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# CHAPTER FIVE SHEAR IN BEAMS 1

## 4 CHAPTER 5: SHEAR IN BEAMS

5.1 Introduction
Loads applied to beams produce bending moments, shearing forces, as shown in Figure 5.1,
and in some cases torques. Design for bending has been elaborately discussed in chapter 4
and torsion will be discussed in chapter 6. In this chapter, design of beams for shear will be
dealt with.

(a) (b)
Figure 5.1: Shear in beams: (a) loaded beam; (b) internal forces at
section A-A
Beams are usually designed for bending moment first; thus cross sectional dimensions are
evaluated along with the required amounts of longitudinal reinforcement. Once this is done,
sections should be checked for shear to determine whether shear reinforcement is required
or not. This by no means indicates that shear is less significant than bending. On the
contrary, shear failure which is usually initiated by diagonal tension is far more dangerous
than flexural failure due to its brittle nature.
5.2 Shear in Homogeneous, Elastic Beams
For good understanding of the subject, consider a simply supported beam subjected to a
uniformly distributed load as shown in Figure 5.2.a. Furthermore, it is assumed that the
beam is made of elastic, homogeneous material.
CHAPTER FIVE SHEAR IN BEAMS 2

Figure 5.2.a : Loaded beam and orientation of cracks
The normal stresses
x
f resulting from bending are given by the following equation as
proved by the classical bending theory;
x
x
x
I
y M
f = ( 5.1)
where
x
M is the bending moment at the section under consideration, y is the distance from
the point under consideration to the neutral axis, and
x
I is the moment of inertia about the
neutral axis.
The shearing stresses
x
are given by the following equation proved in the most classical
mechanics of materials books:
b I
Q V
x
x x
x
= ( 5.2)
where
x
V is the shearing force at the considered section,
x
Q is the moment of the area of
the section located between the point where the shearing stresses are calculated and the
extreme fiber of the section about the neutral axis,
x
I is the moment of inertia about the
neutral axis, and b is the width of the section at the point where shearing stresses are
calculated.
In an attempt to establish the cracking pattern, four elements situated at different distances
from the neutral axis are studied.
Element (1):
It is subjected only to normal, tensile stresses resulting from bending, as shown in Figure
5.2.b. The principal tensile stress is in the same direction as the normal tensile stress. Thus
cracking takes place on a vertical plane due to the weakness of concrete in tension.
CHAPTER FIVE SHEAR IN BEAMS 3

Figure 5.2.b: Principal stresses on elements 1,2,3 and 4
Element (2):
Element (2) that is located below the neutral axis is subjected to normal tensile stresses in
addition to shearing stresses. The cracking surface makes an angle larger than 45 degrees
with the horizontal axis. Cracking takes place on a plane perpendicular to the plane on
which principal tensile stresses occur. See Figure 5.2.b for the orientation of principal
tension stresses along a diagonal crack.
Element (3):
Element (3) that is located at the neutral axis is only subjected to pure shearing stresses.
The cracking surface makes an angle of 45 degrees with the horizontal axis. Thus cracking
takes place on a plane perpendicular to the principal tensile stress, as shown in Figure.5.2.b.
Element (4):
Element (4) that is located above the neutral axis is subjected to normal compressive
stresses in addition to shearing stresses. The cracking surface makes an angle smaller than
45 degrees with the horizontal axis. Cracking takes place on a plane perpendicular to the
CHAPTER FIVE SHEAR IN BEAMS 4
plane on which principal tensile stresses occur. See Figure 5.2.b for the orientation of
principal tension stresses along a diagonal crack.
It is concluded that the shearing force acting on a vertical section in a reinforced concrete
beam does not cause direct rupture of that section. Shear by itself or in combination with
flexure may cause failure indirectly by producing tensile stresses on inclined planes. If
these stresses exceed the relatively low tensile strength of concrete, diagonal cracks
develop. If these cracks are not checked, splitting of the beam or what is known as diagonal
tension failure will take place.
5.3 Types of Shear Cracks
Two types of inclined cracking occur in beams: flexure-shear cracking and web-shear
cracking, shown in Figure 5.3.

