Shear design of beams

Flexural failures are gradual: the steel yields, the beam sags and cracks widen long before collapse. Shear failures are different. A beam without enough shear reinforcement can split along a diagonal crack with almost no warning, and several well-known collapses of warehouses and parking structures were shear failures of this kind. Design codes therefore treat shear conservatively, and the goal of shear design is to make sure that a beam always fails in flexure first. This lesson explains how diagonal cracks form, how stirrups work, and how the codes compute shear resistance, then designs the stirrups for two beams of the operations center.

How diagonal cracks form

Near the supports of a beam, shear and moment act together. In the uncracked web the principal tensile stress is inclined, at about 45 degrees where shear dominates, and when it exceeds the tensile strength of concrete an inclined crack forms. Two types are recognized. Web-shear cracks start in the web near the neutral axis of a beam with a thin web and high shear. Flexure-shear cracks, much more common in ordinary beams, begin as vertical flexural cracks and turn diagonal as they climb toward the compression zone.

Once an inclined crack has formed, shear is carried by several mechanisms: shear in the uncracked compression zone, aggregate interlock across the rough crack faces, dowel action of the longitudinal bars, and, if present, tension in stirrups crossing the crack. Without stirrups, the crack can widen and extend until the compression zone is crushed or the longitudinal bars split away from the concrete. With stirrups, the crack is held closed, aggregate interlock is maintained, and the beam behaves like a truss.

The truss model

Ritter and Mörsch proposed more than a century ago that a cracked beam with stirrups acts as a truss. The compression zone is the top chord, the longitudinal steel is the bottom chord, the concrete between the inclined cracks forms diagonal struts, and the stirrups are vertical ties. If the struts are inclined at angle \(\theta\) to the axis, a length of beam \(d_v\cot\theta\) is crossed by the crack, and every stirrup in that length carries the force \(A_v f_y\). The shear carried by the stirrups is therefore

where \(A_v\) is the area of all the legs of one stirrup and \(s\) is the stirrup spacing. A flatter strut angle means more stirrups are crossed and more shear is carried, but it also means more tension in the longitudinal steel and more compression in the struts. The truss model also explains why the longitudinal bars need extra force near supports and why the web concrete can crush if the shear is too high.

ACI 318 uses the truss with struts at 45 degrees and adds a concrete contribution \(V_c\) equal, roughly, to the shear at which diagonal cracking occurs. The two terms are simply added. This method has a long record of satisfactory service and is easy to apply.

Shear resistance in the code

The design shear strength in ACI 318 is the sum of a concrete term and a steel term, multiplied by \(\phi = 0.75\), with a cap on \(V_s\) that prevents crushing of the web:

Here \(\sqrt{f'_c}\) is limited to 100 psi unless enough stirrups are provided, \(f_{yt}\) is the stirrup yield strength, limited to 60,000 psi for shear, and \(\lambda\) is the lightweight factor. The \(V_c\) expression above applies when the beam has at least the minimum stirrups; ACI 318-19 also allows an alternative that grows with the longitudinal steel ratio:

When the beam has less than the minimum stirrups, ACI 318-19 requires the second form multiplied by a size effect factor \(\lambda_s = \sqrt{2/(1 + d/10)} \le 1\). This change, prompted by tests on deep members without stirrups, means that thick slabs, footings and transfer beams without stirrups have lower shear strength than older editions gave.

The difference between the two codes is philosophical as much as numerical. ACI adds a constant 45 degree truss to a cracking-load concrete term, while the Canadian code lets the strut angle and the concrete term vary with the strain in the member, which captures the effect of moment and axial load on shear strength more directly. For ordinary beams both give similar stirrup designs.

Most US designers use the simple \(2\sqrt{f'_c}\) form for beams with stirrups. The more detailed expressions of older editions, which included the effect of moment and axial load, have been removed or simplified, but the axial load term \(N_u/6A_g\) is still added when there is axial compression.

Where to design for shear and minimum stirrups

Near a support that introduces compression into the beam, a diagonal crack that would cross the support cannot form, so the critical section is taken at \(d\) from the face of the support, and stirrups designed for that shear are continued to the support, provided no concentrated load acts within that distance.

Minimum stirrups are required wherever \(V_u\) exceeds \(0.5\phi V_c\) (with exceptions for shallow members, slabs and joists). They stop a diagonal crack from opening suddenly once it forms. The minimum area and maximum spacing are

The spacing limit makes sure that every potential diagonal crack is crossed by at least one or two stirrups. Stirrups must also be anchored at both ends, with hooks around longitudinal bars, because a stirrup is useless if it pulls out before reaching yield.

