Reinforced

Concrete
Design
chapter Four
Shear in Beams
4-1 Introduction

4-2 Shear Reinforcement Design Requirements

4-3 Shear Analysis Procedure

4-4 Stirrup Design Procedure

4-5 Torsion of Reinforced Concrete Members

4-1 Introduction
In prior chapters we have been concerned primarily with
the bending strength of reinforced concrete beams and
slabs.

The shear forces create additional tensile stresses that

must be considered.

In these members steel reinforcing must be added

specifically to provide additional shear strength if the
shear is in excess of the shear strength of the concrete
itself.
The concepts of bending stresses and shearing stresses in homogeneous elastic beams are
generally discussed at great length in most strength-of-materials texts.

The accepted expressions are

=

=
=

=
All points in the length of the beam, where the shear and bending moment are not equal to zero are
subject to both shearing stresses and bending stresses.

The principal stresses in a beam subjected to shear and bending may be calculated using the following
formula:
The orientation of the principal planes may be calculated using the following formula:
•In Figure 4-1 we isolate a small, square unit element from
the neutral axis of a beam (where f = 0).

•The vertical shear stresses are equal and opposite on the

two vertical faces by reason of equilibrium.

•If these were the only two stresses present, the element
would rotate.
If we consider a set of orthogonal planes that are inclined at
45° with respect to the original element and resolve the shear
stresses into components that are parallel and perpendicular to
these planes, the effect will be as shown in Figure 4-2.

In the beams with which we are concerned the diagonal tension

failure would be the mode of failure in shear. Such a failure is
shown in Figure 4-3.
4-2 Shear Reinforcement Design
Requirements
•ACI 318-11, Chapter 11, addresses shear and torsion design Vu : ultimate shear strength

provisions for both non-prestressed and prestressed concrete Vc : nominal shear strength provided by concrete

members. (λ: 1) normal wt concrete

(λ: 0.85) sand light concrete
•The code allows vertical stirrups and welded wire reinforcement with (λ: 0.75) all light wt concrete
wires located perpendicular to the axis of the member as well as Vs : nominal shear strength provided by steel
spirals, circular ties, or hoops.

For members of normal-weight concrete that are subject to shear and

flexure only,
unreinforced for shear, can resist is Vc:

λ λ=
Vu : ultimate shear strength
Vc : nominal shear strength provided by concrete

Vs : nominal shear strength provided by steel

Av : total cross-sectional area of web reinforcement within a distance s; for single-loop stirrups, Av = 2As,
where As is the cross-sectional area of the stirrup bar (in.2)
bw : web width = b for rectangular sections (in.)
s : center-to-center spacing of shear reinforcement in a direction parallel to the longitudinal reinforcement
(in.)
fyt : yield strength of web reinforcement steel (psi)
Φ: 0.75

The code requires that a minimum area of shear reinforcement be

provided in all reinforced concrete flexural members where Vu exceeds
1/2 фVc except as follows:
Vu : ultimate shear strength
Vc : nominal shear strength provided by concrete

Vs : nominal shear strength provided by steel

Av : total cross-sectional area of web reinforcement within a distance s; for single-loop stirrups, Av = 2As,
where As is the cross-sectional area of the stirrup bar (in.2)
bw : web width = b for rectangular sections (in.)
s : center-to-center spacing of shear reinforcement in a direction parallel to the longitudinal reinforcement
(in.)
fyt : yield strength of web reinforcement steel (psi)
Φ: 0.75

In cases where shear reinforcement is required for strength

or because Vu > 1/2 фVc, the minimum area of shear
reinforcement shall be calculated from
Vu : ultimate shear strength
Vc : nominal shear strength provided by concrete

Vs : nominal shear strength provided by steel

Av : total cross-sectional area of web reinforcement within a distance s; for single-loop stirrups, Av = 2As,
where As is the cross-sectional area of the stirrup bar (in.2)
bw : web width = b for rectangular sections (in.)
s : center-to-center spacing of shear reinforcement in a direction parallel to the longitudinal reinforcement
(in.)
fyt : yield strength of web reinforcement steel (psi)
Φ: 0.75

The ACI Code, Section 11.1.1, states that the basis for shear design must be

f
Vu : ultimate shear strength
Vc : nominal shear strength provided by concrete

Vs : nominal shear strength provided by steel

Av : total cross-sectional area of web reinforcement within a distance s; for single-loop stirrups, Av = 2As,
where As is the cross-sectional area of the stirrup bar (in.2)
bw : web width = b for rectangular sections (in.)
s : center-to-center spacing of shear reinforcement in a direction parallel to the longitudinal reinforcement
(in.)
fyt : yield strength of web reinforcement steel (psi)
Φ: 0.75

For vertical stirrups,

4-3 Shear Analysis Procedure
•The shear analysis procedure involves checking the shear strength in an existing member and
verifying that the various code requirements have been satisfied.