A. Flexure-Shear Cracks
The most common type, develops from the tip of a flexural crack at the tension side of the
beam and propagates towards mid depth until it is checked on the compression side of the
beam. For these cracks to form, the bending moment must exceed the cracking moment of
the cross section and a significant shear must exist.
CHAPTER FIVE SHEAR IN BEAMS 5

Figure 5.3: Types of cracks and associated internal forces: (a) orientation
of cracks; (b) shear force diagram; (c) bending moment diagram
B. Web Shear Cracks
Web shear cracking begins from an interior point in a member at the level of the centroid of
uncracked section and moves on a diagonal path to the tension face when the diagonal
tensile stresses produced by shear exceed the tensile strength of concrete. This type of
cracking is common on beams with thin webs and in regions of high shear and small
moment. This combination exists adjacent to simple supports or at points of inflection in
continuous beams.
5.4 Nominal Shear Stress
The only equation available to relate shear stress to shearing force is derived for a beam of
constant cross section constructed of a homogeneous elastic material. Unfortunately, Eq.
(5.2) can not be applied to reinforced concrete beams for the following reasons:
Reinforced concrete is non-homogeneous material.
(b)
(a)
(c)
CHAPTER FIVE SHEAR IN BEAMS 6
Concrete is not elastic.
Variable extent of cracking along the length of a beam, making it impossible to
determine cross-sectional properties.
Therefore, the ACI Code has adopted a simple procedure for establishing the magnitude of
shear stress v on a cross section
d b
V
v
w
= ( 5.3)
where
v = nominal shear stress
V = shearing force at specified section
w
b = width of web of cross section
d = effective depth of the section.
5.5 Current Shear Design Philosophy
The current ACI Code shear design procedure is based on the assumption that for beams
with no shear reinforcement, failure takes place on a vertical plane when the factored shear
force on this plane exceeds the fictitious shear strength of concrete. The fictitious shear
strength of concrete is evaluated from empirical expressions specified within the ACI Code.
This simplification is done due to the following reasons:
Strength of concrete in tension is highly variable, making it hard to evaluate a
sustainable diagonal tension.
Non-homogeneity of reinforced concrete which makes accurate computation of shear
stresses on a particular section a tough task.
Shear failures occur on diagonal planes as they are usually initiated by diagonal
tension.
According to ACI Code 11.1.1, design of cross sections subject to shear should be based on
the following equation.
u n
V V ( 5.4)
where
u
V = factored shear force at section considered
CHAPTER FIVE SHEAR IN BEAMS 7
n
V = nominal shear strength
= strength reduction factor for shear = 0.75
The nominal shear force is generally resisted by concrete and shear reinforcement or,
s c n
V V V + = ( 5.5)
where
c
V = nominal shear force resisted by concrete
s
V = nominal shear force resisted by shear reinforcement
5.6 Internal Forces in a Beam without Shear Reinforcement
The shear strength of a reinforced concrete beam without shear reinforcement is attributed
to three main sources, shown in Figure 5.4.
z c
V is the shear in the uncracked concrete in the
compression zone,
a
V is the vertical component of the shear transferred across the crack by
interlock of the aggregate particles on the two faces of the crack, and
d
V is the dowel action
of the longitudinal reinforcement. The ACI Code considers the three components combined
as the shear force resisted by concrete
c
V , or
d a z c c
V V V V + + = ( 5.6)

Figure 5.4: Behavior of beams failing in shear
5.7 Strength of Concrete in Shear
Shear strength of concrete
c
Shear stresses are computed by dividing the shearing force resisted by concrete
c
V by
d b
w
. Strength of concrete in shear is directly proportional to the strength of concrete in
CHAPTER FIVE SHEAR IN BEAMS 8
tension, inversely proportional to the magnitude of bending moment at the section under
consideration, and directly proportional to the reinforcement ratio of flexural reinforcement.
For the sake of simplicity
c
V is assumed to be the same for beams with or without shear
reinforcement.
For members subject to shear and bending only, ACI Code 11.2.1.1 gives the following
equation for evaluating
c
V
d b ' f 53 . 0 V
w c c
= ( 5.7)
Eq. (5.7) assumes a constant value of
c
V for all sections along the length of the beam.
A more exact formula is specified by ACI Code 11.2.2.1, given by Eq. (5.8)
d b
M
d V
176 ' f 5 . 0 V
w
u
u
w c c

