Question 2 of 7: Eccentrically loaded round HSS pole column

Question A2: Eccentrically loaded round HSS pole column (10 + 10 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. HSS 406.4×9.53, G40.21 350W Class H (F_y=350 MPa); L=10 m, fixed base/free top (cantilever, K=2.0); load P applied at the free top at eccentricity e=1.2 m.

Find. Maximum factored load P_f the pole can carry as a beam-column.

Approach. A vertical load at the free top offset by e produces a constant axial force P_f and a constant moment M_f=P_fe over the full unbraced cantilever length (zero shear, since there is no lateral load). Compute the tube’s section properties and classification, the compressive resistance C_r and flexural resistance M_rx per CSA S16, then solve the beam-column interaction equation (Cl. 13.8.2) for P_f.

1. Section properties. OD=406.4 mm, wall t=9.53 mm, ID=387.34 mm: $$A=\frac{\pi}{4}(OD^2-ID^2)=\boxed{11\,880\text{ mm}^2},\quad I=\frac{\pi}{64}(OD^4-ID^4)=234.1\times10^6\text{ mm}^4$$ $$Z=\frac{OD^3-ID^3}{6}=1.501\times10^6\text{ mm}^3,\qquad r=\sqrt{I/A}=\boxed{140.4\text{ mm}}$$
2. Classification. $$D/t = 406.4/9.53 = 42.6$$ For a circular HSS with F_y=350 MPa, Class 1 limit is $$13\,000/F_y=37.1$$ and Class 2 is $$18\,000/F_y=51.4$$ – the tube is Class 2, so M_r=φZF_y is permitted, and well below the Class 3 axial limit of $$66\,000/F_y=188.6$$, so no reduced effective area is needed in compression.
3. Compressive resistance. Cantilever fixed-free: $$K=2.0,\quad KL=20\,000\text{ mm},\quad KL/r=142.5$$ With hot-formed (Class H) HSS, n=2.24: $$\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2E}} = 1.897$$ $$C_r=\phi AF_y(1+\lambda^{2n})^{-1/n} = 0.9(11\,880)(350)(1+1.897^{4.48})^{-1/2.24}=\boxed{1014\text{ kN}}$$
4. Flexural resistance and Euler load. $$M_{rx}=\phi ZF_y = 0.9(1.501\times10^6)(350)=\boxed{473\text{ kN}\cdot\text{m}}$$ $$C_{ex}=\frac{\pi^2EI}{(KL)^2}=\boxed{1155\text{ kN}}$$
5. Beam-column interaction. With no transverse load along the unbraced length (uniform end moment), take the amplification factor conservatively as ω₁=1.0. CSA S16 Cl. 13.8.2: $$\frac{C_f}{C_r} + \frac{\omega_1 M_f}{M_{rx}(1-C_f/C_{ex})} \le 1.0,\qquad C_f=P_f,\ \ M_f=P_f(1.2)$$ Substituting and clearing the denominator gives a quadratic in P_f; solving the physically meaningful (smaller) root: $$P_f = \boxed{239\text{ kN}}$$ Check: at P_f=239 kN, U₁=1.26, M_f=287 kN·m, and the interaction sum returns exactly 1.00.