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2.5.1. Failure Criteria
After a soil reaches the critical state, it is no longer contracting or dilating and the shear stress on the failure plane τcrit is determined by the effective normal stress on the failure plane σnand critical state friction angle, φcv:
$\ \mathrm{ \tau_{\text {crit }}=\sigma_{n}^{\prime} \cdot \tan \left(\varphi_{c v}\right)}\tag{2-31}$
The peak strength of the soil may be greater, however, due to the interlocking (dilatancy) contribution. This may be stated:
$\ \tau_{\text {peak }}=\sigma_{\mathrm{n}}^{\prime} \cdot \tan \left(\varphi_{\text {peak }}\right)\tag{2-32}$
Where φpeak > φcv. However, use of a friction angle greater than the critical state value for design requires care. The peak strength will not be mobilized everywhere at the same time in a practical problem such as a foundation, slope or retaining wall. The critical state friction angle is not nearly as variable as the peak friction angle and hence it can be relied upon with confidence. Not recognizing the significance of dilatancy, Coulomb proposed that the shear strength of soil may be expressed as a combination of adhesion and friction components:
$\ \mathrm{\tau=\sigma^{\prime} \cdot \tan (\varphi)+c^{\prime}}\tag{2-33}$
It is now known that the c' and φ parameters in the last equation are not fundamental soil properties. In particular, c' and φ are different depending on the magnitude of effective stress. According to Schofield (2006), the longstanding use of c' in practice has led many engineers to wrongly believe that c' is a fundamental parameter. This assumption that c' and φ are constant can lead to overestimation of peak strengths.
2.5.2. The Phi=0 Concept
When a fast triaxial test is carried out, so an unconsolidated undrained test, it is very well possible that the pore pressures will be equal to the increase of the cell pressure. If a test at high cell pressure is carried out, the only difference with a test with a low cell pressure is the value of the pore pressures. The grain pressures will be almost equal in both cases and the result is, that we will find the same critical Mohr circle. So let’s consider a series of unconsolidated undrained (UU) tests. Three specimens are selected and all are consolidated to 110 kPa. This brings the specimens to the end of step 1 in the UU test program. Now the confining pressures are changed to say 70, 140 and 700 kPa, without allowing further consolidation and the sheared undrained. The result, within experimental scatter, is that the shear stress or radius of the Mohr circle is about 35 kPa for each specimen.
So what happened?
When the confining pressure was changed, the pore pressure in the fully saturated specimens changed just as much as did the confining pressure, and the effective stress remained unchanged and equal in each specimen. Thus the effective stress remained 110 kPa and each specimen behaved during shear just as did the CU specimen. The shear stress and thus the radius of the Mohr circle did not increase and apparently the specimens did not encounter internal friction. This is called the phi=0 concept. In clays with a very low permeability and at a high deformation rate, like during the cutting of clay, the clay behaves like the internal friction angle is zero. So for cutting processes the phi=0 concept will be applied.
2.5.3. Factors Controlling Shear Strength of Soils
The stress-strain relationship of soils, and therefore the shearing strength, is affected by:
1. Soil composition (basic soil material): mineralogy, grain size and grain size distribution, shape of particles, pore fluid type and content, ions on grain and in pore fluid.
2. State (initial): Defined by the initial void ratio, effective normal stress and shear stress (stress history). State can be described by terms such as: loose, dense, over consolidated, normally consolidated, stiff, soft, contractive, dilative, etc.
3. Structure: Refers to the arrangement of particles within the soil mass; the manner the particles are packed or distributed. Features such as layers, joints, fissures, slickensides, voids, pockets, cementation, etc., are part of the structure. Structure of soils is described by terms such as: undisturbed, disturbed, remolded, compacted, cemented; flocculent, honey-combed, single-grained; flocculated, deflocculated; stratified, layered, laminated; isotropic and anisotropic.
4. Loading conditions: Effective stress path, i.e., drained, and undrained; and type of loading, i.e., magnitude, rate (static, dynamic), and time history (monotonic, cyclic).
The shear strength and stiffness of soil determines whether or not soil will be stable or how much it will deform. Knowledge of the strength is necessary to determine if a slope will be stable, if a building or bridge might settle too far into the ground, and the limiting pressures on a retaining wall. It is important to distinguish between failure of a soil element and the failure of a geotechnical structure (e.g., a building foundation, slope or retaining wall); some soil elements may reach their peak strength prior to failure of the structure. Different criteria can be used to define the "shear strength" and the "yield point" for a soil element from a stress-strain curve. One may define the peak shear strength as the peak of a stress strain curve, or the shear strength at critical state as the value after large strains when the shear resistance levels off. If the stress-strain curve does not stabilize before the end of shear strength test, the "strength" is sometimes considered to be the shear resistance at 15% to 20% strain. The shear strength of soil depends on many factors including the effective stress and the void ratio.
The shear stiffness is important, for example, for evaluation of the magnitude of deformations of foundations and slopes prior to failure and because it is related to the shear wave velocity. The slope of the initial, nearly linear, portion of a plot of shear stress as a function of shear strain is called the shear modulus
2.5.4. Friction, Interlocking & Dilation
Soil is an assemblage of particles that have little to no cementation while rock (such as sandstone) may consist of an assembly of particles that are strongly cemented together by chemical bonds. The shear strength of soil is primarily due to inter-particle friction and therefore, the shear resistance on a plane is approximately proportional to the effective normal stress on that plane. [3] But soil also derives significant shear resistance from interlocking of strain. The expansion of the particle matrix due to shearing was called dilatancy by Osborne Reynolds. If one considers the energy required to shear an assembly of particles there is energy input by the shear force, T, moving a distance, x and there is also energy input by the normal force, N, as the sample expands a distance, y. Due to the extra energy required for the particles to dilate against the confining pressures, dilatant soils have greater peak strength than contractive soils. Furthermore, as dilative soil grains dilate, they become looser (their void ratio increases), and their rate of dilation decreases until they reach a critical void ratio. Contractive soils become denser as they shear, and their rate of contraction decreases until they reach a critical void ratio.
The tendency for a soil to dilate or contract depends primarily on the confining pressure and the void ratio of the soil. The rate of dilation is high if the confining pressure is small and the void ratio is small. The rate of contraction is high if the confining pressure is large and the void ratio is large. As a first approximation, the regions of contraction and dilation are separated by the critical state line.
2.5.5. Effective Stress
Karl von Terzaghi (1964) first proposed the relationship for effective stress in 1936. For him, the term ‘effective’ meant the calculated stress that was effective in moving soil, or causing displacements. It represents the average stress carried by the soil skeleton. Effective stress (σ') acting on a soil is calculated from two parameters, total stress (σ) and pore water pressure (u) according to:
$\ \sigma^{\prime}=\sigma-\mathrm{u}\tag{2-34}$
Typically, for simple examples:
$\ \sigma=\gamma_{\mathrm{soil}} \cdot \mathrm{H}_{\mathrm{soil}}\quad\text{ and }\quad\mathrm{u}=\gamma_{\mathrm{w}} \cdot \mathrm{H}_{\mathrm{w}}\tag{2-35}$
Much like the concept of stress itself, the formula is a construct, for the easier visualization of forces acting on a soil mass, especially simple analysis models for slope stability, involving a slip plane. With these models, it is important to know the total weight of the soil above (including water), and the pore water pressure within the slip plane, assuming it is acting as a confined layer.
However, the formula becomes confusing when considering the true behavior of the soil particles under different measurable conditions, since none of the parameters are actually independent actors on the particles.
Consider a grouping of round quartz sand grains, piled loosely, in a classic ‘cannonball’ arrangement. As can be seen, there is a contact stress where the spheres actually touch. Pile on more spheres and the contact stresses increase, to the point of causing frictional instability (dynamic friction), and perhaps failure. The independent parameter affecting the contacts (both normal and shear) is the force of the spheres above. This can be calculated by using the overall average density of the spheres and the height of spheres above.
If we then have these spheres in a beaker and add some water, they will begin to float a little depending on their density (buoyancy). With natural soil materials, the effect can be significant, as anyone who has lifted a large rock out of a lake can attest. The contact stress on the spheres decreases as the beaker is filled to the top of the spheres, but then nothing changes if more water is added. Although the water pressure between the spheres (pore water pressure) is increasing, the effective stress remains the same, because the concept of 'total stress' includes the weight of all the water above. This is where the equation can become confusing, and the effective stress can be calculated using the buoyant density of the spheres (soil), and the height of the soil above.
The concept of effective stress truly becomes interesting when dealing with non-hydrostatic pore water pressure. Under the conditions of a pore pressure gradient, the ground water flows, according to the permeability equation (Darcy's law). Using our spheres as a model, this is the same as injecting (or withdrawing) water between the spheres. If water is being injected, the seepage force acts to separate the spheres and reduces the effective stress. Thus, the soil mass becomes weaker. If water is being withdrawn, the spheres are forced together and the effective stress increases. Two extremes of this effect are quicksand, where the groundwater gradient and seepage force act against gravity; and the 'sandcastle effect', where the water drainage and capillary action act to strengthen the sand. As well, effective stress plays an important role in slope stability, and other geotechnical engineering and engineering geology problems, such as groundwater-related subsidence.
2.5.6. Pore Water Pressure: Hydrostatic Conditions
If there is no pore water flow occurring in the soil, the pore water pressures will be hydrostatic. The water table is located at the depth where the water pressure is equal to the atmospheric pressure. For hydrostatic conditions, the water pressure increases linearly with depth below the water table:
$\ \mathrm{u}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{z}_{\mathrm{w}}\tag{2-36}$
2.5.7. Pore Water Pressure: Capillary Action
Due to surface tension water will rise up in a small capillary tube above a free surface of water. Likewise, water will rise up above the water table into the small pore spaces around the soil particles. In fact the soil may be completely saturated for some distance above the water table. Above the height of capillary saturation, the soil may be wet but the water content will decrease with elevation. If the water in the capillary zone is not moving, the water pressure obeys the equation of hydrostatic equilibrium, = ρw·g·zw, but note that zwis negative above the water table. Hence, hydrostatic water pressures are negative above the water table. The thickness of the zone of capillary saturation depends on the pore size, but typically, the heights vary between a centimeter or so for coarse sand to tens of meters for a silt or clay.
The surface tension of water explains why the water does not drain out of a wet sand castle or a moist ball of clay. Negative water pressures make the water stick to the particles and pull the particles to each other, friction at the particle contacts make a sand castle stable. But as soon as a wet sand castle is submerged below a free water surface, the negative pressures are lost and the castle collapses. Considering the effective stress equation, σ' = σ − u,, if the water pressure is negative, the effective stress may be positive, even on a free surface (a surface where the total normal stress is zero). The negative pore pressure pulls the particles together and causes compressive particle to particle contact forces.
Negative pore pressures in clayey soil can be much more powerful than those in sand. Negative pore pressures explain why clay soils shrink when they dry and swell as they are wetted. The swelling and shrinkage can cause major distress, especially to light structures and roads.
2.5.8. Darcy’s Law
Darcy's law states that the volume of flow of the pore fluid through a porous medium per unit time is proportional to the rate of change of excess fluid pressure with distance. The constant of proportionality includes the viscosity of the fluid and the intrinsic permeability of the soil.
$\ \mathrm{Q}=\frac{-\mathrm{K} \cdot \mathrm{A}}{\mu_{\mathrm{l}}} \cdot \frac{\left(\mathrm{u}_{\mathrm{b}}-\mathrm{u}_{\mathrm{a}}\right)}{\mathrm{L}}\tag{2-37}$
The negative sign is needed because fluids flow from high pressure to low pressure. So if the change in pressure is negative (in the x-direction) then the flow will be positive (in the x-direction). The above equation works well for a horizontal tube, but if the tube was inclined so that point b was a different elevation than point a, the equation would not work. The effect of elevation is accounted for by replacing the pore pressure by excess pore pressureue defined as:
$\ \mathrm{u}_{\mathrm{c}}=\mathrm{u}-\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{z}\tag{2-38}$
Where is the depth measured from an arbitrary elevation reference (datum). Replacing by ue we obtain a more general equation for flow:
$\ \mathrm{Q}=\frac{-\mathrm{K} \cdot \mathrm{A}}{\mu_{\mathrm{l}}} \cdot \frac{\left(\mathrm{u}_{\mathrm{c}, \mathrm{b}}-\mathrm{u}_{\mathrm{c}, \mathrm{a}}\right)}{\mathrm{L}}\tag{2-39}$
Dividing both sides of the equation by A, and expressing the rate of change of excess pore pressure as a derivative, we obtain a more general equation for the apparent velocity in the x-direction:
$\ \mathrm{q_{x}=\frac{-K}{\mu_{l}} \cdot \frac{d u_{c}}{d x}}\tag{2-40}$
Where qx has units of velocity and is called the Darcy velocity, or discharge velocity. The seepage velocity (vsx = average velocity of fluid molecules in the pores) is related to the Darcy velocity, and the porosity, n:
$\ \mathrm{\mathrm{v}_{s, \mathrm{x}}=\frac{\mathrm{q}_{\mathrm{x}}}{\mathrm{n}}}\tag{2-41}$
Civil engineers predominantly work on problems that involve water and predominantly work on problems on earth (in earth’s gravity). For this class of problems, civil engineers will often write Darcy's law in a much simpler form:
$\ \mathrm{q}_{\mathrm{x}}=\mathrm{k} \cdot \mathrm{i}_{\mathrm{x}}\tag{2-42}$
Where k is called permeability, and is defined as:
$\ \mathrm{\mathrm{k}=\mathrm{K} \cdot \frac{\rho_{l} \cdot \mathrm{g}}{\mu_{l}}}\tag{2-43}$
And is called the hydraulic gradient. The hydraulic gradient is the rate of change of total head with distance. Values are for typical fresh groundwater conditions, using standard values of viscosity and specific gravity for water at 20°C and 1 atm.
Table 2-14: Typical values of the permeability k.
Soil
Permeability (m/s)
Degree of permeability
Well sorted gravel
100>k>10-2
Extremely high
Gravel
10-2>k>10-3
Very high
Sandy gravel, clean sand, fine sand
10-3>k>10-5
High to Medium
Sand, dirty sand, silty sand
10-5>k>10-7
Low
Silt, silty clay
10-7>k>10-9
Very low
Clay
<10-9
Vitually impermeable
Highly fractured rocks
100>k>10-3
Very high
Oil reservoir rocks
10-4>k>10-6
Medium to Low
Fresh sandstone
10-7>k>10-8
Very low
Fresh limestone, dolomite
10-9>k>10-10
Vitually impermeable
Fresh granite
<10-11
Vitually impermeable
Table 2-15: Some permeabilities according to Hazen’s equation.
Material
Permeability (m/s)
d10 (mm)
Uniform coarse sand
0.0036
0.6
Uniform medium sand
0.0009
0.3
Clean, well-graded sand
0.0001
0.1
Uniform fine sand
36·10-6
0.06
Well-graded fine sand
4·10-6
0.02
Silty sand
10-6
0.01
Uniform silt
36·10-8
0.006
Sandy clay
4·10-8
0.002
Silty clay
10-8
0.001
Clay
64·10-10
0.0008
Colloidal clay
9·10-11
0.00003
2.5.9. Brittle versus Ductile Failure
The terms ductile failure and brittle failure are often used in literature for the failure of materials with shear strength and tensile strength.
In materials science, ductility is a solid material's ability to deform under tensile stress; this is often characterized by the material's ability to be stretched into a wire. Malleability, a similar property, is a material's ability to deform under compressive stress; this is often characterized by the material's ability to form a thin sheet by hammering or rolling. Both of these mechanical properties are aspects of plasticity, the extent to which a solid material can be plastically deformed without fracture. Ductility and malleability are not always coextensive – for instance, while gold has high ductility and malleability, lead has low ductility but high malleability. The word ductility is sometimes used to embrace both types of plasticity.
A material is brittle if, when subjected to stress, it breaks without significant deformation (strain). Brittle materials absorb relatively little energy prior to fracture, even those of high strength. Breaking is often accompanied by a snapping sound. Brittle materials include most ceramics and glasses (which do not deform plastically) and some polymers, such as PMMA and polystyrene. Many steels become brittle at low temperatures (see ductile-brittle transition temperature), depending on their composition and processing. When used in materials science, it is generally applied to materials that fail when there is little or no evidence of plastic deformation before failure. One proof is to match the broken halves, which should fit exactly since no plastic deformation has occurred. Generally, the brittle strength of a material can be increased by pressure. This happens as an example in the brittle-ductile transition zone at an approximate depth of 10 kilometers in the Earth's crust, at which rock becomes less likely to fracture, and more likely to deform ductile.” (Source Wikipedia).
In rock failure a distinction is made between brittle, brittle ductile and ductile failure. Factors determining those types of failure are the ductility number (ratio compressive strength over tensile strength), the confining pressure and the temperature. During dredging the temperature will have hardly any influence, however when drilling deep oil wells temperature will play an important role. The confining pressure, where the failure transit from brittle to ductile is called σbp.
Brittle failure occurs at relative low confining pressures σ3 < σbp en deviator stress q=σ13 > 1⁄2quThe strength increases with the confining pressure, but decreases after the peak strength to a residual value. The presence of pore water can play an important role.
Brittle failure types are:
• Pure tensile failure with or without a small confining pressure.
• Axial tensile failure
• Shear plane failure
Brittle ductile failure is also called semi brittle. In the transition area where σ3 $\ \approx$ σbp, the deformations are not restricted to local shear planes or fractures but are divided over the whole area. The residual- strength is more or less equal to the peak strength.
Ductile failure. A rock fails ductile when σ3 >> qu and σ3 > σbp while the force stays constant or increases some what with increasing deformation.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.05%3A_Criteria_and_Concepts.txt
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2.6.1. SieveAnalysis
The size distribution of gravel and sand particles are typically measured using sieve analysis. The formal procedure is described in ASTM D6913-04(2009). A stack of sieves with accurately dimensioned holes between a mesh of wires is used to separate the particles into size bins. A known volume of dried soil, with clods broken down to individual particles, is put into the top of a stack of sieves arranged from coarse to fine. The stack of sieves is shaken for a standard period of time so that the particles are sorted into size bins. This method works reasonably well for particles in the sand and gravel size range. Fine particles tend to stick to each other, and hence the sieving process is not an effective method. If there are a lot of fines (silt and clay) present in the soil it may be necessary to run water through the sieves to wash the coarse particles and clods through.
A variety of sieve sizes are available. The boundary between sand and silt is arbitrary. According to the Unified Soil Classification System, a #4 sieve (4 openings per inch) having 4.75mm opening size separates sand from gravel and a #200 sieve with an 0.075 mm opening separates sand from silt and clay. According to the British standard, 0.063 mm is the boundary between sand and silt, and 2 mm is the boundary between sand and gravel.
2.6.2. Hydrometer Analysis
The classification of fine-grained soils, i.e., soils that are finer than sand, is determined primarily by their Atterberg limits, not by their grain size. If it is important to determine the grain size distribution of fine-grained soils, the hydrometer test may be performed. In the hydrometer tests, the soil particles are mixed with water and shaken to produce a dilute suspension in a glass cylinder, and then the cylinder is left to sit. A hydrometer is used to measure the density of the suspension as a function of time. Clay particles may take several hours to settle past the depth of measurement of the hydrometer. Sand particles may take less than a second. Stoke's law provides the theoretical basis to calculate the relationship between sedimentation velocity and particle size. ASTM provides the detailed procedures for performing the Hydrometer test.
Clay particles can be sufficiently small that they never settle because they are kept in suspension by Brownian motion, in which case they may be classified as colloids.
2.6.3. Standard Penetration Test
The standard penetration test (SPT) is an in-situ dynamic penetration test designed to provide information on the geotechnical engineering properties of soil. The test procedure is described in the British Standard BS EN ISO 22476-3, ASTM D1586 and Australian Standards AS 1289.6.3.1.
The test uses a thick-walled sample tube, with an outside diameter of 50 mm and an inside diameter of 35 mm, and a length of around 650 mm. This is driven into the ground at the bottom of a borehole by blows from a slide hammer with a weight of 63.5 kg (140 lb) falling through a distance of 760 mm (30 in). The sample tube is driven 150 mm into the ground and then the number of blows needed for the tube to penetrate each 150 mm (6 in) up to a depth of 450 mm (18 in) is recorded. The sum of the number of blows required for the second and third 6 in. of penetration is termed the "standard penetration resistance" or the "N-value". In cases where 50 blows are insufficient to advance it through a 150 mm (6 in) interval the penetration after 50 blows is recorded. The blow count provides an indication of the density of the ground, and it is used in many empirical geotechnical engineering formulae.
The main purpose of the test is to provide an indication of the relative density of granular deposits, such as sands and gravels from which it is virtually impossible to obtain undisturbed samples. The great merit of the test, and the main reason for its widespread use is that it is simple and inexpensive. The soil strength parameters which can be inferred are approximate, but may give a useful guide in ground conditions where it may not be possible to obtain borehole samples of adequate quality like gravels, sands, silts, clay containing sand or gravel and weak rock. In conditions where the quality of the undisturbed sample is suspect, e.g. very silty or very sandy clays, or hard clays, it is often advantageous to alternate the sampling with standard penetration tests to check the strength. If the samples are found to be unacceptably disturbed, it may be necessary to use a different method for measuring strength like the plate test. When the test is carried out in granular soils below groundwater level, the soil may become loosened. In certain circumstances, it can be useful to continue driving the sampler beyond the distance specified, adding further drilling rods as necessary. Although this is not a standard penetration test, and should not be regarded as such, it may at least give an indication as to whether the deposit is really as loose as the standard test may indicate.
The usefulness of SPT results depends on the soil type, with fine-grained sands giving the most useful results, with coarser sands and silty sands giving reasonably useful results, and clays and gravelly soils yielding results which may be very poorly representative of the true soil conditions. Soils in arid areas, such as the Western United States, may exhibit natural cementation. This condition will often increase the standard penetration value.
The SPT is used to provide results for empirical determination of a sand layer's susceptibility to earthquake liquefaction, based on research performed by Harry Seed, T. Leslie Youd, and others.
Despite its many flaws, it is usual practice to correlate SPT results with soil properties relevant for geotechnical engineering design. The reason being that SPT results are often the only test results available, therefore the use of direct correlations has become common practice in many countries.
Different correlations are proposed for granular and cohesive soils.
2.6.4. Cone Penetration Test
The cone penetration test (CPT) is an in situ testing method used to determine the geotechnical engineering properties of soils and delineating soil stratigraphy. It was initially developed in the 1950s at the Dutch Laboratory for Soil Mechanics in Delft to investigate soft soils. Based on this history it has also been called the "Dutch cone test". Today, the CPT is one of the most used and accepted in situ test methods for soil investigation worldwide. The test method consists of pushing an instrumented cone, with the tip facing down, into the ground at a controlled rate (usually 2 centimeters/second). The resolution of the CPT in delineating stratigraphic layers is related to the size of the cone tip, with typical cone tips having a cross-sectional area of either 10 or 15 cm2, corresponding to diameters of 3.6 and 4.4 cm.
The early applications of CPT mainly determined the soil geotechnical property of bearing capacity. The original cone penetrometers involved simple mechanical measurements of the total penetration resistance to pushing a tool with a conical tip into the soil. Different methods were employed to separate the total measured resistance into components generated by the conical tip (the "tip friction") and friction generated by the rod string. A friction sleeve was added to quantify this component of the friction and aid in determining soil cohesive strength in the 1960s (Begemann, 1965). Electronic measurements began in 1948 and improved further in the early 1970s (de Reister, 1971). Most modern electronic CPT cones now also employ a pressure transducer with a filter to gather pore water pressure data. The filter is usually located either on the cone tip (the so-called U1 position), immediately behind the cone tip (the most common U2 position) or behind the friction sleeve (U3 position). Pore water pressure data aids determining stratigraphy and is primarily used to correct tip friction values for those effects. CPT testing which also gathers this piezometer data is called CPTU testing. CPT and CPTU testing equipment generally advances the cone using hydraulic rams mounted on either a heavily ballasted vehicle or using screwed-in anchors as a counter-force. One advantage of CPT over the Standard Penetration Test (SPT) is a more continuous profile of soil parameters, with CPTU data recorded typically at 2cm intervals.
In addition to the mechanical and electronic cones, a variety of other CPT-deployed tools have been developed over the years to provide additional subsurface information. One common tool advanced during CPT testing is a geophone set to gather seismic shear wave and compression wave velocities. This data helps determine the shear modulus and Poisson's ratio at intervals through the soil column for soil liquefaction analysis and low-strain soil strength analysis. Engineers use the shear wave velocity and shear modulus to determine the soil's behavior under low-strain and vibratory loads. Additional tools such as laser-induced fluorescence, X-ray fluorescence [1], soil conductivity/resistivity, membrane interface probe and cameras for capturing video imagery are also increasingly advanced in conjunction with the CPT probe. An additional CPT deployed tool used in Britain, Netherlands, Germany, Belgium and France is a piezocone combined with a tri-axial magnetometer. This is used to attempt to ensure that tests, boreholes, and piles, do not encounter unexploded ordnance (UXB) or duds. The magnetometer in the cone detects ferrous materials of 50 kg or larger within a radius of up to about 2 m distance from the probe depending on the material, orientation and soil conditions.
CPT for geotechnical applications was standardized in 1986 by ASTM Standard D 3441 (ASTM, 2004). ISSMGE provides international standards on CPT and CPTU. Later ASTM Standards have addressed the use of CPT for various environmental site characterization and groundwater monitoring activities. Particularly for geotechnical soil investigations, CPT is gaining popularity compared to standard penetration testing as a method of geotechnical soil investigation by its increased accuracy, speed of deployment, more continuous soil profile and reduced cost over other soil testing methods. The ability to advance additional in situ testing tools using the CPT direct push drilling rig, including the seismic tools described above, are accelerating this process.
2.6.5. Triaxial Test
A triaxial shear test is a common method to measure the mechanical properties of many deformable solids, especially soil (e.g. sand, clay) and rock, and other granular materials or powders. There are several variations on the test. Although the name triaxial test suggests that the stresses would be different in three directions, this is not true in the test as is usually done. In this test with oil or water as confining medium, the confining pressures are equal in all directions (i.e. in terms of principal stresses: for a compression test: σ1 ≠ σ2 = σ3 and for tensile: σ1 = σ2 ≠ σ3). Only in a true triaxial test the stresses in all three directions can be different (i.e. σ1 ≠ σ2 ≠ σ3). For loose granular materials like sand or gravel, the material is contained in a cylindrical latex sleeve with a flat, circular metal plate or platen closing off the top and bottom ends. This cylinder is placed into a bath of water (mostly water but may be any other fluid) to provide pressure along the sides of the cylinder. The top platen can then be mechanically driven up or down along the axis of the cylinder to squeeze the material. The distance that the upper platen travels is measured as a function of the force required to move it, as the pressure of the surrounding water is carefully controlled. The net change in volume of the material is also measured by how much water moves in or out of the surrounding bath. The test for cohesive (non-loose) materials (e.g. clay, rock) is similar to the test for loose granular materials. For rock testing the sleeve may be a thin metal sheeting rather than latex. Triaxial testing on rock is fairly seldom done because the high forces and pressures required to break a rock sample imply very costly and cumbersome testing equipment available at few laboratories in the world. During the test the pore pressures of fluids (e.g. water, oil) or gasses in the sample may be measured.
The principle behind a triaxial shear test is that the stress applied in the vertical direction (along the axis of the cylindrical sample) can be different from the stresses applied in the horizontal directions perpendicular to the sides of the cylinder, i.e. the confining pressure). In a homogeneous and isotropic material this produces a non- hydrostatic stress state, with shear stress that may lead to failure of the sample in shear. In non-homogeneous and anisotropic samples (e.g. bedded or jointed samples) failure may occur due to bending moments and, hence, failure may be tensile. Also combinations of bending and shear failure may happen in inhomogeneous and anisotropic material.
A solid is defined as a material that can support shear stress without moving. However, every solid has an upper limit to how much shear stress it can support. The triaxial test is designed to measure that limit. The stress on the platens is increased until the material in the cylinder fails and forms sliding regions within itself, known as shear bands. A motion where a material is deformed under shear stress is known as shearing. The geometry of the shearing in a triaxial test typically causes the sample to become shorter while bulging out along the sides. The stress on the platen is then reduced and the water pressure pushes the sides back in, causing the sample to grow taller again. This cycle is usually repeated several times while collecting stress and strain data about the sample. During the shearing, a granular material will typically have a net gain or loss of volume. If it had originally been in a dense state, then it typically gains volume, a characteristic known as Reynolds' dilatancy. If it had originally been in a very loose state, then contraction may occur before the shearing begins or in conjunction with the shearing.
From the triaxial test data, it is possible to extract fundamental material parameters about the sample, including its angle of shearing resistance, apparent cohesion, and dilatancy angle. These parameters are then used in computer models to predict how the material will behave in a larger-scale engineering application. An example would be to predict the stability of the soil on a slope, whether the slope will collapse or whether the soil will support the shear stresses of the slope and remain in place. Triaxial tests are used along with other tests to make such engineering predictions.
The triaxial test can be used to determine the shear strength of a discontinuity. A homogeneous and isotropic sample (see above) fails due to shear stresses in the sample. If a sample with a discontinuity is orientated such that the discontinuity is about parallel to the plane in which maximum shear stress will be developed during the test, the sample will fail due to shear displacement along the discontinuity, and hence, the shear strength of a discontinuity can be calculated.
There are several variations of the triaxial test:
2.6.5.1. Consolidated Drained (CD)
In a consolidated drained test the sample is consolidated and sheared in compression with drainage. The rate of axial deformation is kept constant, i.e. is strain controlled. The idea is that the test allows the sample and the pore pressures to fully consolidate (i.e. adjust) to the surrounding stresses. The test may take a long time to allow the sample to adjust, in particular low permeability samples need a long time to drain and adjust stain to stress levels.
2.6.5.2. Consolidated Undrained (CU)
In a consolidated undrained test the sample is not allowed to drain. The shear characteristics are measured under undrained conditions and the sample is assumed to be fully consolidated under the stresses applied that should be similar to the field conditions. Test in particular used if a change in stress is to happen without time for further consolidation.
2.6.5.3. Unconsolidated Undrained (UU)
In an unconsolidated undrained test the sample is not allowed to drain. The sample is compressed at a constant rate (strain-controlled).
2.6.6. ShearTest
A direct shear test also known as shear box test is a laboratory or field test used by geotechnical engineers to measure the shear strength properties of soil or rock material, or of discontinuities in soil or rock masses. For soil the U.S. and U.K. standards defining how the test should be performed are ASTM D 3080 and BS 1377-7:1990 respectively to establish the shear strength properties of soil. It is also possible to estimate typical values of the shear strength parameters based on the type and classification of the soils. For rock the test is generally restricted to rock with (very) low (shear) strength. The test is, however, standard practice to establish the shear strength properties of discontinuities in rock.
The test is performed on three or four specimens from a relatively undisturbed soil sample. A specimen is placed in a shear box which has two stacked rings to hold the sample; the contact between the two rings is at approximately the mid-height of the sample. A confining stress is applied vertically to the specimen, and the upper ring is pulled laterally until the sample fails, or through a specified strain. The load applied and the strain induced is recorded at frequent intervals to determine a stress-strain curve for the confining stress.
Direct Shear tests can be performed under several conditions. The sample is normally saturated before the test is run, but can be run at the in-situ moisture content. The rate of strain can be varied to create a test of undrained or drained conditions, depending whether the strain is applied slowly enough for water in the sample to prevent pore- water pressure buildup.
Several specimens are tested at varying confining stresses to determine the shear strength parameters, the soil cohesion (c) and the angle of internal friction (commonly friction angle) (φ). The results of the tests on each specimen are plotted on a graph with the peak (or residual) stress on the x-axis and the confining stress on the y- axis. The y-intercept of the curve which fits the test results is the cohesion, and the slope of the line or curve is the friction angle.
2.6.7. PointLoadTest
The Point Load Strength test is intended as an index test for the strength classification of rock materials. It may also be used to predict other strength parameters with which it is correlated, for example the unconfined compressive and the tensile strength. It is measured in accordance with the procedures recommended in ASTM D5731, usually with NX-size core samples. The testing machine consists of a loading frame, which measures the force required to break the sample, and a system for measuring the distance between the two platen contact points. Rock specimens in the form of either core, cut blocks, or irregular lumps are broken by application of concentrated load through a pair of spherically truncated, conical platens. The applied force at failure of the sample and distance between the platen tips are recorded in order to calculate the point load index as follows:
$\ \mathrm{I}_{\mathrm{s}}=\frac{\mathrm{F}}{\mathrm{D}_{\mathrm{e}}^{2}}\tag{2-44}$
Another test that is familiar with the Brazilian splitting test is the point load strength test. This test is executed either axial, diametrical or on irregular pieces. The point load test is frequently used to determine the strength when a large number of samples have to be tested. The tests give for brittle rocks, when tested under diametric loading, values reasonable close to the BTS. Also it is suggested that PLS=0.8*BTS, it is suggested to establish such a relation based on both tests.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.06%3A_Soil_Mechanical_Tests.txt
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Gs
Specific gravity
-
ρs
Density of the soil
kg/m3
ρw
Density of water
kg/m3
g
Gravitational constant (9.81 m/s2)
m/s2
Mt
Mass of the soil, total mass
kg
Ms
Mass of the solids
kg
Mw
Mass of the water
kg
Ma
Mass of the air
kg
Vt
Volume of the soil, total volume
m3
Vs
Volume of the solids
m3
Vw
Volume of the water
m3
Va
Volume of the air
m3
ρt
Density of the soil
kg/m3
$\gamma_{ \mathrm{t}}$
Unit weight of the soil
N/m3
g
Gravitational constant (9.81 m/s2)
m/s2
Dr
Relative density
-
e
Current void ratio of the soil in-situ
-
emax
Void ratio of the soil at its loosest condition
-
emin
Void ratio of the soil at its densest condition
-
n
Porosity of the soil in-situ
-
nmax
Porosity of the soil at its loosest condition
-
nmin
Porosity of the soil in its densest condition
-
Vv
Volume of the voids/pores
m3
Vs
Volume of the solids/grains/particles
m3
n
Porosity
-
e
Void ratio
-
Ct
Sorting coefficient
-
C
Sorting coefficient
-
K
Hydraulic conductivity
m2
k
Permeability
m/s
f(n)
porosity function
-
C
sorting coefficient
de
effective grain diameter
mm
d10
Grain diameter where 10% is smaller
mm
d60
Grain diameter where 60% is smaller
mm
U
Grain uniformity coefficient
-
$\ v$
kinematic viscosity
μ
Dynamic viscosity
Pa.s
ρw
Water density
kg/m3
$\ \gamma_{ \mathrm{w}}$
Unit weight of water
N/m3
Q
units of volume per time
m3/s
K
intrinsic permeability
m2
k
permeability
m/s
A
cross sectional area
m2
L
Length
m
ua
Start excess pore pressure
Pa
ub
End excess pore pressure
Pa
μ
dynamic viscosity of the fluid
Pa.s
c
Cohesion
kPa
UCS
Unconfined Compressive Strength
kPa
V
The total volume of soil
m3
ni
Initial porosity
-
ncv
Porosity at constant volume
-
ε
Dilatation
-
σc
Unconfined Compressive Strength
kPa
F
Maximum Failure Load
kN
A
Cross-sectional area of the core sample
m2
E
Deformation modulus
N/m2
W
Specific work of failure
J/m3
σT
Brazilian Tensile Strength (BTS)
kPa
D
Diameter of the core sample
m
F
Maximum Failure Load
kN
L
Length of the core sample
m
IS
Point load index
kPa
F
Failure load
kN
De
Distance between platen tips
m
De2
= D2 for diametrical test
m2
De2
= 4A/π = for axial, block and lump test
m2
A
= W.D = minimum cross-sectional area of a plane through the platen contact points
m2
ρw Density of water kg/m3
zw Depth below the water table
m
u Hydrostatic pressure
kPa
g Gravitational constant
m/s2
σ1 the major principal stress
kPa
σ3 the minor principal stress kPa
$\ \tau$ the shear strength $\ \tau$ = Su (or somtimes cu) kPa
Su the undrained strength kPa
σ' (σ - u) the effective stress kPa
σ
Total stress applied normal to the shear plane
kPa
u
Pore water pressure acting on the same plane
kPa
φ
Effective stress friction angle or the angle of internal friction after Coulomb friction
deg
c'
Cohesion
kPa
$\ \tau$
The shear strength $\ \tau$ = Su (or sometimes cu)
kPa
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.07%3A_Nomenclature.txt
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In the derivation of the Mohr circle the vertical stress σv and the horizontal stress σh are assumed to be the principal stresses, but in reality these stresses could have any orientation. It should be noted here that the Mohr circle approach is valid for the stress situation in a point in the soil. Now consider an infinitesimal element of soil under plane strain conditions as is shown in Figure 2-44. On the element a vertical stress σv and a horizontal stress σh are acting. On the horizontal and vertical planes the shear stresses are assumed to be zero. Now the question is, what would the normal stress σ and shear stress $\ \tau$ be on a plane with an angle α with the horizontal direction? To solve this problem, the horizontal and vertical equilibriums of forces will be derived. Equilibriums of stresses do not exist. One should consider that the surfaces of the triangle drawn in Figure 2-44 are not equal. If the surface (or length) of the surface under the angle α is considered to be 1, then the surface (or length) of the horizontal side is cos(αand the vertical side sin(α). The stresses have to be multiplied with their surface in order to get forces and forces are required for the equilibriums of forces, see Figure 2-45. The derivation of the Mohr circle is also an exercise for the derivation of many equations in this book where equilibriums of forces and moments are applied.
Since an equilibrium of stresses does not exist, only an equilibrium of forces exists, the forces on the soil element have to be known, or the ratio of the forces has to be known.
These forces are, assuming the length of the side under an angle α is 1:
$\ \mathrm{F}_{\mathrm{h}}=\sigma_{\mathrm{h}} \cdot \sin (\alpha)\quad\text{ and }\quad\mathrm{F}_{\mathrm{v}}=\sigma_{\mathrm{v}} \cdot \cos (\alpha)\tag{2-45}$
And:
$\ \mathrm{F}_{\mathrm{n}}=\sigma \quad\text{ and }\quad \mathrm{F}_{\mathrm{s}}=\tau\tag{2-46}$
The equilibrium of forces in the horizontal direction:
$\ \begin{array}{left}\mathrm{F_{h}=F_{n} \cdot \sin (\alpha)-F_{s} \cdot \cos (\alpha)}\ \mathrm{\sigma_{h} \cdot \sin (\alpha)=\sigma \cdot \sin (\alpha)-\tau \cdot \cos (\alpha)}\end{array}\tag{2-47}$
The equilibrium of forces in the vertical direction:
$\ \begin{array}{left}\mathrm{F_{v}=F_{n} \cdot \cos (\alpha)+F_{s} \cdot \sin (\alpha)}\ \mathrm{\sigma_{v} \cdot \cos (\alpha)=\sigma \cdot \cos (\alpha)+\tau \cdot \sin (\alpha)}\end{array}\tag{2-48}$
Equations (2-47) and (2-48) form a system of two equations with two unknowns σ and $\ \tau$. The normal stresses σh and σv are considered to be known variables. To find a solution for the normal stress σ on the plane considered, equation (2-47) is multiplied with sin(αand equation (2-48) is multiplied with cos(α), this gives:
$\ \mathrm{\sigma_{h} \cdot \sin (\alpha) \cdot \sin (\alpha)=\sigma \cdot \sin (\alpha) \cdot \sin (\alpha)-\tau \cdot \cos (\alpha) \cdot \sin (\alpha)}\tag{2-49}$
$\ \mathrm{\sigma_{v} \cdot \cos (\alpha) \cdot \cos (\alpha)=\sigma \cdot \cos (\alpha) \cdot \cos (\alpha)+\tau \cdot \sin (\alpha) \cdot \cos (\alpha)}\tag{2-50}$
Adding up equations (2-49) and (2-50) eliminates the terms with τ and preserves the terms with σ, giving:
$\ \mathrm{\sigma_{v} \cdot \cos ^{2}(\alpha)+\sigma_{h} \cdot \sin ^{2}(\alpha)=\sigma}\tag{2-51}$
Using some basic rules from trigonometry:
$\ \cos ^{2}(\alpha)=\frac{1+\cos (2 \cdot \alpha)}{2}\tag{2-52}$
$\ \sin ^{2}(\alpha)=\frac{1-\cos (2 \cdot \alpha)}{2}\tag{2-53}$
Giving for the normal stress σ on the plane considered:
$\ \sigma=\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right)+\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right) \cdot \cos (2 \cdot \alpha)\tag{2-54}$
To find a solution for the shear stress τ on the plane considered, equation (2-47) is multiplied with -cos(αand equation (2-48) is multiplied with sin(α), this gives:
$\ \mathrm{-\sigma_{h} \cdot \sin (\alpha) \cdot \cos (\alpha)=-\sigma \cdot \sin (\alpha) \cdot \cos (\alpha)+\tau \cdot \cos (\alpha) \cdot \cos (\alpha)}\tag{2-55}$
$\ \mathrm{\sigma_{v} \cdot \cos (\alpha) \cdot \sin (\alpha)=\sigma \cdot \cos (\alpha) \cdot \sin (\alpha)+\tau \cdot \sin (\alpha) \cdot \sin (\alpha)}\tag{2-56}$
Adding up equations (2-55) and (2-56) eliminates the terms with σ and preserves the terms with $\tau$, giving:
$\ \mathrm{\left(\sigma_{v}-\sigma_{h}\right) \cdot \sin (\alpha) \cdot \cos (\alpha)=\tau}\tag{2-57}$
Using the basic rules from trigonometry, equations (2-52) and (2-53), gives for $\tau$ on the plane considered:
$\ \tau=\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right) \cdot \sin (2 \cdot \alpha)\tag{2-58}$
Squaring equations (2-54) and (2-58) gives:
$\ \left(\sigma-\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right)\right)^{2}=\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)^{2} \cdot \cos ^{2}(2 \cdot \alpha)\tag{2-59}$
And:
$\ \tau^{2}=\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)^{2} \cdot \sin ^{2}(2 \cdot \alpha)\tag{2-60}$
Adding up equations (2-59) and (2-60) gives:
$\ \left(\sigma-\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right)\right)^{2}+\tau^{2}=\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)^{2} \cdot\left(\sin ^{2}(2 \cdot \alpha)+\cos ^{2}(2 \cdot \alpha)\right)\tag{2-61}$
This can be simplified to the following circle equation:
$\ \left(\sigma-\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right)\right)^{2}+\tau^{2}=\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)^{2}\tag{2-62}$
If equation (2-62) is compared with the general circle equation from mathematics, equation (2-63):
$\ \mathrm{\left(x-x_{C}\right)^{2}+\left(y-y_{C}\right)^{2}=R^{2}}\tag{2-63}$
The following is found:
$\ \begin{array}{left}\mathrm{x}=\sigma\ \mathrm{x}_{\mathrm{C}}=\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right)\ \mathrm{y}=\tau\ \mathrm{y}_{\mathrm{C}}=\mathrm{0}\ \mathrm{R}=\left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)\end{array}\tag{2-64}$
Figure 2-46 shows the resulting Mohr circle with the Mohr-Coulomb failure criterion:
$\ \tau=\mathrm{c}+\sigma \cdot \tan (\varphi)\tag{2-65}$
The variable is the cohesion or internal shear strength of the soil. In Figure 2-46 it is assumed that the cohesion c=0, which describes the behavior of a cohesion less soil, sand. Further it is assumed that the vertical stress σv (based on the weight of the soil above the point considered) is bigger than the horizontal stress σh. So in this case the horizontal stress at failure follows the vertical stress. The angle α of the plane considered, appears as an angle of 2·α in the Mohr circle. Figure 2-47: shows how the internal friction angle can be determined from a number of tri-axial tests for a cohesion less soil (sand). The 3 circles in this figure will normally not have the failure line as a tangent exactly, but one circle will be a bit too big and another a bit too small. The failure line found will be a best fit. Figure 2-48 and Figure 2-49 show the Mohr circles for a soil with an internal friction angle and cohesion. In such a soil, the intersection point of the failure line with the vertical axis is considered to be the cohesion.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.08%3A_The_Mohr_Circle.txt
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Active soil failure is failure of the soil where the soil takes action, normally because of gravity. The standard example of active soil failure is illustrated by the retaining wall example. A retaining wall has to withstand the forces exerted on it by the soil, in this case a sand with an internal friction angle φ. The retaining wall has to be strong enough to withstand the maximum possible occurring force. The height of the retaining wall is h. The problem has 4 unknowns; the force on the retaining wall F, the normal force on the shear plane N, the shear force on the shear plane and the angle of the shear plane with the horizontal β. To solve this problem, 4 conditions (equations) have to be defined. The first equation is the relation between the normal force and the shear force S. The second and third equations follow from the horizontal and vertical equilibrium of forces on the triangular wedge that will move downwards when the retaining wall fails to withstand the soil forces. The fourth condition follows from the fact that we search for the maximum possible force, a maximum will occur if the derivative of the force with respect to the angle of the shear plane is zero and the second derivative is negative. It should be mentioned that the direction of the shear force is always opposite to the possible direction of motion of the soil. Since the soil will move downwards because of gravity, the shear force is directed upwards.
To start solving the problem, first the weight of the triangular wedge of soil is determined according to:
$\ \mathrm{G}=\frac{1}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \cot (\beta) \cdot \mathrm{w}\tag{2-66}$
The first relation necessary to solve the problem, the relation between the normal force and the shear force on the shear plane is:
$\ \mathrm{S}=\mathrm{N} \cdot \tan (\varphi)\tag{2-67}$
Further it is assumed that the soil consists of pure sand without cohesion and adhesion and it is assumed that the retaining wall is smooth, so no friction between the sand and the wall.
$\ \begin{array}{left}\text{No cohesion} &\Rightarrow \mathrm{c}=\mathrm{0}\ \text{No adhesion} &\Rightarrow \mathrm{a}=\mathrm{0}\ \text{Smooth wall} &\Rightarrow \delta=\mathrm{0}\end{array}\tag{2-68}$
This gives for the horizontal and vertical equilibrium equations on the triangular wedge:
$\ \begin{array}{left} \text{Horizontal} &\Rightarrow \mathrm{F}+\mathrm{S} \cdot \cos (\beta)-\mathrm{N} \cdot \sin (\beta)=\mathrm{0}\ \text{Vertical} &\Rightarrow \mathrm{G}-\mathrm{N} \cdot \cos (\beta)-\mathrm{S} \cdot \sin (\beta)=\mathrm{0}\end{array}\tag{2-69}$
Substituting equation (2-67) gives:
$\ \begin{array}{left}\mathrm{F}+\mathrm{N} \cdot \tan (\varphi) \cdot \cos (\beta)-\mathrm{N} \cdot \sin (\beta)=\mathrm{0}\ \mathrm{G}-\mathrm{N} \cdot \cos (\beta)-\mathrm{N} \cdot \tan (\varphi) \cdot \sin (\beta)=0\end{array}\tag{2-70}$
Writing the full tangent and multiplying with cos(φgives:
$\ \begin{array}{left} \mathrm{F} \cdot \cos (\varphi)+\mathrm{N} \cdot \sin (\varphi) \cdot \cos (\beta)-\mathrm{N} \cdot \sin (\beta) \cdot \cos (\varphi)=0\ \mathrm{G} \cdot \cos (\varphi)-\mathrm{N} \cdot \cos (\beta) \cdot \cos (\varphi)-\mathrm{N} \cdot \sin (\varphi) \cdot \sin (\beta)=0\end{array}\tag{2-71}$
Now the terms with the normal force can be combined to:
$\ \begin{array}{left}\mathrm{F} \cdot \cos (\varphi)+\mathrm{N} \cdot \sin (\varphi-\beta)=\mathrm{0}\ \mathrm{G} \cdot \cos (\varphi)-\mathrm{N} \cdot \cos (\varphi-\beta)=\mathrm{0}\end{array}\tag{2-72}$
Cross multiplying with sine and cosine to give the normal force the same terms:
$\ \begin{array}{left}\mathrm{F} \cdot \cos (\varphi) \cdot \cos (\varphi-\beta)+\mathrm{N} \cdot \sin (\varphi-\beta) \cdot \cos (\varphi-\beta)=0\ \mathrm{G} \cdot \cos (\varphi) \cdot \sin (\varphi-\beta)-\mathrm{N} \cdot \cos (\varphi-\beta) \cdot \sin (\varphi-\beta)=0\end{array}\tag{2-73}$
Adding up the two equations gives:
$\ \mathrm{F} \cdot \cos (\varphi) \cdot \cos (\varphi-\beta)=-\mathrm{G} \cdot \cos (\varphi) \cdot \sin (\varphi-\beta)\tag{2-74}$
Solving the first 3 equations with the first 3 unknowns gives for the force on the retaining wall:
$\ \mathrm{F}=-\mathrm{G} \cdot \tan (\varphi-\beta)\tag{2-75}$
With the equation for the weight of the sand.
$\ \mathrm{G}=\frac{1}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \cot (\beta) \cdot \mathrm{w}\tag{2-76}$
The equation for the force on the retaining wall is found.
$\ \mathrm{F}=-\frac{1}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \frac{\cos (\beta) \cdot \sin (\varphi-\beta)}{\sin (\beta) \cdot \cos (\varphi-\beta)} \cdot \mathrm{w}\tag{2-77}$
This equation still contains the angle of the shear plane as an unknown. Since we are looking for the maximum possible force, a value for β has to be found where this force reaches a maximum. The derivative of the force and the second derivative have to be determined.
$\ \frac{\mathrm{d} \mathrm{F}}{\mathrm{d} \boldsymbol{\beta}}=\mathrm{0}\tag{2-78}$
$\ \frac{\mathrm{d}^{2} \mathrm{F}}{\mathrm{d} \boldsymbol{\beta}^{2}}<\mathrm{0}\tag{2-79}$
Since the equation of the force on the retaining wall contains this angle both in the nominator and the denominator, determining the derivative may be complicated. It is easier to simplify the equation with the following trick:
$\ \begin{array}{left}-\frac{\cos (\beta) \cdot \sin (\varphi-\beta)}{\sin (\beta) \cdot \cos (\varphi-\beta)}=-\frac{\cos (\beta) \cdot \sin (\varphi-\beta)}{\sin (\beta) \cdot \cos (\varphi-\beta)}-1+1=\ -\frac{\cos (\beta) \cdot \sin (\varphi-\beta)}{\sin (\beta) \cdot \cos (\varphi-\beta)}-\frac{\sin (\beta) \cdot \cos (\varphi-\beta)}{\sin (\beta) \cdot \cos (\varphi-\beta)}+1=1-\frac{\sin (\varphi)}{\sin (\beta) \cdot \cos (\varphi-\beta)}\end{array}\tag{2-80}$
Substituting this result in the equation for the force on the retaining wall gives:
$\ \mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot\left(1-\frac{\sin (\varphi)}{\sin (\beta) \cdot \cos (\varphi-\beta)}\right) \cdot \mathrm{w}\tag{2-81}$
When the denominator in the term between brackets has a maximum, also the whole equation has a maximum. So we have to find the maximum of this denominator.
$\ \mathrm{f}=\sin (\beta) \cdot \cos (\beta-\varphi) \Rightarrow\text{ F maximum if } \mathrm{f} \text{ maximum}\tag{2-82}$
The first derivative of this denominator with respect to the shear angle is:
$\ \frac{\mathrm{d f}}{\mathrm{d} \beta}=\cos (2 \cdot \beta-\varphi)\tag{2-83}$
The second derivative of this denominator with respect to the shear angle is:
$\ \frac{\mathrm{d}^{2} \mathrm{f}}{\mathrm{d} \beta^{2}}=-2 \cdot \sin (2 \cdot \beta-\varphi)\tag{2-84}$
The first derivative is zero when the shear angle equals 45 degrees plus half the internal friction angle:
$\ \frac{\mathrm{d f}}{\mathrm{d} \boldsymbol{\beta}}=\mathrm{0} \Rightarrow \beta=\frac{\pi}{4}+\frac{\mathrm{1}}{2} \cdot \varphi\tag{2-85}$
Substituting this solution in the equation for the second derivative gives a negative second derivative which shows that a maximum has been found.
$\ \frac{\mathrm{d}^{2} \mathrm{f}}{\mathrm{d} \beta^{2}}=-2\text{ for }\beta=\frac{\pi}{4}+\frac{1}{2} \cdot \varphi\tag{2-86}$
Substituting this solution for the shear plane angle in the equation for the force on the retaining wall gives:
$\ \mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot\left(\frac{\mathrm{1}-\sin (\varphi)}{\mathrm{1}+\sin (\varphi)}\right) \cdot \mathrm{w}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \mathrm{w} \cdot \mathrm{K}_{\mathrm{a}}\tag{2-87}$
The factor Ka is often referred to as the coefficient of active failure, which is smaller than 1. In the case of a 30 degrees internal friction angle, the value is 1/3.
$\ \mathrm{K}_{\mathrm{A}}=\frac{1-\sin \varphi}{1+\sin \varphi}=\tan ^{2}(45-\varphi / 2)\tag{2-88}$
The horizontal stresses equal the vertical stresses times the factor of active failure, which means that the horizontal stresses are smaller than the vertical stresses.
$\ \sigma_{\mathrm{h}}=\mathrm{K}_{\mathrm{A}} \cdot \sigma_{\mathrm{v}}\tag{2-89}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.09%3A_Active_Soil_Failure.txt
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Passive soil failure is failure of the soil where the outside world takes action, for example a bulldozer. The standard example of passive soil failure is illustrated by the retaining wall example. A retaining wall has to push to supersede the forces exerted on it by the soil, in this case a sand with an internal friction angle φ. The retaining wall has to push strong enough to overcome the minimum possible occurring force. The height of the retaining wall is h. The problem has 4 unknowns; the force on the retaining wall F, the normal force on the shear plane N, the shear force on the shear plane and the angle of the shear plane with the horizontal β. To solve this problem, 4 conditions (equations) have to be defined. The first equation is the relation between the normal force and the shear force S. The second and third equations follow from the horizontal and vertical equilibrium of forces on the triangular wedge that will move upwards when the retaining wall pushes and the soil fails. The fourth condition follows from the fact that we search for the minimum possible force, a minimum will occur if the derivative of the force with respect to the angle of the shear plane is zero and the second derivative is positive. It should be mentioned that the direction of the shear force is always opposite to the possible direction of motion of the soil. Since the soil will move upwards because of the pushing retaining wall, the shear force is directed downwards.
To start solving the problem, first the weight of the triangular wedge of soil is determined according to:
$\ \mathrm{G}=\frac{1}{2} \cdot \rho_{s} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \cot (\beta) \cdot \mathrm{w}\tag{2-90}$
The first relation necessary to solve the problem, the relation between the normal force and the shear force on the shear plane is:
$\ \mathrm{S}=\mathrm{N} \cdot \tan (\varphi)\tag{2-91}$
Further it is assumed that the soil consists of pure sand without cohesion and adhesion and it is assumed that the retaining wall is smooth, so no friction between the sand and the wall.
$\ \begin{array}{left}\text{No cohesion}& \Rightarrow \mathrm{c}=\mathrm{0}\ \text{No adhesion}& \Rightarrow \mathrm{a}=\mathrm{0}\ \text{Smooth wall}& \Rightarrow \delta=\mathrm{0}\end{array}\tag{2-92}$
This gives for the horizontal and vertical equilibrium equations on the triangular wedge:
$\ \begin{array}{left}\text{Horizontal}& \Rightarrow \mathrm{F}-\mathrm{S} \cdot \cos (\beta)-\mathrm{N} \cdot \sin (\beta)=\mathrm{0}\ \text{Vertical}& \Rightarrow \mathrm{G}-\mathrm{N} \cdot \cos (\beta)+\mathrm{S} \cdot \sin (\beta)=\mathrm{0}\end{array}\tag{2-93}$
Substituting equation (2-91) gives:
$\ \begin{array}{left} \mathrm{F}-\mathrm{N} \cdot \tan (\varphi) \cdot \cos (\beta)-\mathrm{N} \cdot \sin (\beta)=0\ \mathrm{G}-\mathrm{N} \cdot \cos (\beta)+\mathrm{N} \cdot \tan (\varphi) \cdot \sin (\beta)=0\end{array}\tag{2-94}$
Writing the full tangent and multiplying with cos(φgives:
$\ \begin{array}{left}\mathrm{F} \cdot \cos (\varphi)-\mathrm{N} \cdot \sin (\varphi) \cdot \cos (\beta)-\mathrm{N} \cdot \sin (\beta) \cdot \cos (\varphi)=0\ \mathrm{G} \cdot \cos (\varphi)-\mathrm{N} \cdot \cos (\beta) \cdot \cos (\varphi)+\mathrm{N} \cdot \sin (\varphi) \cdot \sin (\beta)=0\end{array}\tag{2-95}$
Now the terms with the normal force can be combined to:
$\ \begin{array}{left}\mathrm{F} \cdot \cos (\varphi)-\mathrm{N} \cdot \sin (\varphi+\beta)=\mathrm{0}\ \mathrm{G} \cdot \cos (\varphi)-\mathrm{N} \cdot \cos (\varphi+\beta)=\mathrm{0}\end{array}\tag{2-96}$
Cross multiplying with sine and cosine to give the normal force the same terms:
$\ \begin{array}{left}\mathrm{F} \cdot \cos (\varphi) \cdot \cos (\varphi+\beta)-\mathrm{N} \cdot \sin (\varphi+\beta) \cdot \cos (\varphi+\beta)=0\ -\mathrm{G} \cdot \cos (\varphi) \cdot \sin (\varphi+\beta)+\mathrm{N} \cdot \cos (\varphi+\beta) \cdot \sin (\varphi+\beta)=0\end{array}\tag{2-97}$
Adding up the two equations gives:
$\ \mathrm{F} \cdot \cos (\varphi) \cdot \cos (\varphi+\beta)=\mathrm{G} \cdot \cos (\varphi) \cdot \sin (\varphi+\beta)\tag{2-98}$
Solving the first 3 equations with the first 3 unknowns gives for the force on the retaining wall:
$\ \mathrm{F}=\mathrm{G} \cdot \tan (\varphi+\beta)\tag{2-99}$
With the equation for the weight of the sand.
$\ \mathrm{G}=\frac{1}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \cot (\beta) \cdot \mathrm{w}\tag{2-100}$
The equation for the force on the retaining wall is found.
$\ \mathrm{F}=\frac{1}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \frac{\cos (\beta) \cdot \sin (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)} \cdot \mathrm{w}\tag{2-101}$
This equation still contains the angle of the shear plane as an unknown. Since we are looking for the minimum possible force, a value for β has to be found where this force reaches a minimum. The derivative of the force and the second derivative have to be determined.
$\ \frac{\mathrm{d} \mathrm{F}}{\mathrm{d} \beta}=\mathrm{0}\tag{2-102}$
$\ \frac{\mathrm{d}^{2} \mathrm{F}}{\mathrm{d} \beta^{2}}>\mathrm{0}\tag{2-103}$
Since the equation of the force on the retaining wall contains this angle both in the nominator and the denominator, determining the derivative may be complicated. It is easier to simplify the equation with the following trick:
$\ \begin{array}{left}\frac{\cos (\beta) \cdot \sin (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)}=\frac{\cos (\beta) \cdot \sin (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)}-1+1\ =\frac{\cos (\beta) \cdot \sin (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)}-\frac{\sin (\beta) \cdot \cos (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)}+1\ =\frac{\cos (-\beta) \cdot \sin (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)}+\frac{\sin (-\beta) \cdot \cos (\varphi+\beta)}{\sin (\beta) \cdot \cos (\varphi+\beta)}+1\ =1+\frac{\sin (\varphi)}{\sin (\beta) \cdot \cos (\varphi+\beta)}\end{array}\tag{2-104}$
Substituting this result in the equation for the force on the retaining wall gives:
$\ \mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot\left(1+\frac{\sin (\varphi)}{\sin (\beta) \cdot \cos (\varphi+\beta)}\right) \cdot \mathrm{w}\tag{2-105}$
When the denominator in the term between brackets has a maximum, also the whole equation has a minimum. So we have to find the maximum of this denominator.
$\ \mathrm{f}=\sin (\beta) \cdot \cos (\beta+\varphi) \Rightarrow\text{ F minimum if } \mathrm{f} \text{ maximum}\tag{2-106}$
The first derivative of this denominator with respect to the shear angle is:
$\ \frac{\mathrm{d f}}{\mathrm{d} \beta}=\cos (2 \cdot \beta+\varphi)\tag{2-107}$
The second derivative of this denominator with respect to the shear angle is:
$\ \frac{\mathrm{d}^{2} \mathrm{f}}{\mathrm{d} \boldsymbol{\beta}^{2}}=-\mathrm{2} \cdot \sin (\mathrm{2} \cdot \boldsymbol{\beta}+\varphi)\tag{2-108}$
The first derivative is zero when the shear angle equals 45 degrees minus half the internal friction angle:
$\ \frac{\mathrm{d f}}{\mathrm{d} \beta}=\mathrm{0} \Rightarrow \beta=\frac{\pi}{4}-\frac{1}{2} \cdot \varphi\tag{2-109}$
Substituting this solution in the equation for the second derivative gives a negative second derivative which shows that a maximum has been found.
$\ \frac{\mathrm{d}^{2} \mathrm{f}}{\mathrm{d} \beta^{2}}=-2\text{ for }\beta=\frac{\pi}{4}-\frac{1}{2} \cdot \varphi\tag{2-110}$
Substituting this solution for the shear plane angle in the equation for the force on the retaining wall gives:
$\ \mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot\left(\frac{\mathrm{1}+\sin (\varphi)}{1-\sin (\varphi)}\right) \cdot \mathrm{w}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \mathrm{w} \cdot \mathrm{K}_{\mathrm{p}}\tag{2-111}$
The factor Kp is often referred to as the coefficient of passive failure, which is larger than 1. In the case of a 30 degrees internal friction angle, the value is 3.
$\ \mathrm{K_{P}=\frac{1+\sin \varphi}{1-\sin \varphi}=\tan ^{2}(45+\varphi / 2)}\tag{2-112}$
The horizontal stresses equal the vertical stresses times the factor of passive failure, which means that the horizontal stresses are larger than the vertical stresses.
$\ \sigma_{\mathrm{h}}=\mathrm{K}_{\mathrm{p}} \cdot \sigma_{\mathrm{v}}\tag{2-113}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.10%3A_Passive_Soil_Failure.txt
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Figure 2-57 gives a summary of the Mohr circles for Active and Passive failure of a cohesion less soil.
Some equations for a cohesion less soil in the active state:
Failure will occur if:
$\ \sin (\varphi)=\frac{\frac{1}{2} \cdot\left(\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}\right)}{\frac{1}{2} \cdot\left(\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}\right)}\tag{2-114}$
This can also be written as:
$\ \left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)-\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right) \cdot \sin (\varphi)=0\tag{2-115}$
Using this equation the value of σh can be expressed into σv:
$\ \sigma_{\mathrm{h}}=\sigma_{\mathrm{v}} \frac{1-\sin (\varphi)}{1+\sin (\varphi)}=\mathrm{K}_{\mathrm{a}} \cdot \sigma_{\mathrm{v}}\tag{2-116}$
On the other hand, the value of σv can also be expressed into σh:
$\ \sigma_{\mathrm{v}}=\sigma_{\mathrm{h}} \frac{1+\sin (\varphi)}{1-\sin (\varphi)}=\mathrm{K}_{\mathrm{p}} \cdot \sigma_{\mathrm{h}}\tag{2-117}$
For the passive state the stresses σv and σh should be reversed.
Figure 2-58 gives a summary of the Mohr circles for Active and Passive failure for a soil with cohesion.
Some equations for a soil with cohesion in the active state:
Failure will occur if:
$\ \sin (\varphi)=\frac{\frac{1}{2} \cdot\left(\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}\right)}{\mathrm{c} \cdot \cot (\varphi)+\frac{1}{2} \cdot\left(\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}\right)}\tag{2-118}$
This can also be written as:
$\ \left(\frac{\sigma_{\mathrm{v}}-\sigma_{\mathrm{h}}}{2}\right)-\left(\frac{\sigma_{\mathrm{v}}+\sigma_{\mathrm{h}}}{2}\right) \cdot \sin (\varphi)-\mathrm{c} \cdot \cos (\varphi)=0\tag{2-119}$
Using this equation the value of σh can be expressed into σv:
$\ \sigma_{\mathrm{h}}=\sigma_{\mathrm{v}} \frac{1-\sin (\varphi)}{1+\sin (\varphi)}-2 \cdot \mathrm{c} \cdot \frac{\cos (\varphi)}{1+\sin (\varphi)}=\mathrm{K}_{\mathrm{a}} \cdot \sigma_{\mathrm{v}}-2 \cdot \mathrm{c} \cdot \sqrt{\mathrm{K}_{\mathrm{a}}}\tag{2-120}$
On the other hand, the value of σv can also be expressed into σh:
$\ \sigma_{\mathrm{v}}=\sigma_{\mathrm{h}} \frac{1+\sin (\varphi)}{1-\sin (\varphi)}+2 \cdot \mathrm{c} \cdot \frac{\cos (\varphi)}{1-\sin (\varphi)}=\mathrm{K}_{\mathrm{p}} \cdot \sigma_{\mathrm{h}}+2 \cdot \mathrm{c} \cdot \sqrt{\mathrm{K}_{\mathrm{p}}}\tag{2-121}$
For the passive state the stresses σv and σh should be reversed.
2.12: Shear Strength versus Friction
To avoid confusion between cohesion and adhesion on one side and internal and external friction on the other side, internal and external friction, also named Coulomb friction, depend linearly on normal stresses, internal friction depends on the normal stress between the sand grains and external friction on the normal stress between the sand grains and another material, for example steel. In civil engineering internal and external friction are denoted by the angle of internal friction and the angle of external friction, also named the soil/interface friction angle. In mechanical engineering the internal and external friction angles are denoted by the internal and external friction coefficient. If there is no normal stress, there is no shear stress resulting from normal stress, so the friction is zero. Adhesion and cohesion are considered to be the sticky effect between two surfaces. Cohesion is the sticky effect between two surfaces of the same material before any failure has occurred and adhesion is the sticky effect between two different materials, for example adhesive tape. Adhesion and cohesion could be named the external and internal shear strength which are independent from normal stresses. The equations for the resulting shear stresses are:
$\ \tau_{\mathrm{in}}=\tau_{\mathrm{c}}+\sigma_{\mathrm{in}} \cdot \tan (\varphi) \quad\text{ or }\quad \tau_{\mathrm{in}}=\tau_{\mathrm{c}}+\sigma_{\mathrm{in}} \cdot \mu_{\mathrm{in}}\tag{2-122}$
$\ \tau_{\mathrm{ex}}=\tau_{\mathrm{a}}+\sigma_{\mathrm{ex}} \cdot \tan (\delta) \quad\text{ or }\quad \tau_{\mathrm{ex}}=\tau_{\mathrm{a}}+\sigma_{\mathrm{ex}} \cdot \mu_{\mathrm{ex}}\tag{2-123}$
Or
$\ \mathrm{\tau_{\mathrm{in}}=c+\sigma_{\mathrm{in}} \cdot \tan (\varphi) \quad\text{ or }\quad \tau_{\mathrm{in}}=c+\sigma_{\mathrm{in}} \cdot \mu_{\mathrm{in}}}\tag{2-124}$
$\ \tau_{\mathrm{ex}}=\mathrm{a}+\sigma_{\mathrm{ex}} \cdot \tan (\delta) \quad\text{ or }\quad \tau_{\mathrm{ex}}=\mathrm{a}+\sigma_{\mathrm{ex}} \cdot \mu_{\mathrm{ex}}\tag{2-125}$
With:
$\ \mu_{\mathrm{in}}=\tan (\varphi)\tag{2-126}$
$\ \mu_{\mathrm{ex}}=\tan (\delta)\tag{2-127}$
The values of the internal friction angle φ and the external friction angle δ not only depend on the soil properties like the density and the shape of the particles, but may also depend on the deformation history.
Figure 2-59, Figure 2-60 and Figure 2-61 show the Ka and Kp coefficients as a function of the internal friction angle.
2.13: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or external shear strength
kPa
c, $\ \tau_\mathrm{c}$
Cohesion or internal shear strength
kPa
f
Function
-
F
Horizontal force
kN
Fh
Horizontal force on soil element
kN
Fv
Verical force on soil element
kN
Fn
Normal force on soil element
kN
Fs
Shear force on soil element
kN
g
Gravitational constant (9.81)
m/s2
G
Gravitational vertical force
kN
h
Height of the dam/soil
m
Ka
Coefficient of active failure
-
Kp
Coefficient of passive failure
-
N
Force normal to the shear plane
kN
S
Shear force on the shear plane
kN
α
Orientation of shear plane (Mohr circle)
rad
β
Angle of the shear plane (active & passive failure)
rad
δ
External friction angle or soil/interface friction angle
rad
φ
Internal friction angle
rad
σ
Normal stress
kPa
σh
Horizontal normal stress (principal stress)
kPa
σv
Vertical normal stress (principal stress)
kPa
σin
Internal normal stress
kPa
σex
External normal stress or soil interface normal stress
kPa
$\ \tau$
Shear stress
kPa
$\ \tau_{\text {in }}$
Internal shear stress
kPa
$\ \tau_{\text {ex }}$
External shear stress or soil interface shear stress
kPa
ρg
Density of the soil
ton/m3
μin
Internal friction coefficient
-
μex
External friction coefficient
-
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/02%3A_Basic_Soil_Mechanics/2.11%3A_Summary.txt
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Hatamura and Chijiiwa (1975), (1976A), (1976B), (1977A) and (1977B) distinguished three failure mechanisms in soil cutting. The Shear Type, the Flow Type and the Tear Type. The Flow Type and the Tear Type occur in materials without an angle of internal friction. The Shear Type occurs in materials with an angle of internal friction like sand.
A fourth failure mechanism can be distinguished (Miedema (1992)), the Curling Type, as is known in metal cutting. Although it seems that the curling of the chip cut is part of the flow of the material, whether the Curling Type or the Flow Type occurs depends on several conditions. The Curling Type in general will occur if the adhesive force on the blade is large with respect to the normal force on the shear plane. Whether the Curling Type results in pure curling or buckling of the layer cut giving obstruction of the flow depends on different parameters. In rock or stone two additional cutting mechanisms may occur, the Crushed Type and the Chip Type. The Crushed Type will occur if a thin layer of rock is scraped or cut like in oil and gas drilling. The mechanism of the Crushed Type is similar to the Shear Type, only first the rock material has to be crushed. The Chip Type will occur when cutting thicker layers of rock or stone. This type is similar to the Tear Type.
Figure 3-1 illustrates the Curling Type, the Flow Type and the Tear Type mechanisms as they might occur when cutting clay, the Shear Type mechanism as it might occur when cutting sand and the Crushed Type and Chip Type as they might occur when cutting rock or stone. Of course also mixed types may occur.
To predict which type of failure mechanism will occur under given conditions with specific soil, a formulation for the cutting forces has to be derived. The derivation is made under the assumption that the stresses on the shear plane and the blade are constant and equal to the average stresses acting on the surfaces. Figure 3-2 gives some definitions regarding the cutting process. The line A-B is considered to be the shear plane, while the line A-C is the contact area between the blade and the soil. The blade angle is named α and the shear angle β. The blade is moving from left to right with a cutting velocity vc. The thickness of the layer cut is hi and the vertical height of the blade hb. The horizontal force on the blade Fh is positive from right to left always opposite to the direction of the cutting velocity vc. The vertical force on the blade Fv is positive downwards.
The shear angle β is determined based on the minimum energy principle. It is assumed that failure will occur at a shear angle where the cutting energy is at a minimum. The cutting power is the cutting energy per unit of time, so the cutting power also has to be at the minimum level.
Since the vertical force is perpendicular to the cutting velocity, the vertical force does not contribute to the cutting power, which is equal to the horizontal cutting force times the cutting velocity:
$\ \mathrm{P}_{\mathrm{c}}=\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}\tag{3-1}$
Whether the minimum energy principle is true and whether the approach of using straight failure planes is right has been validated with experiments. The experimental data, usually measurements of the horizontal and vertical cutting forces and pore pressures, shows that the approach in this book gives a good prediction of the cutting forces.
3.02: Definitions
Definitions:
1. A: The blade tip.
2. B: End of the shear plane.
3. C: The blade top.
4. A-B: The shear plane.
5. A-C: The blade surface.
6. hb: The height of the blade.
7. hi: The thickness of the layer cut.
8. vc: The cutting velocity.
9. α: The blade angle.
10. β: The shear angle.
11. Fh: The horizontal force, the arrow gives the positive direction.
12. Fv: The vertical force, the arrow gives the positive direction.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/03%3A_The_General_Cutting_Process/3.01%3A_Cutting_Mechanisms.txt
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Figure 3-3 and Figure 3-4 show the Flow Type and the Shear Type of cutting process. The Shear Type is modeled as the Flow Type. The difference is that in dry soil the forces calculated for the Flow Type are constant forces because the process is ductile. For the Shear Type the forces are the peak forces, because the process is assumed to be brittle (shear). The average forces can be determined by multiplying the peak forces with a factor of 1⁄4 to 1⁄2.
3.3.1. The Equilibrium of Forces
Figure 3-6 illustrates the forces on the layer of soil cut. The forces shown are valid in general. The forces acting on this layer are:
1. A normal force acting on the shear surface N1 resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A shear force as a result of pure cohesion $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ with the area of the shear plane.
5. A gravity force as a result of the (under water) weight of the layer cut.
6. An inertial force I, resulting from acceleration of the soil.
7. A force normal to the blade N2, resulting from the effective grain stresses.
8. A shear force S2 as a result of the external friction angle N2·tan(δ).
9. A shear force as a result of pure adhesion between the soil and the blade $\ \tau_\mathrm{a}$. This force can be calculated by multiplying the adhesive shear strength $\ \tau_\mathrm{a}$ of the soil with the contact area between the soil and the blade.
10. A force W2 as a result of water under pressure on the blade
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{\mathrm{1}}^{\mathrm{2}}+\mathrm{S}_{\mathrm{1}}^{\mathrm{2}}}\tag{3-2}$
The forces acting on a straight blade when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the external friction angle N2·tan(δ).
3. A shear force as a result of pure adhesion between the soil and the blade $\ \tau_\mathrm{a}$. This force can be calculated by multiplying the adhesive shear strength $\ \tau_\mathrm{a}$ of the soil with the contact area between the soil and the blade.
4. A force W2 as a result of water under pressure on the blade.
These forces are shown in Figure 3-7. If the forces N2 and S2 are combined to a resulting force Kand the adhesive force and the water under pressures forces W1 and W2 are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{3-3}$
The horizontal equilibrium of forces:
$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{h}}=& \mathrm{K}_{1} \cdot \sin (\beta+\varphi)-\mathrm{W}_{1} \cdot \sin (\beta)+\mathrm{C} \cdot \cos (\beta)+\mathrm{I} \cdot \cos (\beta) \ &-\mathrm{A} \cdot \cos (\alpha)+\mathrm{W}_{2} \cdot \sin (\alpha)-\mathrm{K}_{2} \cdot \sin (\alpha+\delta)=0 \end{array}\tag{3-4}$
The vertical equilibrium of forces:
$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{v}}=&-\mathrm{K}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta}+\varphi)+\mathrm{W}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta})+\mathrm{C} \cdot \sin (\boldsymbol{\beta})+\mathrm{I} \cdot \sin (\boldsymbol{\beta}) \ &+\mathrm{G}+\mathrm{A} \cdot \sin (\boldsymbol{\alpha})+\mathrm{W}_{\mathrm{2}} \cdot \mathrm{c o s}(\boldsymbol{\alpha})-\mathrm{K}_{2} \cdot \cos (\boldsymbol{\alpha}+\delta)=\mathrm{0} \end{array}\tag{3-5}$
The force K1 on the shear plane is now:
$\ \begin{array}{left} \mathrm{K}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{3-6}$
The force K2 on the blade is now:
$\ \begin{array}{left} \mathrm{K}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{3-7}$
From equation (3-7) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=-W_{2} \cdot \sin (\alpha)+K_{2} \cdot \sin (\alpha+\delta)+A \cdot \cos (\alpha)}\tag{3-8}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{3-9}$
The normal force on the shear plane is now:
$\ \begin{array}{left} \mathrm{N}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \end{array}\tag{3-10}$
The normal force on the blade is now:
$\ \begin{array}{left} \mathrm{N}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \end{array}\tag{3-11}$
If the equations (3-10) and (3-11) give a positive result, the normal forces are compressive forces. It can be seen from these equations that the normal forces can become negative, meaning that a tensile rupture might occur, depending on values for the adhesion and cohesion and the angles involved. The most critical direction where this might occur can be found from the Mohr circle.
3.3.2. The Individual Forces
If there is no cavitation the water pressures forces W1 and W2 can be written as:
$\ \mathrm{W}_{1}=\frac{\mathrm{p}_{1 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{\varepsilon} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w}}{\left(\mathrm{a}_{\mathrm{1}} \cdot \mathrm{k}_{\mathrm{i}}+\mathrm{a}_{\mathrm{2}} \cdot \mathrm{k}_{\mathrm{m a x}}\right) \cdot \sin (\beta)}=\frac{\mathrm{p}_{1 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}} \cdot \sin (\beta)}\tag{3-12}$
$\ \mathrm{W}_{2}=\frac{\mathrm{p}_{2 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\left(\mathrm{a}_{\mathrm{1}} \cdot \mathrm{k}_{\mathrm{i}}+\mathrm{a}_{\mathrm{2}} \cdot \mathrm{k}_{\mathrm{m a x}}\right) \cdot \sin (\alpha)}=\frac{\mathrm{p}_{2 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}} \cdot \sin (\alpha)}\tag{3-13}$
In case of cavitation W1 and W2 become:
$\ \mathrm{W}_{1}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\tag{3-14}$
$\ \mathrm{W}_{2}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\mathrm{\alpha})}\tag{3-15}$
Wismer and Luth (1972A) and (1972B) investigated the inertia forces term of the total cutting forces. The following equation is derived:
$\ \mathrm{I}=\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{3-16}$
The cohesive and adhesive forces C and A can be determined with soil mechanical experiments. For the cohesive and adhesive forces the following equations are valid:
$\ \mathrm{C=\frac{c \cdot h_{i} \cdot w}{\sin (\beta)}}\tag{3-17}$
$\ \mathrm{A}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{3-18}$
The gravitational force (weight submerged) follows from:
$\ \mathrm{G}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot\left\{\frac{\left(\mathrm{h}_{\mathrm{b}}+\mathrm{h}_{\mathrm{i}} \cdot \sin (\alpha)\right)}{\sin (\alpha)}+\frac{\mathrm{h}_{\mathrm{i}} \cdot \cos (\alpha+\beta)}{2 \cdot \sin (\beta)}\right\}\tag{3-19}$
The gravitational force (dry weight) follows from:
$\ \mathrm{G}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot\left\{\frac{\left(\mathrm{h}_{\mathrm{b}}+\mathrm{h}_{\mathrm{i}} \cdot \sin (\alpha)\right)}{\sin (\alpha)}+\frac{\mathrm{h}_{\mathrm{i}} \cdot \cos (\alpha+\beta)}{2 \cdot \sin (\beta)}\right\}\tag{3-20}$
This is in accordance with the area that is used for the water pore pressure calculations in the case of water saturated sand (see Figure 6-7).
3.04: The Curling Type
In some soils it is possible that the Curling Type of cutting mechanism occurs. This will happen when the layer cut is relatively thin and there is a force on the blade of which the magnitude depends on the blade height, like the adhesive force or the pore pressure force in the case of a cavitating cutting process. In soils like clay and loam, but also in rock under hyperbaric conditions this may occur. Figure 3-8 shows this Curling Type. The question now is, what is the effective blade height hb,m where the soil is in contact with the blade. To solve this problem, an additional equation is required. There is only one equation available and that is the equilibrium equation of moments on the layer cut. Figure 3-9 shows the moments acting on the layer cut. In the case of clay, loam or hyperbaric rock, the contribution of gravity can be neglected.
The equilibrium of moments when the gravity moment is neglected is:
$\ \left(\mathrm{N}_{\mathrm{1}}-\mathrm{W}_{\mathrm{1}}\right) \cdot \mathrm{R}_{\mathrm{1}}=\left(\mathrm{N}_{\mathrm{2}}-\mathrm{W}_{\mathrm{2}}\right) \cdot \mathrm{R}_{\mathrm{2}}\tag{3-21}$
The arms of the 2 moments are:
$\ \mathrm{R_{1}=\frac{\lambda_{1} \cdot h_{\mathrm{i}}}{\sin (\beta)}, R_{2}=\frac{\lambda_{2} \cdot h_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}}\tag{3-22}$
This gives the equilibrium equation of moments on the layer cut:
$\ \left(\begin{array}{c}\frac{\mathrm{W}_{2} \cdot \mathrm{s i n}(\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \ +\frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \ -\mathrm{W}_{1}\end{array}\right) \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}=\left(\begin{array}{c}\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \ +\frac{\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \ -\mathrm{W}_{2}\end{array}\right) \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}\tag{3-23}$
When the equations for W1W2and as mentioned before are substituted, the resulting equation is a second degree equation with hb,m as the variable.
This can be solved using the following set of equations:
$\ \begin{array}{left} \mathrm{A} &\cdot \mathrm{x}^{2}+\mathrm{B} \cdot \mathrm{x}+\mathrm{C}=\mathrm{0} \quad\text { and }\quad \mathrm{h}_{\mathrm{b}, \mathrm{m}}=\mathrm{x}=\frac{-\mathrm{B} \pm \sqrt{\mathrm{B}^{2}-\mathrm{4} \cdot \mathrm{A} \cdot \mathrm{C}}}{\mathrm{2} \cdot \mathrm{A}} \ \mathrm{A}&=\frac{\lambda_{2} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta+\varphi)-\lambda_{2} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\alpha+\beta+\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\alpha)} \ &+\frac{+\mathrm{a} \cdot \lambda_{2} \cdot \cos (\alpha+\beta+\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\alpha)} \ \mathrm{B}&=\frac{\lambda_{1} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\delta)\cdot \mathrm{cos(\varphi)}-\lambda_{2} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \cos (\delta) \cdot \sin (\varphi)}{\sin (\alpha) \cdot \sin (\beta)}\cdot \mathrm{h}_{\mathrm{i}}\ &+\frac{-\mathrm{c} \cdot \lambda_{2} \cdot \cos (\delta) \cdot \cos (\varphi)+\mathrm{a} \cdot \lambda_{1} \cdot \cos (\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \ \mathrm{C}&=\frac{\lambda_{1} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta)\cdot \mathrm{cos(\varphi)}-\lambda_{1} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta+\varphi) }{\sin (\beta) \cdot \sin (\beta)}\cdot \mathrm{h_i \cdot h_i}\ &+\frac{-\mathrm{c} \cdot \lambda_{1} \cdot \cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}\end{array}\tag{3-24}$
The usage is now as follows:
$\ \begin{array}{left}\text{if } \mathrm{h}_{\mathrm{b}, \mathrm{m}}<\mathrm{h}_{\mathrm{b}}\text{ then use } \mathrm{h}_{\mathrm{b}, \mathrm{m}}\ \text{if }\mathrm{h}_{\mathrm{b}, \mathrm{m}} \geq \mathrm{h}_{\mathrm{b}}\text{ then use }\mathrm{h}_{\mathrm{b}}\end{array}\tag{3-25}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/03%3A_The_General_Cutting_Process/3.03%3A_The_Flow_Shear_and_Crushed_Type.txt
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The Tear Type of cutting process has a failure mechanism based on tensile failure. For such a failure mechanism to occur it is required that negative stresses may occur. In sand this is not the case, because in sand the failure lines according to the Mohr-Coulomb criterion will pass through the origin as is shown in Figure 2-46 and Figure 2-47. For the failure lines not to pass through the origin it is required that the soil has a certain cohesion or shear strength like with clay and rock. In clay and rock, normally, the inertial forces and the gravity can be neglected and also the water pore pressures do not play a role. Only with hyperbaric rock cutting the water pore pressures will play a role, but there the Tear Type will not occur. This implies that for the Tear Type and Chip Type a soil with cohesion and adhesion and internal and external friction will be considered.
If clay or rock is considered, the following condition can be derived with respect to tensile rupture:
With the relations for the cohesive force C, the adhesive force and the adhesion/cohesion ratio (the ac ratio r):
$\ \mathrm{C}=\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\tag{3-26}$
$\ \mathrm{A=\frac{\lambda_{s} \cdot a \cdot h_{b} \cdot w}{\sin (\alpha)}}\tag{3-27}$
$\ \mathrm{r}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\tag{3-28}$
The horizontal Fh and vertical Fv cutting forces can be determined according to:
$\ \mathrm{F_{\mathrm{h}}=\lambda_{\mathrm{s}} \cdot c \cdot h_{\mathrm{i}} \cdot w \cdot \frac{\frac{\sin (\alpha+\delta)}{\sin (\beta)} \cdot \cos (\varphi)+r \cdot \frac{\sin (\beta+\varphi)}{\sin (\alpha)} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)}}\tag{3-29}$
$\ \mathrm{F}_{v}=\mathrm{\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\cos (\alpha+\delta)}{\sin (\beta)} \cdot \cos (\varphi)-r \cdot \frac{\cos (\beta+\varphi)}{\sin (\alpha)} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)}}\tag{3-30}$
The shear angle β is determined in the case where the horizontal cutting force Fh is at a minimum, based on the minimum energy principle.
$\ \begin{array}{left} \frac{\partial \mathrm{F}_{\mathrm{h}}}{\partial \boldsymbol{\beta}}&= \frac{\mathrm{r} \cdot \cos (\delta) \cdot \sin (\mathrm{2} \cdot \boldsymbol{\beta}+\varphi) \cdot \sin (\alpha) \cdot \sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)}{\sin ^{2}(\alpha+\beta+\delta+\varphi) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)} \ &+\frac{-\sin (\alpha) \cdot \sin (\alpha+2 \cdot \beta+\delta+\varphi) \cdot(\sin (\alpha) \cdot \sin (\alpha+\delta) \cdot \cos (\varphi))}{\sin ^{2}(\alpha+\beta+\delta+\varphi) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)} =0 \ &+\frac{-\sin (\alpha) \cdot \sin (\alpha+2 \cdot \beta+\delta+\varphi) \cdot(\mathrm{r} \cdot \sin (\beta) \cdot \sin (\beta+\varphi) \cdot \cos (\delta))}{\sin ^{2}(\alpha+\beta+\delta+\varphi) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)} \end{array}\tag{3-31}$
In the special case where there is no adhesion, r = , the shear angle is:
$\ \mathrm{\frac{\partial F_{h}}{\partial \beta}=\frac{-\sin (\alpha+2 \cdot \beta+\delta+\varphi) \cdot \sin (\alpha+\delta) \cdot \cos (\varphi)}{\sin ^{2}(\alpha+\beta+\delta+\varphi) \cdot \sin ^{2}(\beta)}=0}\tag{3-32}$
So:
$\ \sin (\alpha+2 \cdot \beta+\delta+\varphi)=0\text{ for }\alpha+2 \cdot \beta+\delta+\varphi=\pi\text{ giving }\beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}\tag{3-33}$
The cohesion can be determined from the UCS value and the angle of internal friction φ according to (as is shown in Figure 3-12):
$\ \mathrm{c=\frac{\mathrm{U} \mathrm{C S}}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)}\tag{3-34}$
According to the Mohr-Coulomb failure criterion, the following is valid for the shear stress on the shear plane, as is shown in Figure 3-13.
$\ \tau_{\mathrm{S} 1}=\mathrm{c}+\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)\tag{3-35}$
The average stress condition on the shear plane is now σN1, $\ \tau_{\mathrm{S1}}$ as is show in Figure 3-13. A Mohr circle (Mohr circle 1) can be drawn through this point, resulting in a minimum stress σmin which is negative, so tensile. If this minimum normal stress is smaller than the tensile strength σT tensile fracture will occur, as is the case in the figure. Now Mohr circle 1 can never exist, but a smaller circle (Mohr circle 2) can, just touching the tensile strength σT. The question is now, how to get from Mohr circle 1 to Mohr circle 2. To find Mohr circle 2 the following steps have to be taken.
The radius of the Mohr circle 1 can be found from the shear stress $\ \tau_{\mathrm{S1}}$ by:
$\ \mathrm{R=\frac{\tau_{\mathrm{S} 1}}{\cos (\varphi)}}\tag{3-36}$
The center of the Mohr circle 1, σC, now follows from:
$\ \begin{array}{left} \sigma_{\mathrm{C}}&=\mathrm{\sigma_{\mathrm{N} 1}+R \cdot \sin (\varphi)=\sigma_{\mathrm{N} 1}+\tau_{\mathrm{S} 1} \cdot \tan (\varphi)}\ &=\mathrm{\sigma_{\mathrm{N} 1}+\mathrm{c} \cdot \tan (\varphi)+\sigma_{\mathrm{N} 1} \cdot \tan ^{2}(\varphi)}\end{array}\tag{3-37}$
The normal force N1 on the shear plane is now:
\ \begin{aligned} \mathrm{N}_{1}=& \frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \=& \lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{-\frac{\cos (\alpha+\beta+\delta)}{\sin (\beta)}+\mathrm{r} \cdot \frac{\cos (\delta)}{\sin (\alpha)}}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \end{aligned}\tag{3-38}
The normal stress σN1 on the shear plane is:
$\ \begin{array}{left} \sigma_{\mathrm{N} 1}&=\frac{\mathrm{N}_{1} \cdot \sin (\beta)}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}} \&= \lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \frac{-\frac{\sin (\beta) \cdot \cos (\alpha+\beta+\delta)}{\sin (\beta)}+\mathrm{r} \cdot \frac{\sin (\beta) \cdot \cos (\delta)}{\sin (\alpha)}}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \end{array}\tag{3-39}$
The minimum principal stress σmin equals the normal stress in the center of the Mohr circle σC minus the radius of the Mohr circle R:
$\ \sigma_{\min }=\sigma_{\mathrm{C}}-\mathrm{R}=\sigma_{\mathrm{N} 1}+\mathrm{c} \cdot \tan (\varphi)+\sigma_{\mathrm{N} 1} \cdot \tan ^{2}(\varphi)-\frac{\mathrm{c}}{\cos (\varphi)}-\frac{\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)}{\cos (\varphi)}\tag{3-40}$
Rearranging this gives:
$\ \sigma_{\min }=\sigma_{\mathrm{N} 1} \cdot\left(1+\tan ^{2}(\varphi)-\frac{\tan (\varphi)}{\cos (\varphi)}\right)+\mathrm{c} \cdot\left(\tan (\varphi)-\frac{1}{\cos (\varphi)}\right)\tag{3-41}$
$\ \begin{array}{left} \sigma_{\min } &=\frac{\sigma_{\mathrm{N} 1}}{\cos (\varphi)} \cdot\left(\frac{\cos ^{2}(\varphi)+\sin ^{2}(\varphi)-\sin (\varphi)}{\cos (\varphi)}\right)-\mathrm{c} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right) \ &=\left(\frac{\sigma_{\mathrm{N} 1}}{\cos (\varphi)}-\mathrm{c}\right) \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right) \end{array}\tag{3-42}$
Now ductile failure will occur if the minimum principal stress σmin is bigger than then tensile strength σT, thus:
$\ \sigma_{\min }>\sigma_{\mathrm{T}}\tag{3-43}$
If equation (3-43) is true, ductile failure will occur. Keep in mind however, that the tensile strength σT is a negative number. Of course if the minimum normal stress σmin is positive, brittle tensile failure can never occur.
Substituting equation (3-39) for the normal stress on the shear plane gives the following condition for the Tear Type:
$\ \mathrm{c \cdot\left(\frac{r \cdot \frac{\sin (\beta) \cdot \cos (\delta)}{\sin (\alpha)}-\cos (\alpha+\beta+\delta)-\sin (\alpha+\beta+\delta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}\right) \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)>\sigma_{T}}\tag{3-44}$
In clay it is assumed that the internal and external friction angles are zero, while in rock it is assumed that the adhesion is zero. This will be explained in detail in the chapters on clay and rock cutting.
The ratios between the pore pressures and the cohesive shear strength, in the case of hyperbaric rock cutting, can be found according to:
$\ \begin{array}{left}\mathrm{r=\frac{a \cdot h_{b}}{c \cdot h_{i}}, r_{1}=\frac{p_{1 m} \cdot h_{i}}{c \cdot h_{i}}\text{ or }r_{1}=\frac{\rho_{w} \cdot g \cdot(z+10) \cdot h_{i}}{c \cdot h_{i}}}\ \mathrm{r_{2}=\frac{p_{2 m} \cdot h_{b}}{c \cdot h_{i}}\text{ or }r_{2}=\frac{\rho_{w} \cdot g \cdot(z+10) \cdot h_{b}}{c \cdot h_{i}}}\end{array}\tag{3-45}$
Equation (3-46) can be derived for the occurrence of tensile failure under hyperbaric conditions. Under hyperbaric conditions equation (3-46) will almost always be true, because of the terms with r1 and r2 which may become very big (positive). So tensile failure will not be considered for hyperbaric conditions.
$\ \mathrm{c}\cdot \left(\begin{array}{left}\mathrm{r\cdot\frac{sin(\beta)\cdot cos(\delta)}{sin(\alpha)}+r_2 \cdot \frac{sin(\beta)\cdot sin(\delta)}{sin(\alpha)}+r_1 \cdot sin(\alpha + \beta+\delta)}\ + \mathrm{\frac{-cos(\alpha + \beta + \delta)-sin(\alpha + \beta +\delta+\varphi)}{sin(\alpha+\beta+\delta+\varphi)}} \end{array} \right)\cdot \mathrm{\left(\frac{1-sin(\varphi)}{cos(\varphi)} \right)>\sigma_T }\tag{3-46}$
Analyzing equations (3-44) and (3-46) gives the following conclusions:
1. The first term of equations (3-44) and (3-46) is always positive.
2. If the sum of α+β+δ>π/2, in the second term of equation (3-44) and the fourth term of equation (3-46), these terms are positive, which will be the case for normal cutting angles.
3. The second and third terms of equation (3-46) are always positive.
4. The last term in equations (3-44) and (3-46) is always negative.
5. Equation (3-44) may become negative and fulfill the condition for the Tear Type.
6. Equation (3-46) will never become negative under normal conditions, so under hyperbaric conditions the Tear Type will never occur.
7. The Tear Type may occur with clay and rock under atmospheric conditions.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/03%3A_The_General_Cutting_Process/3.05%3A_The_Tear_Type_and_Chip_Type.txt
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3.6.1. The Normal and Friction Forces on the Shear Surface and Blade
On a cutter head, the blades can be divided into small elements, at which a two dimensional cutting process is considered. However, this is correct only when the cutting edge of this element is perpendicular to the direction of the velocity of the element. For most elements this will not be the case. The cutting edge and the absolute velocity of the cutting edge will not be perpendicular. This means the elements can be considered to be deviated with respect to the direction of the cutting velocity. A component of the cutting velocity perpendicular to the cutting edge and a component parallel to the cutting edge can be distinguished. This second component results in a deviation force on the blade element, due to the friction between the soil and the blade. This force is also the cause of the transverse movement of the soil, the snow plough effect.
To predict the deviation force and the direction of motion of the soil on the blade, the equilibrium equations of force will have to be solved in three directions. Since there are four unknowns, three forces and the direction of the velocity of the soil on the blade, one additional equation is required. This equation follows from an equilibrium equation of velocity between the velocity of grains in the shear zone and the velocity of grains on the blade. Since the four equations are partly non-linear and implicit, they have to be solved iteratively. The results of solving these equations have been compared with the results of laboratory tests on sand. The correlation between the two was very satisfactory, with respect to the magnitude of the forces and with respect to the direction of the forces and the flow of the soil on the blade.
Although the normal and friction forces as shown in Figure 3-14 are the basis for the calculation of the horizontal and vertical cutting forces, the approach used, requires the following equations to derive these forces by using equations (3-8) and (3-9). The index 1 points to the shear surface, while the index 2 points to the blade.
$\ \mathrm{F}_{\mathrm{n} 1}=\mathrm{F}_{\mathrm{h}} \cdot \sin (\beta)-\mathrm{F}_{\mathrm{v}} \cdot \cos (\beta)\tag{3-47}$
$\ \mathrm{F}_{\mathrm{f} 1}=\mathrm{F}_{\mathrm{h}} \cdot \cos (\beta)+\mathrm{F}_{\mathrm{v}} \cdot \sin (\beta)\tag{3-48}$
$\ \mathrm{F}_{\mathrm{n} 2}=\mathrm{F}_{\mathrm{h}} \cdot \sin (\alpha)+\mathrm{F}_{\mathrm{v}} \cdot \cos (\alpha)\tag{3-49}$
$\ \mathrm{F_{f 2}=F_{h} \cdot \cos (\alpha)-F_{v} \cdot \sin (\alpha)}\tag{3-50}$
3.6.2. The 3D Cutting Theory
The previous paragraphs summarized the two-dimensional cutting theory. However, as stated in the introduction, in most cases the cutting process is not two-dimensional, because the drag velocity is not perpendicular to the cutting edge of the blade. Figure 3-16 shows this phenomenon. As with snow-ploughs, the soil will flow to one side while the blade is pushed to the opposite side. This will result in a third cutting force, the deviation force Fd. To determine this force, the flow direction of the soil has to be known. Figure 3-17 shows a possible flow direction.
3.6.3. Velocity Conditions
For the velocity component perpendicular to the blade vc, if the blade has a deviation angle $\ \iota$ and a drag velocity vd according to Figure 3-17, it yields:
$\ \mathrm{v}_{\mathrm{c}}=\mathrm{v}_{\mathrm{d}} \cdot \cos (\mathrm{\iota})\tag{3-51}$
The velocity of grains in the shear surface perpendicular to the cutting edge vr1 is now:
$\ \mathrm{v}_{\mathrm{r} 1}=\mathrm{v}_{\mathrm{c}} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)}\tag{3-52}$
The relative velocity of grains with respect to the blade vr2, perpendicular to the cutting edge is:
$\ \mathrm{v}_{\mathrm{r} 2}=\mathrm{v}_{\mathrm{c}} \cdot \frac{\sin (\beta)}{\sin (\alpha+\beta)}\tag{3-53}$
The grains will not only have a velocity perpendicular to the cutting edge, but also parallel to the cutting edge, the deviation velocity components vd1 on the shear surface and vd2 on the blade.
The velocity components of a grain in xand direction can be determined by considering the absolute velocity of grains in the shear surface, this leads to:
$\ \overrightarrow{\mathrm{v}}_{\mathrm{r} 2}+\overrightarrow{\mathrm{v}}_{\mathrm{d} 2}+\overrightarrow{\mathrm{v}}_{\mathrm{d}}=\overrightarrow{\mathrm{v}}_{\mathrm{r} 1}+\overrightarrow{\mathrm{v}}_{\mathrm{d} 1}\tag{3-54}$
$\ \mathrm{v}_{\mathrm{x} 1}=\mathrm{v}_{\mathrm{r} 1} \cdot \cos (\beta) \cdot \cos (\mathfrak{\iota})+\mathrm{v}_{\mathrm{d} 1} \cdot \sin (\iota)\tag{3-55}$
$\ \mathrm{v}_{\mathrm{y} 1}=\mathrm{v}_{\mathrm{r} 1} \cdot \cos (\beta) \cdot \sin (\mathrm{\iota})-\mathrm{v}_{\mathrm{d} 1} \cdot \cos (\mathrm{\iota})\tag{3-56}$
$\ \mathrm{v}_{\mathrm{z 1}}=\mathrm{v}_{\mathrm{r} \mathrm{1}} \cdot \sin (\beta)\tag{3-57}$
The velocity components of a grain can also be determined by a summation of the drag velocity of the blade and the relative velocity between the grains and the blade, this gives:
$\ \mathrm{v}_{\mathrm{x} 2}=\mathrm{v}_{\mathrm{d}}-\mathrm{v}_{\mathrm{r} 2} \cdot \cos (\alpha) \cdot \cos (\iota)-\mathrm{v}_{\mathrm{d} 2} \cdot \sin (\mathfrak{\iota})\tag{3-58}$
$\ \mathrm{v}_{\mathrm{y} 2}=-\mathrm{v}_{\mathrm{r} 2} \cdot \cos (\alpha) \cdot \sin ({\iota})+\mathrm{v}_{\mathrm{d} 2} \cdot \cos (\iota)\tag{3-59}$
$\ \mathrm{v}_{\mathrm{z} 2}=\mathrm{v}_{\mathrm{r} 2} \cdot \sin (\alpha)\tag{3-60}$
Since both approaches will have to give the same resulting velocity components, the following condition for the transverse velocity components can be derived:
$\ \mathrm{v}_{\mathrm{x} 1}=\mathrm{v}_{\mathrm{x} 2} \Longrightarrow \mathrm{v}_{\mathrm{d} 1}+\mathrm{v}_{\mathrm{d} 2}=\mathrm{v}_{\mathrm{d}} \cdot \sin (\mathfrak{\iota})\tag{3-61}$
$\ \mathrm{v}_{\mathrm{y} 1}=\mathrm{v}_{\mathrm{y} 2} \Longrightarrow \mathrm{v}_{\mathrm{d} 1}+\mathrm{v}_{\mathrm{d} 2}=\mathrm{v}_{\mathrm{d}} \cdot \sin (\mathrm{\iota})\tag{3-62}$
$\ \mathrm{v}_{\mathrm{z} \mathrm{1}}=\mathrm{v}_{\mathrm{z} \mathrm{2}}\tag{3-63}$
3.6.4. The Deviation Force
Since a friction force always has a direction matching the direction of the relative velocity between two bodies, the fact that a deviation velocity exists on the shear surface and on the blade, implies that also deviation forces must exist. To match the direction of the relative velocities, the following equations can be derived for the deviation force on the shear surface and on the blade (Figure 3-18):
$\ \mathrm{F}_{\mathrm{d} 1}=\mathrm{F}_{\mathrm{f} 1} \cdot \frac{\mathrm{v}_{\mathrm{d} 1}}{\mathrm{v}_{\mathrm{r} 1}}\tag{3-64}$
$\ \mathrm{F}_{\mathrm{d} 2}=\mathrm{F}_{\mathrm{f} 2} \cdot \frac{\mathrm{v}_{\mathrm{d} 2}}{\mathrm{v}_{\mathrm{r} 2}}\tag{3-65}$
Since perpendicular to the cutting edge, an equilibrium of forces exists, the two deviation forces must be equal in magnitude and have opposite directions.
$\ \mathrm{F}_{\mathrm{d} \mathrm{1}}=\left|\mathrm{F}_{\mathrm{d} 2}\right|\tag{3-66}$
By substituting equations (3-64) and (3-65) in equation (3-66) and then substituting equations (3-48) and (3-50) for the friction forces and equations (3-52) and (3-53) for the relative velocities, the following equation can be derived, giving a second relation between the two deviation velocities:
$\ \lambda=\frac{\mathrm{v}_{\mathrm{d} 1}}{\mathrm{v}_{\mathrm{d} 2}}=\left(\frac{\mathrm{F}_{\mathrm{f} 2}}{\mathrm{F}_{\mathrm{f} 1}}\right) \cdot\left(\frac{\mathrm{v}_{\mathrm{r} 1}}{\mathrm{v}_{\mathrm{r} 2}}\right)=\left(\frac{\mathrm{F}_{\mathrm{h}} \cdot \cos (\alpha)-\mathrm{F}_{\mathrm{v}} \cdot \sin (\alpha)}{\mathrm{F}_{\mathrm{h}} \cdot \cos (\beta)+\mathrm{F}_{\mathrm{v}} \cdot \sin (\beta)}\right) \cdot\left(\frac{\sin (\alpha)}{\sin (\beta)}\right)\tag{3-67}$
To determine Fh and Fv perpendicular to the cutting edge, the angle of internal friction φe and the external friction angle δe mobilized perpendicular to the cutting edge, have to be determined by using the ratio of the transverse velocity and the relative velocity, according to:
$\ \tan \left(\varphi_{\mathrm{e}}\right)=\tan (\varphi) \cdot \cos \left(\operatorname{atn}\left(\frac{\mathrm{v}_{\mathrm{d} 1}}{\mathrm{v}_{\mathrm{r} 1}}\right)\right)\tag{3-68}$
$\ \tan \left(\delta_{\mathrm{e}}\right)=\tan (\delta) \cdot \cos \left(\operatorname{atn}\left(\frac{\mathrm{v}_{\mathrm{d} 2}}{\mathrm{v}_{\mathrm{r}} 2}\right)\right)\tag{3-69}$
For the cohesion and the adhesion this gives:
$\ \mathrm{c}_{\mathrm{e}}=\mathrm{c \cdot cos ( \mathrm { a t n } ( \frac { \mathrm { v } _ { \mathrm { d } 1 } } { \mathrm { v } _ { \mathrm { r } 1 } } ) )}\tag{3-70}$
$\ \mathrm{a}_{\mathrm{e}}=\mathrm{a} \cdot \cos \left(\operatorname{atn}\left(\frac{\mathrm{v}_{\mathrm{d} 2}}{\mathrm{v}_{\mathrm{r} 2}}\right)\right)\tag{3-71}$
3.6.5. The Resulting Cutting Forces
The resulting cutting forces in xand direction can be determined once the deviation velocity components are known. However, it can be seen that the second velocity condition equation (3-67) requires the horizontal and vertical cutting forces perpendicular to the cutting edge, while these forces can only be determined if the mobilized internal and external friction angles and the mobilized cohesion and adhesion (equations (3-68), (3-69), (3-70) and (3-71)) are known. This creates an implicit set of equations that will have to be solved by means of an iteration process. For the cutting forces on the blade the following equation can be derived:
$\ \mathrm{F}_{\mathrm{x} 2}=\mathrm{F}_{\mathrm{h}} \cdot \cos (\mathrm{\iota})+\mathrm{F}_{\mathrm{d} 2} \cdot \sin (\mathrm{\iota})\tag{3-72}$
$\ \mathrm{F}_{\mathrm{y} 2}=\mathrm{F}_{\mathrm{h}} \cdot \sin ({\iota})-\mathrm{F}_{\mathrm{d} 2} \cdot \cos (\iota)\tag{3-73}$
$\ \mathrm{F}_{\mathrm{z} 2}=\mathrm{F}_{\mathrm{v}}\tag{3-74}$
The problem of the model being implicit can be solved in the following way:
A new variable λ is introduced in such a way that:
$\ \mathrm{v_{d 1}=\frac{\lambda}{1+\lambda} \cdot v_{d} \cdot \sin (\iota)}\tag{3-75}$
$\ \mathrm{v}_{\mathrm{d} 2}=\frac{\mathrm{1}}{\mathrm{1}+\lambda} \cdot \mathrm{v}_{\mathrm{d}} \cdot \sin (\iota)\tag{3-76}$
This satisfies the condition from equations (3-61) and (3-62) for the sum of these 2 velocities:
$\ \mathrm{v}_{\mathrm{d} 1}+\mathrm{v}_{\mathrm{d} 2}=\mathrm{v}_{\mathrm{d}} \cdot \sin (\iota)\tag{3-77}$
The procedure starts with a starting value for λ=1. Based on the velocities found with equations (3-75), (3-76), (3-52) and (3-53), the mobilized internal φe and external δfriction angles and the cohesion ce and adhesion ae can be determined using the equations (3-68), (3-69), (3-70) and (3-71). Once these are known, the horizontal Fh and vertical Fv cutting forces in the plane perpendicular to the cutting edge can be determined with equations (3-8) and (3-9). With the equations (3-48), (3-50), (3-64) and (3-65) the friction and deviation forces on the blade and the shear plane can be determined. Now with equation (3-67) the value of the variable λ can be determined and if the starting value is correct, this value should be found. In general this will not be the case after one iteration. But repeating this procedure 3 or 4 times should give enough accuracy.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/03%3A_The_General_Cutting_Process/3.06%3A_The_Snow_Plough_Effect.txt
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The unknown in all the mechanisms is the shear angle β. With the assumption that nature will choose the mechanism configuration that requires the least energy and this energy equals the horizontal force Fh times the cutting velocity vtimes time, the shear angle β should be chosen where the horizontal force Fh is at a minimum. In some cases an analytical solution exists by taking the derivative of the horizontal force Fh with respect to the shear angle β and making it equal to zero. The second derivative has to be positive in this case. In other cases it is more convenient to determine the minimum numerically. This minimum value depends strongly on the blade angle α and the blade height – layer thickness ratio hb/hi. This minimum also depends strongly on the soil properties and thus the type of soil. Different soils will have shear angles in a different range. Different cutting mechanisms will also have shear angles in different ranges. For saturated sand with blade angles α from 30° to 60°, the shear angle β will range from 30° to 20°. For clay, the shear angle depends strongly on the ratio of the adhesion to the cohesion. For very strong clays with a low relative adhesion the shear angle can be in the range of 60° to 75° for blade angles α from 30° to 60°. For soft clays with a high relative adhesion the shear angle is much smaller, from 30° to 40°. In general one can say that the shear angle decreases with increasing blade angle, internal/external friction angle and adhesion.
The criterion of least energy is arbitrary but reasonable. Other criteria may be applied to find the shear angle. Also other mechanisms may be applied leading to slightly different shear angles. In this book it is assumed that the shear plane is a straight line, which is questionable. First of all, the shear plane does not have to be a line without thickness. An area with a certain thickness is also possible. Secondly, the shape of the shear plane could be curved, like a circle segment. The advantage of the approach chosen is, that one can compare the different mechanisms more easily and the models derived give more insight in the basic parameters.
3.09: Specific Cutting Energy Esp
In the dredging industry, the specific cutting energy is described as:
The amount of energy, that has to be added to a volume unit of soil (e.g. sand, clay or rock) to excavate the soil.
The dimension of the specific cutting energy is: kN/m2 or kPa for sand and clay, while for rock often MN/m2 or MPa is used.
For the case as described above, cutting with a straight blade, the specific cutting energy can be written as:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{3-78}$
So the specific cutting energy equals the cutting power divided by the cutting volumetric production. Once the specific cutting energy is known and the installed cutting power is known, this can be used to determine the theoretical cutting production according to:
$\ \mathrm{Q}_{\mathrm{c}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{E}_{\mathrm{s p}}}\tag{3-79}$
It should be noted here that there may be other factors limiting the production, like the hydraulic transport system of a cutter suction dredge, the throughput between the blades of a cutter head or the capacity of the swing winches.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/03%3A_The_General_Cutting_Process/3.08%3A_Finding_the_Shear_Angle.txt
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a1, a2
Coefficients for weighted permeability
-
a, $\ \tau_\mathrm{a}$
Adhesion or external shear strength
kPa
A
Adhesive force on the blade
kN
c, $\ \tau_\mathrm{c}$
Cohesion or internal shear strength
kPa
C, C1
Force due to cohesion in the shear plane
kN
C2
Force due to cohesion on the front of the wedge
kN
C3
Force due to cohesion at the bottom of the wedge
kN
Fh
Horizontal cutting force
kN
Ff1
Friction force on the shear surface
kN
Ff2
Friction force on the blade
kN
Fn1
Normal force on the shear surface
kN
Fn2
Normal force on the blade
kN
Fv
Vertical cutting force
kN
Fd1
Deviation force on the shear surface
kN
Fd, d2
Deviation force on the blade
kN
Fx1, 2
Cutting force in x-direction
kN
Fy1, 2
Cutting force in y-direction
kN
Fz1, 2
Cutting force in z-direction
kN
g
Gravitational constant (9.81)
m/s2
G, G1
Gravitational force on the layer cut
kN
G2
Gravitational force on the wedge
kN
hi
Initial thickness of layer cut
m
hb
Height of blade
m
hb,m
Effective height of the blade in case Curling Type
m
I
Inertial force on the shear plane
kN
ki
Initial permeability
m/s
kmax
Maximum permeability
m/s
km
Average permeability
m/s
K1
Grain force on the shear plane
kN
K2
Grain force on the blade or the front of the wedge
kN
K3
Grain force on the bottom of the wedge
kN
K4
Grain force on the blade (in case a wedge exists)
kN
ni
Initial porosity
%
nmax
Maximum porosity
%
N1
Normal force on the shear plane
kN
N2
Normal force on the blade or the front of the wedge
kN
N3
Normal force on the bottom of the wedge
kN
N4
Normal force on the blade (in case a wedge exists)
kN
p1m
Average pore pressure on the shear surface
kPa
p2m
Average pore pressure on the blade
kPa
Pc
Cutting power
kW
R1
Acting point of resulting forces on the shear plane
m
R2
Acting point of resulting forces on the blade
m
R3
Acting point of resulting forces on the bottom of the wedge
m
R4
Acting point of resulting forces on the blade (in case a wedge exists)
m
S1
Shear force due to friction on the shear plane
kN
S2
Shear force due to friction on the blade or the front of the wedge
kN
S3
Shear force due to friction at the bottom of the wedge
kN
S4
Shear force due to friction on the blade (in case a wedge exists)
kN
vc
Cutting velocity component perpendicular to the blade
m/s
vd
Cutting velocity, drag velocity
m/s
vr1
Velocity of grains in the shear surface
m/s
vr2
Relative velocity of grains on the blade
m/s
vd1
Deviation velocity of grains in the shear surface
m/s
vd2
Deviation velocity of grains on the blade
m/s
vx1,2
Velocity of grains in the x-direction
m/s
vy1,2
Velocity of grains in the y-direction
m/s
vz1,2
Velocity of grains in the z-direction
m/s
W Width of blade m
W1 Force resulting from pore under pressure on the shear plane
kN
W2 Force resulting from pore under pressure on the blade/ front wedge
kN
W3 Force resulting from pore under pressure on the bottom of the wedge kN
W4 Force resulting from pore under pressures on the blade, wedge kN
Z Water depth m
$\ \alpha$
Cutting angle blade
rad
β
Shear angle
rad
ε
Dilatation
-
φ
Angle of internal friction
rad
φe
Angle of internal friction perpendicular to the cutting edge
rad
λs
Strengthening factor
-
λ1
Acting point factor for resulting forces on the shear plane
-
λ2
Acting point factor for resulting forces on the blade or front of wedge
-
λ3
Acting point factor for resulting forces on the bottom of the wedge
-
λ
Acting point factor for resulting forces on the blade
-
δ
External friction angle
rad
δe
External friction angle perpendicular to the cutting edge
rad
$\ \iota$
Deviation angle blade
rad
ρs
Density of the soil
ton/m3
ρl
Density water
ton/m3
θ
Wedge angle
rad
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/03%3A_The_General_Cutting_Process/3.10%3A_Nomenclature.txt
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In dry sand the cutting processes are governed by gravity and by inertial forces. Pore pressure forces, cohesion and adhesion are not present or can be neglected. Internal and external friction are present. The cutting process is of the Shear Type with discrete shear planes, but this can be modeled as the Flow Type, according to Merchant (1944). This approach will give an estimate of the maximum cutting forces. The average cutting forces may be 30%-50% of the maximum cutting forces.
Dry sand cutting is dominated by gravitational and inertial forces and by the internal and external friction angles. The cutting mechanism is the Shear Type. This is covered in Chapter 5: Dry Sand Cutting.