+ = ( 5.8)
where
u
V is the factored shearing force,
u
M is the factored bending moment occurring
simultaneously with
u
V at section considered,
w
is the reinforcement ratio of the web, and
d is the effective depth of the beam. In Eq. (5.8)
u
u
M
d V
should not exceed 1.0, and
c
V
should not exceed d b ' f 93 . 0
w c
.
For members subject to axial compression plus shear, ACI Code 11.2.1.2 gives the
following equation for evaluating
c
V
d b ' f
A 140
N
1 53 . 0 V
w c
g
u
c

+ = ( 5.9)
where
u
N is the factored axial load normal to the cross section occurring simultaneously
with
u
V , and
g
A is the gross area of the cross section. In Eq. (5.9) the shearing force resisted
by concrete
c
V is increased due to the presence of axial compression since the diagonal
tension is decreased.
ACI Code 11.2.2.2 specifies that for members subject to axial compression, it shall be
permitted to compute
c
V using Eq. (5.8) with
m
M substituted for
u
M and
u
u
M
d V
not then
limited to 1.0,
CHAPTER FIVE SHEAR IN BEAMS 9
( )
8
d h 4
N M M
u u m

= (5.10)
where h is overall height of member and
c
V shall not be greater than
g
u
w c c
A 35
N
1 d b ' f 93 . 0 V + = (5.11)
When
m
M as computed by Eq. (5.10) is negative,
c
V shall be computed by Eq. (5.11).
For members subject to axial tension plus shear, ACI Code 11.2.1.3 states that
c
V shall be
taken as zero unless a more detailed analysis is made using ACI 11.2.2.3.
ACI 11.2.2.3 gives the following equation for evaluating
c
V :
d b ' f
A 35
N
1 53 . 0 V
w c
g
u
c

+ = (5.12)
where
u
N is the factored axial load normal to the cross section occurring simultaneously
with
u
V and is taken negative,
g
A is the gross area of the cross section and
c
V shall not be
less than zero.
In Eq. (5.12) the shearing force resisted by concrete
c
V is decreased due to presence of
axial tension which causes widening of cracks.
According to ACI Code 11.2.3, for circular members, the area used to compute
c
V is taken
as the product of the diameter and effective depth of the concrete section. It is permitted to
take d as 0.80 times the diameter of the concrete section.
5.8 Strength Provided by Shear Reinforcement
When the nominal shearing force
n
V exceeds the shearing force that can be resisted by
concrete alone
c
V , shear reinforcement, in any of the forms shown in the following section,
can be used.
5.8.1 Types of Shear Reinforcement
When shear reinforcement is required, the following types of shear reinforcement are
permitted by ACI Code 11.4.1, as shown in Figure 5.5.
a. Vertical Stirrups;
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b. Inclined stirrups making an angle of 45 degree or more with longitudinal tension
reinforcement;
c. Longitudinal reinforcement with bent portion making an angle of 30 degree or
more with the tension reinforcement;
d. Spirals, circular ties, or hoops;
e. Combination of stirrups and bent longitudinal reinforcement.
f. Welded wire reinforcement with wires located perpendicular to axis of member.
Before diagonal cracking occurs, the stirrups remain unstressed. After cracking, the stress
in the stirrups increases as they pick up a portion of the load formerly carried by the
uncracked concrete.
CHAPTER FIVE SHEAR IN BEAMS 11

Figure 5.5: Types of shear reinforcement: (a) vertical stirrups; (b) inclined
stirrups; (c) bent-up bars (two groups); (d) bent-up bars (three groups)
In Figure 5.6, assume an inclined crack making an angle of 45 degree with the longitudinal
reinforcement and extending from the longitudinal reinforcement to the compression zone
of the beam. For the shear reinforcement to be effective, it should intersect this diagonal
crack. Let n be the number of shear reinforcing bars intersecting such crack. The shearing
force resisted by shear reinforcing bars across the crack
s
V is given by
(b)
(a)
(c)
(d)
CHAPTER FIVE SHEAR IN BEAMS 12