•The member may be reinforced or plain.

Example 4-1
Example 4-2
Problems
4-4 Stirrup Design Procedure
In the design of stirrups for shear reinforcement, the end result is a determination of stirrup size
and spacing pattern.

A general procedure may be adopted as follows:

Notes on Stirrup Design
1. Materials and maximum stresses
a) To reduce excessive crack widths in beam webs subject to diagonal tension, the ACI Code, Section
11.4.2, limits the design yield strength of shear reinforcement to 60,000 psi. This increases to 80,000
psi for deformed welded wire reinforcing.
b) The value of Vs must not exceed 8√f′c bw d irrespective of the amount of web reinforcement (ACI Code,
Section 11.4.7.9).
2. Bar sizes for stirrups
a) The most common stirrup size used is a No. 3 bar. Under span and loading conditions where the
shear values are relatively large, it may be necessary to use a No. 4 bar.
b) When conventional single-loop stirrups are used, the web area Av provided by each stirrup is twice the
cross-sectional area of the bar (No. 3 bars, Av = 0.22 in.² ; No. 4 bars, Av = 0.40 in.²) because each
stirrup crosses a diagonal crack twice.
c) If possible, do not vary the stirrup bar sizes; use the same bar sizes unless all other alternatives are
not reasonable.
3. Stirrup spacings
a) When stirrups are required, the maximum spacing for vertical stirrups must not exceed d/2 or 24 in., whichever is
smaller (ACI Code, Section 11.4.5.1).The maximum spacing may also be governed by ACI Equation (11-13), which
gives

b) It is usually undesirable to space vertical stirrups closer than 4 in.

c) It is generally economical and practical to compute the spacing required at several sections and to place stirrups
accordingly in groups of varying spacing.
d) The code permits (Section 11.1.3), when the support reaction introduces a vertical compression into the end region
of a member, no concentrated load occurs between the face of the support and distance d from the face of the
support.
e) The actual stirrup pattern used in the beam is the designer’s choice.
Example 4-3
s (in)

x (ft)
1 3 5 7 9
s (in)

1 3 5 7 9
x (ft)
Problems
A

A
10'
4-5 Torsion of Reinforced
Concrete Members
•The torsion or twisting of reinforced concrete members is
caused by a torsional moment that acts about the
longitudinal axis of the member due to unbalanced loads
applied to the member.
•The torsional moment usually acts in combination with
bending moment and shear force as shown in Figure 4-14.
•A typical example of torsion in concrete members occurs in
a rectangular beam supporting precast hollow-core slabs
(or planks).
•Rectangular and L-beams are more susceptible to torsion
than T-beams.

Figure 4-14.1 Numerical Analysis of Torsional

Behavior of Ultra-High Performance
In the ACI Code, the design for torsion in solid and
hollow concrete beams is based on a thin-walled tube
space truss model (see Figures 4-15 and 4-16).
In the thin-walled space truss model, the outer concrete
cross section that is centered on the stirrups is
assumed to resist the torsion while the concrete in the
core is neglected because after cracking, this core is
relatively ineffective in resisting torsion.
Torsional moments cause additional shear stresses that
result in diagonal tension stresses in the concrete
member.
In the thin-walled tube model, the shear flow q, which is assumed to be constant around the perimeter of
the beam, is equal to the product of the shear stress 𝝉 and the wall thickness, t. Using the thin-walled
tube model, and summing the torques, the equilibrium of torsional moments yields