The forces K1 and K2 on the blade, chisel or pick point are now:
\ \begin{aligned} \mathrm{K}_{1}&= \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \end{aligned}\tag{4-1}
\ \begin{aligned} \mathrm{K}_{2}&= \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \end{aligned}\tag{4-2}
The normal forces N1 on the shear plane and N2 on the blade are:
$\ \mathrm{N}_{1}=\mathrm{K}_{1} \cdot \cos (\varphi) \quad\text{ and }\quad \mathrm{N}_{2}=\mathrm{K}_{2} \cdot \cos (\delta)\tag{4-3}$
The horizontal and vertical forces on the blade, chisel or pick point are:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{2} \cdot \sin (\alpha)+\mathrm{K}_{2} \cdot \sin (\alpha+\delta)+\mathrm{A} \cdot \cos (\alpha)\tag{4-4}$
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{4-5}$
The equilibrium of moments around the blade tip is:
$\ \mathrm{\left(N_{1}-W_{1}\right) \cdot R_{1}-G \cdot R_{3}=\left(N_{2}-W_{2}\right) \cdot R_{2}}\tag{4-6}$
Analyzing these equations results in the following conclusions:
• Since the argument in the cosine of the inertial term in the force K1 is always greater than 90 degrees, the cosine is negative and the term as a whole is positive. This results in positive forces on the blade, chisel or pick point and also positive normal forces.
• There are no forces directly proportional to the (mobilized) blade height or the length of the shear plane, so the equilibrium of moments does not play a role. The Curling Type and the Tear Type will not occur. The acting points of the forces R1 and R2 will be adjusted by nature to form an equilibrium of moments.
• When the argument of the sine in the denominator gets close to 180 degrees, the forces become very large. If the argument is greater than 180 degrees, the forces would become negative. Since both conditions will not happen in nature, nature will find another cutting mechanism, the wedge mechanism.
4.02: Cutting Water Saturated Sand
From literature it is known that, during the cutting process, the sand increases in volume. This increase in volume is accredited to dilatancy. Dilatancy is the change of the pore volume as a result of shear in the sand. This increase of the pore volume has to be filled with water. The flowing water experiences a certain resistance, which causes sub-pressures in the pore water in the sand. As a result the grain stresses increase and therefore the required cutting forces. The rate of the increase of the pore volume in the dilatancy zone, the volume strain rate, is proportional to the cutting velocity. If the volume strain rate is high, there is a chance that the pore pressure reaches the saturated water vapor pressure and cavitation occurs. A further increasing volume strain rate will not be able to cause a further decrease of the pore pressure. This also implies that, with a further increasing cutting velocity, the cutting forces cannot increase as a result of the dilatancy properties of the sand. The cutting forces can, however, still increase with an increasing cutting velocity as a result of the inertia forces and the flow resistance.
The cutting process can be subdivided in 5 areas in relation with the cutting forces:
• Very low cutting velocities, a quasi-static cutting process. The cutting forces are determined by gravitation.
• The volume strain rate is high in relation to the permeability of the sand. The volume strain rate is however so small that inertia forces can be neglected. The cutting forces are dominated by the dilatancy properties of the sand.
• A transition region, with local cavitation. With an increasing volume strain rate, the cavitation area will increase so that the cutting forces increase slightly as a result of dilatancy.
• Cavitation occurs almost everywhere around and on the blade. The cutting forces do not increase anymore as a result of the dilatancy properties of the sand.
• Very high cutting velocities. The inertia forces part in the total cutting forces can no longer be neglected but form a substantial part.
Under normal conditions in dredging, the cutting process in sand will be governed by the effects of dilatation. Gravity, inertia, cohesion and adhesion will not play a role. Internal and external friction are present.
Saturated sand cutting is dominated by pore vacuum pressure forces and by the internal and external friction angles. The cutting mechanism is the Shear Type. This is covered in Chapter 6: Saturated Sand Cutting.
The forces K1 and K2 on the blade, chisel or pick point are now:
$\ \begin{array}{left} \mathrm{K}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{4-7}$
$\ \begin{array}{left} \mathrm{K}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{4-8}$
The normal forces N1 on the shear plane and N2 on the blade are:
$\ \mathrm{N}_{\mathrm{1}}=\mathrm{K}_{1} \cdot \cos (\varphi) \quad\text{ and }\quad \mathrm{N}_{2}=\mathrm{K}_{2} \cdot \cos (\delta)\tag{4-9}$
The horizontal and vertical forces on the blade, chisel or pick point are:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{2} \cdot \sin (\alpha)+\mathrm{K}_{2} \cdot \sin (\alpha+\delta)+\mathrm{A} \cdot \cos (\alpha)\tag{4-10}$
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{4-11}$
The equilibrium of moments around the blade tip is:
$\ \left(\mathrm{N}_{1}-\mathrm{W}_{1}\right) \cdot \mathrm{R}_{1}-\mathrm{G} \cdot \mathrm{R}_{3}=\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{R}_{2}\tag{4-12}$
Analyzing these equations results in the following conclusions:
• The pore pressure forces W1 and W2 are limited by the occurrence of cavitation.
• All the terms are positive, resulting in positive forces on the blade and also positive normal forces.
• In the non-cavitating case the pore pressure forces are related to the (mobilized) blade height or the length of the shear plane. In the cavitating case the pore pressure forces are proportional to the (mobilized) blade height or the length of the shear plane. Theoretically the Curling Type and the Tear Type may occur. This has however never been observed with in dredging normal blade heights and layer thicknesses.
• When the argument of the sine in the denominator gets close to 180 degrees, the forces become very large. If the argument is greater than 180 degrees, the forces would become negative. Since both conditions will not happen in nature, nature will find another cutting mechanism, the wedge mechanism.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/04%3A_Which_Cutting_Mechanism_for_Which_Kind_of_Soil/4.01%3A_Cutting_Dry_Sand.txt
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In clay the cutting processes are dominated by cohesion and adhesion (internal and external shear strength). Because of the φ=0 concept, the internal and external friction angles are set to 0. Gravity, inertial forces and pore pressures are also neglected. This simplifies the cutting equations. Clay however is subject to strengthening, meaning that the internal and external shear strength increase with an increasing strain rate.
The reverse of strengthening is creep, meaning that under a constant load the material will continue deforming with a certain strain rate. Under normal circumstances clay will be cut with the Flow Type mechanism, but under certain circumstances the Curling Type or the Tear Type may occur. The Curling Type will occur when the blade height is large with respect to the layer thickness, hb/hi, the adhesion is high compared to the cohesion a/cand the blade angle α is relatively big. The Tear Type will occur when the blade height is small with respect to the layer thickness, hb/hi, the adhesion is small compared to the cohesion a/c and the blade angle α is relatively small.
Clay cutting is dominated by cohesive (internal shear strength) and adhesive (external shear strength) forces. The basic cutting mechanism is the Flow Type. Cutting a thin layer, combined with a high adhesive force may result in the Curling Type mechanism. Cutting a thick layer combined with a small adhesive force and a low tensile strength may result in the Tear Type mechanism. This is covered in Chapter 7: Clay Cutting.
The forces K1 and K2 on the blade, chisel or pick point are now:
$\ \begin{array}{left} \mathrm{K}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{4-13}$
\ \begin{aligned} \mathrm{K}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \end{aligned}\tag{4-14}
The normal forces N1 on the shear plane and N2 on the blade are:
$\ \mathrm{N}_{1}=\mathrm{K}_{1} \cdot \cos (\varphi) \quad\text{ and }\quad \mathrm{N}_{2}=\mathrm{K}_{2} \cdot \cos (\delta)\tag{4-15}$
The horizontal and vertical forces on the blade, chisel or pick point are:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{2} \cdot \sin (\alpha)+\mathrm{K}_{2} \cdot \sin (\alpha+\delta)+\mathrm{A} \cdot \cos (\alpha)\tag{4-16}$
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{4-17}$
The equilibrium of moments around the blade tip is:
$\ \left(\mathrm{N}_{1}-\mathrm{W}_{1}\right) \cdot \mathrm{R}_{1}-\mathrm{G} \cdot \mathrm{R}_{3}=\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{R}_{2}\tag{4-18}$
Analyzing these equations results in the following conclusions:
• At normal cutting angles in dredging, the argument of the cosine in the cohesive term of Kis greater than 90 degrees, resulting in a small positive term as a whole. Together with the adhesive term, this gives a positive normal stress on the shear plane. The minimum normal stress however equals the normal stress on the shear plane, minus the radius of the Mohr circle, which is the cohesion. The result may be a negative minimum normal stress. If this negative minimum normal stress is smaller than the negative tensile strength, the Tear Type will occur. This occurrence depends on the ratio between the adhesive force to the cohesive force. A large ratio will suppress the Tear Type.
• The adhesive force on the blade is proportional to the (mobilized) length of the blade, so the Curling Type may occur. The cohesive force on the shear plane is proportional to the (mobilized) cohesion, so the Tear Type may occur. The occurrence of the Curling Type or Tear Type depends on the ratio of the adhesive force to the cohesive force. A large ratio results in the Curling Type, a small ratio in the Tear Type.
• When the argument of the sine in the denominator gets close to 180 degrees, the forces become very large. If the argument is greater than 180 degrees, the forces would become negative. Since both conditions will not happen in nature, nature will find another cutting mechanism, the wedge mechanism. In clay this is not likely to occur, since there are only two angles in the argument of the sine in the denominator. It would require very large blade angles to occur.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/04%3A_Which_Cutting_Mechanism_for_Which_Kind_of_Soil/4.03%3A_Cutting_Clay.txt
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Rock is the collection of materials where the grains are bonded chemically from very stiff clay, sandstone to very hard basalt. It is difficult to give one definition of rock or stone and also the composition of the material can differ strongly. Still it is interesting to see if the model used for sand and clay, which is based on the Coulomb model, can be used for rock as well. Typical parameters for rock are the compressive strength UCS and the tensile strength BTS and specifically the ratio between those two, which is a measure for how fractured the rock is. Rock also has shear strength and because it consists of bonded grains it will have an internal friction angle and an external friction angle. It can be assumed that the permeability of the rock is very low, so initially the pore pressures do no play a role under atmospheric conditions. Since the absolute hydrostatic pressure, which would result in a cavitation under pressure of the same magnitude can be neglected with respect to the compressive strength of the rock; the pore pressures are usually neglected. This results in a material where gravity, inertia, pore pressures and adhesion can be neglected.
Rock cutting under atmospheric conditions (normal dredging) is dominated by the internal shear strength and by the internal and external friction angles. The main cutting mechanism is the Chip Type, brittle cuttingCutting a very thin layer or using large blade angles may result in the Crushed Type. This is covered in Chapter 8: Rock Cutting: Atmospheric Conditions.
The forces K1 and K2 on the blade, chisel or pick point are now:
\ \begin{aligned} \mathrm{K}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \end{aligned}\tag{4-19}
$\ \begin{array}{left} \mathrm{K}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{4-20}$
The normal forces N1 on the shear plane and N2 on the blade are:
$\ \mathrm{N}_{\mathrm{1}}=\mathrm{K}_{1} \cdot \cos (\varphi) \quad\text{ and }\quad \mathrm{N}_{2}=\mathrm{K}_{2} \cdot \cos (\delta)\tag{4-21}$
The horizontal and vertical forces on the blade, chisel or pick point are:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{\mathrm{2}} \cdot \sin (\alpha)+\mathrm{K}_{2} \cdot \sin (\alpha+\delta)+\mathrm{A} \cdot \cos (\alpha)\tag{4-22}$
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{4-23}$
The equilibrium of moments around the blade tip is:
$\ \left(\mathrm{N}_{1}-\mathrm{W}_{1}\right) \cdot \mathrm{R}_{1}-\mathrm{G} \cdot \mathrm{R}_{3}=\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{R}_{2}\tag{4-24}$
Analyzing these equations results in the following conclusions:
• Since the argument in the cosine of the cohesive term in the force K1 is always greater than 90 degrees, the cosine is negative and the term as a whole is positive. This results in positive forces on the blade, chisel or pick point and also positive normal forces. The minimum normal stress however equals the normal stress on the shear plane, minus about the radius of the Mohr circle. The result may be a negative minimum normal stress. If this negative minimum normal stress is smaller than the negative tensile strength, brittle tensile failure will occur. Otherwise brittle shear failure will occur. In both cases the forces calculated are peak forces. The average forces are somewhere between 50% and 100% of the peak forces.
• On the blade the normal stresses are always high enough to avoid the occurrence of the Curling Type. In fact the forces on the blade do not depend on the length of the blade. The cohesive force on the shear plane however depends on the (mobilized) cohesion or shear strength, so the Tear Type, here named the Chip Type may occur.
• When the argument of the sine in the denominator gets close to 180 degrees, the forces become very large. If the argument is greater than 180 degrees, the forces would become negative. Since both conditions will not happen in nature, nature will find another cutting mechanism, the wedge mechanism.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/04%3A_Which_Cutting_Mechanism_for_Which_Kind_of_Soil/4.04%3A_Cutting_Rock_Atmospheric.txt
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In the case of hyperbaric rock cutting, the pore pressures cannot be neglected anymore. Gravity and inertial forces can still be neglected. Usually rock has no adhesion. When the hydrostatic pressure is larger than or approaching the UCS value of the rock, the rock tends to fail in a semi-ductile manner, named cataclastic failure. It is almost like the hydrostatic pressure can be added to the tensile strength of the rock.
Rock cutting under hyperbaric conditions (deep sea mining) is dominated by the internal shear strength, the pore vacuum pressure forces and by the internal and external friction angles. The main cutting mechanism is the Crushed Type, cataclastic semi-ductile cuttingThis is covered in Chapter 9: Rock Cutting: Hyperbaric Conditions.
The forces K1 and K2 on the blade, chisel or pick point are now:
$\ \begin{array}{left} \mathrm{K}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)+\mathrm{G} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{4-25}$
$\ \begin{array}{left} \mathrm{K}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G} \cdot \sin (\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \ &+\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \end{array}\tag{4-26}$
The normal forces N1 on the shear plane and N2 on the blade are:
$\ \mathrm{N}_{1}=\mathrm{K}_{1} \cdot \cos (\varphi) \quad\text{ and }\quad \mathrm{N}_{2}=\mathrm{K}_{2} \cdot \cos (\delta)\tag{4-27}$
The horizontal and vertical forces on the blade, chisel or pick point are:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{2} \cdot \sin (\alpha)+\mathrm{K}_{2} \cdot \sin (\alpha+\delta)+\mathrm{A} \cdot \cos (\alpha)\tag{4-28}$
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{4-29}$
The equilibrium of moments around the blade tip is:
$\ \left(\mathrm{N}_{1}-\mathrm{W}_{1}\right) \cdot \mathrm{R}_{1}-\mathrm{G} \cdot \mathrm{R}_{3}=\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{R}_{2}\tag{4-30}$
Analyzing these equations results in the following conclusions:
• Since the argument in the cosine of the cohesive term in the force K1 is always greater than 90 degrees, the cosine is negative and the term as a whole is positive. This results in positive forces on the blade, chisel or pick point and also positive normal forces. The minimum normal stress however equals the normal stress on the shear plane, minus about the radius of the Mohr circle. The result will always be a positive minimum normal stress because of the influence of the large pore under pressure forces.
• On the blade the Curling Type may occur based on the equilibrium of moments, because the pore pressure force depends on the length of the blade. When cutting a very thin layer of rock, compared to the height of the blade, as in oil drilling, this will occur. On the shear plane, both the pore pressure force and the cohesive force depend on the length of the shear plane, which may result in brittle tensile failure, the Tear Type, here named the Chip Type. Usually this is suppressed by the very large pore under pressures in relation to the strength of the rock.
• When the argument of the sine in the denominator gets close to 180 degrees, the forces become very large. If the argument is greater than 180 degrees, the forces would become negative. Since both conditions will not happen in nature, nature will find another cutting mechanism, the wedge mechanism.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/04%3A_Which_Cutting_Mechanism_for_Which_Kind_of_Soil/4.05%3A_Cutting_Rock_Hyperbaric.txt
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The cutting forces for sand, clay and rock can be described by a generic equation, where a number of terms dominate for each individual type of soil. Here dry sand, water saturated sand, clay, atmospheric rock and hyperbaric rock are distinguished. The influences of the different forces for each type of soil are summarized in Table 4-1. The general cutting mechanism is the Flow Type, which is mathematically the same as the Shear Type and the Crushed Type. If the forces on the blade depend on the length of the blade, such as the adhesive force or the pore under pressure force W2, the Curling Type may occur if the layer thickness is very small compared to the blade length. A mobilized blade length (height) is introduced. If the forces on the shear plane depend on the length of the shear plane, such as the cohesive force and the pore under pressure force W1, the Tear Type (or Chip Type) may occur if the layer thickness is large compared to the blade length. A mobilized shear strength is introduced.
However there may also be mixed soils like clay mixed with sand, resulting in a clay with internal friction. Or sand mixed with clay, resulting in a very low permeability. For clay the ratio of the adhesion to the cohesion is very important and little is known about this. Very weak clays may have an adhesion almost equal to the cohesion, but when the cohesion increases the ratio between adhesion and cohesion decreases. A 100 kPa clay may have an adhesion of just 5-10 kPa. For even harder clays the adhesion may drop to zero. The harder clays however seem to have some internal and external friction, increasing with the strength of the clay. A new topic is the cutting of permafrost, frozen clay. From preliminary research it appears that permafrost behaves more like rock, but how exactly is still a question. Future research will give an answer to these questions and hopefully the generic equations will also be applicable for these soils.
Table 4-1: The influences for each type of soil.
Gravity
Inertia
Pore Pressure
Cohesion
Adhesion
Friction
Dry sand
Saturated sand
Clay
Atmospheric rock
Hyperbaric rock
4.07: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or external shear strength
kPa
A
Adhesive force on the blade
kN
c, $\ \tau_\mathrm{c}$
Cohesion or internal shear strength
kPa
cm
Mobilized cohesion in case of Tear Type or Chip Type
m
C
Force due to cohesion in the shear plane
kN
Fh
Horizontal cutting force
kN
Fv
Vertical cutting force
kN
g
Gravitational constant (9.81)
m/s2
G
Gravitational force on the layer cut
kN
hi
Initial thickness of layer cut
m
hb
Height of blade
m
hb,m
Mobilized height of the blade in case Curling Type
m
I
Inertial force on the shear plane
kN
K1
Grain force on the shear plane
kN
K2
Grain force on the blade
kN
N1
Normal force on the shear plane
kN
N2
Normal force on the blade
kN
R1
Acting point forces on the shear plane
m
R2
Acting point forces on the blade
m
R3
Acting point gravity force
m
S1
Shear force due to friction on the shear plane
kN
S2
Shear force due to friction on the blade or the front of the wedge
kN
vc
Cutting velocity
m/s
w
Width of blade
m
W1
Force resulting from pore under pressure on the shear plane
kN
W2
Force resulting from pore under pressure on the blade/ front wedge
kN
z
Water depth
m
$\ \alpha$
Cutting angle blade
rad
β
Shear angle
rad
φ
Angle of internal friction
rad
δ
External friction angle
rad
ρs
Density of the soil
ton/m3
ρl
Density water
ton/m3
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/04%3A_Which_Cutting_Mechanism_for_Which_Kind_of_Soil/4.06%3A_Summary.txt
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In literature most cutting theories are based on one time failure of the sand. Here a continuous cutting process is considered. In dry sand the cutting processes are governed by gravity and by inertial forces. Pore pressure forces, cohesion and adhesion are not present or can be neglected. Internal and external friction are present. The cutting process is of the Shear Type with discrete shear planes (see Figure 5-1), but this can be modeled as the Flow Type (see Figure 5-2), according to Merchant (1944). This approach will give an estimate of the maximum cutting forces. The average cutting forces may be 30%-50% of the maximum cutting forces.
5.02: Definitions
Definitions:
1. A: The blade tip.
2. B: End of the shear plane.
3. C: The blade top.
4. A-B: The shear plane.
5. A-C: The blade surface.
6. hb: The height of the blade.
7. hi: The thickness of the layer cut.
8. vc: The cutting velocity.
9. α: The blade angle.
10. β: The shear angle.
11. Fh: The horizontal force, the arrow gives the positive direction.
12. Fv: The vertical force, the arrow gives the positive direction.
5.03: The Equilibrium of Forces
Figure 5-4 illustrates the forces on the layer of soil cut. The forces shown are valid in general. The forces acting on this layer are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction, N1·tan(φ).
3. A gravity force as a result of the weight of the layer cut.
4. An inertial force I, resulting from acceleration of the soil.
5. A force normal to the blade N2, resulting from the effective grain stresses.
6. A shear force S2 as a result of the soil/steel friction N2·tan(δ).
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{5-1}$
The forces acting on a straight blade when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the soil/steel friction N2·tan(δ).
These forces are shown in Figure 5-5. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{5-2}$
Pure sand is supposed to be cohesion less, meaning it does not have shear strength or the shear strength is zero and the adhesion is also zero. The shear stresses, internal and external, depend completely on the normal stresses. In dry sand the pores between the sand grains are filled with air and although dilatation will occur due to shearing, Miedema (1987 September), there will be hardly any generation of pore under pressures because the permeability for air flowing through the pores is high. This means that the cutting forces do not depend on pore pressure forces, nor on adhesion and cohesion, but only on gravity and inertia, resulting in the following set of equations:
The horizontal equilibrium of forces:
$\ \mathrm{ \sum F_{h}=K_{1} \cdot \sin (\beta+\varphi)+I \cdot \cos (\beta)-K_{2} \cdot \sin (\alpha+\delta)=0}\tag{5-3}$
The vertical equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{\mathrm{1}} \cdot \cos (\beta+\varphi)+\mathrm{I} \cdot \sin (\beta)+\mathrm{G}-\mathrm{K}_{2} \cdot \cos (\alpha+\delta)=\mathrm{0}\tag{5-4}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{G} \cdot \sin (\alpha+\delta)-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{5-5}$
The force K2 on the blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{G} \cdot \sin (\beta+\varphi)+\mathrm{I} \cdot \cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{5-6}$
Wismer and Luth (1972A) and (1972B) researched the inertia forces part of the total cutting forces. The following equation is derived:
$\ \mathrm{I}=\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{2} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{5-7}$
The gravitational force (weight dry) follows, based on Figure 5-2, from:
$\ \mathrm{G}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot\left\{\frac{\left(\mathrm{h}_{\mathrm{b}}+\mathrm{h}_{\mathrm{i}} \cdot \sin (\alpha)\right)}{\sin (\alpha)}+\frac{\mathrm{h}_{\mathrm{i}} \cdot \cos (\alpha+\beta)}{2 \cdot \sin (\beta)}\right\}\tag{5-8}$
In reality the shape of the layer cut may be different since there is no force to keep the sand together and the maximum slope of the sand will be dependent on the angle of natural repose. For the calculations the above equation is applied, since this equation is used for all soil types. Other formulations for the weight of the soil may be used. From equation (5-6) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{2} \cdot \sin (\alpha+\delta)\tag{5-9}$
$\ \mathrm{F}_{v}=\mathrm{K}_{2} \cdot \cos (\alpha+\delta)\tag{5-10}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}=\frac{\mathrm{G} \cdot \sin (\alpha+\delta)-\mathrm{I} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\tag{5-11}$
The normal force on the blade is now:
$\ \mathrm{N}_{2}=\frac{\mathrm{G} \cdot \sin (\beta+\varphi)+\mathrm{I} \cdot \cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\tag{5-12}$
Equations (5-11) and (5-12) show that the normal force on the shear plane N1 can become negative at very high velocities, which are physically impossible, while the normal force on the blade N2 will always be positive. Under normal conditions the sum of α+β+δ will be greater than 90 degrees in which case the cosine of this sum is negative, resulting in a normal force on the shear plane that is always positive. Only in the case of a small blade angle α, shear angle β and angle of external friction δ, the sum of these angles could be smaller than 90°, but still close to 90° degrees. For example a blade angle of 30° would result in a shear angle of about 30°. Loose sand could have an external friction angle of 20°, so the sum would be 80°. But this is a lower limit for α+β+δ. A more realistic example is a blade with an angle of 60°, resulting in a shear angle of about 20° and a medium to hard sand with an external friction angle of 30°, resulting in α+β+δ=110°. So for realistic cases the normal force on the shear plane N1 will always be positive. In dry sand, always the shear type of cutting mechanism will occur.
Based on the weight only of the soil, the forces can also be expressed as:
$\ \begin{array}{left}\mathrm{F_{h}=\rho_{s} \cdot g \cdot h_{i}^{2} \cdot w \cdot \lambda_{H D}}\ \text{With :}\ \lambda_{\mathrm{HD}}=\frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot\left\{\frac{\left(\mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}+\sin (\alpha)\right)}{\sin (\alpha)}+\frac{\cos (\alpha+\beta)}{2 \cdot \sin (\beta)}\right\} \cdot \frac{\sin (\beta+\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\end{array}\tag{5-13}$
$\ \begin{array}{left}\mathrm{F}_{v}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V D}}\ \text{With:}\ \lambda_{\mathrm{VD}}=\frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot\left\{\frac{\left(\mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}+\sin (\alpha)\right)}{\sin (\alpha)}+\frac{\cos (\alpha+\beta)}{2 \cdot \sin (\beta)}\right\} \cdot \frac{\sin (\beta+\varphi) \cdot \cos (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\end{array}\tag{5-14}$
Figure 5-6, Figure 5-7 and Figure 5-8 show the shear angle β, the horizontal cutting force coefficient λHD and the vertical cutting force coefficient λVD. It should be mentioned here that choosing another shape of the layer cut will result in different values for the shear angle and the cutting force coefficients.
For blade angles up to 60°, there is not much influence of the angle of internal friction on the vertical force. The horizontal force and the shear angle however depend strongly on the angle of internal friction. At large blade angles, the horizontal force becomes very large, while the vertical force changes sign and becomes very large negative (upwards directed). The shear angle decreases with increasing blade angle and angle of internal friction.
At large blade angles nature will look for an alternative mechanism, the wedge mechanism, which is discussed in later chapters.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/05%3A_Dry_Sand_Cutting/5.01%3A_Introduction.txt
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The shape of the layer cut will most probably be different with dry sand cutting compared to saturated sand cutting or clay and rock cutting. First of all with dry sand cutting the cutting forces are determined by the weight of the layer cut while with the other types of soil the weight can be neglected. Secondly in dry sand there are no forces to keep the layer cut together, so the sand will move down if possible and the maximum slopes will be under the angle of natural repose φnr (usually about 30°). Figure 5-9 shows this alternative shape of the layer cut. The line D-E-F is the top of the sand, where the two marked areas have the same cross section.
The gravitational force (weight dry) follows, based on Figure 5-9, from:
$\ \mathrm{G}= \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\sin (\boldsymbol{\alpha}+\boldsymbol{\beta})}{\sin (\beta)} \cdot\left\{\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}+\frac{\mathrm{h}_{\mathrm{i}} \cdot \cos (\alpha+\beta)}{2 \cdot \sin (\beta)}-\frac{\mathrm{h}_{\mathrm{i}} \cdot \sin (\alpha+\beta)}{2 \cdot \sin (\beta)} \cdot \frac{\cos \left(\alpha+\varphi_{\mathrm{n r}}\right)}{\sin \left(\alpha+\varphi_{\mathrm{n r}}\right)}\right\} \tag{5-15}$
Based on the weight only of the soil, the forces can now be expressed as:
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{H} \mathrm{D}}\ \text{With :} \quad \lambda_{\mathrm{H D}}=\frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot \frac{\sin (\beta+\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot\left\{\frac{h_{\mathrm{b}} / h_{\mathrm{i}}}{\sin (\alpha)}+\frac{\cos (\alpha+\beta)}{2 \cdot \sin (\beta)}-\frac{\sin (\alpha+\beta)}{2 \cdot \sin (\beta)} \cdot \frac{\cos \left(\alpha+\varphi_{\mathrm{nr}}\right)}{\sin \left(\alpha+\varphi_{\mathrm{nr}}\right)}\right\}\end{array}\tag{5-16}$
$\ \begin{array}{left}\mathrm{F}_{\mathrm{v}}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V D}}\ \text{With : }\quad \lambda_{\mathrm{V D}}=\frac{\sin (\alpha+\beta)}{\sin (\beta)} \cdot \frac{\sin (\beta+\varphi) \cdot \cos (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot\left\{\frac{\mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}}{\sin (\alpha)}+\frac{\cos (\alpha+\beta)}{2 \cdot \sin (\beta)}-\frac{\sin (\alpha+\beta)}{2 \cdot \sin (\beta)} \cdot \frac{\cos \left(\alpha+\varphi_{\mathrm{nr}}\right)}{\sin \left(\alpha+\varphi_{\mathrm{nr}}\right)}\right\}\end{array}\tag{5-17}$
Figure 5-10, Figure 5-11 and Figure 5-12 show the shear angle and the cutting force coefficients for the alternative shape of the layer cut. The difference with the standard configuration is small. Other configurations may exist, but no big differences are expected. The model for dry sand or gravel can also be used for saturated sand, if the cutting process is completely drained and there are no pore vacuum pressures. This only occurs if the permeability is very high, which could be the case in gravel. Of course the dry density of the sand or gravel has to be replaced by the submerged density of the sand or gravel, which is usually close to unity.
The shapes of the curves between the standard configuration and the alternative configuration are very similar. The shear angle first increases with an increasing blade angle up to a maximum after which the shear angle decreases with a further increasing shear angle. The shear angle also decreases with an increasing angle of internal friction. It should be noted here that the external friction angle is assumed to be 2/3 of the internal friction angle.
The cutting forces become very high at large blade angles (close to 90°). Nature will find an alternative cutting mechanism in this case which has been identified as the wedge mechanism. At which blade angle the wedge mechanism will start to occur depends on the internal and external friction angles, but up to a blade angle of 60° the model as described here can be applied. See Chapter 11: A Wedge in Dry Sand Cutting. for detailed information on the wedge mechanism.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/05%3A_Dry_Sand_Cutting/5.04%3A_An_Alternative_Shape_of_the_Layer_Cut.txt
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In the previous chapter the shear angle and the cutting forces are given for the influence of the weight only. This will be appropriate for very small cutting velocities, but the question is of course; what is a very low cutting speed.
Analyzing the equations for the influence of the weight (gravity) and the influence of the inertial forces shows a significant difference. The gravitational forces are proportional to the density of the soil ρs, the gravitational constant g, the thickness of the layer cut hi squared and the width of the blade w. The inertial forces are proportional to the density of the soil ρs, the cutting velocity vc squared, the thickness of the layer cut hi and the width of the blade w. This implies that the ratio between these two forces does not only depends on the geometry, but even stronger on the layer thickness hi and the cutting velocity vc. The thicker the layer cut, the higher the influence of gravity and the higher the cutting velocity, the higher the influence of inertia. One cannot say simply the higher the cutting velocity the higher the influence of inertia.
$\ \text{Gravitation : }\mathrm{F} \propto \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w}\tag{5-18}$
$\ \text{Inertia :} \quad \mathrm{F} \propto \rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{2} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{5-19}$
The contribution of the inertial forces is determined by the following dimensionless parameter:
$\ \lambda_{\mathrm{i}}=\frac{\mathrm{v}_{\mathrm{c}}^{2}}{\mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}}\tag{5-20}$
In dredging a layer thickness of the magnitude of centimeters is common, while for a bulldozer a layer thickness of a magnitude of a meter is not strange. At the same cutting velocity, the relative influence of inertial forces will differ between dredging applications and the operation of bulldozers. If inertial forces dominate the cutting process, the cutting forces can be expressed as:
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}}=\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{H I}}\ \text{With :} \quad \lambda_{\mathrm{H I}}=\frac{\sin (\alpha)}{\sin (\alpha+\beta)} \cdot \frac{\cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \sin (\alpha+\delta)\end{array}\tag{5-21}$
$\ \begin{array}{left}\mathrm{F}_{v}=\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V I}}\ \text{With : }\quad \lambda_{\mathrm{V I}}=\frac{\sin (\alpha)}{\sin (\alpha+\beta)} \cdot \frac{\cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\alpha+\delta)\end{array}\tag{5-22}$
These equations are derived from equations (5-6), (5-7), (5-9) and (5-10). The shear angle β can be derived analytically for the inertial forces, giving:
$\ \beta=\frac{\pi}{2}-\frac{2 \cdot \alpha+\delta+\varphi}{2}\tag{5-23}$
Figure 5-13 shows the percentage of the contribution of the inertial forces to the horizontal cutting force for a layer thickness hi of 1.0 m at a cutting velocity of 0.5 m/sec, giving λi=0.025. Figure 5-14 shows the percentage of the contribution of the inertial forces to the horizontal cutting force for a layer thickness hi of 0.1 m at a cutting velocity of 15.7 m/sec, giving λi=250.
Table 5-1 shows the inertial effect for the dimensionless inertial effect parameter λi ranging from 0.025 to 250. The percentage contribution of the inertial effect on the horizontal force is given as well as the shear angle, both horizontal and vertical cutting force coefficients based on equations (5-21) and (5-22) and both horizontal and vertical cutting force coefficients based on equations (5-13) and (5-14) for the case where the blade height hb equals the layer thickness hi. The table shows that the inertial effect can be neglected at very small values of the dimensionless inertial effect parameter λi, while at large values the gravitational effect can be neglected. The shear angle β decreases with an increasing dimensionless inertial effect parameter λi. Since the inertial forces are not influenced by the blade height hb, the cutting forces are not dependent on the blade height at high cutting velocities. At low cutting velocities there will be an effect of the blade height.
The contribution of the inertial effect only depends on the dimensionless inertial effect parameter λi and not on the cutting velocity or layer thickness individually. The dimensionless inertial effect parameter λi in fact is a Froude number of the cutting process. Figure 5-15, Figure 5-16 and Figure 5-17 show the shear angle and both horizontal and vertical cutting force coefficients at very high values of the dimensionless inertial effect parameter λi (λi=250). The shear angles are considerably smaller than in the case where inertial forces can be neglected. Also in the case where the inertial forces dominate, the cutting forces become very high at large blade angles (close to 90°). Nature will find an alternative cutting mechanism in this case which has been identified as the wedge mechanism. At which blade angle the wedge mechanism will start to occur depends on the internal and external friction angles, but up to a blade angle of 60° the model as described here can be applied. See Chapter 11: A Wedge in Dry SandCutting for detailed information on the wedge mechanism.
Table 5-1: The inertial effect.
λi
%
β
λHI
λVI
λHD
λHV
0.025
0.74
31.6
187.68
27.49
4.78
0.70
0.250
6.65
30.8
20.26
2.97
5.09
0.75
2.500
37.69
26.0
3.12
0.46
7.95
1.16
25.00
78.98
16.9
1.25
0.18
31.40
4.60
250.0
94.98
9.6
0.96
0.14
245.36
35.94
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/05%3A_Dry_Sand_Cutting/5.05%3A_The_Influence_of_Inertial_Forces.txt
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In the dredging industry, the specific cutting energy is described as: The amount of energy, that has to be added to a volume unit of soil (e.g. sand, clay or rock) to excavate the soil. The dimension of the specific cutting energy is: kN/m2 or kPa for sand and clay, while for rock often MN/m2 or MPa is used.
For the case as described above, cutting with a straight blade, the specific cutting energy can be written as:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{5-24}$
At low cutting velocities this gives for the specific cutting energy:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot \lambda_{\mathrm{HD}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot \lambda_{\mathrm{H} \mathrm{D}}\tag{5-25}$
At high cutting velocities this gives for the specific cutting energy:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{H I}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}} \cdot \lambda_{\mathrm{HI}}\tag{5-26}$
At medium cutting velocities a weighted average of both has to be used.
5.07: Usage of the Model for Dry Sand
To use the model for dry sand, first the dry density ρs of the sand and the internal friction angle φ have to be known. The external friction angle δ is assumed to be 2/3 of the internal friction angle φ. Secondly the geometry of the blade, the cutting angle α, the blade height hb and the blade width have to be chosen. Thirdly the operational parameters, the layer thickness hand the cutting velocity vhave to be chosen. Based on the dimensionless inertial effect parameter λi the fraction of the contribution of the inertial force to the total horizontal force can be determined with:
$\ \mathrm{f}_{\mathrm{i}}=\frac{\mathrm{1}}{1+\mathrm{e}^{-\mathrm{2} \cdot \log \left(\lambda_{\mathrm{i}} / \mathrm{5}\right)}}\tag{5-27}$
This equation is empirically derived for a 60° blade and a 40° internal friction angle and may differ for other values of the blade angle and the internal friction angle.
$\ \begin{array}{left} \mathrm{F_{h}=\left(1-f_{i}\right) \cdot \rho_{s} \cdot g \cdot h_{i}^{2} \cdot w \cdot \lambda_{H D}+f_{i} \cdot \rho_{s} \cdot v_{c}^{2} \cdot h_{i} \cdot w \cdot \lambda_{H I}}\ \mathrm{=\rho_{s} \cdot g \cdot h_{i}^{2} \cdot w \cdot\left(\left(1-f_{i}\right) \cdot \lambda_{H D}+f_{i} \cdot \lambda_{i} \cdot \lambda_{H I}\right)}\end{array}\tag{5-28}$
$\ \begin{array}{left}\mathrm{F}_{\mathrm{v}}=\left(\mathrm{1}-\mathrm{f}_{\mathrm{i}}\right) \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V D}}+\mathrm{f}_{\mathrm{i}} \cdot \rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{2} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{VI}}\ =\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{V D}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{VI}}\right)\end{array}\tag{5-29}$
The specific energy is now:
$\ \mathrm{E}_{\mathrm{sp}}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{HD}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{HI}}\right)\tag{5-30}$
In the case of saturated sand or gravel with a very high permeability (in general coarse gravel), the equations change slightly, since the weight of the soil cut is determined by the submerged weight, while the mass of the soil cut also includes the mass of the pore water. The wet density of saturated sand or gravel is usually close to ρs=2 ton/m3, while the submerged weight is close to (ρs-ρw)·g=10 kN/m3 (a porosity of 40% and a quarts density of ρq=2.65 ton/mare assumed). This will double the contribution of the inertial forces as determined by the following dimensionless parameter:
$\ \lambda_{\mathrm{i}}=\frac{\mathrm{v}_{\mathrm{c}}^{2}}{\mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}} \cdot \frac{\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right)}{\rho_{\mathrm{s}}} \approx \frac{\mathrm{2} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}}}{\mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}}\tag{5-31}$
Using this dimensionless inertial effect parameter λi, the cutting forces can be determined by:
$\ \mathrm{F}_{\mathrm{h}}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w} \cdot\left(\left(\mathrm{1}-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{H D}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{HI}}\right)\tag{5-32}$
$\ \mathrm{F}_{\mathrm{v}}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{V D}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{V I}}\right)\tag{5-33}$
The specific energy is now:
$\ \mathrm{E}_{\mathrm{sp}}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{HD}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{HI}}\right)\tag{5-34}$
Under water at high cutting velocities there may also be a drag force which has not been taken into account here.
The horizontal cutting force coefficients λHD and λHI can be found in Figure 5-11 and Figure 5-16. The vertical cutting force coefficients λVD and λVI can be found in Figure 5-12 and Figure 5-17.
The cutting forces calculated are for a plane strain 2D cutting process, so 3D side effects are not included.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/05%3A_Dry_Sand_Cutting/5.06%3A_Specific_Energy.txt
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5.8.1. Hatamura&Chijiiwa(1977B)
Hatamura & Chijiiwa (1977B) carried out very good and extensive research into the cutting of sand, clay and loam. They did not only measure the cutting forces, but also the stresses on the blade, the shear angles and velocity distributions in the sand cut. For their experiments they used a blade with a width of w=0.33 m, a length of L=0.2 m, blade angles of α=30°, 45°, 60°, 75° and 90°, layer thicknesses of hi=0.05 m, 0.10 m and 0.15 m and cutting velocities of vc=0.05 m/sec, 0.10 m/sec and 0.14 m/sec. The sand they used had an internal friction angle φ=38° and an external friction angle δ=26.6° (almost 2/3·φ). The dry density of the sand was ρs=1.46 ton/m3.
Figure 5-19 shows the shear angles measured versus the shear angles calculated with the current model based on the minimum cutting energy criterion. In general there is a good match, especially for the experiments with a layer thickness of 0.1 m. For the experiments with a layer thickness of 0.05 m the theory overestimates the experimental value while for the layer thickness of 0.15 m, the theory underestimates the experimental value. Now the number of experiments is very limited and more experiments are required to get a better validation.
Figure 5-20 shows the total cutting force measured versus the total cutting force calculated. The total cutting force is the vectorial sum of the horizontal and the vertical cutting force. Hatamura & Chijiiwa (1977B) did not give the horizontal and vertical cutting forces, but the total cutting force and the direction of this force. For blade angles up to 60° there is a good match between experiments and theory. However at larger blade angles the theory overestimates the total cutting force strongly. This is most probably caused by the occurrence of a wedge in front of the blade at large blade angles. The occurrence of a wedge will strongly reduce the cutting forces in that case. See also Chapter 11: A Wedge in Dry Sand Cutting.
Figure 5-21 shows the direction of the total cutting force, measured versus calculated. There is an almost perfect match, also for the large blade angles where the forces are overestimated.
The conclusion is that the model developed here matches the experiments well for small blade angles, both in magnitude and direction, for large blade angles the wedge theory has to be applied. Hatamura & Chijiiwa (1977B) also carried out some tests with different cutting velocities, but the velocities were so small that there was hardly any inertial effect.
5.8.2. Wismer & Luth (1972B)
Wismer & Luth (1972B) investigated rate effects in soil cutting in dry sand, clay and loam. They found that in dry quarts sand the cutting forces consist of two components, a static component and a dynamic component. The static component depends on the cutting geometry, like the blade angle and the blade height. The static component also depends on the layer thickness and the soil mechanical parameters, in this case the dry soil density, the internal friction angle and the external friction angle. The dynamic component also depends on the cutting geometry and the soil mechanical properties, but also on the cutting velocity squared. In fact their findings match equations (5-6), (5-7), (5-9) and (5-10), but they use a different formulation for equation (5-8) or (5-15), the cross section of the layer cut. One of the reasons for the latter is that they use a fixed shear angle of β=45-φ/2 resulting in a different weight of the soil cut compared with the theory described here. In the current theory the shear angle depends on the geometry, the operational parameters and the soil mechanical parameters. The test carried out by Wismer & Luth (1972B) were with an α=30° blade with a blade height hb=0.0969 m and a width of w=0.1262 m. The layer thickness was hi=0.098 m. In order to validate the rate effect, first they calibrated the soil mechanical properties, so the cutting forces at zero cutting velocity would match the experiments. This requires an internal friction angle φ=41° and an external friction angle δ=27.3° (δ=2/3·φ), to have the correct ratio between the horizontal and the vertical force. Further, the theoretical cutting forces have to be multiplied by a factor 1.23 in order to match quantitatively. This may be the result of 3D side effects, since the blade used was not very wide compared to the layer thickness and/or the cross section of the layer cut was larger than the here assumed cross section. Both explanations seem to be reasonable. After applying these corrections and calibrations, the cutting forces are determined and plotted in Figure 5-22. The correlation between the theoretical lines and the measured data points is remarkable, resulting in the conclusion that the approach of Wismer & Luth (1972B) to quantify the rate effects for dry sand is a good approach.
Wismer & Luth (1972B) used a fixed shear angle of β=45-φ/2 resulting in β=24.5°. The values found here, based on the minimum energy principle range from β=38.8° at zero cutting velocity to β=32.2° at a cutting velocity vc=3 m/sec, taking into account the effect of the inertial forces on the shear angle.
5.09: Nomenclature
Fh
Horizontal cutting force
kN
Fv
Vertical cutting force
kN
g
Gravitational constant (9.81)
m/s2
G
Gravitational force on the layer cut
kN
hi
Initial thickness of layer cut
m
hb
Height of blade
m
I
Inertial force on the shear plane
kN
K1
Grain force on the shear plane
kN
K2
Grain force on the blade or the front of the wedge
kN
N1
Normal force on the shear plane
kN
N2
Normal force on the blade or the front of the wedge
kN
Pc
Cutting power
kW
S1
Shear force due to friction on the shear plane
kN
S2
Shear force due to friction on the blade or the front of the wedge
kN
vc
Cutting velocity component perpendicular to the blade
m/s
w
Width of blade
m
W1
Force resulting from pore under pressure on the shear plane
kN
W
Force resulting from pore under pressure on the blade or on the front of the wedge
kN
α
Cutting angle blade
rad
β
Shear angle
rad
φ
Angle of internal friction
rad
δ
External friction angle
rad
ρs
Density of the soil
ton/m3
ρw
Density water
ton/m3
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/05%3A_Dry_Sand_Cutting/5.08%3A_Experiments_in_Dry_Sand.txt
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Although calculation models for the determination of the cutting forces for dry soil, based on agriculture, were available for a long time (Hettiaratchi & Reece (1965), (1966), (1967A), (1967B), (1974), (1975) and Hatamura & Chijiiwa (1975), (1976A), (1976B), (1977A) and (1977B) ) it is only since the seventies and the eighties that the cutting process in saturated sand is extensively researched at the Delft Hydraulics Laboratory, at the Delft University of Technology and at the Mineraal Technologisch Instituut (MTI, IHC).
First the process is described, for a good understanding of the terminology used in the literature discussion.
From literature it is known that, during the cutting process, the sand increases in volume (see Figure 6-7). This increase in volume is accredited to dilatancy. Dilatancy is the change of the pore volume as a result of shear in the sand package. This increase of the pore volume has to be filled with water. The flowing water experiences a certain resistance, which causes sub-pressures in the pore water in the sand package. As a result the grain stresses increase and therefore the required cutting forces. The rate of the increase of the pore volume in the dilatancy zone, the volume strain rate, is proportional to the cutting velocity. If the volume strain rate is high, there is a chance that the pore pressure reaches the saturated water vapor pressure and cavitation occurs. A further increasing volume strain rate will not be able to cause a further decrease of the pore pressure. This also implies that, with a further increasing cutting velocity, the cutting forces cannot increase as a result of the dilatancy properties of the sand. The cutting forces can, however, still increase with an increasing cutting velocity as a result of the inertia forces and the flow resistance.
The cutting process can be subdivided in 5 areas in relation with the cutting forces:
• Very low cutting velocities, a quasi-static cutting process. The cutting forces are determined by the gravitation, cohesion and adhesion.
• The volume strain rate is high in relation to the permeability of the sand. The volume strain rate is however so small that inertia forces can be neglected. The cutting forces are dominated by the dilatancy properties of the sand.
• A transition region, with local cavitation. With an increasing volume strain rate, the cavitation area will increase so that the cutting forces increase slightly as a result of dilatancy.
• Cavitation occurs almost everywhere around and on the blade. The cutting forces do not increase anymore as a result of the dilatancy properties of the sand.
• Very high cutting velocities. The inertia forces part in the total cutting forces can no longer be neglected but form a substantial part.
Under normal conditions in dredging, the cutting process in sand will be governed by the effects of dilatation. Gravity, inertia, cohesion and adhesion will not play a role.
6.02: Definitions
Definitions:
1. A: The blade tip.
2. B: End of the shear plane.
3. C: The blade top.
4. A-B: The shear plane.
5. A-C: The blade surface.
6. hb: The height of the blade.
7. hi: The thickness of the layer cut.
8. vc: The cutting velocity.
9. α: The blade angle.
10. β: The shear angle.
11. Fh: The horizontal force, the arrow gives the positive direction.
12. Fv: The vertical force, the arrow gives the positive direction.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.01%3A_Introduction.txt
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In the seventies extensive research is carried out on the forces that occur while cutting sand under water. A conclusive cutting theory has however not been published in this period. However qualitative relations have been derived by several researchers, with which the dependability of the cutting forces with the soil properties and the blade geometry are described (Joanknecht (1974), van Os (1977A), (1976) and (1977B)).
A process that has a lot of similarities with the cutting of sand as far as water pressure development is concerned, is the, with uniform velocity, forward moving breach. Meijer and van Os (1976) and Meijer (1981) and (1985) have transformed the storage equation for the, with the breach, forward moving coordinate system.
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{k}} \cdot\left|\frac{\partial \mathrm{e}}{\partial \mathrm{x}}\right|-\frac{\rho_{\mathrm{w}} \cdot \mathrm{g}}{\mathrm{k}} \cdot\left|\frac{\partial \mathrm{e}}{\partial \mathrm{t}}\right|\tag{6-1}$
In the case of a stationary process, the second term on the right is zero, resulting:
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{k}} \cdot\left|\frac{\partial \mathrm{e}}{\partial \mathrm{x}}\right|\tag{6-2}$
Van Os (1977A), (1976) and (1977B) describes the basic principles of the cutting process, with special attention for the determination of the water sub-pressures and the cavitation. Van Os uses the non-transformed storage equation for the determination of the water sub-pressures.
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g}}{\mathrm{k}} \cdot\left|\frac{\partial \mathrm{e}}{\partial \mathrm{t}}\right|\tag{6-3}$
The average volume strain rate has to be substituted in the term $\ \partial_\mathrm{e}/\partial_\mathrm{t}$ on the right. The average volume strain rate is the product of the average volume strain of the sand package and the cutting velocity and arises from the volume balance over the shear zone. Van Os gives a qualitative relation between the water sub-pressures and the average volume strain rate:
$\ \mathrm{p}:: \frac{\mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \varepsilon}{\mathrm{k}}\tag{6-4}$
The problem of the solution of the storage equation for the cutting of sand under water is a mixed boundary value problem, for which the water sub-pressures along the boundaries are known (hydrostatic).
Joanknecht (1973) and (1974) assumes that the cutting forces are determined by the sub-pressure in the sand package. A distinction is made between the parts of the cutting force caused by the inertia forces, the sub-pressure behind the blade and the soil mechanical properties of the sand. The influence of the geometrical parameters gives the following qualitative relation:
$\ \mathrm{F}_{\mathrm{c i}}:: \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w}\tag{6-5}$
The cutting force is proportional to the cutting velocity, the blade width and the square of the initial layer-thickness. A relation with the pore percentage and the permeability is also mentioned. A relation between the cutting force and these soil mechanical properties is however not given. It is observed that the cutting forces increase with an increasing blade angle.
In the eighties research has led to more quantitative relations. Van Leussen and Nieuwenhuis (1984) discuss the soil mechanical aspects of the cutting process. The forces models of Miedema (1984B), (1985B), (1985A), (1986B) and (1987 September), Steeghs (1985A) and (1985B) and the CSB (Combinatie Speurwerk Baggertechniek) model (van Leussen and van Os (1987 December)) are published in the eighties.
Brakel (1981) derives a relation for the determination of the water sub-pressures based upon, over each other rolling, round grains in the shear zone. The force part resulting from this is added to the model of Hettiaratchi and Reece (1974).
Miedema (1984B) has combined the qualitative relations of Joanknecht (1973) and (1974) and van Os (1976), (1977A) and (1977B) to the following relation:
$\ \mathrm{F}_{\mathrm{ci}}:: \frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot \mathrm{\varepsilon}}{\mathrm{k}_{\mathrm{m}}}\tag{6-6}$
With this basic equation calculation models are developed for a cutter head and for the periodical moving cutter head in the breach. The proportionality constants are determined empirically.
Van Leussen and Nieuwenhuis (1984) discuss the soil mechanical aspects of the cutting process. Important in the cutting process is the way shear takes place and the shape or angle of the shear plane, respectively shear zone. In literature no unambiguous image could be found. Cutting tests along a windowpane gave an image in which the shape of the shear plane was more in accordance with the so-called "stress characteristics" than with the so-called "zero-extension lines". Therefore, for the calculation of the cutting forces, the "stress characteristics method" is used (Mohr-Coulomb failure criterion). For the calculation of the water sub-pressures, however, the "zero- extension lines" are used, which are lines with a zero linear strain. A closer description has not been given for both calculations.
Although the cutting process is considered as being two-dimensional, Van Leussen and Nieuwenhuis found, that the angle of internal friction, measured at low deformation rates in a tri-axial apparatus, proved to be sufficient for dredging processes. Although the cutting process can be considered as a two-dimensional process and therefore it should be expected that the angle of internal friction has to be determined with a "plane deformation test". A sufficient explanation has not been found.
Little is known about the value of the angle of friction between sand and steel. Van Leussen and Nieuwenhuis don't give an unambiguous method to determine this soil mechanical parameter. It is, however, remarked that at low cutting velocities (0.05 mm/s), the soil/steel angle of friction can have a statistical value which is 1.5 to 2 times larger than the dynamic soil/steel angle of friction. The influence of the initial density on the resulting angle of friction is not clearly present, because loosely packed sand moves over the blade. The angles of friction measured on the blades are much larger than the angles of friction measured with an adhesion cell, while also a dependency with the blade angle is observed.
With regard to the permeability of the sand, Van Leussen and Nieuwenhuis found that no large deviations of Darcy's law occur with the water flow through the pores. The found deviations are in general smaller than the accuracy with which the permeability can be determined in situ.
The size of the area where $\ \partial_\mathrm{e}/\partial_\mathrm{t}$ from equation (6-1) is zero can be clarified by the figures published by van Leussen and Nieuwenhuis. The basis is formed by a cutting process where the density of the sand is increased in a shear band with a certain width. The undisturbed sand has the initial density while the sand after passage of the shear band possesses a critical density. This critical density appeared to be in good accordance with the wet critical density of the used types of sand. This implies that outside the shear band the following equation (Biot (1941)) is valid:
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|=\mathrm{0}\tag{6-7}$
Values for the various densities are given for three types of sand. Differentiation of the residual density as a function of the blade angle is not given. A verification of the water pressures calculations is given for a 60o blade with a blade-height/layer-thickness ratio of 1.
Miedema (1984A) and (1984B) gives a formulation for the determination of the water sub-pressures. The deformation rate is determined by taking the volume balance over the shear zone, as van Os (1977A), (1976) and (1977B) did. The deformation rate is modeled as a boundary condition in the shear zone , while the shear zone is modeled as a straight line instead of a shear band as with van Os (1976), (1977A), (1977B), van Leussen and Nieuwenhuis (1984) and Hansen (1958). The influence of the water depth on the cutting forces is clarified. Steeghs (1985A) and (1985B) developed a theory for the determination of the volume strain rate, based upon a cyclic deformation of the sand in a shear band. This implies that not an average value is taken for the volume strain rate but a cyclic, with time varying, value, based upon the dilatancy angle theory.
Miedema (1985A) and (1985B) derives equations for the determination of the water sub-pressures and the cutting forces, based upon Miedema (1982), (1984A) and (1984B). The water sub-pressures are determined with a finite element method. Explained are the influences of the permeability of the disturbed and undisturbed sand and the determination of the shear angle. The derived theory is verified with model tests. On basis of this research nmax is chosen for the residual pore percentage instead of the wet critical density.
Steeghs (1985A) and (1985B) derives equations for the determination of the water sub-pressures according to an analytical approximation method. With this approximation method the water sub-pressures are determined with a modification of equation (6-4) derived by van Os (1976), (1977A), (1977B) and the storage equation (6-7). Explained is how cutting forces can be determined with the force equilibrium on the cut layer. Also included are the gravity force, the inertia forces and the sub-pressure behind the blade. For the last influence factor no formulation is given. Discussed is the determination of the shear angle. Some examples of the cutting forces are given as a function of the cutting velocity, the water depth and the sub-pressure behind the blade. A verification of this theory is not given.
Miedema (1986A) develops a calculation model for the determination of the cutting forces on a cutter-wheel based upon (1985A) and (1985B). This will be discussed in the appropriate section. Also nomograms are published with which the cutting forces and the shear angle can be determined in a simple way. Explained is the determination of the weighted average permeability from the permeability of the disturbed and undisturbed sand. Based upon the calculations it is concluded that the average permeability forms a good estimation.
Miedema (1986B) extends the theory with adhesion, cohesion, inertia forces, gravity, and sub-pressure behind the blade. The method for the calculation of the coefficients for the determination of a weighted average permeability are discussed. It is concluded that the additions to the theory lead to a better correlation with the tests results.
Van Os and van Leussen (1987 December) summarize the publications of van Os (1976), (1977A), (1977B) and of Van Leussen and Nieuwenhuis (1984) and give a formulation of the theory developed in the early seventies at the Waterloopkundig Laboratorium. Discussed are the water pressures calculation, cavitation, the weighted average permeability, the angle of internal friction, the soil/steel angle of friction, the permeability, the volume strain and the cutting forces. Verification is given of a water pressures calculation and the cutting forces. The water sub-pressures are determined with equation (6-4) derived by van Os (1976), (1977A) and (1977B). The water pressures calculation is performed with the finite difference method, in which the height of the shear band is equal to the mesh width of the grid. The size of this mesh width is considered to be arbitrary. From an example, however, it can be seen that the shear band has a width of 13% of the layer-thickness. Discussed is the determination of a weighted average permeability. The forces are determined with Coulomb's method.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.03%3A_Cutting_Theory_Literature.txt
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Figure 6-4 illustrates the forces on the layer of soil cut. The forces shown are valid in general. The forces acting on this layer are:
1. A normal force acting on the shear surface N1.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A force normal to the blade N2.
5. A shear force S2 as a result of the soil/steel friction N2·tan(δ).
6. A force Was a result of water under pressure on the blade.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{6-8}$
The forces acting on a straight blade when cutting soil, can be distinguished as:
1. A force normal to the blade N2.
2. A shear force S2 as a result of the soil/steel friction N2·tan(δ).
3. A force W2 as a result of water under pressure on the blade.
These forces are shown in Figure 6-5. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{6-9}$
Water saturated sand is also cohesionless, although in literature the phenomenon of water under pressures is sometimes referred to as apparent cohesion. It should be stated however that the water under pressures have nothing to do with cohesion or shear strength. The shear stresses still follow the rules of Coulomb friction. Due to dilatation, a volume increase of the pore volume caused by shear stresses, under pressures develop around the shear plane as described by Miedema (1987 September), resulting in a strong increase of the grain stresses. Because the permeability of the flow of water through the pores is very low, the stresses and thus the forces are dominated by the phenomenon of dilatancy and gravitation, inertia, adhesion and cohesion can be neglected.
The horizontal equilibrium of forces is:
$\ \mathrm{\sum F_{h}=K_{1} \cdot \sin (\beta+\varphi)-W_{1} \cdot \sin (\beta)+W_{2} \cdot \sin (\alpha)-K_{2} \cdot \sin (\alpha+\delta)=0}\tag{6-10}$
The vertical equilibrium of forces is:
$\ \mathrm{\sum F_{v}=-K_{1} \cdot \cos (\beta+\varphi)+W_{1} \cdot \cos (\beta)+W_{2} \cdot \cos (\alpha)-K_{2} \cdot \cos (\alpha+\delta)=0}\tag{6-11}$
The force Kon the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{6-12}$
The force K2 on the blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{6-13}$
From equation (6-13) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=-W_{2} \cdot \sin (\alpha)+K_{2} \cdot \sin (\alpha+\delta)}\tag{6-14}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)\tag{6-15}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\tag{6-16}$
The normal force on the blade is now:
$\ \mathrm{N}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\tag{6-17}$
Equations (6-16) and (6-17) show, that the normal forces on the shear plane and the blade are always positive. Positive means compressive stresses. In water saturated sand, always the shear type of cutting mechanism will occur. Figure 6-6 shows these forces on the layer cut.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.04%3A_The_Equilibrium_of_Forces.txt
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The cutting process can be modeled as a two-dimensional process, in which a straight blade cuts a small layer of sand (Figure 6-7). The sand is deformed in the shear zone, also called deformation zone or dilatancy zone. During this deformation the volume of the sand changes as a result of the shear stresses in the shear zone. In soil mechanics this phenomenon is called dilatancy. In hard packed sand the pore volume is increased as a result of the shear stresses in the deformation zone. This increase in the pore volume is thought to be concentrated in the deformation zone, with the deformation zone modeled as a straight line. Water has to flow to the deformation zone to fill up the increase of the pore volume in this zone. As a result of this water flow the grain stresses increase and the water pressures decrease. Therefore there are water under-pressures.
This implies that the forces necessary for cutting hard packed sand under water will be determined for an important part by the dilatancy properties of the sand. At low cutting velocities these cutting forces are also determined by the gravity, the cohesion and the adhesion for as far as these last two soil mechanical parameters are present in the sand. Is the cutting at high velocities, then the inertia forces will have an important part in the total cutting forces especially in dry sand.
If the cutting process is assumed to be stationary, the water flow through the pores of the sand can be described in a blade motions related coordinate system. The determination of the water under-pressures in the sand around the blade is then limited to a mixed boundary conditions problem. The potential theory can be used to solve this problem. For the determination of the water under-pressures it is necessary to have a proper formulation of the boundary condition in the shear zone. Miedema (1984B) derived the basic equation for this boundary condition.
In (1985A) and (1985B) a more extensive derivation is published by Miedema. If it is assumed that no deformations take place outside the deformation zone, then the following equation applies for the sand package around the blade:
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|=\mathrm{0}\tag{6-18}$
The boundary condition is in fact a specific flow rate (Figure 6-8) that can be determined with the following hypothesis. For a sand element in the deformation zone, the increase in the pore volume per unit of blade length is:
$\ \Delta \mathrm{V}=\varepsilon \cdot \Delta \mathrm{A}=\varepsilon \cdot \Delta \mathrm{x} \cdot \Delta \mathrm{h}_{\mathrm{i}}=\varepsilon \cdot \Delta \mathrm{x} \cdot \Delta \mathrm{l} \cdot \sin (\beta)\tag{6-19}$
$\ \varepsilon=\frac{\mathrm{n}_{\mathrm{m a x}}-\mathrm{n}_{\mathrm{i}}}{\mathrm{1}-\mathrm{n}_{\mathrm{m a x}}}\tag{6-20}$
It should be noted that in this book the symbol ε is used for the dilatation, while in previous publications the symbol is often used. This is to avoid confusion with the symbol for the void ratio.
For the residual pore percentage nmax is chosen on the basis of the ability to explain the water under-pressures, measured in laboratory tests. The volume flow rate flowing to the sand element is equal to:
$\ \Delta \mathrm{Q}=\frac{\partial \mathrm{V}}{\partial \mathrm{t}}=\varepsilon \cdot \frac{\partial \mathrm{x}}{\partial \mathrm{t}} \cdot \Delta \mathrm{l} \cdot \sin (\beta)=\varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \Delta \mathrm{l} \cdot \sin (\beta)\tag{6-21}$
With the aid of Darcy's law the next differential equation can be derived for the specific flow rate perpendicular to the deformation zone:
$\ \mathrm{q}=\frac{\partial \mathrm{Q}}{\partial \mathrm{l}}=\mathrm{q}_{1}+\mathrm{q}_{2}=\frac{\mathrm{k}_{\mathrm{i}}}{\rho_{\mathrm{w}} \cdot \mathrm{g}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|_{\mathrm{1}}+\frac{\mathrm{k}_{\mathrm{m} \mathrm{a x}}}{\rho_{\mathrm{w}} \cdot \mathrm{g}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|_{2}=\varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta)\tag{6-22}$
The partial derivative $\ \partial_\mathrm{p}/\partial_\mathrm{n}$ is the derivative of the water under-pressures perpendicular on the boundary of the area, in which the water under-pressures are calculated (in this case the deformation zone). The boundary conditions on the other boundaries of this area are indicated in Figure 6-8. A hydrostatic pressure distribution is assumed on the boundaries between sand and water. This pressure distribution equals zero in the calculation of the water under-pressures, if the height difference over the blade is neglected.
The boundaries that form the edges in the sand package are assumed to be impenetrable. Making equation (6-22) dimensionless is similar to that of the breach equation of Meijer and van Os (1976). In the breach problem the length dimensions are normalized by dividing them by the breach height, while in the cutting of sand they are normalized by dividing them by the cut layer thickness.
Equation (6-22) in normalized format:
$\ \frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}^{\prime}}\right|_{\mathrm{1}}+\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}^{\prime}}\right|_{\mathrm{2}}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \sin (\beta)}{\mathrm{k}_{\mathrm{m a x}}} \quad with: \quad \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\mathrm{h}_{\mathrm{i}}}\tag{6-23}$
This equation is made dimensionless with:
$\ \left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|^{\prime}=\frac{\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}^{\prime}}\right|}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} / \mathrm{k}_{\mathrm{m a x}}}\tag{6-24}$
The accent indicates that a certain variable or partial derivative is dimensionless. The next dimensionless equation is now valid as a boundary condition in the deformation zone:
$\ \frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|_{\mathrm{1}}^{\prime}+\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|_{\mathrm{2}}=\sin (\beta)\tag{6-25}$
The storage equation also has to be made dimensionless, which results in the next equation:
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|^{\prime}+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|^{\prime}=\mathrm{0}\tag{6-26}$
Because this equation equals zero, it is similar to equation (6-18). The water under-pressures distribution in the sand package can now be determined using the storage equation and the boundary conditions. Because the calculation of the water under-pressures is dimensionless the next transformation has to be performed to determine the real water under-pressures. The real water under-pressures can be determined by integrating the derivative of the water under-pressures in the direction of a flow line, along a flow line, so:
$\ \mathrm{P}_{\mathrm{calc}}=\int_{\mathrm{s}^{\prime}}\left|\frac{\partial \mathrm{p}}{\partial \mathrm{s}}\right|^{\prime} \cdot \mathrm{d s}^{\prime}\tag{6-27}$
This is illustrated in Figure 6-9. Using equation (6-30) this is written as:
$\ \mathrm{P}_{\text {real }}=\int_{\mathrm{s}}\left|\frac{\partial \mathrm{p}}{\partial \mathrm{s}}\right| \cdot \mathrm{d s}=\int_{\mathrm{s}^{\prime}} \frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{\varepsilon} \cdot \mathrm{h}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{s}}\right|^{\prime} \cdot \mathrm{d} \mathrm{s}^{\prime} \quad\text{ with: }\quad \mathrm{s}^{\prime}=\frac{\mathrm{s}}{\mathrm{h}_{\mathrm{i}}}\tag{6-28}$
This gives the next relation between the real emerging water under-pressures and the calculated water under-pressures:
$\ \mathrm{P}_{\text {real }}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \boldsymbol{\varepsilon} \cdot \mathrm{h}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot \mathrm{P}_{\mathrm{c a l c}}\tag{6-29}$
To be independent of the ratio between the initial permeability kand the maximum permeability kmax , kmax has to be replaced with the weighted average permeability km before making the measured water under-pressures dimensionless.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.05%3A_Determination_of_the_Pore_Pressures.txt
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The water under-pressures in the sand package on and around the blade are numerically determined using the finite element method. The solution of such a calculation is however not only dependent on the physical model of the problem, but also on the next points:
1. The size of the area in which the calculation takes place.
2. The size and distribution of the elements.
3. The boundary conditions.
The choices for these three points have to be evaluated with the problem that has to be solved in mind. These calculations are about the values and distribution of the water under-pressures in the shear zone and on the blade. A variation of the values for point 1 and 2 may therefore not influence this part of the solution. This is achieved by on the one hand increasing the area in which the calculations take place in steps and on the other hand by decreasing the element size until the variation in the solution was less than 1%. The distribution of the elements is chosen such that a finer mesh is present around the blade tip, the shear zone and on the blade, also because of the blade tip problem. A number of boundary conditions follow from the physical model of the cutting process, these are:
1. The boundary condition in the shear zone. This is described by equation (6-23).
2. The boundary condition along the free sand surface. The hydrostatic pressure at which the process takes place, can be chosen, when neglecting the dimensions of the blade and the layer in relation to the hydrostatic pressure head. Because these calculations are meant to obtain the difference between the water under-pressures and the hydrostatic pressure it is valid to take a zero pressure as the boundary condition.
The boundary conditions, along the boundaries of the area where the calculation takes place that are located in the sand package are not determined by the physical process. For this boundary condition there is a choice between:
1. A hydrostatic pressure along the boundary.
2. A boundary as an impenetrable wall.
3. A combination of a known pressure and a known specific flow rate.
None of these choices complies with the real process. Water from outside the calculation area will flow through the boundary. This also implies, however, that the pressure along this boundary is not hydrostatic. If, however, the boundary is chosen with enough distance from the real cutting process the boundary condition may not have an influence on the solution. The impenetrable wall is chosen although this choice is arbitrary. Figure 6-8 gives an impression of the size of the area and the boundary conditions, while Figure 6-10 shows the element mesh. Figure 6-12 shows the two-dimensional distribution of the water under-pressures. A table with the dimensionless pore pressures can be found in Miedema (1987 September), Miedema & Yi (2001) and in Appendix C: and Appendix R:
The following figures give an impression of how the FEM calculations are carried out:
Figure 6-10 and Figure 6-11: Show how the mesh has been varied in order to get a 1% accuracy.
Figure 6-12: Shows both the equipotential lines and the flow lines (stream function).
Figure 6-14 and Figure 6-15: Show the equipotential lines both as lines and as a color plot. This shows clearly where the largest under pressures occur on the shear plane.
Figure 6-13 shows the pressure distribution on both the shear plane and the blade. From these pressure distributions the average dimensionless pressures p1m and p2m are determined.
Figure 6-16 and Figure 6-17: Show the streamlines both as lines and as a color plot. This shows the paths of the pore water flow.
6.07: The Blade Tip Problem
During the physical modeling of the cutting process it has always been assumed that the blade tip is sharp. In other words, that in the numerical calculation, from the blade tip, a hydrostatic pressure can be introduced as the boundary condition along the free sand surface behind the blade. In practice this is never valid, because of the following reasons:
1. The blade tip always has a certain rounding, so that the blade tip can never be considered really sharp.
2. Through wear of the blade a flat section develops behind the blade tip, which runs against the sand surface (clearance angle $\ \leq$ zero)
3. If there is also dilatancy in the sand underneath the blade tip it is possible that the sand runs against the flank after the blade has passed.
4. There will be a certain under-pressure behind the blade as a result of the blade speed and the cutting process.
A combination of these factors determines the distribution of the water under-pressures, especially around the blade tip. The first three factors can be accounted for in the numerical calculation as an extra boundary condition behind the blade tip. Along the free sand surface behind the blade tip an impenetrable line element is put in, in the calculation. The length of this line element is varied with 0.0·hi0.1·hi and 0.2·hi. It showed from these calculations that especially the water under-pressures on the blade are strongly determined by the choice of this boundary condition as indicated in Figure 6-18 and Figure 6-19.
It is hard to estimate to what degree the influence of the under-pressure behind the blade on the water under- pressures around the blade tip can be taken into account with this extra boundary condition. Since there is no clear formulation for the under-pressure behind the blade available, it will be assumed that the extra boundary condition at the blade tip describes this influence. If there is no cavitation the water pressures forces W1 and W2 can be written as:
$\ \mathrm{W}_{1}=\frac{\mathrm{p}_{1 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m a x}} \cdot \mathrm{s i n}(\beta)}\tag{6-30}$
And
$\ \mathrm{W}_{2}=\frac{\mathrm{p}_{2 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\mathrm{k}_{\max } \cdot \sin (\alpha)}\tag{6-31}$
In case of cavitation W1 and W2 become:
$\ \mathrm{W}_{1}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\tag{6-32}$
And
$\ \mathrm{W}_{2}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{6-33}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.06%3A_Numerical_Water_Pore_Pressure_Calculations.txt
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As is shown in Figure 6-9, the water can flow from 4 directions to the shear zone where the dilatancy takes place. Two of those directions go through the sand which has not yet been deformed and thus have a permeability of ki, while the other two directions go through the deformed sand and thus have a permeability of kmax. Figure 6-12 shows that the flow lines in 3 of the 4 directions have a more or less circular shape, while the flow lines coming from above the blade have the character of a straight line. If a point on the shear zone is considered, then the water will flow to that point along the 4 flow lines as mentioned above. Along each flow line, the water will encounter a certain resistance. One can reason that this resistance is proportional to the length of the flow line and reversibly proportional to the permeability of the sand. Figure 6-20 shows a point on the shear zone and it shows the 4 flow lines. The length of the flow lines can be determined with the equations (6-36), (6-37), (6-38) and (6-39). The variable Lmax in these equations is the length of the shear zone, which is equal to hi/sin(β), while the variable starts at the free surface with a value zero and ends at the blade tip with a value Lmax.
According to the law of Darcy, the specific flow is related to the pressure difference Δaccording to:
$\ \mathrm{q}=\mathrm{k} \cdot \mathrm{i}=\mathrm{k} \cdot \frac{\Delta \mathrm{p}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \Delta \mathrm{s}}\tag{6-34}$
The total specific flow coming through the 4 flow lines equals the total flow caused by the dilatation, so:
$\ \begin{array}{left}\mathrm{q}=\varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta)\ =\mathrm{k}_{\max } \cdot \frac{\Delta \mathrm{p}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{s}_{1}}+\mathrm{k}_{\mathrm{m a x}} \cdot \frac{\Delta \mathrm{p}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{s}_{2}}+\mathrm{k}_{\mathrm{i}} \cdot \frac{\Delta \mathrm{p}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{s}_{3}}+\mathrm{k}_{\mathrm{i}} \cdot \frac{\Delta \mathrm{p}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{s}_{4}}\end{array}\tag{6-35}$
For the lengths of the 4 flow lines, where s2 and s3 have a correction factor of 0.8 based on calibration with the experiments:
$\ \begin{array}{left} \mathrm{s}_{1}=\left(\mathrm{L}_{\mathrm{m} \mathrm{a} \mathrm{x}}-\mathrm{L}\right) \cdot\left(\frac{\pi}{2}+\theta_{1}\right)+\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\ \text{With : }\quad \theta_{1}=\frac{\pi}{2}-(\alpha+\beta)\end{array}\tag{6-36}$
$\ \begin{array}{left} \mathrm{s}_{2}=\mathrm{0.8} \cdot \mathrm{L} \cdot \mathrm{\theta}_{2}\ \text{With : }\quad \theta_{2}=\alpha+\beta\end{array}\tag{6-37}$
$\ \begin{array}{left} \mathrm{s}_{\mathrm{3}}=\mathrm{0 .8 \cdot L \cdot \theta _ { 3 }}\ \text{With : }\quad \theta_{3}=\pi-\beta\end{array}\tag{6-38}$
$\ \begin{array}{left}\mathrm{s}_{4}=\left(\mathrm{L}_{\mathrm{m a x}}-\mathrm{L}\right) \cdot \theta_{4}+\mathrm{0 .9} \cdot \mathrm{h}_{\mathrm{i}} \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\mathrm{h}_{\mathrm{b}}}\right)^{0.5} \cdot(\mathrm{1 .8 5} \cdot \mathrm{\alpha})^{2} \cdot\left(\frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}}\right)^{0.4}\ \mathrm{W i t h}: \quad \theta_{4}=\pi+\beta\end{array}\tag{6-39}$
The equation for the length s4 has been determined by calibrating this equation with the experiments and with the FEM calculations. This length should not be interpreted as a length, but as the influence of the flow of water around the tip of the blade. The total specific flow can also be written as:
$\ \begin{array}{left}\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{q}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{\varepsilon} \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta)\ =\frac{\Delta \mathrm{p}}{\left(\frac{\mathrm{s}_{1}}{\mathrm{k}_{\max }}\right)}+\frac{\Delta \mathrm{p}}{\left(\frac{\mathrm{s}_{2}}{\mathrm{k}_{\max }}\right)}+\frac{\Delta \mathrm{p}}{\left(\frac{\mathrm{s}_{3}}{\mathrm{k}_{\mathrm{i}}}\right)}+\frac{\Delta \mathrm{p}}{\left(\frac{\mathrm{s}_{4}}{\mathrm{k}_{\mathrm{i}}}\right)}=\frac{\Delta \mathrm{p}}{\mathrm{R}_{1}}+\frac{\Delta \mathrm{p}}{\mathrm{R}_{2}}+\frac{\Delta \mathrm{p}}{\mathrm{R}_{3}}+\frac{\Delta \mathrm{p}}{\mathrm{R}_{4}}=\frac{\Delta \mathrm{p}}{\mathrm{R}_{\mathrm{t}}}\end{array}\tag{6-40}$
The total resistance on the flow lines can be determined by dividing the length of a flow line by the permeability of the flow line. The equations (6-41), (6-42), (6-43) and (6-44) give the resistance of each flow line.
$\ \mathrm{R}_{1}=\frac{\mathrm{s}_{1}}{\mathrm{k}_{\mathrm{m} \mathrm{a x}}}\tag{6-41}$
$\ \mathrm{R}_{2}=\frac{\mathrm{s}_{2}}{\mathrm{k}_{\mathrm{m a x}}}\tag{6-42}$
$\ \mathrm{R}_{3}=\frac{\mathrm{s}_{\mathrm{3}}}{\mathrm{k}_{\mathrm{i}}}\tag{6-43}$
$\ \mathrm{R}_{4}=\frac{\mathrm{s}_{4}}{\mathrm{k}_{\mathrm{i}}}\tag{6-44}$
Since the 4 flow lines can be considered as 4 parallel resistors, the total resulting resistance can be determined according to the rule for parallel resistors. Equation (6-45) shows this rule.
$\ \frac{1}{\mathrm{R}_{\mathrm{t}}}=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{4}}}\tag{6-45}$
The resistance Rt in fact replaces the hi/kmax part of the equations (6-23), (6-24), (6-28) and (6-29), resulting in equation (6-46) for the determination of the pore vacuum pressure of the point on the shear zone.
$\ \Delta \mathrm{p}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \sin (\beta) \cdot \mathrm{R}_{\mathrm{t}}\tag{6-46}$
The average pore vacuum pressure on the shear zone can be determined by summation or integration of the pore vacuum pressure of each point on the shear zone. Equation (6-47) gives the average pore vacuum pressure by summation.
$\ \mathrm{p}_{1 \mathrm{m}}=\frac{\mathrm{1}}{\mathrm{n}} \cdot \sum_{\mathrm{i}=\mathrm{0}}^{\mathrm{n}} \Delta \mathrm{p}_{\mathrm{i}}\tag{6-47}$
The determination of the pore pressures on the blade requires a different approach, since there is no dilatation on the blade. However, from the determination of the pore pressures on the shear plane, the pore pressure at the tip of the blade is known. This pore pressure can also be determined directly from:
For the lengths of the 4 flow lines the following is valid at the tip of the blade:
$\ \mathrm{s_{1}=\frac{h_{b}}{\sin (\alpha)}}\tag{6-48}$
$\ \mathrm{s_{2}=0.8 \cdot L_{\max } \cdot \theta_{2} \quad\text{ with : } \quad \theta_{2}=\alpha+\beta}\tag{6-49}$
$\ \mathrm{s_{3}=0.8 \cdot L_{\max } \cdot \theta_{3} \quad\text{ with : }\quad \theta_{3}=\pi-\beta}\tag{6-50}$
$\ \mathrm{s_{4}=0.9 \cdot h_{\mathrm{i}} \cdot\left(\frac{h_{\mathrm{i}}}{h_{\mathrm{b}}}\right)^{0.5} \cdot(1.85 \cdot \alpha)^{2} \cdot\left(\frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{max}}}\right)^{0.4}}\tag{6-51}$
The resistances can be determined with equations (6-41), (6-42), (6-43) and (6-44) and the pore pressure with equation (6-46). Now a linear distribution of the pore pressure on the blade could be assumed, resulting in an average pressure of half the pore pressure at the tip of the blade, but it is not that simple. If the surface of the blade is considered to be a flow line, water will flow from the top of the blade to the tip of the blade. However there will also be some entrainment from the pore water in the sand above the blade, due to the pressure gradient, although the pressure gradient on the blade is considered zero (an impermeable wall). This entrainment flow of water will depend on the ratio of the length of the shear plane to the length of the blade in some way. A high entrainment will result in smaller pore vacuum pressures. When the blade is divided into N intervals, the entrainment per interval will be 1/N times the total entrainment. The two required resistances are now, using i as the counter:
$\ \mathrm{R}_{1, \mathrm{i}}=\frac{\mathrm{s}_{1, \mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left(1-\frac{\mathrm{i}}{\mathrm{N}}\right)\tag{6-52}$
$\ \mathrm{R}_{2}=\frac{\mathrm{s}_{2}}{\mathrm{k}_{\mathrm{m a x}}}\tag{6-53}$
The number of intervals for entrainment and the geometry are taken into account in the constant assumed resistance R2 according to:
$\ \mathrm{R}_{2}^{\prime}=\mathrm{N} \cdot \mathrm{1 .7 5} \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \frac{\sin (\alpha)}{\mathrm{h}_{\mathrm{b}}}\right) \cdot \mathrm{R}_{2}\tag{6-54}$
The total resistance is now:
$\ \frac{1}{\mathrm{R}_{\mathrm{t}, \mathrm{i}}}=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{1}, \mathrm{i}}}+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{2}}^{\prime}}\tag{6-55}$
Now starting from the tip of the blade, the initial flows over the blade are determined.
$\ \mathrm{q}_{0}=\frac{\Delta \mathrm{p}_{\mathrm{tip}}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{R}_{\mathrm{t}, 0}}, \quad \mathrm{q}_{1,0}=\frac{\Delta \mathrm{p}_{\mathrm{tip}}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{R}_{1,0}}, \quad \mathrm{q}_{2,0}=\frac{\Delta \mathrm{p}_{\mathrm{tip}}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{R}_{2}^{\prime}}\tag{6-56}$
However Figure 6-13 (left graph) shows that the pore vacuum pressure distribution is not linear. Going from the tip (edge) of the blade to the top of the blade, first the pore vacuum pressure increases until it reaches a maximum and then it decreases (non-linear) until it reaches zero at the top of the blade. In this graph, the top of the blade is left and the tip of the blade is right. The graph on the right side of Figure 6-13 shows the pore vacuum pressure on the shear zone. In this graph, the tip of the blade is on the left side, while the right side is the point where the shear zone reaches the free water surface. Thus the pore vacuum pressure equals zero at the free water surface (most right point of the graph). Because the distribution of the pore vacuum pressure is non-linear, entrainment used. From the FEM calculations of Miedema (1987 September) and Yi (2000) it is known, that the shape of the pore vacuum pressure distribution on the blade depends strongly on the ratio of the length of the shear zone and the length of the blade, and on the length of the flat wear zone (as shown in Figure 6-18 and Figure 6-19).
The tip effect is taken into account by letting the total flow over the blade increase the first few iteration steps (Int(0.05·N·α)) and then decrease the total flow, so first:
$\ \begin{array}{left} \mathrm{q}_{\mathrm{i}}=\mathrm{q}_{\mathrm{i}-\mathrm{1}}+\mathrm{q}_{\mathrm{2}, \mathrm{i}-1}, \quad \mathrm{q}_{\mathrm{1}, \mathrm{i}}=\mathrm{q}_{\mathrm{i}} \cdot \frac{\mathrm{R}_{\mathrm{t}, \mathrm{i}}}{\mathrm{R}_{1, \mathrm{i}}}, \quad \mathrm{q}_{2, \mathrm{i}}=\mathrm{q}_{\mathrm{i}} \cdot \frac{\mathrm{R}_{\mathrm{t}, \mathrm{i}}}{\mathrm{R}_{2}^{\prime}}\ \Delta \mathrm{p}_{\mathrm{i}}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{q}_{\mathrm{i}} \cdot \mathrm{R}_{\mathrm{t}, \mathrm{i}}\end{array}\tag{6-57}$
In each subsequent iteration step the flow over the blade and the pore vacuum pressure on the blade are determined according to:
$\ \begin{array}{left}\mathrm{q}_{\mathrm{i}}=\mathrm{q}_{\mathrm{i}-1}-\mathrm{q}_{2, \mathrm{i}-1}, \quad \mathrm{q}_{1, \mathrm{i}}=\mathrm{q}_{\mathrm{i}} \cdot \frac{\mathrm{R}_{\mathrm{t}, \mathrm{i}}}{\mathrm{R}_{1, \mathrm{i}}}, \quad \mathrm{q}_{2, \mathrm{i}}=\mathrm{q}_{\mathrm{i}} \cdot \frac{\mathrm{R}_{\mathrm{t}, \mathrm{i}}}{\mathrm{R}_{2}^{\prime}}\ \Delta \mathrm{p}_{\mathrm{i}}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{q}_{\mathrm{i}} \cdot \mathrm{R}_{\mathrm{t}, \mathrm{i}}\end{array}\tag{6-58}$
The average pore vacuum pressure on the blade can be determined by integration or summation.
$\ \mathrm{p}_{2 \mathrm{m}}=\frac{\mathrm{1}}{\mathrm{n}} \cdot \sum_{\mathrm{i}=\mathrm{0}}^{\mathrm{n}} \Delta \mathrm{p}_{\mathrm{i}}\tag{6-59}$
In the past decades many research has been carried out into the different cutting processes. The more fundamental the research, the less the theories can be applied in practice. The analytical method as described here, gives a method to use the basics of the sand cutting theory in a very practical and pragmatic way.
One has to consider that usually the accuracy of the output of a complex calculation is determined by the accuracy of the input of the calculation, in this case the soil mechanical parameters. Usually the accuracy of these parameters is not very accurate and in many cases not available at all. The accuracy of less than 10% of the analytical method described here is small with regard to the accuracy of the input. This does not mean however that the accuracy is not important, but this method can be applied for a quick first estimate.
By introducing some shape factors to the shape of the streamlines, the accuracy of the analytical model has been improved.
Table 6-1: A comparison between the numerical and analytical dimensionless pore vacuum pressures.
ki/kmax=0.25
p1m
p2m
p1m (analytical)
p2m (analytical)
α=30o, β=30o, hb/hi=2
0.294
0.085
0.333
0.072
α=45o, β=25o, hb/hi=2
0.322
0.148
0.339
0.140
α=60o, β=20o, hb/hi=2
0.339
0.196
0.338
0.196
Table 6-1 was determined by Miedema & Yi (2001). Since then the algorithm has been improved, resulting in the program listing of Figure 6-21. With this new program listing also the pore vacuum pressure distribution on the blade can be determined.
Figure 6-21: A small program to determine the pore pressures.
‘Determine the pore vacuum pressure on the shear plane
Teta1 = Pi/2 - Alpha - Beta
Teta2 = Alpha + Beta
Teta3 = Pi - Beta
Teta4 = Pi + Beta
Lmax = Hi / Sin(Beta)
L1 = Hb / Sin(Alpha)
L4 = 0.9 * Hi *(Hi/Hb)^0.5*(1.85*Alpha)^2*(Ki/Kmax)^0.4
N = 100
StepL = Lmax / N
P=0
DPMax = RhoW * G * (Z + 10)
For I = 0 To N
L = I * StepL + 0.0000000001
‘Determine the 4 lengths
S1 = (Lmax - L) * (Pi/2+Teta1) + L1
S2 = 0.8*L * Teta2
S3 = 0.8*L * Teta3
S4 = (Lmax - L) * Teta4 + L4
‘Determine the 4 resistances
R1 = S1 / Kmax
R2 = S2 / Kmax
R3 = S3 / Ki
R4 = S4 / Ki
‘Determine the total resistance
Rt = 1 / (1 / R1 + 1 / R2 + 1 / R3 + 1 / R4)
‘Determine the pore vacuum pressure in point I
DP = RhoW * G * Vc * E * Sin(Beta) * Rt
‘Integrate the pore vacuum pressure
P = P + DP
‘Store the pore vacuum pressure in point I
P1(I)=DP
Next I
‘Store the pore vacuum pressure at the tip of the blade
Ptip=DP
‘Determine the average pore vacuum pressure with correction for integration
P1m = (P - Ptip / 2) / N
‘Determine the pore vacuum pressure on the blade
‘Determine the 2 lengths
S1=L1
S2=0.8*Lmax*Teta2
‘Determine the 2 resistances
R1=S1/Kmax
R2=S2/Kmax
‘Compensate R2 for the number of intervals and the geometry
R2=R2*N*1.75*(Hi*Sin(Alpha)/(Hb*Sin(Beta))
‘Determine the effective resistance
Rt=1/(1/R1+1/R2)
‘Determine the total flow over the blade at the tip of the blade
Q=Ptip/(RhoW*G*Rt)
‘Determine the two flows, Q1 over the blade and Q2 from entrainment
Q1=Ptip/(Rhow*G*R1)
Q2=Ptip/(Rhow*G*R2)
‘Determine the pressure effect near the tip of the blade
TipEffect=Int(0.05*N*Alpha)
‘Now determine the pore vacuum pressure distribution on the blade
P=0
For I = 1 To N
$\ \quad$‘Determine the length of the top of the blade to point I
$\ \quad$S1=L1*(1-I/N)
$\ \quad$‘Determine the resistance of the top of the blade to point I
$\ \quad$R1=S1/Kmax
$\ \quad$‘Determine the effective resistance in point I
$\ \quad$Rt=1/(1/R1+1/R2)
$\ \quad$‘Determine the flow at the tip of the blade
$\ \quad$IF I>TipEffect THEN
$\ \quad\quad$Q=Q-Q2
$\ \quad$ELSE
$\ \quad\quad$Q=Q+Q2
$\ \quad$END IF
$\ \quad$‘Determine the 2 flows
$\ \quad$Q1=Q*Rt/R1
$\ \quad$Q2=Q*Rt/R2
$\ \quad$‘Determine the pore vacuum pressure in point I
$\ \quad$DP=Rhow*G*Q*Rt
$\ \quad$‘Integrate the pore vacuum pressure
$\ \quad$P = P + DP
$\ \quad$‘Store the pore vacuum pressure in point I
$\ \quad$P2(I)=DP
Next I
‘Determine the average pore vacuum pressure with correction for integration
P2m = (P - Ptip / 2) / N
Figure 6-21 shows a program listing to determine the pore pressures with the analytical/numerical method. Figure 6-22, Figure 6-23 and Figure 6-24 show the resulting pore vacuum pressure curves on the shear plane and on the blade for 30, 45 and 60 degree blades with a hi/hb ratio of 1/3 and a ki/kmax ratio of 1/4. The curves match both the FEM calculations and the experiments very well.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.08%3A_Analytical_and_Numerical_Water_Pore_Pressure_Calculations.txt
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The equations are derived with which the forces on a straight blade can be determined according to the method of Coulomb (see Verruyt (1983)). Unknown in these equations is the shear angle β. In literature several methods are used to determine this shear angle.
The oldest is perhaps the method of Coulomb (see Verruyt (1983)). This method is widely used in sheet pile wall calculations. Since passive earth pressure is the cause for failure here, it is necessary to find the shear angle at which the total, on the earth, exerted force by the sheet pile wall is at a minimum.
When the water pressures are not taken into account, an analytical solution for this problem can be found. Another failure criterion is used by Hettiaratchi and Reece (1966), (1967A), (1967B), (1974) and (1975). This principle is based upon the cutting of dry sand. The shear plane is not assumed to be straight as in the method of Coulomb, but the shear plane is composed of a logarithmic spiral from the blade tip that changes into a straight shear plane under an angle of 45oφ/2 with the horizontal to the sand surface. The straight part of the shear plane is part of the so-called passive Rankine zone. The origin of the logarithmic spiral is chosen such that the total force on the blade is minimal.
There are perhaps other failure criterions for sheet pile wall calculations known in literature, but these mechanisms are only suited for a one-time failure of the earth. In the cutting of soil the process of building up stresses and next the collapse of the earth is a continuous process.
Another criterion for the collapse of earth is the determination of those failure conditions for which the total required strain energy is minimal. Rowe (1962) and Josselin de Jong (1976) use this principle for the determination of the angle under which local shear takes place. From this point of view it seems plausible to assume that those failure criterions for the cutting of sand have to be chosen, for which the cutting work is minimal. This implies that the shear angle β has to be chosen for which the cutting work and therefore the horizontal force, exerted by the blade on the soil, is minimal. Miedema (1985B) and (1986B) and Steeghs (1985A) and (1985B) have chosen this method.
Assuming that the water pressures are dominant in the cutting of packed water saturated sand, and thus neglecting adhesion, cohesion, gravity, inertia forces, flow resistance and under-pressure behind the blade, the force Fh (equation (6-14)) becomes for the non-cavitating situation:
$\ \mathrm{F_h}=\left(\begin{array}{left}-\mathrm{p}_{2 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \frac{\sin (\alpha)}{\sin (\alpha)}\ +\mathrm{p}_{\mathrm{2} \mathrm{m}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \frac{\sin (\alpha+\beta+\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi) \cdot \sin (\alpha)}\ +\mathrm{p}_{1 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \frac{\sin (\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi) \cdot \sin (\beta)}\end{array}\right)\cdot \mathrm{\frac{\rho_w \cdot g \cdot v_c \cdot \varepsilon\cdot h_i \cdot w}{(a_1 \cdot k_i + a_2 \cdot k_{max})}}\tag{6-60}$
With the following simplification:
$\ \mathrm{F}_{\mathrm{h}}^{\prime}=\frac{\mathrm{F}_{\mathrm{h}}}{\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\left(\mathrm{a}_{1} \cdot \mathrm{k}_{\mathrm{i}}+\mathrm{a}_{2} \cdot \mathrm{k}_{\mathrm{m a x}}\right)}}\tag{6-61}$
Since the value of the shear angle β, for which the horizontal force is minimal, has to be found, equations (6-62) and (6-65) are set equal to zero. It is clear that this problem has to be solved iterative, because an analytical solution is impossible.
The Newton-Rhapson method works very well for this problem. In Miedema (1987 September) and 0 and 0 the resulting shear angles β, calculated with this method, can be found for several values of δ, φ, α, several ratios of hb/hi and for the non-cavitating and cavitating cutting process.
Interesting are now the results if another method is used. To check this, the shear angles have also been determined according Coulomb’s criterion: there is failure at the shear angle for which the total force, exerted by the blade on the soil, is minimal. The maximum deviation of these shear angles with the shear angles according Miedema (1987 September) has a value of only 3o at a blade angle of 15o. The average deviation is approximately 1.5o for blade angles up to 60o.
The forces have a maximum deviation of less than 1%. It can therefore be concluded that it does not matter if the total force, exerted by the soil on the blade, is minimized, or the horizontal force. Next these calculations showed that the cutting forces, as a function of the shear angle, vary only slightly with the shear angles, found using the above equation. This sensitivity increases with an increasing blade angle. Figure 6-25 shows this for the following conditions:
The forces are determined by minimizing the specific cutting energy and minimizing the total cutting force Ft = 15°30°45° and 60°δ = 24°φ = 42°hb/hi = 1 and a non-cavitating cutting process).
The derivative of the force Fh to the shear angle β becomes:
$\ \begin{array}{left} \frac{\partial \mathrm{F}_{\mathrm{h}}^{\prime}}{\partial \beta}=&-\mathrm{p}_{1 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \frac{\sin (\varphi) \cdot \sin (\alpha+2 \cdot \beta+\delta+\varphi) \cdot \sin (\alpha+\delta)}{\sin ^{2}(\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)^{2}} & \&+\mathrm{p}_{2 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \frac{\sin (\delta) \cdot \sin (\alpha+\delta)}{\sin (\alpha) \cdot \sin (\alpha+\beta+\delta+\varphi)^{2}} & \&+\frac{\partial \mathrm{p}_{1 \mathrm{m}}}{\partial \beta} \cdot \mathrm{h}_{\mathrm{i}} \cdot \frac{\sin (\varphi) \cdot \sin (\alpha+\delta)}{\sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)} & \&+\frac{\partial \mathrm{p}_{2 \mathrm{m}}}{\partial \beta} \cdot \mathrm{h}_{\mathrm{b}} \cdot\left\{\frac{\sin (\alpha+\beta+\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha) \cdot \sin (\alpha+\beta+\delta+\varphi)}-1\right\}=\mathrm{0} \end{array}\tag{6-62}$
For the cavitating situation this gives for the force Fh:
$\ \mathrm{F_h}=\left(\begin{array}{left} -\mathrm{h}_{\mathrm{b}} \cdot \frac{\sin (\alpha)}{\sin (\alpha)}+\mathrm{h}_{\mathrm{b}} \cdot \frac{\sin (\alpha+\beta+\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi) \cdot \sin (\alpha)}\ +\mathrm{h}_{\mathrm{i}} \cdot \frac{\sin (\varphi) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\varphi) \cdot \sin (\beta)}\end{array}\right)\cdot \mathrm{\rho_w \cdot g\cdot (z+10)\cdot w}\tag{6-63}$
With the following simplification:
$\ \mathrm{F}_{\mathrm{h}}^{\prime}=\frac{\mathrm{F}_{\mathrm{h}}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{w}}\tag{6-64}$
The derivative of the force F'h to the shear angle β becomes:
$\ \begin{array}{left}\frac{\partial \mathrm{F_{h}}^{\prime}}{\partial \beta}=-\mathrm{h_{i}} \cdot \frac{\sin (\varphi) \cdot \sin (\alpha+2 \cdot \beta+\delta+\varphi) \cdot \sin (\alpha+\delta)}{\sin ^{2}(\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)^{2}}\ +\mathrm{h_{b}} \cdot \frac{\sin (\delta) \cdot \sin (\alpha+\delta)}{\sin (\alpha) \cdot \sin (\alpha+\beta+\delta+\varphi)^{2}}=0\end{array}\tag{6-65}$
For the cavitating cutting process equation (6-65) can be simplified to:
$\ \mathrm{h}_{\mathrm{b}} \cdot \sin (\delta) \cdot \sin ^{2}(\beta)=\mathrm{h}_{\mathrm{i}} \cdot \sin (\alpha) \cdot \sin (\phi) \cdot \sin (\alpha+2 \cdot \beta+\delta+\varphi)\tag{6-66}$
The iterative results can be approximated by:
$\ \beta=61.29^{\circ}+0.345 \cdot \mathrm{\frac{h_{b}}{h_{i}}}-0.3068 \cdot \alpha-0.4736 \cdot \delta-0.248 \cdot \varphi\tag{6-67}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.09%3A_Determination_of_the_Shear_Angle.txt
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In the derivation of the calculation of the water under-pressures around the blade for the non-cavitating cutting process, resulting in equations (6-30) and (6-31), it already showed that the water under-pressures are determined by the permeability of the undisturbed sand ki and the permeability of the disturbed sand kmax. Equation (6-25) shows this dependence. The water under-pressures are determined for several ratios of the initial permeability of the undisturbed sand to the maximum permeability of the disturbed sand:
ki/kmax = 1
ki/kmax = 0.5
ki/kmax = 0.25
The average water under-pressures p1m and p2m can be put against the ratio ki/kmax, for a certain shear angle β. A hyperbolic relation emerges between the average water under-pressures and the ratio of the permeabilities. If the reciprocal values of the average water under-pressures are put against the ratio of the permeabilities a linear relation emerges.
The derivatives of p1m and p2m to the ratio ki/kmax are, however, not equal to each other. This implies that a relation for the forces as a function of the ratio of permeabilities cannot be directly derived from the found average water under-pressures.
This is in contrast with the method used by Van Leussen and Van Os (1987 December). They assume that the average pore pressure on the blade has the same dependability on the ratio of permeabilities as the average pore pressure in the shear zone. No mathematical background is given for this assumption.
For the several ratios of the permeabilities it is possible with the shear angles determined, to determine the dimensionless forces Fh and Fv. If these dimensionless forces are put against the ratio of the permeabilities, also a hyperbolic relation is found (Miedema (1987 September)), shown in Figure 6-26 and Figure 6-27.
A linear relation can therefore also be found if the reciprocal values of the dimensionless forces are taken. This relation can be represented by:
$\ \frac{1}{\mathrm{F}_{\mathrm{h}}}=\mathrm{a}+\mathrm{b} \cdot \frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}}\tag{6-68}$
With the next transformations an equation can be derived for a weighted average permeability km:
$\ \mathrm{a_{1}=\frac{b}{a+b} \quad\& \quad a_{2}=\frac{a}{a+b}}\tag{6-69}$
So:
$\ \mathrm{k}_{\mathrm{m}}=\mathrm{a}_{\mathrm{1}} \cdot \mathrm{k}_{\mathrm{i}}+\mathrm{a}_{2} \cdot \mathrm{k}_{\mathrm{m a x}}\quad\text{ with: }\quad\mathrm{a}_{1}+\mathrm{a}_{2}=\mathrm{1}\tag{6-70}$
Since the sum of the coefficients a1 and a2 is equal to 1 only coefficient a1 is given in Miedema (1987) and 0. It also has to be remarked that this coefficient is determined on the basis of the linear relation of Fh (dimensionless c1), because the horizontal force gives more or less the same relation as the vertical force, but has besides a much higher value. Only for the 60o blade, where the vertical force is very small and can change direction, differences occur between the linear relations of the horizontal and the vertical force as function of the ratio of the permeabilities.
The influence of the undisturbed soil increases when the blade-height/layer-thickness ratio increases. This can be explained by the fact that the water that flows to the shear zone over the blade has to cover a larger distance with an increasing blade height and therefore has to overcome a higher resistance. Relatively more water will have to flow through the undisturbed sand to the shear zone with an increasing blade height.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.10%3A_The_Coefficients_a1_and_a2.txt
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If only the influence of the water under-pressures on the forces that occur with the cutting of saturated packed sand under water is taken in to account, equations (6-14) and (6-15) can be applied. It will be assumed that the non- cavitating process switches to the cavitating process for that cutting velocity vc, for which the force in the direction of the cutting velocity Fh is equal for both processes. In reality, however, there is a transition region between both processes, where locally cavitation starts in the shear zone. Although this transition region starts at about 65% of the cutting velocity at which, theoretically, full cavitation takes place, it shows from the results of the cutting tests that for the determination of the cutting forces the existence of a transition region can be neglected. In the simplified equations the coefficients c1 and d1 represent the dimensionless horizontal force (or the force in the direction of the cutting velocity) in the non-cavitating and the cavitating cutting process. The coefficients c2 and d2 represent the dimensionless vertical force or the force perpendicular to the direction of the cutting velocity in the non- cavitating and the cavitating cutting process. For the non-cavitating cutting process:
$\ \mathrm{F_{\mathrm{ci}}=\frac{c_{\mathrm{i}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \varepsilon \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}}}}\tag{6-71}$
In which:
$\ \mathrm{c_1}=\left(\begin{array}{left}\frac{\left(\mathrm{p}_{1 \mathrm{m}} \cdot \frac{\mathrm{s i n}(\phi)}{\sin (\beta)}+\mathrm{p}_{\mathrm{2 m}} \cdot \frac{\mathrm{h}_{\mathrm{b}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha+\beta+\phi)}{\sin (\alpha)}\right) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\phi)}\ -\mathrm{p}_{2 \mathrm{m}} \cdot \frac{\mathrm{h}_{\mathrm{b}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha)}{\sin (\alpha)}\end{array}\right)\cdot \mathrm{\frac{(a_1 \cdot k_i +a_2 \cdot k_{max})}{k_{max}}}\tag{6-72}$
And:
$\ \mathrm{c_2}=\left(\begin{array}{left}\frac{\left(\mathrm{p}_{1 \mathrm{m}} \cdot \frac{\mathrm{s i n}(\phi)}{\sin (\beta)}+\mathrm{p}_{\mathrm{2 m}} \cdot \frac{\mathrm{h}_{\mathrm{b}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha+\beta+\phi)}{\sin (\alpha)}\right) \cdot \cos (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\phi)}\ -\mathrm{p}_{2 \mathrm{m}} \cdot \frac{\mathrm{h}_{\mathrm{b}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\cos (\alpha)}{\sin (\alpha)}\end{array}\right)\cdot \mathrm{\frac{(a_1 \cdot k_i +a_2 \cdot k_{max})}{k_{max}}}\tag{6-73}$
And for the cavitating cutting process:
$\ \mathrm{F}_{\mathrm{c i}}=\mathrm{d _ { i }} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{6-74}$
In which:
$\ \mathrm{d}_{1}=\frac{\left(\frac{\sin (\phi)}{\sin (\beta)}+\frac{\mathrm{h}_{\mathrm{b}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha+\beta+\phi)}{\sin (\alpha)}\right) \cdot \sin (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\phi)}-\frac{\mathrm{h}_{\mathrm{b}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha)}{\sin (\alpha)}\tag{6-75}$
And:
$\ \mathrm{\mathrm{d}_{2}=\frac{\left(\frac{\sin (\phi)}{\sin (\beta)}+\frac{h_{\mathrm{b}}}{h_{\mathrm{i}}} \cdot \frac{\sin (\alpha+\beta+\phi)}{\sin (\alpha)}\right) \cdot \cos (\alpha+\delta)}{\sin (\alpha+\beta+\delta+\phi)}-\frac{h_{\mathrm{b}}}{h_{\mathrm{i}}} \cdot \frac{\cos (\alpha)}{\sin (\alpha)}}\tag{6-76}$
The values of the 4 coefficients are determined by minimizing the cutting work that is at that shear angle β where the derivative of the horizontal force to the shear angle is zero. The coefficients c1, c2, d1 and d2 are given in Miedema (1987 September) and in 0 and 0 for the non-cavitating cutting process and 0 and 0 for the cavitating cutting process as functions of αδφ and the ratio hb/hi.
6.11.1. Approximations
Assuming δ=2/3·φ the coefficients can be approximated by:
$\ \begin{array}{left}\alpha=30^{\circ}\quad\text{ and }\quad\mathrm{h_{b} / h_{i}=1:}\ \mathrm{c_{1}=0.0427 \cdot e^{0.0509 \cdot \varphi} \quad\text{ and }\quad c_{2}=0.0343 \cdot e^{0.0341 \cdot \varphi}}\ \mathrm{d_{1}=0.3027 \cdot e^{0.0516 \cdot \varphi} \quad\text{ and }\quad d_{2}=-0.3732+0.0219 \cdot \varphi}\end{array}\tag{6-77}$
$\ \begin{array}{ll}\mathrm{\alpha=30^{\circ} \quad\text { and }\quad h_{b} / h_{i}=2 \text { : }} \ \mathrm{c_{1}=0.0455 \cdot 0^{0.0511 \cdot \varphi} \quad\text { and }\quad c_{2}=0.0304 \cdot e^{0.0356 \cdot \varphi}} \ \mathrm{d_{1}=0.4795 \cdot e^{0.0490 \cdot \varphi} \quad\text { and }\quad d_{2}=-0.5380+0.0159 \cdot \varphi}\end{array}\tag{6-78}$
$\ \begin{array}{ll}\mathrm{\alpha=30^{\circ} \quad\text { and }\quad h_{b} / h_{i}=3:} \ \mathrm{c_{1}=0.0457 \cdot e^{0.0512 \cdot \varphi} \quad\text { and }\quad c_{2}=0.0312 \cdot e^{0.0348 \cdot \varphi} }\ \mathrm{d_{1}=0.6418 \cdot e^{0.0478 \cdot \varphi} \quad\text { and }\quad d_{2}=-0.7332+0.0094 \cdot \varphi}\end{array}\tag{6-79}$
$\ \begin{array}{ll}\mathrm{\alpha=45^{\circ} \quad\text { and }\quad h_{b} / h_{i}=1:} \ \mathrm{c_{1}=0.0485 \cdot e^{0.0577 \cdot \varphi} \quad \text { and } \quad c_{2}=0.0341 \cdot e^{0.0255 \cdot \varphi}} \ \mathrm{d_{1}=0.2618 \cdot e^{0.0603 \cdot \varphi} \quad \text { and } \quad d_{2}=-0.0287+0.0081 \cdot \varphi}\end{array}\tag{6-80}$
$\ \begin{array}{ll}\mathrm{\alpha=45^{\circ} \quad\text { and }\quad h_{b} / h_{i}=2: } \ \mathrm{c_{1}=0.0545 \cdot e^{0.0580 \cdot \varphi} \quad\text { and }\quad c_{2}=0.0281 \cdot e^{0.0238 \cdot \varphi} }\ \mathrm{d_{1}=0.3764 \cdot e^{0.0577 \cdot \varphi} \quad\text { and } \quad d_{2}=-0.0192-0.0017 \cdot \varphi}\end{array}\tag{6-81}$
$\ \begin{array}{ll}\mathrm{\alpha=45^{\circ} \quad\text { and }\quad h_{b} / h_{i}=3:} \ \mathrm{c_{1}=0.0551 \cdot e^{0.0589 \cdot \varphi} \quad\text { and }\quad c_{2}=0.0286 \cdot e^{0.0199 \cdot \varphi}} \ \mathrm{d_{1}=0.4814 \cdot e^{0.0563 \cdot \varphi} \quad\text { and } \quad d_{2}=-0.0295-0.0116 \cdot \varphi}\end{array}\tag{6-82}$
$\ \begin{array}{left}\mathrm{\alpha=60^{\circ}\quad\text{ and }\quad h_{b} / h_{i}=1:}\ \mathrm{c_{1}=0.0474 \cdot e^{0.0688 \cdot \varphi} \quad\text{ and }\quad c_{2}=-0.2902+0.0203 \cdot \varphi-0.000334 \cdot \varphi^{2}}\ \mathrm{d_{1}=0.2342 \cdot e^{0.0722 \cdot \varphi} \quad\text{ and }\quad d_{2}=+1.0548-0.0343 \cdot \varphi}\end{array}\tag{6-83}$
$\ \begin{array}{left}\mathrm{\alpha=60^{\circ}\quad\text{ and }\quad h_{b} / h_{i}=2:}\ \mathrm{c_{1}=0.0562 \cdot e^{0.0686 \cdot \varphi} \quad\text{ and }\quad c_{2}=-0.3550+0.0235 \cdot \varphi-0.000403 \cdot \varphi^{2}}\ \mathrm{d_{1}=0.3148 \cdot e^{0.0695 \cdot \varphi} \quad\text{ and }\quad d_{2}=+1.2737-0.0516 \cdot \varphi}\end{array}\tag{6-84}$
$\ \begin{array}{left}\mathrm{\alpha=60^{\circ}\quad\text{ and }\quad h_{b} / h_{i}=3:}\ \mathrm{c_{1}=0.0593 \cdot e^{0.0692 \cdot \varphi} \quad\text{ and }\quad c_{2}=-0.3785+0.0250 \cdot \varphi-0.000445 \cdot \varphi^{2}}\ \mathrm{d_{1}=0.3889 \cdot e^{0.0680 \cdot \varphi} \quad\text{ and }\quad d_{2}=+1.4708-0.0685 \cdot \varphi}\end{array}\tag{6-85}$
The shear angle β can be approximated by, for the non-cavitating case:
$\ \mathrm{\beta=\frac{\pi-\alpha-\varphi-\delta}{3}-0.0037 \cdot \frac{h_{b}}{h_{i}}}\tag{6-86}$
The shear angle β can be approximated by, for the cavitating case:
$\ \mathrm{\beta=1-\frac{1}{6} \cdot \alpha-\frac{2}{7} \cdot(\varphi+\delta)-0.057 \cdot \frac{h_{b}}{h_{i}}}\tag{6-87}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.11%3A_Determination_of_the_Coefficients_c1_c2_d1_and_d2.txt
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In the dredging industry, the specific cutting energy is described as:
The amount of energy, that has to be added to a volume unit of soil (e.g. sand) to excavate the soil.
The dimension of the specific cutting energy is: kN/m2 or kPa for sand and clay, while for rock often MN/m2 or MPa is used.
Adhesion, cohesion, gravity and the inertia forces will be neglected in the determination of the specific cutting energy. For the case as described above, cutting with a straight blade with the direction of the cutting velocity perpendicular to the blade (edge of the blade) and the specific cutting energy can be written:
$\ \mathrm{E}_{\mathrm{s p}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{6-88}$
The method, with which the shear angle β is determined, is therefore equivalent with minimizing the specific cutting energy, for certain blade geometry and certain soil mechanical parameters. For the specific energy, for the non-cavitating cutting process, it can now be derived from equations (6-71) and (6-88), that:
$\ \mathrm{E}_{\mathrm{n} \mathrm{c}}=\mathrm{c}_{\mathrm{1}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \frac{\varepsilon}{\mathrm{k}_{\mathrm{m}}}\tag{6-89}$
For the specific energy, for the fully cavitating cutting process, can be written from equations (6-74) and (6-88):
$\ \mathrm{E}_{\mathrm{c a}}=\mathrm{d}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0})\tag{6-90}$
From these equations can be derived that the specific cutting energy, for the non-cavitating cutting process is proportional to the cutting velocity, the layer-thickness and the volume strain and inversely proportional to the permeability. For the fully cavitating process the specific cutting energy is only dependent on the water depth.
Therefore it can be posed, that the specific cutting energy, for the fully cavitating cutting process is an upper limit, provided that the inertia forces, etc., can be neglected. At very high cutting velocities, however, the specific cutting energy, also for the cavitating process will increase as a result of the inertia forces and the water resistance.
6.12.1. Specific Energy and Production in Sand
As discussed previously, the cutting process in sand can be distinguished in a non-cavitating and a cavitating process, in which the cavitating process can be considered to be an upper limit to the cutting forces. Assuming that during an SPT test in water-saturated sand, the cavitating process will occur, because of the shock wise behavior during the SPT test, the SPT test will give information about the cavitating cutting process. Whether in practice, the cavitating cutting process will occur, depends on the soil mechanical parameters, the geometry of the cutting process and the operational parameters. The cavitating process gives an upper limit to the forces, power and thus the specific energy and a lower limit to the production and will therefore be used as a starting point for the calculations. For the specific energy of the cavitating cutting process, the following equation can be derived according to Miedema (1987 September):
$\ \mathrm{E}_{\mathrm{s p}}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{d}_{\mathrm{1}}\tag{6-91}$
The production, for an available power Pa, can be determined by:
$\ \mathrm{Q}=\frac{\mathrm{P}_{\mathrm{a}}}{\mathrm{E}_{\mathrm{s p}}}=\frac{\mathrm{P}_{\mathrm{a}}}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{d}_{\mathrm{1}}}\tag{6-92}$
The coefficient d1 is the only unknown in the above equation. A relation between d1 and the SPT value of the sand and between the SPT value and the water depth has to be found. The dependence of d1 on the parameters αhi and hb can be estimated accurately. For normal sands there will be a relation between the angle of internal friction and the soil interface friction. Assume blade angles of 30, 45 and 60 degrees, a ratio of 3 for hb /hi and a soil/interface friction angle of 2/3 times the internal friction angle. For the coefficient d1 the following equations are found by regression:
$\ \mathrm{d}_{1}=\left(\mathrm{0.6 4}+\mathrm{0 .5 6} \cdot \mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}\right)+\left(\mathrm{0 .0 1 6 4}+\mathrm{0 .0 0 8 5} \cdot \mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}\right) \cdot \mathrm{S P T}_{\mathrm{1 0}}(\alpha=30 \text{ degrees})\tag{6-93}$
$\ \mathrm{d}_{1}=\left(\mathrm{0 .8 3}+\mathrm{0 .4 5} \cdot \mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}\right)+\left(\mathrm{0 .0 2 6 8}+\mathrm{0 .0 0 8 5} \cdot \mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}\right) \cdot \mathrm{S P T}_{\mathrm{1 0}}(\alpha=45\text{ degrees })\tag{6-94}$
$\ \mathrm{d}_{1}=\left(\mathrm{0 .9 9}+\mathrm{0 .3 9} \cdot \mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}\right)+\left(\mathrm{0 .0 5 0 3}+\mathrm{0 .0 0 9 9} \cdot \mathrm{h}_{\mathrm{b}} / \mathrm{h}_{\mathrm{i}}\right) \cdot \mathrm{S P T}_{\mathrm{1 0}}(\alpha=60\text{ degrees })\tag{6-95}$
With: SPT10 = the SPT value normalized to 10 m water depth.
