Maximum Load on a Two-Bar Frame Governed by Column Buckling: 04-BS-6, Undated paper, Q5

Question 5: Maximum Load on a Two-Bar Frame Governed by Column Buckling

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
$AB$: length, cross-section	10 ft, 1.5 in × 1.5 in (pin–pin)
$BC$: cross-section, angle to $AB$	0.75 in × 0.75 in, 30°
$E$, $\sigma_Y$ (both members)	$29\times10^3$ ksi, 50 ksi
FS (buckling)	2 — in-plane only; no FS on yield

Find. The largest load $P$ (applied vertically at joint B) the frame can carry.

Frame: strut AB (10 ft, horizontal) pinned at A; diagonal BC at 30 deg to a wall pin C; P applied down at B.

Approach. Resolve the two-force members $AB$, $BC$ at joint B for their forces in terms of $P$; check the compression member ($AB$) against Euler buckling with FS = 2, and both members against yield (no FS) — the smallest resulting $P$ governs.

Joint equilibrium at B. With $BC$ at 30° above horizontal, vertical equilibrium requires $BC$ (tension) to carry the load: $$F_{BC}\sin30^{\circ}=P \Rightarrow F_{BC}=2P\ (\text{tension})$$ Horizontal equilibrium then gives $AB$ in compression: $$F_{AB}=F_{BC}\cos30^{\circ}=\sqrt{3}\,P\ (\text{compression})$$
Euler buckling capacity of AB (pin–pin, $K=1$). $$I_{AB}=\frac{(1.5)^4}{12}=0.4219\text{ in}^4$$ $$P_{cr}=\frac{\pi^2 E I_{AB}}{(KL)^2}=\frac{\pi^2(29{,}000)(0.4219)}{(120)^2}=8.385\text{ kip}$$
Allowable P from buckling (FS = 2). $F_{AB,allow}=P_{cr}/2=4.193$ kip, and $F_{AB}=\sqrt3\,P$, so $$P_{buckling}=\frac{4.193}{\sqrt3}=2.421\text{ kip}$$
Yield checks (no FS) — confirm buckling actually governs. $$P_{yield,AB}=\frac{\sigma_Y A_{AB}}{\sqrt3}=\frac{50(2.25)}{\sqrt3}=64.95\text{ kip},\qquad P_{yield,BC}=\frac{\sigma_Y A_{BC}}{2}=\frac{50(0.5625)}{2}=14.06\text{ kip}$$ Both far exceed 2.421 kip, so: $$\boxed{P_{max}=2.42\text{ kip} \approx 2420\text{ lb (governed by buckling of strut AB)}}$$

Result	Value
$F_{AB}$, $F_{BC}$ (at $P_{max}$)	4.19 kip (compression), 4.84 kip (tension)
$P_{cr}$ (Euler, AB)	8.39 kip
$P_{yield,AB}$, $P_{yield,BC}$	64.95 kip, 14.06 kip (not governing)
$P_{max}$	2.42 kip ≈ 2420 lb

Topic: Two-force-member frame analysis combined with Euler column buckling.

Key relations: Joint equilibrium for pin-connected two-force members; $P_{cr}=\pi^2EI/(KL)^2$; $FS=P_{cr}/P_{allow}$.

Why this works: Slender members in axial compression fail by buckling at a load far below their yield capacity, so a compression member must always be checked against BOTH criteria and the lower value governs. Here $AB$'s buckling load (2.42 kip, at FS 2) is roughly 27× smaller than its yield-based capacity (65 kip) — typical for a slender strut, and the reason column design rarely reduces to a simple stress check.

Common pitfall: Applying the buckling FS to the yield check (or vice versa) — the problem deliberately keeps them independent; also forgetting that only the COMPRESSION member ($AB$) can buckle, not the tension member ($BC$).

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 13 §13.2 (Euler's Formula for Pin-Supported Columns).

Question 5 of 8: Maximum Load on a Two-Bar Frame Governed by Column Buckling

Question 5: Maximum Load on a Two-Bar Frame Governed by Column Buckling