Question 3 of 7: Maximum Factored Load on a Sign Standard Post

Question A3: Maximum Factored Load on a Sign Standard Post (4 + 4 + 12 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 circular hollow section 273.1 mm OD × 12.7 mm wall, G40.21M 350W Class H ($F_y = 350$ MPa). From Figure A3 the post rises 10 m from a rigid concrete foundation and is free at the top. Two horizontal arms project on diametrically opposite sides: the longer arm reaches 3 m and carries $2P_f$, the shorter arm reaches 2 m and carries $P_f$, both acting downwards. $P_f$ is already a factored load.

Find. The greatest factored load $P_f$ the post can carry as a beam-column under CSA S16.

Approach. Reduce the two arm loads to an axial force and a net moment at the base, obtain the section properties and the compressive and bending resistances, then solve the S16 beam-column interaction equation for the largest $P_f$ that keeps the utilisation at unity.

1. Reduce the loads to the post axis. Both arms load the post in compression, while their moments about the post act in opposite senses because the arms are diametrically opposite: $$C_f = 2P_f + P_f = 3P_f, \qquad M_f = 2P_f(3.0) - P_f(2.0) = 4P_f \text{ kN}\cdot\text{m}$$ The moment is constant over the full 10 m height, since no load is applied along the post — a uniform-moment case, which is the most severe distribution for stability.
2. Section properties. With $D = 273.1$ mm and $d = 247.7$ mm, $$A = \tfrac{\pi}{4}(D^2 - d^2) = 10\,389 \text{ mm}^2, \qquad I = \tfrac{\pi}{64}(D^4 - d^4) = 88.27 \times 10^6 \text{ mm}^4$$ $$r = \sqrt{I/A} = 92.2 \text{ mm}, \qquad Z = \frac{D^3 - d^3}{6} = 862 \times 10^3 \text{ mm}^3$$ The wall slenderness $D/t = 21.5$ is well below the Class 1 limit $13\,000/F_y = 37.1$, so the tube is Class 1 in both compression and bending.
3. Effective length and compressive resistance. A column fixed at the base and free at the top has $K = 2.0$, so $KL = 20\,000$ mm and $KL/r = 217$ — a very slender member. The non-dimensional slenderness is $$\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2 E}} = 217\sqrt{\frac{350}{\pi^2(200\,000)}} = 2.889$$ For a Class H hollow section $n = 2.24$, giving $$C_r = \phi A F_y \left(1+\lambda^{2n}\right)^{-1/n} = 0.90(10\,389)(350)(1+2.889^{4.48})^{-1/2.24} = \boxed{391 \text{ kN}}$$ Buckling has cut the squash load of 3273 kN to about 12 % of its value.
4. Bending resistance. A circular hollow section cannot buckle laterally — every axis is a principal axis with the same $I$ — so lateral-torsional buckling does not apply and the Class 1 section reaches its plastic moment: $$M_r = \phi Z F_y = 0.90(862\times10^3)(350) = 271.5 \text{ kN}\cdot\text{m}$$
5. Moment amplification. The Euler load in the plane of bending is $$C_e = \frac{\pi^2 E I}{(KL)^2} = \frac{\pi^2 (200\,000)(88.27\times10^6)}{(20\,000)^2} = 436 \text{ kN}$$ For uniform single-curvature moment $\kappa = -1$ and $\omega_1 = 0.6 - 0.4\kappa = 1.0$, so $U_{1} = \omega_1/(1 - C_f/C_e)$. Because $C_e$ is barely above $C_r$, this amplification will be significant.
6. Solve the interaction equation. For a Class 1 member in uniaxial bending, S16 Cl. 13.8.2 requires $$\frac{C_f}{C_r} + \frac{0.85\,U_{1}M_f}{M_r} \le 1.0$$ Substituting $C_f = 3P_f$ and $M_f = 4P_f$ and solving iteratively for the overall member strength (using $C_r = 391$ kN), $$\frac{3P_f}{391} + \frac{0.85(4P_f)}{271.5\left(1 - 3P_f/436\right)} = 1.0 \;\Rightarrow\; P_f = 40.0 \text{ kN}$$ The cross-sectional strength check, which uses the unreduced $C_r = \phi A F_y = 3273$ kN, yields $P_f = 50.0$ kN and therefore does not govern.
7. Confirm the governing case. At $P_f = 40.0$ kN the actions are $C_f = 120$ kN and $M_f = 160$ kN·m, with $U_1 = 1/(1 - 120/436) = 1.381$. The two interaction terms are 0.31 and 0.69, so the post is moment-dominated: $$\boxed{P_f = 40 \text{ kN}}$$ with the shorter arm carrying 40 kN and the longer arm 80 kN.