Question 9 of 9: Width of a square footing for a short-term factor of safety of 2

Question 9: Width of a square footing for a short-term factor of safety of 2 (24 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. A column load carried on a square footing founded 2.0 m below ground in a normally consolidated soil.

Find. The width B of the square footing at which the short-term (undrained) factor of safety is exactly 2, using Meyerhof's general bearing-capacity analysis.

Approach. Write the general bearing-capacity equation for the undrained case, in which phi_u = 0 makes N_γ = 0 and N_q = 1; note that the depth factor contains B, so the equation for B is implicit and must be iterated; solve it, adopt a practical width, and confirm the drained case does not govern.

1. Reduce the general equation to the undrained case. Meyerhof's general equation is $$q_u = cN_cF_{cs}F_{cd}F_{ci}+qN_qF_{qs}F_{qd}F_{qi} +\tfrac{1}{2}\gamma BN_{\gamma}F_{\gamma s}F_{\gamma d}F_{\gamma i}.$$ With $\phi_u = 0$ the factors are $N_c = 5.14$, $N_q = 1$ and $N_{\gamma} = 0$; the load is vertical so all inclination factors are unity; and $F_{qs} = 1+(B/L)\tan\phi = 1$ and $F_{qd} = 1$ for phi = 0. The whole third term disappears and the second reduces to the surcharge itself.
2. Evaluate the surcharge and the shape factor. The overburden removed at founding level is $$q = \gamma D_f = 19(2.0) = 38.0\ \text{kPa},$$ and for a square footing, B/L = 1, De Beer's shape factor is $$F_{cs} = 1+\left(\frac{B}{L}\right)\left(\frac{N_q}{N_c}\right) = 1+\frac{1}{5.14} = 1.195 .$$
3. Write the depth factor, which is where B enters. Hansen's depth factor for phi = 0 is $$F_{cd} = 1+0.4\left(\frac{D_f}{B}\right)\ \text{for } \frac{D_f}{B}\le 1, \qquad F_{cd} = 1+0.4\tan^{-1}\!\left(\frac{D_f}{B}\right)\ \text{for } \frac{D_f}{B}\gt 1,$$ with the inverse tangent in radians. Because B is unknown, this term makes the design equation implicit.
4. Form the design equation. Working in net terms, which is the convention for a footing that replaces soil it displaces, the net ultimate bearing capacity is $$q_{u(net)} = q_u - q = c_uN_cF_{cs}F_{cd} = 150(5.14)(1.195)F_{cd} = 921.0\,F_{cd} \ \text{kPa},$$ and the net applied pressure is $Q/B^{2}$. Setting the factor of safety to 2, $$FS = \frac{q_{u(net)}}{Q/B^{2}} = 2 \quad\Longrightarrow\quad B^{2} = \frac{2Q}{921.0\,F_{cd}(B)} = \frac{1200}{921.0\,F_{cd}(B)} .$$
5. Iterate. Starting from a trial B = 1.0 m:

   Iteration on the footing width

   Trial B (m)	Df/B	Fcd	qu(net) (kPa)	New B (m)
   1.000	2.000	1.4429	1328.8	0.950
   0.950	2.105	1.4507	1336.1	0.948
   0.948	2.110	1.4511	1336.5	0.948

   The iteration converges after two cycles to $$\boxed{B_{required} = 0.95\ \text{m}}$$
6. Adopt a practical width and confirm the factor of safety. Footings are dimensioned in round increments, so adopt a 1.0 m square footing. At B = 1.0 m, $F_{cd} = 1.443$, $q_{u(net)} = 1328.8$ kPa and the applied net pressure is $600/1.0^{2} = 600$ kPa, giving $$FS = \frac{1328.8}{600} = 2.21 \;\ge\; 2.0 \quad\text{(satisfactory).}$$ On a gross basis the same footing gives $q_u = 1328.8+38.0 = 1366.8$ kPa against a gross applied pressure of 638 kPa, i.e. a gross factor of safety of 2.14, so the conclusion does not depend on which convention is used.
7. Confirm that the long-term case does not govern. Once the clay drains, $c^{\prime} \approx 0$ and $\phi^{\prime} = 36^{\circ}$, for which $$N_q = e^{\pi\tan\phi^{\prime}}\tan^{2}\!\left(45+\frac{\phi^{\prime}}{2}\right) = 37.75, \qquad N_c = (N_q-1)\cot\phi^{\prime} = 50.59,$$ $$N_{\gamma} = (N_q-1)\tan(1.4\phi^{\prime}) = 44.43 .$$ With the corresponding shape and depth factors at B = 1.0 m the gross ultimate capacity is about 3464 kPa, so the drained net factor of safety is roughly 5.7. The short-term case governs, as the question implies.
8. Check the water table. The zone that carries the bearing-capacity mechanism extends about B below founding level, i.e. to 3.0 m depth, whereas the water table is stable at 7.0 to 8.0 m. No groundwater correction to q or to the self-weight term is required, and none has been applied.

A 0.95 m footing under a 600 kN column is small, which is exactly what a 150 kPa undrained strength implies: this is a stiff clay and capacity is generous. That makes it worth stating explicitly that the question asks only for bearing capacity. At an applied pressure of 600 kPa the immediate and consolidation settlement of a 1.0 m footing on this clay would need checking before the design could be issued, and on a stiff clay of this strength it is usually settlement or the structural design of the pad that fixes the final dimension.