Largest Load on a Two-Bar Truss (Buckling): 04-BS-6, December 2016, Q8

Question 8: Largest Load on a Two-Bar Truss (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.

Given. A(0,0) and C(10,0) m are pinned ground supports; B(8,6) m is the apex where load P is applied downward. Members AB (10 m) and BC ($\sqrt{40}=6.32$ m) are pin-ended 120 mm OD × 10 mm wall steel pipes (E=200 GPa, allowable yield 240 MPa, FS=2 on Euler buckling).

Quantity	Value
A, C	(0,0) m, (10,0) m — pinned supports
B	(8,6) m — load point
Pipe	120 mm OD, 10 mm wall (100 mm ID)
Steel	E=200 GPa, allowable σ_y=240 MPa
Buckling factor of safety	2 (Euler load only)

Find. The largest load P the truss can carry.

Two-bar truss: AB and BC meet at the loaded apex B, pinned to ground at A and C.

Approach. Both members are pin-ended two-force members meeting at joint B, so their forces follow directly from joint equilibrium as a multiple of P; each member's allowable load is then the lesser of its Euler-buckling capacity (with FS=2, compression members only) and its yield capacity, and the governing member sets P_max.

Section properties. $A=\tfrac{\pi}{4}(120^2-100^2)=3455.75\ \text{mm}^2$, $I=\tfrac{\pi}{64}(120^4-100^4)=5.2700e+06\ \text{mm}^4$.
Joint equilibrium at B. Unit vectors from B toward A ($(-8,-6)/10$) and toward C ($(2,-6)/6.32$). Solving $\Sigma F_x=0$, $\Sigma F_y=0$ at B for tension-positive member forces per unit P: $$\boxed{F_{AB}=-0.3333\,P}\ (\text{compression}), \qquad \boxed{F_{BC}=-0.8433\,P}\ (\text{compression})$$ both negative (compression), as expected for two struts propping up a downward load at their shared apex.
Euler buckling capacity, each member. $P_{cr}=\pi^2EI/L^2$, then divide by FS=2: $$P_{cr,AB}/2=52013.0\text{ N}, \qquad P_{cr,BC}/2=130032.6\text{ N}$$
Yield capacity, each member. $P_{yield}=\sigma_{allow}\,A=829380.5\text{ N}$ (same for both members, since they share a cross-section).
Governing load. Each member's allowable applied P is $\min(P_{cr}/2,\,P_{yield})$ divided by that member's force coefficient: $$P_{max,AB}=\frac{\min(52013.0,\,829380.5)}{0.3333}=156039.1\text{ N}, \qquad P_{max,BC}=\frac{\min(130032.6,\,829380.5)}{0.8433}=154199.7\text{ N}$$ Buckling governs both members (their Euler-allowable loads are well under their yield loads); the shorter, more steeply-loaded member BC governs overall: $$\boxed{P_{max}=154199.7\text{ N}=154.20\text{ kN}}$$

Quantity	Value
F_AB	-0.3333 P (compression)
F_BC	-0.8433 P (compression)
P_max, AB governs	156039.1 N
P_max, BC governs	154199.7 N
Governing member	BC
P_max	154.20 kN

Topic: Euler column buckling in a two-bar truss.

Key relations: Euler load $P_{cr}=\pi^2EI/L^2$ (pinned-pinned, K=1); allowable load per member $=\min(P_{cr}/\text{FS},\ \sigma_{allow}A)$.

Why this works: Each pin-ended member is a two-force member, so its internal force is exactly the joint-equilibrium coefficient times the applied P; converting that into an allowable P requires checking BOTH failure modes (elastic instability and material yielding) and taking whichever is more restrictive, since a slender long member typically buckles well before it yields.

Common pitfall: applying the buckling factor of safety to the YIELD check too (the problem explicitly says not to), and forgetting that the governing member is the one with the SMALLEST allowable P, not the one with the largest force coefficient alone.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 13 (Buckling of Columns).

Question 8 of 8: Largest Load on a Two-Bar Truss (Buckling)

Question 8: Largest Load on a Two-Bar Truss (Buckling) (20 marks)