Lambe & Whitman (1979), page 78) and Miedema (1995) give the relation between the SPT value, the relative density RD (0-1) and the hydrostatic pressure in two graphs, see Figure 6-29. With some curve-fitting these graphs can be summarized with the following equation:
$\ \operatorname{SPT}=0.243 \cdot\left(82.5+\rho_{\mathrm{l}} \cdot \mathrm{g} \cdot(\mathrm{z}+10)\right) \cdot \mathrm{R D}^{2.52}\tag{6-96}$
And:
$\ \mathrm{RD}=\left(\frac{4.12 \cdot \mathrm{SPT}}{\left(82.5+\rho_{\mathrm{l}} \cdot \mathrm{g} \cdot(\mathrm{z}+10)\right)}\right)^{0.397}\tag{6-97}$
Lambe & Whitman (1979), (page 148) and Miedema (1995) give the relation between the SPT value and the angle of internal friction, also in a graph, see Figure 6-28. This graph is valid up to 12 m in dry soil. With respect to the internal friction, the relation given in the graph has an accuracy of 3 degrees. A load of 12 m dry soil with a density of 1.67 ton/m3 equals a hydrostatic pressure of 20 m.w.c. An absolute hydrostatic pressure of 20 m.w.c. equals 10 m of water depth if cavitation is considered. Measured SPT values at any depth will have to be reduced to the value that would occur at 10 m water depth. This can be accomplished with the following equation (see Figure 6-30):
$\ \operatorname{SPT}_{10}=\frac{282.5}{\left(82.5+\rho_{\mathrm{l}} \cdot g \cdot(\mathrm{z}+10)\right)} \cdot \operatorname{SPT}_{\mathrm{z}}\tag{6-98}$
With the aim of curve-fitting, the relation between the SPT value reduced to 10 m water depth and the angle of internal friction can be summarized to:
$\ \varphi=51.5-25.9 \cdot \mathrm{e}^{-0.01753 \cdot \mathrm{SPT}_{10}}(+/- 3 \text{ degrees })\tag{6-99}$
For water depths of 0, 5, 10, 15, 20, 25 and 30 m and an available power of 100 kW the production is shown graphically for SPT values in the range of 0 to 100 SPT. Figure 6-31 shows the specific energy and Figure 6-32 the production for a 45 degree blade angle.
6.12.2. The Transition Cavitating/Non-Cavitating
Although the SPT value only applies to the cavitating cutting process, it is necessary to have a good understanding of the transition between the non-cavitating and the cavitating cutting process. Based on the theory in Miedema (1987 September), an equation has been derived for this transition. If this equation is valid, the cavitating cutting process will occur.
$\ \mathrm{v}_{\mathrm{c}}>\frac{\mathrm{d}_{1} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{k}_{\mathrm{m}}}{\mathrm{c}_{\mathrm{1}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \varepsilon}\tag{6-100}$
The ratio d1/c1 appears to have an almost constant value for a given blade angle, independent of the soil mechanical properties. For a blade angle of 30 degrees this ratio equals 11.9. For a blade angle of 45 degrees this ratio equals 7.72 and for a blade angle of 60 degrees this ratio equals 6.14. The ratio ε/khas a value in the range of 1000 to 10000 for medium to hard packed sands. At a given layer thickness and water depth, the transition cutting velocity can be determined using the above equation. At a given cutting velocity and water depth, the transition layer thickness can be determined.
6.12.3. Conclusions Specific Energy
To check the validity of the above derived theory, research has been carried out in the laboratory of the chair of Dredging Technology of the Delft University of Technology. The tests are carried out in hard packed water saturated sand, with a blade of 0.3 m by 0.2 m. The blade had cutting angles of 30, 45 and 60 degrees and deviation angles of 0, 15, 30 and 45 degrees. The layer thicknesses were 2.5, 5 and 10 cm and the drag velocities 0.25, 0.5 and 1 m/s. Figure 6-57 shows the results with a deviation angle of 0 degrees, while Figure 6-58 shows the results with a deviation angle of 45 degrees. The lines in this figure show the theoretical forces. As can be seen, the measured forces match the theoretical forces well.
Based on two graphs from Lambe & Whitman (1979) and an equation for the specific energy from Miedema (1987 September) and (1995), relations are derived for the SPT value as a function of the hydrostatic pressure and of the angle of internal friction as a function of the SPT value. With these equations also the influence of water depth on the production can be determined. The specific energies as measured from the tests are shown in Figure 6-57 and Figure 6-58. It can be seen that the deviated blade results in a lower specific energy. These figures also show the upper limit for the cavitating cutting process. For small velocities and/or layer thicknesses, the specific energy ranges from 0 to the cavitating value. The tests are carried out in sand with an angle of internal friction of 40 degrees. According to Figure 6-28 this should give an SPT value of 33. An SPT value of 33 at a water depth of about 0 m, gives according to Figure 6-31, a specific energy of about 450-500 kPa. This matches the specific energy as shown in Figure 6-57.
All derivations are based on a cavitating cutting process. For small SPT values it is however not sure whether cavitation will occur. A non-cavitating cutting process will give smaller forces and power and thus a higher production. At small SPT values however the production will be limited by the bull-dozer effect or by the possible range of the operational parameters such as the cutting velocity.
The calculation method used remains a lower limit approach with respect to the production and can thus be considered conservative. For an exact prediction of the production all of the required soil mechanical properties will have to be known. As stated, limitations following from the hydraulic system are not taken into consideration.
6.12.4. Wear and Side Effects
In the previous chapters the blades are assumed to have a reasonable sharp blade tip and a positive clearance angle. A two dimensional cutting process has also been assumed. In dredging practice these circumstances are hardly encountered. It is however difficult to introduce a concept like wear in the theoretical model, because for every wear stage the water pressures have to be determined numerically again.
Also not clear is, if the assumption that the sand shears along a straight line will also lead to a good correlation with the model tests with worn blades. Only for the case with a sharp blade and a clearance angle of -1o a model test is performed.
It is however possible to introduce the wear effects and the side effects simply in the theory with empirical parameters. To do this the theoretical model is slightly modified. No longer are the horizontal and the vertical forces used, but the total cutting force and its angle with the direction of the velocity component perpendicular to the blade edge are used. Figure 6-33 shows the dimensionless forces c1c2, and cfor the non-cavitating cutting process and the dimensionless forces d1d2 and dt for the cavitating process.
For the total dimensionless cutting forces it can be written:
$\ \begin{array}{left}\quad\quad\quad\text{non-cavitating}\quad\quad\quad\quad\quad\text{cavitating}\ \mathrm{c}_{\mathrm{t}}=\sqrt{\left(\mathrm{c}_{\mathrm{1}} \cdot \mathrm{c}_{\mathrm{1}}+\mathrm{c}_{\mathrm{2}} \cdot \mathrm{c}_{\mathrm{2}}\right)} \quad \quad\mathrm{d}_{\mathrm{t}}=\sqrt{\left(\mathrm{d}_{\mathrm{1}} \cdot \mathrm{d}_{\mathrm{1}}+\mathrm{d}_{\mathrm{2}} \cdot \mathrm{d}_{\mathrm{2}}\right)}\end{array}\tag{6-101}$
For the angle the force makes with the direction of the velocity component perpendicular to the blade edge:
$\ \theta_{\mathrm{t}}=\operatorname{atn}\left(\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}\right) \quad \quad\quad\quad\quad\Theta_{\mathrm{t}}=\operatorname{atn}\left(\frac{\mathrm{d}_{2}}{\mathrm{d}_{1}}\right)\tag{6-102}$
It is proposed to introduce the wear and side effects, introducing a wear factor cs (ds) and a wear angle θs (Θs) according to:
$\ \mathrm{c}_{\mathrm{t s}}=\mathrm{c}_{\mathrm{t}} \cdot \mathrm{c}_{\mathrm{s}} \quad\quad\quad\quad\quad\quad\quad \mathrm{d}_{\mathrm{t} \mathrm{s}}=\mathrm{d}_{\mathrm{t}} \cdot \mathrm{d}_{\mathrm{s}}\tag{6-103}$
And
$\ \theta_{\mathrm{ts}}=\theta_{\mathrm{t}}+\theta_{\mathrm{s}} \quad\quad\quad\quad\quad\quad\quad \Theta_{\mathrm{ts}}=\Theta_{\mathrm{t}}+\Theta_{\mathrm{s}}\tag{6-104}$
For the side effects, introducing a factor cr (dr) and an angle θ(Θr), we can now write:
$\ \mathrm{c}_{\mathrm{tr}}=\mathrm{c}_{\mathrm{t}} \cdot \mathrm{c}_{\mathrm{r}}\quad\quad\quad\quad\quad\quad\quad\mathrm{d_{tr}=d_t\cdot d_r} \tag{6-105}$
And
$\ \theta_{\mathrm{tr}}=\theta_{\mathrm{t}}+\theta_{\mathrm{r}} \quad \quad \Theta_{\mathrm{tr}}=\Theta_{\mathrm{t}}+\Theta_{\mathrm{r}}\tag{6-106}$
In particular the angle of rotation of the total cutting force as a result of wear, has a large influence on the force needed for the haul motion of cutter-suction and cutter-wheel dredgers. Figure 6-34 and Figure 6-35 give an impression of the expected effects of the wear and the side effects.
The angle the forces make with the velocity direction θtΘt, where this angle is positive when directed downward.
The influence of wear on the magnitude and the direction of the dimensionless cutting forces ct or dt for the non-cavitating cutting process.
The influence of side effects on the magnitude and the direction of the dimensionless cutting forces ct or dt for the non-cavitating cutting process.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.12%3A_Specific_Cutting_Energy.txt
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6.13.1. Description of the Test Facility
The tests with the straight blades are performed on two locations:
1. The old laboratory of Dredging Engineering, which will be called the old laboratory DE.
2. The new laboratory of Dredging Engineering, which will be called the new laboratory DE.
The test stand in the old laboratory DE consists of a concrete tank, 30 m long, 2.5 m wide and 1.35 m high, filled with a layer of 0.5 m sand with a d50 of 200 μm and above the sand 0.6 m water. The test stand in the new laboratory DE consists of a concrete tank, 33 m long, 3 m wide and internally 1.5 m high, with a layer of 0.7 m sand with a d50 of 105 μm and above the sand 0.6 m water. In both laboratories a main carriage can ride over the full length of the tank, pulled by two steel cables. These steel cables are winded on the drums of a hydraulic winch, placed in the basement and driven by a squirrel-cage motor of 35 kW in the old laboratory DE and 45 kW in the new laboratory DE.
In the old laboratory DE the velocity of the carriage could be infinitely variable controlled from 0.05 m/s to 2.50 m/s, with a pulling force of 6 kN. In the new laboratory DE the drive is equipped with a hydraulic two-way valve, which allows for the following speed ranges:
1. A range from 0.05 m/s to 1.40 m/s, with a maximum pulling force of 15 kN.
2. A range from 0.05 m/s to 2.50 m/s, with a maximum pulling force of 7.5 kN.
An auxiliary carriage, on which the blades are mounted, can be moved transverse of the longitudinal direction on the main carriage. Hydraulic cylinders are used to adjust the cutting depth and to position the blades in the transverse direction of the tank. Figure 6-36 shows a side view of the concrete tank with the winch drive in the basement and Figure 6-37 shows a cross section with the mounting of cutter heads or the blades underneath the auxiliary carriage (in the new laboratory DE). The main difference between the two laboratories is the side tank, which was added to dump the material excavated. This way the water stays clean and under water video recordings are much brighter. After a test the material excavated is sucked up by a dustpan dredge and put back in the main tank. The old laboratory DE was removed in 1986, when the new laboratory was opened for research. Unfortunately, the new laboratory stopped existing in 2005. Right now there are two such laboratories in the world, one at Texas A&M University in College Station, Texas, USA and one at Hohai University, Changzhou, China. Both laboratories were established around 2005.
Figure 6-38 and Figure 6-39 give an overview of both the old and the new laboratories DE, while Figure 6-40 shows a side view of the carriage, underneath which the blades are mounted.
Removing the spoil tank (3) from this figure gives a good impression of the cutting tank in the old laboratory DE. Instead of a cutter head, blades are mounted under the frame (6) during the cutting tests.
The tests are carried out using a middle blade, flanked on both sides by a side blade, in order to establish a two- dimensional cutting process on the middle blade. The middle blade (center blade) is mounted on a dynamometer, with which the following loads can be measured:
1. The horizontal force
2. The vertical force
3. The transverse force
4. The bending moment
The side blades are mounted in a fork-like construction, attached to some dynamometers, with which the following loads can be measured:
1. The horizontal force
2. The vertical force
Figure 6-41 and Figure 6-42 show the mounting construction of the blades.
In the middle blade, four pore pressure transducers are mounted, with which the pore pressure distribution on the blade can be measured. However no tests are performed in which the forces on the side blades and the pore pressures are measured at the same time. The measuring signals of the dynamometers and the pressure transducers are transmitted to a measurement compartment through pre-amplifiers on the main carriage. In this measurement compartment the measuring signals are suited by 12 bit, 400 Hz A/D converters for processing on a P.C. (personal computer), after which the signals are stored on a flexible disk. Next to the blades, under water, an underwater video camera is mounted to record the cutting process. This also gives a good impression of the shear angles occurring.
Figure 6-44 shows how a blade is mounted under the carriage in the new laboratory DE, in this case for so called snow-plough research. Figure shows the center blade and the two side blades mounted under the carriage in the old laboratory DE. In the center blade the 4 pore pressure transducers can be identified (the white circles) with which the pore pressures are measured.
Figure 6-47 shows the signal processing unit on the carriage, including pre-amplifiers and filters. The pre- amplifiers are used to reduce the noise on the signals that would occur transporting the signals over long distance to the measurement cabin.
Figure 6-46 shows the device used to measure the cone resistance of the sand before every experiment. The cone resistance can be related to the porosity of the sand, where the porosity relates to both the internal and external friction angle and to the permeability.
Figure 6-48 shows the measurement cabin with a PC for data processing and also showing the video screen and the tape recorder to store the video images of all the experiments.
Figure 6-45 shows a side view of the center blades. These blades could also be equipped with a wear flat to measure the influence of worn blades.
6.13.2. Test Program
The theory for the determination of the forces that occur during the cutting of fully water saturated sand with straight blades is verified in two types of sand, sand with a d50 of 200 μm and sand with a d50 of 105 μm. The soil mechanical parameters of these two types of sand can be found in 0 and 0
The research can be subdivided in a number of studies:
1. Research of the water resistance of the blades
2. Research of the accuracy of the assumed two-dimensional character of the cutting process on the middle blade by changing the width of the middle blade with a total width of the middle blade and the side blades of 520 mm. This research is performed in the 200 μm sand.
3. Research of the quantitative character of the side effects in relation to the size and the direction of the cutting forces. This research is performed in the 200 μm sand.
4. Research of the in the theory present scale rules. This research is performed in the 200 μm sand.
5. Research of the accuracy of the theory of the cutting forces and the water sub-pressures in the non-cavitating cutting process. This research is performed in the 200 μm sand.
6. Research of the accuracy of the theory of the forces and the water sub-pressures in the non-cavitating and the partly cavitating cutting process. This research is performed in the 105 μm sand.
From points 4 and 5 it has also been established that the maximum pore percentage of the sand can be chosen for the residual pore percentage. In the 200 μm the dry critical density, the wet critical density and the minimal density are determined, while in the 105 μm sand the wet critical density and the minimal density are determined. These pore values can be found in Appendix K and Appendix L
For both type of sand only the minimal density (maximum pore percentage nmax) gives a large enough increase in volume to explain the measured water sub-pressures. This is in contrast to Van Leussen and Nieuwenhuis (1984) and Van Leussen and Van Os (1987 December), where for the residual density the wet critical density is chosen.
6.13.3. Water Resistance
The water resistance is investigated under circumstances comparable with the cutting tests as far as scale; blade width and cutting velocity are concerned. Since the water resistance during all these tests could be neglected in comparison with the cutting forces, performed under the same conditions (maximum 2%), the water resistance terms are neglected in the further verification. The water resistance could however be more significant at higher cutting velocities above 2 m/s. It should be noted that at higher cutting velocities also the cutting forces will be higher, especially for the non-cavitating cutting process. Further, the inertial force, which is neglected in this research, may also play a role at very high cutting velocities.
6.13.4. The Influence of the Width of the Blade
The blade on which the cutting forces are measured is embedded between two side blades. These side blades have to take care of the three-dimensional side effects, so that on the middle blade a two-dimensional cutting process takes place. The question now is how wide the side blades need to be, at a certain cutting depth, to avoid a significant presence of the side effects on the middle blade. Essential is, that at the deepest cutting depth the side effects on the middle blade are negligible. For this research the following blade configurations are used:
1. A middle blade of 150 mm and two side blades of 185 mm each.
2. A middle blade of 200 mm and two side blades of 160 mm each.
3. A middle blade of 250 mm and two side blades of 135 mm each.
The total blade width in each configuration is therefore 520 mm. The results of this research are, scaled to a middle blade of 200 mm wide, shown in Table 6-2, in which every value is the average of a number of tests. In this table the forces on the 0.20 m and the 0.25 m wide blade are listed in proportion to the 0.15 m wide blade. The change of the direction of the forces in relation to the 0.15 m wide blade is also mentioned. From this table the following conclusions can be drawn:
1. There is no clear tendency to assume that the side effects influence the cutting forces in magnitude.
2. The widening of the middle blade and thus narrowing the side blades, gives slightly more downward aimed forces on the middle blade at a blade angle of 30o. At a blade angle of 45o this tendency can be seen at a blade- height/layer-thickness ratio of 1 and 2, while at a blade-height/ layer-thickness ratio of 3 the forces are just slightly aimed upward. The 60o blade angle gives the same image as the 45o blade angle, however with smaller differences in proportion to the 0.15 m wide blade.
Table 6-2: The influence of the width ratio between the center blade and the side blades.
w=0.20 m (2)
w=0.25 m (3)
α
hb/hi
ct2/ct1
θt2t1
ct3/ct1
θt3t1
30°
1
0.95
+1.0°
1.02
+1.0°
30°
2
1.10
+2.0°
0.93
+4.0°
30°
3
0.96
+5.0°
1.05
+7.0°
45° 1 1.08 +3.0° 1.01 +5.0°
45° 2 0.93 +3.0° 0.93 +5.0°
45° 3 0.93 -8.0° 1.07 -5.0°
60°
1
1.09
+0.0°
1.00
+1.0°
60°
2
0.90
+1.0°
0.92
+2.0°
60°
3
1.04
-5.0°
0.99
-4.0°
The total measured cutting force ct and the force direction θt, at a blade width of 0.20 m (ct2θt2) (2) and a blade width of 0.25 m (ct3θt3) (3) in proportion to the total cutting force and direction at a blade width of 0.15 m (ct1θt1) (1), according the blade configurations mentioned here.
6.13.5. Side Effects
On the outside of the side blades a three-dimensional cutting process acts, in a sense that the shear zone here is three-dimensional, but on top of that the water flows three-dimensional to the shear zone. This makes the cutting forces differ, in magnitude and direction, from the two-dimensional cutting process. Additionally it is imaginable that also forces will act on the blade in the transversal direction (internal forces in the blade). The influence of the side effects is researched by measuring the forces on both the middle blade as on the side blades. Possible present transversal forces are researched by omitting one side blade in order to be able to research the transversal forces due to the three-dimensional side effects. For this research the following blade configurations are used:
1. A middle blade of 150 mm and two side blades of 185 mm each.
2. A middle blade of 200 mm and two side blades of 160 mm each.
3. A middle blade of 250 mm and two side blades of 135 mm each.
4. A middle blade of 200 mm and one side blade of 160 mm
The results of this research can be found in Table 6-3, where every value represents the average of a number of tests. The cutting forces in this table are scaled to the 200 mm blade to simulate a middle blade without side blades.
w=.15 m (1)
w=.20 m (2)
w=.25 m (3)
w=.20 m (4)
α
hb/hi
cr
θr
cr
θr
cr
θr cr θr
30°
1
1.06
+26°
1.23
+14°
1.17
+11°
1.01
+13°
30°
2
0.78
+18°
0.87
+16°
0.83
+10°
1.14
+10°
30°
3
0.74
+22°
0.56
+22°
0.53
+11°
1.45
+ 6°
45°
1
1.13
+23°
1.10
+14°
1.26
+ 9°
1.04
+ 5°
45°
2
0.94
+19°
0.94
+11°
0.93
+ 7°
0.92
+ 7°
45°
3
0.79
+14°
1.10
+17°
0.98
+11°
0.85
+ 6°
60°
1
1.10
+ 8°
1.10
+ 6°
1.10
+ 5°
1.04
+ 2°
60°
2
0.94
+12°
1.10
+ 8°
1.06
+ 6°
0.91
+ 2°
60°
3
0.77
+ 8°
0.99
+15°
1.02
+11°
0.86
+ 3°
The cutting force on the side blades in ratio to the cutting force on the middle blade cr, assuming that the cutting process on the middle blade is two-dimensional. Also shown is the change of direction of the total cutting force θr. The cutting forces are scaled to the width of the middle blade for the blade widths 0.15 m (1), 0.20 m (2) and 0.25 m (3). The second column for w=.20 m (4) contains the results of the tests with only one side blade to measure the side effects on the middle blade. The measured cutting forces are compared to the similar tests where two side blades are used. The blade configurations are according to chapter 6.13.4. From this research the following conclusions can be drawn:
1. For all blade angles the cutting force on the edge is larger than follows from the two-dimensional process, for a blade-height / layer-thickness ratio of 1.
2. A blade-height / layer-thickness ratio of 2 or 3 shows a somewhat smaller cutting force with a tendency to smaller forces with a higher blade-height / layer-thickness ratio.
3. The direction of the cutting force is, for all four blade configurations, aimed more downwards on the sides than in the middle, where the differences with the middle blade decrease with a wider middle blade and therefore less wide side blades. This implies that, with the widening of the middle blade, the influence of the three-dimensional cutting process on the middle blade increases with a constant total blade width. This could be expected. It also explains that the cutting force in the middle blade is directed more downwards with an increasing middle blade width.
4. Blade configuration 4 differs slightly, as far as the magnitude of the forces is concerned, from the tendency seen in the other three configurations with the 30 blade. The direction of the cutting forces match with the other configurations. It has to be remarked that in this blade configuration the side effects occur only on one side of the blade, which explains the small change of the cutting forces.
5. The measured transverse forces for blade configuration 4 are in the magnitude of 1% of the vector sum of the horizontal and the vertical cutting forces and therefore it can be concluded that the transverse forces are negligible for the used sand.
The conclusions found are in principle only valid for the sand used. The influence of the side effects on the magnitude and the direction of the expected cutting forces will depend on the ratio between the internal friction of the sand and the soil/steel friction. This is because the two-dimensional cutting process is dominated by both angles of friction, while the forces that occur on the sides of the blade, as a result of the three-dimensional shear plane, are dominated more by the internal friction of the sand.
6.13.6. Scale Effects
The soil mechanical research showed that the density of the sand increases slightly with the depth. Since both the permeability and the volume strain, and less significant the other soil mechanical parameters, are influenced by the density, it is important to know the size of this influence on the cutting forces (assuming that the two- dimensional cutting theory is a valid description of the process). If the two-dimensional cutting theory is a valid description of the process, the dimensionless cutting forces will have to give the same results for similar geometric ratios, independent of the dimensions and the layer-thickness, according to the equations for the non-cavitating cutting process and the cavitating cutting process. The following blade configurations are used to research the scaling influence:
1. A blade with a width of 150 mm wide and a height of 100 mm.
2. A blade with a width of 150 mm wide and a height of 150 mm.
3. A blade with a width of 150 mm wide and a height of 200 mm.
4. A blade with a width of 150 mm wide and a height of 300 mm.
The results of this research can be found in Table 6-4, where every value represents the average value of a number of tests.
Table 6-4: Influence of the scale factor.
Configuration
1
2
3
4
α
hb/hi
h = 0.10
0.15
0.20
0.30
30°
1
0.93
1.00
0.94
1.18
30°
2
1.23
1.00
1.06
1.13
30°
3
----
1.00
0.89
0.90
45°
1
0.95
1.00
1.13
----
45°
2
0.89
1.00
1.05
1.30
45°
3
----
1.00
1.02
1.13
60°
1
0.91
1.00
---- ----
60°
2
0.90
1.00
1.19
1.04
60°
3
1.02
1.00
1.13
1.21
The total cutting force ct with blade heights of 0.10 m (1), 0.15 m (2), 0.20 m (3) and 0.30 m (4) in proportion to the cutting force at a blade height 0.15 m (2). The blade configurations are according chapter 6.13.4. Because the influences of the gravity and inertia forces can disturb the character of the dimensionless forces compared to 0 to 0, the measured forces are first corrected for these influences. The forces in the table are in proportion to the forces that occurred with blade configuration 2. The following conclusions can be drawn from the table:
1. There is a slight tendency to larger dimensionless forces with increasing dimensions of the blades and the layer-thickness, which could be expected with the slightly increasing density.
2. For a blade angle of 30o and a blade-height / layer-thickness ratio of 2, large dimensionless forces are measured for blade configuration 1. These are the tests with the thinnest layer-thickness of 25 mm. A probable cause can be that the rounding of the blade tip in proportion with the layer-thickness is relatively large, leading to a relatively large influence of this rounding on the cutting forces. This also explains the development of the dimensionless forces at a blade angle of 30o and a blade-height / layer-thickness ratio of 3.
6.13.7. Comparison of Measurements versus Theory
The results of the preceding three investigations are collected in Table 6-5, compared with the theory. Every value is the average of a number of tests. In the table it can be found:
1. The dimensionless forces, the average from the several scales and blade widths.
2. As 1, but corrected for the gravity and inertia forces.
3. The theoretical dimensionless forces according to Appendix D to Appendix J.
measured
calculated
not-corrected
corrected
theoretical
α
hb/hi
ct
θt
ct θt ct θt
30°
1
0.52
+13.3°
0.48
+17.1°
0.39
+28.3°
30°
2
0.56
+17.0°
0.53
+20.1°
0.43
+27.4°
30°
3
0.56
+24.8°
0.53
+28.2°
0.43
+27.3°
45°
1
0.71
+ 4.9°
0.63
+ 7.5°
0.49
+12.9°
45°
2
0.75
+ 6.0°
0.66
+ 8.0°
0.57
+10.7°
45°
3
0.76
+ 5.1°
0.70
+ 6.9°
0.61
+ 9.9°
60°
1
1.06
+ 1.2°
0.88
+ 1.9°
0.69
- 0.7°
60°
2
1.00
- 2.4°
0.84
- 3.4°
0.83
- 3.2°
60°
3
0.99
- 3.4°
0.85
- 4.2°
0.91
- 4.6°
The total cutting force measured (not-corrected and corrected for the gravity and inertia forces) and the theoretical total cutting forces (all dimensionless). The theoretical values for ct and θt are based on an angle of internal friction of 38o, a soil/steel angle of friction of 30o and a weighted average permeability of approximately 0.000242 m/s dependent on the weigh factor a1. The total cutting force ct and the force direction θt are determined according chapter 6.12.4. The following conclusions can be drawn from this table:
1. The measured and corrected cutting forces are larger than the, according to the theory, calculated cutting forces, at blade angles of 30o and 45o. The differences become smaller with an increase in the blade angle and when the blade-height / layer-thickness ratio increases.
2. For a blade angle of 60o the corrected measure forces agree well with the calculated forces.
3. The tendency towards larger forces with a larger blade-height / layer-thickness ratio (theory) is clearly present with blade angles 30o and 45o.
4. At a blade angle of 60o the forces seem to be less dependent of the blade-height / layer-thickness ratio.
5. The direction of the measured cutting forces agrees well with the theoretical determined direction. Only at the blade angle of 30o the forces are slightly aimed more upward for the blade-height / layer-thickness ratios 1 and 2.
6. Neglecting the inertia forces, gravity, etc. introduces an error of at least 15% within the used velocity range. This error occurs with the 60o blade, where the cutting velocity is the lowest of all cutting tests and is mainly due to the gravity.
Considering that the sand, in the course of the execution of the tests, as a result of segregation, has obtained a slightly coarser grain distribution and that the tests are performed with an increasing blade angle, can be concluded that the test results show a good correlation with the theory. It has to be remarked, however, that the scale and side effects can slightly disturb the good correlation between the theory and the measurements.
6.13.8. Location of the Resulting Cutting Force
A quantity that is measured but has not been integrated in the theory is the location of the resulting cutting force. This quantity can be of importance for the determination of the equilibrium of a drag head. The locations, of the in this chapter performed tests, are listed in Table 6-6. Table 6-7 lists the dimensionless locations of the resulting cutting force, in relation with the layer-thickness.
Table 6-6: The location of the resulting cutting force.
Configuration
1
2
3
4
α
hb/hi
h = 0.10
0.15
0.20
0.30
30°
1
51.25
63.1
96.7
157.2
30°
2
76.00
55.7
61.3
84.8
30°
3
----
50.5
54.3
71.5
45°
1
66.38
87.5
128.0
----
45°
2
55.13
56.9
73.4
128.6
45°
3
----
62.0
56.0
82.1
60°
1
69.88
99.5
----
----
60°
2
50.00
68.4
86.1
123.9
60°
3
46.25
55.0
66.3
95.1
The location of the resulting cutting force in mm from the blade tip, for the blade configurations of chapter 6.13.4.
Table 6-7: The location of the resulting cutting force.
Configuration
1
2
3
4
α
hb/hi
h = 0.10
0.15
0.20
0.30
30°
1
0.51
0.42
0.48
0.59
30°
2
1.52
0.75
0.61
0.56
30°
3
----
1.01
0.82
0.71
45°
1
0.67
0.58
0.64
----
45°
2
1.11
0.76
0.63
0.73
45°
3
----
1.25
0.84
0.83
60°
1
0.70
0.66
----
----
60°
2
1.01
0.91
0.86
0.83
60°
3
1.38
1.11
0.99
0.95
The location of the resulting cutting force from the blade tip, along the blade, made dimensionless by dividing with the layer-thickness, for the blade configurations of chapter 6.13.4. From these tables the following conclusions can be drawn:
1. The location of the resulting cutting force is closer to the blade tip with larger blade dimensions.
2. The location of the resulting cutting force is closer to the blade tip with a smaller blade-height / layer-thickness ratio.
The first conclusion can be based upon the fact that a possible present adhesion, on a larger scale (and therefore layer-thickness) causes, in proportion, a smaller part of the cutting force. For the second conclusion this can also be a cause, although the blade-height / layer-thickness ratio must be seen as the main cause.
6.13.9. Verification of the Theory in 200 μm Sand
The linear cutting theory is researched on three points:
1. The distribution of the water sub-pressures on the blade for a blade with a radius of rounding of 1 mm.
2. The distribution of the water sub-pressures on the blade for a blade with a flat wear face of approximately 10 mm and a clearance angle of 1o.
3. The correlation between the measured cutting forces and the theoretical cutting forces.
The dimensions of the blades and the wear faces can be found in Figure 6-45. In Table 6-10 the ratios of the wear face length and the layer-thickness are listed. In the preceding paragraph already a few conclusions are drawn upon the correlation between the measured and the calculated cutting forces. In this research both the forces and the water pressures are measured to increase the knowledge of the accuracy of the theory. Also it has to be mentioned that the soil mechanical parameters are determined during this research.
In Figure 6-56 the results of a test are shown. The results of the whole research of the forces are listed in Table 6-8 for the blade with the radius of rounding of 1 mm and in Table 6-9 for the blade with the wear flat. The dimensionless measured water sub-pressures are shown in Appendix M: Experiments in Water Saturated Sand, in which the theoretical distribution is represented by the solid line. The water sub-pressures are made dimensionless, although the weighted average permeability km is used instead of the permeability kmax used in the equations. From this research the following conclusions can be drawn:
1. The measured forces and water sub-pressures show, in general, a good correlation with the theory.
2. The tendency towards increasing and more upward aimed forces with increasing blade angles can be observed clearly in the Table 6-8 and Table 6-9.
3. The ratio between the measured and calculated forces becomes smaller when the blade angle and the blade-height / layer-thickness ratio increase.
4. The cutting forces on the blade with the wear face are almost equal to the cutting forces on the blade with the radius of rounding, but are slightly aimed more upward.
5. The ratio between the measured and calculated water sub-pressures is, in general, smaller than the ratio between the measured and calculated cutting forces.
6. The measured water sub-pressures on the blade with the wear face and the blade with the radius of rounding differ slightly (Table 6-10) from the water sub-pressures on the blade with the radius of rounding. On the 30o and the 45o blade, the water sub-pressures tends to smaller values for the blade with the wear face, although the differences are very small. On the 60o blade these water sub-pressures are slightly higher. Therefore it can be concluded that, for water pressures calculations, the wear-section-length / layer-thickness ratio w/hi has to be chosen dependent of the blade angle. Which was already clear during the tests because the clearance angle increased with a larger blade angle. For the determination of Appendix H to Appendix J, however, the ratio used was w/hi=0.2, which is a good average value.
measured
calculated
not-corrected
corrected
theoretical
α
hb/hi
ct
θt
ct
θt ct θt
30°
1
0.54
+29.3°
0.49
+29.0°
0.39
+28.3°
30°
2
0.48
+27.5°
0.46
+27.2°
0.43
+27.4°
30°
3
0.49
+27.6°
0.46
+27.3°
0.43
+27.3°
45°
1
0.78
+15.1°
0.58
+13.9°
0.49
+12.9°
45°
2
0.64
+12.3°
0.59
+11.6°
0.57
+10.7°
45°
3
0.60
+11.0°
0.55
+10.5°
0.61
+ 9.9°
60°
1
1.16
+ 0.7°
0.77
- 0.6°
0.69
+ 0.7°
60°
2
0.95
- 1.4°
0.79
- 2.2°
0.83
- 3.2°
60°
3
0.93
- 3.4°
0.82
- 4.0°
0.91
- 4.6°
60°
6
0.70
- 4.8°
0.64
- 5.7°
1.14
- 7.4°
Measured dimensionless forces, not-corrected and corrected for gravity and inertia forces and theoretical values according to Appendix H to Appendix J for the blade with the radius of rounding and the sub-pressure behind the blade. The theoretical values for ct and θt are determined based on values for the angle of internal friction of 38o, a soil/steel angle of friction of 30o and a weighted average permeability of 0.000242 m/s, dependent on the weigh factor a1.
Table 6-9: Measured dimensionless forces.
measured
calculated
not-corrected
corrected
theoretical
α
hb/hi
ct
θt ct θt ct θt
30°
1
0.53
+26.2°
0.48
+25.9°
0.39
+28.3°
30°
2
0.48
+24.0°
0.46
+23.7°
0.43
+27.4°
30°
3
0.49
+24.7°
0.46
+24.3°
0.43
+27.3°
45°
1
0.72
+11.9°
0.57
+11.0°
0.49
+12.9°
45°
2
0.66
+ 8.8°
0.60
+ 8.3°
0.57
+10.7°
45°
3
0.63
+ 7.8°
0.60
+ 7.3°
0.61
+ 9.9°
60°
1
-----
----- ----- ----- ----- -----
60°
2
0.90
- 5.6°
0.80
- 6.2°
0.83
- 3.2°
60°
3
0.95
- 7.3°
0.87
- 8.0°
0.91
- 4.6°
60°
6
0.70
- 9.2°
0.64
-10.1°
1.14
- 7.4°
Measured dimensionless forces, not-corrected and corrected for gravity and inertia forces and theoretical values according to Appendix H to Appendix J for the blade with the flat wear face and the sub-pressure behind the blade. The theoretical values for ct and θt are determined according chapter 6.12.4. They are based on values for the angle of internal friction of 38o, a soil/steel angle of friction of 30o and a weighted average permeability of 0.000242 m/s, dependent on the weigh factor a1.
Table 6-10: Average dimensionless pore pressures on the blade.
α
hb/hi
w
hi
w/hi
p2ma
p2ms
p2m
p2ms/ p2ma
30°
1
10.2
100
0.102
0.076
0.073
0.076
0.96
30°
2
10.2
50
0.204
0.051
0.050
0.049
0.98
30°
3
10.2
33
0.308
0.034
0.030
0.034
0.88
45°
1
11.1
141
0.079
0.090
0.080
0.097
0.89
45°
2
11.1
70
0.159
0.069
0.068
0.082
0.99
45°
3
11.1
47
0.236
0.052
0.051
0.065
0.98
60°
1
13.3
173
0.077
0.107
----
0.091
----
60°
2
13.3
87
0.153
0.083
0.090
0.100
1.08
60°
3
13.3
58
0.229
0.075
0.081
0.094
1.08
60°
6
13.3
30
0.443
0.035
0.038
0.061
1.09
The average dimensionless pore pressures on the blade, on the blade with the radius of rounding p2ma and the blade with the wear face p2ms , the theoretical values p2m and the ratio between the sub-pressures p2ms and p2ma, as a function of the length of the wear face (mm), the layer-thickness hi (mm) and the wear-section-length / layer- thickness ratio.
6.13.10. Verification of the Theory in 105 μm Sand
The linear cutting theory for the 105 μm is investigated on three points:
1. The distribution of the water sub-pressures on the blade in a non-cavitating cutting process.
2. The distribution of the water sub-pressures on the blade in the transition region between the non-cavitating and the cavitating cutting process.
3. The correlation between the measured cutting forces and the theoretical calculated cutting forces.
The dimensions of the blades can be found in Figure 6-45. In this research only a 30o blade with a layer-thickness of 100 mm, a 45o blade with a layer-thickness of 70 mm and a 60o with a layer-thickness of 58 mm, are used, at a blade height h of 200 mm. The soil mechanical parameters of the used sand are listed in Appendix L. The results of the research regarding the cutting forces can be found in Table 6-11.
measured
calculated
α hb/hi
ct
θt ct θt ct θt
no cavitation
not-corrected
corrected
theoretical
30°
1
.45
+16.5°
.45
+25.6°
.41
+25.1°
45°
2
.50
- 3.5°
.47
+ 7.2°
.62
+ 7.6°
60°
3
.60
- 8.8°
.58
- 6.3°
1.02
- 7.5°
cavitation
not-corrected
corrected
theoretical
30°
1
3.4
+13.1°
3.4
+24.2°
3.3
+21.6°
45°
2
4.7
-10.3°
4.2
+ 5.7°
4.6
+ 2.6°
60°
3
4.9
- 9.0°
4.8
- 7.8°
6.8
-12.1°
Measured dimensionless forces, not-corrected and corrected for gravity and inertia forces and the theoretical values according to Appendix C to Appendix G for the non-cavitating cutting process and according to Appendix H to Appendix J for the cavitating cutting process, calculated with a sub-pressure behind the blade. The values of ct and θt are calculated according chapter 6.12.4. They are based on values for the angle of internal friction of 38o, a soil/steel angle of friction of 30o and a weighted average permeability between 0.00011 m/s and 0.00012 m/s, dependent on the weigh factor a1 and the initial pore percentage of the sand bed.
The dimensionless measured water sub-pressures of the non-cavitating cutting process are presented in Appendix M, in which the solid line represents the theoretical distribution. The dimensionless measured water sub-pressures in the transition region are also presented in Appendix M. The figures in Appendix M show the measured horizontal forces Fh, in which the solid line represents the theoretical distribution. Other figures show the measured vertical forces Fv, in which the solid line represents the theoretical distribution. Also shown in is the distribution of the forces, for several water depths, during a fully cavitating cutting process (the almost horizontal lines). From this research the following conclusions can be drawn:
1. The tests with the 30o blade give a good correlation with the theory, both for the forces as for the water sub- pressures. For the 45o blade both the forces and the water sub-pressures are lower than the theoretical calculated values with even larger deviations for the 60o blade. For the 60o blade the forces and the water sub-pressures values are approximately 60% of the calculated values.
2. The direction of the cutting forces agrees reasonably well with the theory for all blade angles, after correction for the gravity and the inertia forces.
3. The figures in Appendix M show that the profile of the water sub-pressures on the blade, clearly changes shape when the peak stress close to the blade tip (sub-pressure) has a value of approximately 65% of the absolute pressure. An increase of the cutting velocity results in a more flattening profile, with a translation of the peak to the middle of the blade. No cavitation is observed but rather an asymptotic approach of the cavitation pressure with an increasing cutting velocity. For the 60o blade the flattening only appears near the blade tip. This can be explained with the large blade-height / layer-thickness ratio. This also explains the low cutting forces in the range where cavitation is expected. There is some cavitation but only locally in the shear zone; the process is not yet fully cavitated.
4. Since, according to the theory, the highest sub-pressures will appear in the shear zone, cavitation will appear there first. The theoretical ratio between the highest sub-pressure in the shear zone and the highest sub-pressure on the blade is approximately 1.6, which is in accordance with conclusion 3. Obviously there is cavitation in the shear zone in these tests, during which the cavitation spot expands to above the blade and higher above the blade with higher cutting velocities.
In Appendix M the pore pressure graphs show this relation between the cavitation spot and the water pressures profile on the blade. The water sub-pressures will become smaller where the cavitation spot ends. This also implies that the measurements give an impression of the size of the cavitation spot.
As soon as cavitation occurs locally in the sand package, it becomes difficult to determine the dimensionless coefficients c1 and c2 or d1 and d2. This is difficult because the cutting process in the transition region varies between a cavitating and a non-cavitating cutting process. The ratio between the average water pressure in the shear zone and the average water pressure on the blade surface changes continuously with an increasing cutting velocity. On top of that the shape and the size of the area where cavitation occurs are unknown. However, to get an impression of the cutting process in the transition region, a number of simplifications regarding the water flow through the pores are carried out.
1. The flow from the free sand surface to the shear zone takes place along circular flow lines (see equations (6-37) and (6-38)), both through the packed sand as through the cut sand. With this assumption the distance from the free sand surface to the cavitation area can be determined, according:
$\ \xi_{0}=\frac{(\mathbf{z}+\mathbf{1 0})}{\mathbf{v}_{\mathbf{c}} \cdot \mathbf{\varepsilon} \cdot \sin (\beta)} \cdot\left(\frac{\mathbf{k}_{\max }}{\alpha+\beta}+\frac{\mathbf{k}_{\mathbf{i}}}{\pi-\beta}\right) \cdot \sin (\alpha+\beta)\tag{6-107}$
2. The flow in the cut sand is perpendicular to the free sand surface, from the breakpoint where the shear plane reaches the free sand surface. This flow fills the water vapor bubbles with water. The distance from the free sand surface to the cavitating area can now be determined, under the assumption that the volume flow rate of the vapor bubbles equals the volume flow rate of the dilatancy, according:
$\ \frac{\mathbf{k}_{\max } \cdot(\mathbf{z}+\mathbf{1 0})}{\xi} \cdot \mathbf{d} \Psi=\mathbf{v}_{\mathbf{c}} \cdot \varepsilon \cdot \frac{\sin (\beta)}{\sin (\alpha+\beta)} \cdot \mathbf{d} \xi\tag{6-108}$
3. In which the right term represents the volume flow rate of the vapor bubbles from the dilatancy zone, while the left term represents the supply of water from the free sand surface. This is shown in Appendix M the pore pressure graphs. With the initial value from equation (6-107) the following solution can be found:
$\ \xi=\sqrt{\xi_{0}^{2}+2 \cdot \frac{\mathbf{k}_{\max } \cdot(\mathbf{z}+\mathbf{1 0})}{\left(\mathbf{v}_{\mathbf{c}} \cdot \frac{\sin (\beta)}{\sin (\alpha+\beta)} \cdot \varepsilon\right)} \cdot \Psi}\tag{6-109}$
The distance from the blade to the cavitation spot is considered to be constant over the blade. The magnitude of this distance is however unknown.
The relation between the dimensions of the cavitation spot, and the water pressure profile on the blade.
The progressive character of the cavitation spot development results from equation (6-109). If, at a certain cutting velocity, cavitation occurs locally in the cavitation zone, then the resulting cavitation spot will always expand immediately over a certain distance above the blade as a result of the fact that a certain time is needed to fill the volume flow rate of the vapor bubbles. The development of the water sub-pressures will, in general, be influenced by the ever in the pore water present dissolved air. As soon as water sub-pressures are developing as a result of the increase in volume in the shear zone, part of the dissolved air will form air bubbles. Since these air bubbles are compressible, a large part of the volume strain will be taken in by the expansion of the air bubbles, which results in a less fast increase of the water sub-pressures with an increasing cutting velocity. The maxima of the water sub- pressures will also be influenced by the present air bubbles. This can be illustrated with the following example: Assume the sand contains 3 volume percent air, which takes up the full volume strain in the dilatancy zone. With a volume strain of 16%, this implies that after expansion, the volume percentage air is 19%. Since it is a quick process, it may be assumed that the expansion is adiabatic, which amounts to maximum water sub-pressures of 0.925 times the present hydrostatic pressure. In an isothermal process the maximum water sub-pressures are 0.842 times the present hydrostatic pressure. From this simple example it can be concluded that the, in the pore water present (either dissolved or not) air, has to be taken into account. In the verification of the water sub-pressures, measured during the cutting tests in the 105 μm sand, the possibility of a presence of dissolved air is recognized but it appeared to be impossible to quantify this influence. It is however possible that the maximum water sub- pressures reached (Appendix M the pore pressure graphs) are limited by the in the pore water present dissolved air.
6.13.11. Determination of φ and δ from Measurements
The soil/steel friction angle δ and the angle of internal friction φ can be determined from cutting tests. Sand without cohesion or adhesion is assumed in the next derivations, while the mass of the cut layer has no influence on the determination of the soil/steel friction angle. In Figure 6-51 it is indicated which forces, acting on the blade, have to be measured to determine the soil/steel friction angle δ.
The forces Fh and Fv can be measured directly. Force W2 results from the integration of the measured water pressures on the blade. From this figure the normal force on the blade, resulting from the grain stresses on the blade, becomes:
$\ \mathrm{F}_{\mathrm{n}}=\mathrm{W}_{\mathrm{2}}-\mathrm{W}_{\mathrm{3}}+\mathrm{F}_{\mathrm{h}} \cdot \sin (\alpha)+\mathrm{F}_{\mathrm{v}} \cdot \cos (\alpha)\tag{6-110}$
The friction force, resulting from the grain stresses on the blade, becomes:
$\ \mathrm{F}_{\mathrm{w}}=\mathrm{F}_{\mathrm{h}} \cdot \cos (\alpha)-\mathrm{F}_{\mathrm{v}} \cdot \sin (\alpha)\tag{6-111}$
The soil/steel angle of friction now becomes:
$\ \delta=\arctan \left(\frac{\mathrm{F}_{\mathrm{w}}}{\mathrm{F}_{\mathrm{n}}}\right)\tag{6-112}$
Determination of the angle of internal friction from the cutting tests is slightly more complicated. In Figure 6-52 it is indicated which forces, acting on the cut layer, have to be measured to determine this angle. Directly known are the measured forces Fh and Fv. The force W1 is unknown and impossible to measure. However from the numerical water pressures calculations the ratio between W1 and W2 is known. By multiplying the measured force W2 with this ratio an estimation of the value of the force W1 can be obtained, so:
$\ \mathrm{W}_{1}=\left(\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}\right)_{\mathrm{c a l c}} \cdot \mathrm{W}_{2 \mathrm{mean}}\tag{6-113}$
For the horizontal and the vertical force equilibrium of the cut layer can now be written:
$\ \mathrm{F}_{\mathrm{h}}-\mathrm{W}_{\mathrm{3}} \cdot \sin (\alpha)=\mathrm{K}_{1} \cdot \sin (\beta+\phi)-\mathrm{W}_{1} \cdot \sin (\beta)+\mathrm{I} \cdot \cos (\beta)\tag{6-114}$
$\ \mathrm{F}_{\mathrm{v}}-\mathrm{W}_{\mathrm{3}} \cdot \cos (\boldsymbol{\alpha})=-\mathrm{K}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta}+\phi)+\mathrm{W}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta})+\mathrm{I} \cdot \sin (\boldsymbol{\beta})+\mathrm{G}\tag{6-115}$
The angle of internal friction:
$\ \mathrm{\phi=\arctan \left(\frac{F_{h}-W_{3} \cdot \sin (\alpha)+W_{1} \cdot \sin (\beta)-I \cdot \cos (\beta)}{-F_{v}+W_{3} \cdot \cos (\alpha)+W_{1} \cdot \cos (\beta)+I \cdot \sin (\beta)+G}\right)-\beta}\tag{6-116}$
The equations derived (6-112) and (6-116) are used to determine the values of φ and δ from the cutting tests carried out. The soil/steel friction angle can quite easily be determined, with the remark that the side and wear effects can influence the results from this equation slightly. The soil/steel friction angle, determined with this method, is therefore a gross value. This value, however, is of great practical importance, because the side and wear effects that occur in practice are included in this value.
The soil/steel friction angle δ, determined with this method, varied between 24o and 35o, with an average of approximately 30o. For both types of sand almost the same results were found for the soil/steel friction angle. A clear tendency towards stress or blade angle dependency of the soil/steel angle of friction is not observed. This in contrast to Van Leussen and Nieuwenhuis (1984), who found a blade angle dependency according Hettiaratchi and Reece (1974).
Harder to determine is the angle of internal friction. The following average values for the angle of internal friction are found, for the 200 μm sand:
• α = 30° » φ = 46.7°
• α = 45° » φ = 45.9°
• α = 60° » φ = 41.0°
These values are high above the angle of internal friction that is determined with soil mechanical research according to Appendix K, for a pore percentage of 38.5%. From equation (6-116) it can be derived that the presence of sub-pressure behind the blade makes the angle of internal friction smaller and also that this reduction is larger when the blade angle is smaller. Within the test program space is created to perform experiments where the sub- pressure is measured both on and behind the blade (Figure 6-53). Pressure transducer p1 is removed from the blade and mounted behind the blade tip. Although the number of measurements was too limited to base a theoretical or empirical model on, these measurements have slightly increased the understanding of the sub-pressure behind the blade. Behind the blade tip sub-pressures are measured, with a value of 30% to 60% of the peak pressure on the blade. The highest sub-pressure behind the blade was measured with the 30o blade. This can be explained by the wedge shaped space behind the blade. The following empirical equation gives an estimate of the force W3 based on these measurements:
$\ \mathrm{W}_{3}=\mathrm{0 .3} \cdot \cot (\alpha) \cdot \mathrm{W}_{2}\tag{6-117}$
The determination of the angle of internal friction corrected for under pressure behind the blade W3 led to thefollowing values:
• α = 30° » φ = 36.6°
• α = 45° » φ = 39.7°
• α = 60° » φ = 36.8°
For the verification of the cutting tests an average value of 38o for the internal angle of friction is assumed. These values are also more in accordance with the values of internal friction mentioned in Appendix K, where a value of approximate 35o can be found with a pore percentage of 38.5%.
The same phenomena are observed in the determination of the angle of internal friction of the 105 μm sand. The assumption of a hydrostatic pressure behind the blade resulted also in too large values for the angle of internal friction, analogously to the calculations of the 200 μm sand. Here the following values are determined:
• α = 30° » φ = 46.2°
• α = 45° » φ = 38.7°
• α = 60° » φ = 40.3°
The determination of the angle of internal friction corrected for under pressure behind the blade W3 led to the following values:
• α = 30° » φ = 38.7°
• α = 45° » φ = 34.0°
• α = 60° » φ = 38.4°
The low value of the angle of internal friction for the 45o blade can be explained by the fact that these tests are performed for the first time in the new laboratory DE in a situation where the sand was not homogenous from top to bottom. For the verification of the cutting forces and the water pressures is, for both sand types, chosen for a soil/steel friction angle of 30o and an angle of internal friction of 38o, as average values.
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From the performed research the following general conclusions can be drawn:
1. Both the measured cutting forces as the measured water sub-pressures agree reasonably with the theory. For both sand types is observed that the cutting forces and the water sub-pressures become smaller in comparison with the theory, when the blade angle becomes larger. For the 30o blade the cutting forces and the water sub- pressures are larger or equal to theoretical derived values, while for the 60o blade the theory can overestimate the measurements with a factor 1.6. This can be explained by assuming that with an increasing blade angle the cutting process becomes more discontinuous and therefore decreases the average volume strain rate. Slices of sand shear off with dilatancy around the shear planes, while the dilatancy is less in the sand between the shear planes. The theory can still be pretty useful since in dredging practice the used blade angles are between 30o and 45o.
2. Side effects can considerably influence the direction of the cutting forces, although the magnitude of the cutting forces is less disturbed. As a result of the side effects the cutting forces are aimed more downward.
3. Wear effects can also influence the direction of the cutting forces considerably, while also the magnitude of the cutting forces is less disturbed. As a result of the wear the cutting forces are, however, aimed more upwards.
6.15: The Snow Plough Effect
To check the validity of the above derived theory, research has been carried out in the new laboratory DE. The tests are carried out in hard packed water saturated sand, with a blade of 0.3 m by 0.2 m. The blade had a cutting angle of 45 degrees and inclination angles of 0, 15, 30 and 45 degrees. The layer thicknesses were 2.5, 5 and 10 cm and the drag velocities 0.25, 0.5 and 1 m/s. Figure 6-57 and Figure 6-58 show the results with and without an inclination angle of 45 degrees. The lines in this figure show the theoretical forces. As can be seen, the measured forces match the theoretical forces well. Since the research is still in progress, further publications on this subject will follow.
More results of measurements can be found in Appendix M and Appendix N
The result of a cutting test graphically. In this figure the horizontal force Fhthe vertical force Fand the water pore-pressures on the blade P1, P2, P3 and P4 are shown. The test is performed with a blade angle α of 45o, a layer thickness hi of 70 mm and a cutting velocity vc of 0.68 m/s in the 200 μm sand.
6.16: Nomenclature
a1,a2
Weight factors k-value (permeability)
-
A
Surface
m2
bpr
Projected width of the blade perpendicular to the velocity direction
m
ci ,c1 ,c2
Coefficients (non-cavitating cutting process)
-
cr
Coefficient side effects
-
cs
Wear coefficient
-
ct
Coefficient total cutting force (non-cavitating cutting process)
-
cts
Coefficient total cutting force including wear effects
-
ctr
Coefficient total cutting force including side effects
-
di ,d1 ,d2
Coefficients (cavitating cutting process)
-
dr
Coefficient side effects
-
ds
Wear coefficient
-
dt
Coefficient total cutting force (cavitating cutting process)
-
dts
Coefficient total cutting force including wear
-
dtr
Coefficient total cutting force including side effects
-
Esp
Specific cutting energy
kN/m2
Egc
Specific cutting energy (no cavitation)
kN/m2
Eca
Specific cutting energy (full cavitation)
kN/m2
Fci
Cutting force (general)
kN
Fcit
Total cutting force (general)
kN
Fh
Horizontal cutting force (parallel to the cutting speed)
kN
Fl
Cutting force parallel to the edge of the blade
kN
Fn
Normal force
kN
Fv
Vertical cutting force (perpendicular to the cutting velocity)
kN
Fw
Friction force
kN
Fx
Cutting force in x-direction (longitudinal)
kN
Fxt
Total cutting force in x-direction (longitudinal)
kN
Fy
Cutting force in y-direction (transversal)
kN
Fyt
Total cutting force in y-direction (transversal)
kN
Fz
Cutting force in z-direction (vertical)
kN
g
Gravitational acceleration
m/s2
hi
Initial layer thickness
m
hb
Blade height
m
k
Permeability
m/s
ki
Initial permeability
m/s
kmax
Maximum permeability
m/s
km
Effective permeability
m/s
K1
Grain force on the shear zone
kN
K2
Grain force on the blade
kN
l
Length of the shear zone
m
n
Normal on an edge
m
n
Porosity
-
ni
Initial pore percentage
%
nmax
Maximum pore percentage
%
N1
Normal force on the shear zone
kN
N2
Normal force on the blade
kN
p
Number of blades excavating element
-
p
Pressure (water pressure)
kPa
patm
Atmospheric pressure
kPa
Pcalc
Calculated dimensionless pressure (water pore pressure)
-
pdamp
Saturated water pore pressure (12 cm.w.c.)
kPa
Preal
Real pore pressure (water pore pressure)
kPa
p1m
Average pore pressure in the shear zone
-
p2m
Average pore pressure on the blade
-
Pc
Drive power excavating element
kW
q, q1 ,q2
Specific flow
m/s
Q
Flow per unit of blade width
m2/s
s
Length of a stream line
m
s
Measure for the layer thickness
m
S1
Shear force on the shear zone
kN
S2
Shear force on the blade
kN
vc
Cutting velocity perpendicular to the edge of the blade
m/s
V
Volume strain per unit of blade width
m2
w
Width of blade of blade element
m
W1
Pore pressure force on the shear zone
kN
W2
Pore pressure force on the blade
kN
x
Coordinate
m
y
Coordinate
m
z
Coordinate
m
z
Water depth
m
α
Blade angle
rad
β
Shear angle
rad
ε
Volume strain
-
φ
Angle of internal friction
rad
δ
Soil/steel interface friction angle
rad
ρg
Wet density of the sand
ton/m3
ρs
Dry density of the sand
ton/m3
ρw
Density of water
ton/m3
θr
Angular displacement force vector as a result of side effects
rad
θs
Angular displacement force vector as a result of wear
rad
θt
Angle force vector angle in relation to cutting velocity vector
rad
θts
Angle force vector angle in relation to velocity vector including wear
rad
θtr
Angle force vector angle in relation to velocity vector including side effects
rad
Θr
Angular displacement force vector as a result of side effects
rad
Θs
Angular displacement force vector as a result of wear
rad
Θt
Angle force vector angle in relation to cutting velocity vector
rad
Θts
Angle force vector angle in relation to velocity vector including wear
rad
Θtr
Angle force vector angle in relation to velocity vector including side effects
rad
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Definitions:
1. A: The blade tip.
2. B: End of the shear plane.
3. C: The blade top.
4. A-C: The blade surface.
5. hb: The height of the blade.
6. hi: The thickness of the layer cut.
7. vc: The cutting velocity.
8. α: The blade angle.
9. β: The shear angle.
10. Fh: The horizontal force, the arrow gives the positive direction.
11. Fv: The vertical force, the arrow gives the positive direction.
7.02: Introduction
Hatamura and Chijiiwa (1975), (1976A), (1976B), (1977A) and (1977B) distinguished three failure mechanisms in soil cutting. The Shear Type, the Flow Type and the Tear Type. The Flow Type and the Tear Type occur in materials without an angle of internal friction. The Shear Type occurs in materials with an angle of internal friction like sand. A fourth failure mechanism can be distinguished (Miedema (1992)), the Curling Type, as is known in metal cutting. Although it seems that the curling of the chip cut is part of the flow of the material, whether the Curling Type or the Flow Type occurs depends on several conditions. The Curling Type in general will occur if the adhesive force on the blade is large with respect to the normal force on the shear plane. Whether the Curling Type results in pure curling or buckling of the layer cut giving obstruction of the flow depends on different parameters.
Figure 7-2 illustrates the Curling Type mechanism, Figure 7-3 the Flow Type mechanism and Figure 7-4 the Tear Type mechanism as they occur when cutting clay or loam. To predict which type of failure mechanism will occur under given conditions with specific soil, a formulation for the cutting forces has to be derived. The derivation is made under the assumption that the stresses on the shear plane and the blade are constant and equal to the average stresses acting on the surfaces. Figure 7-1 gives some definitions regarding the cutting process. The line A-B is considered to be the shear plane, while the line A-C is the contact area between the blade and the soil. The blade angle is named α and the shear angle β. The blade is moving from left to right with a cutting velocity vc. The thickness of the layer cut is hand the vertical height of the blade hb. The horizontal force on the blade Fis positive from right to left always opposite to the direction of the cutting velocity vc. The vertical force on the blade Fis positive downwards.
Since the vertical force is perpendicular to the cutting velocity, the vertical force does not contribute to the cutting power, which is equal to:
$\ \mathrm{P}_{\mathrm{c}}=\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}\tag{7-1}$
In clay the cutting processes are dominated by cohesion and adhesion (internal and external shear strength). Because of the φ=0 concept, the internal and external friction angles are set to 0. Gravity, inertial forces and pore pressures are also neglected. This simplifies the cutting equations. Clay however is subject to strengthening, meaning that the internal and external shear strength increase with an increasing strain rate. The reverse of strengthening is creep, meaning that under a constant load the material will continue deforming with a certain strain rate.
Under normal circumstances clay will be cut with the flow mechanism, but under certain circumstances the Curling Type or the Tear Type may occur.
The Curling Type will occur when the blade height is big with respect to the layer thickness, hb/hi, the adhesion is high compared to the cohesion a/c and the blade angle α is relatively big.
The Tear Type will occur when the blade height is small with respect to the layer thickness, hb/hi, the adhesion is small compared to the cohesion a/c and the blade angle α is relatively small.
This chapter is based on Miedema (1992), (2009) and (2010).
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7.3.1. Introduction
Previous researchers, especially Mitchell (1976), have derived equations for the strain rate dependency of the cohesion based on the "rate process theory". However the resulting equations did not allow pure cohesion and adhesion. In many cases the equations derived resulted in a yield stress of zero or minus infinity for a material at rest. Also empirical equations have been derived giving the same problems.
Based on the "rate process theory" with an adapted Boltzman probability distribution, the Mohr-Coulomb failure criteria will be derived in a form containing the influence of the deformation rate on the parameters involved. The equation derived allows a yield stress for a material at rest and does not contradict the existing equations, but confirms measurements of previous researchers. The equation derived can be used for silt and for clay, giving both materials the same physical background. Based on the equilibrium of forces on the chip of soil cut, as derived by Miedema (1987 September) for soil in general, criteria are formulated to predict the failure mechanism when cutting clay. A third failure mechanism can be distinguished, the "curling type". Combining the equation for the deformation rate dependency of cohesion and adhesion with the derived cutting equations, allows the prediction of the failure mechanism and the cutting forces involved. The theory developed has been verified by using data obtained by Hatamura and Chijiiwa (1975), (1976A), (1976B), (1977A) and (1977B) with respect to the adapted rate process theory and data obtained by Stam (1983) with respect to the cutting forces. However since the theory developed confirms the work carried out by previous researchers its validity has been proven in advance. In this chapter simplifications have been applied to allow a clear description of the phenomena involved.
The theory in this chapter has been published by Miedema (1992) and later by Miedema (2009) and (2010).
7.3.2. The Rate Process Theory
It has been noticed by many researchers that the cohesion and adhesion of clay increase with an increasing deformation rate. It has also been noticed that the failure mechanism of clay can be of the "flow type" or the "tear type", similar to the mechanisms that occur in steel cutting. The rate process theory can be used to describe the phenomena occurring in the processes involved. This theory, developed by Glasstone, Laidler and Eyring (1941) for the modeling of absolute reaction rates, has been made applicable to soil mechanics by Mitchell (1976). Although there is no physical evidence of the validity of this theory it has proved valuable for the modeling of many processes such as chemical reactions. The rate process theory, however, does not allow strain rate independent stresses such as real cohesion and adhesion. This connects with the starting point of the rate process theory that the probability of atoms, molecules or particles, termed flow units having a certain thermal vibration energy is in accordance with the Boltzman distribution (Figure 7-5):
$\ \mathrm{p}(\mathrm{E})=\frac{1}{\mathrm{R} \cdot \mathrm{T}} \cdot \exp \left(\frac{-\mathrm{E}}{\mathrm{R} \cdot \mathrm{T}}\right)\tag{7-2}$
The movement of flow units participating in a time dependent flow is constrained by energy barriers separating adjacent equilibrium positions. To cross such an energy barrier, a flow unit should have an energy level exceeding certain activation energy Ea. The probability of a flow unit having an energy level greater than a certain energy level Ea can be calculated by integrating the Boltzman distribution from the energy level Ea to infinity, as depicted in Figure 7-6, this gives:
$\ \mathrm{p}_{\mathrm{E}>\mathrm{E}_{\mathrm{a}}}=\exp \left(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}\right)\tag{7-3}$
The value of the activation energy Ea depends on the type of material and the process involved. Since thermal vibrations occur at a frequency given by kT/h, the frequency of activation of crossing energy barriers is:
$\ v=\frac{\mathrm{k} \cdot \mathrm{T}}{\mathrm{h}} \cdot \exp \left(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}\right)\tag{7-4}$
In a material at rest the barriers are crossed with equal frequency in all directions. If however a material is subjected to an external force resulting in directional potentials on the flow units, the barrier height in the direction of the force is reduced by (f•λ/2) and raised by the same amount in the opposite direction. Where f represents the force acting on a flow unit and λ represents the distance between two successive equilibrium positions. From this it can be derived that the net frequency of activation in the direction of the force is as illustrated in Figure 7-7:
$\ v=\frac{\mathrm{k} \cdot \mathrm{T}}{\mathrm{h}} \cdot \exp \left(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}\right) \cdot\left\{\exp \left(\frac{+\mathrm{f} \cdot \lambda}{\mathrm{2} \cdot \mathrm{k} \cdot \mathrm{T}}\right)-\exp \left(\frac{-\mathrm{f} \cdot \lambda}{\mathrm{2} \cdot \mathrm{k} \cdot \mathrm{T}}\right)\right\}\tag{7-5}$
If a shear stress $\ \tau$ is distributed uniformly along bonds between flow units per unit area then f=$\ \tau$ /S and if the strain rate is a function of the proportion of successful barrier crossings and the displacement per crossing according to dε/dt=X·$\ v$ then:
$\ \dot{\varepsilon}=2 \cdot \mathrm{X} \cdot \frac{\mathrm{k} \cdot \mathrm{T}}{\mathrm{h}} \cdot \exp \left(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}\right) \cdot \sinh \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{\mathrm{2} \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right) \quad \quad \mathrm{w} \mathrm{i} \mathrm{th}: \mathrm{R}=\mathrm{N} \cdot \mathrm{k}\tag{7-6}$
From this equation, simplified equations can be derived to obtain dashpot coefficients for theological models, to obtain functional forms for the influences of different factors on strength and deformation rate, and to study deformation mechanisms in soils. For example:
$\ \text{if } \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{2 \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)<1\quad\text{ then }\quad\sinh \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{2 \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right) \approx\left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{2 \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)\tag{7-7}$
Resulting in the mathematical description of a Newtonian fluid flow, and:
$\text{if } \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{2 \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)>\mathrm{1}\quad\text{ then }\quad\sinh \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{\mathrm{2} \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right) \approx \frac{\mathrm{1}}{2} \cdot \exp \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{2 \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)\tag{7-8}$
Resulting in a description of the Mohr-Coulomb failure criterion for soils as proposed by Mitchell et al. (1968). Zeng and Yao (1988) and (1991) used the first simplification (7-7) to derive a relation between soil shear strength and shear rate and the second simplification (7-8) to derive a relation between soil-metal friction and sliding speed.
7.3.3. Proposed Rate Process Theory
The rate process theory does not allow shear strength if the deformation rate is zero. This implies that creep will always occur since any material is always exposed to its own weight. This results from the starting point of the rate process theory, the Boltzman distribution of the probability of a flow unit exceeding a certain energy level of thermal vibration. According to the Boltzman distribution there is always a probability that a flow unit exceeds an energy level, between an energy level of zero and infinity, this is illustrated in Figure 7-6.
Since the probability of a flow unit having an infinite energy level is infinitely small, the time-span between the occurrences of flow units having an infinite energy level is also infinite, if a finite number of flow units is considered. From this it can be deduced that the probability that the energy level of a finite number of flow units does not exceed a certain limiting energy level in a finite time-span is close to 1. This validates the assumption that for a finite number of flow units in a finite time-span the energy level of a flow unit cannot exceed a certain limiting energy level El. The resulting adapted Boltzman distribution is illustrated in Figure 7-8. The Boltzman distribution might be a good approximation for atoms and molecules but for particles consisting of many atoms and/or molecules the distribution according to Figure 7-8 seems more reasonable, since it has never been noticed that sand grains in a layer of sand at rest, start moving because of their internal energy. In clay some movement of the clay particles seems probable since the clay particles are much smaller than the sand particles. Since particles consist of many atoms, the net vibration energy in any direction will be small, because the atoms vibrate thermally with equal frequency in all directions.
If a probability distribution according to Figure 7-8 is considered, the probability of a particle exceeding a certain activation energy Ea becomes:
$\ \mathrm{p}_{\mathrm{E}>\mathrm{E}_{\mathrm{a}}}=\frac{\exp \left(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}\right)-\exp \left(\frac{-\mathrm{E}_{\ell}}{\mathrm{R} \cdot \mathrm{T}}\right)}{\mathrm{1 - \operatorname { exp }}\left(\frac{-\mathrm{E}_{\ell}}{\mathrm{R} \cdot \mathrm{T}}\right)} \quad\text{ if }\quad\mathrm{E}_{\mathrm{a}}<\mathrm{E}_{\ell}\quad\text{ and }\quad\mathrm{p}_{\mathrm{E}>\mathrm{E}_{\mathrm{a}}}=\mathrm{0} \quad\text{ if }\quad\mathrm{E}_{\mathrm{a}}>\mathrm{E}_{\ell}\tag{7-9}$
If the material is now subjected to an external shear stress, four cases can be distinguished with respect to the strain rate.
Case 1:
The energy level $\ \mathrm{E}_{\mathrm{a}}+\tau \lambda \mathrm{N} / 2 \mathrm{S}$ is smaller than the limiting energy level El (Figure 7-9). The strain rate equation is now:
$\ \begin{array}{left}\dot{\varepsilon}=2 \cdot \mathrm{X} \cdot \frac{\mathrm{k} \cdot \mathrm{T}}{\mathrm{h} \cdot \mathrm{i}} \cdot \exp \left(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}\right) \cdot \sinh \left(\frac{\tau \cdot \lambda \cdot \mathrm{N}}{\mathrm{2} \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)\\text{with: } \mathrm{i=1-exp\left(\frac{-E_{\ell}}{R \cdot T} \right)}\end{array}\tag{7-10}$
Except for the coefficient i, necessary to ensure that the total probability remains 1, equation (7-10) is identical to equation (7-6).
Case 2:
The activation energy Ea is less than the limiting energy El, but the energy level $\ \mathrm{E}+\tau \lambda \mathrm{N} / 2 \mathrm{S}$ is greater than the limiting energy level El (Figure 7-10).
The strain rate equation is now:
$\ \dot{\varepsilon}=\mathrm{X} \cdot \frac{\mathrm{k} \cdot \mathrm{T}}{\mathrm{h} \cdot \mathrm{i}} \cdot\left\{\exp \left(-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}-\frac{\tau \cdot \lambda \cdot \mathrm{N}}{\mathrm{2} \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)\right)-\exp \left(\frac{-\mathrm{E}_{\ell}}{\mathrm{R} \cdot \mathrm{T}}\right)\right\}\tag{7-11}$
Case 3:
The activation energy Ea is greater than the limiting energy El, but the energy level $\ \mathrm{E}_{\mathrm{a}}-\tau \lambda \mathrm{N} / \mathrm{2} \mathrm{S}$ is less than the limiting energy level El (Figure 7-11). The strain rate equation is now:
$\ \dot{\varepsilon}=\mathrm{X} \cdot \frac{\mathrm{k} \cdot \mathrm{T}}{\mathrm{h} \cdot \mathrm{i}} \cdot\left\{\exp \left(-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R} \cdot \mathrm{T}}-\frac{\tau \cdot \lambda \cdot \mathrm{N}}{\mathrm{2} \cdot \mathrm{S} \cdot \mathrm{R} \cdot \mathrm{T}}\right)\right)-\exp \left(\frac{-\mathrm{E}_{\ell}}{\mathrm{R} \cdot \mathrm{T}}\right)\right\}\tag{7-12}$
Equation (7-12) appears to be identical to equation (7-11), but the boundary conditions differ.
Case 4:
The activation energy Ea is greater than the limiting energy El and the energy level $\ \mathrm{E}_{\mathrm{a}}-\tau \lambda \mathrm{N} / \mathrm{2} \mathrm{S}$ is greater than the limiting energy level E(Figure 7-12). The strain rate will be equal to zero in this case.
The cases 1 and 2 are similar to the case considered by Mitchell (1976) and still do not permit true cohesion and adhesion. Case 4 considers particles at rest without changing position within the particle matrix. Case 3 considers a material on which an external shear stress of certain magnitude must be applied to allow the particles to cross energy barriers, resulting in a yield stress (true cohesion or adhesion).
From equation (7-12) the following equation for the shear stress can be derived:
$\ \begin{array}{left}\tau=\left(\mathrm{E}_{\mathrm{a}}-\mathrm{E}_{\ell}\right) \cdot \frac{\mathrm{2} \cdot \mathrm{S}}{\lambda \cdot \mathrm{N}}+\mathrm{R} \cdot \mathrm{T} \cdot \frac{\mathrm{2} \cdot \mathrm{S}}{\lambda \cdot \mathrm{N}} \cdot \ln \left(\mathrm{1}+\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{\mathrm{0}}}\right)\ \text{with: }\dot{\varepsilon}_{\mathrm{0}}=\frac{\mathrm{X} \cdot \mathrm{k} \cdot \mathrm{T}}{\mathrm{h} \cdot \mathrm{i}} \cdot \exp \left(\frac{-\mathrm{E}_{\ell}}{\mathrm{R} \cdot \mathrm{T}}\right)\end{array}\tag{7-13}$
According to Mitchell (1976), if no shattering of particles occurs, the relation between the number of bonds and the effective stress σe can be described by the following equation:
$\ \mathrm{S}=\mathrm{a}+\mathrm{b} \cdot \sigma_{\mathrm{e}}\tag{7-14}$
Lobanov and Joanknecht (1980) confirmed this relation implicitly for pressures up to 10 bars for clay and paraffin wax. At very high pressures they found an exponential relation that might be caused by internal failure of the particles. For the friction between soil and metal Zeng and Yao (1988) also used equation (7-14), but for the internal friction Zeng and Yao (1991) used a logarithmic relationship, which contradicts Lobanov and Joanknecht and Mitchell, although it can be shown by Taylor series approximation that a logarithmic relation can be transformed into a linear relation for values of the argument of the logarithm close to 1. Since equation (7-14) contains the effective stress it is necessary that the clay used, is fully consolidated. Substituting equation (7-14) in equation (7-13) gives:
$\ \begin{array}{left} \tau=& \mathrm{a} \cdot\left\{\left(\mathrm{E}_{\mathrm{a}}-\mathrm{E}_{\ell}\right) \cdot \frac{\mathrm{2}}{\lambda \cdot \mathrm{N}}+\mathrm{R} \cdot \mathrm{T} \cdot \frac{\mathrm{2}}{\lambda \cdot \mathrm{N}} \cdot \ln \left(\mathrm{1}+\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)\right\} \ &+\mathrm{b} \cdot\left\{\left(\mathrm{E}_{\mathrm{a}}-\mathrm{E}_{\ell}\right) \cdot \frac{\mathrm{2}}{\lambda \cdot \mathrm{N}}+\mathrm{R} \cdot \mathrm{T} \cdot \frac{2}{\lambda \cdot \mathrm{N}} \cdot \ln \left(\mathrm{1}+\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)\right\} \cdot \sigma_{\mathrm{e}} \end{array}\tag{7-15}$
Equation (7-15) is of the same form as the Mohr-Coulomb failure criterion:
$\ \tau=\tau_{\mathrm{c}}+\sigma_{\mathrm{e}} \cdot \tan (\varphi)\tag{7-16}$
Equation (7-15), however, allows the strain rate to become zero, which is not possible in the equation derived by Mitchell (1976). The Mitchell equation and also the equations derived by Zeng and Yao (1988) and (1991) will result in a negative shear strength at small strain rates.
7.3.4. The Proposed Theory versus some other Theories
The proposed new theory is in essence similar to the theory developed by Mitchell (1976) which was based on the "rate process theory" as proposed by Eyring (1941). It was, however, necessary to use simplifications to obtain the equation in a useful form. The following formulation for the shear stress as a function of the strain rate has been derived by Mitchell by simplification of equation (7-6):
$\ \begin{array}{left}\tau=\mathrm{a} \cdot\left\{\mathrm{E}_{\mathrm{a}} \cdot \frac{\mathrm{2}}{\lambda \cdot \mathrm{N}}+\mathrm{R} \cdot \mathrm{T} \cdot \frac{\mathrm{2}}{\lambda \cdot \mathrm{N}} \cdot \ln \left(\frac{\dot{\varepsilon}}{\mathrm{B}}\right)\right\}\ +\mathrm{b} \cdot\left\{\mathrm{E}_{\mathrm{a}} \cdot \frac{\mathrm{2}}{\lambda \cdot \mathrm{N}}+\mathrm{R} \cdot \mathrm{T} \cdot \frac{2}{\lambda \cdot \mathrm{N}} \cdot \ln \left(\frac{\dot{\varepsilon}}{\mathrm{B}}\right)\right\} \cdot \sigma_{\mathrm{e}}\ \text{with: }\mathrm{B}=\frac{\mathrm{X} \cdot \mathrm{k} \cdot \mathrm{T}}{\mathrm{h}}\end{array}\tag{7-17}$
This equation is not valid for very small strain rates, because this would result in a negative shear stress. It should be noted that for very high strain rates the equations (7-15) and (7-17) will have exactly the same form. Zeng and Yao (1991) derived the following equation by simplification of equation (7-6) and by adding some empirical elements:
$\ \begin{array}\ln (\tau)=\mathrm{C_{1}+C_{2} \cdot \ln (\dot{\varepsilon})+C_{3} \cdot \ln \left(1+C_{4} \cdot \sigma_{e}\right)}\end{array}\tag{7-18}$
Rewriting equation (7-18) in a more explicit form gives:
$\ \tau=\exp \left(\mathrm{C}_{1}\right) \cdot(\dot{\varepsilon})^{\mathrm{C}_{2}} \cdot\left(1+\mathrm{C}_{4} \cdot \sigma_{\mathrm{e}}\right)^{\mathrm{C}_{3}}\tag{7-19}$
Equation (7-19) is valid for strain rates down to zero, but not for a yield stress. With respect to the strain rate, equation (7-19) is the equation of a fluid behaving according to the power law named "power law fluids". It should be noted however that equation (7-19) cannot be derived from equation (7-6) directly and thus should be considered as an empirical equation. If the coefficient C3 equals 1, the relation between shear stress and effective stress is similar to the relation found by Mitchell (1976). For the friction between the soil (clay and loam) and metal Zeng and Yao (1988) derived the following equation by simplification of equation (7-6):
$\ \tau_{\mathrm{b}}=\tau_{\mathrm{y a}}+\mathrm{C}_{\mathrm{5}} \cdot \ln (\dot{\varepsilon})+\sigma_{\mathrm{e}} \cdot \tan (\delta)=\tau_{\mathrm{a}}+\sigma_{\mathrm{e}} \cdot \tan (\delta)\tag{7-20}$
Equation (7-20) allows a yield stress, but does not allow the sliding velocity to become zero. An important conclusion of Yao and Zeng is that pasting soil on the metal surface slightly increases the friction meaning that the friction between soil and metal almost equals the shear strength of the soil.
The above-mentioned researchers based their theories on the rate process theory, other researchers derived empirical equations. Turnage and Freitag (1970) observed that for saturated clays the cone resistance varied with the penetration rate according to:
$\ \mathrm{F}=\mathrm{a} . \mathrm{v}^{\mathrm{b}}\tag{7-21}$
With values for the exponent ranging from 0.091 to 0.109 Wismer and Luth (1972B) and (1972A) confirmed this relation and found a value of 0.100 for the exponent, not only for cone penetration tests but also for the relation between the cutting forces and the cutting velocity when cutting clay with straight blades. Hatamura and Chijiiwa (1975), (1976A), (1976B), (1977A) and (1977B) also confirmed this relation for clay and loam cutting and found an exponent of 0.089.
Soydemir (1977) derived an equation similar to the Mitchell equation. From the data measured by Soydemir a relation according to equation (7-21) with an exponent of 0.101 can be derived. This confirms both the Mitchell approach and the power law approach.
7.3.5. Verification of the Theory Developed
The theory developed differs from the other theories mentioned in the previous paragraph, because the resulting equation (7-15) allows a yield strength (cohesion or adhesion). At a certain consolidation pressure level equation (7-15) can be simplified to:
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot \ln \left(1+\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)\tag{7-22}$
If $\ (\mathrm{d} \varepsilon / \mathrm{d t}) /\left(\mathrm{d} \varepsilon_{0} / \mathrm{d} \mathrm{t}\right)$ << 1, equation (7-22) can be approximated by:
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot \frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\tag{7-23}$
This approximation gives the formulation of a Bingham fluid. If the yield strength $\ \tau_{\mathrm{y}}$ is zero, equation represents a Newtonian fluid. If (dε/dt)/(dε0/dt) >> 1, equation (7-22) can be approximated by:
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot \ln \left(\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)\tag{7-24}$
This approximation is similar to equation (7-17) as derived by Mitchell.
If (dε/dt)/(0/dt) >> 1 and $\ \tau-\tau_{\mathrm{y}}<<\tau_{\mathrm{y}}$, equation (7-22) can be approximated by:
$\ \tau=\tau_{\mathrm{y}} \cdot\left(\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)^{\tau_{0} / \tau_{\mathrm{y}}}\tag{7-25}$
This approximation is similar to equation (7-21) as found empirically by Wismer and Luth (1972B) and many other researchers. The equation (7-15) derived in this paper, the equation (7-17) derived by Mitchell and the empirical equation (7-21) as used by many researchers have been fitted to data obtained by Hatamura and Chijiiwa (1975), (1976A), (1976B), (1977A) and (1977B). This is illustrated in Figure 7-13 with a logarithmic horizontal axis. Figure 7-14 gives an illustration with both axis logarithmic. These figures show that the data obtained by Hatamura and Chijiiwa fit well and that the above described approximations are valid.
The values used are $\ \tau$y =28kPa, $\ \tau$0 =4kPa and ε0 =0.03/s.
It is assumed that adhesion and cohesion can both be modeled according to equation (7-22). The research carried out by Zeng and Yao (1991) validates the assumption that this is true for adhesion. In more recent research Kelessidis et al. (2007) and (2008) utilize two rheological models, the Herschel-Bulkley model and the Casson model. The Herschel Bulkley model can be described by the following equation:
$\ \tau=\tau_{\mathrm{y}, \mathrm{H B}}+\mathrm{K} \cdot(\dot \varepsilon)^{\mathrm{n}}\tag{7-26}$
The Casson model can be described with the following equation:
$\ \sqrt{\tau}=\sqrt{\tau_{\mathrm{y}, \mathrm{C a}}}+\sqrt{\mu_{\mathrm{C a}} \cdot \dot{\boldsymbol{\varepsilon}}}\tag{7-27}$
Figure 7-15 compares these models with the model as derived in this paper. It is clear that for the high strain rates the 3 models give similar results. These high strain rates are relevant for cutting processes in dredging and offshore applications.
7.3.6. Abelev & Valent (2010)
Abelev & Valent (2010) investigated the strain rate dependency of the strength of soft marine deposits of the Gulf of Mexico. They used a precision rheometer with rotational rates from 0.25 up to 1000 1/min and water contents of 55% to 95%. They describe several models like an inverse hyperbolic sine:
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot \arcsin \left(\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)\tag{7-28}$
A logarithmic law and a power law:
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot \log _{10}\left(\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right) \quad\text{ and }\quad \tau=\tau_{\mathrm{y}} \cdot\left(\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)^{\beta}\tag{7-29}$
The data of Abelev & Valent (2010) are shown in Figure 7-16, together with a lower limit and an upper limit based on the equation derived in this chapter. Based on their experiments they suggest a modified power law:
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot\left(\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)^{\beta}\tag{7-30}$
The use of the equation derived in this chapter however gives even better results.
$\ \tau=\tau_{\mathrm{y}}+\tau_{0} \cdot \ln \left(1+\frac{\dot{\varepsilon}}{\dot{\varepsilon}_{0}}\right)\tag{7-31}$
One can see some dependency of the strengthening effect on the water content. It seems that the higher the water content, the larger the strengthening effect.
7.3.7. Resulting Equations for the Cutting Process
The strain rate is the rate of change of the strain with respect to time and can be defined as a velocity divided by a characteristic length. For the cutting process it is important to relate the strain rate to the cutting (deformation) velocity vc and the layer thickness hi. Since the deformation velocity is different for the cohesion in the shear plane and the adhesion on the blade, two different equations are found for the strain rate as a function of the cutting velocity.
$\ \dot{\varepsilon}_{\mathrm{c}}=1.4 \cdot \frac{\mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)}\tag{7-32}$
$\ \dot{\varepsilon}_{\mathrm{a}}=1.4 \cdot \frac{\mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\beta)}{\sin (\alpha+\beta)}\tag{7-33}$
This results in the following two equations for the multiplication factor for cohesion (internal shear strength) and adhesion (external shear strength). With $\ \tau_\mathrm{y}$ the cohesion at zero strain rate.
$\ \lambda_{\mathrm{c}}=\mathrm{1 + \frac { \tau _ { 0 } } { \tau _ { \mathrm { y } } } \cdot \operatorname { ln }}\left(1+\frac{1.4 \cdot \frac{\mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)}}{\dot{\varepsilon}_{0}}\right)\tag{7-34}$
$\ \lambda_{\mathrm{a}}=1+\frac{\tau_{0}}{\tau_{\mathrm{y}}} \cdot \ln \left(1+\frac{1.4 \cdot \frac{\mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}}} \cdot \frac{\sin (\beta)}{\sin (\alpha+\beta)}}{\dot{\varepsilon}_{0}}\right)\tag{7-35}$
With:
$\ \tau_{0} / \tau_{\mathrm{y}}=\mathrm{0 .1 4 2 8}, \dot{\varepsilon}_{0}=\mathrm{0 .0 3}\tag{7-36}$
Van der Schrieck (1996) published a graph showing the effect of the deformation rate on the specific energy when cutting clay. Although the shape of the curves found are a bit different from the shape of the curves found with equations (7-34) and (7-35), the multiplication factor for, in dredging common deformation rates, is about 2. This factor matches the factor found with the above equations.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/07%3A_Clay_Cutting/7.03%3A_The_Influence_of_Strain_Rate_on_the_Cutting_Process.txt
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7.4.1. The Forces
The most common failure mechanism in clay is the Flow Type as is shown in Figure 7-18, which will be considered first. The Curling Type and the Tear Type may occur under special circumstances and will be derived from the equations of the Flow Type.
Figure 7-19 illustrates the forces on the layer of soil cut. The forces shown are valid in general. The forces acting on this layer are:
1. A normal force acting on the shear surface N1 resulting from the effective grain stresses.
2. A shear force as a result of pure cohesion $\ \tau_{\mathrm{c}}$. This force can be calculated by multiplying the cohesion c/cohesive shear strength $\ \tau_{\mathrm{c}}$ with the area of the shear plane.
3. A force normal to the blade N2 resulting from the effective grain stresses.
4. A shear force as a result of pure adhesion between the soil and the blade $\ \tau_{\mathrm{a}}$. This force can be calculated by multiplying the adhesion a (adhesive shear strength $\ \tau_{\mathrm{a}}$) of the soil with the contact area between the soil and the blade.
The forces acting on a straight blade when cutting soil, can be distinguished as:
1. A force normal to the blade N2 resulting from the effective grain stresses.
2. A shear force as a result of pure adhesion between the soil and the blade $\ \tau_{\mathrm{a}}$. This force can be calculated by multiplying the adhesive shear strength $\ \tau_{\mathrm{a}}$ of the soil with the contact area between the soil and the blade.
These forces are shown in Figure 7-20.
Pure clay under undrained conditions follows the φ=0 concept, meaning that effectively there is no internal friction and thus there is also no external friction. Under drained conditions clay will have some internal friction, although smaller than sand. The reason for this is the very low permeability of the clay. If the clay is compressed with a high strain rate, the water in the pores cannot flow away resulting in the pore water carrying the extra pressure, the grain stresses do not change. If the grain stresses do not change, the shear stresses according to Coulomb friction do not change and effectively there is no relation between the extra normal stresses and the shear stresses, so apparently φ=0. At very low strain rates the pore water can flow out and the grains have to carry the extra normal stresses, resulting in extra shear stresses. During the cutting of clay, the strain rates, deformation rates, are so big that the internal and external friction angles can be considered to be zero. The adhesive and cohesive forces play a dominant role, so that gravity and inertia can be neglected.
The horizontal equilibrium of forces:
$\ \mathrm{\sum F_{h}=N_{1} \cdot \sin (\beta)+C \cdot \cos (\beta)-A \cdot \cos (\alpha)-N_{2} \cdot \sin (\alpha)=0}\tag{7-37}$
The vertical equilibrium of forces:
$\ \mathrm{\sum F_{v}=-N_{1} \cdot \cos (\beta)+C \cdot \sin (\beta)+A \cdot \sin (\alpha)-N_{2} \cdot \cos (\alpha)=0}\tag{7-38}$
The force K1 on the shear plane is now:
$\ \mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta)+\mathrm{A}}{\sin (\alpha+\beta)}\tag{7-39}$
The force K2 on the blade is now:
$\ \mathrm{N}_{2}=\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\tag{7-40}$
From equation (7-40) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F}_{\mathrm{h}}=\mathrm{N}_{\mathrm{2}} \cdot \sin (\alpha)+\mathrm{A} \cdot \cos (\alpha)\tag{7-41}$
$\ \mathrm{F}_{v}=\mathrm{N}_{\mathrm{2}} \cdot \cos (\alpha)-\mathrm{A} \cdot \sin (\alpha)\tag{7-42}$
Since λc and λa are almost identical, an average value λs is used in the following equations. With the relations for the cohesive force C, the adhesive force and the adhesion/cohesion ratio (the ac ratio r):
$\ \mathrm{C=\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w}{\sin (\beta)}}\tag{7-43}$
$\ \mathrm{A}=\frac{\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{7-44}$
$\ \mathrm{r}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\tag{7-45}$
The horizontal Fh and vertical Fv cutting forces can be determined according to:
$\ \begin{array}{left} \mathrm{F_{h}}&=\mathrm{\frac{C \cdot \sin (\alpha)+A \cdot \sin (\beta)}{\sin (\alpha+\beta)}=\frac{\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w}{\sin (\beta)} \cdot \sin (\alpha)+\frac{\lambda_{s} \cdot a \cdot h_{b} \cdot w}{\sin (\alpha)} \cdot \sin (\beta)}{\sin (\alpha+\beta)}}\ &=\mathrm{\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}+r \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}}\end{array}\tag{7-46}$
$\ \begin{array}{left}\mathrm{F}_{v}&=\mathrm{\frac{C \cdot \cos (\alpha)-A \cdot \cos (\beta)}{\sin (\alpha+\beta)}=\frac{\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w}{\sin (\beta)} \cdot \cos (\alpha)-\frac{\lambda_{s} \cdot a \cdot h_{b} \cdot w}{\sin (\alpha)} \cdot \cos (\beta)}{\sin (\alpha+\beta)}}\ &=\lambda_{\mathrm{s}} \cdot \mathrm{c \cdot h _ { \mathrm { i } } \cdot \mathrm { w } \cdot \frac { \frac { \operatorname { cos } ( \alpha ) } { \operatorname { sin } ( \beta ) } - \mathrm { r } \cdot \frac { \operatorname { cos } ( \beta ) } { \operatorname { sin } ( \alpha ) } } { \operatorname { sin } ( \alpha + \beta ) }}\end{array}\tag{7-47}$
The normal force on the shear plane is now equal to the force K1 as is used in sand cutting, because the internal friction angle φ is zero:
$\ \mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta)+\mathrm{A}}{\sin (\alpha+\beta)}\tag{7-48}$
The normal force on the blade is now equal to the force K2 as is used in sand cutting, because the external friction angle δ is zero:
$\ \mathrm{N}_{2}=\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\tag{7-49}$
Equations (7-48) and (7-49) show that both the normal force on the shear plane N1 and the normal force on the blade N2 may become negative. This depends on the ac ratio between the adhesive and the cohesive forces and on the blade angle α and shear angle β. A negative normal force on the blade will result in the Curling Type of cutting mechanism, while a negative normal force on the shear plane will result in the Tear Type of cutting mechanism. If both normal forces are positive, the Flow Type of cutting mechanism will occur.
7.4.2. Finding the Shear Angle
There is one unknown in the equations and that is the shear angle βThis angle has to be known to determine cutting forces, specific energy and power.
$\ \mathrm{F_{h}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot\left(\frac{\frac{\sin (\alpha)}{\sin (\beta)}+r \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}\right) \quad\text{ with: }r=\frac{a \cdot h_{b}}{c \cdot h_{i}}}\tag{7-50}$
Equation (7-50) for the horizontal cutting force Fh can be rewritten as:
$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot\left(\frac{\sin ^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{HF}}\tag{7-51}$
Equation (7-47) for the vertical cutting force Fv can be rewritten as:
$\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot\left(\frac{\sin (\mathrm{\alpha}) \cdot \cos (\alpha)-\mathrm{r} \cdot \sin (\beta) \cdot \cos (\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V F}}\tag{7-52}$
The strengthening factor λswhich is not very sensitive for β in the range of cutting velocities vc as applied in dredging, can be determined by:
$\ \begin{array}{left}\lambda_{\mathrm{s}}=\left(1+\frac{\tau_{0}}{\tau_{\mathrm{y}}} \cdot \ln \left(1+\frac{1.4 \cdot \frac{\mathrm{v}_{\mathrm{c}}}{h_{\mathrm{i}}} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)}}{\dot{\varepsilon}_{0}}\right) \right)\ \text { With }: \tau_{0} / \tau_{\mathrm{y}}=0.1428 \text { and } \dot{\varepsilon}_{0}=0.03\end{array}\tag{7-53}$
The shear angle β is determined by the case where the horizontal cutting force Fh is at a minimum, based on the minimum energy principle (omitting the strengthening factor λs).
\ \begin{aligned} \frac{\partial \mathrm{F}_{\mathrm{h}}}{\partial \beta}&=\frac{\mathrm{2} \cdot \mathrm{r} \cdot \sin ^{2}(\beta) \cdot \cos (\beta) \cdot \sin (\alpha+\beta) \cdot \sin (\alpha)}{\sin ^{2}(\alpha+\beta) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)} & \ &+\frac{-\sin (\alpha) \cdot \sin (\alpha+2 \cdot \beta) \cdot\left(\sin ^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)\right)}{\sin ^{2}(\alpha+\beta) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)}=0 \end{aligned}\tag{7-54}
In the special case where there is no adhesion a=0r=0, the shear angle β is:
$\ \sin (\alpha+2 \cdot \beta)=0\text{ for }\alpha+2 \cdot \beta=\pi \operatorname{giving} \beta=\frac{\pi}{2}-\frac{\alpha}{2}\tag{7-55}$
An approximation equation for β based on curve fitting on equation (7-54) for the range 0.5<r<2 gives:
$\ \begin{array}{left}\beta=1.26 \cdot \mathrm{e}^{(-0.174 \cdot \alpha-0.3148 \cdot \mathrm{r})}\text{ in radians or}\ \beta=72.2 \cdot \mathrm{e}^{(-0.003 \cdot \alpha-0.3148 \cdot \mathrm{r})}\text{ in degrees}\end{array}\tag{7-56}$
For a clay, with shear strength c=1 kPa, a layer thickness of hi=0.1 m and a blade width of w=1 m Figure 7-21, Figure 7-23 and Figure 7-24 give the values of the shear angle β, the horizontal cutting force Fh and the vertical cutting force Fv for different values of the adhesion/cohesion (ac) ratio and as a function of the blade angle α. The use of the ac ratio makes the graphs independent of individual values of hb and a. In these calculations the strain rate factor λs is set to 1. For different values of the strain rate factor λ, the cohesion c, the blade with and the layer thickness hi, the values found in Figure 7-23 and Figure 7-24 can be multiplied by the corresponding factors.
The horizontal cutting force Fh is at an absolute minimum when:
$\ \alpha+\beta=\frac{\pi}{2}\tag{7-57}$
This is however only useful if the blade angle α can be chosen freely. For a worst case scenario with an ac ratio r=2, meaning a high adhesion, a blade angle α of about 55o is found (see Figure 7-23), which matches blade angles as used in dredging. The fact that this does not give an optimum for weaker clays (clays with less adhesion) is not so relevant.
Figure 7-21, Figure 7-23 and Figure 7-24 show that the shear angle β is decreasing with an increasing blade angle α and an increasing ac ratio r. For practical blade angles between 45 and 60 degrees, the shear angle may vary between 35 and 60-70 degrees, depending on the ac ratio r. The horizontal force first decreases to a minimum with an increasing blade angle, after which it increases. At very large blade angles the horizontal force increases strongly to values that are not reasonable anymore. Nature will find another mechanism with smaller forces, the wedge mechanism, which will be described in Chapter 13: A Wedge in Clay Cutting. The vertical force (positive is downwards directed) is first increasing with an increasing blade angle to a maximum value, after which it is decreasing to very large negative (upwards directed) values at very large blade angles.
Figure 7-22 shows the sum of the blade angle and the shear angle. When this sum is 90 degrees, the minimum of the horizontal force is found. The graph shows clearly that this is the case for a 55 degree blade and an ac ratio r=2.
See Appendix V: Clay Cutting Charts for more and higher resolution charts.
7.4.3. Specific Energy
In the dredging industry, the specific cutting energy Esp is described as:
The amount of energy, that has to be added to a volume unit of soil (e.g. clay) to excavate the soil.
The dimension of the specific cutting energy is: kN/m2 or kPa for sand and clay, while for rock often MN/m2 or MPa is used. For the case as described above, cutting with a straight blade with the direction of the cutting velocity vc perpendicular to the blade (edge of the blade), the specific cutting energy Esp is:
$\ \mathrm{E}_{\mathrm{s p}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{7-58}$
With the following equation for the horizontal cutting force Fh:
$\ \mathrm{F_{h}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot\left(\frac{\sin ^{2}(\alpha)+r \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \lambda_{H F}}\tag{7-59}$
This gives for the specific cutting energy Esp:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot\left(\frac{\mathrm{s i n}^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \lambda_{\mathrm{HF}}\tag{7-60}$
The cohesion is half the UCS value, which can be related to the SPT value of the clay by a factor 12, so the cohesion is related by a factor 6 to the SPT value (see Table 7-1), further, the strengthening λ factor will have a value of about 2 at normal cutting velocities of meters per second, this gives:
$\ \lambda_{\mathrm{s}} \cdot \mathrm{c} \approx \mathrm{2} \cdot \mathrm{6} \cdot \mathrm{S P T}=\mathrm{1 2} \cdot \mathrm{S P T}\tag{7-61}$
Now a simplified equation for the specific energy Esp is found by:
$\ \mathrm{E}_{\mathrm{sp}}=\mathrm{1 2} \cdot \mathrm{S P T} \cdot\left(\frac{\sin ^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=12 \cdot \mathrm{SPT} \cdot \lambda_{\mathrm{HF}}\tag{7-62}$
Figure 7-25 shows the specific energy Esp and the production Pc per 100 kW installed cutting power as a function of the SPT value.
Table 7-1: Guide for Consistency of Fine-Grained Soil (Lambe & Whitman (1979)).
SPT Penetration (blows/ foot)
Estimated Consistency
UCS (kPa)
<2
Very Soft Clay
<24
2 - 4
Soft Clay
24 - 48
4 - 8
Medium Clay
48 - 96
8 - 16
Stiff Clay
96 – 192
16 - 32
Very Stiff Clay
192 – 384
>32
Hard Clay
>384
See Appendix U: Specific Energy in Clay for more graphs on the specific energy in clay.
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7.5.1. Introduction
In the previous chapter, the equations for the cutting forces of the Flow Type cutting mechanism have been derived. These equation however do not take into consideration that normal forces and thus stresses may become negative and may exceed the tensile strength of the clay. If the tensile stresses exceed the tensile strength, tensile failure will occur and the clay will not fail by plastic shear failure, but by tensile failure. The failure mechanism in this case is named the Tear Type mechanism. Based on the Mohr circle, tensile cracks will occur under and angle of 45 degrees downwards with respect of the shear angle as is shown in Figure 7-26. When the blade is progressing with the cutting velocity, after a short while a so called secondary crack will occur under 90 degrees with the first (primary) crack. The model as derived in this chapter, does not assume that the tensile strength is exceeded at the moment of tensile crack forming over the full length of the tensile crack. The model assumes that the tensile strength is exceeded at the start of the tensile crack only. In order to determine whether the tensile strength is exceeded, the average shear stress in the shear plane is used. Of course there may be a stress distribution in the shear plane, leading to locally higher and lower shear stresses and thus normal stresses, but these cannot be determined with the methodology used. Only average stresses can be determined. The methodology applied however gives reasonable and practical tools to determine whether the Tear Type cutting mechanism will occur or not.
7.5.2. The Normal Force on the Shear Plane
In order to determine the normal (possibly tensile) stresses on the shear plane, first the normal force on the shear plane has to be determined.
$\ \mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta)+\mathrm{A}}{\sin (\alpha+\beta)}\tag{7-63}$
Substituting the equations for the cohesive force and the adhesive force gives:
$\ \mathrm{N}_{1}=\frac{-\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)} \cdot \cos (\alpha+\beta)+\frac{\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}}{\sin (\alpha+\beta)}\tag{7-64}$
The average normal stress on the shear plane equals the normal force on the shear plane N1, divided by the cross sectional area of the shear plane, giving:
$\ \sigma_{\mathrm{N} 1}=\frac{\mathrm{N}_{1} \cdot \sin (\beta)}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{7-65}$
Substituting equation (7-64) in equation (7-65) gives for the normal stress on the shear plane:
$\ \begin{array}{left} \sigma_{\mathrm{N} 1}&= \frac{\sin (\beta)}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}} \cdot \frac{-\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)} \cdot \cos (\alpha+\beta)+\frac{\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}}{\sin (\alpha+\beta)} \ &=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \frac{-\cos (\alpha+\beta)+r \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)} \end{array}\tag{7-66}$
Assuming a fixed strain rate factor λs for cohesion and tensile strength, the normal stress minus the shear strength (cohesion) has to be bigger than the tensile strength, where the tensile strength is negative (compressive stresses are positive).
$\ \sigma_{\mathrm{N} 1}-\lambda_{\mathrm{s}} \cdot \mathrm{c} \geq \lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}}\tag{7-67}$
Substituting equation (7-66) into equation (7-67) gives the condition for ductile failure:
$\ \lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \frac{-\cos (\alpha+\beta)+r \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}-\lambda_{\mathrm{s}} \cdot \mathrm{c} \geq \lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}}\tag{7-68}$
The transition from the Flow Type mechanism to the Tear Type mechanism is at the moment where the equal sign is used in the above equation, resulting in a critical ratio between the tensile strength and the shear strength, still also depending on the ac ratio according to:
$\ \mathrm{\frac{\sigma_{T}}{c}}=\mathrm{\left(\frac{r \cdot \frac{\sin (\beta)}{\sin (\alpha)}-\cos (\alpha+\beta)-\sin (\alpha+\beta)}{\sin (\alpha+\beta)}\right)}\tag{7-69}$
Figure 7-27 shows the critical ratio curves of the ratio of the tensile strength to the shear strength (cohesion) of the transition of the Flow Type mechanism to the Tear Type mechanism. Since the tensile strength is considered to be negative, the more negative this ratio, the higher the relative tensile strength. Below a curve the Flow Type may be expected, above a curve the Tear Type. Only negative ratios should be considered, since the tensile strength cannot be positive. The figure shows that for r=1 (high adhesive forces) the curve just touches a ratio of zero, but never becomes negative, meaning the Tear Type will never occur. For smaller values the curves are more negative for a decreasing value. The minimum for is zero (no adhesion). The figure also shows that all curves (except the r=0 curve) start with a positive value, then decrease with an increasing blade angle to a minimum value and with a further increasing blade angle increase again to positive values. For blade angles larger than 90 degrees tensile failure will never occur. Because of the choice of the parameter hb, the blade height, at constant blade height the length of the blade is increasing with a decreasing blade angle. This means that the adhesive force on the blade increases with a decreasing blade angle, resulting in increasing normal stresses on the shear plane. Higher normal stresses suppress tensile failure. On the other hand, an increasing blade angle will increase the normal stress on the shear plane because of the force equilibrium. So we have two effects, the normal stresses on the shear plane will decrease with an increasing blade angle because of the decrease of the adhesive force and the normal stresses will increase with an increases blade angle because of the force equilibrium. The result is a curve with a minimum.
7.5.3. The Mobilized Shear Strength
Assuming a mobilized shear stress cm in the shear plane at the moment of tensile failure, gives:
$\ \mathrm{c_{m} \cdot\left(\frac{r_{m} \cdot \frac{\sin (\beta)}{\sin (\alpha)}-\cos (\alpha+\beta)-\sin (\alpha+\beta)}{\sin (\alpha+\beta)}\right)=\sigma_{T}}\tag{7-70}$
Or:
$\ \mathrm{c}_{\mathrm{m}}=\sigma_{\mathrm{T}} \cdot\left(\frac{\sin (\alpha+\beta)}{\mathrm{r}_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}-\cos (\alpha+\beta)-\sin (\alpha+\beta)}\right)\tag{7-71}$
Since the mobilized shear stress cm is smaller than the shear strength c, also the ac ratio rm will be different from the ac ratio when the shear stress is fully mobilized up to the shear strength. This gives for the mobilized ac ratio rm:
$\ \mathrm{r}_{\mathrm{m}}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c}_{\mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}}}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}}} \cdot\left(\frac{\mathrm{r}_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}-\cos (\alpha+\beta)-\sin (\alpha+\beta)}{\sin (\alpha+\beta)}\right)\tag{7-72}$
The mobilized ac ratio rm is present on both sides of the equal sign. This gives for the mobilized ac ratio rm:
$\ \begin{array}{left}\mathrm{r}_{\mathrm{m}}=\frac{\mathrm{r}_{\mathrm{T}} \cdot\left(\mathrm{1}+\frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}\right)}{\left(\mathrm{r}_{\mathrm{T}} \cdot \frac{\sin (\beta)}{\sin (\alpha) \cdot \sin (\alpha+\beta)}-1\right)}\ \text{With: } \mathrm{r_T}=\mathrm{\frac{a\cdot h_b}{\sigma_t\cdot h_i}}\end{array}\tag{7-73}$
The normal stress on the shear plane is now:
$\ \sigma_{\mathrm{N} 1, \mathrm{m}}=\lambda_{\mathrm{s}} \cdot \mathrm{c}_{\mathrm{m}} \cdot \frac{-\cos (\alpha+\beta)+\mathrm{r}_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}\tag{7-74}$
7.5.4. The Resulting Cutting Forces
Substituting the mobilized shear strength cm and the mobilized ac ratio rm gives the horizontal and vertical forces in the case of brittle failure, the Tear Type cutting mechanism:
$\ \begin{array}{left} \mathrm{F_{\mathrm{h}}}&=\mathrm{\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}+r_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{r_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}-\cos (\alpha+\beta)-\sin (\alpha+\beta)}}\ &=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{r}_{\mathrm{T}} \cdot \frac{\lambda_{\mathrm{HT}}}{\mathrm{r}_{\mathrm{T}}}\end{array}\tag{7-75}$
$\ \begin{array}{left}\mathrm{F}_{\mathrm{v}}&=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\frac{\cos (\alpha)}{\sin (\beta)}-\mathrm{r}_{\mathrm{m}} \cdot \frac{\cos (\beta)}{\sin (\alpha)}}{\mathrm{r}_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}-\cos (\alpha+\beta)-\sin (\alpha+\beta)}\ &=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{r}_{\mathrm{T}} \cdot \frac{\lambda_{\mathrm{V} T}}{\mathrm{r}_{\mathrm{T}}}\end{array}\tag{7-76}$
The cutting forces are not dependent on the shear strength anymore, but completely dependent on the tensile strength and the adhesion.
Figure 7-29, Figure 7-30, Figure 7-31 and Figure 7-32 show the shear angle β, the horizontal cutting force coefficient λHT/rT, the vertical cutting force coefficient λVT/rT and the last one zoomed for the Tear Type of cutting mechanism. The figures show that for large values of rT, the shear angle and the cutting force coefficients hardly depend on the factor rT. It should be mentioned that the graphs show λHT/rT and λVT/rT and not λHT and λVT. A large or very large value of rT means a very small tensile strength compared to the adhesion. Equations (8-112) and (8-113) can be rewritten for the case of a very small relative tensile strength according to:
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{r}_{\mathrm{T}} \cdot \frac{\lambda_{\mathrm{H} \mathrm{T}}}{\mathrm{r}_{\mathrm{T}}}=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}}} \cdot \frac{\lambda_{\mathrm{H} \mathrm{T}}}{\mathrm{r}_{\mathrm{T}}} \ &=\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w} \cdot \frac{\lambda_{\mathrm{H T}}}{\mathrm{r}_{\mathrm{T}}} \end{array}\tag{7-77}$
$\ \begin{array}{left} \mathrm{F}_{\mathrm{v}} &=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{r}_{\mathrm{T}} \cdot \frac{\lambda_{\mathrm{V T}}}{\mathrm{r}_{\mathrm{T}}}=\lambda_{\mathrm{s}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}}} \cdot \frac{\lambda_{\mathrm{V T}}}{\mathrm{r}_{\mathrm{T}}} \ &=\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w} \cdot \frac{\lambda_{\mathrm{V T}}}{\mathrm{r}_{\mathrm{T}}} \end{array}\tag{7-78}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/07%3A_Clay_Cutting/7.05%3A_The_Tear_Type.txt
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7.6.1. Introduction
When the layer thickness becomes very small, two things can happen. The normal force on the blade may become negative or there is no equilibrium of moments. In both cases the contact length between the clay and the blade has to be reduced. There can be different mechanisms for this. In steel cutting the curling of the chip cut is well known, but there could also be buckling or breaking of the layer cut. The result is the same, the clay will have a reduced contact length with the blade. This type of cutting mechanism is named the Curling Type. Both the normal force not becoming negative and the equilibrium of moments will be investigated. The mechanism with the smallest cutting forces is assumed to be the correct mechanism.
7.6.2. The Normal Force on the Blade
From the Flow Type of cutting mechanism the following equation is derived for the normal force on the blade:
$\ \mathrm{N}_{2}=\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\tag{7-79}$
Substituting the equations (7-43) and (7-44) gives:
$\ \begin{array}{left} \mathrm{N}_{2}&= \frac{\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}-\frac{\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)} \&= \lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\frac{1}{\sin (\beta)}-\frac{\mathrm{r}}{\sin (\alpha)} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)} \end{array}\tag{7-80}$
Dividing the normal force by the surface of the blade gives the average normal stress on the blade:
$\ \sigma_{\mathrm{N} 2}=\frac{\mathrm{N}_{2} \cdot \sin (\alpha)}{\mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}\tag{7-81}$
This gives for the normal stress on the blade:
$\ \begin{array}{left} \sigma_{\mathrm{N} 2}&= \frac{\sin (\alpha)}{\mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}} \cdot \frac{\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}-\frac{\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)} \&= \lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \frac{\frac{1}{\mathrm{r}} \cdot \frac{\sin (\alpha)}{\sin (\beta)}-\cos (\alpha+\beta)}{\sin (\alpha+\beta)} \end{array}\tag{7-82}$
As stated before this normal stress should have a value greater than zero, since it is assumed that there is no tensile strength between the clay and the blade.
$\ \sigma_{\mathrm{N} 2} \geq 0\tag{7-83}$
In details this gives for the condition of no negative normal stress on the blade:
$\ \lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \frac{\frac{1}{\mathrm{r}} \cdot \frac{\sin (\alpha)}{\sin (\beta)}-\cos (\alpha+\beta)}{\sin (\alpha+\beta)} \geq 0\tag{7-84}$
At the critical condition where the normal stress equals zero this gives:
$\ \frac{1}{\mathrm{r}} \cdot \frac{\sin (\alpha)}{\sin (\beta)}=\cos (\alpha+\beta)\tag{7-85}$
In the case of the Curling Type, the ac ratio is not fully mobilized giving:
$\ \mathrm{r_{m}}=\frac{\sin (\alpha)}{\sin (\beta)} \cdot \frac{1}{\cos (\alpha+\beta)}\tag{7-86}$
Substituting this mobilized ac ratio rm in equations (7-46) and (7-47) gives for the cutting forces:
$\ \begin{array}{left}\mathrm{F_{\mathrm{h}}}&=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h_{\mathrm{i}}} \cdot \mathrm{w} \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}+\mathrm{r}_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h_{\mathrm{i}}} \cdot \mathrm{w} \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}+\frac{1}{\cos (\alpha+\beta)}}{\sin (\alpha+\beta)}\ &=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\frac{\cos (\alpha)}{\sin (\beta)}}{\cos (\alpha+\beta)}\end{array}\tag{7-87}$
$\ \begin{array}{left}\mathrm{F}_{v}&=\mathrm{\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\cos (\alpha)}{\sin (\beta)}-r_{m} \cdot \frac{\cos (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\cos (\alpha)}{\sin (\beta)}-\frac{\cos (\beta)}{\sin (\beta)} \cdot \frac{1}{\cos (\alpha+\beta)}}{\sin (\alpha+\beta)}}\ &=\mathrm{-\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}}{\cos (\alpha+\beta)}}\end{array}\tag{7-88}$
This method is simple and straightforward, but does not take a normal stress distribution on the blade into account. It does however give a prediction of the cutting forces and the reduced contact length on the blade. The unknown in the equations is the shear angle β. Assuming that the mechanism will choose a shear angle where the cutting energy is at a minimum, a shear angle β is found according to:
$\ \beta=\frac{\pi}{4}-\frac{\alpha}{2}\tag{7-89}$
If we substitute this solution in the cutting force equations we find:
$\ \mathrm{F_{h}=2 \cdot \lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\cos (\alpha)}{1-\sin (\alpha)}}\tag{7-90}$
$\ \mathrm{F}_{v}=\mathrm{-2 \cdot \lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\sin (\alpha)}{1-\sin (\alpha)}}\tag{7-91}$
The horizontal force will increase with an increasing blade angle, the vertical force also, but upwards directed. In the case of the Curling Type, the ac ratio is not fully mobilized giving:
$\ \mathrm{r_{m}=2 \cdot \frac{\sin (\alpha)}{1-\sin (\alpha)}}\tag{7-92}$
The condition of having a normal force of zero on the blade can never fulfill the condition of having an equilibrium of moments on the layer cut, since the normal force on the blade is zero and is therefore rejected. Still this condition gives insight in the behavior of the equations of clay cutting and is therefore mentioned here.
7.6.3. The Equilibrium of Moments
As mentioned in the previous paragraph, the equilibrium of moments on the layer cut has to be fulfilled. If we take the equilibrium of moments around the tip of the blade, there are only two forces participating in the equilibrium of moments, the normal force on the shear plane N1 and the normal force on the blade N2. These forces have acting points R1 and R2 on the shear plane and on the blade. If the normal stresses are uniformly distributed, both acting points will be at the center (half way) the corresponding planes. The acting point of the normal force on the shear plane will be at half the length of the shear plane and the acting point of the normal force on the blade will be at half the (mobilized) length of the blade. Two factors are introduced to give the exact location of these acting points, λ1 on the shear plane and λ2 on the blade. When the moment N2·R2 on the blade is greater than the moment N1·R1 on the blade curling will occur in such a way that both moments are equal. The contact length between the clay and the blade will be reduced to a mobilized contact length hb,m.
