Question 6 of 7: Rectangular footing with a 4 : 1 plan ratio under member A–B–C, and the column-to-pier connection

Question B3: Rectangular footing with a 4 : 1 plan ratio under member A–B–C, 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. Plan dimensions in the 4:1 ratio and a thickness that keep the factored bearing pressure inside $q_r$ with the base in full contact, the flexural and shear reinforcement, and a detailed dowel connection from the column into the square pier.

Approach. Establish the factored bearing resistance, add the pier and footing self weight, transfer the column moment and shear down to founding level, size the plan from the trapezoidal pressure distribution, then design the footing for flexure, one-way shear and punching, check sliding, and develop the column bars into the pier.

1. Convert the bearing capacity into a factored resistance. The paper gives an ultimate capacity, which must be reduced before it can be compared with factored pressures. Canadian limit states practice uses a geotechnical resistance factor of 0.5 for bearing: $$q_r = \phi_g q_{ult} = 0.5(200) = \boxed{100\ \text{kPa}}$$ This is a weak founding stratum for a frame delivering 1621 kN and 1800 kN·m — a fact that dominates the answer.
2. Choose a pier and a trial footing. Interpose a 1500 mm square pier, 1.0 m high, between the 1200 mm column and the footing so the column bars can be developed and the footing top can sit below grade. Try a footing $L = 15.0\ \text{m}$ by $B = L/4 = 3.75\ \text{m}$, thickness $t = 1.2\ \text{m}$, with the long dimension in the plane of the moment — which is the only sensible orientation for a 4:1 rectangle carrying a large uniaxial moment.
3. Accumulate the vertical load at founding level. $$\begin{aligned} A &= 15.0 \times 3.75 = 56.25\ \text{m}^2 \\ W_{footing} &= 1.25(56.25 \times 1.2 \times 24) = 2025\ \text{kN} \\ W_{pier} &= 1.25(1.5^2 \times 1.0 \times 24) = 67.5\ \text{kN} \\ N_f &= 1621 + 2025 + 67.5 = 3713\ \text{kN} \end{aligned}$$ The footing weighs more than the column brings down, which is what a low bearing resistance forces.
4. Transfer the moment down to the underside. The 675 kN thrust acts at A, which is $1.0 + 1.2 = 2.2\ \text{m}$ above the founding plane, so it adds its own moment: $$M_f = 1800 + 675(2.2) = 1800 + 1485 = \boxed{3285\ \text{kN}\cdot\text{m}}$$ Omitting this term would understate the design moment by 45 per cent.
5. Check that the base stays in full contact. $$\begin{aligned} e &= \frac{M_f}{N_f} = \frac{3285}{3713} = 0.885\ \text{m} \\ \frac{L}{6} &= \frac{15.0}{6} = 2.50\ \text{m} \end{aligned}$$ Since $e \le L/6$ the whole base bears and the pressure varies linearly. Had $e$ exceeded $L/6$ the heel would lift and the pressure diagram would become triangular over a reduced length.
6. Evaluate the bearing pressures. $$q = \frac{N_f}{A}\left(1 \pm \frac{6e}{L}\right) = \frac{3713}{56.25}\left(1 \pm \frac{6(0.885)}{15.0}\right) = 66.0(1 \pm 0.354)$$ $$\boxed{\begin{aligned} q_{max} &= 89.4\ \text{kPa} \le q_r = 100\ \text{kPa} \\ q_{min} &= 42.7\ \text{kPa} > 0 \end{aligned}}$$ A working-stress cross-check at $q_{allow} = q_{ult}/3 = 67\ \text{kPa}$ under service loads gives the same plan size, so the two conventions agree here.
7. Work with the net pressure for the structural design. The footing cannot bend under its own weight, because that weight is carried directly by the soil beneath it. Deducting $1.25(1.2 \times 24) = 36.0\ \text{kPa}$ leaves the net design pressure varying from $6.65\ \text{kPa}$ at the toe to $53.4\ \text{kPa}$ at the heel.
