Question 7 of 8: Largest Load Before Buckling in a Two-Rod 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 2017. Closed-book, 3 hours;
any FIVE of the eight questions constitute a complete paper (all eight are answered below as a
complete study resource). Aid sheet permitted; geometric properties of
W-shape sections were supplied as an appendix (not required for any of the eight questions solved
here).
A–B: 4 m horizontal, 3 m vertical; B–C: 2 m horizontal, 3 m vertical (both pin-ended)
Find. The largest load P applicable at B without buckling either member,
using $FS=2$ against the Euler critical load.
Two pin-ended rods AB and BC meeting at B, where load P is applied; both members carry compression.
Approach. AB and BC are pin-ended two-force members meeting only at B, so joint
equilibrium at B gives each member's axial force directly in terms of P. Compare the compressive
force in each (tension members can't buckle) against its own Euler critical load divided by the
safety factor of 2, and take the smaller resulting P as governing.
Geometry. With A at the origin, B at (4, 3) m and C at (6, 0) m (a 3-4-5
triangle for AB):
$$L_{AB}=\sqrt{4^2+3^2}=5.000\text{ m}, \qquad L_{BC}=\sqrt{2^2+3^2}=3.606\text{ m}$$
Joint equilibrium at B. Resolving the two member forces (tension positive)
against the applied downward load P:
$$F_{AB}=-0.5556\,P, \qquad F_{BC}=-0.8012\,P$$
Both coefficients are negative — both rods are in compression under the
downward load, so both are buckling candidates.
Section properties (50 mm diameter rod, both members).
$$I=\frac{\pi}{4}(25)^4=3.068\times10^5\text{ mm}^4$$
Euler critical load, each member (pin–pin, K = 1).
$$P_{cr}=\frac{\pi^2 EI}{L^2}$$
$$P_{cr,AB}=\frac{\pi^2(200{,}000)(3.068\times10^5)}{5000^2}=24{,}224\text{ N}=24.22\text{ kN}$$
$$P_{cr,BC}=\frac{\pi^2(200{,}000)(3.068\times10^5)}{3606^2}=46{,}584\text{ N}=46.58\text{ kN}$$
Apply the safety factor and solve for P from each member. Allowable
compressive force $=P_{cr}/FS$; setting this equal to the actual compressive force
in each member ($|coef|\times P$):
$$P_{AB}=\frac{P_{cr,AB}/2}{0.5556}=21.80\text{ kN}, \qquad P_{BC}=\frac{P_{cr,BC}/2}{0.8012}=29.07\text{ kN}$$
Governing member. The smaller value controls:
$$\boxed{P_{max}=21.8\text{ kN}\quad\text{(member AB governs, by buckling)}}$$
