Structural Design 1
(Reinforced Concrete)
Chapter 5:
Shear and Diagonal Tension
of Beams
1.
Introduction on Shear Failure/Diagonal
Tension Failure
Introduction on Shear Failure/Diagonal
Tension Failure
Beams must also have an adequate safety margin against other types of
failure, some of which may be more dangerous than flexural failure. This may
be so because of the catastrophic nature of some other types of failure, should
they occur.
Shear Failure of reinforced concrete, more properly called Diagonal
Tension Failure, is one example. Shear failure is difficult to predict
accurately. In spite of many decades of experimental research, and the use of
highly sophisticated analytical tools, it is not yet fully understood.
Furthermore, if a beam without properly designed shear reinforcement is
overloaded to failure; shear collapse is likely to occur suddenly, with no
advance warning of distress. This is in strong contrast with the nature of
flexural failure.
Introduction on Shear Failure/Diagonal
Tension Failure
For typically under-reinforced beams, flexural failure is initiated by gradual
yielding of the tension steel, accompanied by obvious cracking of the concrete
and large deflections, giving ample warning and providing the opportunity to
take corrective measures. Because of these differences in behavior, reinforced
concrete beams are generally provided with special shear reinforcement
(Stirrups) to ensure that flexural failure would occur before shear failure if the
member should be severely overloaded.
The shear stresses in most beams are far below the direct shear strength of
the concrete. The real concern is with diagonal tension stress, resulting from
the combination of shear stress and longitudinal flexural stress.
2.
Types of Cracks
Types of Cracks
Cracks produced by
shear/diagonal tension
failure
Types of Cracks
a. Flexural cracks are vertical cracks that extend from the tension sides of
beams up to the region of their neutral axes. Should beams have very deep
webs, the cracks will be very closely spaced, with some of them coming
together above the reinforcing and some disappearing there. These cracks
may be wider up in the middle of the beam than at the bottom.
b. Inclined cracks due to shear can develop in the webs of reinforced concrete
beams either as independent cracks or as extensions of flexural cracks.
Occasionally, inclined cracks will develop independently in a beam, even
though no flexural cracks are in that locality. These cracks are called webshear cracks.
Types of Cracks
c. The usual type of inclined shear cracks are the flexure-shear cracks. They
commonly develop in both prestressed and nonprestressed beams.
d. Torsion cracks are quite similar to shear cracks except that they spiral
around the beam. Should a plain concrete member be subjected to pure
torsion, it will crack and fail along 45° spiral lines due to the diagonal
tension corresponding to the torsional stresses. Although torsion stresses
are very similar to shear stresses, they will occur on all faces of a member.
As a result, they add to the shear stresses on one side and subtract from
them on the other.
Types of Cracks
e. Bond cracks are due to bond stresses between the concrete and the
reinforcement which will lead to a splitting along the bars
f.
Cracks can also occur in concrete members due to shrinkage, temperature
change, settlements, and so on.
3.
Shear Stresses in Concrete Beams
Shear Stresses in Concrete Beams
The average shearing stress, vave in concrete
is calculated from:
(applicable for
rectangular sections.)
where:
vave = average shearing stress (kPa or MPa);
Vu = factored shear force on the critical section
at a distance “d” from the support of the
beam;
bw = width of the web;
d = effective depth.
Shear Stresses in Concrete Beams
This stress is not equal to the diagonal
tension stress.
It merely serves as an
indicator of the magnitude.
Should this
indicator exceeds a certain value, shear or
web reinforcement (such as stirrups) will be
necessary.
The NSCP had presented basic shear
equations that are in terms of shear forces
and not shear stresses. In other words, the
effective beam areas to obtain shear forces
multiply the average shear stresses described
in this section.
4.
Shear Strength for
Non-Prestressed Members
Shear Strength for Non-Prestressed Members
The nominal or theoretical shear strength of a member Vn is provided by the
concrete and the steel reinforcement in accordance to NSCP 422.5.1.1.
Vn = Vc + Vs
The permissible shear strength Vu of a member is
Vu = fVc + fVs
Where:
Vc = nominal shear strength provided by concrete in accordance with NSCP
422.5.5, 422.5.6, and 422.5.7;
Vs = nominal shear strength provided by shear reinforcement in accordance
with Section 422.5.10.