xo and yo are the width and height of the space truss model measured between the centerlines of the tube
walls—that is, the centerlines of the longitudinal corner bars.
•There are two conditions that may occur in the design of
reinforced concrete members for torsion: primary or
equilibrium torsion and secondary or compatibility
torsion.
•Compatibility Torsion
• occurs in statically indeterminate structures, and the design
torque, which cannot be obtained from statics alone, may be
reduced due to redistribution of internal forces to maintain
compatibility of deformations.
• One example of compatibility torsion occurs in spandrel
beams (see Figure 4-17), where the rotation of the slab is
restrained by the spandrel beam.
•Equilibrium Torsion
• For statically determinate structures, the design
torque, which can be obtained from statics
considerations alone, cannot be reduced because
redistribution of internal moments and forces is not
possible in such structures, and in order to
maintain equilibrium, the full design torque has to
be resisted by the beam.
• Examples of concrete members in equilibrium
torsion are shown in Figures 4-18 and 4-19.
Torsion Design of Reinforced Concrete Members (ACI Code
Section, 11.5)

The ACI Code design approach for torsion follows a similar

approach to the design for shear. That is, if
Torsional reinforcement is required to resist the full applied torsional moment as specified in ACI,
Section 11.5.2.1, when

The size of the beam should be increased if these relationships are not satisfied.
Torsion Reinforcement
•The reinforcement required for torsion shall be added to that required for other load effects that act
in combination with the torsional moment, and the most restrictive spacing requirements must be
satisfied (ACI Code, Section 11.5.3.8).
• Vertical Equilibrium of Forces
• the equilibrium of the vertical forces yields V2 can be calculated as

• the ratio of the area of the torsional stirrup to the spacing of the stirrup as
•Horizontal Equilibrium of Forces

•Assuming that xo = x1 and yo = y1 yields the equation for the longitudinal torsion reinforcement as
• Transverse Reinforcement Required for Torsion (Stirrups)
• The required torsional stirrup area,
•The total equivalent transverse reinforcement or stirrups required for combined shear plus torsion
is obtained from the ACI Code, Section 11.5.5.2, as

• Additional Longitudinal Reinforcement Required for Torsion

• Look at page 77 and 78 in your textbook.
Torsion Design Procedure
•The design procedure for torsion is as follows:
Problems

L.L 50 psf L.L 100 psf

15 ft 15ftft
15
• The maximum uniform torsional loading will occur due to
partial loading on the hollow core slabs in which the full design
live load is assumed on the 30-ft-span hollow-core slab and
one-half of the design live load is assumed on the 18-ft-span
hollow core slab.
• This is common practice among some designers and will
generally result in a slightly more conservative design.
• This partial loading is similar to what is prescribed in Section
7.5 of the ASCE 7 Load Standard.
• The maximum torsion will be considered together with the
corresponding maximum uniform vertical load that occurs at
the same time.
Torsion can be neglected if the factored torsional moment
is less than or equal to the concrete torsional strength,
that is, if Tu ≤ 0.25ϕTcr, where the concrete torsional
strength is Because Tu = 22 ft.-kips > 8.78 ft.-kips,
this beam therefore, must be designed for torsion.
Using No. 4 stirrups, Avt (2 legs) = 2 (0.2 in.2) = 0.4 in.2,
the spacing of the stirrups required to resist the maximum
combined shear and torsion is calculated as
If the additional reinforcement is concentrated on the top and bottom
layers, therefore, the total areas of the top and bottom longitudinal • This additional longitudinal reinforcement
reinforcement in the beam are calculated as should be distributed at the corners of the
beam but the spacing between these bars
As,top = As,top(due to bending) + 0.5(1.75 in.2) should be no greater than 12 in.
As,bottom = As,bottom(due to bending) + 0.5(1.75 in.2) • Where the spacing exceeds 12 in., provide
additional longitudinal bars at the midwidth
However, for the beam in this problem, the spacing of the longitudinal or middepth of the beam as required.
reinforcement will exceed the maximum 12 in. because the center-to-
center distance between the top and bottom rebars is approximately 18 in.
The additional longitudinal reinforcement should thus be distributed as
follows:
As,top = As top(due to bending) + (1/3)(1.75 in.2)
= As,top(due to bending) + 0.58 in.2
As,midheight = (1/3 )(1.75 in.2 ) = 0.58 in.2
As,bottom = As,bottom(due to bending) + (1/3)(1.75 in.2)
= As,bottom(due to bending) + 0.58 in.2
Provide additional torsional
longitudinal rebar, in
addition to bending
reinforcement