A-Frame Truss — Buckling of a Compression Strut: 04-BS-6, December 2015, Q5

Question 5: A-Frame Truss — Buckling of a Compression Strut (20 marks)

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
Geometry	Symmetric A-frame, half-base 3 m, height 4 m (member length 5 m each)
Section (each member)	40 mm × 60 mm rectangle, pinned-pinned (K = 1)
Steel	E = 200 GPa, F_y = 240 MPa
Safety factor	2, applied to buckling only

Symmetric two-member A-frame with apex load P and the Section A-A rectangular strut.

Find. The largest P the frame can carry, checking both Euler buckling (in-plane, with the SF) and yielding (no SF), and reporting whichever governs.

Approach. Resolve the apex joint (2-force members only, no other joints) to relate the member axial force to P, then compare the buckling and yield capacities of one member.

Check: the Section A-A view gives no explicit axis callout, so the 60 mm dimension is taken to lie in the plane of the truss (governing in-plane bending resistance with I = 40(60)³/12) — the standard orientation for a compression strut designed against in-plane buckling.

Joint equilibrium at the apex. Each strut makes an angle from vertical with $\cos\alpha = 4/5$ (rise 4 m over a 5 m member); by symmetry the horizontal components cancel and $$2F\cos\alpha = P \;\Rightarrow\; F = \frac{P}{2(4/5)} = 0.625P\ \text{(compression)}$$
Section properties and Euler buckling load. $$A = 40(60) = 2400\ \text{mm}^2\qquad I_{inplane} = \frac{40(60)^3}{12} = 7.20\times10^5\ \text{mm}^4$$ $$P_{cr} = \frac{\pi^2 EI}{L_e^2} = \frac{\pi^2(200000)(7.20\times10^5)}{(5000)^2} = 56{,}849\ \text{N}$$ $$P_{cr,allow} = \frac{P_{cr}}{SF} = \frac{56{,}849}{2} = 28{,}425\ \text{N}$$
Yield capacity (no safety factor) for comparison: $$P_{yield} = F_y A = 240(2400) = 576{,}000\ \text{N} \;\gg\; P_{cr,allow}$$ Buckling clearly governs by a wide margin (this pair of slender struts would buckle at only 5% of the load needed to yield them).
Largest applied load P. $$\boxed{P_{max} = \frac{P_{cr,allow}}{0.625} = \frac{28{,}425}{0.625} = 45{,}479\ \text{N} \approx 45.5\ \text{kN}}$$

Quantity	Value
Member axial force / P	0.625
I (in-plane)	7.20×10⁵ mm⁴
P_cr,allow (per member)	28.43 kN
P_yield (per member)	576.0 kN (does not govern)
P_max	45.5 kN

Topic: Euler buckling of a pin-ended compression strut.

Key relations: $P_{cr}=\pi^2EI/L_e^2$ with $L_e=KL$ ($K=1$ pinned-pinned); governing capacity is $\min(P_{cr}/SF,\ F_yA)$.

Why this works: A slender member in axial compression fails by sudden lateral instability (buckling) at a load far below its material yield capacity; Euler's formula gives that critical load from the member's flexural stiffness $EI$ and its effective length. Because the two struts here are 2-force members meeting only at the apex and each support, joint equilibrium alone (no need for a full method-of-joints solve) gives the force-to-P ratio.

Common pitfall: Checking only yielding and missing that a long slender strut buckles at a small fraction of its yield load — always compute both and report the governing (lower) capacity, never just the more familiar yield check.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 13 (buckling of columns).

Question 5 of 8: A-Frame Truss — Buckling of a Compression Strut

Question 5: A-Frame Truss — Buckling of a Compression Strut (20 marks)