The normal force on the shear plane is now equal to the force N1, because the internal friction angle is zero:
$\ \mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta)+\mathrm{A}}{\sin (\alpha+\beta)}\tag{7-93}$
The normal force on the blade is now equal to the force N2, because the external friction angle is zero:
$\ \mathrm{N}_{2}=\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\tag{7-94}$
This gives for the equilibrium of moments:
$\ \mathrm{N}_{1} \cdot \mathrm{R}_{1}=\mathrm{N}_{2} \cdot \mathrm{R}_{2}\tag{7-95}$
For both acting points we can write:
$\ \mathrm{R_{1}=\frac{\lambda_{1} \cdot h_{\mathrm{i}}}{\sin (\beta)}, R_{2}=\frac{\lambda_{2} \cdot h_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}}\tag{7-96}$
Substituting equations (7-93), (7-94) and (7-96) in equation (7-95) gives:
$\ \left(\frac{\mathrm{A}-\mathrm{C} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\right) \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}=\left(\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\right) \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}\tag{7-97}$
Substituting equations (7-45) and (7-46) for the cohesive and adhesive forces gives:
$\ \left(\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}-\frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \cos (\alpha+\beta)\right) \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}=\left(\frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}-\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \cos (\alpha+\beta)\right) \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}\tag{7-98}$
Rewriting this term by term gives:
$\ \frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}-\frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \cos (\alpha+\beta)=\frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}-\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \cos (\alpha+\beta)\tag{7-99}$
Moving the terms with adhesion to the left side and the terms with cohesion to the right side gives:
$\ \frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}+\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)} \cdot \cos (\alpha+\beta)=\frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}+\frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)} \cdot \cos (\alpha+\beta)\tag{7-100}$
This gives a second degree function of the mobilized blade height according to:
$\ \frac{\lambda_{2} \cdot \mathrm{a} \cdot \cos (\alpha+\beta)}{\sin (\alpha) \cdot \sin (\alpha)} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}+\frac{\lambda_{1} \cdot \mathrm{a}-\lambda_{2} \cdot \mathrm{c}}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}-\frac{\lambda_{1} \cdot \mathrm{c} \cdot \cos (\alpha+\beta)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}=\mathrm{0}\tag{7-101}$
This second degree function can be solved with the A, B, C formula and has two solutions.
$\ \begin{array}{left}\mathrm{A} \cdot \mathrm{x}^{2}+\mathrm{B} \cdot \mathrm{x}+\mathrm{C}=\mathrm{0}\ \mathrm{h}_{\mathrm{b}, \mathrm{m}}=\mathrm{x}=\frac{-\mathrm{B} \pm \sqrt{\mathrm{B}^{2}-\mathrm{4} \cdot \mathrm{A} \cdot \mathrm{C}}}{\mathrm{2A}} \quad\text{with: } \mathrm{r}_{\mathrm{m}}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\ \mathrm{A}=\frac{\lambda_{2} \cdot \mathrm{a} \cdot \cos (\alpha+\beta)}{\sin (\alpha) \cdot \sin (\alpha)}\ \mathrm{B}=\frac{\lambda_{1} \cdot \mathrm{a}-\lambda_{2} \cdot \mathrm{c}}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}}\ \mathrm{C}=-\frac{\lambda_{1} \cdot \mathrm{c} \cdot \cos (\alpha+\beta)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}\end{array}\tag{7-102}$
The following criteria are valid for the use of this method.
$\ \begin{array}{left} \text{if } \mathbf{h}_{\mathbf{b}, \mathbf{m}}<\mathbf{h}_{\mathbf{b}}\text{ then use }\mathbf{h}_{\mathbf{b}, \mathbf{m}}\ \text{if }\mathbf{h}_{\mathbf{b}, \mathbf{m}} \geq \mathbf{h}_{\mathbf{b}}\text{ then use }\mathbf{h}_{\mathbf{b}, \mathbf{m}}\end{array}\tag{7-103}$
To see which solution is valid, the terms of the equation have to be analyzed. For α+β<π/2 the term A>0 and C<0 because of the minus sign. The term is always positive. This will only result in a positive solution if the + sign is applied. For α+β>π/2 the term A<0 and C>0 because of the minus sign. This will only result in a positive solution if the – sign is applied. So at small blade angles the plus sign gives the correct solution, while large blade angles require the minus sign solution.
Figure 7-35, Figure 7-36 and Figure 7-37 show the shear angle and the horizontal cutting force coefficient and the vertical cutting force coefficient for the Curling Type. At large blade angles, both the horizontal and vertical forces become very large. In cases of large blade angles the Curling Type will hardly occur because the Flow Type results in smaller forces.
$\ \mathrm{F_{\mathrm{h}}=\lambda_{\mathrm{s}} \cdot \mathbf{c} \cdot \mathbf{h}_{\mathrm{i}} \cdot \mathbf{w} \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}+r_{\mathrm{m}} \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}=\lambda_{\mathrm{s}} \cdot \mathbf{c} \cdot \mathbf{h}_{\mathrm{i}} \cdot \mathbf{w} \cdot \lambda_{\mathrm{HC}}}\tag{7-104}$
$\ \mathrm{F}_{v}=\mathrm{\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\cos (\alpha)}{\sin (\beta)}-r_{m} \cdot \frac{\cos (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \lambda_{V C}}\tag{7-105}$
Figure 7-35, Figure 7-36 and Figure 7-37 clearly show the transition from the plus root solution to the minus root solution. This transition results in a discontinuity. How exactly this transition will take place in nature is still subject for further research. Confidential tests in clay with blade angles of 20, 30 and 40 degrees have shown that the plus root solution is valid at small blade angles, tests in hyperbaric rock cutting with a blade angle of 110 degrees have shown that the minus root solution is valid at large blade angles (see Chapter 9:). One should consider that the Curling Type only occurs with thin layers. Once the required mobilized blade height exceeds the actual blade height, the Flow Type will occur. So for example, if blade height and layer thickness are equal, the ratio cannot exceed 1 and depending on the a/c ratio, the Flow Type will occur above a certain blade angle.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/07%3A_Clay_Cutting/7.06%3A_The_Curling_Type.txt
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Now the question is, when do we have a Flow TypeCurling Type or Tear Type and how does this depend on the different parameters. This is explained by a number of examples.
Example 1: Cohesion c=1 kPa, adhesion a=1 kPa, tensile strength σT=-0.3 kPa, blade height hb=0.1 m, blade angle α=55°, forces per unit width of the blade.
According to Figure 7-27 (see also Figure 7-40) there will be a transition from the Flow Type to the Tear Type at r=0.3, so a layer thickness hi=0.32 m. But will this really happen? Suppose we investigate the undercutting process of a cutter head, where the layer thickness increases from zero to a maximum during the rotation of a blade. When the blade starts cutting the layer thickness is zero and increases in time. First the cutting process is of the Curling Type up to a layer thickness of about hi=0. 65 m. At this layer thickness the mobilized blade height equals the actual blade height and there is a transition from the Curling Type to the Flow Type. When the layer thickness is increased further, at a layer thickness of about hi=0.32 m the normal stresses on the shear plane result in normal stresses more negative than the tensile strength under an angle of 45° downwards with respect to the direction of the shear plane, so there is a transition from the Flow Type to the Tear Type. However, once the Tear Type of cutting mechanism occurs, this mechanism will search for a shear angle, resulting in a minimum cutting force. This shear angle tends not to be equal to the optimum shear angle of the Flow Type. Figure 7-21 shows the optimum shear angle of the Flow Type, while Figure 7-29 shows the optimum shear angle of the Tear Type. The result is a discontinuity in the cutting force, the cutting force is reduced (the beta real curve) at the moment the Tear Type is the cutting mechanism. Another reduction may occur, because the force calculated is the force at the start of a tensile crack. When the blade continues moving forward, the horizontal force will probably be smaller than the force at the initiation of the tensile crack, resulting in a lower average force.
Now suppose we are overcutting with our cutter head. This means we start with some maximum layer thickness thick enough to cause the Tear Type to occur. When the blade progresses, the layer thickness decreases. But since the curve of the real beta is followed, the Tear Type will continue until a layer thickness of about hi=0.065 m is reached. In fact, each time a block of clay breaks out of the clay and the cutting process starts again. At the layer thickness of about hi=0.065 m there is a transition directly from the Tear Type to the Flow Type.
Figure 7-41 shows the Mohr circles for the Flow Type and the Tear Type for a layer thickness of hi=0.1 m. Both mechanisms are possible. Which one occurs depends on the history, since both only touch one failure criterion. Figure 7-42 shows the Mohr circles for the Flow Type and the Tear Type for a layer thickness of hi=0.5 m. The Mohr circle for shear failure (Flow Type) crosses the tensile failure criterion and thus cannot exist. Only one mechanism is possible, the Tear Type.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/07%3A_Clay_Cutting/7.07%3A_Resulting_Forces.txt
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7.8.1. Experiments of Hatamura & Chijiiwa (1977B)
Hatamura & Chijiiwa (1977B) carried out experiments in sand, clay and loam. The experiments were carried out with blade angles α of 30°, 45°, 60°, 75° and 90°, layer thicknesses hi of 0.05, 0.10 and 0.15 m and cutting velocities vc of 0.05, 0.10 and 0.14 m/sec. The blade had a fixed length L4 of 0.2 m and a fixed width of 0.33 m. The clay/loam had a dynamic cohesion of 27.9 kPa and a dynamic adhesion of 13.95 kPa. Hatamura & Chijiiwa (1977B) only give the dynamic cohesion and adhesion, not the static ones. Based on equation (7-53) an average strengthening factor of about 1.5 can be determined. This factor may however vary with the blade angle and layer thickness. Hatamura & Chijiiwa (1977B) measured the shear angle β, the total cutting force and the direction of the total cutting force. The also determined the location of the acting points of the different forces. In the model they derived they used the horizontal and vertical force equilibrium equations and the equilibrium of moments equation combined with their measured acting points. By solving the 3 equilibrium equations, they solved the horizontal cutting force, the vertical cutting force and the shear angle, based on 3 equations with 3 unknowns. The theory as derived here assumes a shear angle where there is a minimum horizontal force, based on the minimum cutting energy principle. So the two approaches are different. Hatamura & Chijiiwa (1977B) found that cutting tests with cutting angles of 30° and 45° were according to the Tear Type, while the larger cutting angles followed the Flow Type of cutting mechanism. This tells something about the tensile strength of the material. Based on the above an ac ratio of 0.5-0.7 can be derived. Figure 7-27 shows that a tensile strength to cohesion σT/c ratio of about 0.2 may explain this. So it is assumed that the tensile strength is 20% of the cohesion.
Figure 7-44, Figure 7-45 and Figure 7-46 show the results of the experiments and the calcultations. The calculations are carried out for both the Flow Type and the Tear Type. The shear angles predicted are 5°-10° larger than the ones measured, however the tendency is the same.
The measured total cutting forces match the predicted cutting forces very well for the Tear Type for blade angles of 30° and 45° and for the Flow Type for blade angles of 60°, 75° and 90°. The theory does predict the Tear Type and the Flow Type for the corresponding blade angles. The directions measured of the total cutting force also match the theory very well if the correct cutting mechanism is considered. So apparently the total cutting forces and the direction of these forces can be predicted well, but the shear angle gives differences. We should consider that the shear angle as used in the theory here is a straight line, a simplification. In reality the shear plane may be curved, leading to different values of the shear angle measured. For the Tear Type it is not clear what definition Hatamura & Chijiiwa (1977B) used to determine the shear angle. Is it the point where the secondary tensile crack reaches the surface? This explains some of the differences between the measured and calculated shear angles. Overall, the theory as developed here predicts the cutting forces and the direction of these forces very well.
The force for a 60° blade and 0.05 m layer thickness is smaller than expected based on the Flow Type of cutting process. This is caused by the Curling Type as shown below.
Figure 7-45 shows that the experiment with a layer thickness of 0.05 m with a blade angle of 60° gives a smaller cutting force than estimated. Analyzing the 60° experiments as a function of the layer thickness gives Figure 7-47. This figure shows that up to a layer thickness of about 0.08 m there will be a Curling Type of cutting process. Above 0.08 m there will be a Flow Type of cutting process, while above about 0.20 m there will be a Tear Type of cutting process. Once the Tear Type is present, the force will drop to the lower Tear Type curve as is visible in the 30° and 45° experiments. Since all 3 cutting mechanisms were present in the experiments of Hatamura & Chijiiwa (1977B), it is not possible to find just one equation for the cutting forces. Each of the 3 cutting mechanisms has its own model or equation. Figure 7-48 shows the 30° experiment. It is clear from the figure that at 0.10 m layer thickness the cutting mechanism of of the Tear Type.
7.8.2. Wismer & Luth (1972B)
Wismer & Luth (1972B) investigated rate effects in soil cutting in dry sand, clay and loam. For clay and loam they distinguished two rate effects, the inertial forces and the strengthening effect. For cutting velocities as known in dredging (up to 5-6 m/sec), the inertial forces can be neglected compared to the static cutting forces (low cutting velocities) and compared to the strengthening effect. Wismer & Luth (1972B) carried out experiments with blade angles of 30°, 60° and 90°, blades of 0.19·0.29 m, 0.127·0.193 m and 0.0762·0.117 m (7.5·11.45 inch, 5.0·7.6 inch and 3.0·4.59 inch) and layer thicknesses from 0.0225-0.0762 m (0.9-3.0 inch). They did the experiments in two types of clay. Unfortunately they did not mention the cohesion and adhesion, but the mentioned a cone resistance. However, based on their graphs the cohesion could be deducted. The cone index 27 clay should have had a cohesion of about 22.5 kPa and an adhesion of 11.25 kPa, the cone index 42 clay a cohesion of 34 kPa and an adhesion of 17 kPa. The strengthening factor of Wismer & Luth (1972B) can be rewritten in SI Units, using the reference strain rate of 0.03/sec, giving the following equation for the strengthening factor.
$\ \tau=\tau_{\mathrm{y}} \cdot \lambda_{\mathrm{s}} \quad\text{ with: }\quad \lambda_{\mathrm{s}}=\left(\frac{\frac{\mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}}}}{\mathrm{0 . 0 3}}\right)^{\mathrm{0 . 1}}\tag{7-106}$
Figure 7-49 shows the theoretical strengthening factors based on the average of equations (7-34) and (7-35) and for the above equation for the minimum and maximum layer thickness, giving a range for the strengthening factor and comparing the Miedema (1992) equation with the Wismer & Luth (1972B) equation. The figure also shows the results of 5 series of tests as carried out by Wismer & Luth (1972B) with a 30° blade. The two equations match well up to cutting velocities of 1.5 m/sec, but this may differ for other configurations. At high cutting velocities the Wismer & Luth (1972B) equation gives larger strengthening factors. Both equations give a good correlation with the experiments, but of course the number of experiments is limited. A realistic strengthening factor for practical cutting velocities in dredging is a factor 2. In other words, a factor of about 2 should be used to multiply the static measured cohesion, adhesion and tensile strength.
It should be mentioned that the above equation is modified compared with the original Wismer & Luth (1972B) equation. They used the ratio cutting velocity to blade width to get the correct dimension for strain rate, here the ratio cutting velocity to layer thickness is used, which seems to be more appropriate. The constant of 0.03 is the constant found from the experiments of Hatamura & Chijiiwa (1977B).
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/07%3A_Clay_Cutting/7.08%3A_Experiments_in_Clay.txt
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a
Adhesion
kPa
A
Adhesive force on the blade
kN
B
Frequency (material property)
1/s
c
Cohesion
kPa
cm
Mobilized shear strength
kPa
C
Cohesive force on shear plane
kN
E
Energy level
J/kmol
Ea
Activation energy level
J/kmol
El
Limiting (maximum) energy level
J/kmol
Esp
Specific cutting energy
kPa
f
Shear force on flow unit
N
F
Cutting force
kN
Fh
Horizontal cutting force
kN
Fv
Vertical cutting force
kN
G
Gravitational force
kN
h
Planck constant (6.626·10-34 J·s)
J·s
hb
Blade height
m
hb,m
Mobilized blade height
m
hi
Layer thickness
m
k
Boltzman constant (1.3807·10-23 J/K)
J/K
K
Constant Herschel Bulkley equation
-
K1
Grain force on the shear plane
kN
K2
Grain force on the blade
kN
i
Coefficient
-
I
Inertial force on the shear plane
kN
n
Power of strain rate equation
-
N
Avogadro constant (6.02·1026 1/kmol)
-
N1
Normal force on shear plane
kN
N2
Normal force on blade
kN
p
Probability
-
Pc
Cutting power
kW
r
Ratio adhesive force to cohesive force
-
rm
Mobilized ratio adhesive force to cohesive force
-
rT
Ratio adhesive force to tensile force
-
R
Universal gas constant (8314 J/kmol/K)
J/kmol/K
R1
Acting point on shear plane
m
R2
Acting point on blade
m
S
Number of bonds per unit area
1/m2
S1
Shear force due to internal friction on the shear surface
kN
S2
Shear force due to soil/steel friction on the blade
kN
SPT
Standard Penetration Test
Blows/foot
T
Absolute temperature
K
T
Tensile force
kN
vc
Cutting velocity
m/s
w
Blade width
m
W1
Force resulting from pore under pressure on the shear plane
kN
W2
Force resulting from pore under pressure on the blade
kN
X
Function
-
α
Blade angle
rad
β
Angle of the shear plane with the direction of cutting velocity
rad
$\ \dot{\varepsilon}_{\mathbf{a}}$
Strain rate adhesion
1/s
$\ \dot{\varepsilon}_{\mathbf{c}}$
Strain rate cohesion
1/s
$\ \dot{\varepsilon}_{0}$
Strain rate from triaxial test
1/s
$\ v$
frequency of activation
1/s
λ
Distance between equilibrium positions
m
λa
Strain rate factor adhesive force
-
λc
Strain rate factor cohesive force
-
λs
Strain rate factor average adhesion and cohesion (usually 2)
-
λ1
Acting point factor on shear plane
-
λ2
Acting point factor on blade
-
λHF
Horizontal cutting force coefficient Flow Type
-
λVF
Vertical cutting force coefficient Flow Type
-
λHT
Horizontal cutting force coefficient Tear Type
-
λVT
Vertical cutting force coefficient Tear Type
-
λHC
Horizontal cutting force coefficient Curling Type
-
λVC
Vertical cutting force coefficient Curling Type
-
dε/dt
Strain rate
1/s
0/dt
Frequency (material property)
1/s
$\ \tau$
Shear stress
kPa
$\ \tau_{\mathrm{a}}$, a
Adhesive shear strength (strain rate dependent)
kPa
$\ \tau_{\mathrm{c}}$, c
Cohesive shear strength (strain rate dependent)
kPa
$\ \tau_\mathrm{y}$
Shear strength (yield stress, material property)
kPa
$\ \tau_\mathrm{ya}$
Adhesive shear strength (material property)
kPa
$\ \tau_\mathrm{yc}$
Cohesive shear strength (material property)
kPa
$\ \tau_0$
Dynamical shearing resistance factor (material property)
kPa
σe
Effective stress
kPa
σn
Normal stress
kPa
σN1
Normal stress on shear plane
kPa
σN2
Normal stress on blade
kPa
σt
Tensile strength
kPa
φ
Angle of internal friction
rad
δ
Soil/steel friction angle
rad
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/07%3A_Clay_Cutting/7.09%3A_Nomenclature.txt
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As mentioned in chapter 2, rock is a natural occurrence of cohesive organic or inorganic material, which forms a
part of the earth crest. Most rocks are composed of one or more minerals.
Rocks can be classified in different ways. The most used classification is based on their origin, distinguishing the following 3 main classes:
Igneous rock. A rock that consists of solidified molten rock material (magma), which has been generated within the earth. Well known are granite and basalt.
Sedimentary rock. Rock formed by the consolidation of sediment as settled in water, ice or air and accumulated on the earth’s surface, either on dry land or under water. Examples are sandstone, limestone and claystone. Metamorphic rock. Any class of rocks that are the result of partial or complete recrystallization in the solid state of pre-existing rocks under conditions of temperature and pressure that are significantly different from thos obtaining at the surface of the earth.
For the atmospheric cutting of rock models, the unconfined compressive strength (UCS), the unconfined tensile strength (UTS), the Brazilian tensile strength (BTS), the angle of internal friction and the angle of external friction are the dominant material properties.
When cutting rock different types of failure may occur. A distinction is made between brittle, brittle ductile and ductile failure, where brittle can be brittle shear failure, brittle tensile failure or a combination of both. The type of failure is mainly determined by the so called ductility number being the ratio of the compressive strength over the tensile strength (UCS/BTS).
The confining pressure and the temperature may also play a role. Figure 8-1 shows a recording of the cutting forces during brittle and ductile failure, where brittle failure shows strongly fluctuating cutting forces, while ductile failure shows a more constant force. In fact in brittle failure there is a force build up, where failure occurs if the force and thus the stresses exceed a certain limit, after which the rock instantly collapses and the force decreases rapidly. Brittle failure is always destructive, meaning that the structure of the rock changes during failure in an irreversible way. Ductile failure in its pure form is plastic deformation and is reversible. In rock ductile failure is usually cataclastic failure, meaning that the microstructure is destroyed, which is also irreversible. Figure 8-2 shows corresponding stress-strain curves.
8.02: Cutting Process and Failure Criteria
In granular materials a number of failure mechanisms can be distinguished. For clarity of definitions, the following definitions are used:
• Flow Type. Failure is based on plastic shear failure. Non-destructive, continues. Both the stress-strain curve according to Figure 8-2 and the non-destructive plastic deformation show ductile behavior. This type of failure will only occur at very high pressures and/or temperatures. The flow of magma is an example of this.
• Tear TypeUCS/BTS=large. Failure based on 100% tensile failure. This type of failure will occur when the UTS-BTS absolute value is small compared to the UCS value. This is a discontinues mechanism.
• Chip TypeUCS/BTS=medium. Failure based on a combination of shear failure and tensile failure, with a crushed zone near the tool tip. The fractions of shear failure and tensile failure depend on the UCS/BTS ratio. A large ratio results in more tensile failure, a small ratio in more shear failure. This is a discontinues mechanism.
• Shear TypeUCS/BTS=small. Failure based on 100% shear failure. This type of failure occurs when the UTS-BTS value is larger and the normal stresses in the shear plane are high, usually at larger blade angles.
This is a discontinues mechanism.
• Crushed Type: Cataclastic failure based on shear, similar to the Flow Type and the Shear Type like in sand.
The Crushed Type is based on cataclastic failure, disintegration of the grain matrix. This mechanism will be identified as pseudo-ductile since it shows ductile behavior in the stress-strain curve of Figure 8-2, but it is destructive and not plastic.
When cutting in dredging practice, blade or pick point angles of about 60 degrees are used. With these blade angles often the Chip Type of cutting mechanism occurs. Smaller blade angles may show the Tear Type cutting mechanisms, while larger blade angles often show the Shear Type of cutting mechanism. The higher the normal stresses in the rock cut, the less likely the occurrence of tensile failure.
When the pick point starts penetrating the rock, usually very high normal stresses occur in front and below the tip of the pick point, resulting in crushing of the rock. Destroying the grain matrix. In a stress-strain diagram this behavior is ductile, but since its also destructive its named pseudo-ductile. Now if the layer thickness is very small, like in oil drilling, the crushed zone may reach the surface and the whole process is of the Crushed Type. If the layer cut is thicker, like in dredging, the Chip Type cutting mechanism may occur, a combination of mechanisms. In the crushed zone and the intact rock a shear plane can be identified based on the minimum deformation work principle. When the pick point progresses, the shear stress on this shear plane increases. When the shear stress exceeds the shear strength (cohesion) a brittle shear crack will occur. It is not necessary that the shear stress exceeds the shear strength over the full length of the shear plane, it only has to exceed the shear strength at the beginning of the shear crack as in the Nishimatsu (1972) approach. When the pick point progresses, the normal and shear stresses increase, resulting in a Mohr circle with increasing radius. Now if the radius increases faster than the normal stress at the center of the Mohr circle, the minimum principal stress decreases and may even become negative. When it becomes negative it may become smaller than the negative tensile strength, resulting in tensile failure.
So in time it starts with a crushed zone, then a shear plane with possibly shear failure and than possibly tensile failure. If the tensile strength is large, it is possible that only shear failure occurs. If the tensile strength is small, it is possible that only tensile failure occurs. Crushing will start if locally a certain criterion is exceeded. Often the Mogi (1966) criterion is applied, giving a certain ratio between the maximum principal stress and the minimum principal stress. Ratio’s used are 3.4 for sandstone and 4.2 for limestone, while Verhoef (1997) found 6 for limestone. Of course crushing does not start instantly, but gradually, based on the structure of the rock, especially the distribution of the microcracks and the skeleton. With the hypothesis that crushing starts where the rock is the weakest, one may assume that crushing starts at the scale of the microcracks, giving relatively large particles still consisting of many grains. With increasing normal stress these particles will also be fragmented into smaller particles. This process will go on until the smallest possible particles, the rock grains, result. Up to the Mogi (1966) criterion intact rock is assumed, however some fracturing or crushing may already have taken place.
From the perspective of the angle of internal friction, one may assume that the angle of internal friction is based on the internal structure of the rock, and as long as the rock is intact, the angle of internal friction may change slightly based on the stress situation, but not to much. However, when fracturing and crushing starts, the internal structure of the rock is changing and this will result in a decreasing angle of internal friction. Decreasing until the angle of internal friction of the smallest particles, the rock grains is reached at high confining pressures.
Verhoef (1997) shows a complete failure envelope of intact rock, including Mogi’s brittle-ductile transition. Vlasblom (2003-2007) refers to this failure envelope. Figure 8-4 shows this failure envelope, where the maximum normal stress is based on a hydrostatic compression test. So based on hydrostatic pressure, the material is crushed, without the presence of shear. This hydrostatic compressive strength (HCS) is a few times the UCS value of the rock. In the figure HCS+UTS=3.5·UCS. Not all rocks show this kind of behavior however. It is important to know that this envelope is based on tri-axial tests on intact rock.
Verhoef (1997) also shows in figure D3 a different failure envelope beyond the brittle-ductile transition point, which is more related to the cutting process. Beyond this transition point the crushed rock still has a certain internal friction angle, which will be discussed later and is shown in Figure 8-14 and Figure 8-15.
It is thus very important to determine the failure criterion envelope based on tests where shear failure occurs.
Figure 8-5 shows how a failure envelope can be constructed by connecting failure points of different stress situations. The figure shows the UTS, BTS and UCS Mohr circles, the Mohr circle at the Mogi criterion, the Mohr circle of a hydrostatic compression test and three additional Mohr circles. Connecting the failure points gives the failure curve. Surrounding the Mohr circles gives the envelope where Mohr circles have to stay inside to prevent failure. At confining stresses exceeding the Mogi point the two envelopes are slightly different.
As mentioned, the apparent shear strength and the internal friction angle of the intact and the crushed rock may differ. In the case where the Mogi criterion describes the shear strength and the angle of internal friction of the crushed rock, the failure curve for higher normal stresses may be a straight line tangent to the Mogi criterion point. Figure 8-6 shows this type of behavior. The Zijsling (1987) experiments at very high confining pressures show this type of behavior for cutting loads in Mancos Shale. The experiments of Zijsling (1987) will be discussed in chapter 9.
It is however also possible that the shear strength and the internal friction angle of the crushed rock decrease to a certain minimum with increasing normal stresses larger than the Mogi point to a point A or B in Figure 8-7. For higher normal stresses the failure curve will follow a straight line as is shown in the figure. The Zijsling (1987) experiments at very high confining pressures show this type of behavior for cutting loads in Pierre Shale.
When increasing the bottomhole pressure (confining pressure) from 0 MPa to 50 MPa, first the cutting forces and thus the normal stresses and shear stresses increase up to a maximum, after which the cutting forces decrease, but at a certain bottomhole pressure this decrease stops and the cutting forces increase slightly with further increasing bottomhole pressure. So there was still an internal friction angle, but very small. The bottomhole pressure is a good indication of the confining pressure. The Zijsling (1987) experiments did show that the material was crushed.
It should be mentioned that the layer thickness was very small in these experiments, resulting in a crushed zone reaching to the surface. In other words, the rock was crushed completely.
8.2.1. Some Relations
The relation between shear strength (cohesion) c, internal friction angle φ and the minimum and maximum principal stresses can be derived according to, using the basic Mohr-Coulomb relations:
$\ \tau=\mathrm{c}+\sigma \cdot \tan (\varphi)\tag{8-1}$
And:
$\ \tau=\frac{\sigma_{\max }-\sigma_{\min }}{2} \cdot \cos (\varphi) \quad\quad\quad\sigma=\frac{\sigma_{\max }+\sigma_{\min }}{2}-\frac{\sigma_{\max }-\sigma_{\min }}{2} \cdot \sin (\varphi)\tag{8-2}$
This gives:
$\ \frac{\sigma_{\max }-\sigma_{\min }}{2} \cdot \cos (\varphi)=\mathrm{c}+\left(\frac{\sigma_{\max }+\sigma_{\min }}{2}-\frac{\sigma_{\max }-\sigma_{\min }}{2} \cdot \sin (\varphi)\right) \cdot \tan (\varphi)\tag{8-3}$
Multiplying with cos(φand reorganizing gives:
$\ \begin{array}{left}\frac{\sigma_{\max }-\sigma_{\min }}{2} \cdot \sin ^{2}(\varphi)+\frac{\sigma_{\max }-\sigma_{\min }}{2} \cdot \cos ^{2}(\varphi)\ =\mathrm{c} \cdot \cos (\varphi)+\frac{\sigma_{\max }+\sigma_{\min }}{2} \cdot \sin (\varphi)\ \frac{\sigma_{\max }-\sigma_{\min }}{2}=\mathrm{c} \cdot \cos (\varphi)+\frac{\sigma_{\max }+\sigma_{\min }}{2} \cdot \sin (\varphi)\ \sigma_{\max } \cdot(1-\sin (\varphi))=2 \cdot \mathrm{c} \cdot \cos (\varphi)+\sigma_{\min } \cdot(1+\sin (\varphi))\ \sigma_{\max }=\frac{2 \cdot \mathrm{c} \cdot \cos (\varphi)+\sigma_{\min } \cdot(1+\sin (\varphi))}{(1-\sin (\varphi))}\end{array}\tag{8-4}$
This equation can also be written as:
$\ \sigma_{\max }=\sigma_{\min } \cdot \tan ^{2}\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)+2 \cdot \mathrm{c} \cdot \tan \left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\tag{8-5}$
This relation is valid for all linear failure criteria with a cohesion and an internal friction angle φ. Now if two Mohr circles are found with index 1 and 2. Index 1 for the smallest circle and index 2 for the largest circle, the following relation is valid in relation to the failure curve and internal friction angle:
$\ \frac{1+\sin (\varphi)}{1-\sin (\varphi)}=\frac{\sigma_{\max , 2}-\sigma_{\max , 1}}{\sigma_{\min , 2}-\sigma_{\min , 1}}=\mathrm{r}\tag{8-6}$
This gives:
$\ \sin (\varphi)=\frac{\mathrm{r}-1}{\mathrm{r}+1} \quad\text{ and }\quad \cos (\varphi)=\frac{2 \cdot \sqrt{\mathrm{r}}}{\mathrm{r}+1} \quad\text{ and }\quad \tan (\varphi)=\frac{\mathrm{r}-1}{2 \cdot \sqrt{\mathrm{r}}}\tag{8-7}$
Once the internal friction angle is found, the cohesion can be determined as:
$\ \mathrm{c}=\frac{\mathrm{U} \mathrm{C S}}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=\frac{\mathrm{U} \mathrm{C S}}{2 \cdot \sqrt{\mathrm{r}}}\tag{8-8}$
So the Mohr-Coulomb relation is:
$\ \tau=\frac{\mathrm{U C S}}{2 \cdot \sqrt{\mathrm{r}}}+\sigma \cdot \frac{\mathrm{r}-1}{2 \cdot \sqrt{\mathrm{r}}}=\frac{\mathrm{U} \mathrm{C S}+\sigma \cdot(\mathrm{r}-1)}{2 \cdot \sqrt{\mathrm{r}}}\tag{8-9}$
8.2.2. Brittle versus Ductile
The terms ductile failure and brittle failure are often used in literature for the failure of materials with shear strength and tensile strength, but what do the words ductile and brittle mean?
In materials science, ductility is a solid material's ability to deform under tensile stress; this is often characterized by the material's ability to be stretched into a wire. Malleability, a similar property, is a material's ability to deform under compressive stress; this is often characterized by the material's ability to form a thin sheet by hammering or rolling. Both of these mechanical properties are aspects of plasticity, the extent to which a solid material can be plastically deformed without fracture. Ductility and malleability are not always coextensive – for instance, while gold has high ductility and malleability, lead has low ductility but high malleability. The word ductility is sometimes used to embrace both types of plasticity.
A material is brittle if, when subjected to stress, it breaks without significant deformation (strain). Brittle materials absorb relatively little energy prior to fracture, even those of high strength. Breaking is often accompanied by a snapping sound. Brittle materials include most ceramics and glasses (which do not deform plastically) and some polymers, such as PMMA and polystyrene. Many steels become brittle at low temperatures (see ductile-brittle transition temperature), depending on their composition and processing. When used in materials science, it is generally applied to materials that fail when there is little or no evidence of plastic deformation before failure. One proof is to match the broken halves, which should fit exactly since no plastic deformation has occurred. Generally, the brittle strength of a material can be increased by pressure. This happens as an example in the brittle-ductile transition zone at an approximate depth of 10 kilometers in the Earth's crust, at which rock becomes less likely to fracture, and more likely to deform ductile.” (Source Wikipedia).
Rock has both shear strength and tensile strength and normally behaves brittle. If the tensile strength is high the failure is based on brittle shear, but if the tensile strength is low the failure is brittle tensile. In both cases chips break out giving it the name Chip Type. So rock has true brittle behavior. Under hyperbaric conditions however, the pore under pressures will be significant, helping the tensile strength to keep cracks closed. The result is a much thicker crushed zone that may even reach the surface. Crushing the rock is called cataclastic behavior. Since the whole cutting process is dominated by the crushed zone, this is named the Crushed Type. Due to the high pore under pressures the crushed material sticks together and visually looks like a ductile material. That’s the reason why people talk about ductile behavior of hyperbaric rock. In reality it is cataclastic behavior, which could also be named pseudo-ductile behavior.
Now whether the high confining pressure result from a high hyperbaric pressure or from the cutting process itself is not important, in both cases the pseudo-ductile behavior may occur. Figure 8-2 shows the stress-strain behavior typical for brittle and ductile behavior. Based on this stress-strain behavior the term ductile is often used for rock, but as mentioned before this is the result of cataclastic failure.
Gehking (1987) stated that pseudo-ductile behavior will occur when the ratio UCS/BTS<9. Brittle behavior will occur when the ratio UCS/BTS>15. For 9<UCS/BTS<15 there is a transition between brittle and pseudo-ductile. The geometry of the cutting equipment and the operational conditions are not mentioned by Gehking (1987).
Mogi (1966) found a linear relation between the minimum and maximum principal stress at the transition brittle to pseudo-ductile failure. For sandstone he found σmax=3.4·σmin, and for limestone σmax=4.2·σmin. Those values give an indication, since other researchers found σmax>6·σmin (Verhoef, 1997). Now assuming σmax=α·σmin and combining this with Hoek & Brown (1988), gives:
$\ \begin{array}{left}\sigma_{\min }=\mathrm{UCS} \cdot \frac{\mathrm{m}+\sqrt{\mathrm{m}^{2}+4 \cdot(\alpha-1)^{2}}}{2 \cdot(\alpha-1)^{2}}\ \sigma_{\max }=\alpha \cdot \mathrm{UCS} \cdot \frac{\mathrm{m}+\sqrt{\mathrm{m}^{2}+4 \cdot(\alpha-1)^{2}}}{2 \cdot(\alpha-1)^{2}}\end{array}\tag{8-10}$
This gives for the center of the Mohr circle:
$\ \begin{array}{left}\sigma_{\text {center }}=(\alpha+1) \cdot \frac{\text { UCS }}{2} \cdot \frac{\mathrm{m}+\sqrt{\mathrm{m}^{2}+4 \cdot(\alpha-1)^{2}}}{2 \cdot(\alpha-1)^{2}}\ \tau_{\max }=(\alpha-1) \cdot \frac{\mathrm{UCS}}{2} \cdot \frac{\mathrm{m}+\sqrt{\mathrm{m}^{2}+4 \cdot(\alpha-1)^{2}}}{2 \cdot(\alpha-1)^{2}}\end{array}\tag{8-11}$
Figure 8-17 shows the Mogi criterion both for the top of the Mohr circle curve and the failure curve. Left of the Mogi criterion point there will be brittle failure, on the right there will be pseudo-ductile failure. When the coefficient α increases, the Mogi points move to the left.
In the case of a straight failure plane this gives for the normal and shear stress:
$\ \sigma=\frac{\frac{\mathrm{c}}{\cos (\varphi)} \cdot\left(\frac{\alpha+1}{\alpha-1}-\sin (\varphi)\right)}{1-\frac{\tan (\varphi)}{\cos (\varphi)} \cdot\left(\frac{\alpha+1}{\alpha-1}-\sin (\varphi)\right)} \quad\text{ and }\quad \tau=\mathrm{c + \sigma \cdot \operatorname { tan } ( \varphi )}\tag{8-12}$
Which is also shown in Figure 8-17. If the angle of internal friction is to high, the brittle-ductile transition will never be reached. The criterion for this is:
$\ \frac{\alpha-1}{\alpha+1}>\sin (\varphi)\tag{8-13}$
8.2.3. Based on UTS and UCS
Here a linear envelope tangent to the UTS and the UCS Mohr circles is assumed, based on the assumption that the failure curve always has to be tangent to at least two Mohr Coulomb circles. This gives for the principal stresses:
$\ \begin{array}{left}\sigma_{\min , 1}=-\text{UTS}\ \sigma_{\max , 1}=0\ \sigma_{\min , 2}=0\ \sigma_{\max , 2}=\text{UCS}\ \mathrm{r=\frac{U C S-0}{0--U T S}=\frac{U C S}{U T S}=m}\end{array}\tag{8-14}$
This method results in a rather high value for the internal friction angle and consequently a rather low value for the shear strength (cohesion). To find a good estimate for the internal friction angle, there should be two Mohr circles based on shear failure. In this case one circle is based on shear failure, but the other circle is based on tensile failure. So this method is rejected.
Figure 8-9 shows the Mohr circles for UTSBTS and UCS for UCS=100 MPaUTS=BTS=15 MPa. The resulting angle of internal friction φ=47.7o. The transition brittle-ductile according to Mogi (1966) does not exist, the angle of internal friction is too high.
8.2.4. Based on BTS and UCS
Here a linear envelope tangent to the BTS and the UCS Mohr circles is assumed, based on the assumption that the failure curve always has to be tangent to at least two Mohr Coulomb circles. This gives for the principal stresses:
$\ \begin{array}{left}\sigma_{\min , 1}=-\text{BTS}\ \sigma_{\max , 1}=3 \cdot \text{BTS}\ \sigma_{\min , 2}=0\ \sigma_{\max , 2}=\text{UCS}\ \mathrm{r=\frac{U C S-3 \cdot B T S}{0--B T S}=\frac{U C S-3 \cdot B T S}{B T S}=\frac{U C S}{B T S}-3=m-3}\end{array}\tag{8-15}$
This method results in a rather high value for the internal friction angle and consequently a rather low value for the shear strength (cohesion), although the internal friction angle will be lower than from the first method. To find a good estimate for the internal friction angle, there should be two Mohr circles based on shear failure. In this case one circle is based on shear failure, but the other circle is based on tensile failure. So this method is rejected. Figure 8-10 shows the Mohr circles for UTSBTS and UCS for UCS=100 MPaUTS=BTS=15 MPa. The resulting angle of internal friction φ=34.8o. The transition brittle-ductile according to Mogi (1966) is at a normal stress of 316 MPa.
8.2.5. Hoek & Brown (1988)
Over the years Hoek & Brown (1988) developed a failure criterion for rock, based on the UCS and BTS values of the specific rock. The generalised criterion is empirical and yields:
$\ \sigma_{\max }=\sigma_{\min }+\mathrm{U C S} \cdot\left(\mathrm{m} \cdot \frac{\sigma_{\min }}{\mathrm{U C S}}+\mathrm{s}\right)^{\mathrm{a}} \quad\text{ with }\quad\frac{\mathrm{a}=\mathrm{0 .5}}{\mathrm{s}=\mathrm{1 .0}}\text{ for intact rock}\tag{8-16}$
The parameters and are material properties. The parameter is related to the ratio of the UCS value to the BTS value according to:
$\ \mathrm{m}=\frac{\mathrm{U C S}^{2}-\mathrm{B T S}^{2}}{\mathrm{U C S} \cdot \mathrm{B T S}} \quad\text{ for }\frac{\mathrm{B T S}}{\mathrm{U C S}} \ll \mathrm{1} \quad \mathrm{m}=\frac{\mathrm{U C S}}{\mathrm{B T S}}\tag{8-17}$
The parameter is a measure for the amount of fractures in the rock and equals 1 for intact rock. The stresses σmin and σmax are the minimum and maximum principal stresses of the Mohr circle considered. The BTS value can also be represented as a function of and according to:
$\ \mathrm{B T S}=\frac{\mathrm{U C S}}{2} \cdot\left(\mathrm{m}-\sqrt{\mathrm{m}^{2}+4 \cdot \mathrm{s}}\right)\tag{8-18}$
Based on:
$\ \sigma_{\text {center }}=\frac{\sigma_{\max }+\sigma_{\min }}{2}\text{ and }\tau_{\max }=\frac{\sigma_{\max }-\sigma_{\min }}{2}\tag{8-19}$
An equation can be derived relating the maximum shear stress $\ \tau_\text{max}$ (the top of the Mohr circle) to the normal stress at the center of the Mohr circle σcenter.
$\ \tau_{\max }=\frac{1}{8} \cdot\left(-\mathrm{m} \cdot \mathrm{UCS}+\sqrt{(\mathrm{m} \cdot \mathrm{U C S})^{2}+\mathrm{1 6} \cdot\left(\mathrm{m} \cdot \mathrm{U C S} \cdot \sigma_{\text {center }}+\mathrm{U C S}^{2}\right)}\right)\tag{8-20}$
This equation results in a curve through the tops of the Mohr circles and is not yet a failure criterion. For the failure criterion Hoek & Brown (1988) give the following method; First determine a variable according to:
$\ \mathrm{h}=1+\frac{\mathrm{1 6} \cdot(\mathrm{m} \cdot \sigma+\mathrm{s} \cdot \mathrm{U C S})}{\mathrm{3} \cdot \mathrm{m}^{2} \cdot \mathrm{U C S}}\tag{8-21}$
Now an angle θ can be determined:
$\ \theta=\frac{1}{3} \cdot\left(\frac{\pi}{2}+\operatorname{atan}\left(\frac{1}{\sqrt{\mathrm{h}^{3}-1}}\right)\right)\tag{8-22}$
Based on the angle θ the instantaneous internal friction angle can be determined, which is also the tangent to the failure criterion:
$\ \varphi=\operatorname{atan}\left(\frac{1}{\sqrt{4 \cdot \mathrm{h} \cdot \cos ^{2}(\theta)-1}}\right)\tag{8-23}$
Rock Type
Class
Group
Texture
Coarse
Medium
Fine
Very Fine
Sedimentary
Clastic
Conglo- merates (21±3) Breccias (19±5)
Sand-Stones (17±4)
Silt-Stones (7±2) Grey- wackes (18±3)
Clay- Stones (4±2) Shales (6±2) Marls (7±2)
Nonclastic
Carbonates
Crystalline Limestone (12±3)
Sparitic Limestone (10±2)
Micritic Limestone (9±2)
Dolo- mites (9±3)
Evaporites
Gypsum (8±2)
Anhy-drite (12±2)
Organic
Chalk (7±2)
Meta- morphic
Non Foliated
Marble (9±3)
Hornfels (19±4) Meta Sandstone (19±3)
Quartzite (20±3)
Slightly Foliated
Migmatites (29±3)
Amphi- bolites (26±6)
Foliated
Gneiss (28±5)
Schists (12±3)
Phyllites (7±3)
Slates (7±4)
Igneous
Plutonic
Light
Granite (32±3) Grano-diorite (29±3)
Diorite (25±5)
Dark
Gabbro (27±3) Norite (20±5)
Dolerite (16±5)
Hypabyssal
Porphyries (20±5)
Diabase (15±5)
Peridotite (25±5)
Volcanic
Lava
Rhyolite (25±5) Andesite (25±5)
Dacite (25±3) Basalt (25±5)
Obsidian (19±3)
Pyroclastic
Agglomerate (19±3)
Breccia (19±5)
Tuff (13±5)
Last but not least, the shear stress $\ \tau$, matching the normal stress σ can be determined:
$\ \tau=(\cot (\varphi)-\cos (\varphi)) \cdot \frac{\mathrm{m} \cdot \mathrm{U C S}}{\mathrm{8}}\tag{8-24}$
A second way of determining the failure criterion curve is with the following two equations, based on the minimum principal stress:
$\ \sigma=\sigma_{\min }+\frac{\mathrm{U C S}}{2} \cdot \sqrt{\mathrm{m} \cdot \frac{\sigma_{\min }}{\mathrm{U C S}}+\mathrm{s}} \cdot{\left(1-\frac{\mathrm{m}}{\mathrm{m}+4 \cdot \sqrt{\mathrm{m} \cdot \frac{\sigma_{\min }}{\mathrm{U C S}}+\mathrm{s}}}\right)}\tag{8-25}$
$\ \tau=\frac{\mathrm{U C S}}{2} \cdot \sqrt{\mathrm{m} \cdot \frac{\sigma_{\min }}{\mathrm{U C S}}+\mathrm{s}} \cdot \sqrt{ \mathrm{\left(1 - \left( \frac { \mathrm { m } } { \mathrm { m } + 4 \cdot \sqrt { \mathrm { m } \cdot \frac { \sigma _ { \operatorname { min } } } { \mathrm { U C S } } + \mathrm { s } } } \right) ^ { 2 } \right)}}\tag{8-26}$
Figure 8-11 and Figure 8-12 show the Hoek & Brown failure criterion for the top of the Mohr circles (A) and for the real failure condition (B). Although still based on UTS or BTS and UCS and not on two tests with shear failure, the resulting failure curve seems more realistic, which seems logic since it is based on many experiments. The Mohr circles for UTSBTS and UCS are determined for UCS=100 MPaUTS=BTS=15 MPa. The transition brittle-ductile according to Mogi (1966) is at a normal stress of 150 MPa.
Taking an average internal friction angle from a normal stress of zero to a normal stress of 240 MPa gives φ=27.1o.
8.2.6. Parabolic Envelope UTS and UCS
Based on the UTS or BTS and the UCS a parabole can be constructed of the normal stress as a function of the shear stress, with boundary conditions that the parabole goes through the UTS or BTS point (shear stress equals zero, normal stress equals -UTS or -BTS and derivative dσ/d$\ \tau$ equals zero) and touches the UCS Mohr circle as a tangent. With m=UCS/UTS or m=UCS/BTS the equation of this parabole is:
$\ \mathrm{\sigma=\frac{1}{U T S \cdot(\sqrt{m+1}-1)^{2}} \cdot \tau^{2}-U T S}\tag{8-27}$
It is more convenient to write this equation in the form where the shear stress is a function of the normal stress, giving:
$\ \tau^{2}=\mathrm{UTS} \cdot(\sqrt{\mathrm{m}+1}-1)^{2} \cdot(\sigma+\mathrm{UTS})\tag{8-28}$
Figure 8-13 shows the resulting parabole. Although still based on UTS or BTS and UCS and not on two tests with shear failure, the resulting failure curve seems more realistic. The Mohr circles for UTSBTS and UCS are determined for UCS=100 MPaUTS=BTS=15 MPa. The transition brittle-ductile according to Mogi (1966) is at a normal stress of 104 MPa. Taking an average internal friction angle from a normal stress of zero to a normal stress of 240 MPa gives φ=18.6o.
8.2.7. Ellipsoid Envelope UTS and UCS
Instead of a parabole, also an ellipse can be used. The advantage of an ellipse is that it gives more flexibility for the shape of the failure envelope. The general equation for an ellipse is:
$\ \frac{(\sigma-(\mathrm{a}-\mathrm{U T S}))^{2}}{\mathrm{a}^{2}}+\frac{\tau^{2}}{\mathrm{b}^{2}}=1\tag{8-29}$
In order to find an estimate for the radii and b, it is assumed that the ellipse also touches the UCS Mohr circle in the same point as the parabole. With:
$\ \mathrm{f}=\frac{\mathrm{1}}{\mathrm{U T S} \cdot(\sqrt{\mathrm{m}+\mathrm{1}}-\mathrm{1})^{2}}\tag{8-30}$
This gives for the normal stress of the parabole to Mohr circle tangent point:
$\ \mathrm{\sigma_{p}=\frac{-(1-f \cdot U C S)+\sqrt{(1-f \cdot U C S)^{2}+4 \cdot f \cdot U T S}}{2 \cdot f}}\tag{8-31}$
And for the shear stress at the tangent point:
$\ \tau_{\mathrm{p}}^{2}=\mathrm{U T S} \cdot(\sqrt{\mathrm{m}+1}-1)^{2} \cdot\left(\sigma_{\mathrm{p}}+\mathrm{U T S}\right)\tag{8-32}$
Comment: For sandstone a residual internal friction angle of 15 degrees and for limestone 25 degrees have been found at the brittle-ductile transition points.
For a given radius this gives:
$\ \mathrm{b}^{2}=\frac{\tau_{\mathrm{p}}^{2}}{\left(1-\frac{\left(\sigma_{\mathrm{p}}-(\mathrm{a}-\mathrm{U T S})\right)^{2}}{\mathrm{a}^{2}}\right)}\tag{8-33}$
Figure 8-14 shows both the parabolic and the ellipsoid failure envelopes. The ellipsoid failure envelope is determined for a=1.75·UCS. The Mohr circles for UTSBTS and UCS are determined for UCS=100 MPaUTS=BTS=15 MPa. At low normal stresses the parabolic and ellipsoid failure envelopes behave almost identical. Also the Mogi brittle-ductile transition points are very close. Chosing a>10·UCS gives about identical envelopes in the normal stress range considered.
8.2.8. Linear Failure Criterion
The best way to determine the angle of internal friction is to execute at least two tests with different confining pressures in the range of normal stresses the cutting process is expected to operate. Figure 8-16 shows this for a φ=20o internal friction angle. The Mohr circles for UTSBTS and UCS are determined for UCS=100 MPaUTS=BTS=15 MPa. The transition brittle-ductile according to Mogi (1966) is at a normal stress of 95 MPa.
8.2.9. The Griffith (Fairhurst, 1964) Criterion
Griffith (Fairhurst, 1964) has derived a criterion for brittle failure. His hypothesis aasumes that fracture occurs by rapid extension of sub-microscopic. Pre-existing flaws, randomly distributed throughout the material. He defined two criteria. The first criterion is:
$\ \begin{array}{left}3 \cdot \sigma_{\min }+\sigma_{\max } \leq 0\ -3 \cdot \operatorname{UTS}+\sigma_{\max } \leq 0 \quad\text{ or }\quad \sigma_{\max } \leq 3 \cdot \mathrm{UTS}\ -3 \cdot \operatorname{BTS}+\sigma_{\max } \leq 0 \quad\text{ or }\quad \sigma_{\max } \leq 3 \cdot BTS\end{array}\tag{8-34}$
Failure will occur when σmin=-UTS or σmin=-BTS, which is satisfied in the Brazilian split test. However when:
$\ \mathrm{3} \cdot \sigma_{\min }+\sigma_{\max }>0\tag{8-35}$
Failure will occur when:
$\ \left(\sigma_{\max }-\sigma_{\min }\right)^{2}-\mathrm{8} \cdot \mathrm{U T S} \cdot\left(\sigma_{\max }+\sigma_{\min }\right)=\mathrm{0}\tag{8-36}$
With:
$\ \left(\frac{\sigma_{\max }-\sigma_{\min }}{2}\right)^{2}=4 \cdot \mathrm{UTS} \cdot\left(\frac{\sigma_{\max }+\sigma_{\min }}{2}\right)\tag{8-37}$
This can be written as a parabole for the center of the Mohr circles:
$\ \tau_{\mathrm{max}}^{2}=4 \cdot \mathrm{UTS} \cdot \sigma_{\mathrm{center}}\tag{8-38}$
For a UCS test this gives:
$\ \left(\frac{\mathrm{U C S}}{2}\right)^{2}=4 \cdot \mathrm{U T S} \cdot \frac{\mathrm{U C S}}{2} \quad or \quad \frac{\mathrm{U C S}}{\mathrm{U T S}}=\mathrm{8}\tag{8-39}$
If the UCS/UTS or UCS/BTS ratio is larger than 8, brittle failure will occur.
The Griffith criterion as mentioned here is not the failure curve, but the curve connecting the tops of the Mohr circles.
In the original articles tensile is positive and compression negative, resulting in a sign change compared with the equations mentioned here. Als the minimum and maximum principal stresses were reversed.
8.2.10. Conclusions & Discussion
6 concepts for the angle of internal friction and the failure criteria have been discussed. Figure 8-14 and Figure 8-17 show these criteria. To find the best failure criterion curve, many tests should be carried out at different confining pressures, resulting in shear failure and a set of Mohr circles. Since this information is not always available, The Hoek & Brown, Parabole or Ellipse approximations can be used. The preference of the author is the Ellipse Envelope method or the Linear Failure Criterion method, where the internal friction angle is based on the average of the Parabole Envelope method or measured by experiments.
Above the brittle-ductile transition normal stress, the failure curve will decrease according to Verhoef (1997), based on research of van Kesteren (1995). As mentioned before, at higher stress situations there will be fracturing and crushing. This results in a decrease of the angle of internal friction. The higher the normal stresses, the stronger the fracturing and crushing, the smaller the angle of internal friction. When this starts there is a decrease of the angle of internal friction, while the failure curve is still increasing. However at a certain stress situation the failure curve may be at a maximum, since the angle of internal friction decreases to much. This maximum is often close to the Mogi (1966) criterion. Since intact rock and crushed rock are two different materials with different properties, one has to be very carefull with the interpretation of the resulting failure curve. In fact the material has continuously changing properties from the moment is starts fracturing and crushing. First larger particles are formed, consisting of many rock grains. When the stresses increase, these particles will also be fractured or crushed, resulting in smaller particles, until the rock grains are left.
When the angle of internal friction decreases faster than the increase of normal stresses, the failure curve decreases. This does however not mean that there is negative internal friction, normally the tangent to the failure curve. Just that the angle of internal friction decreases faster than the increase of normal stresses and most probably that the shear strength of the crushed rock decreases to zero. Verhoef (1997) and Vlasblom (2003-2007) show a failure curve reducing to zero for very high normal stresses. This seems to be unlikely to happen. It would imply that at very high normal stresses the shear stress equals zero, so no friction at al, which sounds like liquid behavior. It is more likely that the crushed rock, once completely crushed, will have a residual internal friction angle and possibly a residual shear strength. The latter is possible, for example when the particles are so small that van der Waals forces start playing a role. But this will depend completely on the type and composition of the rock.
Figure 8-14 shows a residual internal friction angle for both the ellipse and the parabole, tangent to the failure envelopes at the Mogi brittle-criterion.
For the models derived in this chapter, a constant internal friction angle is assumed, where this constant internal friction angle should match the stress state of the cutting process considered.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/08%3A_Rock_Cutting-_Atmospheric_Conditions/8.01%3A_Introduction.txt
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When cutting rock with a pick point, usually a crushed zone will occur in front of and under the tip of the pick point. If the cutting depth is small, this crushed zone may reach the surface and a sand like cutting process may occur. If the cutting depth is larger, the crushed material cannot escape and the stresses in the crushed zone increase strongly. According to Fairhurst (1964) the cutting forces are transmitted through particle-particle contacts. The stresses are transmitted to the intact rock as discrete point loads this way, causing micro shear cracks and finally a tensile crack. Figure 8-18 and Figure 8-19 sho this cutting mechanism.
As mentioned the type of failure depends on the UCS/BTS ratio. Geking (1987) stated that below a ratio of 9 ductile failure will occur, while above a ratio of 15 brittle failure will occur. In between these limits there is a transition between ductile and brittle failure, which is also in accordance with the findings of Fairhurst (1964). The mechanism as described above is difficult to model. Still a method is desired to predict the cutting forces in rock cutting in order to estimate forces, power and production. In literature some models exist, like the Evans (1964) model based on tensile failure and the Nishimatsu (1972) model based on shear failure. From steel cutting also the Merchant (1944) model is known, based on plastic shear failure. The Evans (1964) model assumes a maximum tensile stress on the entire failure plane, which could match the peak forces, but overestimates the average forces. Nishimatsu (1972) build in a factor for the shear stress distribution on the failure plane, enabling the model to take into account that failure may start when the shear stress is not at a maximum everywhere in the shear plane. Both models are discussed in this chapter.
Based on the Merchant (1944) model for steel cutting and the Miedema (1987 September) model for sand cutting, a new model is developed, both for ductile cutting, ductile cataclastic cutting, brittle shear cutting and brittle tensile cutting. First a model is derived for the Flow Type, which is either ductile shear failure or brittle shear failure. In the case of brittle shear failure, the maximum cutting forces are calculated. For the average cutting forces the maximum cutting forces have to be reduced by 30% to 50%. Based on the Flow Type and the Mohr circle, the shear stress in the shear plane is determined where, on another plane (direction), tensile stresses occur equal to the tensile strength. An equivalent or mobilized shear strength is determined giving this tensile stress, leading to the Tear Type of failure. This approach does not require the tensile stress to be equal to the tensile strength on the whole failure plane, instead it predicts the cutting forces at the start of tensile failure.
This method can also be used for predicting the cutting forces in frozen clay, permafrost.
Roxborough (1987) derived a simple expression for the specific energy based on many experiments in different types of rock. The dimension of this equation is MPa. The two constants in the equation may vary a bit depending on the type of rock. The 0.11 is important at small UCS values, the 0.25 at large values.
$\ \mathrm{E}_{\mathrm{sp}}=\mathrm{0 .2 5} \cdot \mathrm{U.C.S.}+\mathrm{0.11}\tag{8-40}$
The fact that cutting rock is irreversible, compared to the cutting of sand and clay, also means that the 4 standard cutting mechanisms cannot be applied on cutting rock. In fact the Flow Type looks like cataclastic ductile failure from a macroscopic point of view, but the Flow Type (also the Curling Type) are supposed to be real plastic deformation after which the material (clay) is still in tact, while cataclastic ductile failure is much more the crushing of the rock with shear falure in the crushed rock. We will name this the Crushed Type. When the layer cut is thicker, a crushed zone exists but not to the free surface. From the crushed zone first a shear plane is formed from which a tensile crack goes to the free surface. We will name this the Chip Type.
8.3.1. The Model of Evans
For brittle rock the cutting theory of Evans (1964) and (1966) can be used to calculate cutting forces (Figure 8-21). The forces are derived from the geometry of the chisel (width, cutting angle and cutting depth) and the tensile strength (BTS) of the rock. Evans suggested a model on basis of observations on coal breakage by wedges. In this theory it is assumed that:
1. A force is acting under an angle δ (external friction angle) with the normal to the surface A-C of the wedge.
2. A resultant force of the tensile stresses acting at the center of the arc C-D, the line C-D is under an angle β (the shear angle) with the horizontal.
3. A third force is required to maintain equilibrium in the buttock, but does not play a role in the derivation.
4. The penetration of the wedge is small compared to the layer thickness hi.
The action of the wedge tends to split the rock and does rotate it about point D. It is therefore assumed that the force acts through point D. Along the fracture line, it is assumed that a state of plain strain is working and the equilibrium is considered per unit of width of the wedge.
$\ \mathrm{\mathrm{T}=\sigma_{\mathrm{T}} \cdot \mathrm{r} \cdot \int_{-\beta}^{\beta} \cos (\omega) \cdot \mathrm{d} \omega \cdot \mathrm{w}=2 \cdot \sigma_{T} \cdot \mathrm{r} \cdot \sin (\beta) \cdot \mathrm{w}}\tag{8-41}$
Where r·dω is an element of the arc C-D making an angle ω with the symmetry axis of the arc. Let hi be the depth of the cut and assume that the penetration of the edge may be neglected in comparison with hi. This means that the force is acting near point C. Taking moments on the chip cut about point gives:
$\ \mathrm{R} \cdot \cos (\alpha+\beta+\delta) \cdot \frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}=\mathrm{T} \cdot \mathrm{r} \cdot \sin (\beta)=2 \cdot \sigma_{\mathrm{T}} \cdot \mathrm{r} \cdot \sin (\beta) \cdot \mathrm{w} \cdot \mathrm{r} \cdot \sin (\beta)\tag{8-42}$
From the geometric relation it follows:
$\ \mathrm{r} \cdot \sin (\beta)=\frac{\mathrm{h}_{\mathrm{i}}}{2 \cdot \sin (\beta)}\tag{8-43}$
Hence:
$\ \mathrm{R}=\frac{\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{2 \cdot \sin (\beta) \cdot \cos (\alpha+\beta+\delta)}\tag{8-44}$
The horizontal component of is R·sin(α+δand due to the symmetry of the forces acting on the wedge the total cutting force is:
$\ \mathrm{F}_{\mathrm{c}}=\mathrm{2} \cdot \mathrm{R} \cdot \sin (\alpha+\delta)=\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\delta)}{\sin (\beta) \cdot \cos (\alpha+\beta+\delta)}\tag{8-45}$
The normal force ($\ \perp$ on cutting force) is per side:
$\ \mathrm{F}_{\mathrm{n}}=\mathrm{R} \cdot \cos (\alpha+\delta)=\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\cos (\alpha+\delta)}{2 \cdot \sin (\beta) \cdot \cos (\alpha+\beta+\delta)}\tag{8-46}$
The angle β can be determined by using the principle of minimum energy:
$\ \frac{\mathrm{d} \mathrm{F}_{\mathrm{c}}}{\mathrm{d} \beta}=0\tag{8-47}$
Giving:
$\ \begin{array}{left}\cos (\beta) \cdot \cos (\alpha+\beta+\delta)-\sin (\beta) \cdot \sin (\alpha+\beta+\delta)=0\ \Rightarrow \cos (2 \cdot \beta+\alpha+\delta)=0\end{array}\tag{8-48}$
Resulting in:
$\ \beta=\frac{1}{2} \cdot\left(\frac{\pi}{2}-\alpha-\delta\right)=\frac{\pi}{4}-\frac{\alpha+\delta}{2}\tag{8-49}$
With:
$\ \sin (\beta) \cdot \cos (\alpha+\beta+\delta)=\frac{1-\sin (\alpha+\delta)}{2}\tag{8-50}$
This gives for the horizontal cutting force:
$\ \mathrm{F}_{\mathrm{c}}=\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\mathrm{2} \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta)}=\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{H T}}\tag{8-51}$
For each side of the wedge the normal force is now (the total normal/vertical force is zero):
$\ \mathrm{F}_{\mathrm{n}}=\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\cos (\alpha+\delta)}{1-\sin (\alpha+\delta)}=\sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V T}}\tag{8-52}$
Figure 8-20 shows the brittle-tear horizontal force coefficient λHT as a function of the wedge top angle α and the internal friction angle φ. The internal friction angle φ does not play a role directly, but it is assumed that the external friction angle δ is 2/3 of the internal friction angle φ. Comparing Figure 8-20 with Figure 8-42 (the brittle- tear horizontal force coefficient λHT of the Miedema model) shows that the coefficient λHT of Evans is bigger than the λHT coefficient of Miedema. The Miedema model however is based on cutting with a blade, while Evans is based on the penetration with a wedge or chisel, which should give a higher cutting force. The model as is derived in chapter 8.3 assumes sharp blades however.
8.3.2. The Model of Evans under an Angle ε
When it is assumed that the chisel enters the rock under an angle ε and the fracture starts in the same direction as the centerline of the chisel as is shown in Figure 8-22, the following can be derived:
$\ \mathrm{h}=\mathrm{2} \cdot \mathrm{r} \cdot \sin (\beta) \cdot \sin (\beta-\varepsilon)\text{ and }\mathrm{h}_{\mathrm{i}}=2 \cdot \mathrm{r} \cdot \sin ^{2}(\beta)\tag{8-53}$
$\ \mathrm{h_{i}=h \cdot \frac{\sin (\beta)}{\sin (\beta-\varepsilon)}}\tag{8-54}$
Substituting equation (8-53) in equation (8-45) for the cutting force gives:
$\ \begin{array}{left} \mathrm{F}_{\mathrm{c}}&= \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \frac{\mathrm{2} \cdot \mathrm{s i n}(\boldsymbol{\alpha}+\delta)}{\mathrm{1}-\sin (\alpha+\delta)} \&= \sigma_{\mathrm{T}} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\sin (\beta)}{\sin (\beta-\varepsilon)} \cdot \frac{\sin (\alpha+\delta)}{\sin (\beta) \cdot \cos (\alpha+\beta+\delta)} \&= \sigma_{T} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\delta)}{\sin (\beta-\varepsilon) \cdot \cos (\alpha+\beta+\delta)} \end{array}\tag{8-55}$
The horizontal component of the cutting force is now:
$\ \mathrm{F_{\mathrm{ch}}=\sigma_{\mathrm{T}} \cdot h \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\delta)}{\sin (\beta-\varepsilon) \cdot \cos (\alpha+\beta+\delta)} \cdot \cos (\varepsilon)}\tag{8-56}$
The vertical component of this cutting force is now:
$\ \mathrm{F}_{\mathrm{cv}}=\sigma_{\mathrm{T}} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\sin (\alpha+\delta)}{\sin (\beta-\varepsilon) \cdot \cos (\alpha+\beta+\delta)} \cdot \sin (\varepsilon)\tag{8-57}$
Note that the vertical force is not zero anymore, which makes sense since the chisel is not symmetrical with regard to the horizontal anymore. Equation (8-58) can be applied to eliminate the shear angle β from the above equations. When the denominator is at a maximum in these equations, the forces are at a minimum. The denominator is at a maximum when the first derivative of the denominator is zero and the second derivative is negative.
The angle β can be determined by using the principle of minimum energy:
$\ \frac{\mathrm{d} \mathrm{F}_{\mathrm{c}}}{\mathrm{d} \boldsymbol{\beta}}=\mathrm{0}\tag{8-58}$
Giving for the first derivative:
$\ \begin{array}{left} \cos (\beta-\varepsilon) \cdot \cos (\alpha+\beta+\delta)-\sin (\beta-\varepsilon) \cdot \sin (\alpha+\beta+\delta)=0\ \Rightarrow \cos (2 \cdot \beta+\alpha+\delta-\varepsilon)=0\end{array}\tag{8-59}$
Resulting in:
$\ \beta=\frac{1}{2} \cdot\left(\frac{\pi}{2}-\alpha-\delta+\varepsilon\right)=\frac{\pi}{4}-\frac{\alpha+\delta-\varepsilon}{2}\tag{8-60}$
With:
$\ \sin (\beta-\varepsilon) \cdot \cos (\alpha+\beta+\delta)=\frac{1-\sin (\alpha+\delta+\varepsilon)}{2}\tag{8-61}$
Substituting equation (8-61) in equation (8-55) gives for the force Fc:
$\ \mathrm{F_{c}=\sigma_{T} \cdot h \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta+\varepsilon)}}\tag{8-62}$
The horizontal component of the cutting force Fch is now:
$\ \mathrm{F_{c h}=\sigma_{T} \cdot h \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta+\varepsilon)} \cdot \cos (\varepsilon)}\tag{8-63}$
The vertical component of this cutting force Fcv is now:
$\ \mathrm{F_{c v}=\sigma_{T} \cdot h \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta+\varepsilon)} \cdot \sin (\varepsilon)}\tag{8-64}$
8.3.3. The Model of Evans used for a Pick point
In the case where the angle ε equals the angle α, a pick point with blade angle 2·α and a wear flat can be simulated as is shown in Figure 8-23. In this case the equations become:
$\ \mathrm{F_{c}=\sigma_{T} \cdot h \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (2 \cdot \alpha+\delta)}}\tag{8-65}$
The horizontal component of the cutting force Fch is now:
$\ \mathrm{F_{c h}=\sigma_{T} \cdot h \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (2 \cdot \alpha+\delta)} \cdot \cos (\alpha)}\tag{8-66}$
The vertical component of this cutting force Fcv is now:
$\ \mathrm{F_{c v}=\sigma_{T} \cdot h \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (2 \cdot \alpha+\delta)} \cdot \sin (\alpha)}\tag{8-67}$
For the force (see equation (8-45)), acting on both sides of the pick point the following equation can be found:
$\ \mathrm{R}=\frac{\mathrm{F}_{\mathrm{c}}}{\mathrm{2} \cdot \sin (\alpha+\delta)}=\sigma_{\mathrm{T}} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\mathrm{1}}{1-\sin (2 \cdot \alpha+\delta)}\tag{8-68}$
In the case of wear calculations the normal and friction forces on the front side and the wear flat can be interesting. According to Evans the normal and friction forces are the same on both sides, since this was the starting point of the derivation, this gives for the normal force Rn:
$\ \mathrm{R_{n}=\sigma_{T} \cdot h \cdot w \cdot \frac{1}{1-\sin (2 \cdot \alpha+\delta)} \cdot \cos (\delta)}\tag{8-69}$
The friction force Rf is now:
$\ \mathrm{R}_{\mathrm{f}}=\sigma_{\mathrm{T}} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\mathrm{1}}{1-\sin (2 \cdot \alpha+\delta)} \cdot \sin (\delta)\tag{8-70}$
8.3.4. Summary of the Evans Theory
The Evans theory has been derived for 3 cases:
1. The basic case with a horizontal moving chisel and the centerline of the chisel horizontal.
2. A horizontal moving chisel with the centerline under an angle ε.
3. A pick point with the centerline angle ε equal to half the top angle α, horizontally moving.
Once again it should be noted that the angle α as used by Evans is half the top angle of the chisel and not the blade angle as α is used for in most equations in this book. In case 1 the blade angle would be α as used by Evans, in case 2 the blade angle is α+ε and in case 3 the blade angle is 2·α. In all cases it is assumed that the cutting velocity vc is horizontal.
Table 8-2: Summary of the Evans theory.
Case
Cutting forces and specific energy
1
$\ \begin{array}{left}\mathrm{F_{c}=\sigma_{T} \cdot h_{i} \cdot w \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta)}}\ \mathrm{F_{c h}=F_{c}}\ \mathrm{F_{c v}=0}\ \mathrm{E_{s p}=\frac{F_{c h} \cdot v_{c}}{h_{i} \cdot w \cdot v_{c}}=\sigma_{T} \cdot \frac{2 \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta)}}\end{array}\tag{8-71}$
2
$\ \begin{array}{left}\mathrm{F}_{\mathrm{c}}=\sigma_{\mathrm{T}} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\mathrm{2} \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta+\varepsilon)}\ \mathrm{F}_{\mathrm{c h}}=\mathrm{F}_{\mathrm{c}} \cdot \cos (\varepsilon)\ \mathrm{F}_{\mathrm{c v}}=\mathrm{F}_{\mathrm{c}} \cdot \sin (\varepsilon)\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{c h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\sigma_{\mathrm{T}} \cdot \frac{\mathrm{2} \cdot \sin (\alpha+\delta)}{1-\sin (\alpha+\delta+\varepsilon)} \cdot \cos (\varepsilon)\end{array}\tag{8-72}$
3
$\ \begin{array}{left}\mathrm{F}_{\mathrm{c}}=\sigma_{\mathrm{T}} \cdot \mathrm{h} \cdot \mathrm{w} \cdot \frac{\mathrm{2} \cdot \sin (\alpha+\delta)}{1-\sin (\mathrm{2} \cdot \alpha+\delta)}\ \mathrm{F}_{\mathrm{c h}}=\mathrm{F}_{\mathrm{c}} \cdot \cos (\alpha)\ \mathrm{F}_{\mathrm{c v}}=\mathrm{F}_{\mathrm{c}} \cdot \sin (\alpha)\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{c h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\sigma_{\mathrm{T}} \cdot \frac{\mathrm{2} \cdot \sin (\alpha+\delta)}{1-\sin (2 \cdot \alpha+\delta)} \cdot \cos (\alpha)\end{array}\tag{8-73}$
8.3.5. The Nishimatsu Model
For brittle shear rock cutting we may use the equation of Nishimatsu (1972). This theory describes the cutting force of chisels by failure through shear. Figure 8-24 gives the parameters needed to calculate the cutting forces. Nishimatsu (1972) presented a theory similar to Merchant ́s (1944), (1945A) and (1945B) only Nishimatsu’s theory considered the normal and shear stresses acting on the failure plain (A-B) to be proportional to the nth power of the distance λ from point to point B. With being the so called stress distribution factor:
$\ \mathrm{p}=\mathrm{p}_{0} \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}-\lambda\right)^{\mathrm{n}}\tag{8-74}$
Nishumatsu made the following assumptions:
1. The rock cutting is brittle, without any accompanying plastic deformation (no ductile crushing zone).
2. The cutting process is under plain stress condition.
3. The failure is according a linear Mohr envelope.
4. The cutting speed has no effect on the processes.
As a next assumption, let us assume that the direction of the resultant stress is constant along the line A-B. The integration of this resultant stress along the line A-B should be in equilibrium with the resultant cutting force F. Thus, we have:
$\ \mathrm{p}_{0} \cdot \mathrm{w} \cdot \int_{0}^{\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}}\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}-\lambda\right)^{\mathrm{n}} \cdot \mathrm{d} \lambda=\mathrm{F} \Rightarrow \mathrm{p}_{0} \cdot \mathrm{w} \cdot \frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\right)^{\mathrm{n}+1}=\mathrm{F}\tag{8-75}$
Integrating the second term of equation (8-75) allows determining the value of the constant p0.
$\ \mathrm{p}_{0} \cdot \mathrm{w}=(\mathrm{n}+\mathrm{1}) \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\right)^{-(\mathrm{n}+1)} \cdot \mathrm{F}\tag{8-76}$
Substituting this in equation (8-74) gives:
$\ \mathrm{p} \cdot \mathrm{w}=(\mathrm{n}+\mathrm{1}) \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\right)^{-(\mathrm{n}+1)} \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}-\lambda\right)^{\mathrm{n}} \cdot \mathrm{F}\tag{8-77}$
The maximum stress is assumed to occur near the tip of the chisel, so λ=0, giving:
$\ \mathrm{p} \cdot \mathrm{w}=(\mathrm{n}+\mathrm{1}) \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\right)^{-1} \cdot \mathrm{F}\tag{8-78}$
For the normal stress σ and the shear $\ \tau$ stress this gives:
$\ \sigma_{0} \cdot \mathrm{w}=-\mathrm{p} \cdot \mathrm{w} \cdot \cos (\alpha+\beta+\delta)=(\mathrm{n}+1) \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\right)^{-1} \cdot \mathrm{F} \cdot \cos (\alpha+\beta+\delta)\tag{8-79}$
$\ \tau_{0} \cdot \mathrm{w}=\mathrm{p} \cdot \mathrm{w} \cdot \sin (\alpha+\beta+\delta)=(\mathrm{n}+1) \cdot\left(\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\right)^{-1} \cdot \mathrm{F} \cdot \sin (\alpha+\beta+\delta)\tag{8-80}$
Rewriting this gives:
$\ \sigma_{0} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=-\mathrm{p} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\alpha+\beta+\delta)=-(\mathrm{n}+1) \cdot \sin (\beta) \cdot \cos (\alpha+\beta+\delta) \cdot \mathrm{F}\tag{8-81}$
$\ \tau_{0} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{p} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \sin (\alpha+\beta+\delta)=(\mathrm{n}+1) \cdot \sin (\beta) \cdot \sin (\alpha+\beta+\delta) \cdot \mathrm{F}\tag{8-82}$
With the Coulomb-Mohr failure criterion:
$\ \tau_{0}=c+\sigma_{0} \cdot \tan (\varphi)\tag{8-83}$
Substituting equations (8-81) and (8-82) in equation (8-83) gives:
$\ \begin{array}{left}(\mathrm{n}+\mathrm{1}) \cdot \sin (\beta) \cdot \sin (\alpha+\beta+\delta) \cdot \frac{\mathrm{F}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\ =\mathrm{c}-(\mathrm{n}+\mathrm{1}) \cdot \sin (\beta) \cdot \cos (\alpha+\beta+\delta) \cdot \frac{\mathrm{F}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}} \cdot \tan (\varphi)\end{array}\tag{8-84}$
This can be simplified to:
$\ \begin{array}{left} \frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi)}{(\mathrm{n}+1) \cdot \sin (\beta)}=\mathrm{F} \cdot(\sin (\alpha+\beta+\delta) \cdot \cos (\varphi)+\cos (\alpha+\beta+\delta) \cdot \sin (\varphi))\ \quad=\mathrm{F} \cdot \sin (\alpha+\beta+\delta+\varphi)\end{array}\tag{8-85}$
This gives for the force F:
$\ \mathrm{F}=\frac{\mathrm{1}}{(\mathrm{n}+\mathrm{1})} \cdot \frac{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)}\tag{8-86}$
For the horizontal force Fh and the vertical force Fv we find:
$\ \mathrm{F_{h}=\frac{1}{(n+1)} \cdot \frac{c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{\sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)}}\tag{8-87}$
$\ \mathrm{F}_{v}=\mathrm{\frac{1}{(n+1)} \cdot \frac{c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{\sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)}}\tag{8-88}$
To determine the shear angle β where the horizontal force Fh is at the minimum, the denominator of equation (8-86) has to be at a maximum. This will occur when the derivative of Fh with respect to β equals 0 and the second derivative is negative.
$\ \frac{\partial \sin (\alpha+\beta+\delta+\varphi) \cdot \sin (\beta)}{\partial \beta}=\sin (\alpha+2 \cdot \beta+\delta+\varphi)=0\tag{8-89}$
$\ \beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}\tag{8-90}$
Using this, gives for the force F:
$\ \mathrm{F=\frac{1}{(n+1)} \cdot \frac{2 \cdot c \cdot h_{1} \cdot w \cdot \cos (\varphi)}{1+\cos (\alpha+\delta+\varphi)}}\tag{8-91}$
This gives for the horizontal force Fh and the vertical force Fv:
$\ \mathrm{F_{h}=\frac{1}{(n+1)} \cdot \frac{2 \cdot c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\frac{1}{(n+1)} \cdot \lambda_{H F} \cdot c \cdot h_{i} \cdot w}\tag{8-92}$
This solution is the same as the Merchant solution (equations (8-109) and (8-110)) that will be derived in the next chapter, if the value of the stress distribution factor n=0. In fact the stress distribution factor is just a factor to reduce the forces. From tests it appeared that in a type of rock the value of depends on the rake angle. It should be mentioned that for this particular case is about 1 for a large cutting angle. In that case tensile failure may give way to a process of shear failure, which is observed by other researches as well. For cutting angles smaller than 80 degrees is more or less constant with a value of n=0.5. Figure 8-31 and Figure 8-32 show the coefficients λHF and λVF for the horizontal and vertical forces Fh and Fv according to equations (8-109) and (8-110) as a function of the blade angle α and the internal friction angle φ, where the external friction angle δ is assumed to be 2/3·φ. A positive coefficient λVF for the vertical force means that the vertical force Fv is downwards directed. Based on equation (8-97) and (8-109) the specific energy Esp can be determined according to:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\frac{\mathrm{1}}{(\mathrm{n}+\mathrm{1})} \cdot \lambda_{\mathrm{H F}} \cdot \mathrm{c}\tag{8-94}$
The difference between the Nishimatsu and the Merchant approach is that Nishimatsu assumes brittle shear failure, while Merchant assumes plastic deformation as can be seen in steel and clay cutting.
Nishimatsu uses the BTS-UCS method to determine the shear strength and the angle of internal friction. This method gives a high value for the angle of internal friction and a low value for the shear strength. For the factor he found:
$\ \mathrm{n}=-4.9+\mathrm{0 .18} \cdot \alpha\tag{8-95}$
With the blade angle in degrees, for blade angles from 50 to 80 degrees. With this equation is about 0-1 for blade angles around 30 degrees.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/08%3A_Rock_Cutting-_Atmospheric_Conditions/8.03%3A_Cutting_Models.txt
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Rock is the collection of materials where the grains are bonded chemically from very stiff clay, sandstone to very hard basalt. It is difficult to give one definition of rock or stone and also the composition of the material can differ strongly. Still it is interesting to see if the model used for sand and clay, which is based on the Coulomb model, can be used for rock as well. Typical parameters for rock are the compressive strength UCS and the tensile strength BTS and specifically the ratio between those two, which is a measure for how fractured the rock is. Rock also has shear strength and because it consists of bonded grains it will have an internal friction angle and an external friction angle. It can be assumed that the permeability of the rock is very low, so initially the pore pressures do no play a role or cavitation will always occur under atmospheric conditions. But since the absolute hydrostatic pressure, which would result in a cavitation under pressure of the same magnitude can be neglected with respect to the compressive strength of the rock; the pore pressures are usually neglected. This results in a material where gravity, inertia, pore pressures and adhesion can be neglected.
Merchant (1944), (1945A) and (1945B) derived a model for determining the cutting forces when machining steel. The model was based on plastic deformation and a continuous chip formation (ductile cutting). The model included internal and external friction and shear strength, but no adhesion, gravity, inertia and pore pressures. Later Miedema (1987 September) extended this model with adhesion, gravity, inertial forces and pore water pressures.
Definitions:
1. A: The blade tip.
2. B: End of the shear plane.
3. C: The blade top.
4. A-B: The shear plane.
5. A-C: The blade surface.
6. hb: The height of the blade.
7. hi: The thickness of the layer cut.
8. vc: The cutting velocity.
9. α: The blade angle.
10. β: The shear angle.
11. Fh: The horizontal force, the arrow gives the positive direction.
12. Fv: The vertical force, the arrow gives the positive direction.
Figure 8-26 gives some definitions regarding the cutting process. The line A-B is considered to be the shear plane, while the line A-C is the contact area between the blade and the soil. The blade angle is named α and the shear angle β. The blade is moving from left to right with a cutting velocity vc. The thickness of the layer cut is hi and the vertical height of the blade hb. The horizontal force on the blade Fh is positive from right to left always opposite to the direction of the cutting velocity vc. The vertical force on the blade Fv is positive downwards. Since the vertical force is perpendicular to the cutting velocity, the vertical force does not contribute to the cutting power Pc, which is equal to:
$\ \mathrm{P}_{\mathrm{c}}=\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}\tag{8-96}$
The specific energy Esp is defined as the amount of energy used/required to excavate 1 m3 of soil/rock. This can be determined by dividing the cutting power Pc by the production and results in the cutting force Fh in the direction of the cutting velocity vc, divided by the cross section cut hi·w:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{8-97}$
The model for rock cutting under atmospheric conditions is based on the Flow Type of cutting mechanism. Although in general rock will encounter a more brittle failure mechanism and the Flow Type considered represents the ductile failure mechanism, the Flow Type mechanism forms the basis for all cutting processes. The definitions of the Flow Type mechanism are shown in Figure 8-27.
Figure 8-28 illustrates the forces on the layer of rock cut. The forces shown are valid in general. The forces acting on this layer are:
1. A normal force acting on the shear surface N1 resulting from the grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A shear force as a result of the shear strength (cohesion) $\ \tau$c or c. This force can be calculated by multiplying the cohesive shear strength $\ \tau$c with the area of the shear plane.
4. A force normal to the blade N2 resulting from the grain stresses.
5. A shear force S2 as a result of the soil/steel friction N2·tan(δ) or external friction.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1
The forces acting on a straight blade when cutting rock, can be distinguished as:
1. A force normal to the blade N2 resulting from the grain stresses.
2. A shear force S2 as a result of the soil/steel friction N2·tan(δ) or external friction.
These forces are shown in Figure 8-29. If the forces N2 and S2 are combined to a resulting force K2 the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
The horizontal equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{\mathrm{1}} \cdot \sin (\beta+\varphi)+\mathrm{C} \cdot \cos (\beta)-\mathrm{K}_{2} \cdot \sin (\alpha+\delta)=\mathrm{0}\tag{8-98}$
The vertical equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta}+\boldsymbol{\varphi})+\mathrm{C} \cdot \sin (\boldsymbol{\beta})-\mathrm{K}_{\mathrm{2}} \cdot \cos (\boldsymbol{\alpha}+\boldsymbol{\delta})=\mathrm{0}\tag{8-99}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{8-100}$
The force K2 on the blade is now:
$\ \mathrm{K_{2}=\frac{C \cdot \cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}}\tag{8-101}$
The force due to the cohesive shear strength is equal to:
$\ \mathrm{C}=\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\tag{8-102}$
The factor λs in equation (8-102) is the velocity strengthening factor, which causes an increase of the cohesive shear strength. In clay (Miedema (1992) and (2010)) this factor has a value of about 2 under normal cutting conditions. In rock the strengthening effect is not reported, so a value of 1 should be used. From equation (8-101) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=K_{2} \cdot \sin (\alpha+\delta)}\tag{8-103}$
$\ \mathrm{F}_{v}=\mathrm{K}_{\mathrm{2}} \cdot \cos (\alpha+\delta)\tag{8-104}$
Substituting equations (8-102) and (8-101) gives the following equations for the horizontal Fh and vertical Fv cutting forces. It should be remarked that the strengthening factor λs in rock is usually 1.
$\ \mathrm{F_{h}=\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{\sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)}}\tag{8-105}$
$\ \mathrm{F}_{v}=\frac{\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{\sin (\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)}\tag{8-106}$
8.05: Determining the Angle
To determine the shear angle β where the horizontal force Fh is at the minimum, the denominator of equation (8-105) has to be at a maximum. This will occur when the derivative of Fh with respect to β equals 0 and the second derivative is negative.
$\ \frac{\partial \sin (\alpha+\beta+\delta+\varphi) \cdot \sin (\beta)}{\partial \beta}=\sin (\alpha+2 \cdot \beta+\delta+\varphi)=0\tag{8-107}$
$\ \beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}\tag{8-108}$
This gives for the cutting forces:
$\ \mathrm{F_{h}=\frac{2 \cdot c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{H F} \cdot c \cdot h_{i} \cdot w}\tag{8-109}$
$\ \mathrm{F}_{v}=\mathrm{\frac{2 \cdot c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{V F} \cdot c \cdot h_{i} \cdot w}\tag{8-110}$
Equations (8-109) and (8-110) are basically the same as the equations found by Merchant (1944), (1945A) and (1945B). The normal force N1 and the normal stress σN1 on the shear plane are now (with λs=1):
$\ \begin{array}{left}\mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\ \sigma_{\mathrm{N} 1}=\frac{-\mathrm{c} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\end{array}\tag{8-111}$
The normal force N2 and the normal stress σN2 on the blade are now:
$\ \begin{array}{left}\mathrm{N}_{2}&=\frac{\mathrm{C} \cdot \cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\ \sigma_{\mathrm{N} 2}&=\mathrm{c} \cdot \frac{\mathrm{h}_{\mathrm{i}} \cdot \sin (\alpha)}{\mathrm{h}_{\mathrm{b}} \cdot \sin (\beta)} \cdot \frac{\cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\end{array}\tag{8-112}$
Equations (8-111) and (8-112) show that the normal force on the shear plane tends to be negative, unless the sum of the angles α+β+δ is greater than 90°. With the use of equation (8-108) the following condition is found:
$\ \begin{array}{left}\alpha+\beta+\delta=\alpha+\delta+\left(\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}\right)=\frac{\pi}{2}+\frac{\alpha+\delta-\varphi}{2}>\frac{\pi}{2}\ \text{so: }\frac{\alpha+\delta-\varphi}{2}>0\end{array}\tag{8-113}$
Because for normal blade angles this condition is always valid, the normal force is always positive. Figure 8-31 and Figure 8-32 show the coefficients λHF and λVF for the horizontal and vertical forces Fh and Fv according to equations (8-109) and (8-110) as a function of the blade angle α and the internal friction angle φ, where the external friction angle δ is assumed to be 2/3·φ. A positive coefficient λVF for the vertical force means that the vertical force Fis downwards directed.
Based on equation (8-97) and (8-109) the specific energy Esp can be determined according to:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{P}_{\mathrm{c}}}{\mathrm{Q}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\lambda_{\mathrm{H F}} \cdot \mathrm{c}\tag{8-114}$
The cohesive shear strength c is a function of the Unconfined Compressive Strength UCS and the angle of internal friction φ according to (see Figure 8-36):
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)}\tag{8-115}$
This gives for the specific energy Esp:
$\ \mathrm{E}_{\mathrm{sp}}=\lambda_{\mathrm{HF}} \cdot \mathrm{c}=\lambda_{\mathrm{HF}} \cdot \frac{\mathrm{U} \mathrm{C S}}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)\tag{8-116}$
Figure 8-33 shows the specific energy Esp to UCS ratio. In Figure 8-30, Figure 8-31, Figure 8-32 and Figure 8-33 an example is given for an α=60o blade and an internal friction angle of φ=20o.
It should be noted again that the forces and the specific energy are based on peak values. For the average this should be multiplied with a factor between 0.5 and 1.0, but closer to 0.5.
If the forces become to high another mechanism will occur, for example the wedge mechanism.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/08%3A_Rock_Cutting-_Atmospheric_Conditions/8.04%3A_The_Flow_Type_%28Based_on_the_Merchant_Model%29.txt
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Until now only the total normal force on the shear plane Nhas been taken into consideration, but of course this normal force is the result of integration of the normal stresses σN1 on the shear plane. One could consider that cutting is partly bending the material and it is known that with bending a bar, at the inside (the smallest bending radius) compressive stresses will be developed, while at the outside (the biggest bending radius), tensile stresses are developed. So if the normal force N1 equals zero, this must mean that near the edge of the blade tensile stresses (negative) stresses develop, while at the outside compressive (positive) stresses develop. So even when the normal force would be slightly positive, still, tensile stresses develop in front of the edge of the blade. The normal force on the blade however is always positive, meaning that the Curling Type of cutting process will never occur in rock under atmospheric conditions. The previous derivations of the cutting forces are based on the Flow Type, but in reality rock will fail brittle with either the Shear Type or the Tear Type or a combination the Chip Type. For the Shear Type the equations (8-109) and (8-110) can still be used, considering these equations give peak forces. The average forces and thus the average cutting power Pc and the specific energy Esp may be 30%-60% of the peak values. The occurrence of the Tear Type depends on the tensile stress. If somewhere in the rock the tensile stress σmin is smaller than the tensile strength σT, a tensile fracture may occur. One should note here that compressive stresses are positive and tensile stresses are negative. So tensile fracture/rupture will occur if the absolute value of the tensile stress σmin is larger than the tensile strength σT.
If rock is considered, the following condition can be derived with respect to tensile rupture:
The cohesion can be determined from the UCS value and the angle of internal friction according to, as is shown in Figure 8-36:
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)}\tag{8-117}$
According to the Mohr-Coulomb failure criterion, the following is valid for the shear stress on the shear plane, as is shown in Figure 8-37.
$\ \tau_{\mathrm{S} 1}=\mathrm{c}+\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)\tag{8-118}$
The average stress condition on the shear plane is now σN1, $\ \tau$S1 as is shown in Figure 8-37. A Mohr circle (Mohr circle 1) can be drawn through this point, resulting in a minimum stress σmin which is negative, so tensile. If this minimum normal stress is smaller than the tensile strength σT tensile fracture will occur, as is the case in the figure. Now Mohr circle 1 can never exist, but a smaller circle (Mohr circle 2) can, just touching the tensile strength σT. The question is now, how to get from Mohr circle 1 to Mohr circle 2. To find Mohr circle 2 the following steps have to be taken.
The radius of the Mohr circle 1 can be found from the shear stress $\ \tau$S1 by:
$\ \mathrm{R}=\frac{\tau_{\mathrm{S} 1}}{\cos (\varphi)}\tag{8-119}$
The center of the Mohr circle 1, σC, now follows from:
$\ \begin{array}{left} \sigma_{\mathrm{C}} &=\sigma_{\mathrm{N} 1}+\mathrm{R} \cdot \sin (\varphi)=\sigma_{\mathrm{N} 1}+\tau_{\mathrm{S} 1} \cdot \tan (\varphi) \ &=\sigma_{\mathrm{N} 1}+\mathrm{c} \cdot \tan (\varphi)+\sigma_{\mathrm{N} 1} \cdot \tan ^{2}(\varphi) \end{array}\tag{8-120}$
The minimum principal stress σmin equals the normal stress in the center of the Mohr circle σC minus the radius of the Mohr circle R:
$\ \begin{array}{left} \sigma_{\min } &=\sigma_{\mathrm{C}}-\mathrm{R} \ &=\sigma_{\mathrm{N} 1}+\mathrm{c} \cdot \tan (\varphi)+\sigma_{\mathrm{N} 1} \cdot \tan ^{2}(\varphi)-\frac{\mathrm{c}}{\cos (\varphi)}-\frac{\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)}{\cos (\varphi)} \end{array}\tag{8-121}$
Rearranging this gives:
$\ \sigma_{\min }=\sigma_{\mathrm{N} 1} \cdot\left(1+\tan ^{2}(\varphi)-\frac{\tan (\varphi)}{\cos (\varphi)}\right)+\mathrm{c} \cdot\left(\tan (\varphi)-\frac{1}{\cos (\varphi)}\right)\tag{8-122}$
Substituting equation (8-111) for the normal stress on the shear plane gives:
$\ \sigma_{\min }= \frac{-\mathrm{c} \cdot \cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot\left(1+\tan ^{2}(\varphi)-\frac{\tan (\varphi)}{\cos (\varphi)}\right) +\mathrm{c} \cdot\left(\tan (\varphi)-\frac{1}{\cos (\varphi)}\right)>\sigma_{\mathrm{T}} \tag{8-123}$
Now shear failure will occur if the minimum principal stress σmin is larger than the tensile strength σT, thus:
$\ \sigma_{\min }>\sigma_{\mathrm{T}}\tag{8-124}$
If equation (8-124) is true, shear failure will occur. Keep in mind however, that the tensile strength σT is a negative number. Of course if the minimum normal stress σmin or in the graph, Figure 8-38, σT / c is positive, tensile failure can never occur. Equation (8-124) can be transformed to:
$\ \frac{\sigma_{\mathrm{T}}}{\mathrm{c}}<-\frac{\cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot(\cos (\varphi)-\tan (\varphi)+\tan (\varphi) \cdot \sin (\varphi)) +\tan (\varphi)-\frac{1}{\cos (\varphi)}\tag{8-125}$
Substituting equation (8-108) for the shear angle β gives:
$\ \frac{\sigma_{\mathrm{T}}}{\mathrm{c}}<\frac{\sin \left(\frac{\alpha+\delta-\varphi}{2}\right)}{\cos \left(\frac{\alpha+\delta+\varphi}{2}\right)} \cdot(\cos (\varphi)-\tan (\varphi)+\tan (\varphi) \cdot \sin (\varphi)) \quad+\tan (\varphi)-\frac{1}{\cos (\varphi)}\tag{8-126}$
This can be transformed to:
$\ \mathrm{\frac{\sigma_{T}}{c}<\left(\frac{\sin \left(\frac{\alpha+\delta-\varphi}{2}\right)}{\cos \left(\frac{\alpha+\delta+\varphi}{2}\right)}-1\right) \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)}\tag{8-127}$
A mobilized cohesive shear strength cm can be defined, based on the tensile strength σT, by using the equal sign in equation (8-127). With this mobilized cohesive shear strength Mohr circle 2 can be constructed.
$\ \mathrm{c}_{\mathrm{m}}=\frac{\sigma_{\mathrm{T}}}{\left(\frac{\sin \left(\frac{\alpha+\delta-\varphi}{2}\right)}{\cos \left(\frac{\alpha+\delta+\varphi}{2}\right)}-1\right) \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)}\tag{8-128}$
Substituting equation (8-128) in the equations (8-109) and (8-110) gives for the cutting forces:
$\ \mathrm{F_{h}=\frac{2 \cdot c_{m} \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{H T} \cdot \sigma_{T} \cdot h_{i} \cdot w}\tag{8-129}$
$\ \mathrm{F}_{v}=\frac{\mathrm{2} \cdot \mathrm{c}_{\mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi) \cdot \cos (\boldsymbol{\alpha}+\delta)}{\mathrm{1}+\mathrm{c o s}(\boldsymbol{\alpha}+\delta+\varphi)}=\lambda_{\mathrm{V T}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{8-130}$
Figure 8-38 shows the pseudo cohesive shear strength coefficient σT / c from equation (8-127). Below the lines the cutting process is ductile (the Flow Type) or brittle (the Shear Type), while above the lines it is brittle (the Tear Type). It is clear from this figure that an increasing blade angle α and an increasing internal friction angle φ suppresses the occurrence of the Tear Type. The coefficients λHT and λVT are shown in Figure 8-42 and Figure 8-43 for a range of blade angles α and internal friction angles φ.
Equation (8-129) gives for the specific energy Esp:
$\ \mathrm{E}_{\mathrm{s p}}=\lambda_{\mathrm{H T}} \cdot \sigma_{\mathrm{T}}\tag{8-131}$
To determine the cutting forces in rock under atmospheric conditions the following steps have to be taken:
1. Determine whether the cutting process is based on the Flow Type or the Tear Type, using Figure 8-38.
2. If the cutting process is based on the Flow Type, use Figure 8-31 and Figure 8-32 to determine the coefficients λHF and λVF. Use equations (8-109) and (8-110) to calculate the cutting forces. Optionally a factor 0.3-0.5 may be applied in case of brittle shear failure, to account for average forces, power and specific energy.
3. If the cutting process is based on the Tear Type, use Figure 8-42 and Figure 8-43 to determine the coefficients λHT and λVT. Use equations (8-129) and (8-130) to calculate the cutting forces. A factor 0.3-0.6 should be applied to account for average forces, power and specific energy.
For completeness, Figure 8-40 shows the moments on the layer cut.
Based on equation (8-127) and (8-117) the ratio UCS/BTS can also be determined. Gehking (1987) stated that below a ratio of 9 ductile failure will occur, while above a ratio of 15 brittle failure will occur. In between these limits there is a transition between ductile and brittle failure, which is also in accordance with the findings of Fairhurst (1964). Figure 8-39 shows that the ductile limit of 9 is possible for blade angles α between 45o and 60o corresponding with internal friction angles φ of 25o and 15o. For the same blade angles, the corresponding internal friction angles φ are 35o and 25o at the brittle limit of 15. These values match the blade angles as used in dredging and mining and also match the internal friction angle of commonly dredged rock. Figure 8-39 shows that in general a higher internal friction angle φ and a bigger blade angle α suppress tensile failure.
$\ \frac{\mathrm{UCS}}{\mathrm{BTS}}=\frac{2}{\left(\frac{\sin \left(\frac{\alpha+\delta-\varphi}{2}\right)}{\cos \left(\frac{\alpha+\delta+\varphi}{2}\right)}-1\right) \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)^{2}}\tag{8-132}$
Figure 8-41 shows the shear angle with limitations. The limitations occur because at a certain sum of the blade angle, the shear angle and the angle of internal friction, a positive tensile strength would be required to get brittle tensile failure, which is physically impossible. Compressive stresses are defined positive and tensile stresses negative, so a positive tensile stress would in fact be a compressive stress. Beyond this limitation only brittle shear can exist, or if the sum of the angles is to high, probably another mechanism like the wedge mechanism.
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The equations for the Tear Type are derived based on the shear angle β of the Flow Type. It is however a question whether this is correct under all circumstances. At the moment of transition of Flow Type to Tear Type this may be the case, but far away from this transition there may be another optimum shear angle β. Combining equations (7-68), (7-69) and (7-88) with the shear angle β as a variable and determining the minimum horizontal force, gives a different value for the shear angle β.
A shear angle β is found, exactly 22.5° smaller than the shear angle β of the Flow Type (see Figure 8-44).
$\ \beta=\frac{\pi}{2}-\frac{\pi / 4+\alpha+\delta+\varphi}{2}\tag{8-133}$
Figure 8-45 and Figure 8-46 show the horizontal and the vertical cutting force coefficients which are slightly smaller than the horizontal and vertical cutting force coefficients in Figure 8-42 and Figure 8-43. Now there exists a set of parameters where both shear failure and tensile failure give a possible solution. In this range of parameters shear failure will not give tensile stresses that exceed the tensile strength while tensile failure would lead to smaller forces. The occurrence of the Flow Type or the Tear Type will depend on the history of the cutting process.
Substituting the corrected shear angle gives for the mobilized shear strength:
$\ \mathrm{c}_{\mathrm{m}}=\frac{\sigma_{\mathrm{T}}}{\left(\frac{\sin \left(\frac{\alpha+\delta-\varphi-\pi / 4}{2}\right)}{\cos \left(\frac{\alpha+\delta+\varphi-\pi / 4}{2}\right)}-1\right) \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)}\tag{8-134}$
Now the cutting forces can be determined with:
$\ \mathrm{F_{h}=\frac{2 \cdot c_{m} \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{\cos (\pi / 4)+\cos (\alpha+\delta+\varphi)}=\lambda_{H T} \cdot \sigma_{T} \cdot h_{i} \cdot w}\tag{8-135}$
$\ \mathrm{F}_{v}=\frac{\mathrm{2} \cdot \mathrm{c}_{\mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{\cos (\pi / 4)+\cos (\alpha+\delta+\varphi)}=\lambda_{\mathrm{V T}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{8-136}$
8.08: Specific Energy
For the cases as described above, cutting with a straight blade with the direction of the cutting velocity vc perpendicular to the blade (edge of the blade), the specific cutting energy Esp is:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{8-137}$
The specific energy of the Flow Type or Crushed Type of cutting mechanism can be written as:
$\ \mathrm{E}_{\mathrm{s p}}=\lambda_{\mathrm{H F}} \cdot \mathrm{c}\tag{8-138}$
The specific energy of the Tear Type or Chip Type of cutting mechanism can be written as:
$\ \mathrm{E}_{\mathrm{s p}}=\lambda_{\mathrm{H T}} \cdot \sigma_{\mathrm{T}}\tag{8-139}$
Since the specific energy equations are based on the maximum horizontal cutting forces, where the cutting process is most probably either brittle shear or brittle tensile, the average cutting forces will be smaller. How much smaller depends on the type of rock, but literature mentions reductions by 30% to 70%. Since the specific energy is based on the average cutting forces, the values found with the above equations should be multiplied by a factor of 0.3- 0.7.
8.09: Resulting Forces and Mohr Circles
Figure 8-49 and Figure 8-50 show the horizontal and vertical cutting forces. The transition tensile failure/shear failure occurs at a tensile strength of about -8.5 MPa, so 8.5% of the UCS value. From a tensile strength of -8.5 MPa (8.5% UCS) to -20 MPa (20% UCS), both tensile failure and shear failure are possible. Below a tensile strength of -20 MPa (20% UCS) only shear failure is possible. Figure 8-51 and Figure 8-52 show the Mohr circles for tensile strengths of -8.5 MPa (8.5% UCS) and -20 MPa (20% UCS). One can see that with a tensile strength of -8.5 MPa both tensile failure and shear failure are possible, with a tensile strength of -20 MPa also both tensile failure and shear failure are possible. It should be mentioned here that the forces shown are peak forces, so average forces may reduce to 50%-60%. So the two limiting cases are shown.
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In this chapter and in Appendix W many graphs are given with a red or green rectangle giving the value of the different parameters for an α=60o blade and an internal friction angle φ=20o. The external friction angle is assumed to be δ=2/3·φ. Most graphs are dimensionless, but Figure 8-49, Figure 8-50, Figure 8-51 and Figure 8-52 are based on a compressive strength UCS=100 MPa, a blade width w=0.1 m and a layer thickness hi=0.1 m.
8.10.1. Step 1: Brittle Shear
The the shear angle β=43.3o, horizontal force coefficient λHF=1.912, the verticle force coefficient λVF=0.572 and the Esp/UCS ratio=0.669. This gives a horizontal force Fh=0.669 MN, a vertical force Fv=0.200 MN and a specific energy of Esp=66.9 MPa. These values are peak values, but for comparison reasons these values will be used.
8.10.2. Step 2: The Transition Brittle Shear/Brittle Tensile
The transitions brittle shear/brittle tensile occur for UCS/BTS=4.985 and UCS/BTS=11.75. This can also be written as BTS=0.085·UCS and BTS=0.2·UCS, so BTS=8.5 MPa and BTS=20 MPaBTS or UTS are considered positive numbers, while tensile strength is considered to be negative in this book.
This means that below the Lower Limit BTS=8.5 MPa brittle shear failure cannot exist, so there is always brittle tensile failure. Above the Upper Limit BTS=20 MPa brittle tensile failure cannot exist, so there is always brittle shear failure. In between, both can exist , even at the same time, according to Figure 8-19 the Chip Type. Figure 8-49 and Figure 8-50 show the horizontal and vertical cutting forces as a function of the tensile strength for the case considered. Interpolation curves are shown, simulating the simultaneous occurrence of brittle shear and brittle tensile failure according to the Chip Type. For this interpolation the following method is used:
First define a factor according to:
$\ \begin{array}{left}\text{BTS} < \text{LowerLimit BTS} &\Rightarrow \mathrm{f}=\mathrm{1}\ \text{BTS} > \text{UpperLimit BTS}&\Rightarrow \mathrm{f}=\mathrm{0}\ \text{BTS} > \mathrm{Lowerlimit BTS}\ &\Rightarrow \mathrm{f}=\left(\frac{\text { Upperlimit BTS - BTS }}{\text { Upperlimit BTS - LowerLimit BTS }}\right)^\mathrm{p}\ \text{BTS} < \text{UpperLimit BTS}\end{array}\tag{8-140}$
Now the resulting cutting forces can be determined with:
$\ \begin{array}{left}\mathrm{F}_{\mathrm{h}}=\mathrm{F}_{\mathrm{h}, \text { TearType }} \cdot \mathrm{f}+\mathrm{F}_{\mathrm{h}, \text { ShearType }} \cdot(\mathrm{1}-\mathrm{f})\ \mathrm{F}_{\mathrm{v}}=\mathrm{F}_{\mathrm{v}, \text { TearType }} \cdot \mathrm{f}+\mathrm{F}_{\mathrm{v}, \text { ShearType }} \cdot(1-\mathrm{f})\end{array}\tag{8-141}$
The power used in Figure 8-49 and Figure 8-50 is p=1, a linear transition from tensile failure to shear failure, the Chip Type.
8.10.3. Step 3: Applying Tensile Strengths of -5 MPa, -10 MPa and -25 MPa
From Figure 8-49 and Figure 8-50 the horizontal and vertical peak forces can be determined. They are given in the following table. Between brackets estimated average values, based on a 60% ratio between average and peak values.
Table 8-3: Resulting forces and specific energy.
Tensile Strength
Fh (N)
Fv (MN)
Esp (MPa)
Esp/UCS (%)
-5 MPa
0.207 (0.124)
0.062 (0.037)
20.7 (12.4)
20.7 (12.4)
-10 MPa
0.446 (0.268)
0.134 (0.080)
44.6 (26.8)
44.6 (26.8)
-25 MPa
0.669 (0.401)
0.200 (0.120)
66.9 (40.1)
66.9 (40.1)
The UCS/BTS ratio of 10 matches the findings of Roxborough (1987) giving a specific energy of about 25% of the UCS value.
$\ \mathrm{E}_{\mathrm{sp}}=\mathrm{0 .2 5} \cdot \mathrm{U . C . S .}+\mathrm{0 .1 1}\tag{8-142}$
8.11: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesive shear strength
kPa
A
Adhesive force on the blade
kN
BTS
Brazilian Tensile Strength
kPa
c, $\ \tau_\mathrm{c}$
Cohesive shear strength
kPa
cm
Mobilized cohesive shear strength
kPa
C
Cohesive force on shear plane
kN
Esp
Specific energy
kPa
F
Force
kN
Fc
Cutting force on chisel Evans model
kN
Fn
Normal force on chisel Evans model
kN
Fch
Horizontal force component Evans model
kN
Fcv
Vertical force component Evans model
kN
Fh
Horizontal cutting force
kN
Fv
Vertical cutting force
kN
g
Gravitational constant (9.81)
m/s2
G
Gravitational force
kN
hi
Initial thickness of layer cut
m
hb
Height of the blade
m
K1
Grain force on the shear plane
kN
K2
Grain force on the blade
kN
I
Inertial force on the shear plane
kN
n
Power in Nishimatsu model
-
N1
Normal grain force on shear plane
kN
N2
Normal grain force on blade
kN
p
Stress in shear plane Nishimatsu model
kPa
p0
Stress at tip of chisel Nishimatsu model
kPa
Pc
Cutting power
kW
Q
Production
m3
r
Radius in Evans model
m
r
Adhesion/cohesion ratio
-
r1
Pore pressure on shear plane/cohesion ratio
-
r2
Pore pressure on blade/cohesion ratio
-
R
Radius of Mohr circle
kPa
R
Force on chisel Evans model
kN
Rn
Normal force on chisel surface Evans model
kN
Rf
Friction force on chisel surface Evans model
kN
R1
Acting point on the shear plane
m
R2
Acting point on the blade
m
S1
Shear force due to internal friction on the shear plane
kN
S2
Shear force due to external friction on the blade
kN
T
Tensile force
kN
UCS
Unconfined Compressive Strength
kPa
vc
Cutting velocity
m/s
w
Width of the blade
m
W1
Force resulting from pore under pressure on the shear plane
kN
W2
Force resulting from pore under pressure on the blade
kN
α
Blade angle
rad
β
Angle of the shear plane with the direction of cutting velocityrad
rad
ε
Angle of chisel with horizontal Evans model
rad
$\ \tau$
Shear stress
kPa
$\ \tau_\mathrm{a}$, a
Adhesive shear strength (strain rate dependent)
kPa
$\ \tau_\mathrm{c}$, c
Cohesive shear strength (strain rate dependent)
kPa
$\ \tau_{\mathrm{S1}}$
Average shear stress on the shear plane
kPa
$\ \tau_{\mathrm{S2}}$
Average shear stress on the blade
kPa
σ
Normal stress
kPa
σC
Center of Mohr circle
kPa
σT
Tensile strength
kPa
σmin
Minimum principal stress in Mohr circle
kPa
σN1
Average normal stress on the shear plane
kPa
σN2
Average normal stress on the blade
kPa
φ
Angle of internal friction
rad
δ
Angle of external friction
rad
λ
Distance in Nishimatsu model
m
λs
Strengthening factor
-
λ1
Acting point factor on the shear plane
-
λ2
Acting point factor on the blade
-
λHF
Flow Type/Crushed Type horizontal force coefficient
-
λVF
Flow Type/Crushed Type vertical force coefficient
-
λHT
Tear Type/Chip Type horizontal force coefficient
-
λVT
Tear Type/Chip Type vertical force coefficient
-
ω
Angle in Evans model
rad
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For rock cutting in dredging and mining under hyperbaric conditions not much is known yet. The data available are from drilling experiments under very high pressures (Zijsling (1987), Kaitkay and Lei (2005) and Rafatian et al. (2009)). The main difference between dredging and mining applications on one side and drilling experiments on the other side is that in dredging and mining the thickness of the layer cut is relatively big, like 5-10 cm, while in drilling the process is more like scraping with a thickness less than a mm. From the drilling experiments it is known that under high pressures there is a transition from a brittle-shear cutting process to a ductile-flow cutting process. Figure 9-2 and Figure 9-3 from Rafatian et al. (2009) show clearly that with increasing confining pressure, first the specific energy Esp increases with a steep curve, which is the transition brittle-ductile, after which the curve for ductile failure is reached which is less steep. The transition is completed at 690 kPa-1100 kPa, matching a water depth of 69-110 m.
The Carthage Marble has a UCS value of about 100 MPa and the Indiana Limestone a UCS value of 48 MPa. The cutter had a blade angle α of 110o. Figure 9-29 shows the specific energy (according to the theory as developed in this chapter) as a function of the UCS value and the confining pressure (water depth). For the Carthage Marble a specific energy of about 400 MPa is found under atmospheric conditions for the ductile cutting process. For the brittle shear process 25%-50% of this value should be chosen, matching Figure 9-2 at 0 MPa. For a water depth of 65 m, matching 0.65 MPa the graph gives about 500 MPa specific energy, which is a bit lower than the measurements. For the Indiana Limestone a specific energy of about 200 MPa is found under atmospheric conditions for the ductile cutting process. Also here, for the brittle shear process, 25%-50% of this value should be chosen, matching Figure 9-3 at 0 MPa confining pressure. For a water depth of 65 m, matching 0.65 MPa the graph gives about 280 MPa specific energy, which is a bit lower than the measurements.
For deep sea mining applications this is still shallow water. Both graphs show an increase of the Esp by a factor 2-2.5 during the transition brittle-shear to ductile-flow, which matches a reduction factor of 0.25-0.5 for the average versus the maximum cutting forces as mentioned before. Figure 9-22 and Figure 9-23 show the results of Zijsling (1987) in Mancos Shale and Figure 9-1 shows the results of Kaitkay & Lei (2005) in Carthage Marble.
The experiments of Kaitkay & Lei (2005) also show that the transition from brittle-shear to ductile-flow takes place in the first few hundreds of meters of water depth (from 0 to about 2.5 MPa). They also show a multiplication factor of about 3 during this transition. The experiments of Zijsling (1987) are not really suitable for determining the transition brittle-shear to ductile-flow because there are only measurements at 0 MPa and about 10 MPa, so they do not show when the transition is completed, but they do show the increase in forces and Esp.
The explanation for the transition from brittle-shear to ductile-flow is, according to Zijsling (1987), the dilatation due to shear stress in the shear plane resulting in pore under pressures, similar to the cutting process in water saturated sand as has been described by Miedema (1987 September). Zijsling however did not give any mathematical model. Detournay & Atkinson (2000) use the same explanation and use the Merchant (1944) model (equations (8-109) and (8-110) for the flow type cutting process) to quantify the cutting forces and specific energy by adding the pore pressures to the basic equations:
$\ \mathrm{F_{h}=\frac{2 \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)} \cdot\left(c+p_{1 m} \cdot \tan (\varphi)\right)}\tag{9-1}$
The difference between the bottom hole pressure (or hydrostatic pressure) and the average pressure p1m in the shear plane has to be added to the effective stress between the particles in the shear plane A-B. Multiplying this with the tangent of the internal friction angle gives the additional shear stress in the shear plane A-B, see Figure 9-4.
So in the vision of Detournay & Atkinson (2000) the effect of pore water under pressures p1m is like an apparent additional cohesion. Based on this they find a value of the external friction angle which is almost equal to the internal friction angle of 23o for the experiments of Zijsling (1987). Detournay & Atkinson (2000) however forgot that, if there is a very large pore water under pressure in the shear plane, this pore water under pressure has not disappeared when the layer cut moves over the blade or cutter. There will also be a very large pore water under pressures on the blade as has been explained by Miedema (1987 September) for water saturated sand in dredging applications. In the next paragraph this will be explained.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/09%3A_Rock_Cutting-_Hyperbaric_Conditions/9.01%3A_Introduction.txt
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First of all it is assumed that the hyperbaric cutting mechanism is similar to the Flow Type as is shown in Figure 9-5. There may be 3 mechanisms that might explain the influence of large hydrostatic pressures:
1. When a tensile failure occurs, water has to flow into the crack, but the formation of the crack goes so fast that cavitation will occur.
2. A second possible mechanism that might occur is an increase of the pore volume due to the elasticity of the rock and the pore water. If high tensile stresses exist in the rock, then the pore volume will increase due to elasticity. Because of the very low permeability of the rock, the compressibility of the pore water will have to deal with this. Since the pore water is not very compressible, at small volume changes this will already result in large under pressures in the pores. Whether this will lead to full cavitation of the pore water is still a question.
