Question 6 of 7: Rectangular footing at 3:1 for column A–B, and the column-to-pier connection

Question B3: Rectangular footing at 3:1 for column A–B, and the column-to-pier connection (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. Footing plan dimensions and thickness, its flexural and shear design, and a detailed connection between the steel column and the square concrete pier.

Approach. Size the plan area on service loads against the allowable bearing pressure, check the factored pressure against the ultimate bearing capacity, then design the footing for flexure and both shear modes using the net factored pressure, and finally detail a base that behaves as the pin the frame analysis assumed.

1. Collect the loads the base actually sees. The reactions come from the Question A2 frame model, run at service level for sizing and at factored level for strength. Because A is a pin there is no base moment; the horizontal thrust nevertheless produces a moment at the underside of the footing through the lever arm of pier plus footing: $$M = H\left(h_{pier} + t\right)$$ Take a square pier 1000 mm x 1000 mm rising 1.0 m above the footing and a footing 600 mm thick.
2. Size the plan area on service loads. With a 3:1 rectangle, $L = 3B$. Trying $B = 1.5\ \text{m}$, $L = 4.5\ \text{m}$: $$A_f = 6.75\ \text{m}^2, \quad W_{footing} = 6.75(0.6)(24) = 97.2\ \text{kN}, \quad W_{pier} = 24.0\ \text{kN}$$ $$N = 646.8 + 97.2 + 24.0 = 768.0\ \text{kN}, \quad M = 68.3(1.6) = 109.2\ \text{kN}\!\cdot\!\text{m}$$ $$e = \frac{109.2}{768.0} = 0.142\ \text{m} < \frac{L}{6} = 0.75\ \text{m}$$ so the whole base stays in compression and the linear pressure distribution applies: $$\boxed{q_{max} = \frac{N}{A_f}\left(1 + \frac{6e}{L}\right) = 113.8(1.19) = 135.3\ \text{kPa} \le 150\ \text{kPa}\ \checkmark}$$ Orient the 4.5 m length in the plane of the frame so that the eccentricity acts along the long dimension.
3. Check the factored pressure against the ultimate bearing capacity. $$N_f = 883.9 + 1.25(97.2 + 24.0) = 1035.4\ \text{kN}, \quad M_f = 94.1(1.6) = 150.6\ \text{kN}\!\cdot\!\text{m}$$ $$q_{f,max} = 183.1\ \text{kPa} \le 225\ \text{kPa}\ \checkmark, \quad q_{f,min} = 123.6\ \text{kPa}$$
4. Reduce to the net pressure that actually bends the footing. The footing's own weight is carried directly by the soil beneath it and produces no bending, so it is removed from the design pressure: $$q_{sw} = 1.25(0.6)(24) = 18.0\ \text{kPa}$$ $$q_{net} = 165.1\ \text{kPa at the heavily loaded end falling to } 105.6\ \text{kPa at the other}$$
5. Design the flexural reinforcement in the long direction. The critical section is the face of the 1000 mm pier, a cantilever of $(4.5 - 1.0)/2 = 1.75\ \text{m}$. On the heavily loaded side the trapezoidal pressure runs from 142.0 kPa at the face to 165.1 kPa at the tip: $$R = \frac{142.0 + 165.1}{2}(1.75) = 268.7\ \text{kN per metre of width}$$ $$\bar{x} = \frac{1.75}{3}\cdot\frac{2(165.1) + 142.0}{165.1 + 142.0} = 0.897\ \text{m}$$ $$\boxed{M_f = 268.7(0.897) = 241.1\ \text{kN}\!\cdot\!\text{m per metre}}$$
