Question 5 of 7: Footing at E and adequacy of the beam-column CE

Question 5: Footing at E and adequacy of the beam-column CE (12 + 8 = 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. (a) Plan dimensions, thickness and reinforcement of the footing at \(E\); (b) whether the Question 4 section satisfies the CSA S16 Cl. 13.8 beam-column interaction, and if not what section does.

1. Service forces at the base. Bearing pressure is a serviceability check, so use the elastic service analysis of the frame. With uniform \(EI\) and one redundant, the force method gives \(R_A = 184.5\ \text{kN}\), hence
   $$N = 800 - 184.5 = 615.5\ \text{kN}, \quad H = 80\ \text{kN}, \quad M_E = 647.8\ \text{kN}\cdot\text{m}$$
2. Footing plan size. The eccentricity \(M/N \approx 1.05\ \text{m}\) is large, so the plan length is set by keeping the resultant inside the middle third rather than by the bearing pressure. Try \(5.0 \times 2.5\ \text{m}\), 0.9 m thick; its own weight is \(5.0(2.5)(0.9)(24) = 270\ \text{kN}\), and the moment is taken at the underside:
   $$N_{tot} = 885.5\ \text{kN}, \quad M_{tot} = 647.8 + 80(0.9) = 719.8\ \text{kN}\cdot\text{m}, \quad e = 0.813\ \text{m}$$
   $$e = 0.813\ \text{m} \;<\; \frac{L}{6} = 0.833\ \text{m} \quad\text{— full contact retained}$$
3. Bearing pressure check. With the resultant inside the kern the pressure is trapezoidal:
   $$q_{max,min} = \frac{N}{BL}\left(1 \pm \frac{6e}{L}\right) = 70.84(1 \pm 0.976)$$
   $$q_{max} = \boxed{139.9\ \text{kPa}} \le 300\ \text{kPa}, \qquad q_{min} = 1.7\ \text{kPa} > 0$$
   Sliding is checked with a friction coefficient of 0.5: \(\mu N = 442.7\ \text{kN}\) against \(H = 80\ \text{kN}\), a factor of safety of 5.5.
Footing at E under service load: trapezoidal bearing pressure with the resultant just inside the middle third.
4. Factored soil reaction for the structural design. The footing's own weight is carried directly by the soil beneath it and causes no bending, so only the column's factored actions are used:
   $$N_f = 923.2\ \text{kN}, \quad M_f = 1.5(647.8) + 1.5(80)(0.9) = 1079.7\ \text{kN}\cdot\text{m}, \quad e_f = 1.170\ \text{m}$$
   Now \(e_f > L/6\), so the factored pressure block is triangular over a contact length
   $$a = 3\left(\frac{L}{2} - e_f\right) = 3.99\ \text{m}, \qquad q_{f,max} = \frac{2N_f}{Ba} = 185.0\ \text{kPa}$$
5. Flexure in the footing. For a steel column on a base plate, A23.3 Cl. 15.4.1 places the critical section half-way between the column face and the plate edge, i.e. 316 mm from the centreline. Integrating the triangular pressure over the 2.184 m cantilever,
   $$V = 733.9\ \text{kN}, \qquad M = 902.1\ \text{kN}\cdot\text{m}$$
   With \(d = 900 - 75 - 15 = 810\ \text{mm}\), the flexural steel required is 3335 mm², but the shrinkage-and-temperature minimum of Cl. 7.8.1 governs:
   $$A_{s,min} = 0.002 A_g = 0.002(2500)(900) = \boxed{4500\ \text{mm}^2}$$
   Provide 11 – 25M bottom bars at 250 mm centres in the long direction (5500 mm²), and 25M at 300 mm each way elsewhere.
6. Shear in the footing. One-way shear is taken at \(d\) from the critical section, with \(d_v = \max(0.9d,\,0.72h) = 729\ \text{mm}\) and the simplified \(\beta = 0.21\) for a member without stirrups and with \(h > 300\ \text{mm}\):
   $$V_f = 526.2\ \text{kN} \;\le\; V_r = \phi_c\lambda\beta\sqrt{f'_c}\,b_w d_v = 1362.6\ \text{kN}$$
   Two-way (punching) shear around the base plate, \(b_o = 4(700 + 810) = 6040\ \text{mm}\):
   $$V_r = 0.38\phi_c\sqrt{f'_c}\,b_o d = 6619\ \text{kN} \;\gg\; N_f = 923.2\ \text{kN}$$
   Both are satisfied with a large margin; the 900 mm thickness is set by the flexural lever arm, not shear.
7. (b) Beam-column check of the Question 4 section. Under the collapse loading the column carries \(C_f = 930\ \text{kN}\) together with the plastic hinge moment \(M_f = 1080\ \text{kN}\cdot\text{m}\) at \(E\). Bracing at \(C\), \(D\) and \(E\) gives an unbraced length of 5 m, so weak-axis buckling governs with \(r_y = 62.3\ \text{mm}\):
   $$\frac{KL}{r_y} = \frac{5000}{62.3} = 80.3, \quad \lambda = 1.069, \quad C_r = \phi A F_y (1 + \lambda^{2n})^{-1/n} = 2913\ \text{kN}$$
   With \(\omega_2 = 1.42\) the lateral-torsional resistance is \(M_r = 1151.5\ \text{kN}\cdot\text{m}\). The Cl. 13.8.2 interactions are
   $$\text{(a) cross-section: } \frac{930}{5241.6} + 0.85\frac{1080}{1189.5} = 0.949 \;\le\; 1.0 \quad\checkmark$$
   $$\text{(b) overall member: } \frac{930}{2913} + 0.85\frac{1080}{1189.5} = \boxed{1.091} \;>\; 1.0 \quad\times$$
   $$\text{(c) lateral-torsional: } \frac{930}{2913} + 0.85\frac{1080}{1151.5} = \boxed{1.117} \;>\; 1.0 \quad\times$$
   The uniform section from Question 4 is NOT adequate as the beam-column \(CE\). It has just enough plastic moment in pure bending, but no reserve once the 930 kN of axial compression and the 5 m unbraced length are included.
8. Required column section. Increasing the section to flanges \(300 \times 25\ \text{mm}\) and web \(560 \times 12\ \text{mm}\) (\(d = 610\ \text{mm}\), 170.5 kg/m, still Class 1) gives \(r_y = 72.0\ \text{mm}\), \(C_r = 4393\ \text{kN}\) and \(M_r = 1678.4\ \text{kN}\cdot\text{m}\), with lateral-torsional buckling no longer reducing the resistance:
   $$\text{(a) } 0.683, \qquad \text{(b) } \boxed{0.759}, \qquad \text{(c) } 0.759 \;\le\; 1.0 \quad\checkmark$$
   Adopt the 610 mm section for \(CE\) while retaining the 564 mm section for the beam \(AC\) — the frame is then no longer of uniform section, which is the practical consequence of Question 4's "uniform" simplification.