Question 5 of 8: Rigid beam on a pin and two cables — statically indeterminate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-BS-6 Mechanics of Materials. Three-hour, closed-book exam (one double-sided hand-written aid sheet permitted, approved Casio/Sharp calculator only). Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness. No credit is given in the original exam for solutions built by superposing existing standard solutions on Questions 1 and 6 — both are solved here by direct integration/equilibrium as required.
Reference texts: Hibbeler, Mechanics of Materials (10th ed., Pearson) — beam deflection by integration (Ch. 12), stress transformation and Mohr's circle (Ch. 9), columns and buckling (Ch. 13), combined axial & bending loading (Ch. 8), torsion of circular shafts (Ch. 5), transformed-section method for composite beams (Ch. 6), shear and moment diagrams (Ch. 6-7); Beer, Johnston & DeWolf, Mechanics of Materials (7th ed., McGraw-Hill) — cross-check reference for the same topics.
Question 5: Rigid beam on a pin and two cables — statically indeterminate (20 marks)
Rigid beam ABCD: pin at B, cable A (down, 1500 mm), cable C (up, 750 mm), load P at D.
Given.
Quantity
Value
Bay lengths A–B, B–C, C–D
1 m, 2 m, 1.4 m
Cable A length / C length
1500 mm (down) / 750 mm (up)
Cable diameter, $E$, $F_y$
30 mm, 200 GPa, 800 MPa
Load
$P=200$ kN down at D
Pin at B
50 mm diameter, double shear
Find. (a) $T_A$, $T_C$; (b) vertical displacement at D; (c) shear stress in the pin at B.
Approach. This is statically indeterminate to the first degree (one moment equation, two unknown cable tensions): pair the moment-equilibrium equation about B with a compatibility equation from the rigid-body rotation of the beam about the pin, relating each cable's elongation to its distance from B.
Compatibility. Let $\theta$ be the (small) rotation of the rigid beam about B. A point at horizontal distance $x$ from B (measuring B as the origin, D on the positive side) moves vertically by $y=\theta x$. A cable is stretched by exactly the amount its beam end moves away from its own fixed anchor, so with A at $x_A=-1000$ mm (anchor below) and C at $x_C=+2000$ mm (anchor above):
$$\delta_A=\theta\,(-x_A)= -\theta(-1000),\qquad \delta_C=\theta(-x_C)=-\theta(2000)$$
(both cables stretch together for a consistent sense of $\theta$, since A's anchor is below while C's is above). Cable stiffnesses: $k=AE/L$ with $A_c=\pi(15)^2=706.86\ \text{mm}^2$:
$$k_A=\frac{706.86(200\,000)}{1500}=94\,248\ \text{N/mm},\qquad k_C=\frac{706.86(200\,000)}{750}=188\,496\ \text{N/mm}$$
so $T_A=k_A\delta_A$, $T_C=k_C\delta_C$.
Moment equilibrium about B. $T_A$ pulls beam-point A down; $T_C$ pulls beam-point C up; $P$ pulls D down:
$$x_A(-T_A)+x_C(T_C)+x_D(-P)=0\ \Rightarrow\ 1000\,T_A+2000\,T_C=3400(200\,000)$$
Solve simultaneously for $\theta$, then the tensions. Substituting the compatibility relations into the equilibrium equation and solving:
$$\theta=-8.017\times10^{-4}\ \text{rad}\ \Rightarrow\ T_A=\boxed{75.6\ \text{kN}},\qquad T_C=\boxed{302.2\ \text{kN}}$$
Check: $1000(75\,556)+2000(302\,222)=6.80\times10^{8} = 3400(200\,000)$ ✓. Both tensions are well below each cable's yield capacity $F_yA_c=565.5$ kN.
Part (b): displacement at D. $y_D=\theta\,x_D=(-8.017\times10^{-4})(3400)=\boxed{-2.73\ \text{mm}}$ (i.e. D moves down 2.73 mm, consistent with both cables stretching to arrest the rotation caused by $P$).
Part (c): pin shear at B. Vertical equilibrium of the beam: $B_y-T_A+T_C-P=0\Rightarrow B_y=75.6-302.2+200=-26.67$ kN, i.e. the pin carries 26.67 kN. In double shear, each of the two shear planes carries half:
$$\tau_{pin}=\frac{B_y/2}{A_{pin}}=\frac{13\,333}{\pi(25)^2}=\boxed{6.79\ \text{MPa}}$$
