Question 1 of 8: Rigid Bar on a Pin and Two Cables

Question 1: Rigid Bar on a Pin and Two Cables (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) Cable forces T_B, T_C; (b) the vertical displacement at D; (c) the shear stress in the double-shear pin at A.

Approach. The rigid bar is held by one pin and two cables — four reaction unknowns (A_x, A_y, T_B, T_C) against three equilibrium equations, so the bar is indeterminate to one degree; close it with a rigid-body rotation compatibility equation relating the two cable elongations, then finish with statics.

1. Resultant of the triangular load. $$W = \tfrac12(80)(12) = 480\ \text{kN}, \qquad \bar x = \tfrac23(12) = 8\ \text{m from A (i.e. directly over C)}$$
2. Compatibility. The bar is rigid and pinned at A, so it rotates by a single small angle θ about A; the vertical drop at B and C equals the cable elongations $\delta_B=4\theta$, $\delta_C=8\theta$, giving $\delta_C = 2\delta_B$. With $\delta = TL/(AE)$ (same A, E for both cables): $$\frac{T_C(7)}{AE} = 2\,\frac{T_B(3)}{AE} \ \Rightarrow\ T_C = \frac{6}{7}T_B$$
3. Moment equilibrium about A. $$T_B(4) + T_C(8) = W\bar x = 480(8) = 3840\ \text{kN}\cdot\text{m}$$ Substituting $T_C=\tfrac67 T_B$: $$T_B\!\left(4+\tfrac{48}{7}\right)=3840 \ \Rightarrow\ \boxed{T_B = 353.7\ \text{kN}}, \qquad \boxed{T_C = \tfrac67(353.7)=303.2\ \text{kN}}$$
4. Cable stress check. $A_{cable}=\tfrac{\pi}{4}(25)^2=490.9\ \text{mm}^2$: $$\sigma_B=\frac{353{,}700}{490.9}=720.5\ \text{MPa}, \qquad \sigma_C=\frac{303{,}200}{490.9}=617.6\ \text{MPa}\quad(\text{both}<800\ \text{MPa yield, elastic – OK})$$
5. (b) Displacement at D. $$\delta_B=\frac{T_B L_B}{A_{cable}E}= \frac{353{,}700(3000)}{490.9(200{,}000)}=10.81\ \text{mm}, \qquad \theta=\frac{\delta_B}{4000}=2.702\times10^{-3}\ \text{rad}$$ $$\boxed{\delta_D = 12{,}000\,\theta = 32.4\ \text{mm, downward}}$$ (check: $\delta_C=8000\theta=21.6$ mm $=T_C L_C/(A_{cable}E)$, confirming compatibility).
6. (c) Pin shear at A. Vertical equilibrium of the bar (up positive): $$A_y + T_B + T_C - W = 0 \ \Rightarrow\ A_y = 480-353.7-303.2=-176.8\ \text{kN}$$ i.e. the pin must pull DOWN 176.8 kN on the bar (the two cables together over-carry the distributed load, so the pin supplies the balancing downward force). With double shear ($A_{pin}=\tfrac{\pi}{4}(30)^2=706.9\ \text{mm}^2$, two shear planes): $$\boxed{\tau_{pin} = \frac{V}{2A_{pin}} = \frac{176{,}800}{2(706.9)} = 125.1\ \text{MPa}}$$

Final Results.