Question 2 of 7: A2 — Maximum factored load on an eccentrically loaded CHS pole

Question 2: A2 — Maximum factored load on an eccentrically loaded CHS pole (10 + 10 = 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 factored bracket load $P_F$ for which the pole satisfies every applicable S16 13.8.2 combined axial-and-bending check.

Approach. Compute the section properties of the tube, classify it, obtain the compressive and bending resistances for the cantilever effective length $KL = 20$ m, then write the two governing 13.8.2 interaction equations with $M_f = P_F e$ and invert each for $P_F$; the smaller answer governs.

1. Section properties of the round HSS. With inside diameter $d = D - 2t = 387.3$ mm,$$A=\frac{\pi}{4}\left(D^2-d^2\right)=11\,882\ \text{mm}^2,\qquad I=\frac{\pi}{64}\left(D^4-d^4\right)=234.1\times10^{6}\ \text{mm}^4$$ so $r=\sqrt{I/A}=140.4$ mm, and the plastic modulus of a circular tube is $Z=(D^3-d^3)/6=1501\times10^{3}$ mm³.
2. Classify the tube. For a circular hollow section in combined compression and bending the class is set by $D/t = 406.4/9.53 = 42.6$ against the S16 limits $13\,000/F_y = 37.1$ (Class 1) and $18\,000/F_y = 51.4$ (Class 2). The tube is Class 2, so the plastic moment may still be used and $\beta = 0.85$ applies in the interaction equations.
3. Axial resistance for the cantilever. A pole fixed at the base and free at the top has $K = 2.0$, so $KL = 20\,000$ mm and $KL/r = 142.5$ — a very slender column. The non-dimensional slenderness is $\lambda = (KL/r)\sqrt{F_y/(\pi^2 E)} = 1.897$, and with $n = 2.24$ for a class H (stress-relieved) hollow section,$$C_r=\phi A F_y\left(1+\lambda^{2n}\right)^{-1/n}=0.90(11\,882)(350)\left(1+1.897^{4.48}\right)^{-1/2.24}=\boxed{1014\ \text{kN}}$$ against a squash load $\phi A F_y = 3743$ kN: slenderness has removed 73 % of the axial capacity.
4. Bending resistance. A circular tube has no lateral-torsional buckling mode — every axis is a principal axis of equal stiffness, so the section cannot buckle sideways out of the plane of bending. Class 2 therefore gives the full plastic moment,$$M_r=\phi Z F_y = 0.90(1501\times10^{3})(350)=473\ \text{kN}\cdot\text{m}$$ The Euler load in the plane of bending, needed for the second-order amplifier, is $C_e=\pi^2EI/(KL)^2=1155$ kN — only 14 % above $C_r$, which warns in advance that the amplification will be severe.
5. Cross-sectional strength, S16 13.8.2(a). The moment at the base is $M_f = P_F(1.2)$ in kN·m. For a member of an unbraced frame $\omega_1 = 1.0$, so $U_1 = 1/(1-P_F/C_e)$ and the check reads$$\frac{P_F}{\phi A F_y}+\frac{0.85\,U_1 M_f}{M_r}\le 1.0$$ Because $U_1$ itself grows with $P_F$ this must be solved by iteration (a single pass with $U_1 = 1.0$ badly overestimates the answer). Convergence gives $P_F = 311$ kN, at which $U_1 = 1/(1-311/1155) = 1.369$.
6. Overall member strength, S16 13.8.2(b). Here the axial term uses $C_r$ from step 3, and S16 13.8.4 sets $U_1 = 1.0$ for a member of an unbraced (sway) frame, since the maximum moment and the maximum second-order effect already coincide at the base:$$\frac{P_F}{C_r}+\frac{0.85\,M_f}{M_r}\le 1.0\;\Longrightarrow\;\frac{P_F}{1014}+\frac{0.85(1.2P_F)}{473}\le 1.0$$ which inverts directly to $P_F = 318$ kN. Check (c), lateral-torsional buckling, does not apply to a circular tube.
7. Governing answer. The two checks are within 2 % of each other — the paper is calibrated so that neither is obviously redundant — and the smaller controls:$$\boxed{P_F = 311\ \text{kN}}$$ The corresponding base moment is $M_f = 311(1.2) = 373$ kN·m. Note that the moment term, not the axial term, is doing most of the work: at the governing load the axial ratio is only $311/3743 = 0.08$ while the amplified moment ratio is 0.92.