Maximum Load on a Beam Propped by a Buckling Strut: 04-BS-6, May 2014, Q4

Question 4: Maximum Load on a Beam Propped by a Buckling 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.

Beam AB (6 m, UDL w) pinned at A, propped by strut BC (2 m horizontal, 4 m vertical offset), 80×80 mm section, pinned both ends.

Given.

Quantity	Value
Beam length AB	6 m, pinned at A
Strut offsets (B to C)	2 m horizontal, 4 m vertical
Strut section	80 mm × 80 mm, pinned both ends
Strut E, f_y	200 GPa, 350 MPa
Factor of safety on buckling	2 (no safety factor on yield)

Find. The maximum UDL w the beam can carry before the strut buckles.

Approach. The pin-pin strut BC is a two-force member, so its Euler buckling capacity (divided by the FS) sets the maximum axial force it may carry; resolve that force into its vertical component acting on the beam at B, then use moment equilibrium of beam AB about A to convert that into the maximum w.

Strut geometry and section. $$L_{BC} = \sqrt{2000^2+4000^2} = 4472\ \text{mm}, \qquad I = \frac{80(80)^3}{12} = 3.413\times10^6\ \text{mm}^4$$
Euler buckling capacity. Pinned-pinned, K = 1: $$P_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2(200{,}000)(3.413\times10^6)}{4472^2} = 336.9\ \text{kN}$$ $$P_{allow} = \frac{P_{cr}}{FS} = \frac{336.9}{2} = \boxed{168.4\ \text{kN}}$$
Moment equilibrium of beam AB about A. The strut force $F$ acts on the beam at B along the strut's own line, with vertical component fraction $v_{off}/L_{BC}=4000/4472=0.8944$; the UDL resultant $wL_{beam}$ acts at $L_{beam}/2$: $$F\!\left(\frac{v_{off}}{L_{BC}}\right)L_{beam} = wL_{beam}\!\left(\frac{L_{beam}}{2}\right) \ \Rightarrow\ w = \frac{2F(v_{off}/L_{BC})}{L_{beam}}$$
Maximum w at the strut's allowable load. Setting $F=P_{allow}=168.4$ kN: $$\boxed{w_{max} = \frac{2(168.4)(0.8944)}{6} = 50.2\ \text{kN/m}}$$

Final Results.

Quantity	Value
L_BC, I_strut	4472 mm, 3.413 × 10⁶ mm⁴
P_cr (Euler)	336.9 kN
P_allow = P_cr/FS	168.4 kN
Maximum UDL w	50.2 kN/m

Topic: Euler buckling of a pin-ended strut controlling the capacity of a propped beam.

Key relations: $P_{cr}=\pi^2EI/(KL)^2$ (K = 1 for pinned-pinned); a two-force member's line of action fixes how much of its axial capacity converts into a useful vertical support reaction, via simple trigonometry ($v_{off}/L$).

Why this works: Because the strut is loaded only at its pinned ends, it is a two-force member carrying pure axial force regardless of the rest of the structure's stiffness; whichever of buckling or yield is reached first controls, and the question explicitly directs buckling (with its FS) as the sole governing limit.

Common pitfall: Using the full strut force F as if it acted vertically on the beam — only the VERTICAL COMPONENT of the inclined strut force resists the UDL through moment equilibrium of AB; the horizontal component is instead reacted by the pin at A.

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

Question 4 of 8: Maximum Load on a Beam Propped by a Buckling Strut

Question 4: Maximum Load on a Beam Propped by a Buckling Strut (20 marks)