Question 7 of 8: Rigid L-Shaped Link with Two Cables
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2016 — 04-BS-6: Mechanics of Materials (3 hours, closed book, one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete paper on the official exam; every question is answered below. A wide-flange (W-shape) section-property table is attached at the end of the official exam; every question below supplies its own built-up section, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (shear/moment diagrams, deflection by integration, combined axial+bending, transformation of stress via Mohr's circle, torsion of stepped shafts, axial members and pin connections, Euler column buckling).
Question 7: Rigid L-Shaped Link with Two Cables (20 marks)
Given. A rigid L-shaped link, pinned to the wall at D. The vertical arm runs D→C→B→A with C-D=800 mm, B-C=500 mm, A-B=300 mm; the horizontal arm runs D→E=1000 mm. Two horizontal 3 mm dia., 1000 mm long steel cables (E=200 GPa, σy=400 MPa) connect B and C to a fixed wall. A load P=800 N acts downward at E.
Quantity
Value
D to C, C to B, B to A
800, 500, 300 mm
D to E
1000 mm
Cables
3 mm dia, 1000 mm long, E=200 GPa
Load at E
P=800 N, downward
Pin at D
12 mm dia, double shear
Find. Cable tensions TB, TC; horizontal displacement at A; shear stress in the pin at D.
Rigid L-link pinned at D, with horizontal cables at B and C and the load P at E.
Approach. The link is rigid, so a small rotation Δθ about D stretches each horizontal cable by (its height above D) × Δθ. Combine this compatibility relation with moment equilibrium about D to solve for Δθ and hence both tensions (the system is one-degree statically indeterminate against two force unknowns and one moment equation, closed by compatibility, exactly as in a rigid-bar-plus-cables problem).
Compatibility. For a small rotation $\Delta\theta$ about D, cable elongations are $\delta_B=\Delta\theta\cdot y_B$, $\delta_C=\Delta\theta\cdot y_C$ ($y_B=1300$, $y_C=800$ mm), so tensions $T=\dfrac{AE}{L}\delta = \dfrac{AE}{L}\,\Delta\theta\cdot y$, with $A=\pi(1.5)^2=7.0686\ \text{mm}^2$.
Moment equilibrium about D. The load's moment about D, $P\times1000=800000$ N·mm, is resisted by both cable tensions:
$$T_B\,y_B+T_C\,y_C = P\,(1000) \;\Rightarrow\; \frac{AE}{L}\Delta\theta\,(y_B^2+y_C^2) = 800000$$
$$\Delta\theta=\boxed{2.428688e-04\ \text{rad}}$$
Cable tensions.
$$T_B=\frac{AE}{L}\,y_B\,\Delta\theta=\boxed{446.4\text{ N}}, \qquad T_C=\frac{AE}{L}\,y_C\,\Delta\theta=\boxed{274.7\text{ N}}$$
The higher (B) cable, with the larger moment arm, carries more load, as expected. Both stresses ($T/A$) are far below the 400 MPa yield.
Horizontal displacement at A. A ($y_A=1600$ mm from D) moves horizontally by the same small-rotation relation:
$$\boxed{\delta_A=\Delta\theta\cdot y_A=0.38859\text{ mm}}$$
Pin reaction and shear stress at D. Overall equilibrium of the rigid link gives $D_x=T_B+T_C=721.03$ N (balancing the cables' pull toward the wall) and $D_y=P=800$ N, so
$$R_D=\sqrt{D_x^2+D_y^2}=1076.98\text{ N}$$
In double shear the pin carries this resultant across two shear planes, each of area $A_{pin}=\pi(6)^2=113.097\ \text{mm}^2$:
$$\boxed{\tau_{pin}=\frac{R_D/2}{A_{pin}}=4.7613\text{ MPa}}$$
