Question 3 of 7: Reinforced-concrete column AB and its footing

Question 3: Reinforced-concrete column AB and its footing (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.

Find. A square column section with its longitudinal steel and ties, the slenderness/stability check that section must survive, and the plan size and thickness of the spread footing at A.

Approach. Establish the effective length factor from the alignment chart, confirm the slenderness lies in the range where the moment-magnifier method of Clause 10.16 is permitted, build the factored $P$–$M$ interaction diagram by strain compatibility, then proportion the footing for bearing, punching shear, one-way shear and flexure.

1. Part 1 — slenderness and effective length. At B the joint stiffness ratio is$$G_{top} = \frac{\sum(EI/L)_{col}}{\sum(EI/L)_{beam}} = \frac{0.70(3.413\times10^{10})/10\,000}{0.35(1.6875\times10^{11})/15\,000} = 0.607$$and at the pinned base $G_{bot} \to \infty$, taken as 10 in practice. The sway alignment chart then gives $k \approx 2.3$. With $r = 0.3h = 240\ \text{mm}$,$$\boxed{\frac{kl_u}{r} = \frac{2.3(10\,000)}{240} = 95.8}$$This exceeds the 22 of Clause 10.15.2, so slenderness must be accounted for, but it stays below 100, so Clause 10.13.2 permits the magnifier method rather than a full second-order analysis. A 10 m pin-based column is inherently slender, and this is the check that dictates the 800 mm dimension — a 600 mm column would give $kl_u/r = 128$ and would have to be redesigned by Clause 10.14.
2. Reinforcement and the interaction diagram. A23.3 Clause 10.9.1 requires $0.01A_g \le A_{st} \le 0.08A_g$, so at least $0.01(640\,000) = 6400\ \text{mm}^2$. Take $16 - 30\text{M}$, $A_{st} = 11\,200\ \text{mm}^2$, i.e. $\rho = 1.75\ \%$, arranged six per face on two faces and two per side face. Sweeping the neutral-axis depth $c$ and summing $C_c = \alpha_1\phi_cf_c'ba$ with $F_{si} = \phi_sA_{si}f_{si}$, $f_{si} = E_s(0.0035)(c-d_i)/c$ capped at $\pm f_y$, produces the factored interaction curve. Reading it at the two axial loads,$$M_r = 1699.0\ \text{kN}\cdot\text{m at } C_f = 1567.3\ \text{kN}, \qquad M_r = 1662.9\ \text{kN}\cdot\text{m at } C_f = 1387.7\ \text{kN}$$Both exceed their demands, the governing case being C with $1584.7/1699.0 = 0.93$ utilisation. The squash limit for a tied column, $P_{r,max} = 0.80[\alpha_1\phi_cf_c'(A_g - A_{st}) + \phi_sA_{st}f_y] = 10\,943\ \text{kN}$, is never approached.
3. Stability check. The sway magnification is already carried in the design moments: the stability index computed in Question 2 was $Q = 0.109$, so every sway moment was multiplied by $\delta_s = 1.122$ before the column was sized. Non-sway magnification within the member need not be added, because the column is bent in single curvature from $M = 0$ at the pin to the full end moment at B and the critical section is at the restrained end, where the second-order deflection is zero. Ties: 10M at the lesser of 16 bar diameters ($478\ \text{mm}$), 48 tie diameters ($542\ \text{mm}$) and the least column dimension ($800\ \text{mm}$) — use 10M ties at 400 mm, with every corner and alternate bar tied.
4. Part 2 — footing plan size. The base is a pin, so the footing sees axial load and horizontal shear but no moment. At service the axial load at A is $958.7\ \text{kN}$. Trying a $2.4\ \text{m}$ square footing $0.6\ \text{m}$ thick, its own weight is $2.4^2(0.6)(24) = 82.9\ \text{kN}$, so$$\boxed{q = \frac{958.7 + 82.9}{2.4^2} = 180.8\ \text{kPa} < q_a = 200\ \text{kPa}}$$For the structural design the factored soil reaction is $q_f = 1387.7/5.76 = 240.9\ \text{kPa}$, applied as a uniform upward pressure on the footing.
Spread footing at A. With a pinned base there is no moment to resist, so the pressure block is uniform and the footing is governed by minimum steel rather than flexure.
5. Two-way (punching) shear. With 75 mm cover and 20M bars, $d = 600 - 75 - 30 = 495\ \text{mm}$. The critical perimeter lies $d/2$ from the column face, $b_o = 4(800+495) = 5180\ \text{mm}$, and the shear crossing it is$$V_f = q_f\left[A - (0.8+0.495)^2\right] = 240.9(5.76 - 1.677) = 983.7\ \text{kN}$$The resistance takes the least of the three Clause 13.3.4.1 expressions; with $\beta_c = 1$ and $\alpha_s = 4$ the governing coefficient is $0.38$, so $v_c = 0.38\lambda\phi_c\sqrt{f_c'} = 1.353\ \text{MPa}$ and$$V_r = v_cb_od = 3468.9\ \text{kN} \gg V_f$$a very large margin, which is expected: a 600 mm footing under an 800 mm column has a short shear span and punching rarely governs.
6. One-way shear and flexure. The cantilever projection is $(2400-800)/2 = 800\ \text{mm}$. At $d$ from the face, $V_f = 240.9(2.4)(0.305) = 176.4\ \text{kN}$; with $d_v = 445.5\ \text{mm}$ and $\beta = 230/(1000+d_v) = 0.159$ for a member without stirrups, $V_r = \phi_c\beta\sqrt{f_c'}\,b\,d_v = 605.7\ \text{kN}$, so shear is satisfied without stirrups. The cantilever moment is $M_f = q_f b\ell^2/2 = 240.9(2.4)(0.8)^2/2 = 185.0\ \text{kN}\cdot\text{m}$, needing only about $1040\ \text{mm}^2$; but Clause 7.8.1 demands $A_{s,min} = 0.002A_g = 0.002(2400)(600) = 2880\ \text{mm}^2$, so$$\boxed{\text{use }10 - 20\text{M each way}\ (3000\ \text{mm}^2)\ \text{at the bottom of a }2.4 \times 2.4 \times 0.6\ \text{m footing}}$$Complete the detail with dowels matching the column bars, developed into the footing, and a shear key or roughened interface to transfer the $23.8\ \text{kN}$ horizontal reaction into the soil.