5.
Shear Strength
provided by Concrete for NonPrestressed Members, Vc
Shear Strength provided by Concrete for
Non-Prestressed Members, Vc
Based on NSCP 422.5.3.1, the value of !"# used to calculate Vc, Vci, and Vcw for
one-way shear shall not exceed 8.3MPa, unless allowed in Section 422.5.3.2.
NSCP 422.5.5 - Vc for Non-Prestressed Members without Axial Force
1. Simplified Equation (NSCP 422.5.5.1)
!" = 0.17()*′" ,- .
This equation neglects the effect of moment Mu occurring simultaneously
with Vu at the section considered.
Shear Strength provided by Concrete for
Non-Prestressed Members, Vc
2. Detailed / Rational
Equation (NSCP 422.5.5.1)
This value must be
calculated separately for
each
point
being
considered in the beam. The
purpose of his requirement
is to limit Vc near the point
of inflection.
Shear Strength provided by Concrete for
Non-Prestressed Members, Vc
2. Detailed / Rational Equation
(NSCP 422.5.5.1)
Where:
Ø Vud/Mu should not be greater
than 1
Ø λ = modification factor for
lightweight concrete (NSCP
419.2.4)
Ø rw = As/bwd
Ø Mu = moment occurring at the
point where Vu is critical
Shear Strength provided by Concrete for
Non-Prestressed Members, Vc
NSCP 422.5.6 - Vc for Non-Prestressed Members with Axial Compression
1. Simplified Equation (NSCP 422.5.6.1)
*+
!" = 0.17 (1 +
/ 012′" 45 6
14-.
This equation neglects the effect of moment Mu occurring simultaneously
with Vu at the section considered.
Where:
- Nu/Ag shall be expressed in MPa
- Nu = factored axial load normal to cross section occurring simultaneously
with Vu. It is positive for compression.
- Ag = gross area of section.
Shear Strength provided by Concrete for
Non-Prestressed Members, Vc
2. Detailed / Rational
Equation (NSCP 422.5.6.1)
This equation takes into
account the effects of the
longitudinal
reinforcing
bars, moment Mu and
shear Vu magnitudes.
Shear Strength provided by Concrete for
Non-Prestressed Members, Vc
NSCP 422.5.7 - Vc for Non-Prestressed Members with Significant Axial
Tension
1. Detailed / Rational Equation (NSCP 422.5.7.1)
*+
0 123′" 56 7
!" = 0.17 (1 +
3.5./
Where:
- Nu is negative in tension
- Nu/Ag shall be expressed in MPa
- Vc shall not be less than zero
6.
Shear Strength
provided by Shear Reinforcement, Vs
6a.
Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
When the factored shear Vu is high, it shows that serious cracks will develop
unless some type of additional reinforcing is provided. This reinforcement is
called as web reinforcement and takes the form of stirrups that enclose the
longitudinal reinforcing faces of the beam. NSCP 422.5.10.5.1 describes them as
follows:
1. Stirrups, ties, or hoops perpendicular to longitudinal axis of member;
2. Axial welded wire reinforcement with wires located perpendicular to
longitudinal axis of member;
3. Spiral reinforcement.
Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
For non-prestressed members, shear reinforcement shall be permitted to
also consist of in accordance to NSCP 422.5.10.5.2:
1.
Stirrups making an angle of 45 degrees or more with longitudinal tension
reinforcement;
2. Longitudinal reinforcement with bent portion making an angle of 30
degrees or more with the longitudinal tension reinforcement;
3. Combination of stirrups and bent longitudinal reinforcement.
Web Reinforcement
Types of Stirrups
Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
Bars called hangers or stirrup supports are placed on the compression
side of the beam. The stirrups are passed around the tensile steel and to meet
anchorage requirements are run as far into the compression side of the beam
as practical and are hooked around the hangers. Bending the stirrups around
the hangers reduces bearing stresses under the hooks. If these bearing
stresses are too high, the concrete will crush and the stirrups will tear out.
