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# Prestressed Concrete Structures

## 4.1 Design of Members

This section covers the following topics

## Calculation of Demand Design of Sections for Axial Tension

Introduction The design of prestressed concrete members can be done by the limit states method as given in Section 4 of IS:1343 - 1980. First, the force demand in a member under the design loads is determined from a structural analysis. A preliminary size of the member is assumed for analysis. Next, the member is designed to meet the demand. If necessary, another cycle of analysis and design is performed. The following material explains the calculation of the demand in a member under the design loads.

## 4.1.1 Calculation of Demand

Characteristics Loads For dead loads, a characteristic load is defined as the value which has a 95% probability of not being exceeded during the life of the structure. This concept assumes a normal distribution of the values of a particular dead load. In the following figure, the

## Dr. Amlan K Sengupta and Prof. Devdas Menon

shaded area above the characteristic value represents 5% probability of exceedance of the load in the design life of the structure.

## Frequency 5% probability of exceedence Mean Characteristic value

Figure 4-1.1

For live load, wind load and earthquake load, a characteristic load is defined based on an extreme value distribution. For example, the characteristic wind load is defined as the value which has a 98% probability of not being exceeded during a year.

Frequency 2% probability of exceedence Characteristic value Annual maximum mean wind speed

Figure 4-1.2

## Extreme value distribution

The characteristics loads can be obtained from IS:875 - 1987 (Code of Practice for Design Loads for Buildings and Structures) and IS:1893 - 2002 (Criteria for Earthquake Resistant Design of Structures) as follows.

## Dr. Amlan K Sengupta and Prof. Devdas Menon

Codes covering information of loads Code (DL) (LL) (WL) (SL) (EL) IS:875 - 1987 Part 1 IS:875 - 1987 Part 2 IS:875 - 1987 Part 3 IS:875 - 1987 Part 4 IS:1893 - 2002 Part 1

For special loads, there are some guidelines in IS: 875 - 1987, Part 5. In addition, specialised literature may be referred to for these loads. The special loads are listed below. Temperature Hydrostatic Soil pressure Fatigue Accidental load Impact and collision Explosions Fire For special situations, the loads are determined from testing of prototype specimens. Dynamic load tests, wind tunnel tests, shake table tests are some types of tests to determine the loads on a structure. Finite element analysis is used to determine the stresses due to concentrated forces and dynamic loads.

Load Factors and Load Combinations The load factors and the combinations of the various types of loads are given in Table 5 of IS:1343 - 1980. The following are the combinations for the ultimate condition. 1.5 (DL + LL) 1.2 (DL + LL WL) 1.2 (DL + LL EL) 1.5 (DL EL)

## Dr. Amlan K Sengupta and Prof. Devdas Menon

1.5 (DL WL) 0.9 DL 1.5 EL The load combinations for service conditions are as follows. DL + LL DL + 0.8 (LL EL) DL EL DL WL

Analysis of Structures Regarding analysis of structures, IS:1343 - 1980 recommends the same procedure as stated in IS:456 - 2000. A structure can be analysed by the linear elastic theory to calculate the internal forces in a member subjected to a particular type of load.

Design of Members There can be more than one way to design a member. In design, the number of unknown quantities is larger than the number of available equations. Hence, some quantities need to be assumed at the beginning. These quantities are subsequently checked. The member can be designed either for the service loads or, for the ultimate loads. The procedure given here is one of the possible procedures. The design is based on satisfying the allowable stresses under service loads and at transfer. Initially, a lumpsum estimate of the losses is considered under service loads. After the first round of design, detailed computations are done to check the conditions of allowable stresses. Precise values of the losses are computed at this stage. The section is then analysed for the ultimate capacity. The capacity should be greater than the demand under ultimate loads to satisfy the limit state of collapse.

## 4.1.2 Design of Sections for Axial Tension

Introduction Prestressed members under axial loads only, are uncommon. Members such as hangers and ties are subjected to axial tension. Members such as piles may have bending moment along with axial compression or tension.