Figure 5.6: Shear resisted by stirrups
sin T V
s
= (5.13)
where T is the resultant of the forces in the shear reinforcement across the crack and given
by
yt v
f A n T = (5.14)
where
v
A is the total area of shear reinforcement within a distance S ,
and
yt
f is the yield stress of the shear reinforcement.
n is given by
( )
S
d
S
d
S
S
n
) 1 (cot 45 cot cot
1
+
=
+
= =

o
(5.15)
Substituting Eq. (5.14) and Eq. (5.15) into Eq. (5.13), one gets
( ) ( )
S
cos sin d f A
S
f sin A 1 cot d
V
yt v yt v
s
+
=
+
= (5.16)
For vertical stirrups, is equal to 90 degrees, and Eq. (5.16) takes the following form
S
d f A
V
yt v
s
= (5.17)
Eq. (5.18) is valid for inclined stirrups and longitudinal bars bent at more than one point.
When longitudinal reinforcement is bent at a single point,
s
V is given by
d b ' f 8 . 0 sin f A V
w c y v s
= (5.18)
where is the angle between bent-up reinforcement and longitudinal axis of the member.
CHAPTER FIVE SHEAR IN BEAMS 13
5.8.2 Minimum Amount of Shear Reinforcement
The instant diagonal crack forms, the tension carried by the concrete must be transferred to
the stirrups if the beam is not to split into two sections. To ensure that the stirrups will have
sufficient strength to absorb the diagonal tension in the concrete, ACI Code 11.4.6 states
that a minimum area of shear reinforcement is to be provided in concrete members where
the factored shearing force
u
V exceeds half the shear strength provided by concrete
c
V 50 . 0 , except for the following:
Footings and solid slabs.
Concrete joist construction.
Beams with h not greater than 25 cm.
Beams integral with slabs with h not greater than 60 cm and not greater than the lrger
of 2.5 times thickness of flange, and 0.5 times width of web
These exceptions were made because there is a possibility of load sharing between weak
and strong areas.
Where shear reinforcement is required by ACI 11.4.6.1 or for strength and where ACI
11.5.1 allows torsion to be neglected, ACI Code 11.4.6.3 requires that the minimum area of
shear reinforcement,
min , v
A is computed by
yt
w
c min , v
f
S b
' f 2 . 0 A = (5.19)
but not less than be less than
yt
w
min , v
f
S b 5 . 3
A =
where
min , v
A is the minimum area of shear reinforcement within a distance S ,
w
b is the
web width, S is the spacing of shear reinforcement, and
yt
f is the yield stress of the shear
reinforcement.
5.8.3 Maximum Stirrup Spacing
The assumption made in Eq. (5.17) is that one or more stirrups cross each potential
diagonal crack in order to prevent the beam from splitting into two sections between
stirrups. To ensure that, this requirement is satisfied, ACI Code 11.4.5.1 through 11.4.5.3
specifies the following limits for maximum spacing of shear reinforcement.
CHAPTER FIVE SHEAR IN BEAMS 14
A. Vertical Stirrups:
When d b f V
w c s
,
max
S is limited to the smaller of d/2 or 60 cm. This ensures
that at least one stirrup with adequate anchorage will be available to hold the upper
and lower sections together.
When d b f V d b f
w c s w c
> > 10 . 2 ,
max
S is limited to the smaller of d/4 or 30
cm, to ensure that each potential diagonal crack will be crossed by approximately
three stirrups. See Figure 5.7 for schematic representation.
B. Inclines stirrups and Bent-up Bars:
When d b f V
w c s
, inclined stirrups and bent-up longitudinal bars are spaced such that
every 45-degree line, extending toward the reaction from mid depth of member d/2 to
longitudinal tension reinforcement, is to be crossed by at least one stirrup or bent-up bar. If
d b f V d b f
w c s w c
> > 10 . 2 , the line is to be crossed by at least two stirrups or two
groups of bent up bars.
The limit on maximum stirrup spacing serves to limit diagonal crack widths and to provide
a better anchorage for the longitudinal reinforcement against breaking through the concrete
cover or loss of bond with the concrete around the bars.
CHAPTER FIVE SHEAR IN BEAMS 15

