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 8.4.3. Determine the shear stress in the pin.

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 \(P - 2T = 0\).

Combining equilibrium with 8.4.1 gives

\[ \left\lbrace \begin{array}{l} P - 2T = 0\\[4pt] \tau = \dfrac{T}{A} \end{array}\right. \quad\Rightarrow\quad T = \frac{P}{2}, \qquad \tau = \frac{P}{2A} \]

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 8.4.5 (a), and determine the normal and shear stress on that cut. At which angle is the shear stress largest?

The inclined cut has area \(A/\sin\varphi\), which grows without bound as the cut approaches the axis. We take the lower part in Figure 8.4.5 (b), with \(x\) across the bar and \(y\) along it, toward the load. The cut makes the angle \(\varphi\) with the axis, so its outward normal and its tangent are

\[ \bm n = [-\cos\varphi, \sin\varphi]^\mathsf{T}, \qquad \bm t = [\sin\varphi, \cos\varphi]^\mathsf{T} \]

and the internal force on the cut is \(N\bm n + T\bm t\), with \(N\) normal to the cut and \(T\) tangential to it. Equilibrium of the lower part in the two coordinate directions, and the definitions of the two stresses on the cut of area \(A/\sin\varphi\), give four equations for \(N\), \(T\), \(\sigma_\varphi\) and \(\tau_\varphi\),

\[ \begin{aligned} -N\cos\varphi + T\sin\varphi &= 0 \\ N\sin\varphi + T\cos\varphi - P &= 0 \\ \sigma_\varphi &= \frac{N}{A/\sin\varphi} \\ \tau_\varphi &= \frac{T}{A/\sin\varphi} \end{aligned} \]

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

The two stresses on the inclined cut are therefore

\[ \sigma_\varphi = \frac{P\sin^2\varphi}{A}, \qquad \tau_\varphi = \frac{P\sin\varphi\cos\varphi}{A} = \frac{P\sin 2\varphi}{2A} \tag{8.4.2}\]

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.

phi_deg = sp.symbols('phi_deg', real=True)
in_degrees = {varphi: sp.pi*phi_deg/180}          # plot against the angle in degrees

fig, ax = plt.subplots(figsize=(6, 4))
# each force in the colour of the stress it gives, stresses solid and forces dashed
for expr, style, col, lbl in ((sol[sigma_varphi]/P*A, '-', 'C0', r"$\sigma_\varphi A / P$"),
                              (sol[tau_varphi]/P*A, '-', 'C1', r"$\tau_\varphi A / P$"),
                              (sol[N]/P, '--', 'C0', r"$N / P$"),
                              (sol[T]/P, '--', 'C1', r"$T / P$")):
    fplot(sp.simplify(expr).subs(in_degrees), (phi_deg, 0, 90), ax=ax,
          linewidth=2, linestyle=style, color=col, label=lbl)
ax.set_xlabel(r"$\varphi$ [°]")
ax.set_xticks(range(0, 91, 15))
ax.set_ylabel("[-]")
ax.grid(True)
ax.legend()
plt.show()

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.

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 8.4.7 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.

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 \(A\).

With the load \(F_G = 30~\text{kN}\), \(x\) to the right, \(y\) up and counterclockwise moments positive, and with lengths in metres, the free body diagram gives

\[ \begin{aligned} \tfrac{4}{5}F_B \cdot 6 - F_G \cdot 2 &= 0 \\ \tfrac{3}{5}F_B - A_x &= 0 \\ A_y + \tfrac{4}{5}F_B - F_G &= 0 \end{aligned} \]

The pin at \(A\) carries the resultant of the two components,

\[ F_A = \sqrt{A_x^2 + A_y^2} \]

F_B, A_x, A_y = sp.symbols('F_B A_x A_y', real=True)

F_G = 30e3          # applied load [N]

eqs = [sp.Eq(F_B*sp.Rational(4,5)*6 - F_G*2, 0),   # moments about A
       sp.Eq(F_B*sp.Rational(3,5) - A_x, 0),       # horizontal equilibrium
       sp.Eq(A_y - F_G + F_B*sp.Rational(4,5), 0)] # vertical equilibrium

sol = sp.solve(eqs, [F_B, A_x, A_y], dict=True)[0]
F_A = sp.sqrt(sol[A_x]**2 + sol[A_y]**2)

ltx(r"F_B &=", sol[F_B], r"~\text{N}",
    r"\\ A_x &=", sol[A_x], r"~\text{N}",
    r"\\ A_y &=", sol[A_y], r"~\text{N}",
    r"\\ F_A &=", F_A.evalf(), r"~\text{N}", aligned=True, precision=6)

The pin at \(A\) carries \(F_A\), and since it sits in double shear the force splits between two sections, \(T_A = F_A/2\). The pin at \(B\) carries the full link force across a single section, \(T_B = F_B\).

