Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 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 W-shape section-property table is attached at the end of the official exam; it is a red herring here — every question below supplies its own
built-up or standard section, and Question 2 cites the CISC-standard W610×125 properties
directly.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (composite
beams, deflection by integration, combined loading, transformation of stress, columns, torsion,
shear/moment diagrams); CISC Handbook of Steel Construction (W-shape section properties).
A(0,0) and C(4 m,0) pinned to ground; B directly above A at 3 m
Member AB (vertical)
20×20 mm, pin-connected
Member BC (inclined)
60×60 mm, length 5 m (3-4-5 triangle), pin-connected
Steel
E = 200 GPa, Fy = 340 MPa
Safety factor
2, against buckling ONLY (yield unfactored)
A-frame: vertical member AB and inclined strut BC.
Find. The largest horizontal load P the frame can carry.
Approach. AB and BC are two-force (pin-connected) members, so joint equilibrium
at B gives each member's axial force as a multiple of P. Check BC (in compression) for both Euler
buckling (with the SF = 2) and yielding, and AB (in tension) for yielding only (a tension member
cannot buckle); the smallest resulting P governs.
Member forces from joint equilibrium at B. With BC's geometry giving
horizontal/vertical force components in the ratio 4:3 (run 4 m, rise 3 m over the 5 m member),
summing forces at B:
$$\sum F_x=0:\quad P = F_{BC}\!\left(\frac{4}{5}\right)\;\Rightarrow\; F_{BC}=1.25P\ \text{(compression)}$$
$$\sum F_y=0:\quad F_{AB} = F_{BC}\!\left(\frac{3}{5}\right)=0.75P\ \text{(tension)}$$
Euler buckling capacity of BC (pin-pin, $K=1$, $L=5000$ mm, weak axis governs
but the section is square so both axes are identical):
$$I_{BC}=\frac{60^4}{12}=1.080\times10^{6}\ \text{mm}^4\qquad
P_{cr}=\frac{\pi^2 E I_{BC}}{L^2}=\frac{\pi^2(200\,000)(1.080\times10^{6})}{5000^2}=85\,273\ \text{N}$$
$$P_{cr,allow}=\frac{P_{cr}}{SF}=\frac{85\,273}{2}=42\,637\ \text{N}$$
$$\text{From } F_{BC}=1.25P\le42\,637\ \text{N}:\quad P\le34\,109\ \text{N} = 34.11\ \text{kN}$$
Yield capacity of BC (no safety factor):
$$P_{yield,BC}=F_y A_{BC}=340(3600)=1\,224\,000\ \text{N}\;\Rightarrow\;
P\le\frac{1\,224\,000}{1.25}=979.2\ \text{kN}\quad\text{(does not govern)}$$
Yield capacity of AB (tension, no buckling check applies):
$$P_{yield,AB}=F_y A_{AB}=340(400)=136\,000\ \text{N}\;\Rightarrow\;
P\le\frac{136\,000}{0.75}=181.3\ \text{kN}\quad\text{(does not govern)}$$