Figure 5.7: Maximum stirrup spacing
5.8.4 Ensuring Ductile Behavior
To prevent a shear-compression failure caused by diagonal compression stresses in the
compression zone above the tip of a diagonal crack, ACI Code 11.4.7.9 requires that the
maximum force resisted by shear reinforcement
s
V is not to exceed d b ' f 2 . 2
w c
. Since
diagonal tensile stresses develop in the direction perpendicular to the compressive stresses,
the compressive strength of the concrete will be less than that based on the
Uniaxial test. This is done to ensure a ductile mode of failure by forcing the shear
reinforcement to yield before the concrete starts to crush.
CHAPTER FIVE SHEAR IN BEAMS 16
5.9 Critical Section for Shear
According to ACI Code 11.1.3.1, sections located less than a distance d from face of
support are permitted to be designed for the same shear as that calculated at a distance d.
Conditions for the validity of ACI Code 11.1.3.1 are listed in ACI Code 11.1.3 and given
here:
Loads applied near the top of the member, shown in 5.8.a and 5.8.b.
No concentrated loads are applied between the face of support and a distance d from
it, shown in 5.8.c. Otherwise the critical section for shear is taken at the face of the
support.
Support reaction in direction of applied shear, introducing compression into the end
region of the member, shown in 5.8.d. If tension is introduced, the critical section is
taken at the face of the support.

Figure 5.8: Location of critical section for shear
This provision recognizes that a crack adjacent to a support whose reaction induces
compression into a beam will have a horizontal projection of at least d; therefore the
maximum shear force that must be transmitted across the potential failure plane closest to
the support will be equal to the reaction R reduced by any external forces applied to the
beam with a distance d from the support.
(b)
(d) (c)
(a)
CHAPTER FIVE SHEAR IN BEAMS 17
5-10 Punching (Two-way) Shear
When loads are applied over small areas to slabs and footings with no beams, punching
failure may occur. The sloping failure surface takes the shape of a truncated pyramid in
case of rectangular or square columns and a truncated cone in case of circular columns, as
shown in Figure 5.9.

Figure 5.9: Punching shear: (a) punching shear failure of an isolated
footing; (b) actual failure surface; (c) assumed failure surface
The ACI Code R11.11.1.2 assumes that failure takes place on vertical planes located at
distance d/2 from faces of column. In ACI Code 11.11.2.1, the punching shear strength
c
V
is given by the smallest of:
d b
2
1 ' f 53 . 0 V
c c o

+ = (5.20)
d b ' f V
c c o
= (5.21)
d b ' f 2
b
d
27 . 0 V
c
s
c o
o

+ = (5.22)
where
o
b is the perimeter of the critical punching shear section for slabs and footings, d is
the effective depth,
c
is the ratio of long side to short side of column, and
s
is a constant
dependent on the location of the column relative to the slab or footing; is equal to 40 for
interior columns, 30 for side columns, and 20 for corner columns.
(b)
(c)
(a)
CHAPTER FIVE SHEAR IN BEAMS 18
For interior square columns, the perimeter of the critical section ) ( 4 d c b
o
+ = , for
rectangular columns, ) 2 ( 2 d c c b
o
+ + = , and for circular columns, ) ( d D b
o
+ = , where c,
c , and D are column cross sectional dimensions for rectangular and circular columns
respectively. (see Figure 5.10).

(a)

CHAPTER FIVE SHEAR IN BEAMS 19

(b)
Figure 5.10: Punching shear surface for isolated footings:
(a) rectangular column; (b) circular column
The design equation is given here
c u
V V (5.23)
where
u
V is the factored shearing force, is the strength reduction factor for shear, and
c
V
is the nominal punching shear provided by concrete.
5.11 Summary of ACI Shear Design Procedure for Beams
Once the beam is designed for moment, thus establishing the concrete dimensions and the
required longitudinal reinforcement, the beam is designed for shear as explained in the next
steps.
1. Draw the shearing force diagram and establish the critical section for shear.
2. Calculate the nominal capacity of concrete in shear,
c
V .
3. Check whether the chosen concrete dimensions are adequate for ensuring a ductile
mode of failure. If not satisfied, the concrete dimensions should be increased.
4. Classify the factored shearing forces acting on the beam according to the following
categories:
CHAPTER FIVE SHEAR IN BEAMS 20
For
c u
V V 50 . 0 , no shear reinforcement is required.
For
c u c
V V V > > 50 . 0 , minimum shear reinforcement is required.
For
c u
V V shear reinforcement is required. When more than one type of shear
reinforcement is used to reinforce the same portion of a beam, shear strength
s
V is
calculated as the sum of the
s
V values evaluated from each type.
5. Check the spacing between shear reinforcement according to ACI Code limits
discussed earlier in this chapter.
6. Sketch the shear reinforcement along the length of the beam.

