Propped Beam: Strut Buckling Sets the Maximum Load: 04-BS-6, December 2018, Q4

Question 4: Propped Beam: Strut Buckling Sets the Maximum Load (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. Beam $AB$: pinned to the wall at $A$, horizontal, length $L=6\text{ m}$, carrying UDL $w$ (to be found) over its full length. Strut $BC$: pinned at both ends, $60\times60\text{ mm}$ square, $\sigma_Y=500\text{ MPa}$, $E=200\text{ GPa}$; from B, C is $2\text{ m}$ horizontally and $3\text{ m}$ vertically down. Buckling factor of safety $=2$; no yield safety factor applied.

Find. The maximum UDL $w$ the beam can carry, governed by Euler buckling of the strut.

Beam AB (UDL w) propped by inclined strut BC; buckling of the pin-pin strut governs the maximum w.

Approach. Free-body the beam AB alone (pin reactions at A, axial strut force at B); take moments about A to express the strut force in terms of w, then set that force equal to the Euler critical load of the strut divided by the given factor of safety and solve for w.

Strut geometry. $L_{strut}=\sqrt{2^2+3^2}=\boxed{3.606\text{ m}}$.
Moment equilibrium of beam AB about A. The strut is a two-force (pin-pin) member, so its force $S$ (compression) acts on the beam at B along the strut axis; only its VERTICAL component ($S\cdot3/L_{strut}$, since B is at the same height as A) creates moment about A. Setting the strut moment equal to the UDL's moment: $S\left(\dfrac{3}{L_{strut}}\right)(6)=w(6)(3)\Rightarrow S=\dfrac{w\,L\,L_{strut}}{2(3)}=w(3.606)\text{ kN per kN/m of }w$.
Euler buckling load of the strut ($K=1$, pin-pin). $I=\dfrac{60^4}{12}=1.08\times10^{6}\text{ mm}^4$; $P_{cr}=\dfrac{\pi^2EI}{L_{strut}^2}=\dfrac{\pi^2(200{,}000)(1.08\times10^6)}{3{,}605.55^2}=\boxed{163{,}987\text{ N}}=163.99\text{ kN}$.
Allowable strut force and solve for w. $S_{allow}=P_{cr}/\text{FS}=163.99/2=\boxed{81.99\text{ kN}}$. Setting $S=S_{allow}$: $w_{max}=81.99/3.606=\boxed{22.74\text{ kN/m}}$.
Confirm buckling (not yield) governs. At $w_{max}$, the strut stress is $\sigma=S_{allow}/(60\times60)=81{,}990/3{,}600=22.8\text{ MPa}$, far below $\sigma_Y=500\text{ MPa}$ — consistent with the problem statement that no yield safety factor is needed, since buckling of this slender strut governs the design by a wide margin.

Final results
Quantity	Value
L_strut	3.606 m
I (60×60 mm square)	1.08 × 10⁶ mm⁴
P_cr (Euler, K=1)	163.99 kN
Allowable strut force	81.99 kN
w_max	22.74 kN/m

Topic: Euler column buckling governing a propped-beam design

Key relations: Two-force member force from beam moment equilibrium; Euler buckling $P_{cr}=\pi^2EI/(KL)^2$; allowable load $=P_{cr}/\text{FS}$.

Why this works: Because B sits at the same height as A, the strut's horizontal force component contributes no moment about A, which lets the strut force be isolated directly from a single moment equation without ever solving for the pin reactions at A; the strut, being slender ($L/r\approx208$), fails by elastic instability at a small fraction of its yield load, so the buckling check alone governs the whole structure's capacity.

Common pitfall: Forgetting that only the vertical component of an inclined two-force member's force contributes to a moment taken about a point at the SAME height as one of its ends; also applying the buckling factor of safety to the stress instead of to the critical LOAD before comparing with the actual strut force.

Source: Hibbeler, Mechanics of Materials, 10th ed., §13.2–13.3 Euler Column Buckling

Question 4 of 8: Propped Beam: Strut Buckling Sets the Maximum Load

Question 4: Propped Beam: Strut Buckling Sets the Maximum Load (20 marks)