Question 3 of 8: Truss — Buckling of a Compression Member

Question 3: Truss — Buckling of a Compression Member (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.

Find. The largest load P that member 4–6 can carry without buckling.

Approach. Solve the statically determinate truss (method of joints) to express the force in member 4–6 as a multiple of P, then find that member’s allowable compressive force (the lesser of the factored Euler load and the yield load) and equate.

1. Method of joints. Solving the full pin-jointed truss (9 members, 3 reaction components, statically determinate) for the force in the inclined member 4–6 in terms of P gives $$F_{4\text{-}6}=-1.25\,P$$ (negative ⇒ compression); the length of 4-6 is $\sqrt{3^2+4^2}=5$ m.
2. Section properties of the d=40 mm rod: $$A=\dfrac{\pi d^2}{4}=1256.6\text{ mm}^2,\qquad I=\dfrac{\pi d^4}{64}=\dfrac{\pi r^4}{4}=125664\text{ mm}^4.$$
3. Euler buckling load (pin-pin, K=1, L=5000 mm): $$P_{cr}=\dfrac{\pi^2 EI}{L^2}=9922\text{ N}\quad\Rightarrow\quad P_{cr,allow}=\dfrac{P_{cr}}{2}=4961\text{ N}.$$
4. Yield check. $$P_{yield}=\sigma_{yield}A=301593\text{ N}\ \gg\ P_{cr,allow}.$$ At L/r=250 the member is highly slender, so buckling governs, not yielding.
5. Solve for P. Setting the compressive force in 4-6 equal to the allowable buckling load: $$1.25\,P = P_{cr,allow}=4961\text{ N}$$ $$\boxed{P_{max}=3969\text{ N} \approx 3.97\text{ kN}}$$