3. Due to the high effective grain stresses, the particles are removed from the matrix which normally keeps them together and makes it a rock. This will happen near the shear plane. The loose particles will be subject to dilatation, resulting in an increase of the pore volume. This pore volume increase results in water flow to the shear plane, which can only occur if there is an under pressure in the pores in the shear plane. If this under pressure reaches the water vapor pressure, cavitation will occur, which is the lower limit for the absolute pressures and the upper limit for the pressure difference between the bottom hole or hydrostatic pressure and the pore water pressure. The pressure difference is proportional to the cutting velocity and the dilatation, squared proportional to the layer thickness and reversely proportional to the permeability of the rock. If the rock is very impermeable, cavitation will always occur and the cutting forces will match the upper limit.
Now under atmospheric conditions, the compressive strength of the rock will be much bigger than the atmospheric pressure; usually the rock will have a compressive strength of 1 MPa or more while the atmospheric pressure is just 100 kPa. Strong rock may have compressive strengths of 10’s of MPa’s, so the atmospheric pressure and thus the effect of cavitation in the pores or the crack can be neglected. However in oil drilling and deep sea mining at water depths of 3000 m nowadays plus a few 1000’s m into the seafloor (in case of oil drilling), the hydrostatic pressure could easily increase to values higher than 10 MPa up to 100 MPa causing softer rock to behave ductile, where it would behave brittle under low hydrostatic pressures.
It should be noted that brittle-tear failure, which is tensile failure, will only occur under atmospheric conditions and small blade angles as used in dredging and mining. With blade angles larger than 90° brittle-tear will never occur (see Figure 8-38). Brittle-shear may occur in all cases under atmospheric conditions.
Now what is the difference between rock cutting under atmospheric conditions and under hyperbaric conditions? The difference is the extra pore pressure forces W1 and W2 on the shear plane and on the blade as will be explained next.
Figure 9-7 illustrates the forces on the layer of rock cut. The forces acting on this layer are:
1. A normal force acting on the shear surface N1 resulting from the grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A shear force as a result of the cohesive shear strength $\ \tau_\mathrm{c}$ or c. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$/c with the area of the shear plane.
5. A force normal to the blade N2 resulting from the grain stresses.
6. A shear force S2 as a result of the external friction N2·tan(δ).
7. A shear force as a result of pure adhesion between the rock and the blade $\ \tau_\mathrm{a}$ or a. This force can be calculated by multiplying the adhesive shear strength $\ \tau_\mathrm{a}$/a of the rock with the contact area between the rock and the blade. In most rocks this force will be absent.
8. A force Was a result of water under pressure on the blade
The normal force N1 and the shear force S1 on the shear plane can be combined to a resulting grain force K1.
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{9-2}$
The forces acting on a straight blade when cutting rock, can be distinguished as:
1. A force normal to the blade N2 resulting from the grain stresses.
2. A shear force S2 as a result of the external friction N2·tan(δ).
3. A shear force as a result of pure adhesion between the rock and the blade $\ \tau_\mathrm{a}$ or c. This force can be calculated by multiplying the adhesive shear strength $\ \tau_\mathrm{a}$/a of the rock with the contact area between the rock and the blade. In most rocks this force will be absent.
4. A force W2 as a result of water under pressure on the blade
These forces are shown in Figure 9-8. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{\mathrm{2}}^{\mathrm{2}}+\mathrm{S}_{\mathrm{2}}^{\mathrm{2}}}\tag{9-3}$
The horizontal equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{h}}= \mathrm{K}_{1} \cdot \sin (\beta+\varphi)-\mathrm{W}_{1} \cdot \sin (\beta)+\mathrm{C} \cdot \cos (\beta) -\mathrm{A} \cdot \cos (\alpha)+\mathrm{W}_{2} \cdot \sin (\alpha)-\mathrm{K}_{2} \cdot \sin (\alpha+\delta)=0 \tag{9-4}$
The vertical equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{1} \cdot \cos (\beta+\varphi)+\mathrm{W}_{1} \cdot \cos (\beta)+\mathrm{C} \cdot \sin (\beta) +\mathrm{A} \cdot \sin (\alpha)+\mathrm{W}_{2} \cdot \cos (\alpha)-\mathrm{K}_{2} \cdot \cos (\alpha+\delta)=\mathrm{0}\tag{9-5}$
The force Kon the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{9-6}$
The force K2 on the blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{9-7}$
From equation (9-7) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=-W_{2} \cdot \sin (\alpha)+K_{2} \cdot \sin (\alpha+\delta)}\tag{9-8}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)\tag{9-9}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}= \frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) +\frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \tag{9-10}$
The normal force on the blade is now:
$\ \mathrm{N}_{2}= \frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) +\frac{+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \tag{9-11}$
The pore pressure forces can be determined in the case of full-cavitation or the case of no cavitation according to:
$\ \mathrm{W}_{1}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\text{ or }\mathrm{W}_{\mathrm{1}}=\frac{\mathrm{p}_{1 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\tag{9-12}$
$\ \mathrm{W}_{2}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\text{ or }\mathrm{W}_{2}=\frac{\mathrm{p}_{2 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{9-13}$
The forces C and A are determined by the cohesive shear strength c and the adhesive shear strength a according to:
$\ \mathrm{C=\frac{c \cdot h_{i} \cdot w}{\sin (\beta)}}\tag{9-14}$
$\ \mathrm{A}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{9-15}$
The ratio’s between the adhesive shear strength and the pore pressures with the cohesive shear strength can be found according to:
$\ \begin{array}{left}\mathrm{r}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}, \mathrm{r}_{\mathrm{1}}=\frac{\mathrm{p}_{1 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{i}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\text{ or }\mathrm{r}_{\mathrm{1}}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}, \mathrm{r}_{2}=\frac{\mathrm{p}_{2 \mathrm{m}} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\ \text{or } \mathrm{r}_{2}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\end{array}\tag{9-16}$
Finally the horizontal and vertical cutting forces can be written as:
$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{H F}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{9-17}$
$\ \mathrm{F}_{v}=\lambda_{\mathrm{V F}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{9-18}$
Figure 9-9, Figure 9-10 and Figure 9-11 show the horizontal and vertical cutting force coefficients and the shear angle as a function of the ratio of the hydrostatic pressure to the shear strength of the rock rz for a 60 degree blade and full cavitation. If this ratio equals 1, it means the hydrostatic pressure equals the shear strength. At small ratios the resulting values approach atmospheric cutting of rock. Also at small ratios the shear angle approaches the theoretical value for atmospheric cutting. Figure 9-12 shows the Esp/UCS ratio, which is very convenient for production estimation.
The vertical cutting force coefficient λVF is positive downwards directed. From the calculations it appeared that for a 60 degree blade, the Curling Type will already occur with an hb/hi=1. For a 110 degree blade it requires an hb/hi=4-5, depending on the internal friction angle. The transition at small hb/hi ratios, between the Flow Type and the Curling Type, will occur at blade angles between 60 and 90 degrees. So its important to determine the cutting forces for both mechanisms in order to see which of the two should be applied. This is always the mechanism resulting in the smallest horizontal cutting force.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/09%3A_Rock_Cutting-_Hyperbaric_Conditions/9.02%3A_The_Flow_Type_and_the_Crushed_Type.txt
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Similar to the derivation of equation (8-127) for the occurrence of tensile failure under atmospheric conditions, equation (9-19) can be derived for the occurrence of tensile failure under hyperbaric conditions. Under hyperbaric conditions equation (9-19) will almost always be true, because of the terms with rand rwhich may become very big (positive). So tensile failure will not be considered for hyperbaric conditions.
$\ \mathrm{c}\cdot \left( \begin{array}{left}\mathrm{\frac{r\cdot \frac{\sin(\beta)\cdot\cos(\delta)}{\sin(\alpha)}+r_2\cdot\frac{\sin(\beta)\cdot sin(\delta)}{sin(\alpha)}}{sin(\alpha+\beta+\delta+\varphi)}}\ \mathrm{+\frac{+r_1\cdot sin(\alpha+\beta+\delta)}{sin(\alpha+\beta+\delta+\varphi)} }\ +\mathrm{\frac{-cos(\alpha+\beta+\delta)-sin(\alpha+\beta+\delta+\varphi)}{sin(\alpha+\beta+\delta+\varphi)}}\end{array} \right)\cdot\mathrm{\left(\frac{1-sin(\varphi)}{cos(\varphi)} \right)>\sigma_T}\tag{9-19}$
9.04: The Curling Type
When cutting or scraping a very thin layer of rock, the Curling Type may occur. In dredging and mining usually the layer thickness is such that this will not occur, but in drilling practices usually the layer thickness is very small compared with the height of the blade. In the Zijsling (1987) experiments layer thicknesses of 0.15 mm and 0.30 mm were applied with a PDC bit with a height and width of about 10 mm. Under these conditions the Curling Type will occur, which is also named balling. Figure 9-15 shows this type of cutting mechanism.
Now the question is, what is the effective blade height hb,m? In other words, along which distance will the rock cut be in contact with the blade? To solve this problem an additional condition has to be found. This condition is the equilibrium of moments around the blade tip as is shown in Figure 9-16. The only forces that contribute to the equilibrium of moments are the normal forces N1 and N2 and the pore pressure forces W1 and W2. The acting points of these forces are chosen as fractions of the length of the shear plane λ1 and the blade length λ2.
The equilibrium of moments around the blade tip is:
$\ \left(\mathrm{N}_{\mathrm{1}}-\mathrm{W}_{\mathrm{1}}\right) \cdot \mathrm{R}_{\mathrm{1}}=\left(\mathrm{N}_{\mathrm{2}}-\mathrm{W}_{\mathrm{2}}\right) \cdot \mathrm{R}_{\mathrm{2}}\tag{9-20}$
For the acting points the following can be derived:
$\ \mathrm{R_{1}=\frac{\lambda_{1} \cdot h_{\mathrm{i}}}{\sin (\beta)}, R_{2}=\frac{\lambda_{2} \cdot h_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}}\tag{9-21}$
Substituting equations (9-10) and (9-11) into equation (9-20) gives:
$\ \begin{array}{\left}\left(\begin{array}{left}\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \ +\frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\-\mathrm{W}_1\end{array}\right) \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\ =\left(\begin{array}{left}\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \ +\frac{+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\-\mathrm{W_2}\end{array}\right) \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}\end{array}\tag{9-22}$
This can be written as a second degree function of the effective or mobilized blade height hb,m:
$\ \begin{array}{left}\mathrm{A} \cdot \mathrm{x}^{2}+\mathrm{B} \cdot \mathrm{x}+\mathrm{C}=\mathrm{0}\ \mathrm{h}_{\mathrm{b}, \mathrm{m}}=\mathrm{x}=\frac{-\mathrm{B}-\sqrt{\mathrm{B}^{2}-\mathrm{4} \cdot \mathrm{A} \cdot \mathrm{C}}}{\mathrm{2} \cdot \mathrm{A}}\end{array}\tag{9-23}$
With:
$\ \mathrm{A}= \frac{\lambda_{2} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta+\varphi)-\lambda_{2} \cdot \mathrm{p}_{\mathrm{2 m}} \cdot \sin (\alpha+\beta+\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\alpha)} +\frac{+\mathrm{a} \cdot \lambda_{2} \cdot \cos (\alpha+\beta+\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\alpha)} \tag{9-24}$
And:
$\ \mathrm{B}= \frac{\lambda_{1} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\delta) \cdot \cos (\varphi)-\lambda_{2} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \cos (\delta) \cdot \sin (\varphi)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} +\frac{-\mathrm{c} \cdot \lambda_{2} \cdot \cos (\delta) \cdot \cos (\varphi)+\mathrm{a} \cdot \lambda_{1} \cdot \cos (\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \tag{9-25}$
And:
$\ \mathrm{C}= \frac{\lambda_{1} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta) \cdot \cos (\varphi)-\lambda_{1} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta+\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}} +\frac{-\mathrm{c} \cdot \lambda_{1} \cdot \cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}} \tag{9-26}$
If hb,m<hb then the Curling Type will occur, but if hb,m>hthe normal Flow Type will occur.
$\ \begin{array}{left}\text{if } \mathrm{h}_{\mathrm{b}, \mathrm{m}}<\mathrm{h}_{\mathrm{b}}\text{ then use }\mathrm{h}_{\mathrm{b}, \mathrm{m}}\ \text{if }\mathrm{h}_{\mathrm{b}, \mathrm{m}} \geq \mathrm{h}_{\mathrm{b}}\text{ then use }\mathrm{h}_{\mathrm{b}}\end{array}\tag{9-27}$
Now in the case of full cavitation, the adhesion can be neglected and both arms are at 50% of the corresponding length. This simplifies the equations to:
$\ \begin{array}{left}\mathrm{A}=\frac{\mathrm{p}_{\mathrm{m}} \cdot \cos (\alpha+\beta+\varphi) \cdot \sin (\delta)}{\sin (\alpha) \cdot \sin (\alpha)}\ \mathrm{B}=\frac{-\mathrm{p}_{\mathrm{m}} \cdot \sin (\varphi-\delta)-\mathrm{c} \cdot \cos (\delta) \cdot \cos (\varphi)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}}\ \mathrm{C}=\frac{-\mathrm{p}_{\mathrm{m}} \cdot \cos (\alpha+\beta+\delta) \cdot \sin (\varphi)-\mathrm{c} \cdot \cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}\end{array}\tag{9-28}$
Introducing the ratio rz between the absolute hydrostatic pressure and the shear strength c:
$\ \mathrm{r}_{\mathrm{z}}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0})}{\mathrm{c}}\tag{9-29}$
Gives for the A, B and C:
$\ \begin{array}{left}\mathrm{A}=\frac{\mathrm{r_{z}} \cdot \cos (\alpha+\beta+\varphi) \cdot \sin (\delta)}{\sin (\alpha) \cdot \sin (\alpha)}\ \mathrm{B}=\frac{-\mathrm{r}_{\mathrm{z}} \cdot \sin (\varphi-\delta)-\cos (\delta) \cdot \cos (\varphi)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}}\ \mathrm{C}=\frac{-\mathrm{r}_{\mathrm{z}} \cdot \cos (\alpha+\beta+\delta) \cdot \sin (\varphi)-\cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}\end{array}\tag{9-30}$
The term is always negative. The term 4·A·is also always negative. This results in a square root that will always be bigger than |B|. Since the sum of the angles in the arguments of the cosines will always be larger than 90 degrees, the cosines will give a negative result. So will always be negative. This implies that the negative square root gives a positive answer, while the positive square root will give a negative answer. Since the mobilized blade height has to be positive, the negative square root should be used here.
Finally the horizontal and vertical cutting forces can be written as:
$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{H} \mathrm{C}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{9-31}$
$\ \mathrm{F}_{v}=\lambda_{\mathrm{V C}} \cdot \mathrm{c \cdot h}_{\mathrm{i}} \cdot \mathrm{w}\tag{9-32}$
Figure 9-17 and Figure 9-18 show the ratio of the mobilized blade height to the layer thickness hb,m/hi and the shear angle β for a 60 degree blade. From Figure 9-17 it is clear that the Curling Type already occurs at normal hb,m/hi ratios. Especially at small internal friction angles this will be the case. Figure 9-19 and Figure 9-20 show the horizontal and vertical cutting force coefficients, which are not much different from the coefficients of the Flow Type and hb,m/hi=1. . Figure 9-21 shows the Esp/UCS ratio, which is very convenient for production estimation.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/09%3A_Rock_Cutting-_Hyperbaric_Conditions/9.03%3A_The_Tear_Type_and_the_Chip_Type.txt
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The theory developed here, which basically is the theory of Miedema (1987 September) extended with the Curling Type, has been applied on the cutting tests of Zijsling (1987). Zijsling conducted cutting tests with a PDC bit with a width and height of 10 mm in Mancos Shale. This type of rock has a UCS value of about 65 MPa, a cohesive shear strength of about 25 MPa, an internal friction angle φ of 23o, according to Detournay & Atkinson (2000), a layer thickness hof 0.15 mm and 0.30 mm and a blade angle α of 110o. The external friction angle δ is chosen at 2/3 of the internal friction angle φ. Based on the principle of minimum energy a shear angle β of 12o has been derived. Zijsling already concluded that balling would occur. Using equation (9-26) an effective blade height hb,m = 4.04·hhas been foundFigure 9-22 shows the cutting forces as measured by Zijsling compared with the theory derived here. The force FD is the force Fh in the direction of the cutting velocity and the force FN is the force Fv normal to the velocity direction. Figure 9-23 shows the specific energy Esp and the so called drilling strength S. Figure 9-29 and Figure 9-30 show the specific energy Esp as a function of the UCS value of a rock for different UCS/BTS ratio’s and different water depths. Figure 9-29 shows this for a 110o blade as in the experiments of Zijsling (1987). The UCS value of the Mancos Shale is about 65 MPa. It is clear that in this graph the UCS/BTS value has no influence, since there will be no tensile failure at a blade angle of 110o. There could however be brittle shear failure under atmospheric conditions resulting in a specific energy of 30%-50% of the lowest line in the graph. Figure 9-29 gives a good indication of the specific energy for drilling purposes.
Figure 9-30 and Figure 9-31 show this for a 45o and a 60o blade as may be used in dredging and mining. From this figure it is clear that under atmospheric conditions tensile failure may occur. The lines for the UCS/BTS ratios give the specific energy based on the peak forces. This specific energy should be multiplied with 30%-50% to get the average value. Roxborough (1987) found that for all sedimentary rocks and some sandstone, the specific energy is about 25% of the UCS value (both have the dimension kPa or MPa). In Figure 9-30 and Figure 9-31 this would match brittle-shear failure with a factor of 30%-50% (R=2). In dredging and mining the blade angle would normally be in a range of 45o to 60o. Vlasblom (2003-2007) uses a percentage of 40% of the UCS value for the specific energy based on the experience of the dredging industry, which is close to the value found by Roxborough (1987). The percentage used by Vlasblom has the purpose of production estimation and is on the safe side (a bit too high). Both the percentages of Roxborough (1987) and Vlasblom (2003-2007) are based on the brittle shear failure. In the case of brittle tensile failure the specific energy may be much lower.
Resuming it can be stated that the theory developed here matches the measurements of Zijsling (1987) well. It has been proven that the approach of Detournay & Atkinson (2000) misses the pore pressure force on the blade and thus leads to some wrong conclusions. It can further be stated that brittle tensile failure will only occur with relatively small blade angles under atmospheric conditions. Brittle shear failure may also occur with large blade angles under atmospheric conditions. The measurements of Zijsling show clearly that at 0 MPa bottom hole pressure, the average cutting forces are 30%-50% of the forces that would be expected based on the trend. The conclusions are valid for the experiments they are based on. In other types of rock or with other blade angles the theory may have to be adjusted. This can be taken into account by the following equation, where α will have a value of 3-7 depending on the type of material.
$\ \mathrm{F_{h, c}=F_{h} \cdot\left(1-\frac{\alpha}{(z+10)}\right)}\tag{9-33}$
At zero water depth the cutting forces are reduced to α/10, so to 30%-70% depending on the type of rock. At 90 m water depth the reduction is just 3%-7%, matching the Zijsling (1987) experiments, but also the Rafatian et al. (2009) and Kaitkay & Lei (2005) experiments. The equation is empirical and a first attempt, so it needs improvement.
Figure 9-24 and Figure 9-25 show the hb,m/hi ratio and the shear angle β. The Zijsling (1987) experiments match the curves of an internal friction angle of 25 degrees close. Since the blade height in these experiments was about 10 mm, the actual hb,m/hi ratio were 10/.15=66.66 and 10/.3=33.33. In both cases these ratios are much larger than the ones calculated for the Curling Type, leading to the conclusion that the Curling Type always occurs. So in offshore drilling, the Curling Type is the dominant cutting mechanism. On the horizontal axis, a value of 1 matches the shear strength of the rock, being about 25 MPa. A value of 4 matches the maximum hydrostatic pressure of 100 MPa as used in the experiments. The hb,m/hi ratio increases slightly with increasing hydrostatic pressure, the shear angle decreases slightly.
Blade angle α = 110o, blade width = 10 mm, internal friction angle φ = 23.8o, external friction angle δ = 15.87o, shear strength = 24.82 MPa, shear angle β = 12.00o, layer thickness hi = 0.15 mm and 0.30 mm, effective blade height hb = 4.04·hi.
Blade angle α = 110o, blade width = 10 mm, internal friction angle φ = 23.8o, external friction angle δ = 15.87o, shear strength = 24.82 MPa, shear angle β = 12.00o, layer thickness hi = 0.15 mm and 0.30 mm, effective blade height hb = 4.04·hi.
Figure 9-26 and Figure 9-27 show the horizontal and vertical cutting force coefficients. For a hydrostatic pressure of 100 MPa and an internal friction angle of 25 degrees the graphs give a horizontal cutting force coefficient of λHC=45 and a vertical cutting force coefficient of λVC=38 giving cutting forces of Fh=FD=3.25 kN and Fv=FN=2.74 kN, matching the experiments in Figure 9-22. Figure 9-28 shows the Esp/UCS ratio, which is very convenient for production estimation.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/09%3A_Rock_Cutting-_Hyperbaric_Conditions/9.05%3A_Experiments_of_Zijsling_%281987%29.txt
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For the cases as described above, cutting with a straight blade with the direction of the cutting velocity vc perpendicular to the blade (edge of the blade), the specific cutting energy Esp is:
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{9-34}$
The specific energy of the Flow Type of cutting mechanism can be written as:
$\ \mathrm{E}_{\mathrm{s p}}=\lambda_{\mathrm{H F}} \cdot \mathrm{c}\tag{9-35}$
The specific energy of the Curling Type of cutting mechanism can be written as:
$\ \mathrm{E}_{\mathrm{s p}}=\lambda_{\mathrm{H} \mathrm{C}} \cdot \mathrm{c}\tag{9-36}$
Appendix X: Hyperbaric Rock Cutting Charts: Contains graphs for blade angles from 30 degrees up to 120 degrees, covering both dredging and offshore drilling applications.
9.07: Example
In this chapter many graphs are given for an α=60o blade and different internal friction angles. Chosing φ=20o, like in chapter 8, gives the possibility to compare atmospheric and hyperbaric cutting of rock. The external friction angle is assumed to be δ=2/3·φ. Assume a blade width w=0.1 m and a layer thickness hi=0.1 m, similar to chapter 8.
Also choosing UCS=100 MPa gives a specific energy to UCS ratio 0f 0.669 for very small hydrostatic pressure to UCS ratios, which is equal to the peak values found for atmospheric cutting. The atmospheric cutting process however is brittle shear failure in this case, resulting in lower average forces, while the hyperbaric process is supposed to be cataclastic or pseudo ductile. At very small hydrostatic pressures the behavior will still be brittle shear, but at larger water depths pseudo ductile.
Now suppose a rock with a UCS value of 10 MPa and water depths of 100 m, 1000 m and 3000 m. This results in the following forces and specific energies.
Table 9-1: Forces and specific energy example.
Water Depth z (m)
Hydrostatic Pressure/UCS Ratio
β (o)
hb,m/hi (-)
λHC (-) λvC (-)
Esp/UCS (-)
0
0
43.23
0.584
1.94
0.58
0.68
100
0.1
42.33
0.602
2.20
0.60
0.77
1000
1.0
38.51
0.707
4.62
0.81
1.62
3000
3.0
36.50
0.800
10.17
1.12
3.56
The mobilized blade height hb,m is smaller than 1, which means that under normal circumstances the mobilized blade height is smaller than the actual blade height, resulting in the Curling Type. If the mobilized blade height is larger than the actual blade height, the Flow Type or Crushed Type will occur and the numbers in the above table will be different. Figure 9-17, Figure 9-18, Figure 9-19, Figure 9-20 and Figure 9-21 are used to determine the values in the above table.
9.08: Specific Energy Graphs
Blade angle α = 110o, layer thickness hi = 0.00015 m, blade height hb = 0.01 m, angle of internal friction φ =23.80o, angle of external friction δ = 15.87o, shear angle β = 12.00o.
Blade angle α = 45o, layer thickness hi = 0.05 m, blade height hb = 0.1 m, angle of internal friction φ = 20.00o, angle of external friction δ = 13.33o, shear angle β = 40.00o.
Blade angle α = 60o, layer thickness hi = 0.05 m, blade height hb = 0.1 m, angle of internal friction φ = 20.00o, angle of external friction δ = 13.33o, shear angle β = 40.00o.
9.09: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesive shear strength
kPa
A
Adhesive force on the blade
kN
c, $\ \tau_\mathrm{c}$
Cohesive shear strength
kPa
c’
Pseudo cohesive shear strength
kPa
C
Cohesive force on shear plane
kN
Esp
Specific energy
kPa
F
Force
kN
Fh
Horizontal cutting force
kN
FD
Drag force on chisel (horizontal force)
kN
Fv
Vertical cutting force
kN
FN
Normal force on chisel (vertical force)
kN
g
Gravitational constant (9.81)
m/s2
G
Gravitational force
kN
hi
Initial thickness of layer cut
m
hb
Height of the blade
m
hb,m
Mobilized height of the blade in case Curling Type
m
K1
Grain force on the shear plane
kN
K2
Grain force on the blade
kN
I
Inertial force on the shear plane
kN
N1
Normal grain force on shear plane
kN
N2
Normal grain force on blade
kN
p1m
Pore pressure in the shear plane
kPa
p2m
Pore pressure on the blade
kPa
Pc
Cutting power
kW
Q
Production
m3
r
Adhesion/cohesion ratio
-
r1
Pore pressure on shear plane/cohesion ratio
-
r2
Pore pressure on blade/cohesion ratio
-
rz
Ratio hydrostatic pressure to cohesion
-
R
Radius of Mohr circle
kPa
R1
Acting point on the shear plane
m
R2
Acting point on the blade
m
S1
Shear force due to internal friction on the shear plane
kN
S2
Shear force due to external friction on the blade
kN
T
Tensile force
kN
UCS
Unconfined Compressive Stress
kPa
vc
Cutting velocity
m/s
w
Width of the blade
m
W1
Force resulting from pore under pressure on the shear plane
kN
W2
Force resulting from pore under pressure on the blade
kN
α
Blade angle
rad
β
Angle of the shear plane with the direction of cutting velocity
rad
$\ \tau$
Shear stress
kPa
$\ \tau_\mathrm{a}$, a
Adhesive shear strength (strain rate dependent)
kPa
$\ \tau_\mathrm{c}$, c
Cohesive shear strength (strain rate dependent)
kPa
$\ \tau_\mathrm{s1}$
Average shear stress on the shear plane
kPa
$\ \tau_\mathrm{s2}$
Average shear stress on the blade
kPa
σ
Normal stress
kPa
σc
Center of Mohr circle
kPa
σT
Tensile strength
kPa
σmin
Minimum principal stress in Mohr circle
kPa
σN1
Average normal stress on the shear plane
kPa
σN2
Average normal stress on the blade
kPa
φ
Angle of internal friction
rad
δ
Angle of external friction
rad
λ
Strengthening factor
-
λ1
Acting point factor on the shear plane
-
λ2
Acting point factor on the blade
-
λHF
Flow Type/Crushed Type horizontal force coefficient
-
λVF
Flow Type/Crushed Type vertical force coefficient
-
λHT
Tear Type/Chip Type horizontal force coefficient
-
λVT
Tear Type/Chip Type vertical force coefficient
-
λHC
Curling Type horizontal force coefficient
-
λVC
Curling Type vertical force coefficient
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/09%3A_Rock_Cutting-_Hyperbaric_Conditions/9.06%3A_Specific_Energy.txt
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The cutting theories until now works well for small blade angles, however when the blade angle and the other angles involved increase, a problem with the model may occur. The basic equations contain a denominator with the sine of the sum of the blade angle, the shear angle, the internal friction angle and the external friction angle. So if the sum of these angles equals 180 degrees, the denominator is zero, meaning a division by zero giving infinity. Even worse, if the sum of these angles is greater than 180 degrees the sine gives a negative result, meaning the cutting forces become negative. But already if the sum of these angles approach 180 degrees the sine becomes very small and since it is in the denominator, the cutting forces would become very high. Now nature will normally choose the road of least resistance, nature will try to find another mechanism for the cutting process and this mechanism might be the occurrence of a wedge in front of the blade. This wedge will form a pseudo cutting blade A-C with a blade angle much smaller than the angle of the real blade. The probability of the occurrence of a wedge is large for sand and rock since all 4 angles mentioned play a role there. For clay the probability is much smaller, since in clay cutting normally the internal and external friction angles do not play a role.
Now nature may choose another mechanism which will result in even smaller cutting forces, like the model of Hettiaratchi & Reece (1975), but their model is more complicated. The philosophy here is that if a mechanism can be found resulting in smaller cutting forces than the model used for small blade angles, this model will give a better prediction than the model for small blade angles. The wedge mechanism is such a mechanism, with the advantage that it is relatively simple to use and the cutting forces predicted with this model match the cutting forces from the experiments of Hatamura & Chijiiwa (1977B) pretty close. So from a pragmatic point of view this mechanism will be discussed for large blade angles.
Definitions:
1. A: The wedge tip.
2. B: End of the shear plane.
3. C: The blade top.
4. D: The blade tip.
5. A-B: The shear plane.
6. A-C: The wedge surface.
7. A-D: The wedge bottom.
8. D-C: The blade surface.
9. hb: The height of the blade.
10. hi: The thickness of the layer cut.
11. vc: The cutting velocity.
12. α: The blade angle.
13. β: The shear angle.
14. Fh: The horizontal force, the arrow gives the positive direction.
15. Fv: The vertical force, the arrow gives the positive direction.
10.02: TheForceEquilibrium
Figure 10-2 illustrates the forces on the layer of soil cut. The forces shown are valid in general for each type of soil.
The forces acting on the layer A-B are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A shear force C1 as a result of pure cohesion $\ \tau_\mathrm{c}$ or shear strength. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ with the area of the shear plane.
5. A gravity force G1 as a result of the weight of the layer cut.
6. An inertial force I, resulting from acceleration of the soil.
7. A force normal to the pseudo blade N2, resulting from the effective grain stresses.
8. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
9. A shear force C2 as a result of the mobilized cohesion between the soil and the wedge $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the wedge.
10. A force W2 as a result of water under pressure on the wedge.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{10-1}$
The forces acting on the wedge front or pseudo blade A-C when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
3. A shear force C2 as a result of the cohesion between the layer cut and the pseudo blade $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the pseudo blade.
4. A force W2 as a result of water under pressure on the pseudo blade A-C.
These forces are shown in Figure 10-3. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{10-2}$
The forces acting on the wedge bottom A-D when cutting soil, can be distinguished as:
1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.
2. A shear force S3 as a result of the soil/soil friction N3·tan(φ) between the wedge bottom and the undisturbed soil.
3. A shear force C3 as a result of the cohesion between the wedge bottom and the undisturbed soil $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the wedge bottom and the undisturbed soil.
4. A force W3 as a result of water under pressure on the wedge bottom A-D.
The normal force N3 and the shear force S3 can be combined to a resulting grain force K3.
$\ \mathrm{K}_{3}=\sqrt{\mathrm{N}_{3}^{2}+\mathrm{S}_{3}^{2}}\tag{10-3}$
The forces acting on a straight blade C-D when cutting soil (see Figure 10-4), can be distinguished as:
1. A force normal to the blade N4resulting from the effective grain stresses.
2. A shear force Sas a result of the soil/steel friction N4·tan(δ).
3. A shear force as a result of pure adhesion between the soil and the blade $\ \tau_\mathrm{a}$. This force can be calculated by multiplying the adhesive shear strength $\ \tau_\mathrm{a}$ of the soil with the contact area between the soil and the blade.
The normal force N4 and the shear force S4 can be combined to a resulting grain force K4.
$\ \mathrm{K}_{4}=\sqrt{\mathrm{N}_{4}^{2}+\mathrm{S}_{4}^{2}}\tag{10-4}$
The horizontal equilibrium of forces on the layer cut:
$\ \sum \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{1} \cdot \sin (\beta+\varphi)-\mathrm{W}_{1} \cdot \sin (\beta)+\mathrm{C}_{1} \cdot \cos (\beta)+\mathrm{I} \cdot \cos (\beta) -\mathrm{C}_{2} \cdot \cos (\theta)+\mathrm{W}_{2} \cdot \sin (\theta)-\mathrm{K}_{2} \cdot \sin (\theta+\lambda)=\mathrm{0}\tag{10-5}$
The vertical equilibrium of forces on the layer cut:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{1} \cdot \cos (\beta+\varphi)+\mathrm{W}_{1} \cdot \cos (\beta)+\mathrm{C}_{1} \cdot \sin (\beta)+\mathrm{I} \cdot \sin (\beta)+\mathrm{G}_{1}+\mathrm{C}_{2} \cdot \sin (\theta)+\mathrm{W}_{2} \cdot \cos (\theta)-\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0} \tag{10-6}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}= \frac{\mathrm{W}_{2} \cdot \sin (\lambda)+\mathrm{W}_{1} \cdot \sin (\theta+\beta+\lambda)+\mathrm{G}_{1} \cdot \sin (\theta+\lambda)-\mathrm{I} \cdot \cos (\theta+\beta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} +\frac{-\mathrm{C}_{1} \cdot \cos (\theta+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{10-7}$
The force K2 on the pseudo blade is now:
$\ \mathrm{K}_{2}= \frac{\mathrm{W}_{2} \cdot \sin (\theta+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G}_{1} \cdot \sin (\beta+\varphi)+\mathrm{I} \cdot \cos (\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} +\frac{+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\theta+\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \tag{10-8}$
From equation (10-8) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=-W_{2} \cdot \sin (\theta)+K_{2} \cdot \sin (\theta+\lambda)+C_{2} \cdot \cos (\theta)}\tag{10-9}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{2} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)-\mathrm{C}_{2} \cdot \sin (\theta)\tag{10-10}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}= \frac{\mathrm{W}_{2} \cdot \sin (\lambda)+\mathrm{W}_{1} \cdot \sin (\theta+\beta+\lambda)+\mathrm{G}_{1} \cdot \sin (\theta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\varphi) +\frac{-\mathrm{I} \cdot \cos (\theta+\beta+\lambda)-\mathrm{C}_{1} \cdot \cos (\theta+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\varphi)\tag{10-11}$
The normal force on the pseudo blade is now:
$\ \mathrm{N}_{2}= \frac{\mathrm{W}_{2} \cdot \sin (\theta+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{G}_{1} \cdot \sin (\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\lambda) +\frac{+\mathrm{I} \cdot \cos (\varphi)+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\theta+\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\lambda)\tag{10-12}$
Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived. The horizontal equilibrium of forces on the wedge is:
$\ \sum \mathrm{F}_{\mathrm{h}}=-\mathrm{A} \cdot \cos (\alpha)+\mathrm{W}_{4} \cdot \sin (\alpha)-\mathrm{K}_{4} \cdot \sin (\alpha+\delta)+\mathrm{K}_{3} \cdot \sin (\varphi) +\mathrm{C}_{3}-\mathrm{W}_{2} \cdot \sin (\theta)+\mathrm{C}_{2} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda)=\mathrm{0} \tag{10-13}$
The vertical equilibrium of forces on the wedge is:
$\ \sum \mathrm{F}_{\mathrm{v}}=\mathrm{A} \cdot \sin (\alpha)+\mathrm{W}_{4} \cdot \cos (\alpha)-\mathrm{K}_{4} \cdot \cos (\alpha+\delta)+\mathrm{W}_{3}-\mathrm{K}_{3} \cdot \cos (\varphi) -\mathrm{W}_{2} \cdot \cos (\theta)-\mathrm{C}_{2} \cdot \sin (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)+\mathrm{G}_{2}=\mathrm{0}\tag{10-14}$
The unknowns in this equation are K3 and K4, since K2 has already been solved. Three other unknowns are the adhesive force on the blade A, since the adhesion does not have to be mobilized fully if the wedge is static, the external friction angle δ, since also the external friction does not have to be fully mobilized, and the wedge angle θ. These 3 additional unknowns require 3 additional conditions in order to solve the problem. One additional condition is the equilibrium of moments of the wedge, a second condition the principle of minimum required cutting energy. A third condition is found by assuming that the external shear stress (adhesion) and the external shear angle (external friction) are mobilized by the same amount. Depending on whether the soil pushes upwards or downwards against the blade, the mobilization factor is between -1 and +1. Now in practice, sand and rock have no adhesion while clay has no external friction, so in these cases the third condition is not relevant. However in mixed soil both the external shear stress and the external friction may be present.
The force K3 on the bottom of the wedge is now:
$\ \begin{array}{left} \mathrm{K}_{3}&= \frac{-\mathrm{W}_{2} \cdot \sin (\alpha+\delta-\theta)+\mathrm{W}_{3} \cdot \sin (\alpha+\delta)+\mathrm{W}_{4} \cdot \sin (\delta)}{\sin (\alpha+\delta+\varphi)} \&+\frac{\mathrm{K}_{2} \cdot \sin (\alpha+\delta-\theta-\lambda)+\mathrm{G}_{2} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\delta+\varphi)} \&+\frac{\mathrm{A} \cdot \cos (\delta)+\mathrm{C}_{3} \cdot \cos (\alpha+\delta)-\mathrm{C}_{2} \cdot \cos (\alpha+\delta-\theta)}{\sin (\alpha+\delta+\varphi)} & \end{array}\tag{10-15}$
The force K4 on the blade is now:
$\ \begin{array}{left} \mathrm{K}_{4}=& \frac{-\mathrm{W}_{2} \cdot \sin (\theta+\varphi)+\mathrm{W}_{3} \cdot \sin (\varphi)+\mathrm{W}_{4} \cdot \sin (\alpha+\varphi)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda+\varphi)+\mathrm{G}_{2} \cdot \sin (\varphi)}{\sin (\alpha+\delta+\varphi)} \ &+\frac{-\mathrm{A} \cdot \cos (\alpha+\varphi)+\mathrm{C}_{3} \cdot \cos (\varphi)+\mathrm{C}_{2} \cdot \cos (\theta+\varphi)}{\sin (\alpha+\delta+\varphi)} \end{array}\tag{10-16}$
This results in a horizontal force of:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{\mathrm{4}} \cdot \sin (\alpha)+\mathrm{K}_{4} \cdot \sin (\alpha+\delta)+\mathrm{A} \cdot \cos (\alpha)\tag{10-17}$
And in a vertical force of:
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{\mathrm{4}} \cdot \cos (\alpha)+\mathrm{K}_{4} \cdot \cos (\alpha+\delta)-\mathrm{A} \cdot \sin (\alpha)\tag{10-18}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/10%3A_The_Occurrence_of_a_Wedge/10.01%3A_Introduction.txt
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In order to solve the problem, also the equilibrium of moments is required, since the wedge is not subject to rotational acceleration. The equilibrium of moments can be taken around each point of the wedge. Here the tip of the blade is chosen. The advantage of this is that a number of forces do not contribute to the moments on the wedge.
In order to derive the equilibrium of moments equation the arms of all the forces contributing to this equilibrium have to be known. Since these arms depend on the length of all the sides in the cutting process, first these lengths are determined. The length of the shear plane A-B is:
$\ \mathrm{L}_{1}=\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\tag{10-19}$
The length of the pseudo blade A-C is:
$\ \mathrm{L}_{2}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\theta)}\tag{10-20}$
The length of the bottom of the wedge A-D is:
$\ \mathrm{L}_{3}=\mathrm{h}_{\mathrm{b}} \cdot\left(\frac{\mathrm{1}}{\tan (\theta)}-\frac{1}{\tan (\alpha)}\right)\tag{10-21}$
The length of the blade C-D is:
$\ \mathrm{L}_{4}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\tag{10-22}$
The length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side is:
$\ \mathrm{L}_{5}=\mathrm{L}_{3} \cdot \sin (\theta)\tag{10-23}$
The length of the line from point to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{6}=\mathrm{L}_{3} \cdot \cos (\theta)\tag{10-24}$
The distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{7}=\mathrm{L}_{6}-\mathrm{R}_{2}\tag{10-25}$
The values of the acting points R2R3 and R4 follow from calculated or estimated stress distributions.
The equilibrium of moments is now:
$\ \sum \mathrm{M}=\left(\mathrm{N}_{4}-\mathrm{W}_{4}\right) \cdot \mathrm{R}_{4}-\left(\mathrm{N}_{3}-\mathrm{W}_{3}-\mathrm{G}_{2}\right) \cdot \mathrm{R}_{3} +\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{L}_{7}-\left(\mathrm{S}_{2}+\mathrm{C}_{2}\right) \cdot \mathrm{L}_{5}=\mathrm{0} \tag{10-26}$
10.04: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or adhesive shear strength.
kPa
A
Adhesive shear force on the blade.
kN
c, $\ \tau_\mathrm{c}$
Cohesion or cohesive shear strength.
kPa
C1
Cohesive shear force on the shear plane.
kN
C2
Cohesive shear force on the pseudo blade (front of the wedge).
kN
C3
Cohesive shear force on bottom of the wedge.
kN
Fh
Horizontal cutting force.
kN
Fv
Vertical cutting force.
kN
G1
Weight of the layer cut.
kN
G2
Weight of the wedge.
kN
hb
Blade height.
m
hi
Layer thickness.
m
I
Inertial force on the shear plane.
kN
N1
Normal force on the shear plane.
kN
N2
Normal force on the pseudo blade (front of the wedge).
kN
N3
Normal force on bottom of the wedge.
kN
N4
Normal force on the blade.
kN
K1
Sum of N1 and S1 on the shear plane.
kN
K2
Sum of N2 and S2 on the pseudo blade (front of the wedge).
kN
K3
Sum of N3 and S3 on bottom of the wedge.
kN
K4
Sum of N4 and S4 on the blade.
kN
L1
Length of the shear plane.
m
L2
Length of the pseudo blade (front of the wedge).
m
L3
Length of the bottom of the wedge.
m
L4
Length of the blade.
m
L5
Length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side.
m
L6
Length of the line from point to the intersection point of the previous line with side A-C.
m
L7
Distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line L6 with side A-C.
m
R1
Acting point forces on the shear plane.
m
R2
Acting point forces on the pseudo blade (front of the wedge).
m
R3
Acting point forces on the bottom of the wedge.
m
R4
Acting point forces on the blade.
m
S1
Shear (friction) force on the shear plane.
kN
S2
Shear (friction) force on the pseudo blade (front of the wedge).
kN
S3
Shear (friction) force on the bottom of the wedge.
kN
S4
Shear (friction) force on the blade.
kN
W1
Pore pressure force on the shear plane.
kN
W2
Pore pressure force on the pseudo blade (front of the wedge).
kN
W3
Pore pressure force on the bottom of the wedge.
kN
W4
Pore pressure force on the blade.
kN
vc
Cutting velocity.
m/sec
α
Blade angle.
°
β
Shear angle.
°
θ
Wedge angle.
°
φ
Internal friction angle.
°
δ
External friction angle.
°
λ
Internal friction angle on pseudo blade.
°
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/10%3A_The_Occurrence_of_a_Wedge/10.03%3A_The_Equilibrium_of_Moments.txt
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The cutting theories until now works well for small blade angles, however when the blade angle and the other angles involved increase, a problem with the model may occur. The basic equations contain a denominator with the sine of the sum of the blade angle, the shear angle, the internal friction angle and the external friction angle. So if the sum of these angles equals 180 degrees, the denominator is zero, meaning a division by zero giving infinity. Even worse, if the sum of these angles is greater than 180 degrees the sine gives a negative result, meaning the cutting forces become negative. But already if the sum of these angles approach 180 degrees the sine becomes very small and since it is in the denominator, the cutting forces would become very high. Now nature will normally choose the road of least resistance, nature will try to find another mechanism for the cutting process and this mechanism might be the occurrence of a wedge in front of the blade. This wedge will form a pseudo cutting blade A-C with a blade angle much smaller than the angle of the real blade. The probability of the occurrence of a wedge is large for sand and rock since all 4 angles mentioned play a role there. For clay the probability is much smaller, since in clay cutting normally the internal and external friction angles do not play a role.
Now nature may choose another mechanism which will result in even smaller cutting forces, like the model of Hettiaratchi & Reece (1975), but their model is more complicated. The philosophy here is that if a mechanism can be found resulting in smaller cutting forces than the model used for small blade angles, this model will give a better prediction than the model for small blade angles. The wedge mechanism is such a mechanism, with the advantage that it is relatively simple to use and the cutting forces predicted with this model match the cutting forces from the experiments of Hatamura & Chijiiwa (1977B) pretty close. So from a pragmatic point of view this mechanism will be discussed for large blade angles.
Definitions:
1. A: The wedge tip.
2. B: End of the shear plane.
3. C: The blade top.
4. D: The blade tip.
5. A-B: The shear plane.
6. A-C: The wedge surface.
7. A-D: The wedge bottom.
8. D-C: The blade surface.
9. hb: The height of the blade.
10. hi: The thickness of the layer cut.
11. vc: The cutting velocity.
12. α: The blade angle.
13. β: The shear angle.
14. Fh: The horizontal force, the arrow gives the positive direction.
15. Fv: The vertical force, the arrow gives the positive direction.
For the weight of the layer cut G1, see chapter 5: Dry Sand Cutting.
The weight of the wedge G2 is given by:
$\ \mathrm{G}_{2}=\rho_{s} \cdot \mathrm{g} \cdot \frac{\mathrm{h}_{\mathrm{b}}^{2}}{2} \cdot\left(\frac{1}{\tan (\theta)}-\frac{1}{\tan (\alpha)}\right) \cdot \mathrm{w}\tag{11-1}$
11.02: The Force Equilibrium
Figure 11-4 illustrates the forces on the layer of soil cut. The forces shown are valid in general for dry sand. The forces acting on the layer A-B are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A gravity force G1 as a result of the weight of the layer cut.
4. An inertial force I, resulting from acceleration of the soil.
5. A force normal to the pseudo blade N2, resulting from the effective grain stresses.
6. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{11-2}$
The forces acting on the wedge front or pseudo blade A-C when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
These forces are shown in Figure 11-5. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{11-3}$
The forces acting on the wedge bottom A-D when cutting soil, can be distinguished as:
1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.
2. A shear force S3 as a result of the soil/soil friction N3·tan(φ) between the wedge bottom and the undisturbed soil.
The normal force N3 and the shear force S3 can be combined to a resulting grain force K3.
$\ \mathrm{K}_{3}=\sqrt{\mathrm{N}_{\mathrm{3}}^{\mathrm{2}}+\mathrm{S}_{\mathrm{3}}^{\mathrm{2}}}\tag{11-4}$
The forces acting on a straight blade C-D when cutting soil (see Figure 11-6), can be distinguished as:
1. A force normal to the blade N4resulting from the effective grain stresses.
2. A shear force S4 as a result of the soil/steel friction N4·tan(δ).
The normal force N4 and the shear force S4 can be combined to a resulting grain force K4.
$\ \mathrm{K}_{4}=\sqrt{\mathrm{N}_{4}^{2}+\mathrm{S}_{4}^{2}}\tag{11-5}$
The horizontal equilibrium of forces on the layer cut:
$\ \mathrm{\sum F_{h}=K_{1} \cdot \sin (\beta+\varphi)+I \cdot \cos (\beta)-K_{2} \cdot \sin (\theta+\lambda)=0}\tag{11-6}$
The vertical equilibrium of forces on the layer cut:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{1} \cdot \cos (\beta+\varphi)+\mathrm{I} \cdot \sin (\beta)+\mathrm{G}_{1}-\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0}\tag{11-7}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{G}_{1} \cdot \sin (\theta+\lambda)-\mathrm{I} \cdot \cos (\theta+\beta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{11-8}$
The force K2 on the pseudo blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{G}_{1} \cdot \sin (\beta+\varphi)+\mathrm{I} \cdot \cos (\varphi)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{11-9}$
From equation (11-9) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=K_{2} \cdot \sin (\theta+\lambda)}\tag{11-10}$
$\ \mathrm{F}_{v}=\mathrm{K}_{2} \cdot \cos (\theta+\lambda)\tag{11-11}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}=\frac{\mathrm{G}_{1} \cdot \sin (\theta+\lambda)-\mathrm{I} \cdot \cos (\theta+\beta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\varphi)\tag{11-12}$
The normal force on the pseudo blade is now:
$\ \mathrm{N}_{2}=\frac{\mathrm{G}_{1} \cdot \sin (\beta+\varphi)+\mathrm{I} \cdot \cos (\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\lambda)\tag{11-13}$
Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived. The horizontal equilibrium of forces on the wedge is:
$\ \mathrm{\sum F_{h}=-K_{4} \cdot \sin (\alpha+\delta)+K_{3} \cdot \sin (\varphi)+K_{2} \cdot \sin (\theta+\lambda)=0}\tag{11-14}$
The vertical equilibrium of forces on the wedge is:
$\ \mathrm{\sum F_{v}=-K_{4} \cdot \cos (\alpha+\delta)-K_{3} \cdot \cos (\varphi)+K_{2} \cdot \cos (\theta+\lambda)+G_{2}=0}\tag{11-15}$
The unknowns in this equation are K3 and K4, since K2 has already been solved. Two other unknowns are the external friction angle δ, since also the external friction does not have to be fully mobilized, and the wedge angle θ. These 2 additional unknowns require 2 additional conditions in order to solve the problem. One additional condition is the equilibrium of moments of the wedge, a second condition the principle of minimum required cutting energy. Depending on whether the soil pushes upwards or downwards against the blade, the mobilization factor is between -1 and +1.
The force K3 on the bottom of the wedge is now:
$\ \mathrm{K}_{3}=\frac{\mathrm{K}_{2} \cdot \sin (\alpha+\delta-\theta-\lambda)+\mathrm{G}_{2} \cdot \sin (\alpha+\delta)}{\sin (\alpha+\delta+\varphi)}\tag{11-16}$
The force K4 on the blade is now:
$\ \mathrm{K}_{4}=\frac{\mathrm{K}_{2} \cdot \sin (\theta+\lambda+\varphi)+\mathrm{G}_{2} \cdot \sin (\varphi)}{\sin (\alpha+\delta+\varphi)}\tag{11-17}$
This results in a horizontal force on the blade of:
$\ \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{4} \cdot \sin (\alpha+\delta)\tag{11-18}$
And in a vertical force on the blade of:
$\ \mathrm{F}_{\mathrm{v}}=\mathrm{K}_{4} \cdot \cos (\alpha+\delta)\tag{11-19}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/11%3A_A_Wedge_in_Dry_Sand_Cutting/11.01%3A_Introduction.txt
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In order to solve the problem, also the equilibrium of moments is required, since the wedge is not subject to rotational acceleration. The equilibrium of moments can be taken around each point of the wedge. Here the tip of the blade is chosen. The advantage of this is that a number of forces do not contribute to the moments on the wedge.
In order to derive the equilibrium of moments equation the arms of all the forces contributing to this equilibrium have to be known. Since these arms depend on the length of all the sides in the cutting process, first these lengths are determined. The length of the shear plane A-B is:
$\ \mathrm{L}_{1}=\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\tag{11-20}$
The length of the pseudo blade A-C is:
$\ \mathrm{L}_{2}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\theta)}\tag{11-21}$
The length of the bottom of the wedge A-D is:
$\ \mathrm{L}_{3}=\mathrm{h}_{\mathrm{b}} \cdot\left(\frac{\mathrm{1}}{\tan (\theta)}-\frac{\mathrm{1}}{\tan (\alpha)}\right)\tag{11-22}$
The length of the blade C-D is:
$\ \mathrm{L}_{4}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\tag{11-23}$
The length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side is:
$\ \mathrm{L}_{5}=\mathrm{L}_{3} \cdot \sin (\theta)\tag{11-24}$
The length of the line from point to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{6}=\mathrm{L}_{3} \cdot \cos (\theta)\tag{11-25}$
The distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{7}=\mathrm{L}_{6}-\mathrm{R}_{2}\tag{11-26}$
The values of the acting points R2R3 and Rfollow from calculated or estimated stress distributions.
The equilibrium of moments is now:
$\ \sum \mathrm{M}=\mathrm{N}_{4} \cdot \mathrm{R}_{4}-\left(\mathrm{N}_{3}-\mathrm{G}_{2}\right) \cdot \mathrm{R}_{3}+\mathrm{N}_{2} \cdot \mathrm{L}_{7}-\mathrm{S}_{2} \cdot \mathrm{L}_{5}=\mathrm{0}\tag{11-27}$
11.04: Results of some Calculations
Since the wedge model depends on many parameters, some example calculations are carried out with the parameters as used by Hatamura & Chijiiwa (1977B). The calculations are carried out with a blade height hb=0.2 m, a blade width w=0.33 m, an angle of internal friction φ=38°, an angle of internal friction δ=2/3·φ, an angle of internal friction on the pseudo blade of λ=32°, a dry density of ρs=1.59 ton/m3 and a cutting velocity of vc=0.05 m/sec. The difference with the Hatamura & Chijiiwa (1977B) experiments is that here the blade height is constant, while in their experiments the blade length was constant. Further layer thicknesses of hi=0.066 m, 0.10 m and 0.20 m are used in the calculations. Based on these and many more calculations an empirical equation has been found for the wedge angle θ.
$\ \mathrm{\theta=(90-\varphi) \cdot\left(0.73+0.0788 \cdot \frac{h_{\mathrm{b}}}{h_{\mathrm{i}}}\right)}\tag{11-28}$
Figure 11-8, Figure 11-9 and Figure 11-10 show the shear angle, the mobilized external friction angle, the wedge angle, the total cutting force and the direction of the total cutting force.
In the region where the mobilized external friction angle changes from plus the maximum to minus the maximum value, an equilibrium of moments exists. In the case considered this means that a wedge may exist in this region. When the mobilized external friction angle equals minus the maximum value there is no equilibrium of moments. In this region the total cutting force increases rapidly with an increasing blade angle in the calculations, but most probably another mechanism than the wedge mechanism will occur, so the values of the cutting forces in that region are not reliable. In the region of the mobilized external friction angle between plus the maximum to minus the maximum value the total cutting force is almost constant.
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The experiments of Hatamura & Chijiiwa (1977B) were carried out with a blade length L4=0.2 m, a blade width w=0.33 m, an angle of internal friction φ=38°, an angle of internal friction δ=2/3·φ, an angle of internal friction on the pseudo blade of λ=32°, a dry density of ρs=1.46 ton/m3 and a cutting velocity of vc=0.05 m/sec.
Although the number of experiments of Hatamura & Chijiiwa (1977B) is limited, both the shear angles and the total cutting forces tend to follow the wedge theory for blade angles of 75° and 90°. The direction of the total cutting force measured is more upwards directed (negative angle) than predicted with the wedge theory for the 90° blade. This could mean that the real mechanism is different from the wedge mechanism. The cutting forces however match well.
11.06: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or adhesive shear strength.
kPa
A
Adhesive shear force on the blade.
kN
c, $\ \tau_\mathrm{c}$
Cohesion or cohesive shear strength.
kPa
C1
Cohesive shear force on the shear plane.
kN
C2
Cohesive shear force on the pseudo blade (front of the wedge).
kN
C3
Cohesive shear force on bottom of the wedge.
kN
Fh
Horizontal cutting force.
kN
Fv
Vertical cutting force.
kN
G1
Weight of the layer cut.
kN
G2
Weight of the wedge.
kN
hb
Blade height.
m
hi
Layer thickness.
m
I
Inertial force on the shear plane.
kN
N1
Normal force on the shear plane.
kN
N2
Normal force on the pseudo blade (front of the wedge).
kN
N3
Normal force on bottom of the wedge.
kN
N4
Normal force on the blade.
kN
K1
Sum of N1 and S1 on the shear plane.
kN
K2
Sum of N2 and S2 on the pseudo blade (front of the wedge).
kN
K3
Sum of N3 and S3 on bottom of the wedge.
kN
K4
Sum of N4 and S4 on the blade.
kN
L1
Length of the shear plane.
m
L2
Length of the pseudo blade (front of the wedge).
m
L3
Length of the bottom of the wedge.
m
L4
Length of the blade.
m
L5
Length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side.
m
L6
Length of the line from point to the intersection point of the previous line with side A-C.
m
L7
Distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line L6 with side A-C.
m
R1
Acting point forces on the shear plane.
m
R2
Acting point forces on the pseudo blade (front of the wedge).
m
R3
Acting point forces on the bottom of the wedge.
m
R4
Acting point forces on the blade.
m
S1
Shear (friction) force on the shear plane.
kN
S2
Shear (friction) force on the pseudo blade (front of the wedge).
kN
S3
Shear (friction) force on the bottom of the wedge.
kN
S4
Shear (friction) force on the blade.
kN
W1
Pore pressure force on the shear plane.
kN
W2
Pore pressure force on the pseudo blade (front of the wedge).
kN
W3
Pore pressure force on the bottom of the wedge.
kN
W4
Pore pressure force on the blade.
kN
vc
Cutting velocity.
m/sec
α
Blade angle.
°
β
Shear angle.
°
θ
Wedge angle.
°
φ
Internal friction angle.
°
δ
External friction angle.
°
λ
Internal friction angle on pseudo blade.
°
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/11%3A_A_Wedge_in_Dry_Sand_Cutting/11.05%3A_Experiments_of_Hatamura_and_Chijiiwa_%281977B%29.txt
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In the last decennia extensive research has been carried out into the cutting of water saturated sand. In the cutting of water-saturated sand, the phenomenon of dilatation plays an important role. In fact the effects of gravity, inertia, cohesion and adhesion can be neglected at cutting speeds in the range of 0.5 – 10 m/s. In the cutting equations, as published by Miedema (1987 September), there is a division by the sine of the sum of the blade angle, the shear angle, the angle of internal friction and the soil/interface friction angle. When the sum of these angle approaches 180o, a division by zero is the result, resulting in infinite cutting forces. This may occur for example for α=80o, α=30o, α=40o and α=30o. When this sum is greater than 180 degrees, the cutting forces become negative. It is obvious that this cannot be the case in reality and that nature will look for another cutting mechanism.
Hettiaratchi and Reece (1975) found a mechanism, which they called boundary wedges for dry soil. At large cutting angles a triangular wedge will exist in front of the blade, not moving relative to the blade. This wedge acts as a blade with a smaller blade angle. In fact, this reduces the sum of the 4 angles mentioned before to a value much smaller than 180o. The existence of a dead zone (wedge) in front of the blade when cutting at large cutting angles will affect the value and distribution of vacuum water pressure on the interface. He et al. (1998) proved experimentally that also in water saturated sand at large cutting angles a wedge will occur. A series of tests with rake angles 90, 105 and 120 degrees under fully saturated and densely compacted sand condition was performed by He et al. (1998) at the Dredging Technology Laboratory of Delft University of Technology. The experimental results showed that the failure pattern with large rake angles is quite different from that with small rake angles. For large rake angles a dead zone is formed in front of the blade, but not for small rake angles. In the tests he carried out, both a video camera and film camera were used to capture the failure pattern. The video camera was fixed on the frame which is mounted on the main carriage, translates with the same velocity as the testing cutting blade. Shown in the static slide of the video record, as in Figure 12-1, the boundary wedges exist during the cutting test. The assumption of an alternative failure mechanism is based on a small quantity of picture material, see Figure 12-1. It is described as a static wedge in front of the blade, which serves as a new virtual blade over which the sand flows away.
Although the number of experiments published is limited, the research is valuable as a starting point to predict the shape of the wedge. At small cutting angles the cutting forces are determined by the horizontal and vertical force equilibrium equations of the sand cut in front of the blade. These equations contain 3 unknowns, so a third equation/condition had to be found. The principle of minimum energy is used as a third condition to solve the 3 unknowns. This has proved to give very satisfactory results finding the shear angle and the horizontal and vertical cutting forces at small cutting angles. At large cutting angles, a 4th unknown exists, the wedge angle or virtual blade angle. This means that a 4th equation/condition must be found in order to determine the wedge angle. There are 3 possible conditions that can be used: The principle of minimum energy, the circle of Mohr, The equilibrium of moments of the wedge. In fact, there is also a 5th unknown, the mobilized friction on the blade. New research carried out in the Dredging Engineering Laboratory shows that a wedge exists, but not always a static wedge. The sand inside the wedge is still moving, but with a much lower velocity then the sand outside the wedge. This results in fully mobilized friction on the blade. The bottom boundary of the wedge, which is horizontal for a static wedge, may have a small angle with respect to the horizontal in the new case considered.
Definitions:
1. A: The wedge tip.
2. B: End of the shear plane.
3. C: The blade top.
4. D: The blade tip.
5. A-B: The shear plane.
6. A-C: The wedge surface.
7. A-D: The wedge bottom.
8. D-C: The blade surface.
9. hb: The height of the blade.
10. hi: The thickness of the layer cut.
11. vc: The cutting velocity.
12. α: The blade angle.
13. β: The shear angle.
14. Fh: The horizontal force, the arrow gives the positive direction.
15. Fv: The vertical force, the arrow gives the positive direction.
Figure 12-2 shows the definitions of the cutting process with a static wedge. A-B is the shear plane where dilatation occurs. A-C is the front of the static wedge and forms a pseudo cutting blade. A-C-D is the static wedge, where C- D is the blade, A-D the bottom of the wedge and A-C the pseudo blade or the front of the wedge.
The sand wedge theory is based on publications of Hettiaratchi and Reece (1975), Miedema (1987 September), He et al. (1998), Yi (2000), Miedema et al. (2001), Yi et al. (2001), Ma (2001), Miedema et al. (2002A), Miedema et al. (2002B), Yi et al. (2002), Miedema (2003), Miedema et al. (2003), Miedema (2004), Miedema et al. (2004), He et al. (2005), Ma et al. (2006A), Ma et al. (2006B), Miedema (2005), Miedema (2006A), Miedema (2006B).
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Figure 12-4, Figure 12-5 and Figure 12-6 show the forces occurring at the layer cut, the wedge and the blade, while
Figure 12-18 shows the moments occurring on the wedge. The forces are:
The forces acting on the layer A-B are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
5. A force W2 as a result of water under pressure on the wedge.
The forces acting on the wedge front or pseudo blade A-C when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
3. A force W2 as a result of water under pressure on the pseudo blade A-C.
The forces acting on the wedge bottom A-D when cutting soil, can be distinguished as:
1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.
2. A shear force S3 as a result of the soil/soil friction N3·tan(φ) between the wedge bottom and the undisturbed soil.
3. A force W3 as a result of water under pressure on the wedge bottom A-D.
The forces acting on a straight blade C-D when cutting soil, can be distinguished as:
1. A force normal to the blade N4resulting from the effective grain stresses.
2. A shear force S4 as a result of the soil/steel friction N4·tan(δ) between the wedge and the blade.
3. A force W4 as a result of water under pressure on the blade.
To determine the cutting forces on the blade, first the cutting forces on the pseudo blade have to be determined by taking the horizontal and vertical equilibrium of forces on the layer cut B-A-C. The shear angle β is determined by minimizing the cutting energy.
The horizontal equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{\mathrm{1}} \cdot \sin (\boldsymbol{\beta}+\boldsymbol{\varphi})-\mathrm{W}_{\mathrm{1}} \cdot \sin (\boldsymbol{\beta})+\mathrm{W}_{\mathrm{2}} \cdot \sin (\boldsymbol{\theta})-\mathrm{K}_{\mathrm{2}} \cdot \mathrm{\operatorname { sin } ( \boldsymbol { \theta } + \lambda )}=\mathrm{0}\tag{12-1}$
The vertical equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta}+\boldsymbol{\varphi})+\mathrm{W}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta})+\mathrm{W}_{\mathrm{2}} \cdot \mathrm{\operatorname { cos } ( \boldsymbol { \theta } ) - \mathrm { K } _ { \mathrm { 2 } }} \cdot \cos (\boldsymbol{\theta}+\lambda)=\mathrm{0}\tag{12-2}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\lambda)+\mathrm{W}_{1} \cdot \sin (\theta+\beta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{12-3}$
The force K2 on the pseudo blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\theta+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{12-4}$
From equation (12-4) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{2} \cdot \sin (\theta)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda)\tag{12-5}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{\mathrm{2}} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)\tag{12-6}$
The normal force on the shear plane A-B is now:
$\ \mathrm{N}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\lambda)+\mathrm{W}_{1} \cdot \sin (\theta+\beta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\varphi)\tag{12-7}$
The normal force on the pseudo blade A-C is now:
$\ \mathrm{N}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\theta+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\lambda)\tag{12-8}$
Now the force equilibrium on the wedge has to be solved. This is done by first taking the horizontal and vertical force equilibrium on the wedge A-C-D.
The horizontal equilibrium of forces:
$\ \mathrm{\sum F_{h}=+W_{4} \cdot \sin (\alpha)-K_{4} \cdot \sin \left(\alpha+\delta_{e}\right)+K_{3} \cdot \sin (\varphi) -W_{2} \cdot \sin (\theta)+K_{2} \cdot \sin (\theta+\lambda)=0}\tag{12-9}$
The vertical equilibrium of forces:
$\ \sum \mathrm{F}_{\mathrm{v}}=+\mathrm{W}_{4} \cdot \cos (\alpha)-\mathrm{K}_{4} \cdot \cos \left(\alpha+\delta_{\mathrm{e}}\right)+\mathrm{W}_{3}-\mathrm{K}_{3} \cdot \cos (\varphi) -\mathrm{W}_{2} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0}\tag{12-10}$
The grain force K3 on the bottom of the wedge is now:
$\ \mathrm{K}_{3}= \frac{-\mathrm{W}_{2} \cdot \sin \left(\alpha+\delta_{\mathrm{e}}-\theta\right)+\mathrm{K}_{2} \cdot \sin \left(\alpha+\delta_{\mathrm{e}}-\theta-\lambda\right)}{\sin \left(\alpha+\delta_{\mathrm{e}}+\varphi\right)}+\frac{+\mathrm{W}_{3} \cdot \sin \left(\alpha+\delta_{\mathrm{e}}\right)+\mathrm{W}_{4} \cdot \sin \left(\delta_{\mathrm{e}}\right)}{\sin \left(\alpha+\delta_{\mathrm{e}}+\varphi\right)} \tag{12-11}$
The grain force K4 on the blade is now:
\ \begin{aligned} \mathrm{K}_{4}=& \frac{-\mathrm{W}_{2} \cdot \sin (\theta+\varphi)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda+\varphi)}{\sin \left(\alpha+\delta_{\mathrm{e}}+\varphi\right)}+\frac{+\mathrm{W}_{3} \cdot \sin (\varphi)+\mathrm{W}_{4} \cdot \sin (\alpha+\varphi)}{\sin \left(\alpha+\delta_{\mathrm{e}}+\varphi\right)} \end{aligned}\tag{12-12}
From equation (12-12) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{\mathrm{4}} \cdot \sin (\alpha)+\mathrm{K}_{4} \cdot \sin \left(\alpha+\delta_{\mathrm{e}}\right)\tag{12-13}$
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{4} \cdot \cos (\alpha)+\mathrm{K}_{4} \cdot \cos \left(\alpha+\delta_{\mathrm{e}}\right)\tag{12-14}$
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If the cutting process is assumed to be stationary, the water flow through the pores of the sand can be described in a blade motions related coordinate system. The determination of the water vacuum pressures in the sand around the blade is then limited to a mixed boundary conditions problem. The potential theory can be used to solve this problem. For the determination of the water vacuum pressures it is necessary to have a proper formulation of the boundary condition in the shear zone. Miedema (1985A) derived the basic equation for this boundary condition. In later publications a more extensive derivation is published. If it is assumed that no deformations take place outside the deformation zone, then:
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|=\mathrm{0}\tag{12-15}$
Making the boundary condition in the shear plane dimensionless is similar to that of the breach equation of Meijer and Van Os (1976). In the breach problem the length dimensions are normalized by dividing them by the breach height, while in the cutting of sand they are normalized by dividing them by the cut layer thickness. Equation (12-15) is the same as the equation without a wedge. In the shear plane A-B the following equation is valid:
$\ \frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}^{\prime}}\right|_{\mathrm{1}}+\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}^{\prime}}\right|_{\mathrm{2}}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \sin (\beta)}{\mathrm{k}_{\mathrm{m a x}}} \quad\text{ with: }\quad \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\mathrm{h}_{\mathrm{i}}}\tag{12-16}$
This equation is made dimensionless with:
$\ \left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|^{\prime}=\frac{\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}^{\prime}}\right|}{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} / \mathrm{k}_{\mathrm{m a x}}}\tag{12-17}$
The accent indicates that a certain variable or partial derivative is dimensionless. The next dimensionless equation is now valid as a boundary condition in the deformation zone:
$\ \frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|_{\mathrm{1}}+\left|\frac{\partial \mathrm{p}}{\partial \mathrm{n}}\right|_{\mathrm{2}}=\sin (\beta)\tag{12-18}$
The storage equation also has to be made dimensionless, which results in the next equation:
$\ \left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{x}^{2}}\right|+\left|\frac{\partial^{2} \mathrm{p}}{\partial \mathrm{y}^{2}}\right|^{\prime}=\mathrm{0}\tag{12-19}$
Because this equation equals zero, it is similar to equation (12-15). The water under-pressures distribution in the sand package can now be determined using the storage equation and the boundary conditions. Because the calculation of the water under-pressures is dimensionless the next transformation has to be performed to determine the real water under-pressures. The real water under-pressures can be determined by integrating the derivative of the water under-pressures in the direction of a flow line, along a flow line, so:
$\ \mathrm{P}_{\mathrm{c a l c}}=\int_{\mathrm{s^{\prime}}}\left|\frac{\partial \mathrm{p}}{\partial \mathrm{s}}\right|^{\prime} \cdot \mathrm{d s}^{\prime}\tag{12-20}$
This is illustrated in Figure 12-7 and Figure 12-8. Using equation (12-20) this is written as:
$\ \mathrm{P}_{\text {real }}=\int_{\mathrm{s}}\left|\frac{\partial \mathrm{p}}{\partial \mathrm{s}}\right| \cdot \mathrm{d s}=\int_{\mathrm{s}^{\prime}} \frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left|\frac{\partial \mathrm{p}}{\partial \mathrm{s}}\right|^{\prime} \cdot \mathrm{d s}^{\prime} \quad\text{ with: }\quad \mathrm{s}^{\prime}=\frac{\mathrm{s}}{\mathrm{h}_{\mathrm{i}}}\tag{12-21}$
This gives the next relation between the real emerging water under pressures and the calculated water under pressures:
$\ \mathrm{P}_{\text {real }}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{m a x}}} \cdot \mathrm{P}_{\mathrm{c a l c}}\tag{12-22}$
To be independent of the ratio between the initial permeability ki and the maximum permeability kmax , kmax has to be replaced with the weighted average permeability km before making the measured water under pressures dimensionless.
The water vacuum pressures in the sand package on and around the blade are numerically determined using the finite element method. A standard FEM software package is used (Segal (2001)). Within this package and the use of the available "subroutines" a program is written, with which water vacuum pressures can be calculated and be output graphically and numerically. As shown in Figure 12-9, the SEPRAN model is made up of three parts, the original sand layer, the cut sand layer, and the wedge. The solution of such a calculation is however not only dependent on the physical model of the problem, but also on the next points:
1. The size of the area in which the calculation takes place.
2. The size and distribution of the elements
3. The boundary conditions
The choices for these three points have to be evaluated with the problem that has to be solved in mind. These calculations are about the values and distribution of the water under-pressures in the shear zone and on the blade, on the interface between wedge and cut sand, between wedge and the original sand layer. A variation of the values for point 1 and 2 may therefore not influence this part of the solution. This is achieved by on one hand increasing the area in which the calculations take place in steps and on the other hand by decreasing the element size until the variation in the solution was less than 1%. The distribution of the elements is chosen such that a finer mesh is present around the blade tip, the shear zone and on the blade, also because of the blade tip problem. A number of boundary conditions follow from the physical model of the cutting process, these are:
• For the hydrostatic pressure it is valid to take a zero pressure as the boundary condition.
• The boundary conditions along the boundaries of the area where the calculation takes place that are located in the sand package are not determined by the physical process.
• For this boundary condition there is a choice among:
1. A hydrostatic pressure along the boundary.
2. A boundary as an impermeable wall.
3. A combination of a known pressure and a known specific flow rate.
None of these choices complies with the real process. Water from outside the calculation area will flow through the boundary. This also implies, however, that the pressure along this boundary is not hydrostatic. If, however, the boundary is chosen with enough distance from the real cutting process the boundary condition may not have an influence on the solution. The impermeable wall is chosen although this choice is arbitrary. Figure 12-14 and Figure 12-16 give an impression of the equipotential lines and the stream lines in the model area. Figure 12-10 show the dimensionless pore pressure distributions on the lines A-BA-CA-D and D-C. The average dimensionless pore pressures on these lines are named p1mp2mp3m and p4m.
If there is no cavitation the water pressures forces W1W2W3 and W4 can be written as:
$\ \mathrm{W}_{1}=\frac{\mathrm{p}_{1 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w}}{\mathrm{k}_{\max } \cdot \sin (\beta)}\tag{12-23}$
And
$\ \mathrm{W}_{2}=\frac{\mathrm{p}_{2 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{\varepsilon} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m a x}} \cdot \sin (\theta)}\tag{12-24}$
And
$\ \mathrm{W}_{3}=\frac{\mathrm{p}_{3 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m a x}}} \cdot\left(\frac{\cos (\theta)}{\sin (\theta)}-\frac{\cos (\alpha)}{\sin (\alpha)}\right)\tag{12-25}$
And
$\ \mathrm{W}_{4}=\frac{\mathrm{p}_{4 \mathrm{m}} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \varepsilon \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\mathrm{k}_{\max } \cdot \sin (\mathrm{\alpha})}\tag{12-26}$
In case of cavitation W1W2Wand W4 become:
$\ \mathrm{W}_{1}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}{\sin (\beta)}\tag{12-27}$
And
$\ \mathrm{W}_{2}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\theta)}\tag{12-28}$
And
$\ \mathrm{W}_{3}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\mathrm{1}} \cdot\left(\frac{\cos (\theta)}{\sin (\theta)}-\frac{\cos (\alpha)}{\sin (\alpha)}\right)\tag{12-29}$
And
$\ \mathrm{W}_{4}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{12-30}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/12%3A_A_Wedge_in_Saturated_Sand_Cutting/12.03%3A_Pore_Pressures.txt
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Based on the equilibrium of forces on the layer cut B-A-C, FEM calculations of pore water pressures and the minimum of cutting energy the forces N2S2 and W2 are determined; see Miedema (1987 September). To determine the forces on the blade there are still a number of unknowns. W3 and W4 can be determined using FEM calculations of pore water pressures, given the wedge angle θ. Assuming λ=φ as a first estimate, the forces K3 and K4 depend on the wedge angle θ and on the effective external friction angle δe. For a static wedge, meaning that there is no movement between the wedge and the blade, the effective external friction angle can have a value between + and – the maximum external friction angle δ, so –δ<δe. Combining this with the minimum energy principle results in a varying δe and a force N3 being equal to zero for a static wedge. The value of δe follows from the equilibrium of moments. For small values of the blade angle αsmaller than about 60othe effective external friction angle δe=δ and most probably there will not be a wedge. For intermediate values of the blade angle α around 90o, there will be a static wedge and the effective external friction angle δe will decrease from +δ to –δ. For very large values of αlarger than about 120o, the effective external friction angle δe=–δ and N3 will have a positive value, meaning an upwards direction. Probably there will be a movement of soil under the blade. To find the value of the effective external friction angle first the equilibrium of moments has to be solved. Figure 12-18 shows the moments that occur on the wedge as a result of the forces and their acting points.
To determine the moment on the wedge, first the different lengths and distances have to be determined. The length of the shear plane A-B is:
$\ \mathrm{A}-\mathrm{B}=\mathrm{L}_{1}=\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\tag{12-31}$
The length of the pseudo blade or front of the wedge A-C is:
$\ \mathrm{A}-\mathrm{C}=\mathrm{L}_{2}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\theta)}\tag{12-32}$
The length of the bottom of the wedge A-D is:
$\ \mathrm{A}-\mathrm{D}=\mathrm{L}_{3}=\mathrm{h}_{\mathrm{b}} \cdot\left(\frac{1}{\tan (\theta)}-\frac{1}{\tan (\alpha)}\right)\tag{12-33}$
The length of the blade D-C is:
$\ \mathrm{D}-\mathrm{C}=\mathrm{L}_{4}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\tag{12-34}$
The distance between the blade edge and the wedge side A-C (perpendicular) is:
$\ \mathrm{L}_{5}=\mathrm{L}_{3} \cdot \sin (\theta)\tag{12-35}$
The distance from point and the line L5 is:
$\ \mathrm{L}_{6}=\mathrm{L}_{3} \cdot \cos (\theta)\tag{12-36}$
The arm of the acting point of N2 and W2 is now:
$\ \mathrm{L}_{7}=\mathrm{L}_{6}-\mathrm{R}_{2}\tag{12-37}$
The equilibrium of moments can be determined using all those distances:
$\ \sum \mathrm{M}=\left(\mathrm{N}_{4}-\mathrm{W}_{4}\right) \cdot \mathrm{R}_{4}-\left(\mathrm{N}_{3}-\mathrm{W}_{3}\right) \cdot \mathrm{R}_{3}+\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{L}_{7}-\mathrm{S}_{2} \cdot \mathrm{L}_{5}=\mathrm{0}\tag{12-38}$
Equation (12-38) still contains the unknown arms R2R3 and R4. Based on the FEM calculations for the pore pressures, values of 0.35·L20.55·L3 and 0.32·L4 are found, Ma (2001). Figure 12-19 shows the moments on the wedge with respect to the cutting edge as a function of the wedge angle θ for different values of the shear angle β and a blade angle α of 90o. The moment is zero for a wedge angle θ between 50o and 55o.
Figure 12-20 shows the moments as a function of the shear angle β for 4 values of the wedge angle θ. The moment is zero for the wedge angle θ=55o at a shear angle β=18o. It is clear from these figures that the shear angle where the moment is zero is not very sensitive for the shear angle and the wedge angle.
Figure 12-21 shows the force triangles on the 3 sides of the wedges for cutting angles from 60 to 120 degrees. From the calculations it appeared that the pore pressures on interface between the soil cut and the wedge and in the shear plane do not change significantly when the blade angle changes. These pore pressures p1m and p2mresulting in the forces W1 and W2, are determined by the shear angle β, the wedge angle θ and other soil mechanical properties like the permeability.
The fact that the pore pressures do not significantly change, also results in forces K2, acting on the wedge that do not change significantly, according to equations (12-4), (12-5) and (12-6). These forces are shown in Figure 12-21 on the right side of the wedges and the figure shows that these forces are almost equal for all blade angles. These forces are determined by the conventional theory as published by Miedema (1987 September). Figure 12-21 also shows that for the small blade angles the friction force on the wedge is directed downwards, while for the bigger blade angles this friction force is directed upwards.
$\ \mathrm{R}_{2}=\mathrm{e}_{2} \cdot \mathrm{L}_{2}, \quad \mathrm{R}_{3}=\mathrm{e}_{3} \cdot \mathrm{L}_{3}, \quad \mathrm{R}_{4}=\mathrm{e}_{4} \cdot \mathrm{L}_{4}\tag{12-39}$
Now the question is, what is the solution for the cutting of water saturated sand at large cutting angles? From many calculations and an analysis of the laboratory research is described by He (1998), Ma (2001) and Miedema (2005), it appeared that the wedge can be considered a static wedge, although the sand inside the wedge still may have velocity, the sand on the blade is not moving. The main problem in finding acceptable solutions was finding good values for the acting points on the 3 sides of the wedge, e2e3 and e4. If these values are chosen right, solutions exist based on the equilibrium of moments, but if they are chosen wrongly, no solution will be found. So the choice of these parameters is very critical. The statement that the sand on the blade is not moving is based on two things, first of all if the sand is moving with respect to the blade, the soil interface friction is fully mobilized and the bottom of the wedge requires to have a small angle with respect to the horizontal in order to make a flow of sand possible. This results in much bigger cutting forces, while often no solution can be found or unreasonable values for e2e3 and e4 have to be used to find a solution.
So the solution is, using the equilibrium equations for the horizontal force, the vertical force and the moments on the wedge. The recipe to determine the cutting forces seems not to difficult now, but it requires a lot of calculations and understanding of the processes, because one also has to distinguish between the theory for small cutting angles and the wedge theory.
The following steps have to be taken to find the correct solution:
1. Determine the dimensionless pore pressures p1mp2mp3m and p4m using a finite element calculation or the method described by Miedema (2006B), for a variety of shear angles β and wedge angles θ around the expected solution.
2. Determine the shear angle β based on the equilibrium equations for the horizontal and vertical forces, a given wedge angle θ and the principle of minimum energy, which is equivalent to the minimum horizontal force.
This also gives a value for the resulting force K2 acting on the wedge.
3. Determine values of e2e3 and ebased on the results from the pore pressure calculations.
4. Determine the solutions of the equilibrium equations on the wedge and find the solution which has the minimum energy dissipation, resulting in the minimum horizontal force on the blade.
5. Determine the forces without a wedge with the theory for small cutting angles.
6. Determine which horizontal force is the smallest, with or without the wedge.
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To illustrate the results of the calculation method, a non-cavitating case will be discussed. Calculations are carried out for blade angles α of 65o, 70o, 75o, 80o, 85o, 90o, 95o, 100o, 105o, 110o, 115o and 120o, while the smallest angle is around 60o depending on the possible solutions. Also the cutting forces are determined with and without a wedge, so it’s possible to carry out step 6.
The case concerns a sand with an internal friction angle φ of 30o, a soil interface friction angle δ of 20o fully mobilized, a friction angle λ between the soil cut and the wedge equal to the internal friction angle, an initial permeability ki of 6.2*10-5 m/s and a residual permeability kmax of 17*10-5 m/s. The blade dimensions are a width of 0.25 m and a height of 0.2 m, while a layer of sand of 0.05 m is cut with a cutting velocity of 0.3 m/s at a water depth of 0.6 m, matching the laboratory conditions. The values for the acting points of the forces, are e2=0.35e3=0.55 and e4=0.32, based on the finite element calculations carried out by Ma (2001).
Figure 12-22 and Figure 12-23 show the results of the calculations. Figure 12-22 shows the wedge angle θ, the shear angle β, the mobilized internal friction angle λ and the mobilized external friction angle δe as a function of the blade angle α. Figure 12-23 shows the horizontal and vertical cutting forces, with and without a wedge.
The wedge angles found are smaller than 90o, which would match the theory of Hettiaratchi and Reece (1975). The shear angle β is around 20o, but it is obvious that a larger internal friction angle gives a smaller shear angle β. The mobilized external friction angle varies from plus the maximum mobilized external friction angle to minus the maximum mobilized external friction angle as is also shown in the force diagrams in Figure 12-21.
Figure 12-23 shows clearly how the cutting forces become infinite when the sum of the 4 angles involved is 180o and become negative when this sum is larger than 180o. So the transition from the small cutting angle theory to the wedge theory occurs around a cutting angle of 70o, depending on the soil mechanical parameters and the geometry of the cutting process.
12.06: The Cavitating Wedge
Also for the cavitating process, a case will be discussed. The calculations are carried out for blade angles α of 65o, 70o, 75o, 80o, 85o, 90o, 95o, 100o, 105o, 110o, 115o and 120o, while the smallest angle is around 60o depending on the possible solutions. Also the cutting forces are determined with and without a wedge, so it’s possible to carry out step 6.
The case concerns a sand with an internal friction angle φ of 30o, a soil interface friction angle δ of 20o fully mobilized, a friction angle δ between the soil cut and the wedge equal to the internal friction angle, an initial permeability ki of 6.2*10-5 m/s and a residual permeability kmax of 17*10-5 m/s. The blade dimensions are a width of 0.25 m and a height of 0.2 m, while a layer of sand of 0.05 m is cut with a cutting velocity of 0.3 m/s at a water depth of 0.6 m, matching the laboratory conditions. The values for the acting points of the forces, are e2=0.35e3=0.55 and e4=0.32, based on the finite element calculations carried out by Ma (2001).
Figure 12-24 and Figure 12-25 show the results of the calculations. Figure 12-24 shows the wedge angle θ, the shear angle β, the mobilized internal friction angle λ and the mobilized external friction angle δe as a function of the blade angle α. Figure 12-25 shows the horizontal and vertical cutting forces, with and without a wedge.
With the cavitating cutting process, the wedge angle θ always results in an angle of 90o, which matches the theory of Hettiaratchi and Reece (1975).The reason of this is that in the full cavitation situation, the pore pressures are equal on each side of the wedge and form equilibrium in itself. So the pore pressures do not influence the ratio between the grain stresses on the different sides of the wedge. From Figure 12-25 it can be concluded that the transition point between the conventional cutting process and the wedge process occurs at a blade angle of about 77 degrees.
In the non-cavitating cases this angle is about 70 degrees. A smaller angle of internal friction results in a higher transition angle, but in the cavitating case this influence is bigger. In the cavitating case, the horizontal force is a constant as long as the external friction angle is changing from a positive maximum to the negative minimum. Once this minimum is reached, the horizontal force increases a bit. At the transition angle where the horizontal forces with and without the wedge are equal, the vertical forces are not equal, resulting in a jump of the vertical force, when the wedge starts to occur.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/12%3A_A_Wedge_in_Saturated_Sand_Cutting/12.05%3A_The_Non-Cavitating_Wedge.txt
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Instead of carrying out the calculations for each different case, the limits of the occurrence of the wedge can be summarized in a few graphs. Figure 12-26 shows the upper and lower limit of the wedge for the non-cavitating case as a function of the angle of internal friction φ. It can be concluded that the upper and lower limits are not symmetrical around 90o, but a bit lower than that. An increasing angle of internal friction results in a larger bandwidth for the occurrence of the wedge. For blade angles above the upper limit most probably subduction will occur, although there is no scientific evidence for this. The theory developed should not be used for blade angles above the upper limit yet. Further research is required. The lower limit is not necessarily the start of the occurrence of the wedge. This depends on whether the cutting forces with the wedge are smaller than the cutting forces without the wedge. Figure 12-28 shows the blade angle where the wedge will start to occur, based on the minimum of the horizontal cutting forces with and without the wedge. It can be concluded that the blade angle where the wedge starts to occur is larger than the minimum where the wedge can exist, which makes sense. For high angles of internal friction, the starting blade angle is about equal to the lower limit.
For the cavitating case the upper and lower limit are shown in Figure 12-27. In this case the limits are symmetrical around 90o and with an external friction angle of 2/3 of the internal friction angle it can be concluded that these limits are 90o+δ and 90o-δ. The blade angle where the wedge will start to occur is again shown in Figure 12-28.
The methodology applied gives satisfactory results to determine the cutting forces at large cutting angles. The results shown in this paper are valid for the non-cavitating and the cavitating cutting process and for the soils and geometry as used in this paper. The wedge angles found are, in general, a bit smaller then 90o-δ for the non- cavitating case and exactly 90o-δ for the cavitating case, so as a first approach this can be used.
The mobilized external friction angle δe varies from plus the maximum for small blade angles to minus the maximum for large blade angles, depending on the blade angle.
The cutting forces with the wedge do not increase much in the non-cavitating case and not at all in the cavitating case, when the cutting angle increases from 60o to 120o.
If the ratio between the thickness of the layer cut and the blade height changes, also the values of the acting points e2e3 and e4 will change slightly.
It is not possible to find an explicit analytical solution for the wedge problem and it’s even difficult to automate the calculation method, since the solution depends strongly on the values of the acting points.
Figure 12-26, Figure 12-27 and Figure 12-28 are a great help determining whether or not a wedge will occur and at which blade angle it will start to occur.
The theory developed can be applied to cutting processes of bulldozers, in front of the heel of a drag head, ice scour, tunnel boring machines and so on.
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Sand cutting tests have been carried out in the Laboratory of Dredging Engineering at the Delft University.
The cutting tank is a concrete tank with a length of 35 m, a width of 3 m and a depth of 1.5 m. The bottom of the tank is covered with a drainage system. Above the drainage system is a layer of about 0.7 m sand (0.110 mm). On top of the sand is a layer of 0.5 m water. Other soils than the 0.110 mm sand can be used in the tank. On top of the tank rails are mounted on which a carriage can ride with speeds of up to 1.25 m/s with a pulling force of up to 15 kN, or 2.5 m/s with a pulling force of 7.5 kN. On the carriage an auxiliary carriage is mounted that can be moved transverse to the velocity of the main carriage. On this carriage a hydraulic swell simulating system is mounted, thus enabling the cutting tools to be subjected to specific oscillations. Under the carriage dredging equipment such as cutter heads and drag heads can be mounted. The dredging equipment can be instrumented with different types of transducers such as force, speed and density transducers. The signals from these transducers will be conditioned before they go to a computer via an A/D converter. On the carriage a hydraulic system is available, including velocity and density transducers. A 25 kW hydraulic drive is available for cutter heads and dredging wheels. The dredge pump is driven by a 15 kW electric drive with speed control. With the drainage system the pore water pressures can be controlled. Dredged material is dumped in an adjacent hopper tank to keep the water clean for under water video recordings. In the cutting tank research is carried out on cutting processes, mixture forming, offshore dredging, but also jet-cutting, the removal of contaminated silt, etc.
The tests carried out in the Dredging Engineering Laboratory had the objective to find the failure mechanisms of a sand package under large cutting angles of 60 ̊, 75 ̊ and 90 ̊. Main goal of the tests was to visualize the total process in a 2-dimensional view. Besides, the behaviour of sand in front of the blade was to be investigated. As mentioned before, some wedge exists in front of the blade, but it was not clear until now whether this was a kinematic wedge or a dynamic wedge. Visualising the cutting process and visualising the velocity of the sand on the blade has to improve the understanding of the processes involved.
The existing testing facilities have been used to carry out the cutting tests. With these facilities cutting depths from 3 till 7 cm are tested, resulting in an (effective blade height)/(cutting depth) ratio of 2.5 to 6, for the various angles. Cutting velocities of the tests were from 0.1 m/s to 0.4 m/s for smaller and 0.2 m/s for the larger cutting depths. These maximum velocities are limited by the maximum electrical power of the testing facility. In the first series of tests the 2-dimensional cutting process is made visual by doing tests near the window in the cutting tank. The process is not completely 2-dimensional here, because the water pressures and sand friction are influenced by the window, but it gives a good indication of the appearing failure mechanism of the sand package. Figure 12-29 shows a cross-section of the cutting tank and the carriage under which the cutting tools are mounted, while Figure 12-30 shows a front view and Figure 12-31 shows the blades mounted under the carriage.
To visualise the behaviour of the sand package in front of the blade a Perspex window is made in the middle one of the 3 cutting blades. Here we expect the least side influences. The middle blade measures a height of 20 cm and a width of 25 cm. The camera is mounted at the back of the blade, in a cover, as seen in Figure 12-33. In Figure 12-32 you can see an underwater light, which is also mounted in the cover, shining on the camera. This construction gives a view of the process as can be seen in Figure 12-34 and Figure 12-35, at a height of 8 till 9 cm in the blade. The camera records with a frame rate of 25/sec. In the Perspex window, Figure 12-35, a scale of 1 cm is engraved. By tracing sand grains along the window a ratio is determined between the cutting velocity and the velocity along the window at the recorded height, for the angles of 75 ̊ and 90 ̊. These ratios are respectively 0.3 and 0.15. At 60 ̊ this ratio can hardly be determined because it lies in the range of the cutting velocity and out of the range of the recorded frame rate.
With a dynamometer forces on the middle blade are measured. The horizontal cutting forces for the various angles are roughly in a ratio of 1:1.5:2, for 60 ̊, 75 ̊, 90 ̊ respectively. This indicates a changing failure mechanism for the 3 tested angles, which the videos from the tests along the glass also confirm.
Figures 9, 10 and 11 show the horizontal cutting forces as obtained from the experiments.
From the above results two main conclusions can be drawn. First of all, the sand is moving relative to the blade on the blade and secondly the cutting forces at a 90 ̊ blade are much smaller then would be expected from the cutting theory, Miedema (1987 September). As shown in Figure 12-1, He et al. (1998) and also observed according to Figure 12-39, a wedge exists in front of the blade, but apparently this is not a kinematic wedge, but a dynamic wedge.
To determine the flow pattern of the sand in the dynamic wedge, vertical bars of colored sand grains were inserted in the sand. These vertical bars had a length of about 10 cm. Since the maximum cutting depth was 7 cm, the full cutting process was covered by these bars. Figure 12 shows the cutting process with the vertical bars and it shows how the bars are deformed by the cutting process.
12.09: The Dynamic Wedge
As discussed in the above paragraphs, the new research has led to the insight that the wedge in front of the blade is not static but dynamic. The aim of the new research was to get a good understanding of the mechanisms involved in the cutting at large cutting angles. To achieve this, vertical bars of about 10 cm deep with colored sand grains were inserted in the sand as is shown in Figure 12-39. When these bars are cut they will be deformed. If the wedge in front of the blade is a static wedge, meaning that the grains in the wedge have no velocity relative to the blade, then the colored grains from the bars will not enter the wedge. If however the colored grains enter the wedge, this means that the grains in the wedge move with respect to the blade. The research has shown that the colored grains have entered the wedge according to Figure 12-39. In the layer cut, the colored grains show a straight line, which is obvious, because of the velocity distribution in the layer cut. In fact the layer cut moves as a rigid body. In the wedge the colored grains show a curved line. Because of the velocity distribution in the wedge, the grains near the blade move much slower than the grains in the layer cut. Although Figure 12-39 shows a line between the layer cut and the wedge, in reality there does not exist a clear boundary between these two surfaces. The grains on both sides of the drawn boundary line will have (almost) the same velocity, resulting in an internal friction angle λ, which is not fully mobilized. The external friction angle on the blade however is fully mobilized. This contradicts the findings of Miedema et al. (2002A), from previous research. The value of this internal friction angle is between 0<λ<φ. Further research will have to show the value of λ.
12.10: Nomenclature
Blade angle
e2, e3, e4
Acting point of cutting forces
-
F, Fh, Fv
Cutting force (general)
kN
g
Gravitation acceleration
m/s2
hi
Initial layer thickness
m
hb
Blade height
m
ki
Initial permeability
m/s
kmax
Maximum permeability
m/s
K1, 2, 3, 4
Grain force caused by grain stresses
kN
ni
Initial pore percentage
%
nmax
Maximum pore percentage
%
N1, 2, 3, 4
Normal force caused by grain stresses
kN
p1m, 2m, 3m, 4m
Average pore pressure on a surface
-
S1, 2, 3, 4
Force caused by shear stresses
kN
vc
Cutting velocity perpendicular on the blade edge
m/s
w
Width of the blade of blade element
m
W1, 2, 3, 4
Pore pressure forces
kN
z
Water depth
m
α
Blade angle
rad
β
Shear angle
rad
θ
Wedge angle
rad
ε
Volume strain
%
φ
Internal friction angle
rad
δ, δe
External friction angle, mobilized effective external friction angle
rad
ρw
Water density
ton/m3
λ
Angle of internal friction between wedge and layer cut
rad
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Clay cutting is dominated by cohesive and adhesive forces. Pore pressure forces, gravitational forces and inertial forces do not play a role or can be neglected. Clay cutting is regarded to be an undrained process resulting in the φ=0 concept, meaning that the internal and external friction angles can be considered to be zero. Because of the absence of internal and external friction angles, the sine of the sum of the 4 angles in the denominator of the equation for the cutting forces will less likely approach or exceed 180 degrees, resulting in very large or even negative forces. In clay only the blade angle and the shear angle play a role. Now the shear angle will in general be larger in the clay cutting process compared with the sand cutting process, still very large blade angles are required in order to approach the 180 degrees. The shear angle may have values of 30-50 degrees for a blade angle of 90 degrees, still not approaching the total of 180 degrees enough. Blade angles of around 150 degrees will be required to have a sum approaching 180 degrees. In normal dredging the blade angles will be up to about 60 degrees, but the front of a drag head of a trailing suction hopper dredge has an angle larger than 90 degrees, also in the problem of ice berg scour large angles may occur and usually tunnel boring machines have blades with large blade angles. So the problem of having large blade angles is relevant and the transition from the no-wedge mechanism to the wedge mechanism is of interest in engineering practice. Figure 13-1 shows the definitions of the wedge mechanism.
Definitions:
1. A: The wedge tip.
2. B: End of the shear plane.
3. C: The blade top.
4. D: The blade tip.
5. A-B: The shear plane.
6. A-C: The wedge surface.
7. A-D: The wedge bottom.
8. D-C: The blade surface.
9. hb: The height of the blade.
10. hi: The thickness of the layer cut.
11. vc: The cutting velocity.
12. α: The blade angle.
13. β: The shear angle.
14. Fh: The horizontal force, the arrow gives the positive direction.
15. Fv: The vertical force, the arrow gives the positive direction.
13.02: The Equilibrium of Forces
Figure 13-2 illustrates the forces on the layer of soil cut. The forces shown are valid for clay.
The forces acting on the layer A-B are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force C1 as a result of pure cohesion $\ \tau_\mathrm{c}$ or shear strength. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ with the area of the shear plane.
3. A force normal to the pseudo blade N2, resulting from the effective grain stresses.
4. A shear force C2 as a result of the mobilized cohesion between the soil and the wedge $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the wedge.
The forces acting on the wedge front or pseudo blade A-C when cutting clay, can be distinguished as (see Figure 13-3):
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force C2 as a result of the cohesion between the layer cut and the pseudo blade $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the pseudo blade.
The forces acting on the wedge bottom A-D when cutting clay, can be distinguished as:
1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.
2. A shear force C3 as a result of the cohesion between the wedge bottom and the undisturbed soil $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the wedge bottom and the undisturbed soil.
The forces acting on a straight blade C-D when cutting soil (see Figure 13-4), can be distinguished as:
1. A force normal to the blade N4resulting from the effective grain stresses.
2. A shear force as a result of pure adhesion between the soil and the blade $\ \tau_\mathrm{a}$. This force can be calculated by multiplying the adhesive shear strength $\ \tau_\mathrm{a}$ of the soil with the contact area between the soil and the blade.
The horizontal equilibrium of forces on the layer cut:
$\ \sum \mathbf{F}_{\mathbf{h}}=\mathbf{N}_{\mathbf{1}} \cdot \sin (\beta)+\mathbf{C}_{\mathbf{1}} \cdot \cos (\beta)-\mathbf{C}_{2} \cdot \cos (\theta)-\mathbf{N}_{2} \cdot \sin (\theta)=0\tag{13-1}$
The vertical equilibrium of forces on the layer cut:
$\ \sum \mathbf{F}_{\mathbf{v}}=-\mathbf{N}_{\mathbf{1}} \cdot \cos (\beta)+\mathbf{C}_{\mathbf{1}} \cdot \sin (\beta)+\mathbf{C}_{2} \cdot \sin (\theta)-\mathbf{N}_{2} \cdot \cos (\theta)=\mathbf{0}\tag{13-2}$
The force N1 on the shear plane is now:
$\ \mathbf{N}_{1}=\frac{-\mathbf{C}_{1} \cdot \cos (\theta+\beta)+\mathbf{C}_{2}}{\sin (\theta+\beta)}\tag{13-3}$
The force N2 on the pseudo blade is now:
$\ \mathbf{N}_{2}=\frac{+\mathbf{C}_{1}-\mathbf{C}_{2} \cdot \cos (\theta+\beta)}{\sin (\theta+\beta)}\tag{13-4}$
From equation (13-4) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathbf{F}_{\mathbf{h}}=+\mathbf{N}_{\mathbf{2}} \cdot \sin (\theta)+\mathbf{C}_{\mathbf{2}} \cdot \cos (\theta)\tag{13-5}$
$\ \mathbf{F}_{v}=+\mathbf{N}_{2} \cdot \cos (\theta)-\mathbf{C}_{2} \cdot \sin (\theta)\tag{13-6}$
Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived. The adhesive force does not have to be mobilized 100%, while this force could have both directions, depending on the equilibrium of forces and the equilibrium of moments. So for now the mobilized adhesive force Ais used in the equations.
The horizontal equilibrium of forces on the wedge:
$\ \mathrm{\sum F_{h}=N_{4} \cdot \sin (\alpha)+A_{m} \cdot \cos (\alpha)-C_{2} \cdot \cos (\theta)-N_{2} \cdot \sin (\theta)-C_{3}=0}\tag{13-7}$
The vertical equilibrium of forces on the wedge:
$\ \sum \mathbf{F}_{\mathbf{v}}=\mathbf{N}_{4} \cdot \cos (\alpha)-\mathbf{A}_{\mathbf{m}} \cdot \sin (\alpha)-\mathbf{C}_{2} \cdot \sin (\theta)+\mathbf{N}_{2} \cdot \cos (\theta)-\mathbf{N}_{3}=\mathbf{0}\tag{13-8}$
To derive N4:
Multiply the horizontal equilibrium equation with sin(α).
$\ \mathbf{N}_{4} \cdot \sin (\alpha) \cdot \sin (\alpha)+\mathbf{A}_{\mathbf{m}} \cdot \cos (\alpha) \cdot \sin (\alpha)-\mathbf{C}_{2} \cdot \cos (\theta) \cdot \sin (\alpha)-\mathbf{N}_{2} \cdot \sin (\theta) \cdot \sin (\alpha)-\mathbf{C}_{3} \cdot \sin (\alpha)=0\tag{13-9}$
Multiply the vertical equilibrium equation with cos(α).
$\ \mathbf{N}_{4} \cdot \cos (\alpha) \cdot \cos (\alpha)-\mathbf{A}_{\mathbf{m}} \cdot \sin (\alpha) \cdot \cos (\alpha)-\mathbf{C}_{2} \cdot \sin (\theta) \cdot \cos (\alpha) +\mathbf{N}_{2} \cdot \cos (\theta) \cdot \cos (\alpha)-\mathbf{N}_{3} \cdot \cos (\alpha)=\mathbf{0}\tag{13-10}$
Now add up the two resulting equations in order to get an expression for N4.
$\ \mathbf{N}_{4}=\mathbf{C}_{2} \cdot \sin (\alpha+\theta)-\mathbf{N}_{2} \cdot \cos (\alpha+\theta)+\mathbf{C}_{3} \cdot \sin (\alpha)+\mathbf{N}_{3} \cdot \cos (\alpha)\tag{13-11}$
The mobilized adhesive force Am can be derived according to:
First multiply the horizontal equilibrium equation with cos(α).
$\ \mathbf{N}_{4} \cdot \sin (\alpha) \cdot \cos (\alpha)+\mathbf{A}_{\mathbf{m}} \cdot \cos (\alpha) \cdot \cos (\alpha)-\mathbf{C}_{2} \cdot \cos (\theta) \cdot \cos (\alpha) -\mathbf{N}_{2} \cdot \sin (\theta) \cdot \cos (\alpha)-\mathbf{C}_{3} \cdot \cos (\alpha)=0\tag{13-12}$
Now multiply the vertical equilibrium equation with sin(α):
$\ \mathbf{N}_{4} \cdot \cos (\alpha) \cdot \sin (\alpha)-\mathbf{A}_{\mathbf{m}} \cdot \sin (\alpha) \cdot \sin (\alpha)-\mathbf{C}_{2} \cdot \sin (\theta) \cdot \sin (\alpha) +\mathbf{N}_{2} \cdot \cos (\theta) \cdot \sin (\alpha)-\mathbf{N}_{3} \cdot \sin (\alpha)=\mathbf{0}\tag{13-13}$
Subtracting the two resulting equations gives the equation for the mobilized adhesive force.
$\ \mathbf{A}_{\mathbf{m}}=\mathbf{C}_{2} \cdot \cos (\alpha+\theta)+\mathbf{N}_{2} \cdot \sin (\alpha+\theta)+\mathbf{C}_{3} \cdot \cos (\alpha)-\mathbf{N}_{3} \cdot \sin (\alpha)\tag{13-14}$
This can also be rewritten as an equation for the normal force N3 on the bottom of the wedge.
$\ \mathbf{N}_{\mathbf{3}}=\mathbf{N}_{4} \cdot \cos (\alpha)-\mathbf{A}_{\mathbf{m}} \cdot \sin (\alpha)-\mathbf{C}_{2} \cdot \sin (\theta)+\mathbf{N}_{2} \cdot \cos (\theta)\tag{13-15}$
Since both the mobilized adhesive force Am and the normal force on the bottom of the wedge N3 are unknowns, an additional condition has to be found. The wedge angle θ however is also an unknown, requiring an additional condition. Apparently N4 and Am are independent of each other.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/13%3A_A_Wedge_in_Clay_Cutting/13.01%3A_Introduction.txt
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The first additional condition is the equilibrium of moments of the wedge. Since the wedge is not subject to rotational accelerations in a stationary cutting process, the sum of the moments around any point on the wedge has to be zero. Here the tip of the blade is chosen for this point. First the equilibrium of moments is solved in order to find a relation between N3 and N4.
To solve the equilibrium of moments the lengths of the sides of the wedge and arms of the forces have to be determined.
The length of the shear plane A-B is:
$\ \mathrm{L}_{1}=\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\tag{13-16}$
The length of the front of the wedge A-C is:
$\ \mathrm{L}_{2}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\theta)}\tag{13-17}$
The length of the bottom of the wedge A-D is:
$\ \mathrm{L}_{3}=\mathrm{h}_{\mathrm{b}} \cdot\left(\frac{\mathrm{1}}{\tan (\theta)}-\frac{\mathrm{1}}{\tan (\alpha)}\right)\tag{13-18}$
The length of the blade C-D is:
$\ \mathrm{L}_{4}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\tag{13-19}$
The distance of the tip of the blade perpendicular to the front of the wedge is:
$\ \mathrm{L}_{5}=\mathrm{L}_{3} \cdot \sin (\theta)\tag{13-20}$
The distance from point to the intersection point of the line going from the tip of the blade perpendicular to the front of the blade is.
$\ \mathrm{L}_{6}=\mathrm{L}_{3} \cdot \cos (\theta)\tag{13-21}$
The distance of the acting point of the force N2 to the intersection point of the line going from the tip of the blade perpendicular to the front of the blade is:
$\ \mathrm{L}_{7}=\mathrm{L}_{6}-\mathrm{R}_{2}\tag{13-22}$
R2 follows from the equilibrium of moments on the layer cut, assuming the forces on the shear plane act at half the length of the shear plane.
$\ \mathrm{N}_{1} \cdot \mathrm{R}_{1}=\mathrm{N}_{2} \cdot \mathrm{R}_{2}\tag{13-23}$
Now the equilibrium equation of moments can be derived according to:
$\ \sum \mathrm{M}=\frac{\mathrm{N}_{4} \cdot \mathrm{L}_{4}}{2}-\frac{\mathrm{N}_{3} \cdot \mathrm{L}_{3}}{2}+\mathrm{N}_{2} \cdot \mathrm{L}_{7}-\mathrm{C}_{2} \cdot \mathrm{L}_{5}=\mathrm{0}\tag{13-24}$
Both equation (13-11) and equation (13-24) don not contain the mobilized adhesive force Am, giving the possibility to solve the two unknowns N3 and N4. To solve the normal force N3 first an expression for the normal force N4 has to be derived based on the equilibrium of moments.
$\ \mathrm{N}_{4}=\frac{\mathrm{2} \cdot\left(\mathrm{N}_{\mathrm{3}} \cdot \mathrm{L}_{\mathrm{3}} / \mathrm{2}-\mathrm{N}_{\mathrm{2}} \cdot \mathrm{L}_{\mathrm{7}}+\mathrm{C}_{\mathrm{2}} \cdot \mathrm{L}_{\mathrm{5}}\right)}{\mathrm{L}_{\mathrm{4}}}\tag{13-25}$
Equation (13-11) and equation (13-25) should give the same result for the normal force N4, thus:
$\ \begin{array}{left}\frac{2 \cdot\left(\mathrm{N}_{3} \cdot \mathrm{L}_{3} / 2-\mathrm{N}_{2} \cdot \mathrm{L}_{7}+\mathrm{C}_{2} \cdot \mathrm{L}_{5}\right)}{\mathrm{L}_{4}}=\ \mathrm{C}_{2} \cdot \sin (\alpha+\theta)-\mathrm{N}_{2} \cdot \cos (\alpha+\theta)+\mathrm{C}_{3} \cdot \sin (\alpha)+\mathrm{N}_{3} \cdot \cos (\alpha)\end{array}\tag{13-26}$
This can be written as:
$\ \begin{array}{left} \mathrm{N}_{3} \cdot\left(\frac{\mathrm{L}_{3}}{\mathrm{L}_{4}}-\cos (\alpha)\right) &=\mathrm{N}_{2} \cdot\left(\frac{\mathrm{L}_{7}}{2 \cdot \mathrm{L}_{4}}-\cos (\alpha+\theta)\right) \ &+\mathrm{C}_{2} \cdot\left(-\frac{\mathrm{L}_{5}}{\mathrm{2} \cdot \mathrm{L}_{4}}+\sin (\alpha+\theta)\right)+\mathrm{C}_{3} \cdot \sin (\alpha) \end{array}\tag{13-27}$
Now N3 can be expressed in a number of known variables according to:
$\ \begin{array}{left}\mathrm{N}_{3}&=\frac{\mathrm{N}_{2} \cdot\left(\frac{\mathrm{L}_{7}}{\mathrm{2} \cdot \mathrm{L}_{4}}-\cos (\alpha+\theta)\right)+\mathrm{C}_{2} \cdot\left(-\frac{\mathrm{L}_{5}}{2 \cdot \mathrm{L}_{4}}+\sin (\alpha+\theta)\right)}{\left(\frac{\mathrm{L}_{3}}{\mathrm{L}_{4}}-\cos (\alpha)\right)}\ &+\frac{\mathrm{+C_3\cdot\sin(\alpha)}}{\mathrm{\left(\frac{L_3}{L_4}-cos(\alpha) \right)}}\end{array}\tag{13-28}$
Substituting equation (13-28) in equation (13-11) gives a solution for the normal force N4.
$\ \mathrm{N}_{4}=\mathrm{C}_{2} \cdot \sin (\alpha+\theta)-\mathrm{N}_{2} \cdot \cos (\alpha+\theta)+\mathrm{C}_{3} \cdot \sin (\alpha)+\mathrm{N}_{3} \cdot \cos (\alpha)\tag{13-29}$
Substituting equation (13-28) in equation (13-14) gives a solution for the mobilized adhesion Am.
$\ \mathrm{A}_{\mathrm{m}}=\mathrm{C}_{2} \cdot \cos (\alpha+\theta)+\mathrm{N}_{2} \cdot \sin (\alpha+\theta)+\mathrm{C}_{3} \cdot \cos (\alpha)-\mathrm{N}_{3} \cdot \sin (\alpha)\tag{13-30}$
This results in a horizontal force of:
$\ \mathrm{F}_{\mathrm{h}}=\mathrm{N}_{\mathrm{4}} \cdot \sin (\alpha)+\mathrm{A}_{\mathrm{m}} \cdot \cos (\alpha)\tag{13-31}$
And in a vertical force of:
$\ \mathrm{F}_{\mathrm{v}}=\mathrm{N}_{4} \cdot \cos (\alpha)-\mathrm{A}_{\mathrm{m}} \cdot \sin (\alpha)\tag{13-32}$
Based on the experience with sand cutting it is assumed that the wedge angle θ can be determined by assuming that the horizontal force should be at a minimum for the angle chosen. It is very well possible that the mobilized adhesion is negative for large blade angles.
13.04: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or adhesive shear strength.
kPa
A
Adhesive shear force on the blade.
kN
c, $\tau_\mathrm{c}$
Cohesion or cohesive shear strength.
kPa
C1
Cohesive shear force on the shear plane.
kN
C2
Cohesive shear force on the pseudo blade (front of the wedge).
kN
C3
Cohesive shear force on bottom of the wedge.
kN
Fh
Horizontal cutting force.
kN
Fv
Vertical cutting force.
kN
G1
Weight of the layer cut.
kN
G2
Weight of the wedge.
kN
hb
Blade height.
m
hi
Layer thickness.
m
I
Inertial force on the shear plane.
kN
N1
Normal force on the shear plane.
kN
N2
Normal force on the pseudo blade (front of the wedge).
kN
N3
Normal force on bottom of the wedge.
kN
N4
Normal force on the blade.
kN
K1
Sum of N1 and S1 on the shear plane.
kN
K2
Sum of N2 and S2 on the pseudo blade (front of the wedge).
kN
K3
Sum of N3 and S3 on bottom of the wedge.
kN
K4
Sum of N4 and S4 on the blade.
kN
L1
Length of the shear plane.
m
L2
Length of the pseudo blade (front of the wedge).
m
L3
Length of the bottom of the wedge.
m
L4
Length of the blade.
m
L5
Length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side.
m
L6
Length of the line from point to the intersection point of the previous line with side A-C.
m
L7
Distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line L6 with side A-C.
m
R1
Acting point forces on the shear plane.
m
R2
Acting point forces on the pseudo blade (front of the wedge).
m
R3
Acting point forces on the bottom of the wedge.
m
R4
Acting point forces on the blade.
m
S1
Shear (friction) force on the shear plane.
kN
S2
Shear (friction) force on the pseudo blade (front of the wedge).
kN
S3
Shear (friction) force on the bottom of the wedge.
kN
S4
Shear (friction) force on the blade.
kN
W1
Pore pressure force on the shear plane.
kN
W2
Pore pressure force on the pseudo blade (front of the wedge).
kN
W3
Pore pressure force on the bottom of the wedge.
kN
W4
Pore pressure force on the blade.
kN
vc
Cutting velocity.
m/sec
α
Blade angle.
°
β
Shear angle.
°
θ
Wedge angle.
°
φ
Internal friction angle.
°
δ
External friction angle.
°
λ
Internal friction angle on pseudo blade.
°
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/13%3A_A_Wedge_in_Clay_Cutting/13.03%3A_The_Equilibrium_of_Moments.txt
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For completeness of the overview the equations for the cutting of the wedge mechanism for atmospheric rock are given here without further explanation.
Definitions:
1. A: The wedge tip.
2. B: End of the shear plane.
3. C: The blade top.
4. D: The blade tip.
5. A-B: The shear plane.
6. A-C: The wedge surface.
7. A-D: The wedge bottom.
8. D-C: The blade surface.
9. hb: The height of the blade.
10. hi: The thickness of the layer cut.
11. vc: The cutting velocity.
12. α: The blade angle.
13. β: The shear angle.
14. Fh: The horizontal force, the arrow gives the positive direction.
15. Fv: The vertical force, the arrow gives the positive direction.
14.02: The Equilibrium of Forces
Figure 14-2 illustrates the forces on the layer of soil cut. The forces shown are valid in general for each type of soil.
The forces acting on the layer A-B are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A shear force C1 as a result of pure cohesion $\ \tau_\mathrm{c}$ or shear strength. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ with the area of the shear plane.
4. A force normal to the pseudo blade N2, resulting from the effective grain stresses.
5. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
6. A shear force C2 as a result of the mobilized cohesion between the soil and the wedge $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the wedge.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{14-1}$
The forces acting on the wedge front or pseudo blade A-C when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
3. A shear force C2 as a result of the cohesion between the layer cut and the pseudo blade $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the pseudo blade.
These forces are shown in Figure 14-3. If the forces N2 and Sare combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{14-2}$
The forces acting on the wedge bottom A-D when cutting soil, can be distinguished as:
1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.
