Shear stresses in beams

Most beams carry shear force as well as bending moment. The bending moment produces the normal stresses of lesson 5; the shear force produces shear stresses, which are usually smaller but can govern short heavily loaded beams, timber beams (which are weak in shear along the grain), the webs of thin steel girders, and above all the connections that hold built-up members together. This lesson derives the shear stress formula, shows how shear stress is distributed over common sections, and uses the idea of shear flow to space nails, bolts and welds.

Why horizontal shear exists

Stack several loose boards on two supports and load them at midspan. Each board bends on its own, and their ends slide past one another, so the stack is flexible and weak. Glue the boards together and the stack behaves as a single deep beam, far stiffer and stronger, because the glue now prevents the sliding. The glue lines are carrying horizontal shear stress.

The same thing happens inside any solid beam. Where the bending moment changes along the span, the compressive force in the top part of the beam on one side of an element is larger than on the other side. Something must balance the difference, and that something is a shear force on the horizontal plane that separates the top part from the rest. By the equilibrium of a small element, a horizontal shear stress at a point is always accompanied by an equal vertical shear stress. So the vertical shear force \(V\) on a cross-section and the horizontal shear along the beam are two views of the same stresses.

The shear formula

Consider a short length \(dx\) of beam and the part of it above some level \(y_1\). The bending stresses on its two ends are \(-My/I\) and \(-(M + dM)y/I\). Their resultants differ by \((dM/I)\int y\,dA\), taken over the area above \(y_1\). That difference must be balanced by a horizontal shear stress \(\tau\) acting over the width \(t\) of the cut, on an area \(t\,dx\). Setting the forces equal and using \(dM/dx = V\) gives the shear formula:

Here \(V\) is the shear force at the section, \(I\) is the second moment of area of the whole section about the neutral axis, \(t\) is the width of the section at the level where the stress is wanted, and \(Q\) is the first moment of area of the part of the section beyond that level, taken about the neutral axis:

where \(A'\) is the area above (or below) the level and \(\bar{y}'\) is the distance from the neutral axis to its centroid. \(Q\) is zero at the top and bottom surfaces, where there is no area beyond, and it is largest at the neutral axis. So the shear stress is zero at the extreme fibers, exactly where the bending stress is largest, and usually peaks at the neutral axis, where the bending stress is zero.

The formula assumes the shear stress is uniform across the width \(t\). That is accurate for narrow rectangles and thin webs, and less accurate for very wide, shallow sections, where the true stress at the edges can be noticeably higher than the formula predicts. For design purposes it is the standard tool.

Distribution in a rectangular section

For a rectangle of width \(b\) and depth \(h\), the area above a level \(y\) is \(A' = b(h/2 - y)\), with centroid at \(\bar{y}' = (h/2 + y)/2\). So \(Q = (b/2)(h^2/4 - y^2)\), and

The distribution is a parabola, zero at the top and bottom and greatest at the neutral axis. Substituting \(I = bh^3/12\) at \(y = 0\):

The maximum is 50 % higher than the average shear stress \(V/A\). This result is used constantly in timber design, where horizontal shear (sometimes called rolling or longitudinal shear) often governs short, heavily loaded beams.

Distribution in I-sections and thin webs

In an I-shaped section the width changes abruptly at the junction of web and flange. Just inside the flange the width \(t\) is the full flange width, so \(\tau\) is small; just inside the web, \(t\) drops to the web thickness, so \(\tau\) jumps up by the ratio of the widths. Within the web the distribution is a shallow parabola with its peak at the neutral axis.

The web carries about 90 % or more of the vertical shear in a typical wide-flange beam, and the stress over it is nearly uniform. That is why steel design codes check shear on the web alone, using the approximation

where \(d\) is the overall depth and \(t_w\) the web thickness. In slender plate girders, the web may buckle in shear long before it yields, so deep girders are fitted with transverse stiffeners. Steel and bridge design courses cover these checks.

The flanges of an I-beam carry little vertical shear, but they do carry horizontal shear stresses flowing along their width, from the tips toward the web. Those horizontal flange stresses are what lead to the idea of the shear center, discussed at the end of this lesson.