Sofia's program evaluates the shear at several sections and reports the spacing at each, so Owen can choose the steps.

import math
# Shear design to ACI 318 with vertical #3 U-stirrups (two legs), Grade 60
phi, fyt = 0.75, 60000.0
Av = 0.22

def design(name, fc, bw, d, As, wu, P, a, L, xs):
    """Simple span, uniform load wu (kip/ft) plus point load P (kips) at a (ft); xs in ft."""
    rt = min(math.sqrt(fc), 100.0)
    rho_w = As / (bw * d)
    # With at least minimum stirrups: larger of the two Vc expressions (lambda = 1.0)
    Vc = max(2 * rt * bw * d, 8 * rho_w ** (1 / 3) * rt * bw * d) / 1000
    Vc = min(Vc, 5 * rt * bw * d / 1000)
    Vs_cap = 8 * rt * bw * d / 1000
    Av_min_s = max(0.75 * rt, 50) * bw / fyt            # in2 per in.
    RL = wu * L / 2 + P * (L - a) / L
    print(f"{name}: phiVc = {phi*Vc:.1f} kips, Vs limit = {Vs_cap:.0f} kips, RL = {RL:.1f} kips")
    for x in [d / 12] + xs:
        V = abs(RL - wu * x - (P if x > a else 0))
        Vs = max(V / phi - Vc, 0.0)
        s = Av * fyt * d / (Vs * 1000) if Vs > 0 else float("inf")
        smax = min(d / 2, 24) if Vs <= 4 * rt * bw * d / 1000 else min(d / 4, 12)
        need = V > 0.5 * phi * Vc
        s_use = min(s, smax, Av / Av_min_s) if need else float("inf")
        txt = f"{s_use:4.1f} in." if need else "none needed"
        print(f"   x = {x:5.2f} ft  Vu = {V:5.1f}  Vs req = {Vs:5.1f} kips  s = {txt}")

design("office floor beam", 4000, 14, 21.5, 3.16, 2.96, 0.0, 0.0, 25, [5.0, 8.0, 10.0])
design("maintenance bay girder", 5000, 16, 25.5, 3.16, 3.03, 56.0, 5.0, 30, [4.99, 5.01, 12.0])

office floor beam: phiVc = 28.6 kips, Vs limit = 152 kips, RL = 37.0 kips
   x =  1.79 ft  Vu =  31.7  Vs req =   4.2 kips  s = 10.8 in.
   x =  5.00 ft  Vu =  22.2  Vs req =   0.0 kips  s = 10.8 in.
   x =  8.00 ft  Vu =  13.3  Vs req =   0.0 kips  s = none needed
   x = 10.00 ft  Vu =   7.4  Vs req =   0.0 kips  s = none needed
maintenance bay girder: phiVc = 43.3 kips, Vs limit = 231 kips, RL = 92.1 kips
   x =  2.12 ft  Vu =  85.7  Vs req =  56.5 kips  s =  6.0 in.
   x =  4.99 ft  Vu =  77.0  Vs req =  45.0 kips  s =  7.5 in.
   x =  5.01 ft  Vu =  20.9  Vs req =   0.0 kips  s = none needed
   x = 12.00 ft  Vu =   0.2  Vs req =   0.0 kips  s = none needed

The output also shows the effect of the code's alternative \(V_c\) expression: for these beams, with longitudinal steel ratios below about 1.5 percent, the \(8(\rho_w)^{1/3}\) form is smaller than \(2\sqrt{f'_c}\), so the program keeps the familiar value. For heavily reinforced beams the alternative can give more.

Longitudinal steel and other details

The truss model shows that diagonal compression pushes the bottom chord apart near the supports, so the longitudinal tension at a section is larger than \(M_u/d\) alone. ACI 318 covers this through its detailing rules: positive moment bars must extend into simple supports, and bars may not be cut off in a tension zone unless extra stirrups or reduced shear make it safe. Lesson 7 returns to this when it checks bar anchorage.

• Stirrups must be closed or hooked around longitudinal bars in their corners, and extend as close to the compression and tension faces as cover allows.
• Beams that support other beams near their bottom need hanger reinforcement to lift the load of the supported beam to the top of the supporting beam.
• Torsion, as in spandrel beams that support slabs on one side, requires closed stirrups and additional longitudinal bars; it is covered in a later course.
• In seismic regions, beam ends need closely spaced hoops for confinement and shear in plastic hinges.