Question 5 of 8: Statically Indeterminate Cable Frame
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-6: Mechanics of Materials — National Exams, December 2014
3 hours duration. Closed book (one hand-written aid sheet permitted). Any FIVE of the
eight questions constitute a complete paper; all eight are solved below as a complete
study resource.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed.; Beer, Johnston, DeWolf & Mazurek,
Mechanics of Materials, 7th ed.; Gere & Goodno, Mechanics of Materials, 8th ed.
P=1200 N vertical, downward, at D (800 mm horizontal from C)
Find. (a) the force in each cable; (b) the vertical displacement at D; (c) the shear stress in the pin at C.
Rigid frame pivoted at C, cables to A and B (drawn horizontal at their given wall-anchor heights), load P at D.
Approach. The frame is rigid with 4 unknowns (Cx, Cy, FA, FB) but only 3 equilibrium equations — one degree statically indeterminate. Add a compatibility equation from the frame’s small rotation about the pin C, then use equilibrium and Hooke’s law together.
Moment equilibrium about C (both cable forces horizontal, at heights hA, hB above C; load P at horizontal distance xD): $$h_A F_A + h_B F_B = x_D P = 960000\text{ N}\cdot\text{mm}.$$ One equation, two unknowns — the system is indeterminate.
Compatibility. A small rigid-body rotation φ of the frame about C moves point A (directly above C, at height hA) horizontally by φhA, and point B by φhB; since both cables are horizontal, these ARE the cable elongations. So $$\dfrac{\delta_A}{h_A}=\dfrac{\delta_B}{h_B}=\phi \quad\Rightarrow\quad \dfrac{F_A L_A}{h_A}=\dfrac{F_B L_B}{h_B}$$ (both cables share the same AcE). Substituting the geometry gives $F_A = 1.2500\,F_B$.
Solve the two equations. $$\boxed{F_A=1297.3\text{ N},\qquad F_B=1037.8\text{ N}}$$ (both tension; stresses 103.2 MPa and 82.6 MPa, well under 240 MPa yield).
Rotation and displacement at D. $$\phi=\dfrac{F_B L_B}{A_c E\,h_B}=8.259e-04\text{ rad}\qquad\Rightarrow\qquad \boxed{\delta_D=\phi\, x_D = 0.6607\text{ mm (downward)}}$$ (the vertical deflection at any point depends only on its horizontal distance from the pivot C, independent of that point’s own height.)
Pin reaction and shear. $$C_x=F_A+F_B=2335.1\text{ N},\qquad C_y=P=1200.0\text{ N}\qquad\Rightarrow\qquad R_C=\sqrt{C_x^2+C_y^2}=2625.4\text{ N}.$$ In double shear, each of the two shear planes carries half: $$\boxed{\tau_{pin}=\dfrac{R_C/2}{A_{pin}}=6.53\text{ MPa}}$$