With the allowable shear stress \(\tau_{\text{allow}} = 180/1.5 = 120~\text{MPa}\), each diameter follows from setting the actual shear stress on one section equal to the allowable one,

\[ \frac{T}{\pi d^2/4} = \tau_{\text{allow}} \quad\Rightarrow\quad d = \sqrt{\frac{4T}{\pi\,\tau_{\text{allow}}}}, \qquad T_A = \frac{F_A}{2}, \quad T_B = F_B \]

d = sp.symbols('d', positive=True)

tau_allow = 180 / 1.5

T_A = F_A / 2
T_B = sol[F_B]

d_A = sp.solve(sp.Eq(tau_allow, T_A / (sp.pi*(d/2)**2)), d)[0]
d_B = sp.solve(sp.Eq(tau_allow, T_B / (sp.pi*(d/2)**2)), d)[0]

ltx(r"\tau_{\text{allow}} &=", round(tau_allow), r"~\text{MPa}",
    r"\\ d_A &=", d_A.evalf(), r"~\text{mm}",
    r"\\ d_B &=", d_B.evalf(), r"~\text{mm}", aligned=True, precision=4)

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.

Example 4: A pinned lap joint

Two steel plates overlap and are joined by a single pin through the middle of the overlap, as in Figure 8.4.11. The plates are \(w = 40~\text{mm}\) wide and \(t = 5~\text{mm}\) thick, the pin has the diameter \(d = 12~\text{mm}\), and the centre of the hole lies \(e = 20~\text{mm}\) from the end of each plate. The joint carries the tensile load \(P = 20~\text{kN}\). The plates are of S355 steel, \(\sigma_y = 355~\text{MPa}\), and the pin of property class 8.8, \(\sigma_y = 640~\text{MPa}\). Determine the factor of safety against each way the joint can fail, and find the one that governs.

The load passes from plate \(A\) through the pin into plate \(B\), and every part of that path can fail. Plate \(A\) is pulled to the left by \(P\) at its far end and held by the pin, which pushes on the half of the hole toward the plate end. Figure 8.4.12 shows the four ways the plate can fail, and plate \(B\) is loaded the same way in the opposite direction.

Away from the hole the plate carries \(P\) on its whole cross section \(Wt\), the gross section. At the hole the material is missing, and the same force passes through the two ligaments beside it, the net section \((w - d)t\) of Figure 8.4.12 (a),

\[ \sigma_{\text{gross}} = \frac{P}{Wt}, \qquad \sigma_{\text{net}} = \frac{P}{(w - d)t} . \]

The net section stress is an average. The lines of force bend around the hole and crowd together at its edge, so the stress there is higher than the average, as sketched in Figure 8.4.12 (b). The ratio of the peak to the net section stress is the stress concentration factor \(K_t\), which depends only on the geometry. For a round hole in a plate of finite width, a curve fitted to the tabulated values [1] is

\[ \begin{aligned} \sigma_{\max} &= K_t\,\sigma_{\text{net}} \\ K_t &= 3.000 - 3.140\,\frac{d}{w} + 3.667\left(\frac{d}{w}\right)^2 - 1.527\left(\frac{d}{w}\right)^3 \end{aligned} \tag{8.4.3}\]

which runs from \(3\) for a small hole in a wide plate, the classical value for a hole in an infinite plate, down to \(2\) as the hole fills the plate. A hole with a pin pressing on it concentrates the stress more than an open hole, so 8.4.3 is a lower estimate here.

8.4.3 is not a result of theory. Only its left end is: for a hole in an infinite plate, Kirsch solved the elasticity problem in closed form in 1898 and found exactly \(K_t = 3\) at the edge of the hole. For a plate of finite width the values behind the chart were measured, first with photoelastic models and strain gauges, and later computed, with series solutions and with finite element analysis. The polynomial is a curve fitted through those values, so that the chart can be read off a formula. Its coefficients carry no meaning of their own, and the fit is only as good as the data under it, typically within a few percent. Chapter 8.20 checks the value \(3\) against Kirsch’s solution and collects the corresponding fits for notches, grooves and shoulders in flat bars and round shafts.

ratio = sp.symbols('rho', positive=True)         # the hole diameter over the plate width, d/w
K_t_fit = 3 - 3.14*ratio + 3.667*ratio**2 - 1.527*ratio**3

fig, ax = plt.subplots(figsize=(6, 3.6))
fplot(K_t_fit, (ratio, 0, 1), ax=ax, color='C0', linewidth=2, label='fit, Pilkey (2005)')
ax.plot(0, 3, 'ks', label='Kirsch, infinite plate')
ax.plot(0.3, float(K_t_fit.subs(ratio, 0.3)), 'o', color='C3', label='this joint, $d/w = 0.3$')
ax.set_xlabel('$d/w$')
ax.set_ylabel('$K_t$')
ax.set_ylim(1.8, 3.1)
ax.grid(True, alpha=0.4)
ax.legend()
plt.show()

When the chart is not accurate enough, or when no chart exists for the geometry at hand, the stress concentration has to be computed. A finite element model of the plate, with the mesh refined around the hole, gives the peak stress directly, and dividing it by the net section stress gives \(K_t\) for that exact geometry and load. The same model can include what the chart leaves out, such as the pin pressing on one side of the hole instead of a uniform pull far away, which is the case in this joint.

