Question 2 of 7: Beam-column check of member ABC and the footing at A

Question 2: Beam-column check of member ABC and the footing at A (14 + 6 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. Member A–B–C is the windward column of Figure 1, W610×217 from Question 1, 8 m long with lateral support at A, B (the 80 kN load point) and C. From the collapse state of Question 1 it carries $C_f = 932.1$ kN with $M_A = 0$ (pinned base), $M_B = 314.1$ kN·m and $M_C = 862.5$ kN·m. The allowable soil bearing pressure is 500 kPa.

Find. (a) the three Cl 13.8.2 interaction ratios for member ABC and a verdict; (b) plan size, thickness and reinforcement for a spread footing at A.

Approach. Run Cl 13.8.2 in its three parts — cross-sectional strength, overall member strength and lateral-torsional buckling strength — taking each out-of-plane segment separately with its own $\omega_2$, then size the footing on the service axial load and check punching shear, one-way shear and minimum steel.

1. Part (a) — axial resistances. With $n = 1.34$ and $\lambda = \frac{KL}{r}\sqrt{F_y/\pi^2E}$, $$\begin{gathered}\frac{KL}{r_x} = \frac{8000}{262.0} = 30.5 \;\Rightarrow\; C_{rx} = 8160\ \text{kN}, \\ \frac{KL}{r_y} = \frac{5000}{76.9} = 65.1 \;\Rightarrow\; C_{ry} = 5905\ \text{kN},\end{gathered}$$ and over the 3 m segment $C_{ry} = 7723$ kN. The squash load is $C_y = A F_y = 9667$ kN.
2. Moment amplification. The moment ratio over the critical segment B–C is $\kappa = 314.1/(-862.5) = -0.364$ (single curvature), so $\omega_1 = 0.6 - 0.4\kappa = 0.746$. With $C_e = \pi^2 E I_x/(KL)^2 = 58\,488$ kN, $$U_1 = \frac{\omega_1}{1 - C_f/C_e} = \frac{0.746}{1 - 0.0159} = 0.758 \;\Rightarrow\; U_1 = 1.0$$ because Cl 13.8.4 does not permit $U_1 \lt 1.0$ for a member in an unbraced frame.
3. Check (a): cross-sectional strength. $$\frac{C_f}{\phi C_y} + \frac{0.85 U_1 M_f}{\phi M_p} = \frac{932.1}{8700} + \frac{0.85(862.5)}{2144} = 0.107 + 0.342 = 0.449 .$$
4. Check (b): overall member strength. Replace $\phi C_y$ with the in-plane $C_{rx}$: $$\frac{932.1}{8160} + 0.342 = 0.456 .$$
5. Check (c): lateral-torsional buckling strength, segment by segment. Segment A–B ($L = 5$ m, moment $0 \to 314.1$, $\omega_2 = 1.75$) and segment B–C ($L = 3$ m, $\kappa = -0.364$, $\omega_2 = 1.75 + 1.05\kappa + 0.3\kappa^2 = 1.407$) both return $M_u \gg 0.67M_p$, so $M_r = \phi M_p = 2144$ kN·m in each. Then $$\begin{gathered}\text{A--B:}\ \frac{932.1}{5905} + \frac{0.85(314.1)}{2144} = 0.282, \\ \text{B--C:}\ \frac{932.1}{7723} + \frac{0.85(862.5)}{2144} = 0.463 .\end{gathered}$$
6. Verdict on part (a). The governing ratio is $$\boxed{0.46 \;\lt\; 1.0}$$ so member ABC is adequate as a beam-column, with about 54 per cent reserve. The reserve is not waste: the section was sized in Question 1 by the leeward column, which hosts the plastic hinge and carries 9 per cent more axial load, and detailing both columns alike is the practical choice. Had the windward column been sized on its own demand a W610×155 would have sufficed.
7. Part (b) — service loads at A. The base is pinned, so the footing sees axial load and shear but no moment. Dividing out the load factor, $$N = \frac{932.1}{1.5} = 621.4\ \text{kN},\qquad V = \frac{62.8}{1.5} = 41.9\ \text{kN}.$$
8. Plan size on the allowable bearing pressure. Try a 1.2 m square pad 450 mm thick; its own weight is $1.2^2(0.45)(24) = 15.6$ kN, so $$q = \frac{621.4 + 15.6}{1.2^2} = 442\ \text{kPa} \;\le\; 500\ \text{kPa} \quad\checkmark$$ at 88 per cent of the allowable value. The factored net pressure used for the structural checks is $q_f = 932.1/1.44 = 647$ kPa.
9. Two-way (punching) shear. With 75 mm cover and 20M bars, $d = 450 - 75 - 20 = 355$ mm. The critical perimeter at $d/2$ around the 628 × 328 base plate is $b_o = 3332$ mm, and with $\beta_c = 628/328 = 1.91$ the governing stress from Cl 13.3.4 is $0.19\phi_c\lambda\sqrt{f'_c}$ scaled to $0.965$ MPa, so $$V_r = 0.965(3332)(355)/10^3 = 1141\ \text{kN} \;\gt\; V_f = 498\ \text{kN}\quad\checkmark$$
10. One-way shear and flexure. The cantilever from the plate face is $(1200 - 628)/2 = 286$ mm, less than $d = 355$ mm, so the one-way critical section falls outside the footing and that check does not arise. The cantilever moment is $$M_f = \frac{q_f B x^2}{2} = \frac{647(1.2)(0.286)^2}{2} = 31.8\ \text{kN}\cdot\text{m},$$ which needs only 212 mm$^2$ of steel. Minimum reinforcement $A_{s,min} = 0.002 A_g = 0.002(1200)(450) = 1080$ mm$^2$ therefore governs; provide $\boxed{\text{4 – 20M each way}}$ ($1200$ mm$^2$) at the bottom with 75 mm cover.
11. Sliding. The 41.9 kN service shear is resisted by base friction alone: with $\tan\delta = 0.5$ the available resistance is $0.5(621.4 + 15.6) = 319$ kN, a factor of 7.6, so no shear key is needed.