2. A shear force S3 as a result of the soil/soil friction N3·tan(φ) between the wedge bottom and the undisturbed soil.
3. A shear force C3 as a result of the cohesion between the wedge bottom and the undisturbed soil $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the wedge bottom and the undisturbed soil.
The normal force N3 and the shear force S3 can be combined to a resulting grain force K3.
$\ \mathrm{K}_{3}=\sqrt{\mathrm{N}_{\mathrm{3}}^{2}+\mathrm{S}_{\mathrm{3}}^{\mathrm{2}}}\tag{14-3}$
The forces acting on a straight blade C-D when cutting soil (see Figure 14-4), can be distinguished as:
1. A force normal to the blade N4resulting from the effective grain stresses.
2. A shear force S4 as a result of the soil/steel friction N4·tan(δ).
The normal force N4 and the shear force S4 can be combined to a resulting grain force K4.
$\ \mathrm{K}_{4}=\sqrt{\mathrm{N}_{4}^{2}+\mathrm{S}_{4}^{2}}\tag{14-4}$
The horizontal equilibrium of forces on the layer cut:
$\ \sum \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{\mathrm{1}} \cdot \sin (\boldsymbol{\beta}+\boldsymbol{\varphi})+\mathrm{C}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta})-\mathrm{C}_{\mathrm{2}} \cdot \cos (\boldsymbol{\theta})-\mathrm{K}_{\mathrm{2}} \cdot \sin (\boldsymbol{\theta}+\lambda)=\mathrm{0}\tag{14-5}$
The vertical equilibrium of forces on the layer cut:
$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta}+\varphi)+\mathrm{C}_{\mathrm{1}} \cdot \sin (\boldsymbol{\beta})+\mathrm{C}_{\mathrm{2}} \cdot \sin (\boldsymbol{\theta})-\mathrm{K}_{\mathrm{2}} \cdot \mathrm{\operatorname { cos } ( \boldsymbol { \theta } + \lambda )}=\mathrm{0}\tag{14-6}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}=\frac{-\mathrm{C}_{1} \cdot \cos (\theta+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{14-7}$
The force K2 on the pseudo blade is now:
$\ \mathrm{K}_{2}=\frac{+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\theta+\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{14-8}$
From equation (14-8) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{2} \cdot \sin (\theta+\lambda)+\mathrm{C}_{2} \cdot \cos (\theta)\tag{14-9}$
$\ \mathrm{F}_{v}=\mathrm{K}_{2} \cdot \cos (\theta+\lambda)-\mathrm{C}_{2} \cdot \sin (\theta)\tag{14-10}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}=\frac{-\mathrm{C}_{1} \cdot \cos (\alpha+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\alpha+\beta+\lambda+\varphi)} \cdot \cos (\varphi)\tag{14-11}$
The normal force on the pseudo blade is now:
$\ \mathrm{N}_{2}=\frac{+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\lambda+\varphi)} \cdot \cos (\lambda)\tag{14-12}$
Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived. The horizontal equilibrium of forces on the wedge is:
$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{h}}=&-\mathrm{K}_{4} \cdot \sin (\alpha+\delta)+\mathrm{K}_{3} \cdot \sin (\varphi)+\mathrm{C}_{3} \ &+\mathrm{C}_{2} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda)=\mathrm{0} \end{array}\tag{14-13}$
The vertical equilibrium of forces on the wedge is:
$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{v}}=&-\mathrm{K}_{4} \cdot \cos (\alpha+\delta)-\mathrm{K}_{3} \cdot \cos (\varphi) \ &-\mathrm{C}_{2} \cdot \sin (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0} \end{array}\tag{14-14}$
The unknowns in this equation are K3 and K4, since K2 has already been solved. Two other unknowns are, the external friction angle δ, since the external friction does not have to be fully mobilized, and the wedge angle θ. These 2 additional unknowns require 2 additional conditions in order to solve the problem. One additional condition is the equilibrium of moments of the wedge, a second condition the principle of minimum required cutting energy. Depending on whether the soil pushes upwards or downwards against the blade, the mobilization factor is between -1 and +1.
The force K3 on the bottom of the wedge is now:
$\ \mathrm{K}_{3}=\frac{+\mathrm{K}_{2} \cdot \sin (\alpha+\delta-\theta-\lambda)+\mathrm{C}_{3} \cdot \cos (\alpha+\delta)-\mathrm{C}_{2} \cdot \cos (\alpha+\delta-\theta)}{\sin (\alpha+\delta+\varphi)}\tag{14-15}$
The force K4 on the blade is now:
$\ \mathrm{K}_{4}=\frac{+\mathrm{K}_{2} \cdot \sin (\theta+\lambda+\varphi)+\mathrm{C}_{3} \cdot \cos (\varphi)+\mathrm{C}_{2} \cdot \cos (\theta+\varphi)}{\sin (\alpha+\delta+\varphi)}\tag{14-16}$
This results in a horizontal force on the blade of:
$\ \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{4} \cdot \sin (\alpha+\delta)\tag{14-17}$
And in a vertical force on the blade of:
$\ \mathrm{F}_{\mathrm{v}}=\mathrm{K}_{4} \cdot \cos (\alpha+\delta)\tag{14-18}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/14%3A_A_Wedge_in_Atmospheric_Rock_Cutting/14.01%3A_Introduction.txt
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In order to solve the problem, also the equilibrium of moments is required, since the wedge is not subject to rotational acceleration. The equilibrium of moments can be taken around each point of the wedge. Here the tip of the blade is chosen. The advantage of this is that a number of forces do not contribute to the moments on the wedge.
In order to derive the equilibrium of moments equation the arms of all the forces contributing to this equilibrium have to be known. Since these arms depend on the length of all the sides in the cutting process, first these lengths are determined. The length of the shear plane A-B is:
$\ \mathrm{L}_{1}=\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\tag{14-19}$
The length of the pseudo blade A-C is:
$\ \mathrm{L}_{2}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\theta)}\tag{14-20}$
The length of the bottom of the wedge A-D is:
$\ \mathrm{L}_{3}=\mathrm{h}_{\mathrm{b}} \cdot\left(\frac{\mathrm{1}}{\tan (\theta)}-\frac{1}{\tan (\alpha)}\right)\tag{14-21}$
The length of the blade C-D is:
$\ \mathrm{L}_{4}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\tag{14-22}$
The length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side is:
$\ \mathrm{L}_{5}=\mathrm{L}_{3} \cdot \sin (\theta)\tag{14-23}$
The length of the line from point to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{\mathrm{6}}=\mathrm{L}_{\mathrm{3}} \cdot \cos (\theta)\tag{14-24}$
The distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{7}=\mathrm{L}_{6}-\mathrm{R}_{2}\tag{14-25}$
The values of the acting points R2R3 and R4 follow from calculated or estimated stress distributions.
The equilibrium of moments is now:
$\ \sum \mathrm{M}=\mathrm{N}_{4} \cdot \mathrm{R}_{4}-\mathrm{N}_{3} \cdot \mathrm{R}_{3}+\mathrm{N}_{2} \cdot \mathrm{L}_{7}-\left(\mathrm{S}_{2}+\mathrm{C}_{2}\right) \cdot \mathrm{L}_{5}=\mathrm{0}\tag{14-26}$
14.04: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or adhesive shear strength.
kPa
A
Adhesive shear force on the blade.
kN
c, $\ \tau_\mathrm{c}$
Cohesion or cohesive shear strength.
kPa
C1
Cohesive shear force on the shear plane.
kN
C2
Cohesive shear force on the pseudo blade (front of the wedge).
kN
C3
Cohesive shear force on bottom of the wedge.
kN
Fh
Horizontal cutting force.
kN
Fv
Vertical cutting force.
kN
G1
Weight of the layer cut.
kN
G2
Weight of the wedge.
kN
hb
Blade height.
m
hi
Layer thickness.
m
I
Inertial force on the shear plane.
kN
N1
Normal force on the shear plane.
kN
N2
Normal force on the pseudo blade (front of the wedge).
kN
N3
Normal force on bottom of the wedge.
kN
N4
Normal force on the blade.
kN
K1
Sum of N1 and S1 on the shear plane.
kN
K2
Sum of N2 and S2 on the pseudo blade (front of the wedge).
kN
K3
Sum of N3 and S3 on bottom of the wedge.
kN
K4
Sum of N4 and S4 on the blade.
kN
L1
Length of the shear plane.
m
L2
Length of the pseudo blade (front of the wedge).
m
L3
Length of the bottom of the wedge.
m
L4
Length of the blade.
m
L5
Length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side.
m
L6
Length of the line from point to the intersection point of the previous line with side A-C.
m
L7
Distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line L6 with side A-C.
m
R1
Acting point forces on the shear plane.
m
R2
Acting point forces on the pseudo blade (front of the wedge).
m
R3
Acting point forces on the bottom of the wedge.
m
R4
Acting point forces on the blade.
m
S1
Shear (friction) force on the shear plane.
kN
S2
Shear (friction) force on the pseudo blade (front of the wedge).
kN
S3
Shear (friction) force on the bottom of the wedge.
kN
S4
Shear (friction) force on the blade.
kN
W1
Pore pressure force on the shear plane.
kN
W2
Pore pressure force on the pseudo blade (front of the wedge).
kN
W3
Pore pressure force on the bottom of the wedge.
kN
W4
Pore pressure force on the blade.
kN
vc
Cutting velocity.
m/sec
α
Blade angle.
°
β
Shear angle.
°
θ
Wedge angle.
°
φ
Internal friction angle.
°
δ
External friction angle.
°
λ
Internal friction angle on pseudo blade.
°
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/14%3A_A_Wedge_in_Atmospheric_Rock_Cutting/14.03%3A_The_Equilibrium_of_Moments.txt
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For completeness of the overview the equations for the cutting of the wedge mechanism for hyperbaric rock are given here without further explanation.
Definitions:
1. A: The wedge tip.
2. B: End of the shear plane.
3. C: The blade top
4. D: The blade tip.
5. A-B: The shear plane.
6. A-C: The wedge surface.
7. A-D: The wedge bottom.
8. D-C: The blade surface.
9. hb: The height of the blade.
10. hi: The thickness of the layer cut.
11. vc: The cutting velocity.
12. α: The blade angle.
13. β: The shear angle.
14. Fh: The horizontal force, the arrow gives the positive direction.
15. Fv: The vertical force, the arrow gives the positive direction.
15.02: The Equilibrium of Forces
Figure 15-2 illustrates the forces on the layer of soil cut. The forces shown are valid in general for each type of soil.
The forces acting on the layer A-B are:
1. A normal force acting on the shear surface N1, resulting from the effective grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A shear force C1 as a result of pure cohesion $\ \tau_\mathrm{c}$ or shear strength. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ with the area of the shear plane.
5. A force normal to the pseudo blade N2, resulting from the effective grain stresses.
6. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
7. A shear force C2 as a result of the mobilized cohesion between the soil and the wedge $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the wedge.
8. A force W2 as a result of water under pressure on the wedge.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{15-1}$
The forces acting on the wedge front or pseudo blade A-C when cutting soil, can be distinguished as:
1. A force normal to the blade N2, resulting from the effective grain stresses.
2. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.
3. A shear force C2 as a result of the cohesion between the layer cut and the pseudo blade $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the soil and the pseudo blade.
4. A force W2 as a result of water under pressure on the pseudo blade A-C.
These forces are shown in Figure 15-3. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{15-2}$
The forces acting on the wedge bottom A-D when cutting soil, can be distinguished as:
1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.
2. A shear force S3 as a result of the soil/soil friction N3·tan(φ) between the wedge bottom and the undisturbed soil.
3. A shear force C3 as a result of the cohesion between the wedge bottom and the undisturbed soil $\ \tau_\mathrm{c}$. This force can be calculated by multiplying the cohesive shear strength $\ \tau_\mathrm{c}$ of the soil with the contact area between the wedge bottom and the undisturbed soil.
4. A force W3 as a result of water under pressure on the wedge bottom A-D.
The normal force N3 and the shear force S3 can be combined to a resulting grain force K3.
$\ \mathrm{K}_{3}=\sqrt{\mathrm{N}_{3}^{2}+\mathrm{S}_{3}^{2}}\tag{15-3}$
The forces acting on a straight blade C-D when cutting soil (see Figure 15-4), can be distinguished as:
1. A force normal to the blade N4resulting from the effective grain stresses.
2. A shear force S4 as a result of the soil/steel friction N4·tan(δ).
3. A force W4 as a result of water under pressure on the blade.
The normal force N4 and the shear force S4 can be combined to a resulting grain force K4.
$\ \mathrm{K}_{4}=\sqrt{\mathrm{N}_{4}^{2}+\mathrm{S}_{4}^{2}}\tag{15-4}$
The horizontal equilibrium of forces on the layer cut:
$\ \begin{array}{left} \mathrm{\sum F_{h}}=&\mathrm{K_{1} \cdot \sin (\beta+\varphi)-W_{1} \cdot \sin (\beta)+C_{1} \cdot \cos (\beta)}\ &\mathrm{-C_{2} \cdot \cos (\theta)+W_{2} \cdot \sin (\theta)-K_{2} \cdot \sin (\theta+\lambda)=0}\end{array}\tag{15-5}$
The vertical equilibrium of forces on the layer cut:
$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{v}}=&-\mathrm{K}_{1} \cdot \cos (\beta+\varphi)+\mathrm{W}_{1} \cdot \cos (\beta)+\mathrm{C}_{1} \cdot \sin (\beta) \ &+\mathrm{C}_{2} \cdot \sin (\theta)+\mathrm{W}_{2} \cdot \cos (\theta)-\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0} \end{array}\tag{15-6}$
The force K1 on the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\lambda)+\mathrm{W}_{1} \cdot \sin (\theta+\beta+\lambda)-\mathrm{C}_{1} \cdot \cos (\theta+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{15-7}$
The force K2 on the pseudo blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\theta+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\theta+\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{15-8}$
From equation (15-8) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{\mathrm{2}} \cdot \sin (\theta)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda)+\mathrm{C}_{2} \cdot \cos (\theta)\tag{15-9}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{\mathrm{2}} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)-\mathrm{C}_{2} \cdot \sin (\theta)\tag{15-10}$
The normal force on the shear plane is now:
$\ \begin{array}{left} \mathrm{N}_{1}=& \frac{\mathrm{W}_{2} \cdot \sin (\lambda)+\mathrm{W}_{1} \cdot \sin (\theta+\beta+\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\varphi) \ &+\frac{-\mathrm{C}_{1} \cdot \cos (\theta+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\varphi) \end{array}\tag{15-11}$
The normal force on the pseudo blade is now:
$\ \begin{array}{left} \mathrm{N}_{2}=& \frac{\mathrm{W}_{2} \cdot \sin (\theta+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\lambda) \ &+\frac{+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\theta+\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)} \cdot \cos (\lambda) \end{array}\tag{15-12}$
Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived. The horizontal equilibrium of forces on the wedge is:
$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{h}}=& \mathrm{W}_{4} \cdot \sin (\alpha)-\mathrm{K}_{4} \cdot \sin (\alpha+\delta)+\mathrm{K}_{3} \cdot \sin (\varphi) \ &+\mathrm{C}_{3}-\mathrm{W}_{2} \cdot \sin (\theta)+\mathrm{C}_{2} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda)=\mathrm{0} \end{array}\tag{15-13}$
The vertical equilibrium of forces on the wedge is:
$\ \begin{array}{left}\sum \mathrm{F}_{\mathrm{v}}=\mathrm{W}_{4} \cdot \cos (\alpha)-\mathrm{K}_{4} \cdot \cos (\alpha+\delta)+\mathrm{W}_{3}-\mathrm{K}_{3} \cdot \cos (\varphi)\ -\mathrm{W}_{2} \cdot \cos (\theta)-\mathrm{C}_{2} \cdot \sin (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0}\end{array}\tag{15-4}$
The unknowns in this equation are K3 and K4, since K2 has already been solved. Two other unknowns are, the external friction angle δ, since the external friction does not have to be fully mobilized, and the wedge angle θ. These 2 additional unknowns require 2 additional conditions in order to solve the problem. One additional condition is the equilibrium of moments of the wedge, a second condition the principle of minimum required cutting energy. Depending on whether the soil pushes upwards or downwards against the blade, the mobilization factor is between -1 and +1.
The force K3 on the bottom of the wedge is now:
$\ \begin{array}{left} \mathrm{K}_{3}=&\frac{-\mathrm{W}_{2} \cdot \sin (\alpha+\delta-\theta)+\mathrm{W}_{3} \cdot \sin (\alpha+\delta)+\mathrm{W}_{4} \cdot \sin (\delta)}{\sin (\alpha+\delta+\varphi)} \& \frac{+\mathrm{K}_{2} \cdot \sin (\alpha+\delta-\theta-\lambda)+\mathrm{C}_{3} \cdot \cos (\alpha+\delta)-\mathrm{C}_{2} \cdot \cos (\alpha+\delta-\theta)}{\sin (\alpha+\delta+\varphi)} & \end{array}\tag{15-15}$
The force K4 on the blade is now:
$\ \begin{array}{left} \mathrm{K}_{4}=& \frac{-\mathrm{W}_{2} \cdot \sin (\theta+\varphi)+\mathrm{W}_{3} \cdot \sin (\varphi)+\mathrm{W}_{4} \cdot \sin (\alpha+\varphi)}{\sin (\alpha+\delta+\varphi)} \ & \frac{+\mathrm{K}_{2} \cdot \sin (\theta+\lambda+\varphi)+\mathrm{C}_{3} \cdot \cos (\varphi)+\mathrm{C}_{2} \cdot \cos (\theta+\varphi)}{\sin (\alpha+\delta+\varphi)} \end{array}\tag{15-16}$
This results in a horizontal force on the blade of:
$\ \mathrm{F}_{\mathrm{h}}=-\mathrm{W}_{\mathrm{4}} \cdot \sin (\alpha)+\mathrm{K}_{4} \cdot \sin (\alpha+\delta)\tag{15-17}$
And in a vertical force on the blade of:
$\ \mathrm{F}_{\mathrm{v}}=-\mathrm{W}_{\mathrm{4}} \cdot \cos (\alpha)+\mathrm{K}_{4} \cdot \cos (\alpha+\delta)\tag{15-18}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/15%3A_A_Wedge_in_Hyperbaric_Rock_Cutting/15.01%3A_Introduction.txt
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In order to solve the problem, also the equilibrium of moments is required, since the wedge is not subject to rotational acceleration. The equilibrium of moments can be taken around each point of the wedge. Here the tip of the blade is chosen. The advantage of this is that a number of forces do not contribute to the moments on the wedge. In order to derive the equilibrium of moments equation the arms of all the forces contributing to this equilibrium have to be known. Since these arms depend on the length of all the sides in the cutting process, first these lengths are determined. The length of the shear plane A-B is:
$\ \mathrm{L}_{1}=\frac{\mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\tag{15-19}$
The length of the pseudo blade A-C is:
$\ \mathrm{L}_{2}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\theta)}\tag{15-20}$
The length of the bottom of the wedge A-D is:
$\ \mathrm{L}_{3}=\mathrm{h}_{\mathrm{b}} \cdot\left(\frac{\mathrm{1}}{\tan (\theta)}-\frac{1}{\tan (\alpha)}\right)\tag{15-21}$
The length of the blade C-D is:
$\ \mathrm{L}_{4}=\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}\tag{15-22}$
The length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side is:
$\ \mathrm{L}_{5}=\mathrm{L}_{3} \cdot \sin (\theta)\tag{15-23}$
The length of the line from point to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{6}=\mathrm{L}_{3} \cdot \cos (\theta)\tag{15-24}$
The distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line with side A-C is:
$\ \mathrm{L}_{7}=\mathrm{L}_{6}-\mathrm{R}_{2}\tag{15-25}$
The values of the acting points R2R3 and R4 follow from calculated or estimated stress distributions.
The equilibrium of moments is now:
$\ \begin{array}{left} \sum \mathrm{M}=&\left(\mathrm{N}_{4}-\mathrm{W}_{4}\right) \cdot \mathrm{R}_{4}-\left(\mathrm{N}_{3}-\mathrm{W}_{3}\right) \cdot \mathrm{R}_{3} \ &+\left(\mathrm{N}_{2}-\mathrm{W}_{2}\right) \cdot \mathrm{L}_{7}-\left(\mathrm{S}_{2}+\mathrm{C}_{2}\right) \cdot \mathrm{L}_{5}=\mathrm{0} \end{array}\tag{15-26}$
15.04: Nomenclature
a, $\ \tau_\mathrm{a}$
Adhesion or adhesive shear strength.
kPa
A
Adhesive shear force on the blade.
kN
c, $\ \tau_\mathrm{c}$
Cohesion or cohesive shear strength.
kPa
C1
Cohesive shear force on the shear plane.
kN
C2
Cohesive shear force on the pseudo blade (front of the wedge).
kN
C3
Cohesive shear force on bottom of the wedge.
kN
Fh
Horizontal cutting force.
kN
Fv
Vertical cutting force.
kN
G1
Weight of the layer cut.
kN
G2
Weight of the wedge.
kN
hb
Blade height.
m
hi
Layer thickness.
m
I
Inertial force on the shear plane.
kN
N1
Normal force on the shear plane.
kN
N2
Normal force on the pseudo blade (front of the wedge).
kN
N3
Normal force on bottom of the wedge.
kN
N4
Normal force on the blade.
kN
K1
Sum of N1 and S1 on the shear plane.
kN
K2
Sum of N2 and S2 on the pseudo blade (front of the wedge).
kN
K3
Sum of N3 and S3 on bottom of the wedge.
kN
K4
Sum of N4 and S4 on the blade.
kN
L1
Length of the shear plane.
m
L2
Length of the pseudo blade (front of the wedge).
m
L3
Length of the bottom of the wedge.
m
L4
Length of the blade.
m
L5
Length of the line from the tip of the blade to the opposite side of the wedge and perpendicular to this side.
m
L6
Length of the line from point to the intersection point of the previous line with side A-C.
m
L7
Distance from the acting point of the pore pressure force on side A-C to the intersection point of the previous line L6 with side A-C.
m
R1
Acting point forces on the shear plane.
m
R2
Acting point forces on the pseudo blade (front of the wedge).
m
R3
Acting point forces on the bottom of the wedge.
m
R4
Acting point forces on the blade.
m
S1
Shear (friction) force on the shear plane.
kN
S2
Shear (friction) force on the pseudo blade (front of the wedge).
kN
S3
Shear (friction) force on the bottom of the wedge.
kN
S4
Shear (friction) force on the blade.
kN
W1
Pore pressure force on the shear plane.
kN
W2
Pore pressure force on the pseudo blade (front of the wedge)
kN
W3
Pore pressure force on the bottom of the wedge
kN
W4
Pore pressure force on the blade.
kN
vc
Cutting velocity.
m/sec
α
Blade angle.
°
β
Shear angle.
°
θ
Wedge angle.
°
φ
Internal friction angle.
°
δ
External friction angle.
°
λ
Internal friction angle on pseudo blade.
°
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/15%3A_A_Wedge_in_Hyperbaric_Rock_Cutting/15.03%3A_The_Equilibrium_of_Moments.txt
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This book is used for the courses OE4607 Introduction Dredging Engineering and OE4626 Dredging Processes I of the MSc program Offshore & Dredging Engineering of the Delft University of Technology. The exercises are questions of the exams. After each exam, the new questions will be added to this chapter.
16.02: Chapter2- Basic Soil Mechanics
16.2.1. MC: Mohr Circles 1
Which of the following statements are true?
The bold green answers are true.
1. The Mohr circle gives the relation between normal stress and tensile stress.
2. In the τ-σ diagram for soil the positive horizontal axis gives tensile stress.
3. In the τ-σ diagram for soil the positive horizontal axis gives compressive stress.
4. The Mohr circle gives the relation between normal stress and shear stress.
5. In the τ-σ diagram for steel the positive horizontal axis gives compressive stress.
6. In the τ-σ diagram for steel the positive horizontal axis gives tensile stress.
7. In the Mohr circle real angles are shown by a factor 2.
8. In the Mohr circle real angles are shown by a factor 1/2.
16.2.2. MC: Mohr Circles 2
Which of the following statements are true?
The bold green answers are true.
1. The Mohr circle gives the relation between normal stress and shear stress.
2. In the τ-σ diagram for soil the positive horizontal axis gives tensile stress.
3. In the τ-σ diagram for steel the positive horizontal axis gives compressive stress.
4. On the plane of a principle normal stress there is no shear stress.
5. The largest shear stress is always on a plane with an angle of 45 degrees with respect to the principal stresses.
6. In the Mohr circle the angle between the two principal stresses is 180 degrees.
7. Mohr circles can cross the failure line/curve.
8. Tensile failure occurs on a plane with an angle of 90 degrees with the plane with the largest shear stress.
16.2.3. MC: Mohr Circles 3
Which of the following statements are true?
The bold green answers are true.
1. The Mohr circle is based on a force equilibrium.
2. The Mohr circle is based on a stress equilibrium.
3. In the Mohr circle real angles are shown by a factor 2.
4. In the Mohr circle real angles are shown by a factor 1/2.
5. With 1 Mohr circle the angle of internal friction can be determined.
6. At least 2 (different confining pressures) Mohr circles are required to determine the angle of internal friction.
7. Fundamentally the Mohr circle is valid in a point.
8. Fundamentally the Mohr circle is valid in an area.
16.2.4. MC: Active/Passive Soil Failure 1
Which of the following statements are true?
The bold green answers are true.
1. Passive soil failure is the failure where the soil is passive, the outside world is active.
2. Active soil failure is the failure where the outside world is active, the soil is passive.
3. Passive soil failure is the failure where the soil is active, the outside world is passive.
4. Active soil failure is the failure where the outside world is passive, the soil is active.
5. The stresses with passive failure are larger than with active failure.
6. The stresses with active failure are larger than with passive failure.
7. Excavating soil in dredging is a typical example of active failure.
8. Excavating soil in dredging is a typical example of passive failure.
16.2.5. MC: Active/Passive Soil Failure 2
Which of the following statements are true? (Active failure means the horizontal stress is smaller than the vertical stress, passive failure means the horizontal stress is larger than the vertical stress).
The bold green answers are true.
1. The failure of a dike, because it’s too high, is passive failure.
2. At low tide a quay wall is pushed into to the water, this is active failure.
3. A bulldozer pushes a hole in a dike, this is active failure.
4. A very heavy truck drives over a dike. The dike collapses under the weight of the truck. This is active failure.
5. Cutting processes in general are an example of active failure.
6. Forces occurring with active failure are always larger than with passive failure.
7. The cutting process of a cutter head is a typical example of passive failure.
8. The cutting process of a clamshell is a typical example of active failure.
16.2.6. MC: Active/Passive Soil Failure 3
Which of the following statements are true? (Active mode means the horizontal stress is smaller than the vertical stress, passive mode means the horizontal stress is larger than the vertical stress).
The bold green answers are true.
1. Settled or sedimented sand is in active mode.
2. Settled or sedimented sand is in passive mode.
3. Glacial sand, after the ice has melted is in passive mode.
4. Glacial sand, after the ice has melted is in active mode.
5. Glacial sand, when the layer of ice is the thickest is in active mode.
6. Glacial sand, when the layer of ice is the thickest is in passive mode.
7. Sand with a building on top is in passive mode.
8. Sand with a building on top is in active mode.
16.2.7. MC: Active/Passive Soil Failure 4
Which of the following statements are true?
The bold green answers are true.
1. The active soil pressure coefficient increases with increasing internal friction angle.
2. The passive soil pressure coefficient increases with increasing internal friction angle.
3. The active soil pressure coefficient decreases with increasing internal friction angle.
4. The passive soil pressure coefficient decreases with increasing internal friction angle.
5. Passive soil failure is the failure where the soil is passive, the outside world is active.
6. Active soil failure is the failure where the outside world is active, the soil is passive.
7. Passive soil failure is the failure where the soil is active, the outside world is passive.
8. Active soil failure is the failure where the outside world is passive, the soil is active.
16.2.8. Calc.: Bulldozer 1
A bulldozer with a blade height of 0.5 m and a blade width of 3 m and a blade angle of 90 degrees is pushing sand. The internal friction angle of the sand is 45 degrees. The sand has no cohesion or adhesion and the friction between the sand and the blade is assumed to be zero, so a smooth blade. The bulldozer has a maximum forward speed of 1.5 m/sec.
1. What is the coefficient of passive failure for this sand?
$\ \mathrm{K_{p}=\frac{1+\sin (\varphi)}{1-\sin (\varphi)}=\frac{1+\sin (\pi / 4)}{1-\sin (\pi / 4)}=5.826}\quad(-)$
2. What is the pushing force of the bulldozer?
The density of the dry sand ρs with 40% porosity is about 1.6 ton/m3.
$\ \mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \mathrm{w} \cdot \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{1}}{2} \cdot \mathrm{1 .6} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 .5}^{2} \cdot \mathrm{3} \cdot \mathrm{5 .8 2 6}=\mathrm{3 4. 3}\quad(\mathrm{KN})$
3. What is the pushing power of the bulldozer?
$\ \mathrm{P}=\mathrm{F} \cdot \mathrm{v}=34.3 \cdot \mathrm{1 .5}=\mathrm{5 1 .4 5} \quad(\mathrm{kW})$
4. Suppose a total efficiency of 1/3 of the whole drive system of the bulldozer, what is the installed power of the bulldozer.
$\ \mathrm{P_{\text {installed }}=\frac{P}{\eta}=\frac{51.45}{0.3333}=154.35} \quad(\mathrm{kW})$
5. What is the coefficient of active failure of this sand?
$\ \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{1}-\sin (\varphi)}{1+\sin (\varphi)}=\frac{1-\sin (\pi / 4)}{1+\sin (\pi / 4)}=\mathrm{0 .1 7 1 6}\quad(-)$
6. What is the force the bulldozer has to exert on the sand not to make it fail in active mode?
$\ \mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \mathrm{w} \cdot \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{1}}{2} \cdot \mathrm{1 .6} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 .5}^{2} \cdot \mathrm{3} \cdot \mathrm{0 .1 7 1 6}=\mathrm{1 .0 1}\quad(\mathrm{kN})$
16.2.9. Calc.: Bulldozer 2
A bulldozer with a blade height of 1.0 m and a blade angle of 90 degrees is pushing dry sand. The internal friction angle of the sand is 45 degrees. The sand has no cohesion or adhesion and the friction between the sand and the blade is assumed to be zero, so a smooth blade. The bulldozer has a maximum forward speed of 1.0 m/sec. The bulldozer has a maximum pushing force of 100 kN.
1. What is the coefficient of passive failure for this sand?
$\ \mathrm{K_{p}=\frac{1+\sin (\varphi)}{1-\sin (\varphi)}=\frac{1+\sin (\pi / 4)}{1-\sin (\pi / 4)}=5.826}\quad(-)$
2. What is the maximum width of the bulldozer blade?
The density of the dry sand ρs with 40% porosity is about 1.6 ton/m3.
$\ \begin{array}{left}\mathrm{F}=\frac{\mathrm{1}}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \mathrm{w} \cdot \mathrm{K}_{\mathrm{p}}=\mathrm{10 0} \mathrm{k} \mathrm{N}\ \mathrm{w}=\frac{\mathrm{1 0 0}}{\frac{1}{2} \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}^{2} \cdot \mathrm{K}_{\mathrm{p}}}=\frac{\mathrm{1 0 0}}{\mathrm{0 . 5} \cdot \mathrm{1 .6} \cdot \mathrm{9 .8 1} \cdot \mathrm{1}^{2} \cdot \mathrm{5 .8 2 6}}=\mathrm{2. 1 8 7} \mathrm{~ m}\end{array}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/16%3A_Exercises/16.01%3A_Introduction.txt
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16.3.1. MC: Cutting Mechanism
The sine in the denominator of the generic cutting force equation has 4 angles in the argument. What happens if the sum of these 4 angles approaches 180 degrees?
The bold green answers are true.
1. The cutting forces become very high.
2. The cutting forces become negative.
3. Nature will choose another cutting mechanism.
4. The cutting force will be constant above a certain sum of the 4 angles.
5. The cutting process will be cavitating.
6. The adhesion will become zero.
7. The vertical cutting force will become negative.
8. The curling type will occur.
16.3.2. MC: The Snow Plough
Which of the following statements is true?
The bold green answers are true.
1. The snow plough effect occurs when the sum of the 4 angles in the argument of the sine in the denominator of the generic cutting equation is larger than 180 degrees.
2. The snow plough effect occurs when the sum of the 4 angles in the argument of the sine in the denominator of the generic cutting equation is equal to 180 degrees.
3. The snow plough effect will occur when the angle between the cutting velocity and the blade is not 90 degrees.
4. The snow plough effect will occur when the angle between the cutting velocity and the blade edge is smaller than 90 degrees.
5. The snow plough effect will occur when the angle between the cutting velocity and the blade edge is larger than 90 degrees.
6. The snow plough effect will push the blade sideways.
7. The snow plough effect reduces the cutting forces strongly.
8. The snow plough effect increases the cutting forces strongly.
16.04: Chapter 4- Which Cutting Mechanism for Which Kind of Soil
16.4.1. MC: Dry Sand Cutting
Which of the following soil mechanical parameters play a dominant role in the cutting of dry sand?
The bold green answers are true.
1. The density of the sand.
2. The angle of internal friction.
3. The permeability of the sand.
4. The tensile strength of the sand.
5. The porosity of the sand.
6. The adhesion of the sand.
7. The angle of external friction of the sand.
8. The shear strength of the sand.
16.4.2. MC: Water Saturated Sand Cutting
Which of the following soil mechanical parameters play a dominant role in the cutting of saturated sand?
The bold green answers are true.
1. The density of water.
2. The angle of internal friction.
3. The permeability of the sand.
4. The tensile strength of the sand.
5. The porosity of the sand.
6. The adhesion of the sand.
7. The angle of external friction of the sand.
8. The shear strength of the sand.
16.4.3. MC: Clay Cutting
Which soil mechanical parameters dominate the cutting forces in clay?
The bold green answers are true.
1. The gravitational constant.
2. The density of the cutting blade.
3. The internal shear strength or cohesion of the clay.
4. The density of the clay.
5. The permeability of the clay.
6. The external shear strength or adhesion of the clay.
7. The porosity of the clay.
8. The angle of internal friction of the clay.
16.4.4. MC: Atmospheric Rock Cutting
Which material properties play a dominant role in the atmospheric cutting of rock/stone?
The bold green answers are true.
1. The external friction coefficient of the rock.
2. The external shear strength or adhesion of the rock.
3. The external friction angle of the rock.
4. The density of the rock.
5. The permeability of the rock.
6. The tensile strength of the rock.
7. The shear strength or cohesion of the rock.
8. The internal friction angle of the rock.
16.4.5. MC: Hyperbaric Rock Cutting
Which material/environmental properties play a dominant role in the hyperbaric cutting of rock/stone?
The bold green answers are true.
1. The external friction coefficient of the rock.
2. The external shear strength or adhesion of the rock.
3. The external friction angle of the rock.
4. The hydrostatic pressure.
5. The permeability of the rock.
6. The tensile strength of the rock.
7. The shear strength or cohesion of the rock.
8. The internal friction angle of the rock.
16.05: Chapter 5- Dry Sand Cutting
16.5.1. MC: Soil Mechanical Parameters
Which of the following statements is true? The cutting of dry sand is:
The bold green answers are true.
1. Dominated by pore pressures.
2. Dominated by adhesion and cohesion.
3. Dominated by the permeability of the sand.
4. Dominated by the weight of the layer cut.
5. Influenced by the porosity of the sand.
6. Dominated by the inertial forces.
7. Influenced by the angle of external friction of the sand.
8. Dominated by the shear strength of the sand.
16.5.2. MC: The Shear Angle
Which of the following statements are true in the case only the weight of the soil is considered if dry sand is excavated with blade angles above 30o?
The bold green answers are true.
1. If the blade angle α increases, the shear angle β also increases.
2. If the angle of internal friction φ increases, the shear β angle also increases.
3. If the external friction angle δ increases, the shear β angle also increases.
4. If the blade angle α increases, the shear angle β decreases.
5. If the angle of internal friction φ increases, the shear angle β decreases.
6. If the external friction angle δ increases, the shear angle β decreases.
16.5.3. Calc.: The Shear Angle
Suppose the cutting of dry sand is completely dominated by the inertial forces. Derive the analytical solution for the shear angle based on the minimum energy principle.
$\ \mathrm{F_{h}=\rho_{s} \cdot v_{c}^{2} \cdot h_{i} \cdot w \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)} \cdot \frac{\cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \sin (\alpha+\delta)}$
This is at a minimum when the denominator is at a maximum, so:
$\ \begin{array}{left}\frac{\partial}{\partial \beta}(\sin (\alpha+\beta) \cdot \sin (\alpha+\beta+\delta+\varphi))=0\ \cos (\alpha+\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)+\sin (\alpha+\beta) \cdot \cos (\alpha+\beta+\delta+\varphi)=\sin (2 \cdot \alpha+2 \cdot \beta+\delta+\varphi)=0\ 2 \cdot \alpha+2 \cdot \beta+\delta+\varphi=\pi\ \beta=\frac{\pi}{2}-\frac{2 \cdot \alpha+\delta+\varphi}{2}\end{array}$
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/16%3A_Exercises/16.03%3A_Chapter_3-_The_General_Cutting_Process.txt
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16.6.1. MC: Soil Mechanical Parameters
Which of the following soil mechanical parameters play a dominant role in the cutting of saturated sand?
The bold green answers are true.
1. The density of water.
2. The angle of internal friction.
3. The permeability of the sand.
4. The tensile strength of the sand.
5. The porosity of the sand.
6. The adhesion of the sand.
7. The angle of external friction of the sand.
8. The shear strength of the sand.
16.6.2. MC: Dilatation
What is the definition of dilatation?
The bold green answers are true.
1. Dilatation is the increase of the pore volume of sand caused by gravitation.
2. Dilatation is the decrease of the pore volume of sand caused by shear stress.
3. Dilatation is the decrease of the pore volume of sand caused by dredging.
4. Dilatation is the increase of the pore volume of sand caused by shear stress.
5. Dilatation is the increase of the pore volume of sand caused by dredging.
6. Dilatation is the decrease of the pore volume of sand caused by gravitation.
16.6.3. MC: Cavitation
Which statements are true?
The bold green answers are true.
1. Cavitation is in fact the boiling of water.
2. Cavitation at 10 degrees centigrade occurs at about 0.1 bar absolute pressure.
3. At 100 degrees centigrade cavitation will occur at about 100 kPa.
4. Cavitation is the vaporization of water.
5. Cavitation at 10 degrees centigrade occurs at about 0.01 bar absolute pressure.
6. High in the mountains cavitation will occur at a temperature lower than 100 degrees centigrade.
7. At an atmospheric pressure of 100 kPa cavitation occurs at 10 degrees centigrade.
8. High in the mountains cavitation will occur at a temperature higher than 100 degrees centigrade.
16.6.4. Calc.: Porosity
If the porosity of sand is 42% before cutting (in situ) and 50% after cutting, what is the volume increase of the sand (grains + pores) as a fraction?
$\ \varepsilon=\frac{\mathrm{n}_{\max }-\mathrm{n}_{\mathrm{i}}}{\mathrm{1}-\mathrm{n}_{\max }}=\frac{\mathrm{0 .5}-\mathrm{0 . 4 2}}{\mathrm{1 - 0 .5}}=\mathrm{0 .1 6}$
16.6.5. Calc. : Density
Assume a water density of 1.000 ton/m3 and a quarts density of 2.650 ton/m3 . What is the density of a sand with a porosity of 40% (saturated sand)?
$\ \rho_{\mathrm{s}}=(1-\mathrm{n}) \cdot \rho_{\mathrm{q}}+\mathrm{n} \cdot \rho_{\mathrm{w}}=(1-\mathrm{0 .4}) \cdot \mathrm{2 .6 5}+\mathrm{0 .4} \cdot \mathrm{1 . 0 0}=\mathrm{1 .9 9}\quad (\mathrm{ton/m^3})$
16.6.6. Calc.: Permeability
What is the value of the mean permeability km as used in the cutting equations for water saturated sand? Consider a sand with the following properties:
Initial permeability: ki=0.0001 m/sec
Maximum permeability: kmax=0.0004 m/sec
$\ \mathrm{k}_{\mathrm{m}}=\frac{\mathrm{k}_{\mathrm{i}}+\mathrm{k}_{\mathrm{max}}}{2}=\frac{\mathrm{0 . 0 0 0 1 + 0 . 0 0 0 4}}{2}=\mathrm{0 .0 0 0 2 5} \quad(\mathrm{m} / \mathrm{sec})$
16.6.7. Calc.: Dilatancy
What is the value of the dilatancy ε?
Consider a sand with the following properties:
initial porosity: ni=40%
Maximum porosity: nmax=50%
$\ \varepsilon=\frac{\mathrm{n}_{\mathrm{max}}-\mathrm{n}_{\mathrm{i}}}{1-\mathrm{n}_{\mathrm{max}}}=\frac{\mathrm{0 .5 - 0 .4}}{\mathrm{1 - 0 .5}}=\mathrm{0 .2 0}\quad (-)$
16.6.8. Calc.: Transition Velocity
Give the equation for the transition velocity from non-cavitating cutting to cavitating cutting.
$\ \mathrm{v}_{\mathrm{c}}=\frac{\mathrm{d}_{1} \cdot(\mathrm{z}+\mathrm{1} \mathrm{0}) \cdot \mathrm{k}_{\mathrm{m}}}{\mathrm{c}_{\mathrm{1}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{\varepsilon}}$
16.6.9. Calc.: Cutting Forces & Specific Energy
Consider a sand with the following properties:
Angle of internal friction: φ=36o
Angle of external friction: δ=24o
Initial permeability: ki=0.00005 m/sec
Maximum permeability: kmax=0.00025 m/sec
initial porosity: ni=42%
Maximum porosity: nmax=50%
The cutting blade has the following properties:
The cutting angle: α=60o
The shear angle: β=20o
The blade height: hb=0.2 m
The thickness of the layer to be cut: hi=0.1 m
The width of the blade: w=1 m
Coefficients for the cutting equations:
The coefficient for non-cavitating cutting: c1=0.35
The coefficient for cavitating cutting: d1=4.5
General constants:
The density of water: ρw=1.025 tons/m3
The gravitational constant: g=9.81 m/sec2
A: What are the horizontal cutting forces at 0 m, 15 m and 30 m water depth for the non-cavitational cutting process at a cutting velocity of 0.5 m/s?
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\frac{\mathrm{c}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \varepsilon \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}}}=\frac{\mathrm{0 .3 5} \cdot \mathrm{1 .0 2 5} \cdot \mathrm{9 .8 1 \cdot 0 .} \mathrm{1}^{2} \cdot \mathrm{0 .1 6} \cdot \mathrm{1}}{\mathrm{0 .0 0 0 1 5}} \cdot \mathrm{v}_{\mathrm{c}} \ &=\mathrm{3 7 .5 4} \cdot \mathrm{v}_{\mathrm{c}}=\mathrm{3 7 .5 4} \cdot \mathrm{0 .5}=\mathrm{1 8 .7 7} \end{array}\quad(\mathrm{kN})$
The non-cavitational force does not depend on the water depth.
B: What are the horizontal cutting forces at 0 m, 15 m and 30 m water depths for the cavitational cutting process?
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\mathrm{d}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{4 .5} \cdot \mathrm{1 .0 2 5} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 .1} \cdot \mathrm{1} \cdot(\mathrm{z}+\mathrm{1 0}) \ &=\mathrm{4 .5 2} \cdot(\mathrm{z}+\mathrm{1 0}) \end{array}\quad(\mathrm{kN})$
This gives for 0 m water depth a force of 45.2 kN, for 15 m water depth a force of 113 kN and for 30 m water depth a force of 180.8 kN.
C: At which velocities are the transitions between the cavitational and the non-cavitational cutting process at 0 m, 15 m and 30 m water depths?
$\ \mathrm{37.54 \cdot v_{c}=4.52 \cdot(z+10) \quad\Rightarrow \quad v_{c}=0.12 \cdot(z+10)\quad(m / s e c)}$
This gives for 0 m water depth a transition velocity of 1.2 m/sec, for 15 m water depth a transition velocity of 3.0 m/sec and for 30 m water depth a transition velocity of 4.8 m/sec.
D: What is the specific energy at 0 m, 15 m and 30 m water depth at a cutting velocity of 1 m/s?
$\ \mathrm{\mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{0 . 1} \cdot \mathrm{1}}=\mathrm{1 0} \cdot \mathrm{F}_{\mathrm{h}}}\quad (\mathrm{kPa})$
At a cutting velocity of 1 m/sec the cutting process is non-cavitational at all 3 water depths, so in all 3 cases the specific energy is 375.4 kPa.
E: What is the specific energy at a water depth of 0 m, 15 m and 30 m at a cutting velocity of 2 m/s?
At a cutting velocity of 2 m/sec, the cutting process at 0 m water depth is cavitating, giving a specific energy of 452 kPa. At a water depth of 15 m the cutting process is non-cavitational giving a specific energy of 750.8 kPa. At a water depth of 30 m the cutting process is also non-cavitational giving a specific energy of 750.8 kPa.
F: Determine the pore pressure at the centre of the shear plane using the parallel resistor method for a cutting velocity of 0.5 m/sec.
First the 4 lengths have to be determined:
$\ \begin{array}{left} \mathrm{s_{1}=\left(L_{\max }-L\right) \cdot\left(\frac{\pi}{2}+\frac{\pi}{2}-(\alpha+\beta)\right)+\frac{h_{b}}{\sin (\alpha)}=0.486}\ \mathrm{s_{2}=0.8 \cdot L \cdot(\alpha+\beta)=0.163}\ \mathrm{s_{3}=0.8 \cdot L \cdot(\pi-\beta)=0.327}\ \mathrm{s_{4}=\left(L_{\max }-L\right) \cdot(\pi+\beta)+0.9 \cdot h_{\mathrm{i}} \cdot\left(\frac{h_{\mathrm{i}}}{h_{\mathrm{b}}}\right)^{0.5} \cdot(1.85 \cdot \alpha)^{2} \cdot\left(\frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{max}}}\right)^{0.4}=0.636}\end{array}$
Secondly the total resistance has to be determined:
$\ \mathrm{\frac{1}{R_{t}}=\left(\frac{s_{1}}{k_{\max }}\right)+\left(\frac{s_{2}}{k_{\max }}\right)+\left(\frac{s_{3}}{k_{i}}\right)+\left(\frac{s_{4}}{k_{i}}\right) \Rightarrow R_{t}=439}$
$\ \mathrm{\Delta \mathrm{p}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta) \cdot \mathrm{R}_{\mathrm{t}}=120.8(\mathrm{k} \mathrm{P} \mathrm{a})}$
Below which water depth will we have cavitation at this point?
An absolute pressure of 120.8 kPa is reached at a water depth of 2.07 m, so the point on the shear plane considered will cavitated for water depths below 2.07 m.
16.6.10. Calc.: Cutting Forces & Specific Energy
Consider a sand with the following properties:
Angle of internal friction: φ=36o
Angle of external friction: δ=24o
Initial permeability: ki=0.000025 m/sec
Maximum permeability: kmax=0.000125 m/sec
initial porosity: ni=42%
Maximum porosity: nmax=50%
The cutting blade has the following properties:
The cutting angle: α=60o
The shear angle: β=20o
The blade height: hb=0.2 m
The thickness of the layer to be cut: hi=0.1 m
The width of the blade: w=1 m
Coefficients for the cutting equations:
The coefficient for non-cavitating cutting: c1=0.45
The coefficient for cavitating cutting: d1=5.5
General constants:
The density of water: ρw=1.025 tons/m3
The gravitational constant: g=9.81 m/sec2
A: What are the horizontal cutting forces at 0 m, 10 m and 20 m water depth for the non-cavitational cutting process at a cutting velocity of 1.5 m/s?
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\frac{\mathrm{c}_{\mathrm{1}} \cdot \mathrm{\rho}_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{\varepsilon} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}}}=\frac{\mathrm{0} . \mathrm{45} \cdot \mathrm{1 . 0 2 5} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 . 1}^{2} \cdot \mathrm{0 . 1 6} \cdot \mathrm{1}}{\mathrm{0.} \mathrm{0 0 0 0 7 5}} \cdot \mathrm{v}_{\mathrm{c}} \ &=\mathrm{96 .5 3} \cdot \mathrm{v}_{\mathrm{c}}=\mathrm{9 6 .5 3} \cdot \mathrm{1 .5}=\mathrm{1 4 4 .8} \end{array}\quad(\mathrm{kN})$
The non-cavitational force does not depend on the water depth.
B: What are the horizontal cutting forces at 0 m, 10 m and 20 m water depths for the cavitational cutting process?
$\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\mathrm{d}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{5.5} \cdot \mathrm{1 .0 2 5} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 .1} \cdot \mathrm{1} \cdot(\mathrm{z}+\mathrm{1 0}) \ &=\mathrm{5 .5 3} \cdot(\mathrm{z}+\mathrm{1 0}) \end{array}\quad (\mathrm{kN})$
This gives for 0 m water depth a force of 55.3 kN, for 10 m water depth a force of 110.6 kN and for 20 m water depth a force of 165.9 kN.
C: At which velocities are the transitions between the cavitational and the non-cavitational cutting process at 0 m, 10 m and 20 m water depths?
$\ \mathrm{96.53\cdot v_c = 5.53 \cdot (z+10) \quad\Rightarrow \quad v_{c}=0.0573 \cdot(z+10)(m / s e c)}$
This gives for 0 m water depth a transition velocity of 0.573 m/sec, for 10 m water depth a transition velocity of 1.146 m/sec and for 20 m water depth a transition velocity of 1.719 m/sec.
D: What is the specific energy at 0 m, 10 m and 20 m water depth at a cutting velocity of 1 m/s?
$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{0 .1} \cdot \mathrm{1}}=\mathrm{1 0} \cdot \mathrm{F}_{\mathrm{h}} \quad (\mathrm{kPa})$
At a cutting velocity of 1 m/sec the cutting process is non-cavitational at all z=0 m and non-cavitational at z=10 m and z=20 m, so in the specific energy is 553 kPa at z=0 m and 965.3 kPa at z=10 m and z=20 m.
E: What is the specific energy at a water depth of 0 m, 10 m and 20 m at a cutting velocity of 2 m/s?
At a cutting velocity of 2 m/sec, the cutting process at 0 m water depth is cavitating, giving a specific energy of 553 kPa. At a water depth of 10 m the cutting process is cavitational giving a specific energy of 1106 kPa. At a water depth of 20 m the cutting process is also cavitational giving a specific energy of 1659 kPa.
F: Determine the pore pressure at the centre of the shear plane using the parallel resistor method for a cutting velocity of 1.5 m/sec.
First the 4 lengths have to be determined:
$\ \begin{array}{left}\mathrm{s}_{1}=\left(\mathrm{L}_{\max }-\mathrm{L}\right) \cdot\left(\frac{\pi}{2}+\frac{\pi}{2}-(\alpha+\beta)\right)+\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}=\mathrm{0 .4 8 6}\ \mathrm{s_{2}=0.8 \cdot L \cdot(\alpha+\beta)=0.163}\ \mathrm{s_{3}=0.8 \cdot L \cdot(\pi-\beta)=0.327}\ \mathrm{s_{4}=\left(L_{\max }-L\right) \cdot(\pi+\beta)+0.9 \cdot h_{\mathrm{i}} \cdot\left(\frac{h_{\mathrm{i}}}{h_{\mathrm{b}}}\right)^{0.5} \cdot(1.85 \cdot \alpha)^{2} \cdot\left(\frac{\mathrm{k}_{\mathrm{i}}}{k_{\max }}\right)^{0.4}=0.636}\end{array}$
Secondly the total resistance has to be determined:
$\ \mathrm{\frac{1}{\mathrm{R}_{t}}=\left(\frac{s_{1}}{k_{\max }}\right)+\left(\frac{s_{2}}{k_{\max }}\right)+\left(\frac{s_{3}}{k_{i}}\right)+\left(\frac{s_{4}}{k_{i}}\right) \Rightarrow R_{t}=878}$
$\ \Delta \mathrm{p}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta) \cdot \mathrm{R}_{\mathrm{t}}=72 \mathrm{5}(\mathrm{k} \mathrm{P} \mathrm{a})$
Below which water depth will we have cavitation at this point?
An absolute hydrostatic pressure of 725 kPa is reached at a water depth of 62.15 m, so the point on the shear plane considered will cavitated for water depths below 62.15 m.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/16%3A_Exercises/16.06%3A_Chapter_6-_Water_Saturated_Sand_Cutting.txt
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16.7.1. Calc.: Cutting Forces
Consider clay with cohesion of 200 kPa and an adhesion of 50 kPa. The strengthening factor is 2. A blade angle of 55 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m. Assume the Flow Type.
What is the ac ratio r?
$\ \mathrm{r=\frac{a \cdot h_{b}}{c \cdot h_{i}}=\frac{50 \cdot 0.1}{200 \cdot 0.1}=\frac{1}{4}\quad(-)}$
What is the shear angle?
See Figure 7-21 (Figure 7.20, 1st edition), blade angle α=55 degrees and r=0.25 gives a shear angle β of about 57 degrees.
What are the horizontal and the vertical cutting forces?
Figure 7-23 (Figure 7.22, 1st edition) gives a horizontal cutting force coefficient λHF of about 1.3 and Figure 7-24 (Figure 7.23, first edition) gives a vertical cutting force coefficient λVF of 0.6. This gives for the Flow Type:
$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{HF}}=\mathrm{2} \cdot \mathrm{2} \mathrm{0} \mathrm{0} \cdot \mathrm{0.1} \cdot \mathrm{1} \cdot \mathrm{1 .3}=\mathrm{5} \mathrm{2} \mathrm{k} \mathrm{N}$
$\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V F}}=\mathrm{2} \cdot \mathrm{2} \mathrm{0} \mathrm{0} \cdot \mathrm{0.1} \cdot \mathrm{1} \cdot \mathrm{0.6}=\mathrm{2} \mathrm{4} \mathrm{k} \mathrm{N}$
16.7.2. Calc.: Cutting Forces & Mechanisms
Consider clay with cohesion of 200 kPa and an adhesion of 10 kPa. The strengthening factor is 2. A blade angle of 55 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m. Assume the Flow Type.
What is the ac ratio r?
$\ \mathrm{r}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}=\frac{\mathrm{1 0} \cdot \mathrm{0.1}}{\mathrm{2 0 0} \cdot \mathrm{0 . 1}}=\frac{\mathrm{1}}{\mathrm{2 0}} \quad\mathrm{( - )}$
What is the shear angle?
See Figure 7-21 (Figure 7.20, 1st edition), blade angle α=55 degrees and r=0.05 gives a shear angle β of about 62 degrees.
What are the horizontal and the vertical cutting forces?
Figure 7-23 (Figure 7.22, 1st edition) gives a horizontal cutting force coefficient λHF of about 1.1 and Figure 7-24 (Figure 7.23, first edition) gives a vertical cutting force coefficient λVF of 0.7. This gives for the Flow Type:
$\ \mathrm{F_{h}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \lambda_{H F}=2 \cdot 200 \cdot 0.1 \cdot 1 \cdot 1.1=44 \mathrm{kN}}$
$\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V F}}=\mathrm{2} \cdot \mathrm{2 0 0} \cdot \mathrm{0 .1} \cdot \mathrm{1} \cdot \mathrm{0 .7}=\mathrm{2 8} \mathrm{k N}$
If the tensile strength is -20 kPa, will we have the Tear Type or the Flow Type?
A tensile strength of -20 kPa gives a σT/c ratio of -0.1. With an ac ratio of r=0.05 this ratio should be below -0.5 according to Figure 7-27 (Figure 7-26, 1st edition) for the Flow Type, it is not, so we have the Tear Type.
16.08: Chapter 8- Atmospheric Rock Cutting
16.8.1. Calc.: Cutting Forces & Mechanisms
Consider a rock with a compressive strength of 30 MPa and a tensile strength of -2 MPa. The angle of internal friction is 18 degrees, the angle of external friction is 12 degrees. A blade angle of 55 degrees is used and a blade height of 0.2 m and blade width w=0.1 m. The layer thickness is 0.1 m.
Is the cutting process brittle shear or brittle tensile?
The cohesion or shear strength is:
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=10.9 M P a}$
According to Figure 8-38 (Figure 8.22, 1st edition) & (φ=18°), the BTS/Cohesion ratio should be above -0.3 for tensile failure, the ratio is -2/10.9=-0.183 which is above -0.3, so the process is brittle tensile failure.
If the tensile strength is -10 MPa, is the cutting process brittle shear or brittle tensile?
Now the ratio is -10/10.9=-0.92, which is below -0.3, so the process is shear failure, which is often related to ductile failure, but it is brittle shear failure.
What are the horizontal and the vertical cutting forces?
According to Figure 8-45 (Figure 8.28, 1st edition) the brittle horizontal coefficient λHT is about 3.07 and according to Figure 8-46 (Figure 8.29, 1st edition) the brittle vertical coefficient λVT about 1.30. This gives for a tensile strength of -2 MPa:
$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{H T}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{3.0 7} \cdot \mathrm{2 0 0 0} \cdot \mathrm{0 .1} \cdot \mathrm{0 .1}=\mathrm{6 1} \mathrm{k N}$
$\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{V T}} \cdot \sigma_{\mathrm{T}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{1 .3 0} \cdot \mathrm{2 0 0 0} \cdot \mathrm{0 .1} \cdot \mathrm{0 .1}=\mathrm{2 6} \mathrm{k N}$
For the case with a tensile strength of -10 MPa, the following is found:
According to Figure 8-31 (Figure 8.16, 1st edition) the brittle shear horizontal coefficient λHF is about 1.61 and according to Figure 8-32 (Figure 8.17, 1st edition) the brittle shear vertical coefficient λVF about 0.68. This gives for a compressive strength of 30 MPa (cohesion about 10.9 MPa):
$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{HF}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{1 .6 1} \cdot \mathrm{1 0 9 0 0} \cdot \mathrm{0 .1} \cdot \mathrm{0 .1}=\mathrm{17 6} \mathrm{k N}$
$\ \mathrm{F_{v}=\lambda_{V F} \cdot \mathrm{c} \cdot h_{i} \cdot \mathrm{w}=0.68 \cdot 10900 \cdot 0.1 \cdot 0.1=75 \mathrm{kN}}$
16.8.2. Calc.: Cutting Forces & Mechanisms
Consider a rock with a compressive strength of 100 MPa and a tensile strength of -10 MPa. The angle of internal friction is 20 degrees, the angle of external friction is 13.33 degrees. A blade angle of 60 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m.
The shear angle is:
$\ \beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}=43.33\text{ degrees}$
The cohesion or shear strength is:
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=35 M P a}$
The normal stress on the shear plane is:
$\ \sigma_{\mathrm{N} 1}=\frac{-\mathrm{c} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)=21.52 \mathrm{M} \mathrm{Pa}$
The shear stress on the shear plane is:
$\ \tau_{\mathrm{S} 1}=\mathrm{c}+\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)=42.84 \mathrm{M} \mathrm{P} \mathrm{a}$
The normal stress in the center of the Mohr circle is:
$\ \sigma_{\mathrm{C}}=\sigma_{\mathrm{N} 1}+\tau_{\mathrm{S} 1} \cdot \tan (\varphi)=37.1 \mathrm{M} \mathrm{P} \mathrm{a}$
The radius of this Mohr circle is now:
$\ \mathrm{R}=\frac{\tau_{\mathrm{S} 1}}{\cos (\varphi)}=45.59 \mathrm{M P a}$
The minimum principle stress of this Mohr circle is:
$\ \sigma_{\min }=\sigma_{\mathrm{C}}-\mathrm{R}=-\mathrm{8 .4 8} \mathrm{M P a}$
Since -8.48 MPa>-10 MPa there is no tensile failure but shear failure.
The horizontal force is now:
$\ \mathrm{F_{h}=\frac{2 \cdot c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{H F} \cdot c \cdot h_{i} \cdot w=1.912 \cdot 35 \cdot 0.1 \cdot 1=6.691 \ M N}$
The vertical force is now:
$\ \mathrm{F}_{v}=\frac{\mathrm{2} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{\mathrm{1}+\cos (\alpha+\delta+\varphi)}=\lambda_{\mathrm{V F}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{0 .5 7 2} \cdot \mathrm{3 5} \cdot \mathrm{0 .1} \cdot \mathrm{1}=\mathrm{2.0} \mathrm{\ M N}$
16.8.3. Calc.: Cutting Forces & Mechanisms
Consider a rock with a compressive strength of 100 MPa and a tensile strength of -5 MPa. The angle of internal friction is 20 degrees, the angle of external friction is 13.33 degrees. A blade angle of 60 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m.
The shear angle is:
$\ \beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}=43.33\text{ degrees}$
The cohesion or shear strength is:
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=35 M P a}$
The normal stress on the shear plane is:
$\ \sigma_{\mathrm{N} 1}=\frac{-\mathrm{c} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)=21.52 \text{ MPa}$
The shear stress on the shear plane is:
$\ \tau_{\mathrm{S} 1}=\mathrm{c}+\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)=42.84 \text{ MPa}$
The normal stress in the center of the Mohr circle is:
$\ \sigma_{\mathrm{C}}=\sigma_{\mathrm{N} 1}+\tau_{\mathrm{S} 1} \cdot \tan (\varphi)=37.1 \text{ MPa}$
The radius of this Mohr circle is now:
$\ \mathrm{R}=\frac{\tau_{\mathrm{S} 1}}{\cos (\varphi)}=45.59 \text{ MPa}$
The minimum principle stress of this Mohr circle is:
$\ \mathrm{\sigma_{\min }=\sigma_{\mathrm{C}}-R=-8.48 \text{ MPa}}$
Since -8.48 MPa<-5 MPa there is tensile failure but no shear failure. This results in another shear angle of 25.8°.
The horizontal force is now, Figure 8-45 (Figure 8.28, 1st edition):
$\ \mathrm{F_{h}=\frac{2 \cdot c_{m} \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{H T} \cdot \sigma_{T} \cdot h_{i} \cdot w=4 \cdot 5 \cdot 0.1 \cdot 1=2} \text{ MN}$
The vertical force is now, Figure 8-46 (Figure 8.29, 1st edition):
$\ \mathrm{F}_{v}=\mathrm{\frac{2 \cdot c_{m} \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{V T} \cdot \sigma_{T} \cdot h_{i} \cdot w=1.25 \cdot 5 \cdot 0.1 \cdot 1=0.625}\text{ MN}$
16.8.4. Calc.: Cutting Forces & Mechanisms
Consider a rock with a compressive strength of 60 MPa and a tensile strength of -10 MPa. The angle of internal friction is 20 degrees, the angle of external friction is 13.33 degrees. A blade angle of 60 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m.
The shear angle is:
$\ \beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}=43.33\text{ degree}$
The cohesion or shear strength is:
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=21}\text{ MPa}$
The normal stress on the shear plane is:
$\ \sigma_{\mathrm{N} 1}=\frac{-\mathrm{c} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)=12.9 \text{ MPa}$
The shear stress on the shear plane is:
$\ \mathrm{\tau_{\mathrm{S} 1}=c+\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)=25.7 \text{ MPa}}$
The normal stress in the center of the Mohr circle is:
$\ \sigma_{\mathrm{C}}=\sigma_{\mathrm{N} 1}+\tau_{\mathrm{S} 1} \cdot \tan (\varphi)=22.27 \text{ MPa}$
The radius of this Mohr circle is now:
$\ \mathrm{R}=\frac{\tau_{\mathrm{S} 1}}{\cos (\varphi)}=27.35 \text{ MPa}$
The minimum principle stress of this Mohr circle is:
$\ \sigma_{\min }=\sigma_{\mathrm{C}}-\mathrm{R}=-\mathrm{5.0 9} \text{ MPa}$
Since -5.09 MPa>-10 MPa there is no tensile failure but shear failure.
The horizontal force is now:
$\ \mathrm{F_{h}}=\mathrm{\frac{2 \cdot c \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{H F} \cdot c \cdot h_{i} \cdot w=1.912 \cdot 21 \cdot 0.1 \cdot 1=4.02} \text{ MN}$
The vertical force is now:
$\ \mathrm{F}_{v}=\frac{\mathrm{2} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{\mathrm{1}+\cos (\alpha+\delta+\varphi)}=\lambda_{\mathrm{V F}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{0 .5 7 2} \cdot \mathrm{2 1} \cdot \mathrm{0 .1} \cdot \mathrm{1}=\mathrm{1 .2 0} \text{ MN}$
16.8.5. Calc.: Cutting Forces & Mechanisms
Consider a rock with a compressive strength of 60 MPa and a tensile strength of -3 MPa. The angle of internal friction is 20 degrees, the angle of external friction is 13.33 degrees. A blade angle of 60 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m.
The shear angle is:
$\ \beta=\frac{\pi}{2}-\frac{\alpha+\delta+\varphi}{2}=43.33\text{ degrees}$
The cohesion or shear strength is:
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=21 \text{ MPa}}$
The normal stress on the shear plane is:
$\ \sigma_{\mathrm{N} 1}=\frac{-\mathrm{c} \cdot \cos (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)=12.9 \text{ MPa}$
The shear stress on the shear plane is:
$\ \tau_{\mathrm{S} 1}=\mathrm{c}+\sigma_{\mathrm{N} 1} \cdot \tan (\varphi)=25.7 \text{ MPa}$
The normal stress in the center of the Mohr circle is:
$\ \sigma_{\mathrm{C}}=\sigma_{\mathrm{N} 1}+\tau_{\mathrm{S} 1} \cdot \tan (\varphi)=22.27 \text{ MPa}$
The radius of this Mohr circle is now:
$\ \mathrm{R}=\frac{\tau_{\mathrm{S} 1}}{\cos (\varphi)}=27.35 \text{ MPa}$
The minimum principle stress of this Mohr circle is:
$\ \sigma_{\min }=\sigma_{\mathrm{C}}-\mathrm{R}=-5.09 \text{ MPa}$
Since -5.09 MPa<-3 MPa there is tensile failure but no shear failure. This results in another shear angle of 25.8°.
The horizontal force is now, Figure 8-45 (Figure 8.28, 1st edition):
$\ \mathrm{F_{h}=\frac{2 \cdot c_{m} \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \sin (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{H T} \cdot \sigma_{T} \cdot h_{i} \cdot w=4.1 \cdot 3 \cdot 0.1 \cdot 1=1.24} \text{ MN}$
The vertical force is now, Figure 8-46 (Figure 8.29, 1st edition):
$\ \mathrm{F}_{v}=\mathrm{\frac{2 \cdot c_{m} \cdot h_{i} \cdot w \cdot \cos (\varphi) \cdot \cos (\alpha+\delta)}{1+\cos (\alpha+\delta+\varphi)}=\lambda_{V T} \cdot \sigma_{T} \cdot h_{i} \cdot w=1.24 \cdot 3 \cdot 0.1 \cdot 1=0.371} \text{ MN}$
16.09: Chapter 9- Hyperbaric Rock Cutting
16.9.1. MC: Soil Mechanical Parameters
Which material/environmental properties play a dominant role in the hyperbaric cutting of rock/stone at high cutting velocities?
The bold green answers are true.
1. The ratio of the hydrostatic pressure to the cohesion of the rock.
2. The external shear strength or adhesion of the rock.
3. The external friction angle of the rock.
4. The hydrostatic pressure to adhesion ratio.
5. The permeability of the rock.
6. The tensile strength of the rock.
7. The shear strength or cohesion of the rock.
8. The internal friction angle of the rock.
1.1.1 Exercise 2
Consider a rock with a compressive strength of 30 MPa and a tensile strength of -2 MPa. The angle of internal friction is 20 degrees, the angle of external friction is 13 degrees. A blade angle of 60 degrees is used and a blade height of 0.1 m and blade width w=0.05 m. The layer thickness is 0.01 m. The water depth is 1000 m.
What is the hydrostatic pressure to cohesion ratio rz?
The cohesion or shear strength is according to equation (8-117) (eqn 8.77, 1st edition):
$\ \mathrm{c=\frac{U C S}{2} \cdot\left(\frac{1-\sin (\varphi)}{\cos (\varphi)}\right)=10.5} \text{ MPa}$
The hydrostatic pressure to cohesion ratio rz is according to equation (9-29) (eqn 9-27, 1st edition):
$\ \mathrm{r}_{\mathrm{z}}=\frac{\rho_{\mathrm{l}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0})}{\mathrm{c}}=\frac{\mathrm{1 .0 2 5} \cdot \mathrm{9 . 8 1} \cdot(\mathrm{1 0 0 0}+\mathrm{1 0})}{\mathrm{1 0 5 0 0}}=\mathrm{0 .9 6 7}$
What is the mobilized blade height hb,m?
The ratio hb,m/hi=0.65, see Figure 9-17 (Figure 9-16, 1st edition).
$\ \mathrm{h}_{\mathrm{b}, \mathrm{m}}=\mathrm{0 .6 5} \cdot \mathrm{h}_{\mathrm{i}}=\mathrm{0. 6 5} \cdot \mathrm{0 .0 1}=\mathrm{0 .0 0 6 5}(\mathrm{m})$
So what is the cutting mechanism with the original blade height of 0.1 m?
Since the mobilized blade height hb,m is smaller than the blade height hb, the Curling Type will occur.
What are the horizontal and the vertical cutting forces?
The horizontal force coefficient is about 2.9 according to Figure 9-19 (Figure 9-18, 1st edition) and the vertical force coefficient is about 0.68 Figure 9-20 (Figure 9-19, 1st edition).
$\ \begin{array}{left} \text{Horizontal }& \mathrm{F_{h}=\lambda_{H C} \cdot c \cdot h_{i} \cdot w=2.9 \cdot 10500 \cdot 0.01 \cdot 0.05=15.23} \quad(\mathrm{kN})\ \text{Vertical } & \mathrm{F_{v}=\lambda_{V C} \cdot c \cdot h_{i} \cdot w=0.68 \cdot 10500 \cdot 0.01 \cdot 0.05=3.57}\quad(\mathrm{kN})\end{array}$
What is the shear angle β?
The shear angle is β=41°, according to Figure 9-18 (Figure 9-17, 1st edition).
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/16%3A_Exercises/16.07%3A_Chapter_7-_Clay_Cutting.txt
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Table C-1: The dimensionless pore pressures.
hb/hi
ki/kmax=1
ki/kmax=0.25
β =30o
37.5o
45o
30o
37.5o
45o
α =15o
1 (s)
0.156
0.168
0.177
0.235
0.262
0.286
2 (s)
0.157
0.168
0.177
0.236
0.262
0.286
3 (s)
0.158
0.168
0.177
0.237
0.262
0.286
1 (b)
0.031
0.033
0.035
0.054
0.059
0.063
2 (b)
0.016
0.017
0.018
0.028
0.030
0.032
3 (b)
0.011
0.011
0.012
0.019
0.020
0.021
β =25o
30o
35o
25o
30o
35o
α =30o
1 (s)
0.178
0.186
0.193
0.274
0.291
0.308
2 (s)
0.179
0.187
0.193
0.276
0.294
0.310
3 (s)
0.179
0.187
0.193
0.277
0.294
0.310
1 (b)
0.073
0.076
0.078
0.126
0.133
0.139
2 (b)
0.049
0.049
0.049
0.084
0.085
0.086
3 (b)
0.034
0.034
0.033
0.059
0.059
0.059
β =20o
25o
30o
20o
25o
30o
α =45o
1 (s)
0.185
0.193
0.200
0.289
0.306
0.325
2 (s)
0.190
0.198
0.204
0.304
0.322
0.339
3 (s)
0.192
0.200
0.205
0.308
0.325
0.340
1 (b)
0.091
0.097
0.104
0.161
0.174
0.187
2 (b)
0.081
0.082
0.083
0.146
0.148
0.151
3 (b)
0.067
0.065
0.063
0.120
0.116
0.114
β =15o
20o
25o
15o
20o
25o
α =60o
1 (s)
0.182
0.192
0.200
0.278
0.303
0.324
2 (s)
0.195
0.204
0.211
0.314
0.339
0.359
3 (s)
0.199
0.208
0.214
0.327
0.350
0.368
1 (b)
0.091
0.103
0.112
0.158
0.184
0.205
2 (b)
0.100
0.106
0.109
0.182
0.196
0.204
3 (b)
0.094
0.095
0.093
0.174
0.176
0.174
The dimensionless pore pressures p1m in the shear zone (s) and p2m on the blade surface (b) as a function of the blade angle α, de shear angle β, the ratio between the blade height hb and the layer thickness hi and the ratio between the permeability of the situ sand kand the permeability of the sand cut kmax, with a wear zone behind the edge of the blade of 0.2·hi.
17.04: Appendix D- The Shear Angle Non-Cavitating
Table D-1: β for hb/hi=1, non-cavitating.
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15
15o
40.892
40.152
39.169
38.012
36.727
18o
39.024
38.380
37.483
36.402
35.184
21o
37.355
36.781
35.947
34.924
33.756
24o
35.847
35.321
34.534
33.552
32.423
27o
34.468
33.975
33.220
32.269
31.166
30o
33.196
32.723
31.989
31.058
29.973
30
15o
37.967
36.937
35.707
34.334
32.854
18o
36.187
35.250
34.100
32.795
31.372
21o
34.564
33.696
32.606
31.353
29.974
24o
33.072
32.255
31.209
29.994
28.648
27o
31.690
30.907
29.893
28.705
27.382
30o
30.401
29.640
28.646
27.476
26.166
45
15o
33.389
32.254
30.936
29.481
27.919
18o
31.792
30.726
29.467
28.061
26.539
21o
30.326
29.310
28.092
26.720
25.224
24o
28.969
27.984
26.793
25.442
23.963
27o
27.700
26.733
25.557
24.218
22.745
30o
26.503
25.543
24.373
23.036
21.562
60
15o
28.220
26.928
25.482
23.917
22.253
18o
26.813
25.569
24.160
22.623
20.978
21o
25.500
24.287
22.901
21.379
19.742
24o
24.264
23.067
21.692
20.174
18.535
27o
23.091
21.897
20.522
18.999
17.350
30o
21.967
20.767
19.382
17.845
16.177
The shear angle β as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for the non-cavitating cutting process, for hb/hi=1.
Table D-2: β for hb/hi=2, non-cavitating.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
41.128
40.402
39.427
38.273
36.986
18o
39.239
38.609
37.720
36.643
35.424
21o
37.554
36.993
36.167
35.147
33.979
24o
36.030
35.517
34.738
33.760
32.630
27o
34.638
34.158
33.410
32.462
31.358
30o
33.354
32.893
32.167
31.238
30.152
30o
15o
39.129
37.939
36.562
35.056
33.457
18o
37.223
36.144
34.859
33.429
31.894
21o
35.458
34.468
33.258
31.891
30.408
24o
33.820
32.899
31.748
30.432
28.992
27o
32.293
31.425
30.320
29.043
27.637
30o
30.864
30.035
28.965
27.718
26.336
45o
15o
33.483
32.334
30.991
29.508
27.918
18o
31.743
30.679
29.408
27.985
26.444
21o
30.142
29.141
27.925
26.547
25.043
24o
28.660
27.704
26.527
25.182
23.705
27o
27.278
26.353
25.202
23.879
22.420
30 o
25.982
25.074
23.939
22.630
21.179
60o
15o
27.692
26.533
25.186
23.694
22.085
18o
26.156
25.057
23.759
22.307
20.729
21o
24.744
23.683
22.418
20.991
19.432
24o
23.432
22.394
21.147
19.733
18.180
27o
22.203
21.173
19.932
18.520
16.965
30o
21.039
20.008
18.763
17.344
15.776
The shear angle β as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for the non-cavitating cutting process, for hb/hi=2.
Table D-3: β for hb/hi=3, non-cavitating.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
42.346
41.502
40.418
39.164
37.786
18o
40.414
39.674
38.681
37.507
36.198
21o
38.673
38.010
37.086
35.973
34.718
24o
37.087
36.481
35.609
34.542
33.328
27o
35.631
35.064
34.230
33.197
32.013
30 o
34.283
33.742
32.934
31.926
30.763
30o
15o
40.176
38.793
37.257
35.619
33.909
18o
38.242
36.978
35.537
33.977
32.331
21o
36.421
35.258
33.900
32.407
30.817
24o
34.711
33.631
32.341
30.906
29.364
27o
33.103
32.090
30.858
29.470
27.968
30o
31.590
30.631
29.444
28.095
26.625
45o
15o
35.406
33.895
32.248
30.509
28.703
18o
33.548
32.142
30.578
28.907
27.156
21o
31.788
30.472
28.981
27.368
25.665
24o
30.126
28.885
27.455
25.891
24.230
27o
28.557
27.376
25.996
24.474
22.845
30o
27.075
25.941
24.600
23.111
21.509
60o
15o
28.252
26.972
25.516
23.930
22.241
18o
26.613
25.406
24.010
22.472
20.823
21o
25.094
23.940
22.588
21.086
19.464
24o
23.677
22.560
21.238
19.760
18.156
27o
22.348
21.253
19.950
18.485
16.890
30o
21.092
20.008
18.713
17.254
15.600
The shear angle β as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for the non-cavitating cutting process, for hb/hi=3.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.03%3A_Appendix_C-_Dimensionless_Pore_Pressures_p1m_and_p2m.txt
|
Table E-1: c1 for hb/hi=1.
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.104
0.118
0.132
0.146
0.162
18o
0.119
0.134
0.150
0.167
0.186
21o
0.133
0.150
0.169
0.189
0.210
24o
0.147
0.167
0.188
0.211
0.236
27o
0.162
0.184
0.209
0.235
0.264
30o
0.177
0.202
0.229
0.259
0.292
30o
15o
0.175
0.203
0.234
0.268
0.306
18o
0.195
0.227
0.261
0.300
0.343
21o
0.215
0.251
0.290
0.334
0.384
24o
0.236
0.276
0.320
0.370
0.427
27o
0.257
0.302
0.352
0.409
0.474
30o
0.279
0.329
0.385
0.450
0.525
45o
15o
0.254
0.304
0.360
0.425
0.502
18o
0.279
0.334
0.398
0.472
0.560
21o
0.305
0.367
0.438
0.523
0.624
24o
0.332
0.401
0.482
0.578
0.695
27o
0.360
0.437
0.529
0.639
0.774
30o
0.390
0.477
0.580
0.706
0.863
60o
15o
0.360
0.445
0.547
0.671
0.826
18o
0.393
0.488
0.604
0.746
0.928
21o
0.428
0.535
0.666
0.831
1.045
24o
0.466
0.587
0.736
0.928
1.180
27o
0.507
0.643
0.815
1.039
1.341
30o
0.553
0.707
0.905
1.169
1.534
The dimensionless force c1, in the direction of the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=1.
Table E-2: c1 for hb/hi=2.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.106
0.119
0.133
0.148
0.163
18o
0.120
0.135
0.152
0.169
0.187
21o
0.135
0.152
0.171
0.191
0.213
24o
0.149
0.169
0.191
0.214
0.239
27o
0.164
0.187
0.211
0.237
0.267
30o
0.179
0.205
0.232
0.262
0.296
30o
15o
0.185
0.214
0.246
0.281
0.320
18o
0.207
0.240
0.276
0.317
0.362
21o
0.230
0.267
0.308
0.354
0.407
24o
0.254
0.296
0.342
0.395
0.455
27o
0.278
0.325
0.378
0.437
0.507
30o
0.303
0.356
0.415
0.483
0.563
45o
15o
0.282
0.335
0.396
0.466
0.547
18o
0.313
0.373
0.441
0.521
0.616
21o
0.345
0.412
0.490
0.582
0.692
24o
0.379
0.454
0.543
0.648
0.775
27o
0.414
0.499
0.600
0.721
0.869
30o
0.452
0.547
0.662
0.801
0.974
60o
15o
0.415
0.509
0.622
0.760
0.932
18o
0.458
0.565
0.693
0.853
1.056
21o
0.504
0.625
0.772
0.958
1.197
24o
0.554
0.690
0.860
1.077
1.362
27o
0.607
0.762
0.958
1.213
1.556
30o
0.665
0.843
1.070
1.372
1.787
The dimensionless force c1 in the direction of the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=2.
Table E-3: c1 for hb/hi=3.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.105
0.119
0.133
0.148
0.164
18o
0.120
0.135
0.152
0.169
0.188
21o
0.135
0.152
0.171
0.192
0.214
24o
0.150
0.170
0.191
0.215
0.240
27o
0.165
0.188
0.212
0.239
0.268
30o
0.180
0.206
0.234
0.264
0.298
30o
15o
0.185
0.215
0.247
0.282
0.322
18o
0.208
0.241
0.278
0.318
0.364
21o
0.232
0.269
0.310
0.357
0.410
24o
0.256
0.298
0.345
0.398
0.459
27o
0.280
0.328
0.381
0.441
0.511
30o
0.306
0.359
0.419
0.488
0.569
45o
15o
0.290
0.345
0.408
0.480
0.565
18o
0.324
0.386
0.457
0.541
0.640
21o
0.359
0.429
0.511
0.607
0.722
24o
0.396
0.476
0.568
0.679
0.813
27o
0.436
0.525
0.631
0.758
0.914
30o
0.478
0.579
0.699
0.846
1.029
60o
15o
0.439
0.538
0.657
0.802
0.983
18o
0.489
0.601
0.737
0.906
1.120
21o
0.542
0.670
0.826
1.024
1.278
24o
0.599
0.744
0.926
1.157
1.461
27o
0.660
0.827
1.037
1.310
1.676
30o
0.728
0.918
1.163
1.487
1.933
The dimensionless force c1 in the direction of the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=3.
|
textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.05%3A_Appendix_E-_The_Coefficient_c1.txt
|
Table F-1: c2 for hb/hi=1.
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.113
0.137
0.161
0.187
0.215
18o
0.110
0.134
0.159
0.186
0.215
21o
0.106
0.130
0.156
0.184
0.214
24o
0.101
0.126
0.152
0.181
0.213
27o
0.096
0.121
0.148
0.178
0.211
30o
0.090
0.116
0.143
0.174
0.208
30o
15o
0.117
0.146
0.177
0.211
0.249
18o
0.110
0.139
0.171
0.206
0.206
21o
0.103
0.132
0.164
0.200
0.241
24o
0.094
0.123
0.156
0.193
0.235
27o
0.084
0.114
0.147
0.184
0.228
30o
0.074
0.103
0.136
0.174
0.218
45o
15o
0.101
0.130
0.164
0.202
0.247
18o
0.090
0.119
0.152
0.191
0.237
21o
0.078
0.106
0.139
0.178
0.224
24o
0.064
0.092
0.124
0.162
0.208
27o
0.049
0.075
0.106
0.143
0.188
30o
0.032
0.056
0.085
0.120
0.164
60o
15o
0.060
0.084
0.112
0.146
0.189
18o
0.041
0.063
0.088
0.120
0.160
21o
0.021
0.039
0.061
0.088
0.124
24o
-0.003
0.011
0.028
0.050
0.078
27o
-0.030
-0.021
-0.011
0.003
0.021
30o
-0.061
-0.059
-0.057
-0.055
-0.053
The dimensionless force c2 perpendicular to the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=1.
Table F-2: c2 for hb/hi=2.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.113
0.136
0.161
0.187
0.215
18o
0.109
0.133
0.159
0.186
0.216
21o
0.105
0.130
0.156
0.184
0.215
24o
0.101
0.126
0.153
0.182
0.214
27o
0.095
0.121
0.148
0.178
0.212
30o
0.089
0.115
0.143
0.174
0.209
30o
15o
0.113
0.143
0.174
0.209
0.249
18o
0.105
0.135
0.168
0.204
0.245
21o
0.096
0.126
0.160
0.197
0.239
24o
0.086
0.116
0.150
0.188
0.232
27o
0.075
0.105
0.139
0.178
0.223
30o
0.062
0.092
0.127
0.166
0.212
45o
15o
0.092
0.123
0.158
0.199
0.247
18o
0.078
0.109
0.144
0.185
0.234
21o
0.062
0.092
0.127
0.168
0.217
24o
0.044
0.073
0.107
0.148
0.197
27o
0.023
0.051
0.084
0.124
0.173
30o
0.001
0.027
0.058
0.096
0.143
60o
15o
0.042
0.068
0.099
0.137
0.184
18o
0.017
0.040
0.069
0.104
0.148
21o
-0.012
0.008
0.033
0.063
0.103
24o
-0.044
-0.029
-0.010
0.015
0.046
27o
-0.081
-0.071
-0.060
-0.045
-0.025
30o
-0.123
-0.121
-0.120
-0.118
-0.116
The dimensionless force c2 perpendicular to the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=2.
Table F-3: c2 for hb/hi=3.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.113
0.137
0.161
0.188
0.216
18o
0.110
0.134
0.159
0.187
0.216
21o
0.105
0.130
0.156
0.185
0.216
24o
0.101
0.126
0.153
0.182
0.214
27o
0.096
0.121
0.149
0.179
0.212
30o
0.090
0.116
0.144
0.175
0.210
30o
15o
0.113
0.142
0.174
0.209
0.248
18o
0.105
0.135
0.167
0.204
0.244
21o
0.096
0.126
0.159
0.196
0.239
24o
0.085
0.116
0.149
0.188
0.231
27o
0.074
0.104
0.138
0.177
0.222
30o
0.061
0.091
0.125
0.165
0.211
45o
15o
0.089
0.121
0.156
0.197
0.246
18o
0.073
0.105
0.140
0.182
0.232
21o
0.056
0.086
0.122
0.163
0.214
24o
0.035
0.065
0.100
0.141
0.192
27o
0.012
0.041
0.074
0.115
0.164
30o
-0.013
0.013
0.045
0.083
0.131
60o
15o
0.032
0.058
0.090
0.129
0.177
18o
0.002
0.026
0.055
0.091
0.136
21o
-0.031
-0.011
0.014
0.045
0.085
24o
-0.069
-0.054
-0.035
-0.011
0.021
27o
-0.112
-0.104
-0.093
-0.079
-0.059
30o
-0.162
-0.162
-0.162
-0.162
-0.162
The dimensionless force c2 perpendicular to the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=3.
|
textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.06%3A_Appendix_F-_The_Coefficient_c2.txt
|
Table G-1: a1 for hb/hi=1.
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.525
0.520
0.515
0.509
0.503
18o
0.520
0.516
0.510
0.505
0.498
21o
0.516
0.511
0.506
0.500
0.494
24o
0.511
0.507
0.502
0.496
0.490
27o
0.507
0.503
0.498
0.492
0.485
30o
0.503
0.498
0.493
0.487
0.481
30o
15o
0.526
0.522
0.517
0.512
0.506
18o
0.523
0.519
0.514
0.509
0.503
21o
0.520
0.516
0.511
0.506
0.500
24o
0.517
0.512
0.508
0.502
0.497
27o
0.514
0.509
0.504
0.499
0.493
30o
0.510
0.506
0.501
0.496
0.490
45o
15o
0.534
0.530
0.525
0.520
0.514
18o
0.531
0.527
0.522
0.517
0.511
21o
0.528
0.524
0.519
0.514
0.508
24o
0.525
0.521
0.516
0.511
0.505
27o
0.523
0.518
0.513
0.508
0.501
30o
0.520
0.515
0.510
0.504
0.498
60o
15o
0.535
0.528
0.521
0.513
0.505
18o
0.530
0.524
0.517
0.509
0.500
21o
0.526
0.519
0.512
0.504
0.494
24o
0.521
0.515
0.507
0.498
0.489
27o
0.517
0.510
0.502
0.493
0.483
30o
0.512
0.505
0.497
0.487
0.477
The weigh factor a1, for the determination of the weighted average permeability km, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=1.
Table G-2: a1 for hb/hi=2.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.522
0.518
0.513
0.507
0.501
18o
0.518
0.514
0.509
0.503
0.497
21o
0.514
0.510
0.505
0.499
0.493
24o
0.510
0.506
0.501
0.495
0.489
27o
0.506
0.502
0.497
0.491
0.485
30o
0.502
0.498
0.493
0.487
0.481
30o
15o
0.531
0.526
0.521
0.516
0.511
18o
0.527
0.523
0.518
0.513
0.508
21o
0.524
0.520
0.515
0.510
0.505
24o
0.521
0.517
0.512
0.507
0.501
27o
0.518
0.514
0.509
0.504
0.498
30o
0.514
0.510
0.506
0.500
0.495
45o
15o
0.554
0.550
0.546
0.541
0.536
18o
0.552
0.548
0.544
0.539
0.534
21o
0.550
0.546
0.542
0.537
0.532
24o
0.548
0.544
0.539
0.535
0.529
27o
0.546
0.542
0.537
0.532
0.527
30o
0.544
0.540
0.535
0.530
0.524
60o
15o
0.575
0.569
0.563
0.556
0.549
18o
0.571
0.566
0.559
0.552
0.545
21o
0.568
0.562
0.556
0.549
0.541
24o
0.565
0.559
0.552
0.545
0.536
27o
0.561
0.555
0.548
0.541
0.532
30o
0.558
0.552
0.544
0.536
0.527
The weigh factor a1, for the determination of the weighted average permeability km, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=2.
Table G-3: a1 for hb/hi=3.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.522
0.517
0.512
0.507
0.501
18o
0.518
0.513
0.508
0.503
0.497
21o
0.514
0.509
0.504
0.499
0.493
24o
0.510
0.505
0.500
0.495
0.489
27o
0.506
0.501
0.497
0.491
0.485
30o
0.502
0.498
0.493
0.487
0.480
30o
15o
0.534
0.529
0.524
0.519
0.514
18o
0.531
0.526
0.521
0.516
0.511
21o
0.528
0.523
0.519
0.513
0.508
24o
0.525
0.520
0.516
0.511
0.505
27o
0.522
0.517
0.513
0.508
0.502
30o
0.519
0.514
0.510
0.504
0.499
45o
15o
0.552
0.548
0.544
0.540
0.536
18o
0.550
0.547
0.543
0.539
0.534
21o
0.549
0.545
0.541
0.537
0.532
24o
0.547
0.543
0.539
0.535
0.531
27o
0.545
0.542
0.538
0.533
0.529
30o
0.544
0.540
0.536
0.531
0.527
60o
15o
0.580
0.575
0.570
0.565
0.559
18o
0.578
0.573
0.568
0.563
0.557
21o
0.576
0.571
0.566
0.560
0.554
24o
0.573
0.569
0.564
0.558
0.551
27o
0.571
0.566
0.561
0.555
0.548
30o
0.569
0.564
0.558
0.552
0.545
The weigh factor a1, for the determination of the weighted average permeability km, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=3.
|
textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.07%3A_Appendix_G-_The_Coefficient_a1.txt
|
Table H-1: β for hb/hi=1, cavitating.
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15o
15o
37.217
37.520
37.355
36.831
36.026
18o
34.461
34.854
34.790
34.370
33.669
21o
32.163
32.598
32.594
32.243
31.613
24o
30.212
30.661
30.689
30.379
29.796
27o
28.530
28.973
29.012
28.726
28.173
30o
27.060
27.483
27.520
27.243
26.707
30o
15o
39.766
39.060
38.014
36.718
35.232
18o
37.341
36.757
35.823
34.628
33.233
21o
35.196
34.696
33.844
32.725
31.399
24o
33.280
32.837
32.041
30.977
29.704
27o
31.554
31.145
30.387
29.363
28.127
30o
29.985
29.593
28.859
27.860
26.650
45o
15o
36.853
35.599
34.097
32.412
30.591
18o
34.768
33.616
32.202
30.594
28.839
21o
32.866
31.789
30.441
28.892
27.188
24o
31.119
30.094
28.794
27.288
25.623
27o
29.502
28.512
27.246
25.770
24.132
30o
27.996
27.026
25.781
24.325
22.705
60o
15o
31.992
30.395
28.608
26.683
24.654
18o
30.155
28.634
26.911
25.039
23.055
21o
28.444
26.979
25.303
23.471
21.520
24o
26.841
25.414
23.772
21.968
20.040
27o
25.330
23.927
22.306
20.520
18.605
30o
23.897
22.506
20.896
19.118
17.208
The shear angle β as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for the cavitating cutting process, for hb/hi=1.
Table H-2: β for hb/hi=2, cavitating.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
28.724
29.560
29.957
29.994
29.733
18o
26.332
27.162
27.586
27.670
27.472
21o
24.420
25.221
25.643
25.747
25.582
24 o
22.849
23.608
24.014
24.120
23.968
27o
21.528
22.240
22.621
22.716
22.566
30o
20.396
21.059
21.407
21.485
21.329
30o
15o
33.398
33.367
32.937
32.198
31.215
18o
30.972
31.019
30.677
30.027
29.134
21o
28.922
29.011
28.721
28.131
27.299
24o
27.161
27.265
27.004
26.451
25.659
27o
25.622
25.725
25.476
24.944
24.177
30o
24.259
24.349
24.101
23.576
22.823
45o
15o
32.378
31.721
30.741
29.516
28.100
18o
30.207
29.642
28.751
27.610
26.271
21o
28.308
27.801
26.970
25.887
24.605
24o
26.624
26.149
25.357
24.314
23.070
27 o
25.110
24.652
23.881
22.862
21.643
30o
23.736
23.280
22.518
21.512
20.306
60o
15o
28.906
27.806
26.445
24.886
23.174
18o
26.993
25.974
24.686
23.194
21.540
21o
25.276
24.309
23.072
21.626
20.014
24o
23.716
22.781
21.576
20.159
18.574
27o
22.283
21.364
20.176
18.776
17.204
30o
20.955
20.038
18.855
17.461
15.892
The shear angle β as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for the cavitating cutting process, for hb/hi=2.
Table H-3: β for hb/hi=3, cavitating.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
24.046
25.019
25.609
25.872
25.856
18o
21.976
22.900
23.476
23.751
23.765
21o
20.350
21.217
21.763
22.030
22.053
24o
19.031
19.838
20.348
20.596
20.615
27o
17.932
18.680
19.150
19.374
19.381
30o
16.996
17.687
18.117
18.313
18.303
30o
15o
29.286
29.575
29.466
29.038
28.353
18o
26.992
27.319
27.267
26.908
26.297
21o
25.100
25.435
25.410
25.090
24.525
24o
23.504
23.828
23.811
23.511
22.973
27o
22.130
22.433
22.410
22.116
21.592
30o
20.928
21.202
21.165
20.867
20.346
45o
15o
29.236
28.919
28.257
27.325
26.179
18o
27.101
26.853
26.266
25.411
24.339
21o
25.277
25.065
24.524
23.719
22.699
24o
23.690
23.493
22.977
22.203
21.215
27o
22.288
22.091
21.584
20.825
19.857
30o
21.031
20.823
20.315
19.561
18.600
60o
15o
26.619
25.832
24.754
23.450
21.967
18o
24.711
23.995
22.987
21.750
20.329
21o
23.037
22.362
21.398
20.206
18.826
24o
21.543
20.889
19.951
18.785
17.431
27o
20.193
19.545
18.617
17.464
16.121
30o
18.958
18.303
17.374
16.222
14.880
The shear angle β as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for the cavitating cutting process, for hb/hi=3.
|
textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.08%3A_Appendix_H-_The_Shear_Angle__Cavitating.txt
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Table I-1: d1 for hb/hi=1
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15o
15o
1.390
1.505
1.625
1.753
1.890
18o
1.626
1.766
1.913
2.069
2.238
21o
1.860
2.028
2.205
2.393
2.597
24o
2.092
2.291
2.501
2.726
2.970
27o
2.324
2.557
2.803
3.068
3.358
30o
2.556
2.826
3.112
3.423
3.764
30o
15o
1.206
1.374
1.559
1.766
2.000
18o
1.381
1.575
1.791
2.033
2.309
21o
1.559
1.783
2.033
2.315
2.638
24o
1.741
1.998
2.286
2.613
2.991
27o
1.928
2.222
2.552
2.930
3.370
30o
2.121
2.455
2.833
3.269
3.781
45o
15o
1.419
1.688
2.000
2.365
2.800
18o
1.598
1.905
2.262
2.685
3.192
21o
1.784
2.133
2.543
3.032
3.625
24o
1.980
2.376
2.846
3.411
4.105
27o
2.186
2.636
3.174
3.829
4.642
30o
2.404
2.916
3.533
4.292
5.249
60o
15o
1.879
2.331
2.883
3.570
4.444
18o
2.099
2.615
3.252
4.054
5.090
21o
2.336
2.925
3.661
4.602
5.837
24o
2.593
3.267
4.120
5.228
6.711
27o
2.872
3.645
4.639
5.952
7.746
30o
3.179
4.069
5.232
6.798
8.991
The dimensionless force d1, in the direction of the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=1.
Table I-2: d1 for hb/hi=2.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
2.295
2.460
2.627
2.801
2.984
18o
2.683
2.889
3.098
3.315
3.545
21o
3.062
3.313
3.569
3.836
4.119
24o
3.435
3.735
4.042
4.364
4.707
27o
3.803
4.156
4.520
4.903
5.313
30o
4.169
4.579
5.005
5.455
5.941
30o
15o
1.729
1.934
2.156
2.401
2.674
18o
1.997
2.239
2.503
2.794
3.122
21o
2.267
2.550
2.860
3.205
3.593
24o
2.539
2.868
3.230
3.634
4.093
27o
2.815
3.195
3.614
4.085
4.625
30o
3.097
3.532
4.015
4.563
5.195
45o
15o
1.836
2.142
2.492
2.898
3.377
18o
2.093
2.447
2.854
3.330
3.897
21o
2.357
2.765
3.238
3.794
4.462
24o
2.631
3.100
3.646
4.296
5.084
27o
2.917
3.454
4.085
4.843
5.772
30o
3.217
3.830
4.558
5.442
6.541
60o
15o
2.269
2.764
3.364
4.104
5.038
18o
2.567
3.139
3.837
4.710
5.827
21o
2.883
3.543
4.357
5.388
6.728
24o
3.221
3.982
4.933
6.154
7.771
27o
3.586
4.464
5.578
7.031
8.995
30o
3.982
4.998
6.306
8.047
10.453
The dimensionless force d1, in the direction of the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=2.
Table I-3: d1 for hb/hi=3.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
3.145
3.362
3.578
3.799
4.028
18o
3.672
3.945
4.218
4.497
4.789
21o
4.185
4.519
4.855
5.200
5.562
24o
4.687
5.087
5.492
5.910
6.351
27o
5.180
5.652
6.132
6.631
7.159
30o
5.667
6.216
6.778
7.366
7.993
30o
15o
2.216
2.458
2.717
3.000
3.312
18o
2.567
2.858
3.169
3.510
3.889
21o
2.919
3.262
3.632
4.038
4.492
24o
3.272
3.673
4.107
4.587
5.127
27o
3.629
4.093
4.599
5.162
5.799
30o
3.991
4.525
5.110
5.766
6.515
45o
15o
2.222
2.566
2.954
3.402
3.925
18o
2.549
2.951
3.408
3.938
4.562
21o
2.883
3.350
3.885
4.509
5.252
24o
3.228
3.768
4.391
5.123
6.004
27o
3.585
4.207
4.929
5.788
6.831
30o
3.958
4.671
5.508
6.513
7.750
60o
15o
2.632
3.170
3.817
4.610
5.605
18o
2.999
3.627
4.387
5.329
6.526
21o
3.387
4.116
5.008
6.128
7.572
24o
3.799
4.645
5.692
7.025
8.774
27o
4.240
5.222
6.453
8.044
10.175
30o
4.717
5.856
7.307
9.217
11.833
The dimensionless force d1, in the direction of the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=3.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.09%3A_Appendix_I-_The_Coefficient_d1.txt
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Table J-1: d2 for hb/hi=1.
hb/hi=1
φ
32o
37o
42o
47o
52o
α δ
15o
15o
0.409
0.608
0.816
1.037
1.274
18o
0.312
0.528
0.754
0.995
1.255
21o
0.205
0.436
0.680
0.939
1.220
24o
0.087
0.333
0.592
0.870
1.172
27o
-0.040
0.219
0.493
0.788
1.110
30o
-0.175
0.095
0.382
0.692
1.034
30o
15o
0.474
0.642
0.828
1.035
1.269
18o
0.412
0.588
0.782
1.000
1.249
21o
0.341
0.523
0.725
0.954
1.216
24o
0.261
0.447
0.657
0.895
1.169
27o
0.171
0.361
0.576
0.822
1.108
30o
0.071
0.264
0.483
0.735
1.031
45o
15 o
0.398
0.553
0.733
0.945
1.196
18o
0.325
0.481
0.664
0.879
1.138
21o
0.241
0.396
0.579
0.797
1.061
24o
0.145
0.298
0.478
0.696
0.962
27o
0.037
0.183
0.358
0.572
0.836
30o
-0.086
0.051
0.217
0.421
0.678
60o
15o
0.195
0.317
0.465
0.650
0.885
18o
0.083
0.193
0.329
0.500
0.721
21o
-0.047
0.047
0.164
0.313
0.510
24o
-0.198
-0.126
-0.036
0.081
0.238
27o
-0.372
-0.331
-0.278
-0.208
-0.113
30o
-0.575
-0.574
-0.573
-0.572
-0.570
The dimensionless force d2, perpendicular to the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=1.
Table J-2: d2 for hb/hi=2.
hb/hi=2
φ
32o
37o
42o
47o
52o
α δ
15o
15o
-0.024
0.262
0.552
0.853
1.170
18o
-0.253
0.064
0.387
0.722
1.076
21o
-0.496
-0.151
0.202
0.569
0.959
24o
-0.752
-0.381
-0.001
0.396
0.820
27o
-1.018
-0.626
-0.221
0.204
0.660
30o
-1.294
-0.884
-0.458
-0.007
0.479
30o
15o
0.266
0.471
0.693
0.938
1.211
18o
0.136
0.354
0.592
0.854
1.149
21o
-0.008
0.222
0.473
0.752
1.067
24o
-0.165
0.074
0.337
0.631
0.965
27o
-0.336
-0.089
0.183
0.490
0.841
30o
-0.520
-0.268
0.011
0.327
0.693
45o
15o
0.216
0.393
0.595
0.830
1.107
18o
0.087
0.267
0.475
0.718
1.007
21o
-0.059
0.123
0.334
0.582
0.880
24o
-0.221
-0.040
0.170
0.420
0.723
27o
-0.401
-0.226
-0.020
0.227
0.529
30 o
-0.600
-0.435
-0.240
-0.002
0.293
60o
15o
-0.009
0.124
0.285
0.484
0.735
18o
-0.182
-0.060
0.089
0.275
0.513
21o
-0.379
-0.274
-0.145
0.019
0.233
24o
-0.603
-0.523
-0.422
-0.293
-0.122
27o
-0.859
-0.812
-0.753
-0.676
-0.571
30o
-1.151
-1.151
-1.150
-1.148
-1.146
The dimensionless force d2, perpendicular to the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=2.
Table J-3: d2 for hb/hi=3.
hb/hi=3
φ
32o
37o
42o
47o
52o
α δ
15o
15o
-0.552
-0.177
0.198
0.581
0.979
18o
-0.921
-0.501
-0.080
0.350
0.800
21o
-1.306
-0.846
-0.384
0.092
0.590
24o
-1.703
-1.208
-0.708
-0.191
0.353
27o
-2.111
-1.586
-1.053
-0.498
0.090
30o
-2.528
-1.979
-1.417
-0.828
-0.201
30o
15o
0.020
0.263
0.522
0.805
1.118
18o
-0.182
0.079
0.360
0.667
1.009
21o
-0.402
-0.124
0.176
0.505
0.873
24o
-0.638
-0.346
-0.030
0.319
0.711
27o
-0.890
-0.588
-0.259
0.107
0.521
30o
-1.158
-0.850
-0.511
-0.132
0.301
45o
15o
0.017
0.215
0.440
0.698
1.001
18o
-0.171
0.034
0.267
0.537
0.856
21o
-0.379
-0.171
0.068
0.346
0.677
24o
-0.608
-0.400
-0.160
0.122
0.460
27o
-0.858
-0.656
-0.420
-0.141
0.199
30o
-1.133
-0.941
-0.717
-0.447
-0.114
60o
15o
-0.221
-0.076
0.097
0.310
0.578
18o
-0.455
-0.321
-0.159
0.042
0.298
21o
-0.718
-0.602
-0.460
-0.282
-0.052
24o
-1.014
-0.925
-0.814
-0.673
-0.488
27o
-1.349
-1.297
-1.231
-1.147
-1.034
30o
-1.728
-1.727
-1.726
-1.724
-1.722
The dimensionless force d2, perpendicular to the cutting velocity, as a function of the blade angle α, the angle of internal friction φ, the soil/interface friction angle δ, for hb/hi=3.
|
textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.10%3A_Appendix_J-_The_Coefficient_d2.txt
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The sand in the old laboratory DE, with a d50 of 200 μm, is examined for the following soil mechanical parameters:
1. The minimum and the maximum density, Table K-1: Pore percentages.
2. The dry critical density, Table K-1: Pore percentages.
3. The saturated critical density, Table K-1: Pore percentages. The permeability as a function of the density, Table K-2: Permeability as a function of the porosity.
4. The angle of internal friction as a function of the density, Table K-4: The angle of internal friction as function of the pore percentage.
5. The d50 as a function of the time, Table K-3: The d50 of the sand as function of the time.
6. The cone resistance per experiment.
7. The density in the test stand in combination with the cone resistance.
The points 7 and 8 need some explanation. With the aid of a Troxler density measuring set density measurements are performed in situ, that is in the test stand. During each measurement the cone resistance is determined at the same position. In this way it is possible to formulate a calibration formula for the density as a function of the cone resistance. The result is:
$\ \mathrm{n=\frac{65.6}{C_{p}^{0.082}} \quad\text{ with: n in %, } \mathrm{C_{p}}\text{ in kPa}} \tag{k-1}$
In which the cone resistance is determined in a top layer of 18 cm, where the cone resistance was continuously increasing and almost proportional with the depth. The value to be used in this equation is the cone resistance for the 18 cm depth.
With the aid of this equation it was possible to determine the density for each cutting test from the cone resistance measurements. The result was an average pore percentage of 38.53% over 367 tests.
By interpolating in Table K-2 it can be derived that a pore percentage of 38.53% corresponds to a permeability of 0.000165 m/s. By extrapolating in this table it can also be derived that the maximum pore percentage of 43.8% corresponds to a permeability of approximately 0.00032 m/s. At the start of the cutting tests the pore percentage was averaged 38%, which corresponds to a permeability of 0.00012 m/s.
Table K-1: Pore percentages.
Minimum density
43.8%
Maximum density
32.7%
Dry critical density
39.9%
Saturated critical density
40.7%-41.7%
Initial density
38.5%
Table K-2: Permeability as a function of the porosity.
Pore percentage
Permeability (m/s)
36.97%
0.000077
38.48%
0.000165
38.98%
0.000206
39.95%
0.000240
40.88%
0.000297
41.84%
41.84%
43.07%
0.000289
43.09%
0.000322
Table K-3: The d50 of the sand as function of the time.
Date
d50 (mm)
22-09-1982
0.175
17-12-1984
0.180
02-01-1985
0.170
08-01-1985
0.200
14-01-1985
0.200
21-01-1985
0.200
28-01-1985
0.195
04-02-1985
0.205
26-02-1985
0.210
Table K-4: The angle of internal friction as function of the pore percentage.
Pore percentage
Cell pressure kPa
Angle of internal friction
Dry
43.8%
50
35.1o
41.2%
50
36.0o
39.9%
50
38.3o
Saturated undrained
43.8%
100
30.9o
42.1%
10
31.2o
42.1%
50
31.2o
42.1%
100
31.6o
42.2%
100
32.0o
41.8%
10
33.1o
41.3%
10
31.9o
41.2%
50
32.2o
41.1%
50
30.1o
41.1%
100
31.3o
41.1%
100
33.7o
41.0%
100
35.2o
40.5%
10
33.8o
40.3%
50
33.7o
40.4%
100
33.1o
39.8%
10
34.1o
39.2%
10
33.8o
39.2%
50
33.8o
39.2%
100
33.9o
38.2%
10
35.2o
38.1%
50
35.3o
38.1%
100
35.0o
37.3%
10
37.4o
37.0%
10
38.6o
37.0%
50
37.3o
36.9%
100
36.8o
36.2%
100
38.0o
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.11%3A_Appendix_K-_The_Properties_of_the_200_m_Sand.txt
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The sand in the new laboratory DE, with a d50 of 105 μm, is examined for the following soil mechanical parameters:
1. The minimum and the maximum density, Table L-1: Pore percentages, indicated are the average measured densities for the various blade angles.
2. The saturated critical density, Table L-1: Pore percentages, indicated are the average measured densities for the various blade angles.
3. The permeability as a function of the density, Table L-2: Permeabilities, indicated are the average permeabilities for the various blade angles.
4. The angle of internal friction as a function of the density, Table L-4: The angle of internal friction as a function of the pore percentage.
5. The d50 as a function of the time, Table L-3: The d50 of the sand as a function of time.
6. The cone resistance per test.
7. The density in the test stand in combination with the cone resistance.
The points 6 and 7 need some explanation. As with the 200 μm sand density measurements are performed in situ with the aid of a Troxler density measuring set. The calibration formula for the 105 μm sand is:
$\ \mathrm{n=\frac{69.9}{C_{p}^{0.068}} \quad\text{ with: n in %, }C_{p}\text{ in kPa}}\tag{L-1}$
In which the cone resistance is determined in a top layer of 12 cm, where the cone resistance was continuously increasing and almost proportional with the depth. The value to be used in this equation is the cone resistance for the 12 cm depth.
With the aid of this equation it was possible to determine the density for each cutting test from the cone resistance measurements. As, however, new sand was used, the density showed changed in time. The sand was looser in the first tests than in the last tests. This resulted in different average initial densities for the different test series. The tests with a 45o blade were performed first with an average pore percentage of 44.9%. The tests with the 60o blade were performed with an average pore percentage of 44.2%. The tests with the 30o blade were performed with an average pore percentage of 43.6%. Because of the consolidation of the sand a relatively large spread was found in the first tests.
Table L-2 lists the permeabilities corresponding to the mentioned pore percentages. By extrapolation in Table L-2 a permeability of 0.00017 m/s is derived for the maximum pore percentage of 51.6%.
The sand bed is flushed after the linear tests because of the visibility in the water above the sand. In the tables it is indicated which soil mechanical parameters are determined after the flushing of the sand bed.
Table L-1: Pore percentages, indicated are the average measured densities for the various blade angles.
Minimum density
51.6%
Maximum density
38.3%
Initial density 30 o
43.6%
Initial density 45 o
44.9%
Initial density 60 o
44.2%
After the flushing
Minimum density
50.6%
Maximum density
37.7%
Saturated critical density
44.5%
Table L-2: Permeabilities, indicated are the average permeabilities for the various blade angles.
Pore percentage
Permeability (m/s)
42.2%
0.000051
45.6%
0.000082
47.4%
0.000096
49.4%
0.000129
Initial
43.6%
0.000062
44.2%
0.000067
44.9%
0.000075
After the flushing
39.6%
0.000019
40.7%
0.000021
41.8%
0.000039
43.8%
0.000063
45.7%
0.000093
48.3%
0.000128
Table L-3: The d50 of the sand as a function of time.
Date
d50 (mm)
06-08-1986
0.102
06-08-1986
0.097
06-08-1986
0.104
06-08-1986
0.129
06-08-1986
0.125
06-08-1986
0.123
29-08-1986
0.105
29-08-1986
0.106
29-08-1986
0.102
16-09-1986
0.111
16-09-1986
0.105
16-09-1986
0.107
Table L-4: The angle of internal friction as a function of the pore percentage.
Pore percentage
Cell pressure kPa
Angle of internal friction
Saturated undrained
After the flushing
44.7%
100
33.5o
44.9%
200
33.3o
44.5%
400
32.8o
42.6%
100
35.0o
42.1%
200
35.5o
42.2%
400
34.8o
39.8%
100
38.6o
39.9%
200
38.3o
39.6%
400
37.9o
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.12%3A_Appendix_L-_The_Properties_of_the_105_m_Sand.txt
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M.1 Pore pressures and cutting forces in 105 μm Sand
The cutting forces on the blade. Experiments in 105 μm sand, with α=30°, β=30o, φ=41o, δ=27o, ni=43.6%, nmax=51.6%, ki=0.000062 m/s, kmax=0.000170 m/s, hi=100 mm, hb=100 mm, w=0.2 m, z=0.6 m and a partial cavitating cutting process.
The cutting forces on the blade. Experiments in 105 μm sand, with α=45°, β=30o, φ=38o, δ=25o, ni=45.0%, nmax=51.6%, ki=0.000075 m/s, kmax=0.000170 m/s, hi=70 mm, hb=100 mm, w=0.2 m, z=0.6 m and a partial cavitating cutting process.
The cutting forces on the blade. Experiments in 105 μm sand, with α=60°, β=30o, φ=36o, δ=24o, ni=44.3%, nmax=51.6%, ki=0.000067 m/s, kmax=0.000170 m/s, hi=58 mm, hb=100 mm, w=0.2 m, z=0.6 m and a partial cavitating cutting process.
M.2 Pore Pressures in 200 μm Sand
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=30°, β=30o, φ=38o, δ=30o, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=33 mm, hb=100 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=30°, β=29°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=50 mm, hb=100 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=30°, β=29°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=100 mm, hb=100 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=45°, β=25°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=47 mm, hb=141 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=45°, β=24oφ=38oδ=30oni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=70 mm, hb=141 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=45°, β=25°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=141 mm, hb=141 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=60°, β=19oφ=38oδ=30oni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=30 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=60°, β=19oφ=38oδ=30oni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=58 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=60°, β=19°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=87 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The dimensionless water pore pressures on the blade. Experiments in 200 μm sand, with α=60°, β=20°, φ=38o, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=173 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
M.3 Cutting Forces in 200 μm Sand
The cutting forces Fh and Fv on the blade. Experiments in 200 μm sand, with α=30°, β=30°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=33 mm, hb=100 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fon the blade. Experiments in 200 μm sand, with α=30°, β=30°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=50 mm, hb=100 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fv on the blade. Experiments in 200 μm sand, with α=30°, β=30oφ=38oδ=30oni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=100 mm, hb=100 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fv on the blade. Experiments in 200 μm sand, with α=45°, β=30o, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=47 mm, hb=141 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fon the blade. Experiments in 200 μm sand, with α=45°, β=30°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=70 mm, hb=141 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fv on the blade. Experiments in 200 μm sand, with α=45°, β=30°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=141 mm, hb=141 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fon the blade. Experiments in 200 μm sand, with α=45°, β=30oφ=38oδ=30oni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=58 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fv on the blade. Experiments in 200 μm sand, with α=45°, β=30°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=87 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
The cutting forces Fh and Fv on the blade. Experiments in 200 μm sand, with α=45°, β=30°, φ=38°, δ=30°, ni=38.53%, nmax=43.88%, ki=0.000165 m/s, kmax=0.000320 m/s, hi=173 mm, hb=173 mm, w=0.2 m, z=0.6 m and a non-cavitating cutting process.
17.14: Appendix N- The Snow Plough Effect
The 105 μm sand properties from Appendix L were used.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.13%3A_Appendix_M-_Experiments_in_Water_Saturated_Sand.txt
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Table R-1: The average water pore pressure and total pressure along the four sides.