8. Design for flexure about the pier face. The cantilever length is $(15.0 - 1.5)/2 = 6.75\ \text{m}$, and the net pressure at the critical face is $32.35\ \text{kPa}$ rising to $53.4\ \text{kPa}$ at the heel. Treating the trapezoid as a uniform part plus a triangular part, $$M_f = \left[\frac{q_1 c^2}{2} + \frac{(q_2 - q_1)c^2}{3}\right] B = \left[\frac{32.35(6.75)^2}{2} + \frac{21.02(6.75)^2}{3}\right] 3.75 = \boxed{3961\ \text{kN}\cdot\text{m}}$$
9. Provide the bottom steel. With 75 mm cover to a 30M bar, $d = 1200 - 75 - 12.5 = 1112.5\ \text{mm}$. Trying 24–30M ($A_s = 16\,800\ \text{mm}^2$) across the 3.75 m width, $$\begin{aligned} a &= \frac{0.85(16\,800)(400)}{0.7975(0.65)(25)(3750)} = 115.4\ \text{mm} \\ M_r &= 0.85(16\,800)(400)(1112.5 - 57.7) = 6025\ \text{kN}\cdot\text{m} \end{aligned}$$ giving a utilisation of $3961/6025 = 0.657$. The shrinkage and temperature minimum, $0.002A_g = 9000\ \text{mm}^2$, is comfortably exceeded. Place the bars at 155 mm centres in the long direction and provide 20–25M in the short direction, which is minimum-governed.
10. Check one-way shear at $d$ from the pier face. Without transverse reinforcement the general method gives $\beta = 230/(1000 + d_v)$ with $d_v = \max(0.9d,\ 0.72t) = 1001\ \text{mm}$, so $\beta = 0.1149$: $$V_r = \phi_c \beta \sqrt{f'_c}\, b d_v = 0.65(0.1149)\sqrt{25}\,(3750)(1001) = 1402\ \text{kN}$$ against a demand of $943\ \text{kN}$ from the net pressure outboard of the critical section, a ratio of $\boxed{0.672}$. Thick footings pay a size-effect penalty through $\beta$, which is why the check is not trivial despite the modest pressures.
11. Check punching around the pier. The critical perimeter is at $d/2$ from the pier face, so $b_o = 4(1500 + 1112.5) = 10\,450\ \text{mm}$. For a square loaded area, $\beta_c = 1.0$ and $$v_c = \min\left[\phi_c\left(1 + \frac{2}{\beta_c}\right)0.19\sqrt{f'_c},\ \phi_c(0.38)\sqrt{f'_c}\right] = 1.235\ \text{MPa}$$ so $V_r = 1.235(10\,450)(1112.5)/10^3 = 14\,358\ \text{kN}$ against about 3349 kN acting outside the perimeter — a ratio of 0.233. A footing this thick is never punching-critical.
12. Check sliding. The 675 kN thrust must be resisted at the base. With a conservative friction coefficient of 0.5 on the founding plane, $$\frac{V_f}{\mu N_f} = \frac{675}{0.5(3713)} = 0.364\ \ \checkmark$$ Passive resistance on the 1.2 m buried face is available as reserve and has been ignored.
13. Detail the column-to-pier connection. Every column bar must be able to deliver its full tensile force into the pier, because the joint sees the same moment as the column base plus the thrust over the pier height, $1800 + 675(1.0) = 2475\ \text{kN}\cdot\text{m}$. Match the column reinforcement with 32–30M dowels cast into the pier and footing in the same pattern as the column bars. In 25 MPa concrete the straight tension development length for a 30M bar is $$l_d = \frac{0.45 k_1k_2k_3k_4 f_y d_b}{\sqrt{f'_c}} = \frac{0.45(400)(29.9)}{\sqrt{25}} = 1076\ \text{mm}$$ and the compression development length is 574 mm. Project the dowels 1100 mm above the pier top and lap them with the column bars over a Class B tension lap; hook the lower ends into the footing bottom mat so the 1076 mm can be accommodated within the 1.2 m thickness. Confine the lap zone with 10M ties at 150 mm, and roughen and clean the construction joint at the pier top to at least 5 mm amplitude so the 675 kN horizontal shear transfers by shear friction across it.

Final Results