When a significant amount of torsion is present in a member, it will be
necessary to use closed stirrups.
Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
NSCP 420.2.2.4. Types of non-prestressed bars and wires to be
specified for particular structural applications shall be in accordance
with Table 420.2.2.4(a) for deformed reinforcement and Table
420.2.2.4(b) for plain reinforcement.
Web Reinforcement
6b.
Minimum Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
NSCP 409.6.3.1 states that the
minimum
area
of
shear
reinforcement, Av(min), shall be
provided in all regions where Vu
exceeds 0.5ϕVc except for the cases
in Table 409.6.3.1, where at least
Av(min) shall be provided where Vu >
ϕVc·
Minimum Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
When shear reinforcing is required, the amount provided must fall between
certain specified lower and upper limits. If the amount of reinforcing is too
low, it may yield or even snap immediately after the formation of an inclined
crack. As soon as diagonal cracks develop, the tension previously carried by
the concrete is transferred to the web reinforcement. To prevent the stirrups
from snapping at that time, their area is limited to the minimum value. The
least amount of web reinforcement for any section of the beam is
Minimum Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
NSCP
409.6.3.3.
If
shear
reinforcement is required and
torsional effects can be neglected
according to Section 409.5.4.1,
Av(min) shall be in accordance with
Table 409.6.3.3.
Minimum Web Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
Bent-up Bar Web Reinforcing
Minimum Web Reinforcement
6c.
Spacing Limits for Shear Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
The maximum shear Vu must not exceed the permissible shear capacity of
the section fVn
Vu ≤ fVn
The purpose of stirrups is to minimize the size of the diagonal tension
cracks or to carry the diagonal tension stress from one side on the crack to the
other. Very little tension is carried by the stirrups until after a crack begins to
form. Before the inclined cracks begin to form, the strain in the stirrups is
equal to the strain in the adjacent concrete. Because this concrete cracks at
very low diagonal tensile stresses, the stresses in the stirrups at that time are
very small. Stirrups do not prevent inclined cracks until the cracks develop.
Spacing Limits for Shear Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
The nominal shear strength of the stirrups or shear reinforcement Vs
crossing the crack can be calculated from the following expression
Vs = Avfyn
where
- Vs = nominal shear strength provided by the shear reinforcement;
- Av = cross sectional area of the stirrup bar;
- n = number of stirrups crossing the crack.
If a U stirrup is used, Av = 2 times the cross sectional area of the stirrup bar.
If a UU stirrup is used, Av = 4 times the cross sectional area of the stirrup bar.
Spacing Limits for Shear Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
If it is conservatively assumed that the horizontal projection of the crack
equals the effective depth d of the section (thus a 45° crack), the number of
stirrups crossing the crack can be determined from the following:
n = d/s
where
- s = center-to-center spacing of the stirrups.
- d = effective depth.
From the given equations:
(
!" = $% &'
"
#$ %& '
!=
(!
(NSCP 422.5.10.5.3)
Spacing Limits for Shear Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
Going through a similar derivation, the following can be determined for the
required area for inclined stirrups (NSCP 422.5.10.5.4):
#$ %& '
(!*+, + ./!,)
!=
(!
Where:
- a = angle between the stirrups and the longitudinal axis. It shall not be less
than 30° (NSCP 422.5.10.6.1).
Spacing Limits for Shear Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
NSCP 422.5.10.6.2 If shear reinforcement consists of a single bar or a single
group of parallel bars having an area Av, all bent the same distance from the
support, Vs shall be the lesser of (a) and (b):
where a is the angle between bent-up reinforcement and longitudinal axis of
the member.
Spacing Limits for Shear Reinforcement
Shear Strength provided by Shear
Reinforcement, Vs
Stirrups
cannot
resist
appreciable shear unless they are
crossed at an inclined crack. In
order to make sure that each 45°
crack is intercepted by at least one
stirrup, the maximum spacing of
vertical stirrups according to NSCP
409.7.6.2.2.
But the desirable
stirrups is 100 mm.
spacing
for
Spacing Limits for Shear Reinforcement
7.