θ=30o
β
p1m
p2m
p3m
p4m
15o
0.2489
0.0727
0.1132
0.0313
20o
0.2675
0.0713
0.1133
0.0290
25o
0.2852
0.0702
0.1139
0.0268
30o
0.3014
0.0695
0.1149
0.0249
θ=40o
β
p1m
p2m
p3m
p4m
15o
0.2798
0.1040
0.1728
0.0688
20o
0.2980
0.1047
0.1788
0.0672
25o
0.3145
0.1036
0.1827
0.0640
30o
0.3291
0.1022
0.1859
0.0607
θ=50o
β
p1m
p2m
p3m
p4m
15o
0.3043
0.1338
0.2357
0.1141
20o
0.3240
0.1377
0.2523
0.1158
25o
0.3404
0.1373
0.2635
0.1134
30o
0.3544
0.1353
0.2722
0.1096
θ=55o
β
p1m
p2m
p3m
p4m
15o
0.3152
0.1492
0.2720
0.1392
20o
0.3367
0.1549
0.2967
0.1435
25o
0.3540
0.1549
0.3143
0.1422
30o
0.3684
0.1526
0.3284
0.1388
θ=59o
β
p1m
p2m
p3m
p4m
15o
0.3242
0.1626
0.3089
0.1607
20o
0.3481
0.1699
0.3436
0.1676
25o
0.3675
0.1705
0.3707
0.1679
30o
0.3838
0.1683
0.3922
0.1654
For α=60o; hi=1; hb=3; ki/kmax=0.25.
Table R-2: The average water pore pressure and total pressure along the four sides.
θ=30o
β
p1m
p2m
p3m
p4m
15o
0.2489
0.0727
0.1132
0.0313
20o
0.2675
0.0713
0.1133
0.0290
25o
0.2852
0.0702
0.1139
0.0268
30o
0.3014
0.0695
0.1149
0.0249
θ=40o
β
p1m
p2m
p3m
p4m
15o
0.2798
0.1040
0.1728
0.0688
20o
0.2980
0.1047
0.1788
0.0672
25o
0.3145
0.1036
0.1827
0.0640
30o
0.3291
0.1022
0.1859
0.0607
θ=50o
β
p1m
p2m
p3m
p4m
15o
0.3043
0.1338
0.2357
0.1141
20o
0.3240
0.1377
0.2523
0.1158
25o
0.3404
0.1373
0.2635
0.1134
30o
0.3544
0.1353
0.2722
0.1096
θ=55o
β
p1m
p2m
p3m
p4m
15o
0.3152
0.1492
0.2720
0.1392
20o
0.3367
0.1549
0.2967
0.1435
25o
0.3540
0.1549
0.3143
0.1422
30o
0.3684
0.1526
0.3284
0.1388
θ=59o
β
p1m
p2m
p3m
p4m
15o
0.3242
0.1626
0.3089
0.1607
20o
0.3481
0.1699
0.3436
0.1676
25o
0.3675
0.1705
0.3707
0.1679
30o
0.3838
0.1683
0.3922
0.1654
For α=60o; hi=1; hb=3; ki/kmax=0.25.
Table R-2: The average water pore pressure and total pressure along the four sides.
θ=30o
β
p1m
p2m
p3m
p4m
15o
0.2499
0.0773
0.1071
0.0339
20o
0.2679 0.0735
0.1048
0.0292
25o
0.2854
0.0715
0.1041
0.0261
30o
0.3015
0.0704
0.1043
0.0240
θ=40o
β
p1m
p2m
p3m
p4m
15o
0.2825
0.1127
0.1622
0.0712
20o
0.2992 0.1088 0.1626 0.0651
25o
0.3152
0.1060
0.1634
0.0603
30o
0.3297
0.1039
0.1646
0.0564
θ=50o
β
p1m
p2m
p3m
p4m
15o
0.3088
0.1438
0.2160
0.1129
20o
0.3259 0.1422 0.2230 0.1086
25o
0.3414
0.1399
0.2283
0.1038
30o
0.3549
0.1373
0.2325
0.0992
θ=55o
β
p1m
p2m
p3m
p4m
15o
0.3175
0.1496
0.1994
0.1104
20o
0.3382 0.1556 0.2103 0.1128
25o
0.3547
0.1563
0.2156
0.1110
30o
0.3682
0.1548
0.2184
0.1076
θ=60o
β
p1m
p2m
p3m
p4m
15o
0.3300
0.1719
0.2720
0.1562
20o
0.3498 0.1745 0.2907 0.1567
25o
0.3664
0.1736
0.3043
0.1540
30o
0.3803
0.1710
0.3150
0.1497
θ=60o
β
p1m
p2m
p3m
p4m
15o
0.3474
0.1984
0.3369
0.1970
20o
0.3737
0.2066
0.3760
0.2050
25o 0.3953 0.2081 0.4060 0.2063
30o
0.4134
0.2062
0.4306
0.2041
For α=70o; hi=1; hb=3; ki/kmax=0.25.
Table R-3: The average water pore pressure and total pressure along the four sides.
θ=30o
β
p1m
p2m
p3m
p4m
15o
0.2493
0.0738
0.0973
0.0279
20o
0.2679 0.0723 0.0966 0.0260
25o
0.2856
0.0712
0.0962
0.0242
30o
0.3018
0.0705
0.0964
0.0226
θ=40o
β
p1m
p2m
p3m
p4m
15o
0.2810
0.1058
0.1450
0.0595
20o
0.2992 0.1065 0.1481 0.0581
25o
0.3156
0.1055
0.1493
0.0555
30o
0.3302
0.1042
0.1501
0.0527
θ=50o
β
p1m
p2m
p3m
p4m
15o
0.3062
0.1352
0.1917
0.0967
20o
0.3257 0.1393 0.2010 0.0978
25o
0.3420
0.1393
0.2010 0.0978
30o
0.3557
0.1378
0.2085
0.0919
θ=55o
β
p1m
p2m
p3m
p4m
15o
0.3170
0.1495
0.2153
0.1167
20o
0.3378 0.1554 0.2284 0.1195
25o
0.3542
0.1560
0.2355
0.1176
30o
0.3678
0.1544
0.2400
0.1140
θ =60o
β
p1m
p2m
p3m
p4m
15o
0.3271
0.1639
0.2398
0.1375
20o
0.3493 0.1716 0.2572 0.1422
25o
0.3663
0.1728
0.2672
0.1411
30o
0.3799
0.1712
0.2739
0.1375
θ =70o
β
p1m
p2m
p3m
p4m
15o
0.3465
0.1944
0.2946
0.1820
20o
0.3727 0.2057 0.3231 0.1914
25o
0.3922
0.2082
0.3419
0.1923
30o
0.4070
0.2062
0.3549
0.1890
For α=80o; hi=1; hb=3; ki/kmax=0.25.
Table R-4: The average water pore pressure and total pressure along the four sides.
θ=30o
β
p1m
p2m
p3m
p4m
15o
0.2494
0.0740
0.0917
0.0270
20o
0.2680 0.0726 0.0908 0.0252
25o
0.2857
0.0715
0.0902
0.0235
30o
0.3018
0.0708
0.0901
0.0220
θ=40o
β
p1m
p2m
p3m
p4m
15o
0.2813
0.1062
0.1358
0.0569
20o
0.2995 0.1070 0.1381 0.0556
25o
0.3159
0.1060
0.1387
0.0530
30o
0.3305
0.1047
0.1389
0.0504
θ=50o
β
p1m
p2m
p3m
p4m
15o
0.3067
0.1355
0.1782
0.0917
20o
0.3262 0.1397 0.1860 0.0926
25o
0.3424
0.1397
0.1893
0.0904
30o
0.3561
0.1383
0.1910
0.0871
θ=55o
β
p1m
p2m
p3m
p4m
15o
0.3175
0.1496
0.1994
0.1104
20o
0.3382 0.1556 0.2103 0.1128
25o
0.3547
0.1563
0.2156
0.1110
30o
0.3682
0.1548
0.2184
0.1076
θ=60o
β
p1m
p2m
p3m
p4m
15o
0.3276
0.1637
0.2209
0.1296
20o
0.3497 0.1713 0.2353 0.1338
25o
0.3666
0.1727
0.2428
0.1327
30o
0.3800
0.1713
0.2471
0.1292
θ=70o
β
p1m
p2m
p3m
p4m
15o
0.3464
0.1927
0.2670
0.1706
20o
0.3719 0.2038 0.2894 0.1780
25o
0.3907
0.2065
0.3027
0.1793
30o
0.4047
0.2049
0.3110
0.1759
θ=80o
β
p1m
p2m
p3m
p4m
15o
0.3658
0.2253
0.3216
0.2157
20o
0.3965
0.2400
0.3556
0.2289
25o 0.4185 0.2435 0.3776 0.2311
30o
0.4347
0.2411
0.3930
0.2277
For θ=90o; hi=1; hb=3; ki/kmax=0.25.
Table R-5: Acting points for α=90o; hi=1; hb=3; ki/kmax=0.25.
θ
β
E2
E3
E4
60o
15o
0.36
0.56
0.40
60o
20o
0.36
0.56
0.40
60o
25o
0.35
0.56
0.39
60o
30o
0.34
0.57
0.39
55o
15o
0.35
0.56
0.40
55o
20o
0.34
0.56
0.40
55o
25o
0.33
0.57
0.39
55o
30o
0.33
0.57
0.39
50o
15o
0.34
0.58
0.40
50o
20o
0.34
0.58
0.40
50o
25o
0.33
0.59
0.40
50o
30o
0.33
0.59
0.40
40o
15o
0.32
0.61
0.40
40o
20o
0.31
0.62
0.40
40o
25o
0.30
0.62
0.40
40o
30o
0.29
0.63
0.39
Table R-6: Acting points for α=80o; hi=1; hb=3; ki/kmax=0.25.
θ
β
E2
E3
E4
60o
15o
0.36
0.54
0.39
60o
20o
0.35
0.54
0.39
60o
25o
0.35
0.55
0.38
60o
30o
0.34
0.56
0.37
55o
15o
0.35
0.55
0.40
55o
20o
0.34
0.55
0.39
55o
25o
0.33
0.56
0.39
55o
30o
0.33
0.57
0.38
50o
15o
0.35
0.57
0.40
50o
20o
0.34
0.57
0.40
50o
25o
0.33
0.58
0.39
50o
30o
0.32
0.58
0.39
Table R-7: Acting points for α=70o; hi=1; hb=3; ki/kmax=0.25.
θ
β
E2
E3
E4
60o
15o
0.36
0.53
0.36
60o
20o
0.35
0.53
0.35
60o
25o
0.34
0.54
0.34
60o
30o
0.34
0.54
0.34
55o
15o
0.35
0.54
0.37
55o
20o
0.34
0.54
0.37
55o
25o
0.33
0.54
0.36
55o
30o
0.33
0.54
0.35
50o
15o
0.34
0.55
0.38
50o
20o
0.33
0.55
0.38
50o
25o
0.32
0.56
0.37
50o
30o
0.31
0.56
0.36
Table R-8: Acting points for α=60o; hi=1; hb=3; ki/kmax=0.25.
θ
β
E2
E3
E4
55o
15o
0.34
0.52
0.34
55o
20o
0.32
0.52
0.33
55o
25o
0.31
0.52
0.32
55o
30o
0.30
0.52
0.31
50o
15o
0.33
0.54
0.35
50o
20o
0.32
0.54
0.34
50o
25o
0.31
0.54
0.34
50o
30o
0.31
0.54
0.33
17.19: Appendix S- FEM Calculations with Wedge
S.2 The 60 Degree Blade
The wedge angle in these calculations is 59 degrees. The pore pressures on the blade C-D are almost equal to the pore pressures on the front of the wedge A-C, which they should be with a blade angle of 60 degrees and a wedge angle of 59 degrees. The pore pressures on the front of the wedge C-A are drawn in red on top of the pore pressures on the blade C-A and match almost exactly.
17.22: Appendix V- Clay Cutting Charts
V.1 The Flow Type
Shear angle and cutting forces for a layer thickness hi=0.1 m, a blade width w=1 m and a strain rate factor λc=1.
17.23: Appendix W- Rock Cutting Charts
W.1 Brittle Shear
In all figures an example is given for a 60o blade and an internal friction angle of 20o.
W.2 The Transition Brittle Shear/Brittle Tensile A
Below the lines the cutting process is subject to shear failure, above the lines to tensile failure. The curves are based on the shear angle resulting from shear failure.
In all figures an example is given for a 60o blade and an internal friction angle of 20o.
W.3 The Transition Brittle Shear/Brittle Tensile B
Below the lines the cutting process is subject to shear failure, above the lines to tensile failure. The curves are based on the shear angle resulting from tensile failure.
In all figures an example is given for a 60o blade and an internal friction angle of 20o.
In all figures an example is given for a 60o blade and an internal friction angle of 20o.
W.5 Brittle Tensile Failure based on Brittle Shear Shear Angle
The Evans approach gives much higher values, since it is based on reaching the tensile strength in the whole failure plane. The DSCRCM model assumes reaching the tensile strength only at the start of the tensile crack.
17.25: Appendix Y- Applications and Equipment
Y.13 PDC Cutters (Oil & Gas Drilling)
PDC cutters (Polycrystalline Diamond Composite) are widely used in petroleum/oil field PDC bit, eological PDC exploration bits, gas exploration bits, PCD/PDC coal mining drill bits, ilfield drilling bits, currently, we developed new type of PDC cutters, and the length of the cutter is 32mm, with a round radius of the tungsten carbide substrate of the PDC cutter.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.18%3A_Appendix_R-_Pore_Pressures_with_Wedge.txt
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1. Miedema, S.A., "The soil reaction forces on a crown cutter head on a swell compensated ladder". LaO/81/97, Delft University of Technology, 1981, 36 pages.
2. Miedema, S.A., "Computer program for the determination of the reaction forces on a cutter head, resulting from the motions of the cutter head". Delft Hydraulics, 1981, 82 pages.
3. Miedema, S.A., "The mathematical modeling of the soil reaction forces on a cutter head and the development of the computer program DREDMO". CO/82/125, Delft University of Technology, 1982, with appendices 600 pages.
4. Miedema, S.A., "The Interaction between Cutter head and Soil at Sea" (In Dutch). Proc. Dredging Day November 19th , Delft University of Technology 1982.
5. Koning, J. de, Miedema, S.A., & Zwartbol, A., "Soil/Cutter head Interaction under Wave Conditions ". Proc. WODCON X, Singapore 1983.
6. Miedema, S.A., "Mathematical Modeling of a Seagoing Cutter Suction Dredge" (In Dutch). Published: The Hague, 18-9-1984, KIVI Lectures, Section Under Water Technology.
7. Miedema, S.A., "The Cutting of Densely Compacted Sand under Water ". Terra et Aqua No. 28, October 1984 pp. 4-10.
8. Miedema, S.A., "Mathematical Modeling of the Cutting of Densely Compacted Sand Under Water". Dredging & Port Construction, July 1985, pp. 22-26.
9. Miedema, S.A., "Derivation of the Differential Equation for Sand Pore Pressures". Dredging & Port Construction, September 1985, pp. 35.
10. Miedema, S.A., "The Application of a Cutting Theory on a Dredging Wheel ". Proc. WODCON XI, Brighton 1986.
11. Miedema, S.A., "Underwater Soil Cutting: a Study in Continuity". Dredging & Port Construction, June 1986, pp. 47-53.
12. Miedema, S.A., "The cutting of water saturated sand, laboratory research" (In Dutch). Delft University of Technology, 1986, 17 pages.
13. Miedema, S.A., "The forces on a trenching wheel, a feasibility study" (In Dutch). Delft, 1986, 57 pages + software.
14. Miedema, S.A., "Calculation of the Cutting Forces when Cutting Water Saturated Sand ". Basic Theory and Applications for 3-D Blade Movements and Periodically Varying Velocities for, in Dredging Commonly used Excavating Means. Ph.D. Thesis, Delft University of Technology, September 15th 1987.
15. Bakker, A. & Miedema, S.A., "The Specific Energy of the Dredging Process of a Grab Dredge". Delft University of Technology, 1988, 30 pages.
16. Miedema, S.A., "On the Cutting Forces in Saturated Sand of a Seagoing Cutter Suction Dredge". Proc. WODCON XII, Orlando, Florida, USA, April 1989. This paper was given the IADC Award for the best technical paper on the subject of dredging in 1989.
17. Miedema, S.A., "On the Cutting Forces in Saturated Sand of a Seagoing Cutter Suction Dredge". Terra et Aqua No. 41, December 1989, Elsevier’s Scientific Publishers.
18. Miedema, S.A., "New Developments of Cutting Theories with respect to Dredging, the Cutting of Clay". Proc. WODCON XIII, Bombay, India, 1992.
19. Miedema, S.A. & Journee, J.M.J. & Schuurmans, S., "On the Motions of a Seagoing Cutter Dredge, a Study in Continuity". Proc. WODCON XIII, Bombay, India, 1992.
20. Becker, S. & Miedema, S.A. & Jong, P.S. de & Wittekoek, S., "On the Closing Process of Clamshell Dredges in Water Saturated Sand". Proc. WODCON XIII, Bombay, India, 1992. This paper was given the IADC Award for the best technical paper on the subject of dredging in 1992.
21. Becker,S.&Miedema,S.A.&Jong,P.S.de&Wittekoek,S.,"TheClosingProcessofClamshellDredges in Water Saturated Sand". Terra et Aqua No. 49, September 1992, IADC, The Hague.
22. Miedema, S.A., "Modeling and Simulation of Dredging Processes and Systems". Symposium "Zicht op Bagger processen", Delft University of Technology, Delft, The Netherlands, 29 October 1992.
23. Miedema, S.A. & Becker, S., "The Use of Modeling and Simulation in the Dredging Industry, in Particular the Closing Process of Clamshell Dredges", CEDA Dredging Days 1993, Amsterdam, Holland, 1993.
24. Miedema, S.A., "On the Snow-Plough Effect when Cutting Water Saturated Sand with Inclined Straight Blades". ASCE Proc. Dredging 94, Orlando, Florida, USA, November 1994.
25. Miedema, S.A., "Production Estimation Based on Cutting Theories for Cutting Water Saturated Sand". Proc. WODCON IV, November 1995, Amsterdam, The Netherlands 1995.
26. Miedema, S.A. & Zhao Yi, "An Analytical Method of Pore Pressure Calculations when Cutting Water Saturated Sand". Texas A&M 33nd Annual Dredging Seminar, June 2001, Houston, USA 2001.
27. ZhaoYi,&Miedema,S.A.,"Finite Element Calculations To Determine The Pore Pressures When Cutting Water Saturated Sand At Large Cutting Angles". CEDA Dredging Day 2001, November 2001, Amsterdam, the Netherlands.
28. Miedema, S.A., & Ma, Y., "The Cutting of Water Saturated Sand at Large Cutting Angles". Proc. Dredging 02, May 5-8, Orlando, Florida, USA.
29. Miedema,S.A.,&He,Y.,"The Existence of Kinematic Wedgesat Large Cutting Angles". Proc.WEDA XXII Technical Conference & 34th Texas A&M Dredging Seminar, June 12-15, Denver, Colorado, USA.
30. Miedema, S.A., Frijters, D., "The Mechanism of Kinematic Wedges at Large Cutting Angles - Velocity and Friction Measurements". 23rd WEDA Technical Conference & 35th TAMU Dredging Seminar, Chicago, USA, June 2003.
31. Miedema, S.A.,"The Existence of Kinematic Wedgesat Large Cutting Angles".CHIDA Dredging Days, Shanghai, China, November 2003.
32. Miedema, S.A. & Frijters, D.D.J., "The wedge mechanism for cutting of water saturated sand at large cutting angles". WODCON XVII, September 2004, Hamburg Germany.
33. Miedema, S.A., "THE CUTTING MECHANISMS OF WATER SATURATED SAND AT SMALL AND LARGE CUTTING ANGLES". International Conference on Coastal Infrastructure Development - Challenges in the 21st Century. Hong Kong, November 2004.
34. He,J., Miedema, S.A. & Vlasblom, W.J., "FEM Analyses Of Cutting Of Anisotropic Densely Compacted and Saturated Sand", WEDAXXV & TAMU37, New Orleans, USA, June 2005.
35. Miedema,S.A.,"The Cutting of Water Saturated Sand, the FINAL Solution". WEDAXXV & TAMU37, New Orleans, USA, June 2005.
36. Miedema, S.A., "THE CUTTING OF WATER SATURATED SAND, THE SOLUTION". CEDA African Section: Dredging Days 2006 - Protection of the coastline, dredging sustainable development, Nov. 1-3, Tangiers, Morocco.
37. Miedema, S.A. & Vlasblom, W.J., "THE CLOSING PROCESS OF CLAMSHELL DREDGES IN WATER-SATURATED SAND". CEDA African Section: Dredging Days 2006 - Protection of the coastline, dredging sustainable development, Nov. 1-3, Tangiers, Morocco.
38. Miedema,S.A."THE CUTTING OF WATER SATURATED SAND, THE SOLUTION". The 2nd China Dredging Association International Conference & Exhibition, themed 'Dredging and Sustainable Development' and in Guangzhou, China, May 17-18 2006.
39. Ma, Y, Ni, F. & Miedema, S.A., "Calculation of the Blade Cutting Force for small Cutting Angles based on MATLAB". The 2nd China Dredging Association International Conference & Exhibition, themed 'Dredging and Sustainable Development' and in Guangzhou, China, May 17-18 2006.
40. Miedema, S.A. , Kerkvliet, J., Strijbis, D., Jonkman, B., Hatert, M. v/d, "THE DIGGING AND HOLDING CAPACITY OF ANCHORS". WEDA XXVI AND TAMU 38, San Diego, California, June 25-28, 2006.
41. Ma Yasheng, Ni Fusheng, S.A. Miedema, "Mechanical Model of Water Saturated Sand Cutting at Blade Large Cutting Angles", Journal of Hohai University Changzhou, ISSN 1009-1130, CN 32-1591, 2006.
42. Miedema,S.A.,Lager,G.H.G.,Kerkvliet,J.,“An Overview of Drag Embedded Anchor Holding Capacity for Dredging and Offshore Applications”. WODCON, Orlando, USA, 2007.
43. Miedema, S.A., "A Sensitivity Analysis of the Production of Clamshells". WEDA XXVIII & Texas A&M 39. St. Louis, USA, June 8-11, 2008.
44. Miedema, S.A., "A Sensitivity Analysis of the Production of Clamshells". WEDA Journal of Dredging Engineering, December 2008.
45. Miedema, S.A., "New Developments Of Cutting Theories With Respect To Dredging, The Cutting Of Clay And Rock". WEDA XXIX & Texas A&M 40. Phoenix Arizona, USA, June 14-17 2009.
46. Miedema, S.A., “New developments of cutting theories with respect to offshore applications, the cutting of sand, clay and rock”. ISOPE 2010, Beijing China, June 2010
47. Miedema, S.A.,“The influence of the strain rate on cutting processes”. ISOPE 2010, Beijing China, June 2010.
48. Abdeli, M., Miedema, S.A., Schott, D., AlvarezGrima, M., “The application of discrete elementm odeling in dredging”. WODCON XIX, Beijing China, September 2010.
49. Rahman, M., Schott, D.L., Miedema, S.A., Lodewijks, G., "Simulation of cutting process by hybrid granular and multi-body dynamics software". 3rd International Conference on Bulk solids. Glasgow, Scotland, September 9-10, 2010.
50. Rahman, M., Abdeli, M., Miedema, S.A., Schott, D., "Simulation of passive soil failure & cutting processes in sand. OMAE 2011 ASME, June 19-24, Rotterdam, the Netherlands.
51. Miedema, S.A., "Soil cutting processes, the cutting of water saturated sand". OMAE 2011 ASME, June 19-24, Rotterdam, the Netherlands.
52. Miedema, S.A., “THE BULLDOZER EFFECT WHEN CUTTING WATER SATURATED SAND”. OMAE 2012 ASME, June 10-15, Rio de Janeiro, Brazil.
53. Miedema, S.A., Zijsling, D., “HYPERBARIC ROCK CUTTING”. OMAE 2012 ASME, June10-15,Rio de Janeiro, Brazil.
54. Kuiper, R.J., Miedema, S.A., Frumeau, J.C.L. & van Rhee, C., “Influence of the Hyperbaric Effect on Apparrent Material Strength of Fully Saturated Porous Rock”. OTC 2013, Houston, Texas, U.S.A., May 2013.
55. Helmons, R.I.J. & Miedema, S.A., “Cutting Through Hard Rock-Like Materials, A Review of the Process”. WODCON XX, Brussels, Belgium, June 2013.
56. Chen, X. & Miedema, S.A., “Porosity Calculation in Discrete Element Modeling of Sand Cutting Process”. WODCON XX, Brussels, Belgium, June 2013.
57. Chen, X. & Miedema, S.A., “Influence of Particle Geometry on the Simulation of Sand Cutting Process”. OMAE 2013, Nantes, France, June 2013.
58. Helmons, R.I.J. & Miedema, S.A., “Rock Cutting for Deep Sea Mining: an Extension into Poromechanics”. Poromechanics V © ASCE 2013.
59. Liefferink, D.M., Alvarez Grima, M., Miedema, S.A., Plat, R., Rhee, C. van, “Failure mechanism of cutting submerged frozen clay in an arctic trenching process”. OTC 2014. Houston, Texas, U.S.A., May 2014.
60. Chen, X., Miedema, S.A., “Numerical methods for modeling the rock cutting process in deep sea mining”. ASME-OMAE 2014, San Francisco, USA, June 2014.
61. Helmons, R.L.J., Miedema, S.A., Rhee, C. van, “A new approach to model hyperbaric rock cutting processes”. ASME-OMAE 2014, San Francisco, USA, June 2014.
62. Miedema, S.A., “A NEW APPROACH TO DETERMINE CUTTING FORCES IN BRITTLE ROCK UNDER HYPERBARIC CONDITIONS”. ASME-OMAE 2014, San Francisco, USA, June 2014.
63. Chen, X., Miedema, S.A. & Rhee, C. van, “Numerical modeling of excavation process in dredging engineering”. 7th World Congress on Particle Technology. Beijing, China, May 19-22, 2014.
64. Miedema, S.A., “The Delft Sand, Clay & Rock Cutting Model”. IOS Press, Delft University Press, Delft, The Netherlands, 2015.
65. Chen, X., Miedema, S.A. & Rhee, C. van, “Numerical modeling of excavation process in dredging engineering”. Procedia Engineering, Vol. 102, pp. 804-814, 2015.
66. Miedema, S.A., “The Delft Sand, Clay & Rock Cutting Model”. WEDA TAMU, Houston, Texas, USA, June, 2015.
67. Combe, Th., & Miedema, S.A., “The influence of adhesion on cutting processes in dredging”. WEDA TAMU, Houston, Texas, USA, June, 2015.
68. Alvarez Grima, M., Miedema, S.A., Ketterij, R.G. van de, Yenigul, N.B., Rhee, C. van, Ëffect of high hyperbaric pressure on rock cutting process”. Engineering Geology, Vol. 196, pp. 24-36, 2015.
69. Chen, X., Miedema, S.A., Rhee, C. van, “Sensitivity study of bond radius and energy dissipation in parallel bond method”. Joint Conference of 5th UK-China and 13th UK Particle Technology Forum, Leeds, UK, 2015.
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textbooks/eng/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/17%3A_Appendices/17.26%3A_Appendix_Z-_Publications.txt
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Transport is the movement of humans,animals and goods from one location to another.Transportation moves people and goods from one place to another using a variety of vehicles across different infrastructure systems. It does this using not only technology (namely vehicles, energy, and infrastructure), but also people’s time and effort; producing not only the desired outputs of passenger trips and freight shipments, but also adverse outcomes such as air pollution, noise, congestion, crashes, injuries, and fatalities.If agriculture and industries are supposed to be the body of country,transport may be said to be the nerves and veins of the economy.
Figure 1 illustrates the inputs, outputs, and outcomes of transportation. In the upper left are traditional inputs (infrastructure including pavements, bridges, etc.), labor required to produce transportation, land consumed by infrastructure, energy inputs, and vehicles). Infrastructure is the traditional preserve of civil engineering, while vehicles are anchored in mechanical engineering. Energy, to the extent it is powering existing vehicles is a mechanical engineering question, but the design of systems to reduce or minimize energy consumption require thinking beyond traditional disciplinary boundaries.
On the top of the figure are Information, Operations, and Management, and Travelers’ Time and Effort. Transportation systems serve people, and are created by people, both the system owners and operators, who run, manage, and maintain the system and travelers who use it. Travelers’ time depends both on freeflow time, which is a product of the infrastructure design and on delay due to congestion, which is an interaction of system capacity and its use. On the upper right side of the figure are the adverse outcomes of transportation, in particular its negative externalities:
• by polluting, systems consume health and increase morbidity and mortality;
• by being dangerous, they consume safety and produce injuries and fatalities;
• by being loud they consume quiet and produce noise (decreasing quality of life and property values); and
• by emitting carbon and other pollutants, they harm the environment.
All of these factors are increasingly being recognized as costs of transportation, but the most notable are the environmental effects, particularly with concerns about global climate change. The bottom of the figure shows the outputs of transportation. Transportation is central to economic activity and to people’s lives, it enables them to engage in work, attend school, shop for food and other goods, and participate in all of the activities that comprise human existence. More transportation, by increasing accessibility to more destinations, enables people to better meet their personal objectives, but entails higher costs both individually and socially. While the “transportation problem” is often posed in terms of congestion, that delay is but one cost of a system that has many costs and even more benefits. Further, by changing accessibility, transportation gives shape to the development of land.
Modalism and Intermodalism
Transportation is often divided into infrastructure modes: e.g. highway, rail, water, pipeline and air. These can be further divided. Highways include different vehicle types: cars, buses, trucks, motorcycles, bicycles, and pedestrians. Transportation can be further separated into freight and passenger, and urban and inter-city. Passenger transportation is divided in public (or mass) transit (bus, rail, commercial air) and private transportation (car, taxi, general aviation).
These modes of course intersect and interconnect. At-grade crossings of railroads and highways, inter-modal transfer facilities (ports, airports, terminals, stations).
Different combinations of modes are often used on the same trip. I may walk to my car, drive to a parking lot, walk to a shuttle bus, ride the shuttle bus to a stop near my building, and walk into the building where I take an elevator.
Transportation is usually considered to be between buildings (or from one address to another), although many of the same concepts apply within buildings. The operations of an elevator and bus have a lot in common, as do a forklift in a warehouse and a crane at a port.
Motivation
Transportation engineering is usually taken by undergraduate Civil Engineering students. Not all aim to become transportation professionals, though some do. Loosely, students in this course may consider themselves in one of two categories: Students who intend to specialize in transportation (or are considering it), and students who don't. The remainder of civil engineering often divides into two groups: "Wet" and "Dry". Wets include those studying water resources, hydrology, and environmental engineering, Drys are those involved in structures and geotechnical engineering.
Transportation students
Transportation students have an obvious motivation in the course above and beyond the fact that it is required for graduation. Transportation Engineering is a pre-requisite to further study of Highway Design, Traffic Engineering, Transportation Policy and Planning, and Transportation Materials. It is our hope, that by the end of the semester, many of you will consider yourselves Transportation Students. However not all will.
"Wet Students"
I am studying Environmental Engineering or Water Resources, why should I care about Transportation Engineering?
Transportation systems have major environmental impacts (air, land, water), both in their construction and utilization. By understanding how transportation systems are designed and operate, those impacts can be measured, managed, and mitigated.
"Dry Students"
I am studying Structures or Geomechanics, why should I care about Transportation Engineering?
Transportation systems are huge structures of themselves, with very specialized needs and constraints. Only by understanding the systems can the structures (bridges, footings, pavements) be properly designed. Vehicle traffic is the dynamic structural load on these structures.
Citizens and Taxpayers
Everyone participates in society and uses transportation systems. Almost everyone complains about transportation systems. In developed countries you seldom hear similar levels of complaints about water quality or bridges falling down. Why do transportation systems engender such complaints, why do they fail on a daily basis? Are transportation engineers just incompetent? Or is something more fundamental going on? By understanding the systems as citizens, you can work toward their improvement. Or at least you can entertain your friends at parties.
Goal
It is often said that the goal of Transportation Engineering is "The Safe and Efficient Movement of People and Goods."
But that goal (safe and efficient movement of people and goods) doesn’t answer:
Who, What, When, Where, How, Why?
Overview
This libretext is broken into 4 major units
• Transportation Planning: Forecasting, determining needs and standards.
• Transit: Transit demand, service planning, and operations.
• Traffic Engineering (Operations): Queueing, Traffic Flow Highway Capacity and Level of Service (LOS)
• Highway Engineering (Design): Vehicle Performance/Human Factors, Geometric Design
Thought Questions
• What constraints keeps us from achieving the goal of transportation systems?
• What is the "Transportation Problem"?
Sample Problem
• Identify a transportation problem (local, regional, national, or global) and consider solutions. Research the efficacy of various solutions. Write a one-page memo documenting the problem and solutions, documenting your references.
Abbreviations/Key Terms
• LOS - Level of Service
• ITE - Institute of Transportation Engineers
• TRB - Transportation Research Board
• TLA - Three letter abbreviation
• Hierarchy of Roads
• Functional Classification
• Modes
• Vehicles
• Freight, Passenger
• Urban, Intercity
• Public, Private
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/01%3A_Introduction_and_Planning/1.01%3A_Fundamentals_of_Transportation_Introduction.txt
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Decision Making is the process by which one alternative is selected over another. Decision making generally occurs in the planning phases of transportation projects, but last minute decision making has been shown to occur, sometimes successfully. Several procedures for making decisions have been outlined in effort to minimize inefficiencies or redundancies. These are idealized (or normative) processes, and describe how decisions might be made in an ideal world, and how they are described in official documents. Real-world processes are not as orderly.
Applied systems analysis is the use of rigorous methods to assist in determining optimal plans, designs and solutions to large scale problems through the application of analytical methods. Applied systems analysis focuses upon the use of methods, concepts and relationships between problems and the range of techniques available. Any problem can have multiple solutions. The optimal solution will depend upon technical feasibility (engineering) and costs and valuation (economics). Applied systems analysis is an attempt to move away from the engineering practice of design detail and to integrate feasible engineering solutions with desirable economic solutions. The systems designer faces the same problem as the economist, "efficient resource allocation" for a given objective function.
Systems analysis emerged during World War II, especially with the deployment of radar in a coordinated way. It spread to other fields such as fighter tactics, mission planning and weapons evaluation. Ultimately the use of mathematical techniques in such problems came to be known as operations research, while other statistical and econometric techniques are being applied. Optimization applies to cases where data is under-determined (fewer observations than dependent variables) and statistics where data is over-determined (more observations than dependent variables). After World War II, techniques spread to universities. Systems analysis saw further mathematical development and application to a broad variety of problems.
It has been said of Systems Analysis, that it is:
• "A coordinated set of procedures which addresses the fundamental issues of design and management: that of specifying how men, money and materials should be combined to achieve a higher purpose" - De Neufville
• "... primarily a methodology, a philosophical approach to solving problems for and for planning innovative advances" - Baker
• "Professionals who endeavor to analyze systematically the choices available to public and private agencies in making changes in the transportation system and services in a particular region" - Manheim
• "Systems analysis is not easy to write about: brief, one sentence definitions frequently are trivial" - Thomas
The most prominent decision-making process to emerge from systems analysis is rational planning, which will be discussed next, followed by some critiques and alternatives.
How does one (rationally) decide what to do?
The figure identifies three layers of abstraction. The first layer (top row) describes the high level process, which we can summarize in six steps. A second layer details many of the components of the first layer. A third layer, identified by the blue box, "abstract into model or framework" depends on the problem at hand.
Overview data
The first step is observational, review and gather data about the system under consideration. An understanding of the world around is required, including specifying the system.
The problem (defined in the next step) lies within a larger system, that comprises
1. Objectives - measure the effectiveness or performance
2. Environment - things which affect the system but are not affected by it
3. Resources - factor inputs to do the work
4. Components - set of activities or tasks of the system
5. Management - sets goals, allocates resources and exercises control over components
6. Model of how variables in 1-5 relate to each other
the detailed objectives are identified in the following step, and the detailed model for analysis of the problem is specified in the step after that.
Note
For instance in the case of intercity transportation in California, data about existing demand conditions, existing supply conditions, future demand expectations, and proposed changes to supply would be important inputs. Changes in technology and environmental conditions are important considerations for long-term projects. We would also want to know the certainty of the forecasts, not just a central tendency, and the potential for alternative scenarios which may be widely different.
Define the problem
The second step is to define the problem more narrowly, in a sense to identify needs.
Note
Rather than an amorphous issue (intercity transportation), we might be interested in a more detailed question, e.g. how to serve existing and future demands between two cities (say metropolitan Los Angeles and San Francisco). The problem might be that demand is expected to grow and outstrip supply.
Formulate goal
The third step is to formulate a goal. For major transportation projects, or projects with intense community interest, this may involve the public. For instance
Note
To serve future passenger demand between Los Angeles and San Francisco, quickly, safely, cleanly, and inexpensively.
The goal will need to be testable, the process below "formulate goal" in the flowchart suggests this process in more detail.
The first aspect is to operationalize the goal. We need to measure the adverbs in the goal (e.g. how do we measure "quickly", "safely", "cleanly", or "inexpensively"). Some are straight-forward. "Quickly" is a measure of travel time or speed. But it needs to account for both the access and egress time, the waiting time, and the travel time, and these may not be weighted the same.
The second step is identifying the decision criteria. Each adverb may have a certain value, but it might be that an alternative has not merely have the most points in one area, but establish at least minimum satisfactory points in all areas. So a very fast mode must meet a specific safety test, and going faster does not necessarily mean it can also be more dangerous (despite what a rational economist might think about trade-offs).
The third is to weight those criteria. E.g. how important is speed vs. safety? This is in many ways a value question, though economics can try to value each of these aspects in monetary form, enabling Evaluation. For instance, many Negative externalities have been monetized, giving a value of time in delay, a value of pollution damages, and a value of life.
Generate alternatives
Examining, evaluating, and recommending alternatives is often the job of professionals, engineers, planners, and economists. Final selection is generally the job of elected or appointed officials for important projects.
There are several sub-problems here, the first is to generate alternatives. This may require significant creativity. Within major alternatives, there may be many sub-alternatives, e.g. the main alternative may be mode of travel, the sub-alternatives may be different alignments. For network problems there may be many combinations of alternative alignments. If the analyst is lucky, these are separable problems, that is, the choice of one sub-alignment is independent of the choice of alternative sub-alignments.
1. Algorithms-systematic search over available alternatives
1. Analytical
2. Exact numerical
3. Heuristic numerical
2. Generate alternatives selectively, evaluate subjectively
1. Fatal flaw analysis
2. Simple rating schemes
3. Delphi exercises
3. Generate alternatives judgmentally, evaluate scientifically using system model
A critical issue is how many alternatives to consider. In principle, an infinite number of more or less similar alternatives may be generated, not all are practical, and some may be minor variations. In practice a stopping rule to consider a reasonable number of alternatives is used. Major exemplars of the alternatives may be used, with fine-tuning awaiting a later step after the first set of alternatives is analyzed. The process may be iterative, winnowing down alternatives and detailing alternatives as more information is gained throughout the analysis.
Note
Several major alternatives may be suggested, expand highways, expand air travel, or construct new high-speed rail line, along with a no-build alternative.
Abstract into model or framework
"All Models are Wrong, Some Models are Less Wrong than Others"—Anonymous
"All Models are Wrong, Some Models are Useful"—George Box
The term Model refers here to a mathematical representation of a system, while a Framework is a qualitative organizing principle for analyzing a system. The terms are sometimes used interchangeably.
Framework Example: Porter's Diamond of Advantage
To illustrate the idea of a framework, consider Porter's Diamond of Advantage
Michael Porter proposes four key determinants of competitiveness, which he calls the "Diamond of Advantage," based on cases from around the world:
1. factor conditions, such as a specialized labor pool, specialized infrastructure and sometimes selective disadvantages that drive innovation;
2. home demand, or local customers who push companies to innovate, especially if their tastes or needs anticipate global demand;
3. related and supporting industries, specifically internationally competitive local supplier industries, creating a high quality, supportive business infrastructure, and spurring innovation and spin-off industries; and
4. industry strategy/rivalry, involving both intense local rivalry among area industries that is more motivating than foreign competition and as well as a local "culture" which influences individual industries' attitudes toward innovation and competition.
Model Example: The Four-Step Urban Transportation Planning System
Within the rational planning framework, transportation forecasts have traditionally followed the sequential four-step model or urban transportation planning (UTP) procedure, first implemented on mainframe computers in the 1950s at the Detroit Area Transportation Study and Chicago Area Transportation Study (CATS).
Land use forecasting sets the stage for the process. Typically, forecasts are made for the region as a whole, e.g., of population growth. Such forecasts provide control totals for the local land use analysis. Typically, the region is divided into zones and by trend or regression analysis, the population and employment are determined for each.
The four steps of the classical urban transportation planning system model are:
• Trip generation determines the frequency of origins or destinations of trips in each zone by trip purpose, as a function of land uses and household demographics, and other socio-economic factors.
• Destination choice matches origins with destinations, often using a gravity model function, equivalent to an entropy maximizing model. Older models include the fratar model.
• Mode choice computes the proportion of trips between each origin and destination that use a particular transportation mode. This model is often of the logit form, developed by Nobel Prize winner Daniel McFadden.
• Route choice allocates trips between an origin and destination by a particular mode to a route. Often (for highway route assignment) Wardrop's principle of user equilibrium is applied, wherein each traveler chooses the shortest (travel time) path, subject to every other driver doing the same. The difficulty is that travel times are a function of demand, while demand is a function of travel time.
Ascertain performance
This is either an output of the analytical model, or the result of subjective judgment.
Sherden identifies a number of major techniques for technological forecasting that can be used to ascertain expected performance of particular technologies, but that can be used within a technology to ascertain the performance of individual projects. These are listed in the following box:
"Major techniques for technological forecasting [3]
• Delphi method: a brain-storming session with a panel of experts.
• Nominal group process: a variant of the Delphi method with a group leader.
• Case study method: an analysis of analogous developments in other technologies.
• Trend analysis: the use of statistical analysis to extend past trends into the future.
• S-curve: a form of trend analysis using an s-shaped curve to extend past trends into the future.
• Correlation analysis: the projection of development of a new technology past developments in similar technologies.
• Lead-user analysis: the analysis of leading-edge users of a new technology predict how the technology will develop.
• Analytic hierarchy process: the projection of a new technology by analyzing a hierarchy of forces influencing its development.
• Systems dynamics: the use of a detailed model to assess the dynamic relationships among the major forces influencing the development of the technology.
• Cross-impact analysis: the analysis of potentially interrelated future events that may affect the future development of a technology.
• Relevance trees: the breakdown of goals for a technology into more detailed goals and then assigning probabilities that the technology will achieve these detail goals.
• Scenario writing: the development of alternative future views on how the new technology could be used."
Rate alternatives
The performance of each of the alternatives is compared across decision criteria, and weighted depending on the importance of those criteria. The alternative with the highest ranking would be identified, and this information would be brought forward to decision-makers.
Compute optimal decision
The analyst is generally not the decision maker. The actual influence of the results of the analysis in actual decisions will depend on:
1. Determinacy of evaluation
2. Confidence in the results on the part of the decision maker
3. Consistency of rating among alternatives
Implement alternatives
A decision is made. A project is constructed or a program implemented.
Evaluate outcome
Evaluating outcomes of a project includes comparing outcome against goals, but also against predictions, so that forecasting procedures can be improved. Analysis and implementation experience lead to revisions in systems definition, and may affect the values that underlay that definition. The output from this "last" step in is used as input to earlier steps in subsequent analyses. See e.g. Parthasarathi, Pavithra and David Levinson (2010) Post-Construction Evaluation of Traffic Forecast Accuracy. Transport Policy
Relationship to other models
We need a tool to "Identify Needs" and "Evaluate Options". This may be the transportation forecasting model.
Problem PRT: Skyweb Express
The Metropolitan Council of Governments (the region's main transportation planning agency) is examining whether the Twin Cities should build a new Personal Rapid Transit system in downtown Minneapolis, and they have asked you to recommend how it should be analyzed
1. What kind of model should be used. Why?
2. What data should be collected.
Form groups of 3 and take 15 minutes and think about what kinds of models you want to run and what data you want to collect, what questions you would ask, and how it should be collected. Each group should have a note-taker, but all members of the group should be able to present findings to the class.
Thought Questions
• Is the "rational planning" process rational?
• Compare and contrast the rational planning process with the scientific method?
Some Issues with Rational Planning
Nevertheless, some issues remain with the rational planning model:
Problems of incomplete information
• Limited Computational Capacity
• Limited Solution Generating Capacity
• Limited input data
• Cost of Analysis
Problems of incompatible desires
• Conflicting Goals
• Conflicting Evaluation Criteria
• Reliance on Experts (What about the People?)
Alternative Planning Decision Making Paradigms: Are They Irrational?
No one really believes the rational planning process is a good description of most decision making, as it is highly idealized. Alternatives normative and positive paradigms to the rational planning process include:
Several strategies normatively address the problems associated with incomplete information:
• Satisficing
• Decomposition hierarchically into Strategy/Tactics/Operations.
Other strategies describe how organizations and political systems work:
• Organizational Process
• Political Bargaining
Some do both:
• Incrementalism
The paper Montes de Oca, Norah and David Levinson (2006) Network Expansion Decision-making in the Twin Cities. Journal of the Transportation Research Board: Transportation Research Record #1981 pp 1-11 describes the actual decision making process for road construction in the Twin Cities.
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A benefit-cost analysis (BCA)[1] is often required in determining whether a project should be approved and is useful for comparing similar projects. It determines the stream of quantifiable economic benefits and costs that are associated with a project or policy. If the benefits exceed the costs, the project is worth doing; if the benefits fall short of the costs, the project is not. Benefit-cost analysis is appropriate where the technology is known and well understood or a minor change from existing technologies is being performed. BCA is not appropriate when the technology is new and untried because the effects of the technology cannot be easily measured or predicted. However, just because something is new in one place does not necessarily make it new, so benefit-cost analysis would be appropriate, e.g., for a light-rail or commuter rail line in a city without rail, or for any road project, but would not be appropriate (at the time of this writing) for something truly radical like teleportation.
The identification of the costs, and more particularly the benefits, is the chief component of the “art” of Benefit-Cost Analysis. This component of the analysis is different for every project. Furthermore, care should be taken to avoid double counting; especially counting cost savings in both the cost and the benefit columns. However, a number of benefits and costs should be included at a minimum. In transportation these costs should be separated for users, transportation agencies, and the public at large. Consumer benefits are measured by consumer’s surplus. It is important to recognize that the demand curve is downward sloping, so there a project may produce benefits both to existing users in terms of a reduction in cost and to new users by making travel worthwhile where previously it was too expensive.
Agency benefits come from profits. But since most agencies are non-profit, they receive no direct profits. Agency construction, operating, maintenance, or demolition costs may be reduced (or increased) by a new project; these cost savings (or increases) can either be considered in the cost column, or the benefit column, but not both.
Society is impacted by transportation project by an increase or reduction of negative and positive externalities. Negative externalities, or social costs, include air and noise pollution and accidents. Accidents can be considered either a social cost or a private cost, or divided into two parts, but cannot be considered in total in both columns.
If there are network externalities (i.e. the benefits to consumers are themselves a function of the level of demand), then consumers’ surplus for each different demand level should be computed. Of course this is easier said than done. In practice, positive network externalities are ignored in Benefit Cost Analysis.
Background
Early Beginnings
When Benjamin Franklin was confronted with difficult decisions, he often recorded the pros and cons on two separate columns and attempted to assign weights to them. While not mathematically precise, this “moral or prudential algebra”, as he put it, allowed for careful consideration of each “cost” and “benefit” as well as the determination of a course of action that provided the greatest benefit. While Franklin was certainly a proponent of this technique, he was certainly not the first. Western European governments, in particular, had been employing similar methods for the construction of waterway and shipyard improvements.
Ekelund and Hebert (1999) credit the French as pioneers in the development of benefit-cost analyses for government projects. The first formal benefit-cost analysis in France occurred in 1708. Abbe de Saint-Pierre attempted to measure and compare the incremental benefit of road improvements (utility gained through reduced transport costs and increased trade), with the additional construction and maintenance costs. Over the next century, French economists and engineers applied their analysis efforts to canals (Ekelund and Hebert, 1999). During this time, The École Polytechnique had established itself as France’s premier educational institution, and in 1837 sought to create a new course in “social arithmetic”: “…the execution of public works will in many cases tend to be handled by a system of concessions and private enterprise. Therefore our engineers must henceforth be able to evaluate the utility or inconvenience, whether local or general, or each enterprise; consequently they must have true and precise knowledge of the elements of such investments.” (Ekelund and Hebert, 1999, p. 47). The school also wanted to ensure their students were aware of the effects of currencies, loans, insurance, amortization and how they affected the probable benefits and costs to enterprises.
In the 1840s French engineer and economist Jules Dupuit (1844, tr. 1952) published an article “On Measurement of the Utility of Public Works”, where he posited that benefits to society from public projects were not the revenues taken in by the government (Aruna, 1980). Rather the benefits were the difference between the public’s willingness to pay and the actual payments the public made (which he theorized would be smaller). This “relative utility” concept was what Alfred Marshall would later rename with the more familiar term, “consumer surplus” (Ekelund and Hebert, 1999).
Vilfredo Pareto (1906) developed what became known as Pareto improvement and Pareto efficiency (optimal) criteria. Simply put, a policy is a Pareto improvement if it provides a benefit to at least one person without making anyone else worse off (Boardman, 1996). A policy is Pareto efficient (optimal) if no one else can be made better off without making someone else worse off. British economists Kaldor and Hicks (Hicks, 1941; Kaldor, 1939) expanded on this idea, stating that a project should proceed if the losers could be compensated in some way. It is important to note that the Kaldor-Hicks criteria states it is sufficient if the winners could potentially compensate the project losers. It does not require that they be compensated.
Benefit-cost Analysis in the United States
Much of the early development of benefit-cost analysis in the United States is rooted in water related infrastructure projects. The US Flood Control Act of 1936 was the first instance of a systematic effort to incorporate benefit-cost analysis to public decision-making. The act stated that the federal government should engage in flood control activities if “the benefits to whomsoever they may accrue [be] in excess of the estimated costs,” but did not provide guidance on how to define benefits and costs (Aruna, 1980, Persky, 2001). Early Tennessee Valley Authority (TVA) projects also employed basic forms of benefit-cost analysis (US Army Corp of Engineers, 1999). Due to the lack of clarity in measuring benefits and costs, many of the various public agencies developed a wide variety of criteria. Not long after, attempts were made to set uniform standards.
The U.S. Army Corp of Engineers “Green Book” was created in 1950 to align practice with theory. Government economists used the Kaldor-Hicks criteria as their theoretical foundation for the restructuring of economic analysis. This report was amended and expanded in 1958 under the title of “The Proposed Practices for Economic Analysis of River Basin Projects” (Persky, 2001).
The Bureau of the Budget adopted similar criteria with 1952’s Circular A-47 - “Reports and Budget Estimates Relating to Federal Programs and Projects for Conservation, Development, or Use of Water and Related Land Resources”.
Modern Benefit-cost Analysis
During the 1960s and 1970s the more modern forms of benefit-cost analysis were developed. Most analyses required evaluation of:
1. The present value of the benefits and costs of the proposed project at the time they occurred
2. The present value of the benefits and costs of alternatives occurring at various points in time (opportunity costs)
3. Determination of risky outcomes (sensitivity analysis)
4. The value of benefits and costs to people with different incomes (distribution effects/equity issues) (Layard and Glaister, 1994)
The Planning Programming Budgeting System (PPBS) - 1965
The Planning Programming Budgeting System (PPBS) developed by the Johnson administration in 1965 was created as a means of identifying and sorting priorities. This grew out of a system Robert McNamara created for the Department of Defense a few years earlier (Gramlich, 1981). The PPBS featured five main elements:
1. A careful specification of basic program objectives in each major area of governmental activity.
2. An attempt to analyze the outputs of each governmental program.
3. An attempt to measure the costs of the program, not for one year but over the next several years (“several” was not explicitly defined).
4. An attempt to compare alternative activities.
5. An attempt to establish common analytic techniques throughout the government.
Office of Management and Budget (OMB) – 1977
Throughout the next few decades, the federal government continued to demand improved benefit-cost analysis with the aim of encouraging transparency and accountability. Approximately 12 years after the adoption of the PPBS system, the Bureau of the Budget was renamed the Office of Management and Budget (OMB). The OMB formally adopted a system that attempts to incorporate benefit-cost logic into budgetary decisions. This came from the Zero-Based Budgeting system set up by Jimmy Carter when he was governor of Georgia (Gramlich, 1981).
Executive Order 12292, issued by President Reagan in 1981, required a regulatory impact analysis (RIA) for every major governmental regulatory initiative over $100 million. The RIA is basically a benefit-cost analysis that identifies how various groups are affected by the policy and attempts to address issues of equity (Boardman, 1996). According to Robert Dorfman, (Dorfman, 1997) most modern-day benefit-cost analyses suffer from several deficiencies. The first is their attempt “to measure the social value of all the consequences of a governmental policy or undertaking by a sum of dollars and cents”. Specifically, Dorfman mentions the inherent difficultly in assigning monetary values to human life, the worth of endangered species, clean air, and noise pollution. The second shortcoming is that many benefit-cost analyses exclude information most useful to decision makers: the distribution of benefits and costs among various segments of the population. Government officials need this sort of information and are often forced to rely on other sources that provide it, namely, self-seeking interest groups. Finally, benefit-cost reports are often written as though the estimates are precise, and the readers are not informed of the range and/or likelihood of error present. The Clinton Administration sought proposals to address this problem in revising Federal benefit-cost analyses. The proposal required numerical estimates of benefits and costs to be made in the most appropriate unit of measurement, and “specify the ranges of predictions and shall explain the margins of error involved in the quantification methods and in the estimates used” (Dorfman, 1997). Executive Order 12898 formally established the concept of Environmental Justice with regards to the development of new laws and policies, stating they must consider the “fair treatment for people of all races, cultures, and incomes.” The order requires each federal agency to identify and address “disproportionately high and adverse human health or environmental effects of its programs, policies and activities on minority and low-income populations.” Probabilistic Benefit-Cost Analysis In recent years there has been a push for the integration of sensitivity analyses of possible outcomes of public investment projects with open discussions of the merits of assumptions used. This “risk analysis” process has been suggested by Flyvbjerg (2003) in the spirit of encouraging more transparency and public involvement in decision-making. The Treasury Board of Canada’s Benefit-Cost Analysis Guide recognizes that implementation of a project has a probable range of benefits and costs. It posits that the “effective sensitivity” of an outcome to a particular variable is determined by four factors: • the responsiveness of the Net Present Value (NPV) to changes in the variable; • the magnitude of the variable's range of plausible values; • the volatility of the value of the variable (that is, the probability that the value of the variable will move within that range of plausible values); and • the degree to which the range or volatility of the values of the variable can be controlled. It is helpful to think of the range of probable outcomes in a graphical sense, as depicted in Figure 1 (probability versus NPV). Once these probability curves are generated, a comparison of different alternatives can also be performed by plotting each one on the same set of ordinates. Consider for example, a comparison between alternative A and B (Figure 2). In Figure 2, the probability that any specified positive outcome will be exceeded is always higher for project B than it is for project A. The decision maker should, therefore, always prefer project B over project A. In other cases, an alternative may have a much broader or narrower range of NPVs compared to other alternatives (Figure 3). Some decision-makers might be attracted by the possibility of a higher return (despite the possibility of greater loss) and therefore might choose project B. Risk-averse decision-makers will be attracted by the possibility of lower loss and will therefore be inclined to choose project A. Discount rate Both the costs and benefits flowing from an investment are spread over time. While some costs are one-time and borne up front, other benefits or operating costs may be paid at some point in the future, and still others received as a stream of payments collected over a long period of time. Because of inflation, risk, and uncertainty, a dollar received now is worth more than a dollar received at some time in the future. Similarly, a dollar spent today is more onerous than a dollar spent tomorrow. This reflects the concept of time preference that we observe when people pay bills later rather than sooner. The existence of real interest rates reflects this time preference. The appropriate discount rate depends on what other opportunities are available for the capital. If simply putting the money in a government insured bank account earned 10% per year, then at a minimum, no investment earning less than 10% would be worthwhile. In general, projects are undertaken with those with the highest rate of return first, and then so on until the cost of raising capital exceeds the benefit from using that capital. Applying this efficiency argument, no project should be undertaken on cost-benefit grounds if another feasible project is sitting there with a higher rate of return. Three alternative bases for the setting the government’s test discount rate have been proposed: 1. The social rate of time preference recognizes that a dollar's consumption today will be more valued than a dollar's consumption at some future time for, in the latter case, the dollar will be subtracted from a higher income level. The amount of this difference per dollar over a year gives the annual rate. By this method, a project should not be undertaken unless its rate of return exceeds the social rate of time preference. 2. The opportunity cost of capital basis uses the rate of return of private sector investment, a government project should not be undertaken if it earns less than a private sector investment. This is generally higher than social time preference. 3. The cost of funds basis uses the cost of government borrowing, which for various reasons related to government insurance and its ability to print money to back bonds, may not equal exactly the opportunity cost of capital. Typical estimates of social time preference rates are around 2 to 4 percent while estimates of the social opportunity costs are around 7 to 10 percent. Generally, for Benefit-Cost studies an acceptable rate of return (the government’s test rate) will already have been established. An alternative is to compute the analysis over a range of interest rates, to see to what extent the analysis is sensitive to variations in this factor. In the absence of knowing what this rate is, we can compute the rate of return (internal rate of return) for which the project breaks even, where the net present value is zero. Projects with high internal rates of return are preferred to those with low rates. Determine a present value The basic math underlying the idea of determining a present value is explained using a simple compound interest rate problem as the starting point. Suppose the sum of$100 is invested at 7 percent for 2 years. At the end of the first year the initial $100 will have earned$7 interest and the augmented sum ($107) will earn a further 7 percent (or$7.49) in the second year. Thus at the end of 2 years the $100 invested now will be worth$114.49.
The discounting problem is simply the converse of this compound interest problem. Thus, $114.49 receivable in 2 years time, and discounted by 7 per cent, has a present value of$100.
Present values can be calculated by the following equation:
(1) $P=\dfrac{F}{(1+i)^n}$
where:
• F = future money sum
• P = present value
• i = discount rate per time period (i.e. years) in decimal form (e.g. 0.07)
• n = number of time periods before the sum is received (or cost paid, e.g. 2 years)
Illustrating our example with equations we have:
$P=\dfrac{F}{(1+i)^n}=\dfrac{114.49}{(1+0.07)^2}=100.00$
The present value, in year 0, of a stream of equal annual payments of A starting year 1, is given by the reciprocal of the equivalent annual cost. That is, by:
(2) $P=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]$
where:
• A = Annual Payment
For example: 12 annual payments of $500, starting in year 1, have a present value at the middle of year 0 when discounted at 7% of:$3971
$P=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]=500\left[\dfrac{(1+0.07)^{12}-1}{0.07(1+0.07)^{12}}\right]=3971$
The present value, in year 0, of m annual payments of A, starting in year n + 1, can be calculated by combining discount factors for a payment in year n and the factor for the present value of m annual payments. For example: 12 annual mid-year payments of $250 in years 5 to 16 have a present value in year 4 of$1986 when discounted at 7%. Therefore in year 0, 4 years earlier, they have a present value of $1515. $P_{Y=4}=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]=250\left[\dfrac{(1+0.07)^{12}-1}{0.07(1+0.07)^{12}}\right]=1986$ $P_{Y=0}=\dfrac{F}{(1+i)^n}=\dfrac{P_{Y=4}}{(1+i)^n}=\dfrac{1986}{(1+0.07)^4}=1515$ Evaluation criterion Three equivalent conditions can tell us if a project is worthwhile 1. The discounted present value of the benefits exceeds the discounted present value of the costs 2. The present value of the net benefit must be positive. 3. The ratio of the present value of the benefits to the present value of the costs must be greater than one. However, that is not the entire story. More than one project may have a positive net benefit. From the set of mutually exclusive projects, the one selected should have the highest net present value. We might note that if there are insufficient funds to carry out all mutually exclusive projects with a positive net present value, then the discount used in computing present values does not reflect the true cost of capital. Rather it is too low. There are problems with using the internal rate of return or the benefit/cost ratio methods for project selection, though they provide useful information. The ratio of benefits to costs depends on how particular items (for instance, cost savings) are ascribed to either the benefit or cost column. While this does not affect net present value, it will change the ratio of benefits to costs (though it cannot move a project from a ratio of greater than one to less than one). Examples Example 1: Benefit Cost Application This problem, adapted from Watkins (1996), illustrates how a Benefit Cost Analysis might be applied to a project such as a highway widening. The improvement of the highway saves travel time and increases safety (by bringing the road to modern standards). But there will almost certainly be more total traffic than was carried by the old highway. This example excludes external costs and benefits, though their addition is a straightforward extension. The data for the “No Expansion” can be collected from off-the-shelf sources. However the “Expansion” column’s data requires the use of forecasting and modeling. Assume there are 250 weekdays (excluding holidays) each year and four rush hours per weekday. Table 1: Data No Expansion Expansion Peak Passenger Trips (per hour) 18,000 24,000 Trip Time (minutes) 50 30 Off-peak Passenger Trips (per hour) 9,000 10,000 Trip Time (minutes) 35 25 Traffic Fatalaties (per year) 2 1 Note: the operating cost for a vehicle is unaffected by the project, and is$4.
Table 2: Model Parameters
Peak Value of Time ($/minute)$0.15
Off-Peak Value of Time ($/minute)$0.10
Value of Life ($/life)$3,000,000
What is the benefit-cost relationship?
Solution
A 50 minute trip at $0.15/minute is$7.50, while a 30 minute trip is only $4.50. So for existing users, the expansion saves$3.00/trip. Similarly in the off-peak, the cost of the trip drops from $3.50 to$2.50, saving $1.00/trip. Consumers’ surplus increases both for the trips which would have been taken without the project and for the trips which are stimulated by the project (so-called “induced demand”), as illustrated above in Figure 1. Our analysis is divided into Old and New Trips, the benefits are given in Table 3. Table 3: Hourly Benefits TYPE Old trips New Trips Total Peak$54,000 $9000$63,000
Off-Peak $9,000$500 $9,500 Note: Old Trips: For trips which would have been taken anyway the benefit of the project equals the value of the time saved multiplied by the number of trips. New Trips: The project lowers the cost of a trip and public responds by increasing the number of trips taken. The benefit to new trips is equal to one half of the value of the time saved multiplied by the increase in the number of trips. There are 1000 peak hours per year. With 8760 hours per year, we get 7760 offpeak hours per year. These numbers permit the calculation of annual benefits (shown in Table 4). Table 4: Annual Travel Time Benefits TYPE Old trips New Trips Total Peak$54,000,000 $9,000,000$63,000,000
Off-Peak $69,840,000$3,880,000 $73,720,000 Total$123,840,000 $12,880,000$136,720,000
The safety benefits of the project are the product of the number of lives saved multiplied by the value of life. Typical values of life are on the order of $3,000,000 in US transportation analyses. We need to value life to determine how to trade off between safety investments and other investments. While your life is invaluable to you (that is, I could not pay you enough to allow me to kill you), you don’t act that way when considering chance of death rather than certainty. You take risks that have small probabilities of very bad consequences. You do not invest all of your resources in reducing risk, and neither does society. If the project is expected to save one life per year, it has a safety benefit of$3,000,000. In a more complete analysis, we would need to include safety benefits from non-fatal accidents.
The annual benefits of the project are given in Table 5. We assume that this level of benefits continues at a constant rate over the life of the project.
Table 5: Total Annual Benefits
Type of Benefit Value of Benefits Per Year
Time Saving $136,720,000 Reduced Risk$3,000,000
Total $139,720,000 Costs Highway costs consist of right-of-way, construction, and maintenance. Right-of-way includes the cost of the land and buildings that must be acquired prior to construction. It does not consider the opportunity cost of the right-of-way serving a different purpose. Let the cost of right-of-way be$100 million, which must be paid before construction starts. In principle, part of the right-of-way cost can be recouped if the highway is not rebuilt in place (for instance, a new parallel route is constructed and the old highway can be sold for development). Assume that all of the right-of-way cost is recoverable at the end of the thirty-year lifetime of the project. The $1 billion construction cost is spread uniformly over the first four-years. Maintenance costs$2 million per year once the highway is completed.
The schedule of benefits and costs for the project is given in Table 6.
Table 6: Schedule of Benefits And Costs ($millions) Time (year) Benefits Right-of-way costs Construction Costs Maintenance costs 0 0 100 0 0 1-4 0 0 250 0 5-29 139,72 0 0 2 30 139.72 -100 0 2 Conversion to Present Value The benefits and costs are in constant value dollars. Assume the real interest rate (excluding inflation) is 2%. The following equations provide the present value of the streams of benefits and costs. To compute the Present Value of Benefits in Year 5, we apply equation (2) from above. $P=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]=139.72\left[\dfrac{(1+0.02)^{26}-1}{0.02(1+0.02)^{26}}\right]=2811.31$ To convert that Year 5 value to a Year 1 value, we apply equation (1) $P=\dfrac{F}{(1+i)^n}=\dfrac{2811.31}{(1+0.02)^4}=2597.21$ The present value of right-of-way costs is computed as today’s right of way cost ($100 M) minus the present value of the recovery of those costs in Year 30, computed with equation (1):
$P=\dfrac{F}{(1+i)^n}=\dfrac{100}{(1+0.02)^{30}}=55.21$
$100-55.21=44.79$
The present value of the construction costs is computed as the stream of $250M outlays over four years is computed with equation (2): $P=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]=250\left[\dfrac{(1+0.02)^4-1}{0.02(1+0.02)^4}\right]=951.93$ Maintenance Costs are similar to benefits, in that they fall in the same time periods. They are computed the same way, as follows: To compute the Present Value of$2M in Maintenance Costs in Year 5, we apply equation (2) from above.
$P=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]=2\left[\dfrac{(1+0.02)^{26}-1}{0.02(1+0.02)^{26}}\right]=40.24$
To convert that Year 5 value to a Year 1 value, we apply equation (1)
$P=\dfrac{F}{(1+i)^n}=\dfrac{40.24}{(1+0.02)^4}=37.18$
As Table 7 shows, the benefit/cost ratio of 2.5 and the positive net present value of $1563.31 million indicate that the project is worthwhile under these assumptions (value of time, value of life, discount rate, life of the road). Under a different set of assumptions, (e.g. a higher discount rate), the outcome may differ. Table 7: Present Value of Benefits and Costs ($ millions)
Present Value
Benefits 2,597.21
Costs
Right-of-Way 44.79
Construction 951.93
Maintenance 37.18
Costs SubTotal 1,033.90
Net Benefit(B-C) 1,563.31
Benefit/Cost Ratio 2.5
Thought Questions
Decision Criteria
Which is a more appropriate decision criteria: Benefit/Cost or Benefit - Cost? Why?
Is it only money that matters?
Problem
Is money the only thing that matters in Benefit-Cost Analysis? Is "converted" money the only thing that matters? For example, the value of human life in dollars?
Solution
Absolutely not. A lot of benefits and costs can be converted to monetary value, but not all. For example, you can put a price on human safety, but how can you put a price on, say, aesthetics—something that everyone agrees is beneficial. What else can you think of?
Can small units of time be given the same value of time as larger units of time?
In other words, do 60 improvements each saving a traveler 1 minute equal 1 improvement saving a traveler 60 minutes? Similarly, does 1 improvement saving a 1000 travelers 1 minute equal the value of time of a single traveler of 1000 minutes. These are different problems, one is intra-traveler and one is inter-traveler, but related.
Several issues arise.
A. Is value of time linear or non-linear? To this we must conclude the value of time is surely non-linear. I am much more agitated waiting 3 minutes at a red light than 2, and I begin to suspect the light is broken. Studies of ramp meters show a similar phenomena.[2]
B. How do we apply this in a benefit-cost analysis? If we break one project into 60 smaller projects, each with a smaller value of travel time saved, and then we added the gains, we would get a different result than the what obtains with a single large project. For analytical convenience, we would like our analyses to be additive, not sub-additive, otherwise arbitrarily dividing the project changes the result. In particular many smaller projects will produce an undercount that is quite significant, and result in a much lower benefit than if the projects were bundled.
As a practical matter, every Benefit/Cost Analysis assumes a single value of time, rather than assuming non-linear value of time. This also helps avoiding biasing public investments towards areas with people who have a high value of time (the rich)
On the other hand, mode choice analyses do however weight different components of travel time differently, especially transit time (i.e. in-vehicle time is less onerous than waiting time). The implicit value of time for travelers does depend on the type of time (though generally not the amount of time). Using the log-sum of the mode choice model as a measure of benefit would implicitly account for this.
Are sunk costs sunk, is salvage value salvageable? A paradox in engineering economics analysis.
Salvage value is defined as "The estimated value of an asset at the end of its useful life." Sunk cost is defined as "Cost already incurred which cannot be recovered regardless of future events."
It is often said in economics that "sunk costs are sunk", meaning they should not be considered a cost in economic analysis, because the money has already been spent.
Now consider two cases
In case 1, we have a road project that costs $10.00 today, and at the end of 10 years has some economic value remaining, let's say a salvage value of$5.00, which when discounted back to the present is $1.93 (at 10% interest). This value is the residual value of the road. Thus, the total present cost of the project$10.00 - $1.93 =$8.07. Clearly the road cannot be moved. However, its presence makes it easier to build future roads ... the land has been acquired and graded, some useful material for aggregate is on-site perhaps, and can be thought of as the amount that it reduces the cost of future generations to build the road. Alternatively, the land could be sold for development if the road is no longer needed, or turned into a park.
Assume the present value of the benefit of the road is $10.00. The benefit/cost ratio is$10.00 over $8.07 or 1.23. If we treat the salvage value as a benefit rather than cost, the benefit is$10.00 + $1.93 =$11.93 and the cost is $10, and the B/C is 1.193. In 10 years time, the community decides to replace the old worn out road with a new road. This is a new project. The salvage value from the previous project is now the sunk cost of the current project (after all the road is there and could not be moved, and so does not cost the current project anything to exploit). So the cost of the project in 10 years time would be$10.00 - $5.00 =$5.00. Discounting that to the present is $1.93. The benefit in 10 years time is also$10.00, but the cost in 10 years time was $5.00, and the benefit/cost ratio they perceive is$10.00/$5.00 = 2.00 Aggregating the two projects • the benefits are$10 + $3.86 =$13.86
• the costs are $8.07 +$1.93 = $10.00 • the collective benefit/cost ratio is 1.386 • the NPV is benefits - costs =$3.86
One might argue the salvage value is a benefit, rather than a cost reduction. In that case
• the benefits are $10.00 +$1.93 + $3.86 =$15.79
• the costs are $10.00 +$1.93 = $11.93 • the collective benefit/cost ratio is 1.32 • the NPV remains$3.86
Case 2 is an identical road, but now the community has a 20 year time horizon to start. The initial cost is $10, and the cost in 10 years time is$5.00 (discounted to $1.93). The benefits are$10 now and $10 in 10 years time (discounted to$3.86). There is no salvage value at the end of the first period, nor sunk costs at the beginning of the second period. What is the benefit cost ratio?
• the costs are $11.93 • the benefits are still$13.86
• the benefit/cost ratio is 1.16
• the NPV is $1.93. If you are the community, which will you invest in? Case 1 has an initial B/C of 1.23 (or 1.193), Case 2 has a B/C of 1.16. But the real benefits and real costs of the roads are identical. The salvage value in this example is, like so much in economics (think Pareto optimality), an accounting fiction. In this case no transaction takes place to realize that salvage value. On the other hand, excluding the salvage value over-estimates the net cost of the project, as it ignores potential future uses of the project. Time horizons on projects must be comparable to correctly assess relative B/C ratio, yet not all projects do have the same benefit/cost ratio. Software Tools for Impact Analysis The majority of economic impact studies for highway capacity projects are undertaken using conventional methods. These methods tend to focus on the direct user impacts of individual projects in terms of travel costs and outcomes, and compare sums of quantifiable, discounted benefits and costs. Inputs to benefit-cost analyses can typically be obtained from readily available data sources or model outputs (such as construction and maintenance costs, and before and after estimates of travel demand, by vehicle class, along with associated travel times). Valuation of changes in external, somewhat intangible costs of travel (e.g., air pollution and crash injury) can usually be accommodated by using shadow price estimates, such as obtained from FHWA-suggested values, based on recent empirical studies. The primary benefits included in such studies are those related to reductions in user cost, such as travel time savings and vehicle operating costs (e.g. fuel costs, vehicle depreciation, etc.). Additional benefits may stem from reductions in crash rates, vehicle emissions, noise, and other costs associated with vehicle travel. Project costs are typically confined to expenditures on capital investment, along with ongoing operations and maintenance costs. A number of economic analysis tools have been developed under the auspices of the United States Federal Highway Administration (FHWA) permitting different forms of benefit-cost analysis for different types of projects, at different levels of evaluation. Several of these tools are prevalent in past impact analyses, and are described here. However, none identifies the effects of infrastructure on the economy and development. MicroBENCOST MicroBENCOST is a sketch planning tool for estimating basic benefits and costs of a range of highway improvement projects, including capacity addition projects. In each type of project, attention is focused on corridor traffic conditions and their resulting impact on motorist costs with and without a proposed improvement. This type of approach may be appropriate for situations where projects have relatively isolated impacts and do not require regional modeling. SPASM The Sketch Planning Analysis Spreadsheet Model (SPASM) is a benefit-cost tool designed for screening level analysis. It outputs estimates of project costs, cost-effectiveness, benefits, and energy and air quality impacts. SPASM is designed to allow for comparison among multiple modes and non-modal alternatives, such as travel demand management scenarios. The model is comprised of three modules (worksheets) relating to public agency costs, characteristics of facilities and trips, and a travel demand component. Induced traffic is dealt with through the use of elasticity-based methods, where an elasticity of vehicle-miles of travel (VMT) with respect to travel time is defined and applied. Vehicle emissions are estimated based on calcuations of VMT, trip length and speeds, and assumed shares of travel occuring in cold start, hot start, and hot stabilized conditions. Analysis is confined to a corridor level, with all trips having the same origin, destination and length. This feature is appropriate for analysis of linear transportation corridors, but also greatly limits the ability to deal with traffic drawn to or diverted from outside the corridor. DeCorla-Souza et al. (1996) describe the model and its application to a freeway corridor in Salt Lake City, Utah. STEAM The Surface Transportation Efficiency Analysis Model (STEAM) is a planning-level extension of the SPASM model, designed for a fuller evaluation of cross-modal and demand management policies. STEAM was designed to overcome the most important limitations of its predecessor, namely the assumption of average trip lengths within a single corridor and the inability to analyze systemwide effects. The enhanced modeling capabilities of STEAM feature greater compatibility with existing four-step travel demand models, including a trip table module that is used to calculate user benefits and emissions estimates based on changes in network conditions and travel behavior. Also, the package features a risk analysis component to its evaluation summary module, which calculates the likelihood of various outcomes such as benefit-cost ratios. An overview of STEAM and a hypothetical application are given by DeCorla-Souza et al. (1998). SMITE The Spreadsheet Model for Induced Travel Estimation (SMITE) is a sketch planning application that was designed for inclusion with STEAM in order to account for the effects of induced travel in traffic forecasting. SMITE's design as a simple spreadsheet application allows it to be used in cases where a conventional, four-step travel demand model is unavailable or cannot account for induced travel effects in its structure. SMITE applies elasticity measures that describe the response in demand (VMT) to changes in travel time and the response in supply (travel time) to changes in demand levels. SCRITS As a practical matter, highway corridor improvements involving intelligent transportation systems (ITS) applications to smooth traffic flow can be considered capacity enhancements, at least in the short term. The FHWA's SCRITS (SCReening for ITS) is a sketch planning tool that offers rough estimates of ITS benefits, for screening-level analysis. SCRITS utilizes aggregate relationships between average weekday traffic levels and capacity to estimate travel speed impacts and vehicle-hours of travel (VHT). Like many other FHWA sketch planning tools, it is organized in spreadsheet format and can be used in situations where more sophisticated modeling systems are unavailable or insufficient. HERS In addition to helping states plan and manage their highway systems, the FHWA's Highway Economic Requirements System for states (HERS-ST) offers a model for economic impacts evaluation. In one case, Luskin (2005) use HERS-ST to conclude that Texas is under-invested in highways – particularly urban systems and lower-order functional classes – by 50 percent. Combining economic priniciples with engineering criteria, HERS evaluates competing projects via benefit-cost ratios. Recognizing user benefits, emissions levels, and construction and maintenance costs, HERS operates within a GIS environment and will be evaluated under this project, for discussion in project deliverables. Well established software like HERS offer states and regions an oportunity to readily pursue standardized economic impact evaluations on all projects, a key advantage for many users, as well as the greater community. Summary of Software Tools Many analytical tools, like those described above, are favored due to their relative ease of use and employment of readily available or easily acquired data. However, several characteristics limit their effectiveness in evaluating the effects of new highway capacity. First, they are almost always insufficient to describe the full range of impacts of new highway capacity. Such methods deliberately reduce economic analysis to the most important components, resorting to several simplifying assumptions. If a project adds capacity to a particularly important link in the transportation network, its effects on travel patterns may be felt outside the immediate area. Also, the effects of induced travel, in terms of either route switching or longer trips, may not be accounted for in travel models based on a static, equilibrum assignment of traffic. In the longer term, added highway capacity may lead to the spatial reorganization of activities as a result of changes in regional accessibility. These types of changes cannot typically be accounted for in analysis methods. Second, there is the general criticism of methods based on benefit-cost analysis that they cannot account for all possible impacts of a project. Benefit-cost methods deliberately reduce economic analysis to the most important components and often must make simplifying assumptions. The project-based methods described here generally do not describe the economic effects of a project on different user or non-user groups. Winners and losers from a new capacity project cannot be effectively identified and differentiated. Third, a significant amount of uncertainty and risk is involved in the employment of project-based methods. Methods that use benefit-cost techniques to calculate B/C ratios, rates of return, and/or net present values are often sensitive to certain assumptions and inputs. With transportation infrastructure projects, the choice of discount rate is often critical, due to the long life of projects and large, up-front costs. Also, the presumed value of travel time savings is often pivotal, since it typically reflects the majority of project benefits. Valuations of travel time savings vary dramatically across the traveler population, as a function of trip purpose, traveler wage, household income, and time of day. It is useful to test several plausible values. Assessment procedures in the UK and other parts of Europe have moved towards a multi-criteria approach, where economic development is only one of several appraisal criteria. Environmental, equity, safety, and the overall integration with other policy sectors are examined in a transparent framework for decision makers. In the UK, the Guidance on the Methodologies for Multi-Modal Studies (2000) provides such a framework. These procedures require a clear definition of project goals and objectives, so that actual effects can be tied to project objectives, as part of the assessment procedure. This is critical for understanding induced travel effects. Noland (2007) has argued that this implies that comprehensive economic assessment, including estimation of land valuation effects, is the only way to fully assess the potential beneficial impacts of projects. Sample Problems Problem 1: A new transportation project is proposed to the city. This project is a form of "guide wire", where cars can hook to these moving, below-ground wires and be transported for free around town. This project is proven to reduce gas consumption by$5 million in its completion year. The city's preferred contractor says that it will take 10 years to build the thing and cost $500,000 a year, which is "perfect" because costs would add up to the benefit. Knowing that inflation is 3 percent, as an expert evaluator, is this a wise decision? Answer Convert everything to Present Value and see just how great a deal this is. For the present value of the$5,000,000 benefit (gas reduction):
$P=\dfrac{F}{(1+i)^n}=\dfrac{5,000,000}{(1+0.03)^{10}}=3,720,469.57$
The benefit in present-day value is $3,720,469.57. For the money (cost) being sent to the contractor, a payment of$470,000 per year, the present value would be:
$P=A\left[\dfrac{(1+n)^n-1}{i(1+i)^n}\right]=470,000\left[\dfrac{(1+0.03)^{10}-1}{0.03(1+0.03)^{10}}\right]=4,009,195.33$
The cost in present-day value is $4,009,195.33. Therefore, this detail, while shiny in appearance at first, is NOT a wise decision, since costs exceed benefits. Problem 2: A new Southstar rail line is proposed. This project is expected to reduce travel time for 2,000 commuters by 30 minutes per day, in its completion year. The line only operates on weekdays (Monday-Friday). It will take 2 years to build and cost$320,000,000 total (Net Present Value). If the interest rate is 3 percent, above what value of time must average value of time for SouthStar Passengers be in order for the benefit/cost ratio to exceed 1.
• Assume a 30 year lifespan. The interest rate is annual*
Answer
Benefits considered are only Travel Time Savings.
Travel Time Savings = 2,000 Commuters x 0.5 Hours/day (30 minutes/day)
Travel Time Savings = 1000 Commuters-Hours/day x 5 days/week 52 weeks/year
Travel Time Savings = 260,000 Commuters-Hours/year
Travel Time Savings start after two years (given in the problem statement). Other assumptions are: Present Year is 0, no growth (constant commuters ev- ery year, and thus constant savings), and constant Value of Travel Time (VOT). Therefore, Benefits must be discounted to Present value for each year and added for a total during the lifespan of the project considered (30 years).
Adding up all Present Value of Travel Time
Present Value of Travel Time = $(VOT)(260,000)(\dfrac{1}{[1+0.03]^2}+\dfrac{1}{[1+0.03]^3}+...+\dfrac{1}{[1+0.03]^{29}}+\dfrac{1}{[1+0.03]^{30}})$
You can sum it up in an excel spreadsheet or recognize that this is a geometric series.
The sum is 18.63 inside the parentheses.
Total Present Value of Travel Time = (4,843,687.57)(VOT)
Costs are given by Total Present Value of $320,000,000. Benefits/Costs = 1 Thus $\dfrac{(4,843,687.57)(VOT)}{320,000,000}=1$ VOT = USD$66/Hr.
The VOT must be at least 66 US dollars.
Key Terms
• Benefit-Cost Analysis
• Profits
• Costs
• Discount Rate
• Present Value
• Future Value
External Exercises
Use the SAND software at the STREET website to learn how to evaluate network performance given a changing network scenario.
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/01%3A_Introduction_and_Planning/1.03%3A_Evaluation.txt
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All forecasts are wrong; some forecasts are more wrong than others. - anonymous
Modeling is a means for representing reality in an abstracted way. Your mental models are your world view: your outlook on life, and the world. The world view is your internal model of how the world works; it is employed every time you make a prediction: what do you expect, what is a surprise. The expression “Where you stand depends on where you sit” epitomizes this idea. Your world view is shaped by your experience and your position.
When modeling, the issue of Point of View should be considered. It must be clear who (and what) the results are for. If you are modeling for personal pleasure, it will naturally reflect your own worldview, but if you are working for an employer or client, their point of view must also be considered, if the inputs or results deviate significantly from their worldview, they may adapt their worldview, but more likely will dismiss the model.
Modeling can be conducted both for subjective advocacy and for objective analysis. The same methods may be employed in either, and the ethical modeler will produce the same results in either case, but they may be used differently.
Why Model?
There are a variety of reasons to model. Modeling helps
• gain insight into complex situations by understanding simpler situations resembling them
• optimize the use of resources in building or maintaining systems
• operate system, particularly by testing alternative operational scenarios
• educate and provide experience for model-builders
• provide a platform for testing contending ideas and use in negotiations.
Particular applications in transportation include:
• Forecasting traffic
• Testing scenarios (alternative land uses, networks, policies)
• Planning projects/corridor studies
• Regulating land use: Growth management/public facility adequacy
• Managing complexity, when eyeballs are insufficient, (different people have different intuitions)
• Understanding travel behavior
• Influencing decisions
• Estimating traffic in the absence of data
Developing Models
As an engineer, economist, or planner you may be given a model to use. But that model was not spontaneously generated, it came from other engineers, economists, or planners who undertook a series of steps to translate raw data into a tool that could be used to generate useful information.
The first step, specification, tells us what depends on what (the number of trips leaving a zone may depend on the number of households). The estimation step tells us how strong these relationships mathematically are, (each household generates two trips in the peak hour). Implementation takes those relationships and puts them into a computer program. Calibration looks at the output of the computer program and compares it to other data, if they don't match exactly, adjustments to the models may be made. Validation compares the results to data at another point in time. Finally the model is applied to look at a project (e.g. how much traffic will use Granary Road after Washington Avenue is closed to traffic).
• Specification:
• \[y=f(X)\]
• Estimation:
• \[y=mX+b;m=1,b=2\]
• Implementation
• If Z > W, Then \[y=mX+b\]
• Calibration
• \[y_{predicted}+k=y_{observed}\]
• Validation
• \[y_{predicted1990}+k=y_{observed1990}\]
• Application
Specification
When building a model system, numerous decisions must be made. These are discussed below:
Types of Models
There are numerous types of models, a short list is below. Each has different applicability, multiple methods may be used in pursuit of the same question, sometimes they are complementary, and sometimes competitive techniques.
• Network analysis
• Linear Programming
• Nonlinear Programming
• Simulation
• Deterministic queuing
• Probabilistic queuing
• Regression
• Neural Nets
• Genetic Algorithm
• Cost/ Benefit Analysis
• Life-cycle costing
• System Dynamics
• Control Theory
• Difference Equations
• Differential Equations
• Probabilistic Risk Assessment
• Supply/Demand Equilibrium
• Game Theory
• Statistical Decision Theory
• Markov Models
• Cellular Automata
• Etc.
Model Trade-offs
Building a model requires trading-off time and resource constraints. One could always be more detailed, more accurate, or more comprehensive if resources were not constrained. However, the following must also be considered.
• Money,
• Data,
• Computation,
• Labor,
• Ease of Use,
• Convincing (e.g. Graphic Displays),
• Extendable,
• Evidence of Model Benefits,
• Measuring Model Success
Organization of Model System
• Hierarchy of Models
• Centralized vs. Decentralized (Optimization (Global) vs. Agent, Local Optimization)
Time
• Time Frame
• Static vs. Dynamic
• Real Time vs. Offline
• Short Term vs. Long Term (Partial vs. General Equilibrium)
• Proactive vs. Reactive (Predictive vs. Responsive)
Space
• Scale/Detail
• Spatial Extent
• Boundaries (Boundary Effects)
• Macroscopic vs. Microscopic (Zones, Flows vs. Individuals, Vehicles)
Process
• Stochastic vs. Deterministic
• Linear vs. Nonlinear
• Continuous vs. Discrete
• Numerical Simulation vs. Closed Form Solution
• Equilibrium vs. Disequilibrium
Type
• Behavioral vs. Aggregate Model
• Physical vs. Mathematical Models
Solution Techniques
When solving the model, the system as a whole must be understood. Several questions arise:
• Does the solution exist?
• Is the solution unique?
• Is the solution feasible?
Solution techniques often trade-off accuracy vs. speed. Some solution techniques may only guarantee a local optima, while others (such as brute force techniques) can guarantee a global optimum, but may be much slower.
“Four-Step” Urban Transportation Planning Models
We want to answer a number of related questions (who, what, where, when, why, and how):
• Who is traveling or what is being shipped?
• Where are the origin and destination of those trips, shipments?
• When do those trips begin and end (how long do they take, how far away are the trip ends)?
• Why are the trips being made, what is their purpose?
• How are the trips getting there, what routes or modes are they taking?
If we know the answers to those questions, we want to know what are the answers to those questions a function of?
• Cost: Money, Time spent on the trip,
• Cost: Money and Time of alternatives.
• Benefit (utility) of trip (e.g. the activity at the destination)
• Benefit of alternatives
The reason for this is to understand what will happen under various circumstances:
• How much “induced demand” will be generated if a roadway is expanded?
• How many passengers will be lost if bus services are cut back?
• How many people will be “priced off” if tolls are implemented?
• How much traffic will a new development generate?
• How much demand will I lose if I raise shipping costs to my customers?
In short, for urban passenger travel, we are trying to predict the number of trips by:
• Origin Activity,
• Destination Activity,
• Origin Zone,
• Destination Zone,
• Mode,
• Time of Day, and
• Route.
This is clearly a multidimensional problem.
In practice, the mechanism to do this is to employ a "four-step" urban transportation planning model, each step will be detailed in subsequent modules. These steps are executed in turn, though there may be feedback between steps:
• Trip Generation - How many trips \(T_i\) or \(T_j\) are entering or leaving zone \(i\) or \(j\)
• Trip Distribution or Destination Choice - How many trips \(T_{ij}\) are going from zone \(i\) to zone \(j\)
• Mode Choice - How many trips \(T_{ijm}\) from \(i\) to \(j\) are using mode \(m\)
• Route Choice - Which links are trips \(T_{ijmr}\) from \(i\) to \(j\) by mode \(m\) using route \(r\)
Thought Questions
• Is past behavior reflective of future behavior?
• Can the future be predicted?
• Is the future independent of decisions, or are prophecies self-fulfilling?
• How do we know if forecasts were successful?
• Against what standard are they to be judged?
• What values are embedded in the planning process?
• What happens when values change?
Additional Problems
Homework
1. Explain the four-step transportation planning model?
2. What are the outputs of the planning model and what they are used for?
3. What are the strength and weaknesses of this type of the forecasting procedure?
4. List 4 applications of transportation planning models.
5. List 4 typical data sources used in estimating, calibrating, and validating transportation planning models.
6. What is the rational planning model, and how does the four-step transportation planning model relate to it?
7. Identify 5 ways in which the conventional 4-step transportation planning model is imperfect.
Additional Problems
1. List five inputs to transportation planning models – are they measured or forecast?
2. Why is modeling important, (give reasons, purposes of modeling)?
3. What would be a possible complication or problem that would arise if in the planning of a project no forecasting was used?
4. What are the steps in the rational planning model?
5. Why is the rational planning model not always used (examples when it is not used)?
6. What are some of the factors preventing accurate forecasting?
7. List the steps in the four-step urban transportation planning model?
8. Explain the advantages and disadvantages of public vs. privately owned transportation?
9. What types of matrices are there in the transportation model, how do they differ?
10. How does the effect of population increase effect transportation planning?
11. What are the step in the generic modeling process?
12. Are all transportation modes publicly owned? Give examples of public and private.
13. When would it be reasonable to use the rational planning model?
14. What significance does the equilibrium point in transportation economics have in design?
15. What makes forecasting with one method better than another, and how are differences resolved?
16. How do measured inputs affect the forecasting inputs?
17. What is the idealized model to plan for a transportation project?
18. On what type of economic model does transportation depend?
19. Who, other than planners and engineers, should be involved in defining the needs and objectives for future projects?
20. Can we steer development and growth trends through transportation planning decision?
21. Give three examples of intermodalism?
22. Are actual transportation routes planned using the Rational Planning model?
23. List at least 10 things that must be considered before undertaking a major transit project?
24. How is the rational planning model used?
25. Why do we combine vector matrices to a full matrix?
26. Is household income measured or forecast?
27. What statistical methods play a role in transportation engineering/planning?
28. Who measures transit use or travel times?
29. What is a zone centroid?
30. What is forecasting?
31. How are various transportation needs provided?
32. What is considered a carrier?
33. How does vehicle occupancy influence our design process?
34. Is rational planning process more likely to be used on major projects or small ones?
35. What are the axes labels on the supply / demand curve?
36. Explain why all forecasts are wrong?
37. How does supply and demand work? How do they relate?
38. How does intermodealism affect forecasting?
39. What is the difference between modalism and intermodalism?
Key Terms
• Rational Planning
• Transportation planning model
• Matrix, Full Matrix, Vector Matrix, Scalar Matrix
• Trip table
• Travel time matrix
• Origin, Destination
• Purpose
• Network
• Zone (Traffic Analysis Zone or Transportation Analysis Zone, or TAZ)
• External station or external zone
• Centroid
• Node
• Link
• Turn
• Route
• Path
• Mode
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/02%3A_Planning_Models/2.01%3A_Modeling.txt
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There are a variety of types of transportation data used in analysis. Some are listed below:
• Infrastructure Status
• Traffic Counts
• Travel Behavior Inventory
• Land Use Inventory
• Truck/Freight Demand
• External/Internal Demand (by Vehicle Type)
• Special Generators
Revealed Preference
Household travel surveys which ask people what they actually did are a type of Revealed Preference survey data that have been obtained by direct observation of the choice that individuals make with respect to travel behavior. Travel Cost Analysis uses the prices of market goods to evaluate the value of goods that are not traded in the market.
Hedonic Pricing uses the market price of a traded good and measures of its component attributes to calculate value. There are other methods to attain Revealed Preference information, but surveys are the most common in travel behavior.
Travel Behavior Inventory
While the Cleveland Regional Area Traffic Study in 1927 was the first metropolitan planning attempt sponsored by the US federal government, the lack of comprehensive survey methods and standards at that time precluded the systematic collection of information such as travel time, origin and destination, and traffic counts. The first US travel surveys appeared in urban areas after the Federal-Aid Highway Act of 1944 permitted the spending of federal funds on urban highways.[1] A new home- interview origin-destination survey method was developed in which households were asked about the number of trips, purpose, mode choice, origin and destination of the trips conducted on a daily basis. In 1944, the US Bureau of Public Roads printed the Manual of Procedures for Home Interview Traffic Studies.[2] This new procedure was first implemented in several small to mid-size areas. Highway engineers and urban planners made use of the new data collected after the 1944 Highway Act extended federally sponsored planning to travel surveys as well as traffic counts, highway capacity studies, pavement condition studies and cost-benefit analysis.
Attributes of a household travel survey, or travel behavior inventory include:
• Travel Diary of ~ 1% sample of population (all trips made on one day) every 10 years
• Socioeconomic/demographic data of survey respondents
• Collection methodology:
• Phone,
• Mail,
• In-Person at Home,
• In-Person at Work,
• Roadside
Many such surveys are available online at: Metropolitan Travel Survey Archive
Thought Question
What are the advantages and disadvantages of Revealed Preference surveys?
Stated Preference
In contrast with revealed preference, Stated preference is a group of techniques used to calculate the utility functions of transport options based on the response of an individual decision-maker to certain given options. The options generally are based on descriptions of the transport scenario or are constructed by the researcher
• Contingent valuation is based on the assumption that the best way to find out the value that an individual places on something is known by asking.
• Compensating variation is the compensating payment that leaves an individual as well off as before the economic change.
• Equivalent variation for a benefit is the minimum amount of money one would have to be compensated to leave the person as well as they would be after the change.
• Conjoint analysis refers to the application of the design of experiments to obtain the preferences of the individual, breaking the task into a list of choices or ratings that enable us to compute the relative importance of each of the attributes studied
Thought Question
What are the advantages and disadvantages of Stated Preference surveys?
Pavement conditions
adapted from Xie, Feng and David Levinson (2008) The Use of Road Infrastructure Data for Urban Transportation Planning: Issues and Opportunities. Published in Infrastructure Reporting and Asset Management Edited by Adjo Amekudzi and Sue McNeil. pp- 93-98. Publisher: American Society of Civil Engineers, Reston, Virginia.
Road infrastructure represents the supply side of an urban transportation system. Pavement condition is a critical indicator to the quality of road infrastructure in terms of providing a smooth and reliable driving environment on roads. A series of indices have been developed to evaluate the pavement conditions of road segments in their respective jurisdictions: Pavement Condition Index (PCI) is scored as a perfect roadway (100 points) minus point deductions for “distresses” that are observed; Present Serviceability Rating (PSR) is measured as vertical movement per unit horizontal movement (e.g. millimeters of vertical displacement per meter of horizontal displacement) as one drives along the road; (SR) Surface Rating is calculated by reviewing images of the roadway based on the frequency and severity of defects; Pavement Quality Index (PQI) is calculated using the PSR and SR to evaluate the general condition of the road. A high PQI (up to 4.5) means a road will most likely not need maintenance soon, whereas a low PQI means it can be selected for maintenance.
These indices of pavement quality are basic measures for road maintenance and preservation, for which each county develops its own performance standards to evaluate pavement conditions and make decisions on maintenance and preservation projects. Typically, pavement preservation projects are prioritized based on PCI of road segments: the lower the PCI, the higher the likelihood of selection. Taking Washington County, Minnesota as an example, the county has determined that a reasonable standard to maintain is an average PCI of 72. Thus any road segment with its PCI below 72 has a chance to be selected for preservation. Dakota County, Minnesota on the other hand, scores its preservation projects according to the measure of PQI: a road segment will be allocated 17 points (out of a possible 100) if its PQI falls lower than 3.1.
The pavement data structure is incompatible with the link-node structure of the planning road network used by the Metropolitan Council and other planning agencies. Typically, the measures of PCI, PSR, and PQI are stored in records indexed by “highway segment numbers” along each highway route. Highway sections with the same highway segment number are differentiated by their starting and ending stations. There is no exact match between highway segments in the actual road network and links in the planning network, as stationing is a position along the curved centerline of a highway while a planning network is a simplified structure consisting of only straight lines intersecting at points. Historic pavement data is generally unavailable in electronic format, although the information on pavement history such as pavement life and the duration since last repaving are important to estimate the cost of a preservation project, also affecting the decision whether a specific project will get selected and how much funding will be allocated.
Traffic flow
Traffic conditions reflect the travel demand loaded on a given road infrastructure. Traffic flows on roads, together with road capacity, can be used to calculate the volume/capacity (V/C) ratio, which is an approximate indicator for the level of service of road infrastructure, and is commonly adopted by the jurisdictions in their respective decision making processes. The traffic flows on the planning road network are predicted by the transportation planning model, but the results have to be calibrated with actual traffic data.
Loop detectors are the primary technology currently employed in many US metropolitan areas to collect actual traffic data. E.g. In the Twin Cities of Minneapolis-St. Paul, about one thousand detector stations have been buried on major highways, through which Mn/DOT’s Traffic Management Center collects, stores, and makes public traffic data every 30 seconds, including measured volume (flow) and occupancy, and calculated speed data for each detector station. Although the estimates of Annual Average Daily Traffic (AADT) for the planning road network are readily available, loop detectors provide more accurate measures of traffic volume, since they are collecting real-time data on a continuous basis. It also allows for calibrating models to hourly rather than daily conditions.
Due to the limited ability to convert raw data collected by loop detectors, however, most forecasting models rely on AADT data. The raw data are stored in a 30- second interval in binary codes. For planning uses they have to be converted and aggregated into desired measures, such as peak hour average, average for a particular month or a particular season, etc, in a systematic way.
Another issue in integrating loop detector data into a planning road network is to match the detector stations with the links in planning networks. Similar to the problem encountered in translating pavement data, detectors are located along major highways and mapped on the actual geometry of the network, while the planning road network is a simplified structure with only straight lines.
Sampling Issues (Statistics)
• Sample Size,
• Population of Interest
• Sampling Method,
• Error:
• Measurement,
• Sampling,
• Computational,
• Specification,
• Transfer,
• Aggregation
• Bias,
• Oversampling
• Extent of Collection
• Spatial
• Temporal
• Span of Data
• Cross-section,
• Time Series, and
• Panel
Metadata
Adapted from Levinson, D. and Zofka, Ewa. (2006) “The Metropolitan Travel Survey Archive: A Case Study in Archiving” in Travel Survey Methods: Quality and Future Directions, Proceedings of 5th International Conference on Travel Survey Methods (Peter Stopher and Cheryl Stecher, editors)
Metadata allows data to function together. Simply put, metadata is information about information – labeling, cataloging and descriptive information structured to permit data to be processed. Ryssevik and Musgrave (1999) argue that high quality metadata standards are essential as metadata is the launch pad for any resource discovery, maps complex data, bridges the gap between data producers and consumers, and links data with its resultant reports and scientific studies produced about it. To meet the increasing needs for the proper data formats and encoding standards, the World Wide Web Consortium (W3C) has developed the generic Resource Description Framework (RDF) (W3C 2002). RDF treats metadata more generally, providing a standard way to use Extended Markup Language (XML) to “represent metadata in the form of statements about properties and relationships of items” (W3C 2002). Resources can be almost any type of file, including of course, travel surveys. RDF delivers detailed and unified data description vocabulary.
Applying these tools specifically to databases, the Data Documentation Initiative (DDI) for Document Type Definitions (DTD) applies metadata standards used for documenting datasets. DDI was first developed by European and North American data archives, libraries and official statistics agencies. “The Data Documentation Initiative (DDI) is an effort to establish an international XML-based standard for the content, presentation, transport, and preservation of documentation for datasets in the social and behavioral sciences” (Data Documentation Initiative 2004). As this international standardization effort gathers momentum it is expected more and more datasets to be documented using DDI as the primary metadata format. With DDI, searching data archives on the Internet no longer depends on an archivist's skill at capturing the information that is important to researchers. The standard of data description provides sufficient detail sorted in a user-friendly manner.
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/02%3A_Planning_Models/2.02%3A_Data.txt
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Transportation systems have specific structure. Roads have length, width, and depth. The characteristics of roads depends on their purpose.
Roads
A road is a path connecting two points. The English word ‘road’ comes from the same root as the word ‘ride’ –the Middle English ‘rood’ and Old English ‘rad’ –meaning the act of riding. Thus a road refers foremost to the right of way between an origin and destination. In an urban context, the word street is often used rather than road, which dates to the Latin word ‘strata’, meaning pavement (the additional layer or stratum that might be on top of a path).
Modern roads are generally paved, and unpaved routes are considered trails in some countries. The pavement of roads began early in history. Approximately 2600 BCE, the Egyptians constructed a paved road out of sandstone and limestone slabs to assist with the movement of stones on rollers between the quarry and the site of construction of the pyramids. The Romans and others used brick or stone pavers to provide a more level, and smoother surface, especially in urban areas, which allows faster travel, especially of wheeled vehicles. The innovations of Thomas Telford and John McAdam reinvented roads in the early nineteenth century, by using less expensive smaller and broken stones, or aggregate, to maintain a smooth ride and allow for drainage. Later in the nineteenth century, application of tar (asphalt) further smoothed the ride. In 1824, asphalt blocks were used on the Champs-Élysées in Paris. In 1872, the first asphalt street (Fifth Avenue) was paved in New York (due to Edward de Smedt), but it wasn’t until bicycles became popular in the late nineteenth century that the “Good Roads Movement” took off. Bicycle travel, more so than travel by other vehicles at the time, was sensitive to rough roads. Demands for higher quality roads really took off with the widespread adoption of the automobile in the United States in the early twentieth century.
The first good roads in the twentieth century were constructed of Portland cement concrete (PCC). The material is stiffer than asphalt (or asphalt concrete) and provides a smoother ride. Concrete lasts slightly longer than asphalt between major repairs, and can carry a heavier load, but is more expensive to build and repair. While urban streets had been paved with concrete in the US as early as 1889, the first rural concrete road was in Wayne County, Michigan, near to Detroit in 1909, and the first concrete highway in 1913 in Pine Bluff, Arkansas. By the next year over 2300 miles of concrete pavement had been laid nationally. However over the remainder of the twentieth century, the vast majority of roadways were paved with asphalt. In general only the most important roads, carrying the heaviest loads, would be built with concrete.
Roads are generally classified into a hierarchy. At the top of the hierarchy are freeways, which serve entirely a function of moving vehicles between other roads. Freeways are grade-separated and limited access, have high speeds and carry heavy flows. Below freeways are arterials. These may not be grade-separated, and while access is still generally limited, it is not limited to the same extent as freeways, particularly on older roads. These serve both a movement and an access function. Next are collector/distributor roads. These serve more of an access function, allowing vehicles to access the network from origins and destinations, as well as connecting with smaller, local roads, that have only an access function, and are not intended for the movement of vehicles with neither a local origin nor destination. Local roads are designed to be low speed and carry relatively little traffic.
The class of the road determines which level of government administers it. The highest roads will generally be owned, operated, or at least regulated (if privately owned) by the higher level of government involved in road operations; in the United States, these roads are operated by the individual states. As one moves down the hierarchy of roads, the level of government is generally more and more local (counties may control collector/distributor roads, towns may control local streets). In some countries freeways and other roads near the top of the hierarchy are privately owned and regulated as utilities, these are generally operated as toll roads. Even publicly owned freeways are operated as toll roads under a toll authority in other countries, and some US states. Local roads are often owned by adjoining property owners and neighborhood associations.
The design of roads is specified in a number of design manual, including the AASHTO Policy on the Geometric Design of Streets and Highways (or Green Book). Relevant concerns include the alignment of the road, its horizontal and vertical curvature, its super-elevation or banking around curves, its thickness and pavement material, its cross-slope, and its width.
Freeways
A motorway or freeway (sometimes called an expressway or thruway) is a multi-lane divided road that is designed to be high-speed free flowing, access-controlled, built to high standards, with no traffic lights on the mainline. Some motorways or freeways are financed with tolls, and so may have tollbooths, either across the entrance ramp or across the mainline. However in the United States and Great Britain, most are financed with gas or other tax revenue.
Though of course there were major road networks during the Roman Empire and before, the history of motorways and freeways dates at least as early as 1907, when the first limited access automobile highway, the Bronx River Parkway began construction in Westchester County, New York (opening in 1908). In this same period, William Vanderbilt constructed the Long Island Parkway as a toll road in Queens County, New York. The Long Island Parkway was built for racing and speeds of 60 miles per hour (96 km/hr) were accommodated. Users however had to pay a then expensive \$2.00 toll (later reduced) to recover the construction costs of \$2 million. These parkways were paved when most roads were not. In 1919 General John Pershing assigned Dwight Eisenhower to discover how quickly troops could be moved from Fort Meade between Baltimore and Washington to the Presidio in San Francisco by road. The answer was 62 days, for an average speed of 3.5 miles per hour (5.6 km/hr). While using segments of the Lincoln Highway, most of that road was still unpaved. In response, in 1922 Pershing drafted a plan for an 8,000 mile (13,000 km) interstate system which was ignored at the time.
The US Highway System was a set of paved and consistently numbered highways sponsored by the states, with limited federal support. First built in 1924, they succeeded some previous major highways such as the Dixie Highway, Lincoln Highway and Jefferson Highway that were multi-state and were constructed with the aid of private support. These roads however were not in general access-controlled, and soon became congested as development along the side of the road degraded highway speeds.
In parallel with the US Highway system, limited access parkways were developed in the 1920s and 1930s in several US cities. Robert Moses built a number of these parkways in and around New York City. A number of these parkways were grade separated, though they were intentionally designed with low bridges to discourage trucks and buses from using them. German Chancellor Adolf Hitler appointed a German engineer Fritz Todt Inspector General for German Roads. He managed the construction of the German Autobahns, the first limited access high-speed road network in the world. In 1935, the first section from Frankfurt am Main to Darmstadt opened, the total system today has a length of 11,400 km. The Federal-Aid Highway Act of 1938 called on the Bureau of Public Roads to study the feasibility of a toll-financed superhighway system (three east-west and three north-south routes). Their report Toll Roads and Free Roads declared such a system would not be self-supporting, advocating instead a 43,500 km (27,000 mile) free system of interregional highways, the effect of this report was to set back the interstate program nearly twenty years in the US.
The German autobahn system proved its utility during World War II, as the German army could shift relatively quickly back and forth between two fronts. Its value in military operations was not lost on the American Generals, including Dwight Eisenhower.
On October 1, 1940, a new toll highway using the old, unutilized South Pennsylvania Railroad right-of-way and tunnels opened. It was the first of a new generation of limited access highways, generally called superhighways or freeways that transformed the American landscape. This was considered the first freeway in the US, as it, unlike the earlier parkways, was a multi-lane route as well as being limited access. The Arroyo Seco Parkway, now the Pasadena Freeway, opened December 30, 1940. Unlike the Pennsylvania Turnpike, the Arroyo Seco parkway had no toll barriers.
A new National Interregional Highway Committee was appointed in 1941, and reported in 1944 in favor of a 33,900 mile system. The system was designated in the Federal Aid Highway Act of 1933, and the routes began to be selected by 1947, yet no funding was provided at the time. The 1952 highway act only authorized a token amount for construction, increased to \$175 million annually in 1956 and 1957.
The US Interstate Highway System was established in 1956 following a decade and half of discussion. Much of the network had been proposed in the 1940s, but it took time to authorize funding. In the end, a system supported by gas taxes (rather than tolls), paid for 90% by the federal government with a 10% local contribution, on a pay-as-you-go” system, was established. The Federal Aid Highway Act of 1956 had authorized the expenditure of \$27.5 billion over 13 years for the construction of a 41,000 mile interstate highway system. As early as 1958 the cost estimate for completing the system came in at \$39.9 billion and the end date slipped into the 1980s. By 1991, the final cost estimate was \$128.9 billion. While the freeways were seen as positives in most parts of the US, in urban areas opposition grew quickly into a series of freeway revolts. As soon as 1959, (three years after the Interstate act), the San Francisco Board of Supervisors removed seven of ten freeways from the city’s master plan, leaving the Golden Gate bridge unconnected to the freeway system. In New York, Jane Jacobs led a successful freeway revolt against the Lower Manhattan Expressway sponsored by business interests and Robert Moses among others. In Baltimore, I-70, I-83, and I-95 all remain unconnected thanks to highway revolts led by now Senator Barbara Mikulski. In Washington, I-95 was rerouted onto the Capital Beltway. The pattern repeated itself elsewhere, and many urban freeways were removed from Master Plans.
In 1936, the Trunk Roads Act ensured that Great Britain’s Minister of Transport controlled about 30 major roads, of 7,100 km (4,500 miles) in length. The first Motorway in Britain, the Preston by-pass, now part of the M-6, opened in 1958. In 1959, the first stretch of the M1 opened. Today there are about 10,500 km (6300 miles) of trunk roads and motorways in England.
Australia has 790 km of motorways, though a much larger network of roads. However the motorway network is not truly national in scope (in contrast with Germany, the United States, Britain, and France), rather it is a series of local networks in and around metropolitan areas, with many intercity connection being on undivided and non-grade separated highways. Outside the Anglo-Saxon world, tolls were more widely used. In Japan, when the Meishin Expressway opened in 1963, the roads in Japan were in far worse shape than Europe or North American prior to this. Today there are over 6,100 km of expressways (3,800 miles), many of which are private toll roads. France has about 10,300 km of expressways (6,200 miles) of motorways, many of which are toll roads. The French motorway system developed through a series of franchise agreements with private operators, many of which were later nationalized. Beginning in the late 1980s with the wind-down of the US interstate system (regarded as complete in 1990), as well as intercity motorway programs in other countries, new sources of financing needed to be developed. New (generally suburban) toll roads were developed in several metropolitan areas.
An exception to the dearth of urban freeways is the case of the Big Dig in Boston, which relocates the Central Artery from an elevated highway to a subterranean one, largely on the same right-of-way, while keeping the elevated highway operating. This project is estimated to be completed for some \$14 billion; which is half the estimate of the original complete US Interstate Highway System.
As mature systems in the developed countries, improvements in today’s freeways are not so much widening segments or constructing new facilities, but better managing the road space that exists. That improved management, takes a variety of forms. For instance, Japan has advanced its highways with application of Intelligent Transportation Systems, in particular traveler information systems, both in and out of vehicles, as well as traffic control systems. The US and Great Britain also have traffic management centers in most major cities that assess traffic conditions on motorways, deploy emergency vehicles, and control systems like ramp meters and variable message signs. These systems are beneficial, but cannot be seen as revolutionizing freeway travel. Speculation about future automated highway systems has taken place almost as long as highways have been around. The Futurama exhibit at the New York 1939 World’s Fair posited a system for 1960. Yet this technology has been twenty years away for over sixty years, and difficulties remain.
Layers of Networks
Data Unit Layer Function
Host Layers Data 7. Application Network process to application
Host Layers Data 6. Presentation Data representation,encryption and decryption
Host Layers Data 5. Session Interhost communication
Host Layers Segments 4. Transport End-to-end connections and reliability,Flow control
Media Layers Packet 3. Network Path determination and logical addressing
Media Layers Frame 2. Data Link Physical addressing
Media Layers Bit 1. Physical Media, signal and binary transmission
All networks come in layers. The OSI Reference Model for the Internet is well-defined. Roads too are part of a layer of subsystems of which the pavement surface is only one part. We can think of a hierarchy of systems.
• Places
• Trip Ends
• End to End Trip
• Driver/Passenger
• Service (Vehicle & Schedule)
• Signs and Signals
• Markings
• Pavement Surface
• Structure (Earth & Pavement and Bridges)
• Alignment (Vertical and Horizontal)
• Right-Of-Way
• Space
At the base is space. On space, a specific right-of-way is designated, which is property where the road goes. Originally right-of-way simply meant legal permission for travelers to cross someone's property. Prior to the construction of roads, this might simply be a well-worn dirt path.
On top of the right-of-way is the alignment, the specific path a transportation facility takes within the right-of-way. The path has both vertical and horizontal elements, as the road rises or falls with the topography and turns as needed.
Structures are built on the alignment. These include the roadbed as well as bridges or tunnels that carry the road.
Pavement surface is the gravel or asphalt or concrete surface that vehicles actually ride upon and is the top layer of the structure. That surface may have markings to help guide drivers to stay to the right (or left), delineate lanes, regulate which vehicles can use which lanes (bicycles-only, high occupancy vehicles, buses, trucks) and provide additional information. In addition to marking, signs and signals to the side or above the road provide additional regulatory and navigation information.
Services use roads. Buses may provide scheduled services between points with stops along the way. Coaches provide scheduled point-to-point without stops. Taxis handle irregular passenger trips.
Drivers and passengers use services or drive their own vehicle (producing their own transportation services) to create an end-to-end trip, between an origin and destination. Each origin and destination comprises a trip end and those trip ends are only important because of the places at the ends and the activity that can be engaged in. As transportation is a derived demand, if not for those activities, essentially no passenger travel would be undertaken.
With modern information technologies, we may need to consider additional systems, such as Global Positioning Systems (GPS), differential GPS, beacons, transponders, and so on that may aide the steering or navigation processes. Cameras, in-pavement detectors, cell phones, and other systems monitor the use of the road and may be important in providing feedback for real-time control of signals or vehicles.
Each layer has rules of behavior:
• some rules are physical and never violated, others are physical but probabilistic
• some are legal rules or social norms which are occasionally violated
Hierarchy of Roads
Even within each layer of the system of systems described above, there is differentiation.
Transportation facilities have two distinct functions: through movement and land access. This differentiation:
• permits the aggregation of traffic to achieve economies of scale in construction and operation (high speeds);
• reduces the number of conflicts;
• helps maintain the desired quiet character of residential neighborhoods by keeping through traffic away from homes;
• contains less redundancy, and so may be less costly to build.
Functional Classification Types of Connections Relation to Abutting Property Minnesota Examples
Limited Access (highway) Through traffic movement between cities and across cities Limited or controlled access highways with ramps and/or curb cut controls. I-94, Mn280
Linking (arterial:principal and minor) Traffic movement between limited access and local streets. Direct access to abutting property. University Avenue, Washington Avenue
Local (collector and distributor roads) Traffic movement in and between residential areas Direct access to abutting property. Pillsbury Drive, 17th Avenue
Network
• Zone Centroid - special node whose number identifies a zone, located by an "x" "y" coordinate representing longitude and latitude (sometimes "x" and "y" are identified using planar coordinate systems).
• Node (vertices) - intersection of links , located by x and y coordinates
• Links (arcs) - short road segments indexed by from and to nodes (including centroid connnectors), attributes include lanes, capacity per lane, allowable modes
• Turns - indexed by at, from, and to nodes
• Routes, (paths) - indexed by a series of nodes from origin to destination. (e.g. a bus route)
• Modes - car, bus, HOV, truck, bike, walk etc.