Shear flow in built-up members

Built-up beams are made by joining pieces: boards nailed into a box beam, plates welded into a girder, a steel beam connected to a concrete slab by shear studs. The connectors must carry the horizontal shear that would otherwise make the pieces slide. The useful quantity is the shear flow \(q\), the shear force per unit length along the beam:

Here \(Q\) is the first moment of the part being held on (for example, one flange) about the neutral axis of the whole section. If connectors each able to carry a force \(F\) are spaced at a pitch \(s\) along the beam, they can carry a shear flow of \(F/s\) (or \(nF/s\) if there are \(n\) connectors in each row). The required spacing is therefore

Because \(q\) is proportional to \(V\), connectors can be spaced more widely near midspan, where shear is small, and must be closer near the supports. Welds are designed the same way: the shear flow at the web–flange junction divided by the strength of the weld per unit length gives the weld size or the length of intermittent welds.

Thin-walled open sections and the shear center

In thin-walled open sections, shear flow runs along the walls. In a channel loaded through its web, the horizontal shear flows in the two flanges point in opposite directions and form a couple. That couple twists the channel unless the load is applied at a particular point outside the web, the shear center. Only loads through the shear center bend a beam without twisting it.

For sections with two axes of symmetry, such as I-beams and rectangular tubes, the shear center coincides with the centroid, so no special care is needed. For channels, angles and Z-sections it does not. That is why channels used as lintels or purlins are often installed in back-to-back pairs or braced, and why a load hung from the toe of an angle produces twisting. Light-gauge steel purlins and girts, which are usually C- or Z-shaped, are designed with the shear center in mind.

Shear stress through the depth by code

The program below evaluates \(VQ/(It)\) at many levels through a symmetric I-section, showing the jump at the flange–web junction and the nearly uniform stress in the web. It also compares the peak with the simple \(V/(d\,t_w)\) approximation.

# Shear stress through the depth of a symmetric I-section (mm, N)
bf, tf, d, tw = 200.0, 15.0, 400.0, 9.0
V = 250e3
hw = d - 2 * tf
I = bf * d**3 / 12 - (bf - tw) * hw**3 / 12

def Q_above(y):                     # first moment of the area above level y (y from neutral axis)
    if y >= d / 2 - tf:             # inside the top flange
        return bf * (d / 2 - y) * (d / 2 + y) / 2
    Qf = bf * tf * (d / 2 - tf / 2)
    return Qf + tw * (d / 2 - tf - y) * (d / 2 - tf + y) / 2

def width(y):
    return bf if abs(y) > d / 2 - tf else tw

print(f"I = {I/1e6:.1f} x10^6 mm4")
for y in [200, 190, 185.01, 184.99, 150, 100, 50, 0]:
    tau = V * Q_above(y) / (I * width(y))
    where = "flange" if width(y) == bf else "web"
    print(f"y = {y:7.2f} mm ({where:6s})  tau = {tau:6.1f} MPa")
print(f"Approximation V/(d*tw) = {V / (d * tw):.1f} MPa")
web_force = sum(V * Q_above(y) / I * 0.5 for y in [i * 0.5 - (hw / 2) + 0.25 for i in range(int(hw / 0.5))])
print(f"Share of V carried by the web = {web_force / V:.1%}")

I = 260.4 x10^6 mm4
y =  200.00 mm (flange)  tau =    0.0 MPa
y =  190.00 mm (flange)  tau =    1.9 MPa
y =  185.01 mm (flange)  tau =    2.8 MPa
y =  184.99 mm (web   )  tau =   61.6 MPa
y =  150.00 mm (web   )  tau =   67.2 MPa
y =  100.00 mm (web   )  tau =   73.2 MPa
y =   50.00 mm (web   )  tau =   76.8 MPa
y =    0.00 mm (web   )  tau =   78.0 MPa
Approximation V/(d*tw) = 69.4 MPa
Share of V carried by the web = 96.6%

The stress jumps from a few megapascals in the flange to over 60 MPa in the web at the junction, and the web carries the vast majority of the shear. The simple approximation is about 11 % below the true peak at the neutral axis, but it is close to the average stress over the web, which carries about 97 % of the shear here. Design codes calibrate their shear resistances to the simple formula, so it is the one to use in a code check.