1
Example (5.1):
A rectangular beam has the dimensions shown in the figure and is loaded with a 40 ton
concentrated factored load at the middle of the span. Design the necessary shear
reinforcement given that
2
/ 250 cm kg f
c
= and
2
/ 4200 cm kg f
y
= . Width of support is
equal to 30 cm.
Solution:
Assuming 8 mm stirrups,
cm d 95 . 63 25 . 1 80 . 0 4 70 = =
1- Draw shearing force diagram:
Shearing force diagram is shown in the figure.
2- Calculate the shear capacity of concrete:
( ) ( ) tons 06 . 12
1000
95 . 63
30 250 53 . 0 75 . 0 V
c
=

=

(a) Beam and loads; (b) shear force diagram; (c) bending moment
diagram
(a)
(b)
(c)

2
3- Check whether the dimensions of the cross section are adequate for ensuring a
ductile mode of failure:
s c
u
V V
V
+ =

tons 59 . 10
75 . 0
06 . 12
75 . 0
0 . 20
V
s
= =
( ) tons 59 . 10 tons 74 . 66
1000
95 . 63
30 250 2 . 2 d b ' f 2 . 2
w c
> =

=
i.e. cross sectional dimensions are adequate in terms of preventing brittle mode of failure.
4- Classify the factored shear force:
c u
V V >
Trying two-legged 8 mm vertical stirrups, its spacing is evaluated from Eq. (5.17)
( )
( ) ( )
cm / cm 0394 . 0
95 . 63 4200
1000 59 . 10
d f
V
S
A
2
yt
s v
= = =
The required shear reinforcement is not to be less than the larger of:
( )
cm / cm 0226 . 0
4200
30 250 20 . 0
f
b ' f 20 . 0
S
A
2
yt
w c
min
v
= = =

and
( )
cm / cm 0025 . 0
4200
30 5 . 3
f
b 5 . 3
S
A
2
yt
w
min
v
= = =

cm / cm 0394 . 0
2
< O.K.
5- Check stirrups spacing:
( ) tons 59 . 10 tons 37 . 33
1000
95 . 63
30 250 1 . 1 d b ' f 1 . 1
w c
> =

=
Maximum stirrup spacing is not to exceed the smaller of d/2 = 31.97 cm or 60 cm.
Use 8 mm vertical stirrups spaced at 25 cm.
6- The shear reinforcement is drawn in the figure.

3

Shear reinforcement
Example (5.2):
A rectangular beam has the dimensions shown in the figure and is loaded with a uniform
service dead load of 4.0 t/m (including own weight of beam) and a service live load of 2.5
t/m. Design the necessary shear reinforcement given that
2
/ 280 cm kg f
c
= and
2
/ 4200 cm kg f
y
= . Width of support is equal to 30 cm.

(a) Beam and loads; (b) shearing force diagram
Solution:
1- Draw shearing force diagram:
The shearing force diagram is shown in the figure.
Shear force at the centerline of the support is given as:
(b)
(a)

4
( ) ( ) [ ]( ) tons 8 . 30 2 / 7 5 . 2 6 . 1 4 2 . 1 V
u
= + =
Critical section for shear is located at a distance of 54.2 cm from the face of support. From
similar triangles, design shear force is equal to 24.71 tons.
2- Calculate the shear capacity of concrete:
( ) ( ) tons 82 . 10
1000
2 . 54
30 280 53 . 0 75 . 0 V
c
=

=
tons 41 . 5
2
V
c
=

s c
u
V V
V
+ =

tons 52 . 18
75 . 0
82 . 10
75 . 0
71 . 24
V
s
= =
3- Check whether the dimensions of the cross section are adequate for ensuring a
ductile mode of failure:
( ) tons 52 . 18 tons 86 . 59
1000
2 . 54
30 280 2 . 2 d b ' f 2 . 2
w c
> =

=
i.e. cross sectional dimensions are adequate in terms of preventing brittle mode of failure.
4- Classified the factored shear forces:
The forces are shown in the figure.

