Factor of safety against elastic buckling of a strut: 22-Mec-B9 Advanced Engineering Structures, May 2016, Q4

Question 4: Factor of safety against elastic buckling of a 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.

Given. A pin-jointed frame in which a two-force strut props the tip of a uniformly loaded cantilevered beam.

Given data — Question 4
Quantity	Symbol	Value
Beam span, A to B	$L_{AB}$	3.0 m
Vertical offset, A to C	$h$	1.5 m
Uniformly distributed load on AB	$w$	12.5 kN/m
Strut outside diameter	$D$	55 mm
Strut wall thickness	$t$	5 mm
Young’s modulus	$E$	200 GPa
Yield strength	$\sigma_{Y}$	320 MPa

Find. The factor of safety of strut BC against elastic (Euler) buckling.

Figure 4.1 — the frame ABC. The beam AB carries the distributed load; the inclined member BC is pinned at both ends and therefore carries axial force only.

Approach. Take moments about the pin at A to get the axial force in the two-force strut, compute the Euler load of the same strut treated as pin-ended over its true inclined length, and divide.

Establish the strut geometry. B lies 3.0 m along the beam from A and C lies 1.5 m directly below A, so the strut runs the diagonal $$L_{BC} = \sqrt{3.0^{2}+1.5^{2}} = \sqrt{11.25} = 3.3541\ \text{m}$$ Its direction cosines are $3.0/3.3541 = 0.8944$ horizontally and $1.5/3.3541 = 0.4472$ vertically.
Resolve the applied load. The uniformly distributed load has a resultant $$W = wL_{AB} = 12.5\times3.0 = 37.5\ \text{kN}$$ acting at mid-span, 1.5 m from A.
Take moments about A to find the strut force. BC is pinned at both ends and carries no transverse load, so it is a two-force member and its force acts along its own axis. Only the vertical component of that force has a lever arm about A: $$\sum M_{A}=0:\quad F_{BC}\left(\frac{1.5}{3.3541}\right)(3.0) = 37.5\times1.5$$ $$F_{BC} = \frac{56.25}{1.34164} = \boxed{41.93\ \text{kN (compression)}}$$ The strut is in compression because it must push the tip of the beam upward; its horizontal component, 37.50 kN, is reacted at the pin A.
Compute the section properties of the tube. With $D = 55$ mm and $t = 5$ mm the bore is $d = 45$ mm, so $$I = \frac{\pi\left(D^{4}-d^{4}\right)}{64} = \frac{\pi\left(9\,150\,625-4\,100\,625\right)}{64} = 2.4789\times10^{5}\ \text{mm}^{4}$$ $$A = \frac{\pi\left(D^{2}-d^{2}\right)}{4} = \frac{\pi\left(3025-2025\right)}{4} = 785.40\ \text{mm}^{2}$$ and the radius of gyration is $r = \sqrt{I/A} = 17.766$ mm.
Confirm that elastic buckling is the governing mode. The slenderness ratio of a pin-ended strut of this length is $$\frac{L_{BC}}{r} = \frac{3354.1}{17.766} = 188.8$$ against a transition slenderness $\sqrt{2\pi^{2}E/\sigma_{Y}} = \sqrt{2\pi^{2}(200\,000)/320} = 111.1$. The strut is well past the transition, so the Euler formula applies without any inelastic correction — which is exactly why the question supplies $\sigma_{Y}$ and asks specifically for the elastic buckling factor.
Compute the Euler critical load. Both ends are pinned, so the effective length factor is $K = 1$ and the effective length is the full 3.3541 m: $$P_{\text{cr}} = \frac{\pi^{2}EI}{\left(KL\right)^{2}} = \frac{\pi^{2}\left(200\,000\right)\left(2.4789\times10^{5}\right)}{3354.1^{2}} = \boxed{43.49\ \text{kN}}$$ The corresponding critical stress is $P_{\text{cr}}/A = 55.38$ MPa, only 17 per cent of the yield strength, which confirms the slenderness check above.
Form the factor of safety. $$\text{FoS} = \frac{P_{\text{cr}}}{F_{BC}} = \frac{43.49}{41.93} = \boxed{1.04}$$ The strut is on the point of buckling. For comparison, the factor against squashing is $A\sigma_{Y}/F_{BC} = 251.3/41.93 = 6.0$, so stability, not strength, controls this member by a factor of nearly six.
State what the answer means for the design. A factor of 1.04 against elastic buckling leaves no margin at all for initial crookedness, end-fixity that is less than ideal, or load overshoot, and real columns fail below the Euler load because of exactly those effects. If a working factor of 2.0 were required the strut would need $P_{\text{cr}} = 83.9$ kN, which at the same length means $I \ge 4.78\times10^{5}$ mm⁴ — for instance a 70 mm outside diameter with the same 5 mm wall.

Check — end fixity. The circles at A, B and C on the figure are read as frictionless pins, which makes BC a genuine two-force member and gives $K = 1$. If the connection at C were treated as fully fixed in the plane and B as pinned, $K$ would fall to about 0.7 and the critical load would rise to roughly 89 kN, doubling the factor of safety. The pinned reading is the conservative one and is what the figure shows.

Question 4 — results
Quantity	Result
Strut length	3.3541 m
Total distributed load on AB	37.5 kN
Axial force in BC	41.93 kN compression
Second moment of area of the tube	$2.4789\times10^{5}$ mm⁴
Cross-sectional area	785.40 mm²
Radius of gyration and slenderness	17.77 mm; $L/r = 188.8$
Euler critical load (pin–pin)	43.49 kN
Critical stress	55.38 MPa
Factor of safety against elastic buckling	1.04

Concept & Theory

Topic. Euler buckling of a pin-ended tubular strut in a statically determinate pin-jointed frame.

Key relations. Two-force member equilibrium and $\sum M_{A}=0$ for the strut force; tube properties $I=\pi\left(D^{4}-d^{4}\right)/64$ and $A=\pi\left(D^{2}-d^{2}\right)/4$; the Euler load $P_{\text{cr}}=\pi^{2}EI/\left(KL\right)^{2}$; the elastic/inelastic transition at $L/r=\sqrt{2\pi^{2}E/\sigma_{Y}}$.

Why this works. Because BC is pinned at both ends and unloaded along its length, its internal force must lie along the line joining the pins, which collapses the whole frame analysis to a single moment equation. The Euler formula is then exact for an ideal, initially straight, perfectly elastic column, and it applies here because the critical stress of 55 MPa is far below the proportional limit; had the strut been stocky, the critical stress would have exceeded yield and a tangent-modulus or Johnson parabola treatment would have been needed instead. Checking $L/r$ against the transition value is the one-line test that decides which regime you are in, and it costs nothing.

Common pitfall. Using the horizontal span, 3.0 m, as the buckling length instead of the true inclined length of 3.354 m; that alone would overstate the critical load by 25 per cent and turn a marginal strut into an apparently safe one. The second trap is comparing $F_{BC}$ with the squash load $A\sigma_{Y}$, which returns a comfortable factor of 6.0 and completely misses the real failure mode.

Source. Hibbeler, Mechanics of Materials, 10th ed., Ch. 13 (buckling of columns); Boresi and Schmidt, Advanced Mechanics of Materials, 6th ed., Ch. 12.

Question 4 of 8: Factor of safety against elastic buckling of a strut

Question 4: Factor of safety against elastic buckling of a strut (20 marks)