Question

3.12 for the joint shown in the figure, calculate (1) the largest bearing stress between the pin and the members; (2) the average shear stress in the pin; and (3) the largest average stress in the members.

Explanation:

Step1: Calculate bearing - area for pin - member contact

The bearing area between the pin and the members is calculated for the thinnest member. Here, the thinnest member has a thickness \(t = 7.5\mathrm{mm}\) and the diameter of the pin \(d=12.5\mathrm{mm}\). The bearing area \(A_b\) for a cylindrical - to - flat contact is \(A_b = td\).
\[A_b=7.5\times10^{- 3}\mathrm{m}\times12.5\times10^{-3}\mathrm{m}=93.75\times10^{-6}\mathrm{m}^2\]
The force \(P = 25\times10^{3}\mathrm{N}\). The bearing stress \(\sigma_b=\frac{P}{A_b}\).
\[ \sigma_b=\frac{25\times10^{3}\mathrm{N}}{93.75\times10^{-6}\mathrm{m}^2}\approx266.67\times10^{6}\mathrm{Pa}=266.67\mathrm{MPa}\]

Step2: Calculate shear - area of the pin

The pin is in double - shear. The cross - sectional area of the pin \(A_s=\frac{\pi d^{2}}{4}\), and since it is in double - shear, the shear area \(A_{total}=2\times\frac{\pi d^{2}}{4}=\frac{\pi d^{2}}{2}\), where \(d = 12.5\times10^{-3}\mathrm{m}\).
\[A_{total}=\frac{\pi\times(12.5\times10^{-3}\mathrm{m})^{2}}{2}\approx122.72\times10^{-6}\mathrm{m}^2\]
The shear stress \(\tau=\frac{P}{A_{total}}\), with \(P = 25\times10^{3}\mathrm{N}\).
\[ \tau=\frac{25\times10^{3}\mathrm{N}}{122.72\times10^{-6}\mathrm{m}^2}\approx203.71\times10^{6}\mathrm{Pa}=203.71\mathrm{MPa}\]

Step3: Calculate the net - area of the members

The net area of the member is calculated by subtracting the area of the hole from the gross area. The gross area of the member with the hole \(A_{gross}=18\times10^{-3}\mathrm{m}\times25\times10^{-3}\mathrm{m}=450\times10^{-6}\mathrm{m}^2\). The area of the hole \(A_{hole}=\pi\times(\frac{12.5\times10^{-3}\mathrm{m}}{2})^{2}\approx122.72\times10^{-6}\mathrm{m}^2\). The net area \(A_{net}=A_{gross}-A_{hole}\)
\[A_{net}=450\times10^{-6}\mathrm{m}^2 - 122.72\times10^{-6}\mathrm{m}^2=327.28\times10^{-6}\mathrm{m}^2\]
The average stress in the members \(\sigma=\frac{P}{A_{net}}\), with \(P = 25\times10^{3}\mathrm{N}\).
\[ \sigma=\frac{25\times10^{3}\mathrm{N}}{327.28\times10^{-6}\mathrm{m}^2}\approx76.4\times10^{6}\mathrm{Pa}=76.4\mathrm{MPa}\]

Answer:

(1) The largest bearing stress between the pin and the members is approximately \(266.67\mathrm{MPa}\).
(2) The average shear stress in the pin is approximately \(203.71\mathrm{MPa}\).
(3) The largest average stress in the members is approximately \(76.4\mathrm{MPa}\).