6. Solve for the steel. With 75 mm cover to the bottom mat and 20M bars, $d = 600 - 75 - 20 - 10 = 495\ \text{mm}$. Solving the same rectangular-block quadratic used in B1, now with $b = 1000\ \text{mm}$: $$A_{s,req} = 1490\ \text{mm}^2/\text{m}, \quad A_{s,min} = 0.002A_g = 1200\ \text{mm}^2/\text{m}$$ Provide 20M at 175 mm ($1714\ \text{mm}^2/\text{m}$): $$M_r = 276.1\ \text{kN}\!\cdot\!\text{m per metre} \ \Rightarrow\ \frac{241.1}{276.1} = 0.873\ \checkmark$$ In the short direction the cantilever is only 0.25 m and minimum steel governs; Clause 15.4.4 requires the fraction $2/(\beta + 1) = 0.5$ of the short-direction steel to lie in a band of width $B$ centred on the pier, so use 20M at 160 mm within the 1.5 m band and 20M at 250 mm outside it.
7. Check one-way shear. The critical section is at $d$ from the pier face: $$V_f = 188.6\ \text{kN per metre}, \quad d_v = \max(0.9d,\ 0.72h) = 445.5\ \text{mm}$$ $$\beta = \frac{230}{1000 + d_v} = 0.159, \quad V_c = 0.65(0.159)(5)(1000)(445.5)/10^{3} = 230.4\ \text{kN/m}$$ $$\frac{188.6}{230.4} = 0.819\ \checkmark$$ This is the check that sets the 600 mm thickness; a 500 mm footing would fail it.
8. Check two-way (punching) shear. The critical perimeter lies $d/2$ from the pier face, giving a 1.495 m square, $b_o = 5980\ \text{mm}$: $$V_f = 883.9 + 1.25(24.0) - 135.4(2.235) = 611.3\ \text{kN}$$ $$v_c = \min\left[\left(1 + \tfrac{2}{\beta_c}\right)0.19,\ \tfrac{\alpha_s d}{b_o} + 0.19,\ 0.38\right]\lambda\phi_c\sqrt{f_c'} = 0.38(0.65)(5) = 1.235\ \text{MPa}$$ $$V_r = 1.235(5980)(445.5)/10^{3} = 3290\ \text{kN} \ \Rightarrow\ \frac{611.3}{3290} = 0.186\ \checkmark$$ Punching is nowhere near critical because the 1.0 m pier spreads the load over a long perimeter.
9. Detail the column-to-pier connection as a genuine pin. The frame analysis assumed zero moment at A, and the base must be built so that it delivers that. (i) Base plate 900 x 400 x 25 mm, 300W. The bearing pressure is $883.9\times10^{3}/(900 \times 400) = 2.46\ \text{MPa}$; concrete bearing resistance under Clause 10.8 of A23.3 is $0.85\phi_cf_c'A_1\sqrt{A_2/A_1} = 8288\ \text{kN}$, so bearing is not critical, and the plate cantilever check $t \ge \max(m, n)\sqrt{2f_p/\phi F_y} = 11.2\ \text{mm}$ is met with margin by 25 mm. (ii) Two 30 mm anchor rods (F1554 Gr. 300) on the centre-line transverse to the plane of the frame, so that the base cannot develop a couple about the axis of bending. Their shear resistance is $0.6\phi_{br}A_bF_u = 113.7\ \text{kN}$ each, giving 227 kN against the 94.1 kN thrust, a ratio of 0.41. (iii) A 150 x 150 x 20 mm shear lug welded to the underside of the plate and set in a keyed pocket, so that the horizontal thrust is delivered in bearing on the pier rather than in shear-friction across the anchor rods. (iv) The pier is reinforced with 8-25M vertical dowels lapped into the footing mat, with 15M ties at 300 mm, and 50 mm of non-shrink grout under the plate.
10. State the residual check. A base plate this stiff will always attract some moment. Provided the anchor rods stay on the axis of bending, the couple they can generate is negligible compared with the 1191 kN·m at the column head, so the pinned model is safe. If the base were built with four rods outside the flanges instead, it would attract perhaps 15 per cent of the column moment, the footing would need to be designed for it, and the frame's sway would reduce — a change that helps the frame and hurts the foundation, and one that must not be made accidentally on site.