The pin presses on the half of the hole toward the plate end, Figure 8.4.12 (c). The contact pressure varies around the hole, and its average over the projected area \(dt\), the rectangle the pin covers when seen along the load, is the bearing stress. The same push drives the material between the hole and the plate end out of the plate, Figure 8.4.12 (d), by shearing it along two planes of length \(e - d/2\) each,

\[ \sigma_b = \frac{P}{dt}, \qquad \tau_{\text{out}} = \frac{P}{2(e - d/2)t} . \]

Finally the pin itself is cut at the one plane between the plates, in single shear like the pin at \(B\) in Figure 8.4.10, \(\tau_{\text{pin}} = P/(\pi d^2/4)\).

P, w, t, d, e = sp.symbols('P w t d e', positive=True)
K_t = 3 - 3.14*(d/w) + 3.667*(d/w)**2 - 1.527*(d/w)**3     # Pilkey (2005)

stresses = {r"\sigma_{\text{gross}}": P/(w*t),
            r"\sigma_{\text{net}}": P/((w - d)*t),
            r"\sigma_{\max}": K_t*P/((w - d)*t),
            r"\sigma_b": P/(d*t),
            r"\tau_{\text{out}}": P/(2*(e - d/2)*t),
            r"\tau_{\text{pin}}": P/(sp.pi*d**2/4)}

values = {P: 20e3, w: 40, t: 5, d: 12, e: 20}              # N and mm, so stresses in MPa
sigma_y_plate, sigma_y_pin = 355, 640
# shear yields at half the tensile yield stress, a conservative estimate
limits = {r"\sigma_{\text{gross}}": sigma_y_plate, r"\sigma_{\text{net}}": sigma_y_plate,
          r"\sigma_{\max}": sigma_y_plate, r"\sigma_b": sigma_y_plate,
          r"\tau_{\text{out}}": sigma_y_plate/2, r"\tau_{\text{pin}}": sigma_y_pin/2}

rows = []
for name, expr in stresses.items():
    value = float(expr.subs(values))
    rows += [("" if not rows else r"\\ ") + name + " &=", sp.Float(value),
             r"~\text{MPa}, \qquad \eta =", sp.Float(limits[name]/value)]
ltx(*rows, aligned=True, precision=4)

The shear limits in the code take a material to yield in shear at half its tensile yield stress. That is a conservative estimate, and Chapter 8.7 shows where it comes from. The factor of safety of each mode is then its limit over its stress, as in 8.3.3.

The two lowest factors of safety are almost equal, \(\eta = 1.06\) for the peak stress at the hole edge and \(\eta = 1.07\) in bearing, and they mean different things. The peak stress is reached only in a thin band of material at the edge of the hole. In a ductile steel under a static load that band yields a little, the stress spreads over the rest of the net section, and the joint carries on; the net section itself has a factor of safety of \(2.5\). The peak matters when the load varies over time, because fatigue cracks start where the stress is highest, and in brittle materials, which cannot redistribute it. The bearing stress, in contrast, is an average over the whole contact, and exceeding it crushes the edge of the hole and lets the pin move, so under a static load bearing is what limits this joint. Design codes allow a bearing stress somewhat above \(\sigma_y\), because the material around the contact is confined, so comparing with \(\sigma_y\) as we did errs on the safe side.

Each mode points to its own remedy. A thicker plate lowers every plate stress at once. A larger edge distance \(e\) helps only tear-out, a wider plate only the tension modes, and a larger pin lowers the bearing and pin stresses but removes more of the net section. Designing a joint means balancing these against each other until no single mode is much weaker than the rest.

A lap joint has one more weakness that none of these formulas contains. The two plates do not lie in one plane, so the forces \(P\) are offset by the thickness \(t\) and form a couple that bends the plates and the pin. A double-shear joint, with one plate held between two others as in Example 1, is symmetric and avoids it.

Computing the stress concentration

The notebook below solves the plate of Figure 8.4.13 with the finite element method, for any width, hole, thickness, load and material. It meshes a quarter of the plate with six-node triangles, uses the two symmetry lines through the hole, and compares the computed \(K_t\) with the chart. Refining the mesh along the hole brings the two together.