Two-rod truss — Euler buckling capacity: 04-BS-6, May 2013, Q3

Question 3: Two-rod truss — Euler buckling capacity (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.

Truss geometry: A (wall pin), B (free joint), C (ground pin); load P at 45° from vertical.

Given.

Quantity	Value
$L_{AB}$ (4 m horiz., 3 m vert.)	5 m
$L_{BC}$ (vertical)	4 m
Rod diameter	120 mm (both rods, pinned both ends)
$F_y$, $E$	350 MPa, 200 GPa
Euler safety factor	2 (no SF on yielding)

Find. The largest load $P$ the truss can support.

Approach. Solve joint B for the member forces in terms of $P$, then compare each member's actual compressive force against its lesser of (i) yield capacity $F_yA$ and (ii) allowable Euler load $\pi^2EI/(KL)^2\big/2$ (pin-ended, $K=1$); the governing member sets $P_{max}$.

Joint equilibrium at B. Unit vectors: $\hat u_{BA}=(-0.8,0.6)$, $\hat u_{BC}=(0,-1)$. With $P$ applied at $45^\circ$ from vertical, down-left: $P_x=-P\sin45^\circ$, $P_z=-P\cos45^\circ$. Solving $T_{AB}\hat u_{BA}+T_{BC}\hat u_{BC}+\vec P=0$: $$T_{AB}=-0.8839\,P,\qquad T_{BC}=-1.2374\,P$$ Both are negative — both rods are in compression for any positive $P$, so both must be checked for buckling.
Section and Euler loads. $A=\pi(60)^2=11\,309.7\ \text{mm}^2$, $I=\pi(60)^4/4=1.01788\times10^{7}\ \text{mm}^4$. $$P_{cr,AB}=\frac{\pi^2EI}{L_{AB}^2}=\frac{\pi^2(200\,000)(1.01788\times10^7)}{5000^2}=803.7\ \text{kN},\qquad P_{cr,BC}=\frac{\pi^2EI}{L_{BC}^2}=1255.8\ \text{kN}$$
Allowable compressive load per member. Yield capacity $F_yA=350(11\,309.7)=3958.4$ kN is far above either Euler load, so buckling governs both members: $$P_{allow,AB}=\frac{P_{cr,AB}}{2}=401.8\ \text{kN},\qquad P_{allow,BC}=\frac{P_{cr,BC}}{2}=627.9\ \text{kN}$$
Back out the governing applied load. Using the force coefficients from Step 1: $$P\big|_{AB\ \text{limit}}=\frac{401.8}{0.8839}=454.6\ \text{kN},\qquad P\big|_{BC\ \text{limit}}=\frac{627.9}{1.2374}=507.4\ \text{kN}$$ Member AB reaches its Euler-buckling limit first, so $$P_{max}=\boxed{454.6\ \text{kN}}$$

Quantity	Result
$T_{AB}$, $T_{BC}$ (at $P_{max}$)	−401.8 kN, −562.6 kN (both compression)
Governing mode	Euler buckling of AB (SF = 2)
$P_{max}$	454.6 kN

Concept & Theory

Topic. Pin-jointed two-force members governed by Euler column buckling.

Key relations. $P_{cr}=\pi^2EI/(KL)^2$ (pinned-pinned, $K=1$); allowable load with a specified safety factor $= P_{cr}/SF$; failure governed by $\min(F_yA,\ P_{cr}/SF)$ for a compression member.

Why this works. Both rods are slender (length/diameter ≈ 42 and 33), so elastic (Euler) buckling occurs at a stress far below yield — confirmed by comparing $P_{cr}/2$ against $F_yA$ before deciding which failure mode governs.

Common pitfall. Checking only the more heavily loaded member (BC, larger force coefficient) without checking whether the shorter, stiffer member (AB) might still govern once its own (higher) Euler capacity is divided by its own (lower) force coefficient — here AB governs despite carrying a smaller fraction of $P$.

Source. Hibbeler, Mechanics of Materials, Ch. 13 (Buckling of Columns, Euler's Formula).

Question 3 of 8: Two-rod truss — Euler buckling capacity

Question 3: Two-rod truss — Euler buckling capacity (20 marks)