Matrices
Scalar
A scalar is a single value that applies model-wide; e.g. the price of gas or total trips.
Total Trips
Variable T
Vectors
Vectors are values that apply to particular zones in the model system, such as trips produced or trips attracted or number of households. They are arrayed separately when treating a zone as an origin or as a destination so that they can be combined into full matrices.
• vector (origin) - a column of numbers indexed by traffic zones, describing attributes at the origin of the trip (e.g. the number of households in a zone)
Trips Produced at Origin Zone
Origin Zone 1 \[T_{i1}\]
Origin Zone 2 \[T_{i2}\]
Origin Zone 3 \[T_{i3}\]
• vector (destination) - a row of numbers indexed by traffic zones, describing attributes at the destination
Destination Zone 1 Destination Zone 2 Destination Zone 3
Trips Attracted to Destination Zone \[T_{j1}\] \[T_{j2}\] \[T_{j3}\]
Full Matrices
A full or interaction matrix is a table of numbers, describing attributes of the origin-destination pair
Destination Zone 1 Destination Zone 2 Destination Zone 3
Origin Zone 1 \[T_{11}\] \[T_{12}\] \[T_{13}\]
Origin Zone 2 \[T_{21}\] \[T_{22}\] \[T_{23}\]
Origin Zone 3 \[T_{31}\] \[T_{32}\] \[T_{33}\]
Thought Questions
• Identify the rules associated with each layer?
• Why aren’t all roads the same?
• How might we abstract the real transportation system when representing it in a model for analysis?
• Why is abstraction useful?
Abbreviations
• SOV - single occupant vehicle
• HOV - high occupancy vehicle (2+, 3+, etc.)
• TAZ - transportation analysis zone or traffic analysis zone
Variables
• msXX - scalar matrix
• moXX - origin vector matrix
• mdXX - destination vector matrix
• mfXX - full vector matrix
• - Total Trips
• - Trips Produced from Origin Zone
• - Trips Attracted to Destination Zone
• - Trips Going Between Origin Zone and Destination Zone
Key Terms
• Zone Centroid
• Node
• Links
• Turns
• Routes
• Modes
• Matrices
• Right-of-way
• Alignment
• Structures
• Pavement Surface
• Markings
• Signs and Signals
• Services
• Driver
• Passenger
• End to End Trip
• Trip Ends
• Places
External Exercises
Use the ADAM software at the STREET website and examine the network structure. Familiarize yourself with the software, and edit the network, adding at least two nodes and four one-way links (two two-way links), and deleting nodes and links. What are the consequences of such network adjustments? Are some adjustments better than others?
2.04: Modes
Wikipedia’s article gives a list of a variety of modes largely defined by their technology (are they powered by animals or engines, are the engines on the vehicle or is the vehicle powered by a cable, do the vehicles travel on land, sea, or water, etc. While this is a reasonably comprehensive list and a reasonable organization of the subject for Wikipedia, it does not really get at the transportation aspect of modes, focusing instead on their mechanical aspects.
What characteristics describe and differentiate modes? Every mode must differ from every other mode on at least one dimension (otherwise they would be the same mode). This is analogous to the idea of speciation in biology. The Figure is a first cut at this for surface passenger transportation.
This distinguishes primarily on the non-mechanical (non-propulsion) characteristics of the service. Of course not every possible dimension is identified, and a few of the circles contain multiple modes which are otherwise obviously distinct (e.g. gondolas and subways are much the same from a transportation service perspective but for one is underground and uses a train and the other is suspended by a cable which moves it). This taxonomy differentiates things that were qualitatively different rather than quantitatively different.
So the first cut is about time, is a reservation required or not (i.e. does it need some advance planning).
The second cut is about time as well, is the service scheduled or dynamic.
The third cut is about space, are the routes fixed or dynamic.
If the route is fixed, are stops fixed (i.e. does the vehicle stop at every stop, or only when called, like a bus).
Otherwise if the routes are dynamic, things get a bit more ad-hoc, as the key question changes.
Some traditional distinctions (access mode vs. primary mode, such as walk to transit vs. drive to transit) are not distinguished here, rather that would be thought of as at least two trips, one where you walk or drive to some place (with the purpose of changing modes), and second where you take some form of transit.
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```
03: Modeling Methods
Agent-based modeling
Transportation engineers and planners rely on transportation forecasting models to address a wide range of increasingly complicated issues, from congestion and air quality, to social equity concerns. Two major strands of travel demand models have emerged over the past several decades, trip-based and activity-based approaches.
The traditional four-step travel demand model, often referred to as the trip-based approach, takes individual trips as the elementary subjects and considers aggregate travel choices in four steps: trip generation, trip distribution, modal split, and route assignment. This sequential travel demand modeling paradigm, which originated in the 1950s when limited data, computational power, and algorithms were available, ignores the diversity across individuals and does not have solid foundation in travel behavior theory. Discrete choice analysis, describes travel demand as a multi-dimensional hierarchical choice process, including residential and business location choice, trip origin, trip destination, travel modes and etc. Although discrete choice models could improve travel demand prediction by classifying travelers according to certain attributes such as age, gender and household incomes, it still in the end focuses on aggregate travel behaviors and ignores individual decision-making processes. Another flaw of four-step model lies in the fact that this sequential modeling process ignores the interaction between steps and could not predicts certain phenomenon such as induced travel or demand, which can be thought of as a feedback from traffic assignment to trip generation, distribution, and mode split. Although introducing feedback and iteratively applying four-step approach could mitigate this problem, researchers believe that a coherent framework should be introduced to address four steps simultaneously.
To overcome these inadequacies of conventional four-step modeling, activity-based models have been applied in travel demand analysis since the 1970s. Activity-based models predict activities and related travel choices by considering time and space constraints as well as individual characteristics. Individuals will follow a sequence of activities and make corresponding trips connecting those activities to maximize their utilities. Macroscopic travel patterns are predicted through aggregation of individual travel choice.
Although activity-based models have the potential to bridge the gap between individual decision-making processes and macroscopic travel demand, these models require solving many optimization problems simultaneously, which is computationally difficult and behaviorally unrealistic. Therefore, some models employs external aggregate methods such as User Equilibrium (Deterministic (DUE) or Stochastic (SUE)) to address route choice, which compromises their claims as microscopic decision-making models.
The agent-based travel demand model has emerged as a new generation of transportation forecasting tools and provides an alternative to address the topic of travel demand modeling. This modeling approach is flexible and capable to model individual decision making process. There have been many applications of agent-based model in transportation (Transportation Research Part C (2002) dedicated a special issue to this topic). This modeling strategy, however, has not yet been widely adopted in travel demand modeling practice.
To build a pedagogically appropriate model, this chapter introduces an Agent-based Demand and Assignment Model (ADAM), extending Zhang and Levinson (2004), which addresses the destination choice and route choice problems with consideration of congestion. Students have the opportunity to work with the ADAM model for several exercises.
Introduction to agent-based models for transport
While agent-based models are not commonly used in travel demand forecasting as such, many activity-based models are agent-based models of a sort, at least in part, though the behaviors of the agents are typically very complex. Historically, agent-based models come from different fields such as genetics, artificial intelligence, cognitive science, social science. The advantage of using them in transportation begins first with the intuition they provide. It makes more sense to people to think of individual travelers behaving rather than flows. This is in part because it is also more realistic, in that it can be formulated to capture the process by which travelers make decisions, and because it is tracking individuals, can be internally consistent (so that a given traveler has a particular set of constraints (like income, obligations, and time available)
There are several elements in an agent-based model:
• Agents are like people who have characteristics, goals and behavioral rules. The actions of agents depend on the environment they inhabit.
• The environment provides a space where agents live. The environment is shaped by the actions of agents.
• Interaction rules describe how agents and the environment interact
An agent-based model evolves by itself once those micro-level elements are specified. Macro-level properties emerge from this evolutionary process.
An exploratory agent-based model is presented below. The advantage of this model is its simplicity. Clearly, it will lose some predictive detail, but hopefully gives you a flavor of the kinds of modeling approaches and things that can be modeled with agent-based models in the realm of travel demand.
Agent-based Demand and Assignment Model (ADAM)
The agent-based modeling approach assumes that aggregate urban travel demand patterns emerge from multi-dimensional choice process of individuals. All agents have individual characteristics, goals, and rules of travel behaviors. Agents exchange information with the environment on their travel experiences and adjust their travel choices according to available information. In ADAM travelers are active agents and nodes are fixed point agents, while links comprise the environment.
ADAM can be thought of as modeling the AM commute. As shown in Figure 1, ADAM examines the status of each traveler after updating turning matrices at nodes. If a traveler has not found a satisfactory job (status = 1), that traveler will continue the random process of job searching following the rules presented later in this paper. The process will repeat until either all travelers have found jobs (chosen a destination) or some maximum number of iterations are reached. The key components of the agent-based model are introduced in turn below.
Agents
Travelers aim to find a job on the network and a route leading from their origin to this destination with the lowest cost. In the searching process, each traveler visits a node and decides to either accept or reject a job available at that node according to rules discussed later in this paper. If they reject a job at that node, they proceed to another node. Travelers learn current link travel times in the neighborhood of the node when they visit a node through this link and they will only proceed through one link at each step. By accumulating link travel time information during the trip, travelers could derive travel cost between any two nodes they visited.
Nodes are geographic locations where links intersect in the real world. In this model, they also represent the abstract centroids of traffic zone where travelers originate from and are destined to. Furthermore, nodes are carriers of pooled, collective knowledge, including both shortest path information and attractiveness of adjacent nodes. Travelers would exchange knowledge with nodes once they arrived at a new node. The knowledge and exchanging behavior is an abstraction of information spread in a community and communications among travelers in the real world. Links represents roads in the real world and have attributes such as length, free flow travel time, and capacity. Links also provide information about traffic flow and travel time to travelers passing by, which abstracts travelers’ observation of traffic condition in the real world. Links impose geographic constraints to travelers since they are only able to visit adjacent nodes directly connected by a link with the node they are currently visiting.
Rules
Rules are the most important attributes of an agent-based model, which drives the evolution of the model given initial condition. There are two fundamental rules in ADAM: turning rules for finding a destination and information exchange rules for improving paths.
Destination selection rules: Network Origin-Destination Exploration
The first element of ADAM is for each traveler, the discovery of a destination. The model which does this Network Origin-Destination Exploration (NODE) is described below.
Nodes provide turning guidance matrices to travelers, which decide the probability for each traveler to accept a job or proceed to the next node and which direction to go in the later case. Each node () has a set of supply nodes and a set of demand nodes . Therefore, a matrix will be provided and each term, (for simplicity, is omitted), represents the probability to move from supply node to demand node .
(1)
The probability is determined by many factors, including travelers’ characteristics ($\Omega_t$), the opportunity (or attractiveness) at the current node ($b_i$), the opportunity at demand nodes ($b_d$) and the ease of reaching those opportunities ($A$).
(2) $P=f(\Omega_t,b_i,b_d,A)$
Different definitions of turning probability reflect assumptions of different underlying decision-making processes of travelers regarding where to work and may lead to very different travel demand patterns on the network. Zhang and Levinson (2004) assumed that this probability is proportional to jobs available at each node and ignored the ease of reaching them (travel cost). Another disadvantage of this assumption is that if a node does not have available jobs, travelers will never search in this direction even though more jobs may be available via this node.
Extending Zhang and Levinson (2004), a Logit-form probability is used, where $c_d$ represents travel cost to a destination while $c_i$ is the corresponding intrazonal travel cost. The parameter $\theta$ indicates the importance of travel cost when travelers evaluate possible destinations, while $\beta$ is related to people’s relative willingness to travel. A larger $\beta$ implies that travelers are more likely to accept jobs at current node, thus have a shorter travel length.
(3) $p_{s,d}=0 {if} s=d$
(4) $p_{s,d}=\dfrac{b_de^{-\theta c_d}{\beta b_ie^\theta c_d+\displaystyle \sum_{d\epsilon,d\cancel=i}b_de^{-\thetac_d}} (5) if and The variable $b_d$ reflects the opportunity or attractiveness of a node and can be further generalized beyond the number of jobs. We could define it as the summation of jobs on all nodes adjacent to node d, which abstracts the regional accessibility discussed in many previous studies (Handy, 1993). This definition could mitigate the aforementioned problem of search direction. Using accessibility to the whole network is another possibility. However, this may lead to essentially random search since accessibility to the whole network of nearby nodes may be quite similar. In this study we adopt regional accessibility as the indicator of attractiveness to next node, while willingness to accept a job (to stay) is proportional to jobs available at current node. Path learning rule: Agent-based route choice The other important rule in ADAM is the path learning rule. Travelers will learn travel cost of links on their travel route, while nodes keep information about the shortest path from itself to all other nodes which have been visited by travelers to that node. Once a traveler arrives at a new node, that traveler compares their knowledge about travel cost from the current node to each node on the traveler’s travel route. Both of them will keep the shorter "shortest path" after knowledge exchange. Although nodes originally have very limited knowledge about the routes in the remaining network, information spreads rapidly on the network. With the congested link travel time, which can be simply defined as any available travel time-flow relationship, each traveler’s choice will change the link travel time on the network and thus affect destination and route choice of other travelers. Travelers’ route adjustments will trigger more significant change on the network thus other travelers’ behavior. This mechanism reflects the complexity of the real world. The initial route choice can be either given or generated by a random-walk route searching process at iteration 0. In the random walk scenario, travelers set off from their origins and travel in a randomly chosen direction, updating directions after arriving at each node. However, directed cycles and U-turns are prevented. Once travelers arrive at the destination, their travel routes become the initial travel route and will be updated in subsequent iterations. The randomness of searching direction and the large number of travelers will ensure the diversity of initial route choices, which comprises the knowledge based on subsequent iterations. On subsequent iterations, each traveler follows a fixed route chosen at the end of the previous iteration. Once arriving at a destination centroid, travelers will enrich the information set with their individual knowledge while benefiting from the pooled knowledge at the same time by exchanging both shortest path and toll information with centroids. Those travelers will also bring that updated information back to their origin and repeat the exchange process. The information exchange mechanism is illustrated by Figure 1. As illustrated in Figure 1, suppose that the traveler originating at node 1 is traveling to node 5, initially via node 4. His initial shortest path knowledge is 1-3-4-5. Suppose the shortest path information stored at node 5 is 4-5, 3-5, 2-3-5 and 1-2-3-5, respectively from nodes 4, 3, 2 and 1. The comparison starts from the node closest to the current node along the path chain in traveler's memory and repeats for each node on this chain until reaching the origin. After comparing the path from node 3 to 5, the traveler's path information is updated to 1-3-5 since the shortest path for this path segment proposed by the node is shorter than that held by the traveler. Notice that this improvement has also changed the shortest path from node 1 to 5 in the traveler's memory. Consequently, the node will adopt the path from node 1 proposed by the traveler since 1-3-5 is better than 1-2-3-5. The updated path from node 1 to 5 then becomes part of the traveler's shortest path information. This information exchange mechanism will naturally mutate the path chain and generate the most efficient route, sometimes better than all known existing routes. Since nodes store K alternative paths, nodes will insert the path proposed by the visitor in their information pool as long as this path is better than the longest path stored. This information will also be shared with those travelers visiting node 5 at subsequent steps. After stopping at the destination node, travelers compare their travel route determined at the end of previous iterations and shortest path learned during the currently iteration. The path length is evaluated in dollar value by each traveler, considering their individual value of time and the toll charged by each link segment. Since travelers have different values of time, the cost of K alternatives should be reevaluated and sorted for each traveler. If the path suggested by the destination node is better than their current route, the travelers have a probability to switch to the better route that iteration. In general, \[P=f(\sigma,\Delta,T$
To apply this model, we choose a specific form:
T}}\\end{array}}{\rm {}}}\\end{array}}{\rm {}}\{\begin{array}{*{20}c}{P=0}&
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Where:
• $\Delta$ represents the potential benefit by switching routes, which is defined as the time or money saving by choosing route proposed by the destination node instead of sticking to the current route.
• $T$ is the threshold of benefit perception, which reflects both the incapability to perceive small benefit and the inertia for people to change route.
• $\sigma$ denotes the probability of perceiving an existing better route in a given day, and captures the differentiation in the effectiveness of social networks defines the shape of the probability curve.
ARC simulates the day-to-day route choice behavior of travelers and this probability curve must account for two factors:
1. the probability a traveler perceives this better path once its information is available and
2. the probability a traveler takes this path once it is learned. It should be noted that information spreading takes time and not everyone learns immediately.
Travelers with more effective social networks are more likely to be exposed to such information and thus have a higher probability of learning the better path. Once a new road opens, it takes weeks or even months before the flow reaches a stable level. Even when people learn a better alternative, route change involves a certain switching cost preventing travelers from changing routes immediately. Or travelers may just resist changing because of inertia. Considering these factors, this curve should increase as benefits increase and reach some upper limit predicted by the willingness to learn. Estimation of this curve through survey or other psychological studies will enhance the empirical foundation of the model.
Figure 2 illustrates the flow chart of ARC. After travelers choose their routes according to the aforementioned probability, link flow and link travel time will be updated. Consequently, the cost of all possible paths stored both at nodes and travelers will be updated without changing the choice set. Then travelers will follow their new route and repeat the described process until an equilibrium pattern is reached (equilibrium is defined here as link flow variance smaller than a pre-determined threshold $\epsilon$, we arbitrarily choose $\epsilon=5$. Once this equilibrium is reached, no traveler has the incentive to change their travel route according to their behavioral rules and available information. Thus a link flow pattern would be reached and could be provided to other model components under a more comprehensive framework.
Iterations
Traditional travel demand models disentangled this complexity by formulating an optimization problem, using either Deterministic or Stochastic User Equilibrium. However, algorithms employed to solve such optimization problems are computationally cumbersome and behaviorally unrealistic. Instead, ADAM introduces a heuristic learning process to address this challenge. Under this framework, travelers will reenter the network and choose their destination and route again according to the link travel time resulting from their previous choices. Updated shortest path information will be learned and spread by travelers. This process mimics people’s job change and route change behavior. Given the initial condition, ADAM evolves with previously defined rules and a pattern may be achieved according to certain convergence rules, from which macroscopic information such as trip distribution and traffic assignment can be extracted by summing up individual choices.
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Mode choice analysis is the third step in the conventional four-step transportation forecasting model, following Trip Generation and Destination Choice but before Route Choice. While trip distribution's zonal interchange analysis yields a set of origin destination (OD) tables which tells where the trips will be made, mode choice analysis allows the modeler to determine what mode of transport will be used.
Disaggregate Travel Demand models
Travel demand theory was introduced in the appendix on traffic generation. The core of the field is the set of models developed following work by Stan Warner in 1962 (Strategic Choice of Mode in Urban Travel: A Study of Binary Choice). Using data from the CATS, Warner investigated classification techniques using models from biology and psychology. Building from Warner and other early investigators, disaggregate demand models emerged. Analysis is disaggregate in that individuals are the basic units of observation, yet aggregate because models yield a single set of parameters describing the choice behavior of the population. Behavior enters because the theory made use of consumer behavior concepts from economics and parts of choice behavior concepts from psychology. Researchers at the University of California, Berkeley (especially Daniel McFadden, who won a Nobel Prize in Economics for his efforts) and the Massachusetts Institute of Technology (Moshe Ben-Akiva) (and in MIT associated consulting firms, especially Cambridge Systematics) developed what has become known as choice models, direct demand models (DDM), Random Utility Models (RUM) or, in its most used form, the multinomial logit model (MNL).
Choice models have attracted a lot of attention and work; the Proceedings of the International Association for Travel Behavior Research chronicles the evolution of the models. The models are treated in modern transportation planning and transportation engineering textbooks.
One reason for rapid model development was a felt need. Systems were being proposed (especially transit systems) where no empirical experience of the type used in diversion curves was available. Choice models permit comparison of more than two alternatives and the importance of attributes of alternatives. There was the general desire for an analysis technique that depended less on aggregate analysis and with a greater behavioral content. And, there was attraction too, because choice models have logical and behavioral roots extended back to the 1920s as well as roots in Kelvin Lancaster’s consumer behavior theory, in utility theory, and in modern statistical methods.
The Logit Model
The Logit Model, widely used for transportation forecasting in various forms, was first theorized by Daniel McFadden. The Logit model says, the probability that a certain mode choice will be taken is proportional to raised to the utility over the sum of $e$ raised to the utility.
$P_m=\dfrac{e^{u_{ijm}}}{\sum{} e^{u_{ijm}}}$
For any Logit Model the sum of the probability of all modes will equal 1.
$1=\sum{} P_m$
The Logit Model also says that if a new mode of transportation is added to a system (or taken away) then the original modes will lose (or gain) an amount of travels proportional to their share originally.
Steps for the Logit Model:
• Compute the Utility for each OD pair and mode
• Compute Exponentiated utilities for each OD pair and mode
• Sum Exponentiated utilities for each OD pair
• Compute Probability for each mode by OD pair
• Multiply Probability for OD pair by number of trips for each OD pair
Psychological roots
Early psychology work involved the typical experiment: Here are two objects with weights, $w_1$ and $w_2$, which is heavier? The finding from such an experiment would be that the greater the difference in weight, the greater the probability of choosing correctly. Graphs similar to the one on the right result.
Louis Leon Thurstone proposed (in the 1920s) that perceived weight,
$w=v+e$
where $v$ is the true weight and $e$ is random with $E(e)=0$
The assumption that is normally and identically distributed (NID) yields the binary probit model.
Econometric formulation
Economists deal with utility rather than physical weights, and say that
observed utility = mean utility + random term.
Utility in this context refers to the total satisfaction (or happiness) received from making a particular choice or consuming a good or service.
The characteristics of the object, x, must be considered, so we have
$u(x)=v(x)+e(x)$
If we follow Thurston's assumption, we again have a probit model.
An alternative is to assume that the error terms are independently and identically distributed with a Weibull, Gumbel Type I, or double exponential distribution (They are much the same, and differ slightly in their tails (thicker) from the normal distribution). This yields the multinomial logit model (MNL). Daniel McFadden argued that the Weibull had desirable properties compared to other distributions that might be used. Among other things, the error terms are normally and identically distributed. The logit model is simply a log ratio of the probability of choosing a mode to the probability of not choosing a mode.
$log\left(\dfrac{P_i}{1-P_i}\right)=v(x_i)$
Observe the mathematical similarity between the logit model and the S-curves we estimated earlier, although here share increases with utility rather than time. With a choice model we are explaining the share of travelers using a mode (or the probability that an individual traveler uses a mode multiplied by the number of travelers).
The comparison with S-curves is suggestive that modes (or technologies) get adopted as their utility increases, which happens over time for several reasons. First, because the utility itself is a function of network effects, the more users, the more valuable the service, higher the utility associated with joining the network. Second, because utility increases as user costs drop, which happens when fixed costs can be spread over more users (another network effect). Third, technological advances, which occur over time and as the number of users increases, drive down relative cost.
An illustration of a utility expression is given:
$log\left(\dfrac{P_A}{1-P_A}\right)=\beta_0+\beta_1(c_A-c_T)+\beta_2(t_A-t_T)+\beta_3I+\beta_4N=v_A$
where
• Pi = Probability of choosing mode i.
• PA = Probability of taking auto
• cA,cT = cost of auto, transit
• tA,tT = travel time of auto, transit
• I = income
• N = Number of travelers
With algebra, the model can be translated to its most widely used form:
$\dfrac{P_A}{1-P_A}=e^{v_A}$
$P_A=e^{v_A}-P_Ae^{v_A}$
$P_A(1+e^{v_A})=e^{v_A}$
$P_A=\dfrac{e^{v_A}}{1+e^{v_A}}$
It is fair to make two conflicting statements about the estimation and use of this model:
1. It's a "house of cards", and
2. Used by a technically competent and thoughtful analyst, it's useful.
The "house of cards" problem largely arises from the utility theory basis of the model specification. Broadly, utility theory assumes that (1) users and suppliers have perfect information about the market; (2) they have deterministic functions (faced with the same options, they will always make the same choices); and (3) switching between alternatives is costless. These assumptions don’t fit very well with what is known about behavior. Furthermore, the aggregation of utility across the population is impossible since there is no universal utility scale.
Suppose an option has a net utility ujk (option k, person j). We can imagine that having a systematic part vjk that is a function of the characteristics of an object and person j, plus a random part ejk, which represents tastes, observational errors, and a bunch of other things (it gets murky here). (An object such as a vehicle does not have utility, it is characteristics of a vehicle that have utility.) The introduction of e lets us do some aggregation. As noted above, we think of observable utility as being a function:
$v_A=\beta_0+\beta_1(c_A-c_T)+\beta_2(t_A-t_T)+\beta_3I+\beta_4N$
where each variable represents a characteristic of the auto trip. The value β0 is termed an alternative specific constant. Most modelers say it represents characteristics left out of the equation (e.g., the political correctness of a mode, if I take transit I feel morally righteous, so β0 may be negative for the automobile), but it includes whatever is needed to make error terms NID.
Econometric Estimation
Turning now to some technical matters, how do we estimate v(x)? Utility (v(x)) isn’t observable. All we can observe are choices (say, measured as 0 or 1), and we want to talk about probabilities of choices that range from 0 to 1. (If we do a regression on 0s and 1s we might measure for j a probability of 1.4 or -0.2 of taking an auto.) Further, the distribution of the error terms wouldn’t have appropriate statistical characteristics.
The MNL approach is to make a maximum likelihood estimate of this functional form. The likelihood function is:
$L^*=\displaystyle \prod_{n=1}^N f(y_n|x_n, \theta)$
we solve for the estimated parameters
$\hat \theta$
that max L*. This happens when:
$\dfrac{\partial L}{\partial \hat \theta_N}=0$
The log-likelihood is easier to work with, as the products turn to sums:
$lnL^*=\displaystyle \prod_{n=1}^N lnf(y_n|x_n, \theta)$
Consider an example adopted from John Bitzan’s Transportation Economics Notes. Let X be a binary variable that is gamma and 0 with probability (1- gamma). Then f(0) = (1- gamma) and f(1) = gamma. Suppose that we have 5 observations of X, giving the sample {1,1,1,0,1}. To find the maximum likelihood estimator of gamma examine various values of gamma, and for these values determine the probability of drawing the sample {1,1,1,0,1} If gamma takes the value 0, the probability of drawing our sample is 0. If gamma is 0.1, then the probability of getting our sample is: f(1,1,1,0,1) = f(1)f(1)f(1)f(0)f(1) = 0.1*0.1*0.1*0.9*0.1=0.00009. We can compute the probability of obtaining our sample over a range of gamma – this is our likelihood function. The likelihood function for n independent observations in a logit model is
$L^*=\displaystyle \prod_{n=1}^N P_i^{Y_i}(1-P_i)^{1-Y_i}$
where: Yi = 1 or 0 (choosing e.g. auto or not-auto) and Pi = the probability of observing Yi=1
The log likelihood is thus:
$\ell=lnL^*=\displaystyle \prod_{i=1}^n[Y_ilnP_i+(1-Y_i)ln(1-P_i)]$
In the binomial (two alternative) logit model,
$P_auto=\dfrac{e^{v(x_{auto})}{1+e^{v(x_{auto})}$, so
$\ell=lnL^*=\displaystyle \prod_{i=1}^n[Y_iv(x_{auto})-ln(1+e^{v(x_{auto})}]$
The log-likelihood function is maximized setting the partial derivatives to zero:
$\dfrac{\partial L}{\partial \beta}=\displaystyle \prod_{i=1}^n(Y_i-\hat P_i)=0$
The above gives the essence of modern MNL choice modeling.
Independence of Irrelevant Alternatives (IIA)
Independence of Irrelevant Alternatives is a property of Logit, but not all Discrete Choice models. In brief, the implication of IIA is that if you add a mode, it will draw from present modes in proportion to their existing shares. (And similarly, if you remove a mode, its users will switch to other modes in proportion to their previous share). To see why this property may cause problems, consider the following example: Imagine we have seven modes in our logit mode choice model (drive alone, carpool 2 passenger, carpool 3+ passenger, walk to transit, auto driver to transit (park and ride), auto passenger to transit (kiss and ride), and walk or bike). If we eliminated Kiss and Ride, a disproportionate number may use Park and Ride or carpool.
Consider another example. Imagine there is a mode choice between driving and taking a red bus, and currently each has 50% share. If we introduce another mode, let's call it a blue bus with identical attributes to the red bus, the logit mode choice model would give each mode 33.3% of the market, or in other words, buses will collectively have 66.7% market share. Logically, if the mode is truly identical, it would not attract any additional passengers (though one can imagine scenarios where adding capacity would increase bus mode share, particularly if the bus was capacity constrained.
There are several strategies that help with the IIA problem. Nesting of choices allows us to reduce this problem. However, there is an issue of the proper Nesting structure. Other alternatives include more complex models (e.g. Mixed Logit) which are more difficult to estimate.
Returning to roots
The discussion above is based on the economist’s utility formulation. At the time MNL modeling was developed there was some attention to psychologist's choice work (e.g., Luce’s choice axioms discussed in his Individual Choice Behavior, 1959). It has an analytic side in computational process modeling. Emphasis is on how people think when they make choices or solve problems (see Newell and Simon 1972). Put another way, in contrast to utility theory, it stresses not the choice but the way the choice was made. It provides a conceptual framework for travel choices and agendas of activities involving considerations of long and short term memory, effectors, and other aspects of thought and decision processes. It takes the form of rules dealing with the way information is searched and acted on. Although there is a lot of attention to behavioral analysis in transportation work, the best of modern psychological ideas are only beginning to enter the field. (e.g. Golledge, Kwan and Garling 1984; Garling, Kwan, and Golledge 1994).
Modal Split
This page describes historical, but no longer standard, practice in Mode Choice models.
The early transportation planning model developed by the Chicago Area Transportation Study (CATS) focused on transit, it wanted to know how much travel would continue by transit. The CATS divided transit trips into two classes: trips to the CBD (mainly by subway/elevated transit, express buses, and commuter trains) and other (mainly on the local bus system). For the latter, increases in auto ownership and use were trade off against bus use; trend data were used. CBD travel was analyzed using historic mode choice data together with projections of CBD land uses. Somewhat similar techniques were used in many studies. Two decades after CATS, for example, the London study followed essentially the same procedure, but first dividing trips into those made in inner part of the city and those in the outer part. This procedure was followed because it was thought that income (resulting in the purchase and use of automobiles) drove mode choice.
Diversion Curve techniques
The CATS had diversion curve techniques available and used them for some tasks. At first, the CATS studied the diversion of auto traffic from streets and arterial to proposed expressways. Diversion curves were also used as bypasses were built around cities to establish what percentage of the traffic would use the bypass. The mode choice version of diversion curve analysis proceeds this way: one forms a ratio, say:
$\dfrac{c_{transit}}{c_{auto}}=R$
where:
cm = travel time by mode m and
R is empirical data in the form:
Given the R that we have calculated, the graph tells us the percent of users in the market that will choose transit. A variation on the technique is to use costs rather than time in the diversion ratio. The decision to use a time or cost ratio turns on the problem at hand. Transit agencies developed diversion curves for different kinds of situations, so variables like income and population density entered implicitly.
Diversion curves are based on empirical observations, and their improvement has resulted from better (more and more pointed) data. Curves are available for many markets. It is not difficult to obtain data and array results. Expansion of transit has motivated data development by operators and planners. Yacov Zahavi's UMOT studies contain many examples of diversion curves.
In a sense, diversion curve analysis is expert system analysis. Planners could "eyeball" neighborhoods and estimate transit ridership by routes and time of day. Instead, diversion is observed empirically and charts can be drawn.
Examples
Example 1: Mode Choice Model
You are given this mode choice model
$U_{ijm}=-0.412(C_c/w)-0.0201*C_{ivt}-0.0531*C_{ovt}-0.89*D_1-1.783D_3-2.15D_4$
Where:
• $C_c/w$ = cost of mode (cents) / wage rate (in cents per minute)
• $C_{ivt}$ = travel time in-vehicle (min)
• $C_{ovt}$ = travel time out-of-vehicle (min)
• $D$ = mode specific dummies: (dummies take the value of 1 or 0)
• $D_1$ = driving,
• $D_2$ = transit with walk access, [base mode]
• $D_3$ = transit with auto access,
• $D_4$ = carpool
With these inputs:
Driving
Walk
Connect
Transit
Auto
Connect
Transit
Carpool
t = travel time in-vehicle (min) 10 30 15 12
t0 = travel time out-of-vehicle (min) 0 15 10 3
$D_1$ = driving,
1 0 0 0
$D_2$ = transit with walk access, [base mode] 0 1 0 0
$D_3$ = transit with auto access, 0 0 1 0
$D_4$ = carpool 0 0 0 1
COST 25 100 100 150
WAGE 60 60 60 60
What are the resultant mode shares?
Solution
Outputs 1: Driving
2: Walk
Connect
Transit
3:
Auto
Connect
Transit
4: Carpool Sum
Utilities -1.26 -2.09 -3.30 -3.58
EXP(V) 0.28 0.12 0.04 0.03
P(V) 59.96% 26.31% 7.82% 5.90% 100%
Interpretation
Value of Time:
$.0411/2.24=0.0183/min=1.10/hour$
(in 1967 $, when the wage rate was about$2.85/hour)
implication, if you can improve the travel time (by more buses, less bottlenecks, e.g.) for less than $1.10/hour/person, then it is socially worthwhile. Example 2: Mode Choice Model Interpretation What mode would a perfectly rational, perfectly informed traveler choose in a deterministic world given these facts: Case 1 ' Bus Car Parameter Tw 10 min 5 min -0.147 Tt 40 min 20 min -0.0411 C$2 $1 -2.24 Car always wins (independent of parameters as long as all are < 0) Case 2 ' Bus Car Parameter Tw 5 min 5 min -0.147 Tt 40 min 20 min -0.0411 C$2 \$4 -2.24
Results -6.86 -10.51
Sample Problem
Problem 1
You are given the following mode choice model.
$U_{ijm}=-1C_{ijm}+5D_T$
Where:
• $C_{ijm}$ = travel cost between $i$ and $j$ by mode $m$
• $D_T$ = dummy variable (alternative specific constant) for transit
and Travel Times
Auto Travel Times
Origin/Destination Dakotopolis New Fargo
Dakotopolis 5 7
New Fargo 7 5
Transit Travel Times
Origin/Destination Dakotopolis New Fargo
Dakotopolis 10 15
New Fargo 15 8
A. Using a logit model, determine the probability of a traveler driving.
B. Using the results from the previous problem (#2), how many car trips will there be?
Answer
A. Using a logit model, determine the probability of a traveler driving.
Solution Steps
1. Compute Utility for Each Mode for Each Cell
2. Compute Exponentiated Utilities for Each Cell
3. Sum Exponentiated Utilities
4. Compute Probability for Each Mode for Each Cell
5. Multiply Probability in Each Cell by Number of Trips in Each Cell
Auto Utility: $U_{auto}$
Origin\Destination Dakotopolis New Fargo
Dakotopolis -5 -7
New Fargo -7 -5
Transit Utility: $U_{transit}$
Origin\Destination Dakotopolis New Fargo
Dakotopolis -5 -10
New Fargo -10 -3
$e^{U_{auto}}$
Origin\Destination Dakotopolis New Fargo
Dakotopolis 0.0067 0.0009
New Fargo 0.0009 0.0067
$e^{U_{transit}}$
Origin\Destination Dakotopolis New Fargo
Dakotopolis 0.0067 0.0000454
New Fargo 0.0000454 0.0565
Sum: $e^{U_{transit}}+e^{U_{auto}}$
Origin\Destination Dakotopolis New Fargo
Dakotopolis 0.0134 0.0009454
New Fargo 0.0009454 0.0498
P(Auto) = $e^{U_{auto}}/(e^{U_{auto}}+e^{U_{transit}})$
Origin\Destination Dakotopolis New Fargo
Dakotopolis 0.5 0.953
New Fargo 0.953 0.12
P(Transit) = $e^{U_{transit}}/(e^{U_{auto}}+e^{U_{transit}})$
Origin\Destination Dakotopolis New Fargo
Dakotopolis 0.5 0.047
New Fargo 0.047 0.88
Part B
B. Using the results from the previous problem (#2), how many car trips will there be?
Recall
Total Trips
Origin\Destination Dakotopolis New Fargo
Dakotopolis 9395 5606
New Fargo 6385 15665
Total Trips by Auto = $T_{ij}*P(Auto)$
Origin\Destination Dakotopolis New Fargo
Dakotopolis 4697 5339
New Fargo 6511 1867
Problem 2
Prior to the collapse, there were two modes serving the Marcytown-Rivertown corridor: driving alone (d) and carpool (c), which takes advantage of an uncongested carpool lane. The utilities of the modes are as given below.
$U_d=-t_d$
$U_c=-12-t_c$
where t is the travel time
Assuming a multinomial logit model, and
A) That the congested time by driving alone was 15 minutes and time by carpool was 5 minutes. What was the modeshare prior to the collapse?
B) How would you interpret the constant of -12 in the expression for Uc?
C) After the collapse, because of a shift in travelers from other bridges, the travel time by both modes increased by 12 minutes. What is the post-collapse modeshare?
Answer
A) That the congested time by driving alone was 15 minutes and time by carpool was 5 minutes. What was the modeshare prior to the collapse?
Ud = -15
Uc = -17
Pd = 0.88
Pc = 0.12
B) How would you interpret the constant of -12 in the expression for Uc?
The constant -12 is an alternative specific constant. In this example, even if the travel time for the drive and carpool modes were the same, the utility of the carpool mode is lesser than the drive alone due to the negative constant -12. This indicates that there is a lesser probability of an individual choosing the carpool mode due to its lower utility.
C) After the collapse, because of a shift in travelers from other bridges, the travel time by both modes increased by 12 minutes. What is the post-collapse modeshare?
In this question since the travel time for both modes increase by the same 12 minutes, the post-collapse mode share will be the same as before
D) The transit agency decides to run a bus to help out the commuters after the collapse. Again assuming the multinomial logit model holds, without knowing how many travelers take the bus, what proportion of travelers on the bus previously took the car? Why? Comment on this result. Does it seem plausible?
The logit model will indicate that 88% of the bus riders previously drove alone due to the underlying Independence of Irrelevant Alternatives (IIA) property. The brief implication of IIA is that when you add a new mode, it will draw from the existing modes in proportion to their existing shares. This doesnt seem plausible since the bus is more likely to draw from the carpool mode than the drive alone mode.
Additional Questions
Homework
1. Identify five independent variables that you believe affect mode choice. Pose hypotheses about how each variable affects share of travel by each mode.
2. Explain Independence from Irrelevant Alternatives
3. You are given the following mode choice model.
$U_{ij}=-2C_{ijm}+5D_T-5D_W$
Where: $C_{ijm}$ = trabel time between i and j by mode m
$D_T$ = dummy variable (alternative specific constant) for transit
$D_W$ = dummy variable (alternative specific constant) for walking
Here auto is assumed the baseline mode.
The auto travel time between zones (in minutes) is given by the following matrix:
Origin \ Destination To Dakotopolis To New Fargo
From Dakotopolis 5 9
From New Fargo 9 4
The transit travel time between zones (in minutes) is given by the following matrix:
Origin \ Destination To Dakotopolis To New Fargo
From Dakotopolis 10 15
From New Fargo 15 8
The walk travel time between zones (in minutes) is given by the following matrix:
Origin \ Destination To Dakotopolis To New Fargo
From Dakotopolis 10 30
From New Fargo 30 10
Using a logit model, what is the probability of a traveler taking transit?
Additional Questions
1. What factors determine mode choice, utility $U_{mij}$
2. What role does socio-economics play in mode choice?
3. What is an aggregate model, disaggregate model?
4. How is probability ($P_m$) calculated?
5. What happens to probability when utility increases or decreases?
6. Why is it hard for transportation engineers to determine how many people will be taking one mode?
7. Who developed the logit model for transportation mode choice?
8. What is “kiss and ride”
9. What happens when a new mode is introduced?
10. Explain the red bus/blue bus paradox, and Independence of Irrelevant Alternatives feature of multinomial logit models.
11. Compare the logit model with the gravity model.
12. What is an alternative specific constant?
Variables
• $U_{ijm}$- Utility of traveling from i to j by mode m
• $D_n$ = mode specific dummies: (dummies take the value of 1 or 0)
• $P_m$ = Probability of mode m
• $C_c/w$ = cost of mode (cents) / wage rate (in cents per minute)
• $C_{ivt}$ = travel time in-vehicle (min)
• $C_{ovt}$ = travel time out-of-vehicle (min)
• $D_n$ = mode specific dummies: (dummies take the value of 1 or 0)
Abbreviations
• WCT - walk connected transit
• ADT - auto connect transit (drive alone/park and ride)
• APT - auto connect transit (auto passenger/kiss and ride)
• AU1 - auto driver (no passenger)
• AU2 - auto 2 occupants
• AU3+ - auto 3+ occupants
• WK/BK - walk/bike
• IIA - Independence of Irrelevant Alternatives
Key Terms
• Mode choice
• Logit
• Probability
• Independence of Irrelevant Alternatives (IIA)
• Dummy Variable (takes value of 1 or 0)
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/03%3A_Modeling_Methods/3.02%3A_Choice_Modeling.txt
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Land use forecasting undertakes to project the distribution and intensity of trip generating activities in the urban area. In practice, land use models are demand driven, using as inputs the aggregate information on growth produced by an aggregate economic forecasting activity. Land use estimates are inputs to the transportation planning process.
The discussion of land use forecasting to follow begins with a review of the Chicago Area Transportation Study (CATS) effort. CATS researchers did interesting work, but did not produce a transferable forecasting model, and researchers elsewhere worked to develop models. After reviewing the CATS work, the discussion will turn to the first model to be widely known and emulated: the Lowry model developed by Ira S. Lowry when he was working for the Pittsburgh Regional Economic Study. Second and third generation Lowry models are now available and widely used, as well as interesting features incorporated in models that are not widely used.
Today, the transportation planning activities attached to metropolitan planning organizations are the loci for the care and feeding of regional land use models. In the US, interest in and use of models is spotty, for most agencies are concerned with short run planning and day-to-day decisions. Interest is higher in Europe and elsewhere.
Even though there isn’t much use of full blown land use modeling in the US today, we need to understand the subject: the concepts and analytic tools pretty much define how land use-transportation matters are thought about and handled; there is a good bit of interest in the research community where there have been important developments; and when the next upturn in infrastructure development comes along the present models will form the starting place for work.
Land Use Analysis at the Chicago Area Transportation Study
In brief, the CATS analysis of the 1950s was “by mind and hand” distribute growth. The product was maps developed with a rule-based process. The rules by which land use was allocated were based on state-of-the art knowledge and concepts, and it hard to fault CATS on those grounds. The CATS took advantage of Colin Clark’s extensive work on the distribution of population densities around city centers. Theories of city form were available, sector and concentric circle concepts, in particular. Urban ecology notions were important at the University of Chicago and University of Michigan. Sociologists and demographers at the University of Chicago had begun its series of neighborhood surveys with an ecological flavor. Douglas Carroll, the CATS director, had studied with Amos Hawley, an urban ecologist at Michigan.
Colin Clark studied the population densities of many cities, and he found traces similar to those in the figure. Historic data show how the density line has changed over the years. To project the future, one uses changes in the parameters as a function of time to project the shape of density in the future, say in 20 years. The city spreads glacier-like. The area under the curve is given by population forecasts.
The CATS did extensive land use and activity surveys, taking advantage of the City work done by the Chicago Planning Commission. Hock’s work forecasting activities said what the land uses-activities were that would be accommodated under the density curve. Existing land use data were arrayed in cross section. Land uses were allocated in a manner consistent with the existing pattern.
The study area was divided into transportation analysis zones: small zones where there was a lot of activity, larger zones elsewhere. The original CATS scheme reflected its Illinois State connections. Zones extended well away from the city. The zones were defined to take advantage of Census data at the block and minor civil division levels. They also strove for homogeneous land use and urban ecology attributes.
The first land use forecasts at CATS arrayed developments using “by hand” techniques, as stated. We do not fault the “by hand” technique – the then state of computers and data systems forced it. It was a rule based land use allocation. Growth was the forcing function, as were inputs from the economic study. Growth said that the population density envelope would have to shift. The land uses implied by the mix of activities were allocated from “Where is the land available?” and “What’s the use now?” Considerations. Certain types of activities allocate easily: steel mills, warehouses, etc.
Conceptually, the allocation rules seem important. There is lot of spatial autocorrelation in urban land uses; it’s driven by historical path dependence: this sort of thing got started here and seeds more of the same. This autocorrelation was lost somewhat in the step from “by hand” to analytic models.
The CATS procedure was not viewed with favor by the emerging Urban Transportation Planning professional peer group, and in the late 1950s there was interest in the development of analytic forecasting procedures. At about the same time, similar interests emerged to meet urban redevelopment and sewer planning needs, and interest in analytic urban analysis emerged in political science, economics, and geography.
Lowry Model
Hard on the heels of the CATS work, several agencies and investigators began to explore analytic forecasting techniques, and between 1956 and the early 1960s a number of modeling techniques evolved. Irwin (1965) provides a review of the status of emerging models. One of the models, the Lowry model, was widely adopted.
Supported at first by local organizations and later by a Ford Foundation grant to the RAND Corporation, Ira S. Lowry undertook a three-year study in the Pittsburgh metropolitan area. (Work at RAND will be discussed later.) The environment was data rich, and there were good professional relationships available in the emerging emphasis on location and regional economies in the Economics Department at the University of Pittsburgh under the leadership of Edgar M. Hoover. The structure of the Lowry model is shown on the flow chart.
The flow chart gives the logic of the Lowry model. It is demand driven. First, the model responds to an increase in basic employment. It then responds to the consequent impacts on service activities. As Lowry treated his model and as the flow chart indicates, the model is solved by iteration. But the structure of the model is such that iteration is not necessary.
Although the language giving justification for the model specification is an economic language and Lowry is an economist, the model is not an economic model. Prices, markets, and the like do not enter.
A review of Lowry’s publication will suggest reasons why his approach has been widely adopted. The publication was the first full elaboration of a model, data analysis and handling problems, and computations. Lowry’s writing is excellent. He is candid and discusses his reasoning in a clear fashion. One can imagine an analyst elsewhere reading Lowry and thinking, “Yes, I can do that.”
The diffusion of innovations of the model is interesting. Lowry was not involved in consulting, and his word of mouth contacts with transportation professionals were quite limited. His interest was and is in housing economics. Lowry did little or no “selling.” We learn that people will pay attention to good writing and an idea whose time has come.
The model makes extensive use of gravity or interaction decaying with distance functions. Use of “gravity model” ideas was common at the time Lowry developed his model; indeed, the idea of the gravity model was at least 100 years old at the time. It was under much refinement at the time of Lowry’s work; persons such as Alan Voorhees, Mort Schneider, John Hamburg, Roger Creighon, and Walter Hansen made important contributions. (See Carrothers 1956).
The Lowry Model provided a point of departure for work in a number of places. Goldner (1971) traces its impact and modifications made. Steven Putnam at the University of Pennsylvania used it to develop PLUM (Projective Land Use Model) and I(incremental)PLUM. We estimate that Lowry derivatives are used in most MPO studies, but most of today’s workers do not recognize the Lowry heritage, the derivatives are one or two steps away from the mother logic.
Penn-Jersey Model
The P-J (Penn-Jersey, greater Philadelphia area) analysis had little impact on planning practice. It will now be discussed, even so, because it illustrates what planners might have done, given available knowledge building blocks. It is an introduction to some of the work by researchers who are not practicing planners.
The P-J study scoped widely for concepts and techniques. It scoped well beyond the CATS and Lowry efforts, especially taking advantage of things that had come along in the late 1950s. It was well funded and viewed by the State and the Bureau of Public Roads as a research and a practical planning effort. Its Director’s background was in public administration, and leading personnel were associated with the urban planning department at the University of Pennsylvania. The P-J study was planning and policy oriented.
The P-J study drew on several factors "in the air". First, there was a lot of excitement about economic activity analysis and the applied math that it used, at first, linear programming. T. J. Koopmans, the developer of activity analysis, had worked in transportation. There was pull for transportation (and communications) applications, and the tools and interested professionals were available.
There was work on flows on networks, through nodes, and activity location. Orden (1956) had suggested the use of conservation equations when networks involved intermediate modes; flows from raw material sources through manufacturing plants to market were treated by Beckmann and Jacob Marschak (1955) and Goldman (1958) had treated commodity flows and the management of empty vehicles.
Maximal flow and synthesis problems were also treated (Boldreff 1955, Gomory and Hu 1962, Ford and Fulkerson 1956, Kalaba and Juncosa 1956, Pollack 1964). Balinski (1960) considered the problem of fixed cost. Finally, Cooper (1963) considered the problem of optimal location of nodes. The problem of investment in link capacity was treated by Garrison and Marble (1958) and the issue of the relationship between the length of the planning time-unit and investment decisions was raised by Quandt (1960) and Pearman (1974).
A second set of building blocks was evolving in location economics, regional science, and geography. Edgar Dunn (1954) undertook an extension of the classic von Thünen analysis of the location of rural land uses. Also, there had been a good bit of work in Europe on the interrelations of economic activity and transportation, especially during the railroad deployment era, by German and Scandinavian economists. That work was synthesized and augmented in the 1930’s by August Lösch, and his The Location of Economic Activities was translated into English during the late 1940s. Edgar Hoover’s work with the same title was also published in the late 1940s. Dunn’s analysis was mainly graphical; static equilibrium was claimed by counting equations and unknowns. There was no empirical work (unlike Garrison 1958). For its time, Dunn’s was a rather elegant work.
William Alonso’s (1964) work soon followed. It was modeled closely on Dunn’s and also was a University of Pennsylvania product. Although Alonso’s book was not published until 1964, its content was fairly widely known earlier, having been the subject of papers at professional meetings and Committee on Urban Economics (CUE) seminars. Alonso’s work became much more widely known than Dunn’s, perhaps because it focused on “new” urban problems. It introduced the notion of bid rent and treated the question of the amount of land consumed as a function of land rent.
Wingo (1961) was also available. It was different in style and thrust from Alonso and Dunn’s books and touched more on policy and planning issues. Dunn’s important, but little noted, book undertook analysis of location rent, the rent referred to by Marshall as situation rent. Its key equation was:
$R-Y(P-c)-Ytd$
where: R = rent per unit of land, P = market price per unit of product, c = cost of production per unit of product, d = distance to market, and t = unit transportation cost.
In addition, there were also demand and supply schedules.
This formulation by Dunn is very useful, for it indicates how land rent ties to transportation cost. Alonso’s urban analysis starting point was similar to Dunn’s, though he gave more attention to market clearing by actors bidding for space.
The question of exactly how rents tied to transportation was sharpened by those who took advantage of the duality properties of linear programming. First, there was a spatial price equilibrium perspective, as in Henderson (1957, 1958) Next, Stevens (1961) merged rent and transportation concepts in a simple, interesting paper. In addition, Stevens showed some optimality characteristics and discussed decentralized decision-making. This simple paper is worth studying for its own sake and because the model in the P-J study took the analysis into the urban area, a considerable step.
Stevens 1961 paper used the linear programming version of the transportation, assignment, translocation of masses problem of Koopmans, Hitchcock, and Kantorovich. His analysis provided an explicit link between transportation and location rent. It was quite transparent, and it can be extended simply. In response to the initiation of the P-J study, Herbert and Stevens (1960) developed the core model of the P-J Study. Note that this paper was published before the 1961 paper. Even so, the 1961 paper came first in Stevens’ thinking.
The Herbert-Stevens model was housing centered, and the overall study had the view that the purpose of transportation investments and related policy choices was to make Philadelphia a good place to live. Similar to the 1961 Stevens paper, the model assumed that individual choices would lead to overall optimization.
The P-J region was divided into u small areas recognizing n household groups and m residential bundles. Each residential bundle was defined on the house of apartment, the amenity level in the neighborhood (parks, schools, etc.), and the trip set associated with the site. There is an objective function:
$max Z = \displaystyle \sum_{k=1}^u \displaystyle \sum_{i=1}^n \displaystyle \sum_{h=1}^m x_{ih}^k(b_{ih}-c_{ih}^k)$
$x_{ih}^k \ge 0$
wherein xihk is the number of households in group i selecting residential bundle h in area k. The items in brackets are bih (the budget allocated by i to bundle h) and cihk, the purchase cost of h in area k. In short, the sum of the differences between what households are willing to pay and what they have to pay is maximized; a surplus is maximized. The equation says nothing about who gets the surplus: it is divided between households and those who supply housing in some unknown way. There is a constraint equation for each area limiting the land use for housing to the land supply available.
$\displaystyle \sum_{i=1}^n \displaystyle \sum_{h=1}^m s_{ih}x_{ih}^k \le L^k$
where: sih = land used for bundle h Lk = land supply in area k
And there is a constraint equation for each household group assuring that all folks can find housing.
$\displaystyle \sum_{k=1}^u \displaystyle \sum_{h=1}^m x_{ih}^k = N_i$
where: Ni = number of households in group i
A policy variable is explicit, the land available in areas. Land can be made available by changing zoning and land redevelopment. Another policy variable is explicit when we write the dual of the maximization problem, namely:
$min Z'= \displaystyle \sum_{k=1}^u r^kL^k+ \displaystyle \sum_{i=1}^n v_i(-N_i)$
Subject to:
$s_{ih}r^k-v_i \ge b_{ih}-c_{ih}^k$
$r^k \ge 0$
The variables are rk (rent in area k) and vi an unrestricted subsidy variable specific to each household group. Common sense says that a policy will be better for some than others, and that is reasoning behind the subsidy variable. The subsidy variable is also a policy variable because society may choose to subsidize housing budgets for some groups. The constraint equations may force such policy actions.
It is apparent that the Herbert-Stevens scheme is a very interesting one. Its also apparent that it is housing centered, and the tie to transportation planning is weak. That question is answered when we examine the overall scheme for study, the flow chart of a single iteration of the model. How the scheme works requires little study. The chart doesn’t say much about transportation. Changes in the transportation system are displayed on the chart as if they are a policy matter.
The word “simulate” appears in boxes five, eight, and nine. The P-J modelers would say, “We are making choices about transportation improvements by examining the ways improvements work their way through urban development. The measure of merit is the economic surplus created in housing.”
Academics paid attention to the P-J study. The Committee on Urban Economics was active at the time. The committee was funded by the Ford Foundation to assist in the development of the nascent urban economics field. It often met in Philadelphia for review of the P-J work. Stevens and Herbert were less involved as the study went along. Harris gave intellectual leadership, and he published a fair amount about the study (1961, 1962). However, the P-J influence on planning practice was nil. The study didn’t put transportation up front. There were unsolvable data problems. Much was promised but never delivered. The Lowry model was already available.
Kain Model
About 1960, the Ford Foundation made a grant to the RAND Corporation to support work on urban transportation problems (Lowry’s work was supported in part by that grant). The work was housed in the logistics division of RAND, where the economists at RAND were housed. The head of that division was then Charles Zwick, who had worked on transportation topics previously.
The RAND work ranged from new technology and the cost of tunneling to urban planning models and analyses with policy implications. Some of the researchers at RAND were regular employees. Most, however, were imported for short periods of time. The work was published in several formats: first in the RAND P series and RM series and then in professional publications or in book form. Often, a single piece of work is available in differing forms at different places in the literature.
In spite of the diversity of topics and styles of work, one theme runs through the RAND work – the search for economic policy guides. We see that theme in Kain (1962), which is discussed by de Neufville and Stafford, and the figure is adapted from their book.
Kain’s model dealt with direct and indirect affects. Suppose income increases. The increase has a direct effect on travel time and indirect affects through the use of land, auto ownership, and choice of mode. Work supported at RAND also resulted in Meyer, Kain and Wohl (1964). These parts of the work at RAND had considerable influence on subsequent analysis (but not so much on practice as on policy). John Meyer became President of the National Bureau of Economic Research and worked to refocus its lines of work. Urban analysis Kain-style formed the core of a several-year effort and yielded book length publications (see, e.g., G. Ingram, et al., The NBER Urban Simulation Model, Columbia Univ. Press, 1972). After serving in the Air Force, Kain moved to Harvard, first to redirect the Urban Planning Department. After a time, he relocated at the Kennedy School, and he, along with José A. Gómez-Ibáñez, John Meyer, and C. Ingram, lead much work in an economic-policy analysis style. Martin Wohl moved on from RAND, eventually, to Carnegie-Mellon University, where he continued his style of work (e.g. Wohl 1984).
Policy Oriented Gaming
The notion that the impact of policy on urban development might be simulated was the theme for a conference at Cornell in the early 1960s; collegiums were formed, several streams of work emerged. Several persons developed rather simple (from today’s view) simulation games. Land use development was the outcome of gravitational type forces and the issue faced was that of conflicts between developers and planners when planners intervened in growth. CLUG and METROPOLIS are two rather well known products from this stream of work (they were the SimCity of their day); there must be twenty or thirty other similar planner vs. developer in the political context games. There seems to have been little serious attempt to analyze use of these games for policy formulation and decision-making, except for work at the firm Environmetrics.
Peter House, one of the Cornell Conference veterans, established Environmetrics early in the 1960s. It, too, started with relatively simple gaming ideas. Over about a ten-year period, the comprehensiveness of gaming devices was gradually improved and, unlike the other gaming approaches, transportation played a role in their formulation. Environmetrics’ work moved into the Environmental Protection Agency and was continued for a time at the EPA Washington Environmental Studies Center.
A model known as River Basin was generalized to GEM (General Environmental Assessment Model) and then birthed SEAS (Strategic Environmental Assessment Model) and SOS (Son of SEAS). There was quite a bit of development as the models were generalized, too much to be discussed here.
The most interesting thing to be noted is change in the way the use of the models evolved. Use shifted from a “playing games” stance to an “evaluate the impact of federal policy” stance. The model (both equations and data) is viewed as a generalized city or cities. It responds to the question: What would be the impact of proposed policies on cities?
An example of generalized question answering is LaBelle and Moses (1983) La Belle and Moses implement the UTP process on typical cities to assess the impact of several policies. There is no mystery why this approach was used. House had moved from the EPA to the DOE, and the study was prepared for his office.
University of North Carolina
A group at Chapel Hill, mainly under the leadership of Stuart Chapin, began its work with simple analysis devices somewhat similar to those used in games. Results include Chapin (1965), Chapin and H. C. Hightower (1966) and Chapin and Weiss (1968). That group subsequently focused on (1) the ways in which individuals make tradeoffs in selecting residential property, (2) the roles of developers and developer decisions in the urban development process, and (3) information about choices obtained from survey research. Lansing and Muller (1964 and 1967) at the Survey Research Center worked in cooperation with the Chapel Hill Group in developing some of this latter information.
The first work was on simple, probabilistic growth models. It quickly moved from this style to game-like interviews to investigate preferences for housing. Persons interviewed would be given “money” and a set of housing attributes – sidewalks, garage, numbers of rooms, lot size, etc. How do they spend their money? This is an early version of the game The Sims. The work also began to examine developer behavior, as mentioned. (See: Kaiser 1972).
Reviews and Surveys
In addition to reviews at CUE meetings and sessions at professional meetings, there have been a number of organized efforts to review progress in land use modeling. An early effort was the May 1965 issue of the Journal of the American Institute of Planners edited by B. Harris. The next major effort was a Highway Research Board Conference in June, 1967 (HRB 1968) and this was most constructive. This reference contains a review paper by Lowry, comments by Chapin, Alonso, and others. Of special interest is Appendix A, which listed several ways that analysis devices had been adapted for use. Robinson (1972) gives the flavor of urban redevelopment oriented modeling. And there have been critical reviews (e.g. Brewer 1973, Lee 1974). Pack (1978) addresses agency practice; it reviews four models and a number of case studies of applications. (See also Zettel and Carll 1962 and Pack and Pack 1977). The discussion above has been limited to models that most affected practice (Lowry) and theory (P-J, etc.) there are a dozen more that are noted in reviews. Several of those deal with retail and industry location. There are several that were oriented to urban redevelopment projects where transportation was not at issue.
Discussion
Lowry-derived land use analysis tools reside in the MPOs. The MPOs also have a considerable data capability including census tapes and programs, land use information of varied quality, and survey experiences and survey-based data. Although large model work continues, fine detail analysis dominates agency and consultant work in the US. One reason is the requirement for environmental impact statements. Energy, noise, and air pollution have been of concern, and techniques special to the analysis of these topics have been developed. Recently, interest has increased in the uses of developer fees and/or other developer transportation related actions. Perceived shortages for funds for highways and transit are one motive for extracting resources or actions from developers. There’s also the long-standing ethic that those who occasion costs should pay. Finally, there is a small amount of theoretical or academic work. Small is the operative word. There are few researchers and the literature is limited.
The discussion to follow will first emphasize the latter, theory-oriented work. It will then turn to a renewed interest in planning models in the international arena. Modern behavioral, academic, or theory-based analysis of transportation and land use date from about 1965. By modern we mean analysis that derives aggregate results from micro behavior. First models were Herbert-Stevens in character. Similar to the P-J model, they:
• Treated land as the constraining resource and land use choices given land rent variations as the critical behavior.
• Imagined roles for policy makers.
• Emphasized residential land uses and ignored interdependencies in land uses.
• Used closed system, comparative statics ways of thinking.
• And gave no special attention to transportation.
There have been three major developments subsequently:
1. Consideration of transportation activities and labor and capital inputs in addition to land inputs,
2. Efforts to use dynamic, open system ways of thinking, and
3. Inquiry into how micro choice behavior yields macro results.
The Herbert-Stevens model was not a behavioral model in the sense that it did not try to map from micro to macro behavior. It did assume rational, maximizing behavior by locators. But that was attached to macro behavior and policy by assumed some centralized authority that provided subsidies. Wheaton (1974) and Anderson (1982) modified the Herbert-Stevens approach in different, but fairly simple, ways to deal with the artificiality of the Herbert-Stevens formulation.
An alternative to the P-J, Herbert-Stevens tradition was seeded when Edwin S. Mills, who is known as the father of modern urban economics, took on the problem of scoping more widely. Beginning with Mills (1972), Mills has developed a line of work yielding more publications and follow on work by others, especially his students.
Using a Manhattan geometry, Mills incorporated a transportation component in his analysis. Homogeneous zones defined by the transportation system were analyzed as positioned x integer steps away from the central zone via the Manhattan geometry. Mills treated congestion by assigning integer measures to levels of service, and he considered the costs of increasing capacity. To organize flows, Mills assumed a single export facility in the central node. He allowed capital-land rent trade offs yielding the tallest buildings in the central zones.
Stating this in a rather long but not difficult to understand linear programming format, Mills’ system minimizes land, capital, labor, and congestion costs, subject to a series of constraints on the quantities affecting the system. One set of these is the exogenously gives vector of export levels. Mills (1974a,b) permitted exports from non-central zones, and other modifications shifted the ways congestion is measured and allowed for more than one mode of transport.
With respect to activities, Mills introduced an input-output type coefficient for activities; aqrs, denotes land input q per unit of output r using production technique s. T.J. Kim (1979) has followed the Mills tradition through the addition of articulating sectors. The work briefly reviewed above adheres to a closed form, comparative statics manner of thinking. This note now will turn to dynamics.
The literature gives rather varied statements on what consideration of dynamics means. Most often, there is the comment that time is considered in an explicit fashion, and analysis becomes dynamic when results are run out over time. In that sense, the P-J model was a dynamic model. Sometimes, dynamics are operationalized by allowing things that were assumed static to change with time. Capital gets attention. Most of the models of the type discussed previously assume that capital is malleable, and one considers dynamics if capital is taken as durable yet subject to ageing – e.g., a building once built stays there but gets older and less effective. On the people side, intra-urban migration is considered. Sometimes too, there is an information context. Models assume perfect information and foresight. Let’s relax that assumption.
Anas (1978) is an example of a paper that is “dynamic” because it considers durable capital and limited information about the future. Residents were mobile; some housing stock was durable (outlying), but central city housing stock was subject to obsolescence and abandonment.
Persons working in other traditions tend to emphasize feedbacks and stability (or the lack of stability) when they think “dynamics,” and there is some literature reflecting those modes of thought. The best known is Forester (1968), which set off an enormous amount of critique and some follow on thoughtful extensions (e.g., Chen (ed), 1972)
Robert Crosby in the University Research Office of the US DOT was very much interested in the applications of dynamics to urban analysis, and when the DOT program was active some work was sponsored (Kahn (ed) 1981). The funding for that work ended, and we doubt if any new work was seeded.
The analyses discussed use land rent ideas. The direct relation between transportation and land rent is assumed, e.g., as per Stevens. There is some work that takes a less simple view of land rent. An interesting example is Thrall (1987). Thrall introduces a consumption theory of land rent that includes income effects; utility is broadly considered. Thrall manages both to simplify analytic treatment making the theory readily accessible and develop insights about policy and transportation.
Wachs summarizes Lee's (1973) "Requiem for Large Scale Models."
Note
Lee couched his argument, as many of you will recall, in terms of what he called the seven deadly sins of modeling. The seven sins were:
1) Hypercomprehensiveness: Meaning that the models tried to replicate too complex a system in a single shot, and were expected to serve too many different purposes at the same time.
2) Grossness: In a way, the converse of hypercomprehensiveness. Even though they tried to do too much and serve too many purposes, their results or outputs were too coarse and aggregate, too simplistic to be useful for complicated and sophisticated policy requirements.
3) Data Hungriness: Even to produce, gross outputs (a few variables), the models required us to input many variables for many geographic units, and from at least several time periods in order to produce approximate projections, and very often we could not afford the data collection efforts needed to run the models. In other instances, data simply didn't exist at the levels of specificity which would be appropriate to run them.
4) Wrongheadedness: Lee meant that the models suffered from substantial and largely unrecognized deviations between the behavior claimed for them and the variables and equations which actually determined their behavior. As an example, when regional averages were used to calibrate models, but forecasts were made for local areas, the models deviated from reality because of specification errors which were often not even recognized by their users.
5) Complicatedness: Even though when you looked at them through one set of lenses the models seemed terribly simplistic, when looked at through another set of lenses they were outrageously complex. Too simplistic in replicating urban economic and social processes, the models were too complex in their computational algorithms. Errors were multiplied because there were so many equations, spatial units, and time periods. Even the theoretical notion of the model or its representation of an urban process was grossly simplistic compared with reality. Often, the user didn't know how the errors were propagated through series of sequential operations; and sometimes we needed to use systematic adjustments or "correction factors" to make the models more realistic even though we did not completely comprehend the sources of all the errors and could not interpret the correction factors in real-world terms.
6) Mechanicalness: Lee meant that we routinely went through many steps in a modeling process without completely understanding why we did so, and without fully comprehending the consequences in terms of validity or error magnification. He stated, for example, that even rounding errors could be compounded beyond reasonable bounds by mechanical steps taken to calibrate and apply many models without the user's knowledge.
7) Expensiveness: The costs of the models, derived from their grossness, data hungriness, complicatedness, and so on, placed them beyond the financial means of many agencies, or depleted the resources of agencies so much that the very use of models precluded having the resources available to improve them or to fine tune them to make them appropriate to their applications.
Lee argued in 1973 that the models should be improved in four ways:
1) Models should be made more transparent to users and policymakers.
2) Models should combine strong theoretical foundations, objectives information, and wisdom or good judgment. Without these elements, they remain exercises in empty-headed empiricism, abstract theorizing, or false consciousness of what is actually going on in our urban areas.
3) We should start with problems and match our methods to the needs of particular situations, gathering no more information and using no more modeling complexity than is really needed.
4) We should build the simplest models possible, since complex models do not work well, and certainly are unlikely to be understood by those who are asked to act on the basis of the model outputs.
Martin Wachs, Keynote Address: Evolution and Objectives of the Travel Model Improvement Program, in Travel Model Improvement Program Conference Proceedings August 14–17, 1994, edited by Shunk, Gordon and Bass, Patricia
John Landis, has responded about seven challenges facing large scale models:
1. Models - microbehavioral (actors and agents) ... Social Benefit/Social Action
2. Simulation - multiple movies/scenarios
3. Respond to constraints and investments
4. Nonlinearity - path dependence in non-artifactual way (structure and outcomes, network effects)
5. spatial vs. real autocorrelation, emergence - new dynamics, threshold network effects
6. preference utility diversity and change over time
7. Useful beyond calibration periods. Embed innovators and norming agents. Strategic and response function.
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Trip Generation is the first step in the conventional four-step transportation forecasting process (followed by Destination Choice, Mode Choice, and Route Choice), widely used for forecasting travel demands. It predicts the number of trips originating in or destined for a particular traffic analysis zone.
Every trip has two ends, and we need to know where both of them are. The first part is determining how many trips originate in a zone and the second part is how many trips are destined for a zone. Because land use can be divided into two broad category (residential and non-residential) we have models that are household based and non-household based (e.g. a function of number of jobs or retail activity).
For the residential side of things, trip generation is thought of as a function of the social and economic attributes of households (households and housing units are very similar measures, but sometimes housing units have no households, and sometimes they contain multiple households, clearly housing units are easier to measure, and those are often used instead for models, it is important to be clear which assumption you are using).
At the level of the traffic analysis zone, the language is that of land uses "producing" or attracting trips, where by assumption trips are "produced" by households and "attracted" to non-households. Production and attractions differ from origins and destinations. Trips are produced by households even when they are returning home (that is, when the household is a destination). Again it is important to be clear what assumptions you are using.
Activities
People engage in activities, these activities are the "purpose" of the trip. Major activities are home, work, shop, school, eating out, socializing, recreating, and serving passengers (picking up and dropping off). There are numerous other activities that people engage on a less than daily or even weekly basis, such as going to the doctor, banking, etc. Often less frequent categories are dropped and lumped into the catchall "Other".
Every trip has two ends, an origin and a destination. Trips are categorized by purposes, the activity undertaken at a destination location.
Observed trip making from the Twin Cities (2000-2001) Travel Behavior Inventory by Gender
Trip Purpose Males Females Total
Work 4008 3691 7691
Work related 1325 698 2023
Attending school 495 465 960
Other school activities 108 134 242
Childcare, daycare, after school care 111 115 226
Quickstop 45 51 96
Shopping 2972 4347 7319
Visit friends or relatives 856 1086 1942
Personal business 3174 3928 7102
Eat meal outside of home 1465 1754 3219
Entertainment, recreation, fitness 1394 1399 2793
Civic or religious 307 462 769
Pick up or drop off passengers 1612 2490 4102
With another person at their activities 64 48 112
Some observations:
• Men and women behave differently on average, splitting responsibilities within households, and engaging in different activities,
• Most trips are not work trips, though work trips are important because of their peaked nature (and because they tend to be longer in both distance and travel time),
• The vast majority of trips are not people going to (or from) work.
People engage in activities in sequence, and may chain their trips. In the Figure below, the trip-maker is traveling from home to work to shop to eating out and then returning home.
Specifying Models
How do we predict how many trips will be generated by a zone? The number of trips originating from or destined to a purpose in a zone are described by trip rates (a cross-classification by age or demographics is often used) or equations. First, we need to identify what we think the relevant variables are.
Home-end
The total number of trips leaving or returning to homes in a zone may be described as a function of:
$T_h = f(housing \text{ }units, household \text{ }size, age, income, accessibility, vehicle \text{ }ownership)$
Home-End Trips are sometimes functions of:
• Housing Units
• Household Size
• Age
• Income
• Accessibility
• Vehicle Ownership
• Other Home-Based Elements
Work-end
At the work-end of work trips, the number of trips generated might be a function as below:
$T_w=f(jobs(area \text{ }of \text{ } space \text{ } by \text{ } type, occupancy \text{ } rate$
Work-End Trips are sometimes functions of:
• Jobs
• Area of Workspace
• Occupancy Rate
• Other Job-Related Elements
Shop-end
Similarly shopping trips depend on a number of factors:
$T_s = f(number \text{ }of \text{ }retail \text{ }workers, type \text{ }of \text{ }retail, area, location, competition)$
Shop-End Trips are sometimes functions of:
• Number of Retail Workers
• Type of Retail Available
• Area of Retail Available
• Location
• Competition
• Other Retail-Related Elements
Input Data
A forecasting activity conducted by planners or economists, such as one based on the concept of economic base analysis, provides aggregate measures of population and activity growth. Land use forecasting distributes forecast changes in activities across traffic zones.
Estimating Models
Which is more accurate: the data or the average? The problem with averages (or aggregates) is that every individual’s trip-making pattern is different.
Home-end
To estimate trip generation at the home end, a cross-classification model can be used. This is basically constructing a table where the rows and columns have different attributes, and each cell in the table shows a predicted number of trips, this is generally derived directly from data.
In the example cross-classification model: The dependent variable is trips per person. The independent variables are dwelling type (single or multiple family), household size (1, 2, 3, 4, or 5+ persons per household), and person age.
The figure below shows a typical example of how trips vary by age in both single-family and multi-family residence types.
The figure below shows a moving average.
Non-home-end
The trip generation rates for both “work” and “other” trip ends can be developed using Ordinary Least Squares (OLS) regression (a statistical technique for fitting curves to minimize the sum of squared errors (the difference between predicted and actual value) relating trips to employment by type and population characteristics.
The variables used in estimating trip rates for the work-end are Employment in Offices ($E_{off}$), Retail ($E_{ret}$), and Other ($E_{oth}$)
A typical form of the equation can be expressed as:
$T_{D,k}=a_1E_{off,k}+a_2E_{oth,k}+a_3E_{ret,k}$
Where:
• $T_{D,k}$ - Person trips attracted per worker in Zone k
• $E_{off,i}$ - office employment in the ith zone
• $E_{oth,i}$ - other employment in the ith zone
• $E_{ret,i}$- retail employment in the ith zone
• $a_1,a_2,a_3$ - model coefficients
Normalization
For each trip purpose (e.g. home to work trips), the number of trips originating at home must equal the number of trips destined for work. Two distinct models may give two results. There are several techniques for dealing with this problem. One can either assume one model is correct and adjust the other, or split the difference.
It is necessary to ensure that the total number of trip origins equals the total number of trip destinations, since each trip interchange by definition must have two trip ends.
The rates developed for the home end are assumed to be most accurate,
The basic equation for normalization:
$T'_{D,j}=T_{D,j} \dfrac{ \displaystyle \sum{i=1}^I T_{O,i}}{\displaystyle \sum{j=1}^J T_{TD,j}}$
Sample Problems
Problem 1:
Planners have estimated the following models for the AM Peak Hour
$T_{O,i}=1.5*H_i$
$T_{D,j}=(1.5*E_{off,j})+(1*E_{oth,j})+(0.5*E_{ret,j})$
Where:
$T_{O,i}$ = Person Trips Originating in Zone $i$
$T_{D,j}$ = Person Trips Destined for Zone $j$
$H_i$ = Number of Households in Zone $i$
You are also given the following data
Data
Variable Dakotopolis New Fargo
$H$ 10000 15000
$E_{off}$ 8000 10000
$E_{oth}$ 3000 5000
$E_{ret}$ 2000 1500
A. What are the number of person trips originating in and destined for each city?
B. Normalize the number of person trips so that the number of person trip origins = the number of person trip destinations. Assume the model for person trip origins is more accurate.
Answer
A. What are the number of person trips originating in and destined for each city?
Solution to Trip Generation Problem Part A
Households Office Employees Other Employees Retail Employees Origins Destinations
Dakotopolis 10000 8000 3000 2000 15000 16000
New Fargo 15000 10000 5000 1500 22500 20750
Total 25000 18000 8000 3000 37500 36750
B. Normalize the number of person trips so that the number of person trip origins = the number of person trip destinations. Assume the model for person trip origins is more accurate.
Use:
$T'_{D,j}=T_{D,j} \dfrac{ \displaystyle \sum{i=1}^I T_{O,i}}{\displaystyle \sum{j=1}^J T_{TD,j}}=>T_{D,j} \dfrac{37500}{36750}=T_{D,j}*1.0204$
Solution to Trip Generation Problem Part B
Origins Destinations Adjustment Factor Normalized Destinations Rounded
Dakotopolis 15000 16000 1.0204 16326.53 16327
New Fargo 22500 20750 1.0204 21173.47 21173
Total 37500 36750 1.0204 37500 37500
Problem 2:
Modelers have estimated that the number of trips leaving Rivertown ($T_O$) is a function of the number of households (H) and the number of jobs (J), and the number of trips arriving in Marcytown ($T_D$) is also a function of the number of households and number of jobs.
$T_O=1H+0.1J;R^2=0.9$
$T_D=0.1H+1J;R^2=0.5$
Assuming all trips originate in Rivertown and are destined for Marcytown and:
Rivertown: 30000 H, 5000 J
Marcytown: 6000 H, 29000 J
Determine the number of trips originating in Rivertown and the number destined for Marcytown according to the model.
Which number of origins or destinations is more accurate? Why?
Answer
Determine the number of trips originating in Rivertown and the number destined for Marcytown according to the model.
T_Rivertown =T_O ; T_O= 1(30000) + 0.1(5000) = 30500 trips
T_(MarcyTown)=T_D ; T_D= 0.1(6000) + 1(29000) = 29600 trips
Which number of origins or destinations is more accurate? Why?
Origins(T_{Rivertown}) because of the goodness of fit measure of the Statistical model (R^2=0.9).
Problem 3
Modelers have estimated that in the AM peak hour, the number of trip origins (T_O) is a function of the number of households (H) and the number of jobs (J), and the number of trip destinations (T_D) is also a function of the number of households and number of jobs.
$T_O=1.0H+0.1J;R^2=0.9$
$T_D=0.1H+1J;R^2=0.5$
Suburbia: 30000 H, 5000 J
Urbia: 6000 H, 29000 J
1) Determine the number of trips originating in and destined for Suburbia and for Urbia according to the model.
2) Does this result make sense? Normalize the result to improve its accuracy and sensibility?
3) If the interzonal travel cost (from Suburbia to Urbia or Urbia to Suburbia) is 10 minutes, and the intrazonal travel cost (trips within Suburbia or within Urbia) is 5 minutes, use a (doubly-constrained) gravity model wherein the impedance is$f(t_{ij})=t_{ij}^{-2}$, calculate the impedance, balance the matrix to match trip generation, and determine the trip table (within 2 percent accuracy)
Variables
• $T_{O,i}$ - Person trips originating in Zone i
• $T_{D,j}$ - Person Trips destined for Zone j
• $T_{O,i'}$ - Normalized Person trips originating in Zone i
• $T_{D,j'}$ - Normalized Person Trips destined for Zone j
• $T_h$ - Person trips generated at home end (typically morning origins, afternoon destinations)
• $T_w$ - Person trips generated at work end (typically afternoon origins, morning destinations)
• $T_s$ - Person trips generated at shop end
• $H_i$ - Number of Households in Zone i
• $E_{off,k}$ - office employment in Zone k
• $E_{ret,k}$ - retail employment in Zone k
• $E_{oth,k}$ - other employment in Zone k
• $B_n$ - model coefficients
Abbreviations
• H2W - Home to work
• W2H - Work to home
• W2O - Work to other
• O2W - Other to work
• H2O - Home to other
• O2H - Other to home
• O2O - Other to other
• HBO - Home based other (includes H2O, O2H)
• HBW - Home based work (H2W, W2H)
• NHB - Non-home based (O2W, W2O, O2O)
External Exercises
Use the ADAM software at the STREET website and try Assignment #1 to learn how changes in analysis zone characteristics generate additional trips on the network.
Additional Problems
Homework
1. Keep a travel diary for the next week. Analyze your personal travel diary. Summarize the number of trips by mode, by purpose, by time of day, etc. To do so, for each trip record the following information
1. the start and end time (to the nearest minute)
2. start and end location of each trip,
3. primary mode you took (drive alone, car driver with passenger, car passenger, bus, LRT, walk, bike, motorcycle, taxi, Zipcar, other). (use the codes provided)
4. purpose (to work, return home, work related business, shopping, family/personal business, school, church, medical/dental, vacation, visit friends or relatives, other social recreational, other) (use the codes provided)
5. if you traveled with anyone else, and if so whether they lived in your household or not.
Bonus: Email your professor at the end of everyday with a detailed log of your travel diary. (+5 points on the first exam)
Additional Problems
1. Are number of destinations always less than origins?
2. Pose 5 hypotheses about factors that affect work, non-work trips? How do these factors affect accuracy, and thus normalization?
3. What is the acceptable level of error?
4. Describe one variable used in trip generation and how it affects the model.
5. What is the basic equation for normalization?
6. Which of these models (home-end, work-end) are assumed to be more accurate? Why is it important to normalize trip generation models
7. What are the different trip purposes/types trip generation?
8. Why is it difficult to know who is traveling when?
9. What share of trips during peak afternoon peak periods are work to home (>50%, <50%?), why?
10. What does ORIO abbreviate?
11. What types of employees (ORIO) are more likely to travel from work to home in the evening peak
12. What does the trip rate tell us about various parts of the population?
13. What does the “T-statistic” value tell us about the trip rate estimation?
14. Why might afternoon work to home trips be more or less than morning home to work trips? Why might the percent of trips be different?
15. Define frequency.
16. Why do individuals > 65 years of age make fewer work to home trips?
17. Solve the following problem. You have the following trip generation model:
$Trips=B_1Off+B_2Ind+B_3Ret$
And you are given the following coefficients derived from a regression model.
B_1 = 0.61
B_2 = 0.15
B_3 = 0.123
If there are 600 office employees, 300 industrial employees, and 200 retail employees, how many trips are going from work to home?
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Everything is related to everything else, but near things are more related than distant things. - Waldo Tobler's 'First Law of Geography’
Destination Choice (or trip distribution or zonal interchange analysis), is the second component (after Trip Generation, but before Mode Choice and Route Choice) in the traditional four-step transportation forecasting model. This step matches tripmakers’ origins and destinations to develop a “trip table”, a matrix that displays the number of trips going from each origin to each destination. Historically, trip distribution has been the least developed component of the transportation planning model.
Table: Illustrative Trip Table
Origin \ Destination 1 2 3 Z
1 T11 T12 T13 T1Z
2 T21
3 T31
Z TZ1 TZZ
Where: $T_{ij}$ = Trips from origin i to destination j.
Work trip distribution is the way that travel demand models understand how people take jobs. There are trip distribution models for other (non-work) activities, which follow the same structure.
Fratar Models
The simplest trip distribution models (Fratar or Growth models) simply extrapolate a base year trip table to the future based on growth,
$T_{ij,y+1}=g*T_{ij,y}$
where:
• $T_{ij,y}$ - Trips from $i$ to $j$ in year $y$
• $g$ - growth factor
Fratar Model takes no account of changing spatial accessibility due to increased supply or changes in travel patterns and congestion.
Gravity Model
The gravity model illustrates the macroscopic relationships between places (say homes and workplaces). It has long been posited that the interaction between two locations declines with increasing (distance, time, and cost) between them, but is positively associated with the amount of activity at each location (Isard, 1956). In analogy with physics, Reilly (1929) formulated Reilly's law of retail gravitation, and J. Q. Stewart (1948) formulated definitions of demographic gravitation, force, energy, and potential, now called accessibility (Hansen, 1959). The distance decay factor of has been updated to a more comprehensive function of generalized cost, which is not necessarily linear - a negative exponential tends to be the preferred form. In analogy with Newton’s law of gravity, a gravity model is often used in transportation planning.
The gravity model has been corroborated many times as a basic underlying aggregate relationship (Scott 1988, Cervero 1989, Levinson and Kumar 1995). The rate of decline of the interaction (called alternatively, the impedance or friction factor, or the utility or propensity function) has to be empirically measured, and varies by context.
Limiting the usefulness of the gravity model is its aggregate nature. Though policy also operates at an aggregate level, more accurate analyses will retain the most detailed level of information as long as possible. While the gravity model is very successful in explaining the choice of a large number of individuals, the choice of any given individual varies greatly from the predicted value. As applied in an urban travel demand context, the disutilities are primarily time, distance, and cost, although discrete choice models with the application of more expansive utility expressions are sometimes used, as is stratification by income or auto ownership.
Mathematically, the gravity model often takes the form:
$T_{ij}=r_is_jT_{O,i}T_{D,j}F(C_{ij})$
$\displaystyle \sum{j}T_{ij}=T_{O,i}, \displaystyle \sum{i} T_{ij} = T_{D,j}$
$r_i = (\displaystyle \sum{j} s_jT_{D,j}f(C_{ij}))^{-1}$
$s_j=(\displaystyle \sum{i} r_iT_{O,i}f(C_{ij}))^{-1}$
where
• $T_{ij}$ = Trips between origin and destination
• $T_{O,i}$ = Trips originating at
• $T_{D,j}$ = Trips destined for
• $C_{ij}$ = travel cost between and
• $r_i,s_j$ = balancing factors solved iteratively.
• $f$ = impedance or distance decay factor
It is doubly constrained so that Trips from to equal number of origins and destinations.
Balancing a matrix
Balancing a matrix can be done using what is called the Furness Method, summarized and generalized below.
1. Assess Data, you have $T_{O,i}$,$T_{D,j}$, $C_{ij}$
2. Compute $f(C_{ij})$, e.g.
$f(C_{ij})=C_{ij}^{-2}$
$f(C_{ij})=e^{\beta C_{ij}}$
3. Iterate to Balance Matrix
(a) Multiply Trips from Zone $i(T_i)$ by Trips to Zone $j(T_j)$ by Impedance in Cell $ij(f(C_{ij})$ for all $ij$
(b) Sum Row Totals $T'_{O,i}$, Sum Column Totals $T'_{D,j}$
(c) Multiply Rows by $N{O,i}=T_{O,i}/T'_{O,i}$
(d) Sum Row Totals $T'_{O,i}$, Sum Column Totals $T'_{D,j}$
(e) Compare $T_{O,i}$ and $T'_{O,i}$, $T_{D,j}$ $T'_{D,j}$ if within tolerance stop, Otherwise go to (f)
(f) Multiply Columns by $N_{D,j}=T_{D,j}/T'_{D,j}$
(g) Sum Row Totals $T'_{O,i}$, Sum Column Totals $T'_{D,j}$
(h) Compare $T_{O,i}$ and $T'_{O,i}$, $T_{D,j}$ and $T'_{D,j}$ if within tolerance stop, Otherwise go to (b)
Issues
Feedback
One of the key drawbacks to the application of many early models was the inability to take account of congested travel time on the road network in determining the probability of making a trip between two locations. Although Wohl noted as early as 1963 research into the feedback mechanism or the “interdependencies among assigned or distributed volume, travel time (or travel ‘resistance’) and route or system capacity”, this work has yet to be widely adopted with rigorous tests of convergence or with a so-called “equilibrium” or “combined” solution (Boyce et al. 1994). Haney (1972) suggests internal assumptions about travel time used to develop demand should be consistent with the output travel times of the route assignment of that demand. While small methodological inconsistencies are necessarily a problem for estimating base year conditions, forecasting becomes even more tenuous without an understanding of the feedback between supply and demand. Initially heuristic methods were developed by Irwin and Von Cube (as quoted in Florian et al. (1975) ) and others, and later formal mathematical programming techniques were established by Evans (1976).
Feedback and time budgets
A key point in analyzing feedback is the finding in earlier research by Levinson and Kumar (1994) that commuting times have remained stable over the past thirty years in the Washington Metropolitan Region, despite significant changes in household income, land use pattern, family structure, and labor force participation. Similar results have been found in the Twin Cities by Barnes and Davis (2000).
The stability of travel times and distribution curves over the past three decades gives a good basis for the application of aggregate trip distribution models for relatively long term forecasting. This is not to suggest that there exists a constant travel time budget.
In terms of time budgets:
• 1440 Minutes in a Day
• Time Spent Traveling: ~ 100 minutes + or -
• Time Spent Traveling Home to Work: 20 – 30 minutes + or -
Research has found that auto commuting times have remained largely stable over the past forty years, despite significant changes in transportation networks, congestion, household income, land use pattern, family structure, and labor force participation. The stability of travel times and distribution curves gives a good basis for the application of trip distribution models for relatively long term forecasting.
Examples
Example 1: Solving for impedance
You are given the travel times between zones, compute the impedance matrix $f(C_{ij})$, assuming $f(C_{ij})=C_{ij}^{-2}$.
Travel Time OD Matrix ()
Origin Zone Destination Zone 1 Destination Zone 2
1 2 5
2 5 2
Compute impedances ($f(C_{ij})$)
Solution
Impedance Matrix ($f(C_{ij})$)
Origin Zone Destination Zone 1 Destination Zone 2
1
2
Example 2:
You are given the travel times between zones, trips originating at each zone (zone1 =15, zone 2=15) trips destined for each zone (zone 1=10, zone 2 = 20) and asked to use the classic gravity model $f(C_{ij})=C_{ij}^{-2}$
Travel Time OD Matrix ()
Origin Zone Destination Zone 1 Destination Zone 2
1 2 5
2 5 2
Solution
(a) Compute impedances ($f(C_{ij})$)
Impedance Matrix ($f(C_{ij})$)
Origin Zone Destination Zone 1 Destination Zone 2
1 0.25 0.04
2 0.04 0.25
(b) Find the trip table
Balancing Iteration 0 (Set-up)
Origin Zone Trips Originating Destination Zone 1 Destination Zone 2
Trips Destined 10 20
1 15 0.25 0.04
2 15 0.04 0.25
Balancing Iteration 1 ($T_{ij,iteration1}=T_{O,i}T_{D,j}f(C_{ij})$)
Origin Zone Trips Originating Destination Zone 1 Destination Zone 2 Row Total Normalizing Factor
Trips Destined 10 20
1 15 37.50 12 49.50 0.303
2 15 6 75 81 0.185
Column Total 43.50 87
Balancing Iteration 2 ($T_{ij,iteration2}=T_{ij,iteration1}*N_{O,i,iteration1$)
Origin Zone Trips Originating Destination Zone 1 Destination Zone 2 Row Total Normalizing Factor
Trips Destined 10 20
1 15 11.36 3.64 15.00 1.00
2 15 1.11 13.89 15.00 1.00
Column Total 12.47 17.53
Normalizing Factor 0.802 1.141
Balancing Iteration 3 ($T_{ij,iteration3}=T_{ij,iteration2}*N_{D,j,iteration2$)
Origin Zone Trips Originating Destination Zone 1 Destination Zone 2 Row Total Normalizing Factor
Trips Destined 10 20
1 15 9.11 4.15 13.26 1.13
2 15 0.89 15.85 16.74 0.90
Column Total 10.00 20.00
Normalizing Factor = 1.00 1.00
Balancing Iteration 4 ($T_{ij,iteration4}=T_{ij,iteration3}*N_{O,i,iteration3$)
Origin Zone Trips Originating Destination Zone 1 Destination Zone 2 Row Total Normalizing Factor
Trips Destined 10 20
1 15 10.31 4.69 15.00 1.00
2 15 0.80 14.20 15.00 1.00
Column Total 11.10 18.90
Normalizing Factor = 0.90 1.06
Balancing Iteration 16 ($T_{ij,iteration16}=T_{ij,iteration15}*N_{D,i,iteration5$)
Origin Zone Trips Originating Destination Zone 1 Destination Zone 2 Row Total Normalizing Factor
Trips Destined 10 20
1 15 9.39 5.61 15.00 1.00
1 15 9.39 5.61 15.00 1.00
Column Total 10.01 19.99
Normalizing Factor = 1.00 1.00
So while the matrix is not strictly balanced, it is very close, well within a 1% threshold, after 16 iterations. The threshold refers to the proximity of the normalizing factor to 1.0.
Additional Questions
Homework
1. Identify five independent variables that you believe affect trip generation. Pose hypotheses about how each variable affects number of trips generated.
2. Identify three different types of trip distribution models. Which one includes the most information? Which one is most common?
3. You are given the following situation: The towns of Saint Cloud and Minneapolis, separated by 110 km as the crow flies, are to be connected by a railroad, a freeway, and a rural highway. Answer the following questions related to this problem
Trip Generation and Distribution
Your planners have estimated the following models for the AM Peak Hour
$T_{O,i}=500+0.5*{HH}_i$
$T_{D,j}=250+0.5*{OFFEMP_j}+0.355*{OTHEMP_j}+0.094*{RETEMP_j}$
Where: $T_{O,i}=$ Person Trips Originating in Zone i
$T_{D,j}=$Person Trips Destined for Zone j
$HH_i=$ Number of Households in Zone i
${OFFEMP}_j=$Office Employees in Zone j
${OTHEMP}_j=$Other Employees in Zone j
${RETEMP}_j$Retail Employees in Zone j
Your are also given the following data
Saint Cloud Minneapolis
47,604 476,040
33,675 336,750
14,500 145,000
22,000 220,000
The travel time between zones (in minutes) is given by the following matrix:
To Minneapolis To Saint Cloud
From Minneapolis 5 70
From Saint Cloud 70 5
(a) (10) What are the number of AM peak hour person trips originating in and destined for Saint Cloud and Minneapolis.
(b) (10) Assuming the origins are more accurate, normalize the number of destination trips for Saint Cloud and Minneapolis.
(c) (10) Assume a gravity model where the impedance $f(C_{ij})=C_{ij}^{-2}$. Estimate the proportion of trips that go from Saint Cloud to Minneapolis. Solve your matrix within 5 percent of a balanced solution.
Additional Questions
1. What are the different kinds of models for trip distribution? How do they differ?
2. What factors do conventional Trip Distribution models neglect?
3. In a day, how many minutes are spent traveling for the average person, traveling to work? Why might this be stable, not stable?
4. Do travelers have a travel time tolerance or a travel time budget?
5. What affects travel impedance?
6. If impedance increases, will willingness to travel increase or decrease? Is a person more likely to travel to a closer place or a farther place?
7. Why does willingness to travel have a negative exponential form?
8. Briefly describe the gravity model? How might the gravity model be extended to depend on more than just size (opportunities) and distance (impedance)?
9. Why is balancing a matrix an iterative procedure?
10. What importance does impedance play in balancing iterations
11. Give two functions of impedance (f(Cij)) used in gravity models.
12. How do you calculate impedance between zones?
13. What is congested travel time?
14. Conventional trip distribution models are estimated for which mode?
15. How is trip distribution applied?
16. What can a trip distribution curve tell you about maximum and average trip times and willingness to take trips?
17. What factors are multiplied to give a resulting trip distribution curve
Variables
• $T_{O,i}$ - Trips leaving origin
• $T_{D,j}$ - Trips arriving at destination {\displaystyle j}
• $T'_{D,j}$ - Effective Trips arriving at destination , computed as a result for calibration to the next iteration
• $T_{ij}$ - Total number of trips between origin and destination
• $r_i$ - Calibration parameter for Origins
• $s_j$ - Calibration parameter for Destinations
• $f(C_{ij})$ - Cost function between origin and destination
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/03%3A_Modeling_Methods/3.05%3A_Destination_Choice.txt
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Route assignment, route choice, or traffic assignment concerns the selection of routes (alternative called paths) between origins and destinations in transportation networks. It is the fourth step in the conventional transportation forecasting model, following Trip Generation, Destination Choice, and Mode Choice. The zonal interchange analysis of trip distribution provides origin-destination trip tables. Mode choice analysis tells which travelers will use which mode. To determine facility needs and costs and benefits, we need to know the number of travelers on each route and link of the network (a route is simply a chain of links between an origin and destination). We need to undertake traffic (or trip) assignment. Suppose there is a network of highways and transit systems and a proposed addition. We first want to know the present pattern of travel times and flows and then what would happen if the addition were made.
Link Performance Function
The cost that a driver imposes on others is called the marginal cost. However, when making decisions, a driver only faces his own cost (the average cost) and ignores any costs imposed on others (the marginal cost).
• $AverageCost=\dfrac{S_T}{Q}$
• $MarginalCost=\dfrac{\delta S_T}{\delta Q}$
where $S_T$ is the total cost, and $Q$ is the flow.
BPR Link Performance Function
Suppose we are considering a highway network. For each link there is a function stating the relationship between resistance and volume of traffic. The Bureau of Public Roads (BPR) developed a link (arc) congestion (or volume-delay, or link performance) function, which we will term Sa(Qa)
$S_a(Q_a)=t_a(1+0.15\dfrac ({Q_a}{c_a})^4)$
ta = free-flow travel time on link a per unit of time
Qa = flow (or volume) of traffic on link a per unit of time (somewhat more accurately: flow attempting to use link a)
ca = capacity of link a per unit of time
Sa(Qa) is the average travel time for a vehicle on link a
There are other congestion functions. The CATS has long used a function different from that used by the BPR, but there seems to be little difference between results when the CATS and BPR functions are compared.
Can Flow Exceed Capacity?
On a link, the capacity is thought of as “outflow.” Demand is inflow.
If inflow > outflow for a period of time, there is queueing (and delay).
For Example, for a 1 hour period, if 2100 cars arrive and 2000 depart, 100 are still there. The link performance function tries to represent that phenomenon in a simple way.
Wardrop's Principles of Equilibrium
User Equilibrium
Each user acts to minimize his/her own cost, subject to every other user doing the same. Travel times are equal on all used routes and lower than on any unused route.
System optimal
Each user acts to minimize the total travel time on the system.
Price of Anarchy
The reason we have congestion is that people are selfish. The cost of that selfishness (when people behave according to their own interest rather than society's) is the price of anarchy.
The ratio of system-wide travel time under User Equilibrium and System Optimal conditions.
Price of Anarchy = 1}" aria-hidden="true" src="wikimedia.org/api/rest_v1/me...d15c6e66395909">
For a two-link network with linear link performance functions (latency functions), Price of Anarchy is < 4/3.
Is this too much? Should something be done, or is 33% waste acceptable? [The loss may be larger/smaller in other cases, under different assumptions, etc.]
Conservation of Flow
An important factor in road assignment is the conservation of flow. This means that the number of vehicles entering the intersection (link segment) equals the number of vehicles exiting the intersection for a given period of time (except for sources and sinks).
Similarly, the number of vehicles entering the back of the link equals the number exiting the front (over a long period of time).
Auto assignment
Long-standing techniques
The above examples are adequate for a problem of two links, however real networks are much more complicated. The problem of estimating how many users are on each route is long standing. Planners started looking hard at it as freeways and expressways (motorways) began to be developed. The freeway offered a superior level of service over the local street system and diverted traffic from the local system. At first, diversion was the technique. Ratios of travel time were used, tempered by considerations of costs, comfort, and level of service.
The Chicago Area Transportation Study (CATS) researchers developed diversion curves for freeways versus local streets. There was much work in California also, for California had early experiences with freeway planning. In addition to work of a diversion sort, the CATS attacked some technical problems that arise when one works with complex networks. One result was the Moore algorithm for finding shortest paths on networks.
The issue the diversion approach didn’t handle was the feedback from the quantity of traffic on links and routes. If a lot of vehicles try to use a facility, the facility becomes congested and travel time increases. Absent some way to consider feedback, early planning studies (actually, most in the period 1960-1975) ignored feedback. They used the Moore algorithm to determine shortest paths and assigned all traffic to shortest paths. That’s called all or nothing assignment because either all of the traffic from i to j moves along a route or it does not.
The all-or-nothing or shortest path assignment is not trivial from a technical-computational view. Each traffic zone is connected to n - 1 zones, so there are numerous paths to be considered. In addition, we are ultimately interested in traffic on links. A link may be a part of several paths, and traffic along paths has to be summed link by link.
An argument can be made favoring the all-or-nothing approach. It goes this way: The planning study is to support investments so that a good level of service is available on all links. Using the travel times associated with the planned level of service, calculations indicate how traffic will flow once improvements are in place. Knowing the quantities of traffic on links, the capacity to be supplied to meet the desired level of service can be calculated.
Heuristic procedures
To take account of the affect of traffic loading on travel times and traffic equilibria, several heuristic calculation procedures were developed. One heuristic proceeds incrementally. The traffic to be assigned is divided into parts (usually 4). Assign the first part of the traffic. Compute new travel times and assign the next part of the traffic. The last step is repeated until all the traffic is assigned. The CATS used a variation on this; it assigned row by row in the O-D table.
The heuristic included in the FHWA collection of computer programs proceeds another way.
• Step 0: Start by loading all traffic using an all or nothing procedure.
• Step 1: Compute the resulting travel times and reassign traffic.
• Step 2: Now, begin to reassign using weights. Compute the weighted travel times in the previous two loadings and use those for the next assignment. The latest iteration gets a weight of 0.25 and the previous gets a weight of 0.75.
• Step 3. Continue.
These procedures seem to work “pretty well,” but they are not exact.
Frank-Wolfe algorithm
Dafermos (1968) applied the Frank-Wolfe algorithm (1956, Florian 1976), which can be used to deal with the traffic equilibrium problem.
Equilibrium Assignment
To assign traffic to paths and links we have to have rules, and there are the well-known Wardrop equilibrium (1952) conditions. The essence of these is that travelers will strive to find the shortest (least resistance) path from origin to destination, and network equilibrium occurs when no traveler can decrease travel effort by shifting to a new path. These are termed user optimal conditions, for no user will gain from changing travel paths once the system is in equilibrium.
The user optimum equilibrium can be found by solving the following nonlinear programming problem
$min \displaystyle \sum_{a} \displaystyle\int\limits_{0}^{v_a}S_a(Q_a)\, dx$
subject to:
$Q_a=\displaystyle\sum_{i}\displaystyle\sum_{j}\displaystyle\sum_{r}\alpha_{ij}^{ar}Q_{ij}^r$
$sum_{r}Q_{ij}^r=Q_{ij}$
$Q_a\ge 0, Q_{ij}^r\ge 0$
where $Q_{ij}^r$ is the number of vehicles on path r from origin i to destination j. So constraint (2) says that all travel must take place: i = 1 ... n; j = 1 ... n
$\alpha_{ij}^{ar}$= 1 if link a is on path r from i to j ; zero otherwise.
So constraint (1) sums traffic on each link. There is a constraint for each link on the network. Constraint (3) assures no negative traffic.
Transit assignment
There are also methods that have been developed to assign passengers to transit vehicles. In an effort to increase the accuracy of transit assignment estimates, a number of assumptions are generally made. Examples of these include the following:
• All transit trips are run on a set and predefined schedule that is known or readily available to the users.
• There is a fixed capacity associated with the transit service (car/trolley/bus capacity).
Examples
Example 1
Solve for the flows on Links a and b in the Simple Network of two parallel links just shown if the link performance function on link a:
$S_a=5+2*Q_a$
and the function on link b:
$S_b=10+Q_b$
where total flow between the origin and destination is 1000 trips.
Solution
Time (Cost) is equal on all used routes so $S_a=S_b$
And we have Conservation of flow so, $Q_a+Q_b=Q_o=Q_d=1000$
$5+2*(1000-Q_b)=10+Q_b$
$1995=3Q_b$
$Q_b=665;Q_a=335$
Example 2
An example from Eash, Janson, and Boyce (1979) will illustrate the solution to the nonlinear program problem. There are two links from node 1 to node 2, and there is a resistance function for each link (see Figure 1). Areas under the curves in Figure 2 correspond to the integration from 0 to a in equation 1, they sum to 220,674. Note that the function for link b is plotted in the reverse direction.
$S_a=15(1+0.15(\dfrac{Q_a}{1000})^4)$
$S_b=20(1+0.15(\dfrac{Q_a}{3000})^4)$
$Q_a+Q_b=8000$
Show graphically the equilibrium result.
Solution
At equilibrium there are 2,152 vehicles on link a and 5,847 on link b. Travel time is the same on each route: about 63.
Figure 3 illustrates an allocation of vehicles that is not consistent with the equilibrium solution. The curves are unchanged, but with the new allocation of vehicles to routes the shaded area has to be included in the solution, so the Figure 3 solution is larger than the solution in Figure 2 by the area of the shaded area.
Example 3
Assume the traffic flow from Milwaukee to Chicago, is 15000 vehicles per hour. The flow is divided between two parallel facilities, a freeway and an arterial. Flow on the freeway is denoted $Q_f$, and flow on the two-lane arterial is denoted $Q_a$.
The travel time (in minutes) on the freeway ($C_f$) is given by:
$C_f=10+Q_f/1500$
$C_a=15+Q_a/1000$
Apply Wardrop's User Equilibrium Principle, and determine the flow and travel time on both routes.
Solution
The travel times are set equal to one another
$C_f=C_a$
$10+Q_f/1500=15+Q_a/1000$
The total traffic flow is equal to 15000
$Q_f+Q_a=15000$
$Q_a=15000-Q_f$
$10+Q_f/1500=15+(15000-Q_f)/1000$
Solve for $Q_f$
$Q_f=60000/5=12000$
$Q_a=15000-Q_f=3000$
$C_f=18$
$C_a=18$
Thought Questions
• How can we get drivers to consider their marginal cost?
• Alternatively: How can we get drivers to behave in a “System Optimal” way?
Sample Problems
Problem 1
Given a flow of six (6) units from origin “o” to destination “r”. Flow on each route ab is designated with Qab in the Time Function. Apply Wardrop's Network Equilibrium Principle (Users Equalize Travel Times on all used routes)
A. What is the flow and travel time on each link? (complete the table below) for Network A
Link Attributes
Link Link Performance Function Flow Time
o-p $C_{op}=5*Q_{op}$
p-r $C_{pr}=25+Q_{pr}$
o-q $C_{oq}=20+2*Q_{oq}$
q-r $C_{qr}=5*Q_{qr}$
B. What is the system optimal assignment?
C. What is the Price of Anarchy?
Answer
What is the flow and travel time on each link? Complete the table below for Network A:
Link Attributes
Link Link Performance Function Flow Time
o-p $C_{op}=5*Q_{op}$
p-r $C_{pr}=25+Q_{pr}$
o-q $C_{oq}=20+2*Q_{oq}$
q-r $C_{qr}=5*Q_{qr}$
These four links are really 2 links O-P-R and O-Q-R, because by conservation of flow Qop = Qpr and Qoq = Qqr.
Link Attributes
Link Link Performance Function Flow Time
o-p-r $C_{opr}=25+6*Q_{opr}$
o-q-r $C_{oqr}=20+7*Q_{oqr}$
By Wardrop's Equilibrium Principle, the travel time (cost) on each used route must be equal. Therefore $C_{opr}=C_{oqr}$
OR
$25+6*Q_{opr}=20+7*Q_{oqr}$
$5+6*Q_{opr}=7*Q_{oqr}$
$Q_{oqr}=5/7+6*Q_{opr}/7$
By the conservation of flow principle
$Q_{oqr}+Q_{opr}=6$
$Q_{opr}=6-Q_{oqr}$
By substitution
\Q_{oqr}=5/7+6/7(6-Q_{oqr})=41/7-6*Q_{oqr}/7\)
$13*Q_{oqr}=41$
$Q_{oqr}=41/13=3.15$
$Q_{opr}=2.84$
Check
$42.01=25+6(2.84)$
$42.05=20+7(3.15)$
Check (within rounding error)
Link Attributes
Link Link Performance Function Flow Time
o-p-r $C_{opr}=25+6*Q_{opr}$ 2.84 42.01
o-q-r $C_{oqr}=20+7*Q_{oqr}$ 3.15 42.01
or expanding back to the original table:
Link Attributes
Link Link Performance Function Flow Time
o-p $C_{op}=5*Q_{op}$ 2.84 14.2
p-r $C_{pr}=25+Q_{pr}$ 2.84 27.84
o-q $C_{oq}=20+2*Q_{oq}$ 3.15 26.3
q-r $C_{qr}=5*Q_{qr}$ 3.15 15.75
User Equilibrium: Total Delay = 42.01 * 6 = 252.06
Part B
What is the system optimal assignment?
Conservation of Flow:
$Q_{opr}+Q_{oqr}=6$
$TotalDelay=Q_{opr}(25+6*Q_{oqr})+Q_{oqr}(20+7*Q_{oqr})$
$25Q_{opr}+6Q_{opr}^2+(6_Q_{opr})(20+7(6-Q_{opr}))$
$25Q_{opr}+6Q_{opr}^2+(6_Q_{opr})(62-7Q_{opr}))$
$25Q_{opr}+6Q_{opr}^2+372-62Q_{opr}-42Q_{opr}+7Q_{opr}^2$
$13Q_{opr}^2-79Q_{opr}+372$
Analytic Solution requires minimizing total delay
$\deltaC/\deltaQ=26Q_{opr}-79=0$
$Q_{opr}=79/26-3.04$
$Q_{oqr}=6-Q_{opr}=2.96$
And we can compute the SO travel times on each path
$C_{opr,SO}=25+6*3.04=43.24$
$C_{opr,SO}=20+7*2.96=40.72$
Note that unlike the UE solution, $C_{opr,SO}\g C_{oqr,SO}$
Total Delay = 3.04(25+ 6*3.04) + 2.96(20+7*2.96) = 131.45+120.53= 251.98
Note: one could also use software such as a "Solver" algorithm to find this solution.
Part C
What is the Price of Anarchy?
User Equilibrium: Total Delay =252.06 System Optimal: Total Delay = 251.98
Price of Anarchy = 252.06/251.98 = 1.0003 < 4/3
Problem 2
The Marcytown - Rivertown corridor was served by 3 bridges, according to the attached map. The bridge over the River on the route directly connecting Marcytown and Citytown collapsed, leaving two alternatives, via Donkeytown and a direct. Assume the travel time functions Cij in minutes, Qij in vehicles/hour, on the five links routes are as given.
Marcytown - Rivertown Cmr = 5 + Qmr/1000
Marcytown - Citytown (prior to collapse) Cmc = 5 + Qmc/1000
Marcytown - Citytown (after collapse) Cmr = ∞
Citytown - Rivertown Ccr = 1 + Qcr/500
Marcytown - Donkeytown Cmd = 7 + Qmd/500
Donkeytown - Rivertown Cdr = 9 + Qdr/1000
Also assume there are 10000 vehicles per hour that want to make the trip. If travelers behave according to Wardrops user equilibrium principle.
Answer
A) Prior to the collapse, how many vehicles used each route?
Route A (Marcytown-Rivertown) = Ca = 5 + Qa/1000
Route B (Marcytown-Citytown-Rivertown) = Cb = 5 + Qb/1000 + 1 + Qb/500 = 6 + 3Qb/1000
Route C (Marcytown-Donkeytown-Rivertown)= Cc = 7 + Qc/500 + 9 + Qc/1000 = 16 + 3Qc/1000
At equilibrium the travel time on all three used routes will be the same: Ca = Cb = Cc
We also know that Qa + Qb + Qc = 10000
Solving the above set of equations will provide the following results:
Qa = 8467;Qb = 2267;Qc = −867
We know that flow cannot be negative. By looking at the travel time equations we can see a pattern.
Even with a flow of 0 vehicles the travel time on route C(16 minutes) is higher than A or B. This indicates that vehicles will choose route A or B and we can ignore Route C.
Solving the following equations:
Route A (Marcytown-Rivertown) = Ca = 5 + Qa /1000
Route B (Marcytown-Citytown-Rivertown) = Cb = 6 + 3Qb /1000
Ca = Cb
Qa + Qb = 10000
We can the following values:
Qa = 7750; Qb = 2250; Qc = 0
B) After the collapse, how many vehicles used each route?
We now have only two routes, route A and C since Route B is no longer possible. We could solve the following equations:
Route A (Marcytown-Rivertown) = Ca = 5 + Qa /1000
Route C (Marcytown- Donkeytown-Rivertown) = Cc = 16 + 3Qc /1000
Ca = Cc
Qa+ Qc= 10000
But we know from above table that Route C is going to be more expensive in terms of travel time even with zero vehicles using that route. We can therefore assume that Route A is the only option and allocate all the 10,000 vehicles to Route A.
If we actually solve the problem using the above set of equations, you will get the following results:
Qa = 10250; Qc = -250
which again indicates that route C is not an option since flow cannot be negative.
C) After the collapse, public officials want to reduce inefficiencies in the system, how many vehicles would have to be shifted between routes? What is the “price of anarchy” in this case?
User Equilibrium
TotalDelayUE =(15)(10,000)=150,000
System Optimal
TotalDelaySO =(Qa)(5+Qa/1000)+(Qc)(16+3Qc/1000)
Using Qa + Qc = 10,000
TotalDelaySO =(Qa2)/250−71Qa+460000
Minimize total delay ∂((Qa2)/250 − 71Qa + 460000)/∂Qa = 0
Qa/125−7 → Qa = 8875 Qc = 1125 Ca = 13,875 Cc = 19,375
TotalDelaySO =144938
Price of Anarchy = 150,000/144,938 = 1.035
Variables
• $C_T$ - total cost
• $C_k$ - travel cost on link $k$
• $Q_k$ - flow (volume) on link $k$
Abbreviations
• VDF - Volume Delay Function
• LPF - Link Performance Function
• BPR - Bureau of Public Roads
• UE - User Equilbrium
• SO - System Optimal
• DTA - Dynamic Traffic Assignment
• DUE - Deterministic User Equilibrium
• SUE - Stochastic User Equilibrium
• AC - Average Cost
• MC - Marginal Cost
Key Terms
• Route assignment, route choice, auto assignment
• Volume-delay function, link performance function
• User equilibrium
• System optimal
• Conservation of flow
• Average cost
• Marginal cost
External Exercises
Use the ADAM software at the STREET website and try Assignment #3 to learn how changes in network characteristics impact route choice.
Additional Questions
Homework
1. If trip distribution depends on travel times, and travel times depend on the trip table (resulting from trip distribution) that is assigned to the road network, how do we solve this problem (conceptually)?
2. Do drivers behave in a system optimal or a user optimal way? How can you get them to move from one to the other.
3. Identify a mechanism that can ensure the system optimal outcome is achieved in route assignment, rather than the user equilibrium. Why would we want such an outcome? What are the drawbacks to the mechanism you identified?
4. Assume the flow from Dakotopolis to New Fargo, is 5300 vehicles per hour. The flow is divided between two parallel facilities, a freeway and an arterial. Flow on the freeway is denoted $Q_f$, and flow on the two-lane arterial is denoted $Q_r$. The travel time on the freeway $C_f$ is given by:
$C_f=5+Q_f/1000$
The travel time on the arterial (Cr) is given by
$C_r=7+Q_r/500$
(a) Apply Wardrop's User Equilibrium Principle, and determine the flow and travel time on both routes from Dakotopolis to New Fargo.
(b) Solve for the System Optimal Solution and determine the flow and travel time on both routes.
5. Given a flow of 10,000 vehicles from origin to destination traveling on three parallel routes. Flow on each route A, B, or C is designated with $Q_a$, $Q_b$, $Q_c$ in the Time Function Respectively. Apply Wardrop's Network Equilibrium Principle (Users Equalize Travel Times on all used routes), and determine the flow on each route.
$T_A=500+20Q_A$
$T_B=1000+10Q_B$
$T_C=2000+30Q_C$
Additional Questions
1. How does average cost differ from marginal cost?
2. How do System Optimal and User Equilibrium travel time differ?
3. Why do we want people to behave in an SO way?
4. How can you get people to behave in an SO way?
5. Who was John Glen Wardrop?
6. What are Wardrop’s Two Principles?
7. What does conservation of flow require in route assignment?
8. Can Variable Message Signs be used to encourage System Optimal behavior?
9. What is freeflow travel time?
10. If a problem has more than two routes, where does the extra equation come from?
11. How can you determine if a route is unused?
12. What is the difference between capacity and flow
13. Draw a typical volume-delay function for a deterministic, static user equilibrium assignment.
14. Can Q be negative?
15. What is route assignment?
16. Is it important that the output travel times from route choice be consistent with the input travel times for destination choice and mode choice? Why?
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textbooks/eng/Civil_Engineering/Fundamentals_of_Transportation/03%3A_Modeling_Methods/3.06%3A_3-6_Route_Choice.txt
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