Shear force zone classification
Zone (A):
2
c
u
V
V

5
No shear reinforcement is required, but it is recommended to use minimum area of shear
reinforcement.
Trying 8 mm vertical stirrups
( ) ( )
( )
cm 83 . 41
30 280 2 . 0
4200 50 . 0 2
S
1
= =
( ) ( )
( )
cm 0 . 40
30 5 . 3
4200 50 . 0 2
b 5 . 3
f A
S
w
yt min , v
= = =
Maximum stirrup spacing:
Maximum stirrup spacing is not to exceed the smaller of d/2 = 27.1 cm or 60 cm. So, use
8 mm vertical stirrups spaced at 25 cm.
Zone (B):
2
c
u c
V
V V

> > , minimum shear reinforcement is required.
Use two-legged 8 mm vertical stirrups spaced at 25 cm similar to zone A.
Zone (C):
c u
V V >
s c
u
V V
V
+ =

tons 52 . 18 V
s
=
Trying two-legged 8 mm vertical stirrups, its spacing is evaluated from Eq. (4.15)
( ) ( ) ( )
( )
cm 29 . 12
1000 52 . 18
2 . 54 4200 50 . 0 2
V
d f A
S
s
yt v
= = =
( ) tons 52 . 18 tons 93 . 29
1000
2 . 54
30 280 1 . 1 d b ' f 1 . 1
w c
> =

=
Maximum stirrup spacing is not to exceed the smaller of d/2 = 27.1 cm or 60 cm. Use 8
mm vertical stirrups spaced at 12 cm.
5- Check stirrups spacing
Stirrups spacing is checked for all the zones.

6

Shear reinforcement
6- The shear reinforcement is shown in the figure.

7
Problems
P.1 Design shear reinforcement required for the beam shown in Figure P.1. The beam
carries a uniformly distributed service dead load of 2.0 t/m, including own weight of beam,
and a service live load of 2.5 t/m, in addition to two concentrated service live loads, 2.5 tons
each, distanced at 1.50 m from the supports.
Use
2
/ 250 cm kg f
c
= and
2
/ 4200 cm kg f
y
= .

Figure P4.1

P.2 The 3.0 m-long cantilever beam shown in Figure P.2 carries service loads consisting of
0.80 t/m (own weight is not included), and 2.0 ton concentrated live load at the free end.
Design the required shear reinforcement.
Use
2
/ 250 cm kg f
c
= and
2
/ 4200 cm kg f
y
= .

Figure P.2
P.3 The beam shown in Figure P.3 supports a factored dead load of 0.75 t/m (own weight of
beam is included) and a factored live load of 4.25 t/m. Design the required shear
reinforcement.
Use
2
/ 280 cm kg f
c
= and
2
/ 4200 cm kg f
y
= .

8

Figure P.3

P.4 For the beam shown in Figure P.4, design the required shear reinforcement to resist the
Use
2
/ 280 cm kg f
c
= and
2
/ 4200 cm kg f
y
= .

Figure P.4

P.5 For the beam shown in Figure P.5, design the required shear reinforcement to resist the
Use
2
/ 250 cm kg f
c
= and
2
/ 4200 cm kg f
y
= .

9

Figure P.5

P.6 The beam shown in Figure P.6 contains four bars each is 25 mm in diameter out of
which two bars are bent up at 45 degree angle near the support. Design the required shear
reinforcement to resist the given load.
Use
2
/ 300 cm kg f
c
= and
2
/ 4200 cm kg f
y
= .

Figure P.6

P.7 For the two isolated footings shown in Figure P.7, check whether the provided depths
are adequate for resisting punching shear for the given service loads. The allowable net soil
pressure is equal to 1.5 kg/cm
2
. Use
2
/ 250 cm kg f
c
= .

10

(a) (b)
Figure P.7
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