The normal stress of the previous chapter came from cutting a bar perpendicular to the load, so that the internal force pointed straight out of the cut surface. Nothing forces us to cut that way. If we instead cut parallel to the load, the internal force lies in the plane of the cut and tries to slide one face of the cut past the other. The internal force per unit area is then called the shear stress, and it is what fails a bolt, a rivet, a weld or a glued lap joint.
Figure 8.4.1: Two plates joined by a pin, with the load trying to slide one face past the other.
Denoting the internal force tangential to the cut by \(T\) and the area of the cut by \(A\), the average shear stress is
\[
\boxed{\tau = \frac{T}{A}}
\tag{8.4.1}\]
which has the same form and the same units as 8.3.2. The difference is entirely in the direction: \(\sigma\) acts along the surface normal and \(\tau\) acts in the surface. Since the two are distinguished only by the orientation of the cut, we should already suspect that they are two faces of one object rather than two separate quantities, and Example 2 below makes that suspicion concrete.
Two arrangements occur so often that they have names. In single shear the joint presents one cut surface, so the pin carries the whole load across a single section. In double shear the joint presents two, and each carries half the load, which halves the stress for the same pin.
Figure 8.4.2: Single shear
Figure 8.4.3: Double shear
Figure 8.4.4: The shear plane
Counting shear planes correctly is the most common source of error in joint design, and the free body diagram of the pin settles it every time.
Example 1: A pin in double shear
A cylindrical pin of cross-sectional area \(A\) joins a clevis to a tongue and carries a force \(P\), as shown in Figure 8.4.5. Determine the shear stress in the pin.
Figure 8.4.5: A pin in a clevis joint.
We isolate the pin. The tongue pushes on it with \(P\) in the middle, and the two arms of the clevis push back with \(T\) each, so equilibrium reads \(P - 2T = 0\).
The two shear planes halve the stress compared to a single-shear joint carrying the same load, which is exactly why clevis joints are preferred wherever the geometry allows them.
Example 2: Stress on an inclined cut
A prismatic bar of cross-sectional area \(A\) is pulled by an axial force \(P\). Cut it at an angle \(\varphi\) to the bar axis, as in Figure 8.4.7, and determine the normal and shear stress on that cut. At which angle is the shear stress largest?
Figure 8.4.7: A bar cut at an angle to its axis.
The inclined cut has area \(A/\sin\varphi\), which grows without bound as the cut approaches the axis. Resolving the internal force into a component \(N\) normal to the cut and a component \(T\) tangential to it, equilibrium of the free body in the two coordinate directions gives two equations, and the definitions of the two stresses give two more.
Figure 8.4.8: Free body diagram of the inclined cut.
N, T, varphi, P, A, sigma_varphi, tau_varphi = sp.symbols('N T varphi P A sigma_varphi tau_varphi', real=True)eqs = [sp.Eq(N*sp.sin(varphi) + T*sp.cos(varphi) - P, 0), # equilibrium along the load sp.Eq(-N*sp.cos(varphi) + T*sp.sin(varphi), 0), # equilibrium across the load sp.Eq(sigma_varphi, N/(A/sp.sin(varphi))), sp.Eq(tau_varphi, T/(A/sp.sin(varphi)))]sol = sp.solve(eqs, [N, T, sigma_varphi, tau_varphi], dict=True)[0]sol = {k: sp.simplify(v) for k, v in sol.items()}sol | la
\[ \begin{aligned}
N &= P \sin{\left(\varphi \right)} \\[9.0pt]
T &= P \cos{\left(\varphi \right)} \\[9.0pt]
\sigma_{\varphi} &= \dfrac{P \sin^{2}{\left(\varphi \right)}}{A} \\[9.0pt]
\tau_{\varphi} &= \dfrac{P \sin{\left(2 \varphi \right)}}{2 A}
\end{aligned} \]
The two stresses on the inclined cut are therefore
and the double angle in the second expression tells us immediately where the maximum lies. Plotting both over the full range of cut angles makes the behaviour clear.
Figure 8.4.9: Normal and shear stress on a cut inclined at \(\varphi\) to the axis of a bar in uniaxial tension.
The shear stress peaks at \(\varphi = \pi/4\), that is at \(45^\circ\), where it reaches \(P/2A\), half the maximum normal stress. The normal stress meanwhile grows monotonically and reaches its own maximum \(P/A\) on the perpendicular cut at \(\varphi = \pi/2\), where the shear stress has fallen to zero.
This deserves more attention than a worked example usually gets. We applied a single axial load and made no change whatsoever to the bar or to the loading, yet by choosing a different cut we obtained a different pair of stresses. A cut at \(90^\circ\) sees pure tension, a cut at \(45^\circ\) sees the largest shear the bar will ever experience, and every angle in between sees a mixture. The stress at a point is therefore not one number but a rule that assigns a traction to every possible cut direction, and the mathematical object encoding such a rule is a tensor. We construct it in Chapter 8.7, where 8.4.2 reappears as a special case of the transformation of the stress tensor under rotation.
The physical consequence is visible in any tensile test of a ductile metal. The specimen is pulled axially but yields along planes inclined at roughly \(45^\circ\), because plastic slip in metals is driven by shear rather than by tension. A brittle material such as chalk instead fails on the plane of maximum normal stress, straight across.
Example 3: Dimensioning two pins
The rigid beam in Figure 8.4.10 is pinned at \(A\) and supported by a link at \(B\) inclined so that its direction has components \((3, 4)/5\). A load of \(30~\text{kN}\) acts \(2~\text{m}\) from \(A\), and the link attaches \(6~\text{m}\) from \(A\). The maximum allowable shear stress in the pins is \(180~\text{MPa}\) and the factor of safety is \(1.5\). The pin at \(A\) is in double shear and the pin at \(B\) is in single shear. Determine the required pin diameters.
Figure 8.4.10: The beam and its supports.
The free body diagram of the beam gives three equilibrium equations for the three unknown reaction components. Taking moments about \(A\) isolates the link force \(F_B\) immediately, after which the two force equations give the components at \(A\).
The pin at \(A\) carries the resultant \(F_A = \lvert (A_x, A_y) \rvert\), and since it sits in double shear the force splits between two sections, \(V_A = F_A/2\). The pin at \(B\) carries the full link force across a single section, \(V_B = F_B\).
Figure 8.4.12: Pin at A, double shear
Figure 8.4.13: Pin at B, single shear
With the allowable shear stress \(\tau_{\text{allow}} = 180/1.5 = 120~\text{MPa}\), each diameter follows from setting the actual shear stress equal to the allowable one.
Rounding upward, the pin at \(A\) needs at least \(11~\text{mm}\) and the pin at \(B\) at least \(12~\text{mm}\). In practice we would fit \(12~\text{mm}\) pins in both positions. The slightly heavier pin at \(A\) costs almost nothing, whereas two different pin sizes in one assembly cost a second tool, a second stock item and a chance to assemble the joint wrongly.
Note that the pin at \(B\) ends up larger despite carrying a smaller force, \(12.5~\text{kN}\) against \(21.4~\text{kN}\). The double shear at \(A\) is what makes the difference, and it illustrates that the load path through a joint matters as much as the magnitude of the load itself.