Design of Reinforced Concrete
Beam for Shear
Design of Reinforced Concrete
Beam for Shear
The following steps summarize the design of vertical shear reinforcement
(Stirrups):
1.
Determine the design shear Vu that the member can sustain;
2.
Draw the Vu diagram;
3.
Calculate Vu at a distance d from the support;
4.
Calculate fVc either considering the effects of Mu or not, depending on the
given;
5.
Determine the need for stirrups
If Vu ≥ 0.5fVc then provide stirrups
Design of Reinforced Concrete
Beam for Shear
The following steps summarize the design of vertical shear reinforcement
(Stirrups):
6. Calculate the maximum theoretical spacing of the stirrups (fy ≤ 415 MPa)
#$ %& '
!=
(!
Design of Reinforced Concrete
Beam for Shear
The following steps summarize the design of vertical shear reinforcement
(Stirrups):
7. Determine the maximum spacing to provide minimum area of shear
reinforcement.
8.
Compute maximum spacing.
9.
For convenience, calculate the maximum theoretical spacing at 1-meter
points along the span. Normally, the first stirrup is placed at a distance of
d or s/2 from the face of the support. In practice, the practical minimum
spacing is 75-100 mm.
8.
Design Considerations
Design Considerations
In practice 10-12mm f
stirrups are used. However,
if the calculated design
spacing are less than ¼ d,
larger diameter stirrups
can be used.
Another alternative is to
use UU stirrups instead of U
stirrups. Different diameter
stirrups should not be used
in the same beam so as not
to produce confusion.
Design Considerations
It is convenient to draw the Vudiagram and carefully label it with
values of items such as fVc, ½ fVc, Vu,
and Vu at a distance d from the
support and to show the dimensions
involved.
The first stirrup may be placed
Ø at a distance d from the face of the
support.
Ø at ½ s from the face of the support.
Ø at 75 mm to 100 mm from the
support.
Examples
(Shear and Diagonal Tension of
Beams)
Examples – Shear and Diagonal Tension of Beams
1. A rectangular reinforced concrete beam has the dimensions and
reinforcement shown. If f’c = 20.70 MPa and fy = 415 MPa, what should be
the ultimate shear capacity, Vu of the beam? Use the rational method of
computing Vc. with Mu = 225 kNm.
Examples – Shear and Diagonal Tension of Beams
2. A simply supported beam has a span of 6m and carries a uniformly
distributed service dead load of 21.2 kN/m and a service live load of 51.10
kN/m. The weight of the beam is included in the service dead load. The beam
is reinforced with a constant area of flexural steel equal to 3910 mm2. f‘c =
17.24 MPa, fy = 344.70 MPa, Width of beam is 400 mm with an effective depth
of 560 mm.
a. Compute the nominal shear strength of the concrete at a distance “d”
from the support of the beam;
b. Compute the maximum spacing of the 10mm stirrups and show the
actual positions of the vertical stirrups; and
c. Determine the location where stirrups are no longer required on either
side of the midspan section.
Examples – Shear and Diagonal Tension of Beams
3. A simply supported rectangular beam 400 mm wide having an effective
depth of 550 mm carries a total factored load of 116 kN/m on a 6 m clear
span. It is reinforced with 6360 mm2 of tensile steel, which continues
uninterrupted into the supports. f‘c = 20.73 MPa
a. Compute the design shear force which results from the application of
the factored loads;
b. Compute the shear capacity provided by concrete; and
c. At what point is the web reinforcement no longer required measured
from the support.
Examples – Shear and Diagonal Tension of Beams
4. A rectangular beam has a width of 250 mm and an effective depth of
437.50 mm. It is reinforced with a tensile reinforcement having an area of
1875 mm2 placed at 62.5 mm above the bottom of the beam. The beam is
subjected to a factored shear force Vu = 178 kN and moment Mu = 40 kN.m
occurring at the point where Vu is critical. f‘c = 24.82 MPa.
a. Compute the nominal shear strength provided by concrete (rational
method);
b. Compute the nominal shear strength provided by concrete with a tensile
force of 44 kN (rational method); and
c. Compute the nominal shear strength provided by concrete with an axial
compression force of 44 kN.
Any questions?