Maximum Loads Against Buckling of a Two-Force Strut: 04-BS-6, May 2015, Q3

Question 3: Maximum Loads Against Buckling of a Two-Force 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 (pinned at A) propped by pin-ended strut BC (0.5 m horizontal, 1.5 m vertical offset), loads P at 1 m and 2P at 2 m from A.

Given.

Quantity	Value
Beam AB	3 m, pinned at A; P at x=1 m, 2P at x=2 m (both downward)
Strut BC offsets	0.5 m horizontal, 1.5 m vertical (pin-pin)
Strut section	40 mm × 40 mm square, E = 200 GPa, f_y = 280 MPa
Safety factor	2 against buckling; none against yielding

Find. The largest P (and 2P) the beam can carry without the strut BC buckling.

Approach. BC is pinned at both ends with no load along its length, so it is a two-force (axial-only) member. Express its compressive force F in terms of P from moment equilibrium of beam AB about A, then set F equal to the SMALLER of (i) the Euler buckling load divided by the safety factor and (ii) the yield load (no safety factor) to find the governing P.

Strut geometry. $$L_{BC}=\sqrt{0.5^2+1.5^2}=1.581\ \text{m}, \qquad I=\frac{(40)^4}{12}=213{,}333\ \text{mm}^4=2.133\times10^{-7}\ \text{m}^4$$
Strut force F in terms of P (moment equilibrium of AB about A). The strut pushes up on the beam at B with vertical component fraction $1.5/1.581=0.9487$; loads P (x=1) and 2P (x=2) act downward: $$3(0.9487)F = 1(P)+2(2P)=5P \ \Rightarrow\ \boxed{F=1.757P}\ \text{(compression)}$$
Euler buckling capacity (pin–pin, K=1). $$P_{cr}=\frac{\pi^2EI}{L_{BC}^2}=\frac{\pi^2(200{,}000)(213{,}333)}{1581.1^2}= 168.4\ \text{kN}$$ $$F_{allow,buckling}=\frac{P_{cr}}{2}=\boxed{84.22\ \text{kN}} \ \Rightarrow\ P_{buckling}=\frac{84.22}{1.757}=\boxed{47.94\ \text{kN}}$$
Yield capacity (no safety factor). $$F_{yield}=\sigma_yA=280(1600)=448.0\ \text{kN} \ \Rightarrow\ P_{yield}=\frac{448.0}{1.757}=255.0\ \text{kN}$$
Governing case. $P_{buckling}=47.94\ \text{kN} \ll P_{yield}=255.0\ \text{kN}$, so buckling governs by a wide margin: $$\boxed{P_{max}=47.9\ \text{kN}, \qquad 2P_{max}=95.9\ \text{kN}}$$

Final Results.

Quantity	Value
L_BC, I_strut	1.581 m, 213,333 mm⁴
F = 1.757P	relation between strut force and P
P_cr (Euler), F_allow	168.4 kN, 84.22 kN
P_buckling, P_yield	47.94 kN, 255.0 kN
Governing P, 2P	47.9 kN, 95.9 kN (buckling governs)

Topic: Euler buckling of a pin-ended two-force strut controlling the capacity of a propped beam, checked against yielding.

Key relations: $P_{cr}=\pi^2EI/(KL)^2$ (K=1 pinned-pinned); $F_{yield}=\sigma_yA$; the true capacity is the SMALLER of the two, each with its own specified factor of safety.

Why this works: A member pinned at both ends with no transverse load between the pins carries pure axial force by equilibrium alone; converting that axial force to a moment about A (via its vertical component) is all that is needed to connect the strut's buckling capacity to the beam's applied loads.

Common pitfall: Applying the safety factor to BOTH failure modes, or assuming yielding always governs a short, thick-looking strut — here the strut is slender enough (L/depth ≈ 40) that buckling governs by more than a factor of 5 over yielding.

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

Question 3 of 8: Maximum Loads Against Buckling of a Two-Force Strut

Question 3: Maximum Loads Against Buckling of a Two-Force Strut (20 marks)