Question: Each of the four vertical links has an $8 \times 36$-mm uniform rectangular cross section and each of the four pins has a $16$-mm diameter. Take $P = 22$ kN. $P$ $0.25$ m $0.4$ m $0.2$ m $A$ $B$ $C$ $D$ $E$ $O$ Determine the average shearing stress in the pin at $B$. The average shearing stress in the pin at $B$ is ____ MPa. Graph shape: a mechanical linkage diagram in the center showing a horizontal bar with pins at $B$ and $C$ connected by two vertical links to a lower support, an upward force $P$ applied at the left end $O/A$, and dimension arrows labeled $0.25$ m, $0.4$ m, and $0.2$ m; position_hint = center.

1. **Problem Statement:** Determine the average shearing stress in the pin at $B$ given: - Four vertical links with $8 \times 36$ mm rectangular cross section. - Pins have diameter $16$ mm. - Applied force $P = 22$ kN. - Dimensions: $0.25$ m, $0.4$ m, $0.2$ m. 2. **Understanding the problem:** The pin at $B$ experiences shear due to the applied load $P$. The average shearing stress $\tau$ is given by: $$\tau = \frac{F}{A}$$ where $F$ is the shear force on the pin and $A$ is the cross-sectional area of the pin. 3. **Calculate the shear force on pin $B$: ** The force $P$ is applied at point $O/A$ and transmitted through the linkage. We assume the load is equally shared by the two pins $B$ and $C$ because of symmetry. Therefore, shear force on pin $B$: $$F_B = \frac{P}{2} = \frac{22,000}{2} = 11,000 \text{ N}$$ 4. **Calculate the cross-sectional area of the pin:** Pin diameter $d = 16$ mm = $0.016$ m. Cross-sectional area $A$ of the pin (circular cross section): $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.016}{2}\right)^2 = \pi (0.008)^2 = \pi \times 6.4 \times 10^{-5} = 2.01 \times 10^{-4} \text{ m}^2$$ 5. **Calculate the average shearing stress:** $$\tau = \frac{F_B}{A} = \frac{11,000}{2.01 \times 10^{-4}} = 54,726,368 \text{ Pa} = 54.7 \text{ MPa}$$ 6. **Final answer:** The average shearing stress in the pin at $B$ is approximately **54.7 MPa**.