Largest Load on a Two-Rod Truss (Euler Buckling): 04-BS-6, December 2013, Q3

Question 3: Largest Load on a Two-Rod Truss (Euler Buckling) (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.

Two-rod truss: A(−5,0), B(0,4), C(1,0) m; P applied downward at B.

Given.

Quantity	Value
Geometry (A→B→C offsets)	5 m horiz. / 4 m vert. (AB); 1 m horiz. / 4 m vert. (BC)
Rod section	100 mm OD, 5 mm wall (90 mm ID), both members
Buckling factor of safety	2 (Euler load only; no FS on yield)
Yield strength / E (steel)	240 MPa / 200 GPa

Find. The largest P that keeps both two-force members within their governing (buckling or yield) capacity.

Approach. Resolve joint B for the member forces in terms of P, compute each member's Euler buckling capacity (pinned-pinned, K = 1) and yield capacity, then find the P that first exhausts the more critical member.

Member lengths and joint-B equilibrium. $$L_{AB}=\sqrt{5^2+4^2}=6.403\ \text{m}, \qquad L_{BC}=\sqrt{1^2+4^2}=4.123\ \text{m}$$ Resolving the two axial member forces against the applied P at B (both rods carry the load down into their supports, so both come out in compression): $$F_{AB} = -0.2668\,P, \qquad F_{BC} = -0.8590\,P$$
Section properties (both rods identical). $$A = \frac{\pi}{4}\left(100^2-90^2\right) = 1492.3\ \text{mm}^2, \qquad I = \frac{\pi}{64}\left(100^4-90^4\right) = 1.688\times10^6\ \text{mm}^4$$
Buckling and yield capacity of each member. $$P_{cr}=\frac{\pi^2 EI}{L^2}\ (K=1),\qquad P_{cr,allow}=\frac{P_{cr}}{2},\qquad P_{yield}=F_y A = 240(1492.3)=358.1\ \text{kN}$$ Member AB ($L=6403$ mm): $P_{cr}=81.27$ kN $\Rightarrow P_{cr,allow}=40.64$ kN (governs over 358.1 kN yield).
Member BC ($L=4123$ mm): $P_{cr}=196.0$ kN $\Rightarrow P_{cr,allow}=98.01$ kN (governs over yield). Both members are buckling-governed, as expected for such slender rods.
Largest P from each member, and the controlling value. $$P_{AB} = \frac{40.64}{0.2668} = 152.3\ \text{kN}, \qquad P_{BC} = \frac{98.01}{0.8590} = 114.1\ \text{kN}$$ Member BC reaches its buckling capacity first, so $$\boxed{P_{allow} = 114.1\ \text{kN}}$$ (At this load, $F_{AB}=0.2668(114.1)=30.4$ kN compression — comfortably below AB's own 40.64 kN capacity, confirming BC governs.)

Final Results.

Quantity	Value
A, I (each rod)	1492 mm², 1.688×10⁶ mm⁴
P_cr,allow (AB / BC)	40.64 kN / 98.01 kN
F_AB / F_BC at P_allow	30.4 kN (C) / 98.0 kN (C)
Largest load P	114.1 kN, governed by buckling of BC

Topic: Euler column buckling of pin-ended compression members in a truss.

Key relations: $P_{cr}=\pi^2EI/(KL)^2$ (K = 1 for pinned-pinned); a two-force member's axial force is fixed by joint equilibrium, independent of its own stiffness.

Why this works: Both AB and BC are two-force members (loaded only at their pinned ends), so each carries pure axial force set by statics at joint B; whichever member's buckling (or yield) capacity is reached first, at the smallest P, controls the whole structure.

Common pitfall: Forgetting that the SHORTER member is not automatically safer — capacity depends on the interaction of L, I and the force ratio together, so both members must be checked against their own force fraction of P, not just compared by length.

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

Question 3 of 8: Largest Load on a Two-Rod Truss (Euler Buckling)

Question 3: Largest Load on a Two-Rod Truss (Euler Buckling) (20 marks)