Design of Prestressing Force First, a preliminary dimension of the member is selected based on the architectural requirement. The prestressing force at transfer (P0) should be such that the compressive stress in concrete is limited to the allowable value. should be within the allowable value. At service, the designed prestressing force (Pe) should be such that the tensile stress in concrete The amount of prestressing steel (Ap) is determined from the designed prestressing force based on the allowable stress in steel. At transfer, in absence of non-prestressed reinforcement, the stress in concrete (fc) is given as follows.
fc = P0 Ac

(4-1.1)

Here, Ac = net area of concrete P0 = prestress at transfer after short-term losses. In presence of non-prestressed reinforcement, the stress in the concrete (fc) can be calculated as follows.
fc = P0 Ac + (Es /Ec )As

(4-1.2)

Here, As = area of non-prestressed reinforcement Es = modulus of elasticity of steel Ec = modulus of elasticity of concrete. At service, the stress in concrete (fc) can be calculated as follows.
fc = Pe P Ac At

(4-1.3)

## Dr. Amlan K Sengupta and Prof. Devdas Menon

Here, At = transformed area of section P = external axial force Pe = effective prestress. The external axial force is considered positive if it is tension and negative if it is compression. In the above expression, non-prestressed reinforcement is not considered. If there is non-prestressed reinforcement, Ac is to be substituted by (Ac + (Es/Ec) As) and At is to be calculated including As. Analysis of Ultimate Strength The ultimate tensile strength of a section (PuR) is calculated as per Clause 22.3, IS:1343 - 1980. The ultimate strength should be greater than the demand due to factored loads. In absence of non-prestressed reinforcement, the ultimate tensile strength of a section (PuR) is given as follows.
PuR = 0.87fpk Ap

(4-1.4)

## In presence of non-prestressed reinforcement,

PuR = 0.87fy As + 0.87fpk Ap

(4-1.5)

In the previous equations, fy = characteristic yield stress for non-prestressed reinforcement with mild steel bars = characteristic 0.2% proof stress for non-prestressed reinforcement with high yield strength deformed bars. fpk = characteristic tensile strength of prestressing tendons. The following example shows the design of a post-tensioned hanger for tension.

## Dr. Amlan K Sengupta and Prof. Devdas Menon

Example 4-1.1
Design a post-tensioned hanger to carry an axial tension of PDL = 300 kN (dead load including self-weight) and PLL = 130 kN. The dimension of the hanger is 250 250 mm2.

Design the section without considering non-prestressed reinforcement. Tension is not allowed under service loads.

The grade of concrete is M 35. The age at transfer is 28 days. Assume 15% long term losses in the prestress.

The following properties of the prestressing strands are available from tests. Type of prestressing tendon : 7 wire strand Nominal diameter Nominal area Tensile strength fpk Modulus of elasticity = 12.8 mm = 99.3 mm2 = 1860 N/mm2 = 195 kN/mm2.

Solution Preliminary calculations at transfer Ac A = 250 250 = 62,500 mm2 Allowable stress for M35 concrete under direct compression at transfer
fcc,all = 0.8 0.51fci = 0.8 0.5135 = 14.3 N/mm2

fc = -

Pe P + Ac At

## Allowable stress at service Considering 15% loss

fct,all = 0 N/mm2

Pe = 0.85P0
Substituting the values
0= 0.85P0 P + A A

## fp 0 = 0.8fpk = 0.81860 = 1488 N/mm2

Required area of tendon
Ap = 506,000 1488 = 340 mm2

## Ap = 499.3 = 397.2 mm2

Prestress at transfer

P0 = 397.21488 N = 591kN

## Dr. Amlan K Sengupta and Prof. Devdas Menon

Final calculations at transfer Ac = 62,500 - 397 = 62103 mm2 Stress in concrete fc = =P0 Ac 591,000 62,103 OK

= - 9.5 N/mm2
fc < fcc,all

## Final calculations at service

E p = 195 kN/mm2
Ec = 5,000 35 = 29,580 N/mm2

At = 62,103 +

## 195 397.2 29.6 = 64,720 mm2

Stress in concrete
fc = Pe P + Ac At

fc = -

## 0.85 591,000 (300 +130)103 + 62,103 64,720

= - 1.4 N/mm2
No tensile stress in concrete. OK. Final calculations for ultimate strength

## Pu = 1.5(300 +130) = 645.0 kN

PuR Pu Designed cross-section OK

250

250

## (4) 7-wire strands with P0 = 591 kN

Nominal non-prestressed reinforcement is provided for resisting thermal and shrinkage cracks.
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