Question 6: Width of a square footing on normally consolidated clay (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 square footing carries a column load Q = 300 kN at a founding depth D_f = 2.0 m in a normally consolidated clay, and two sets of strength parameters are supplied — one for each drainage condition.

Find. The square footing width B that delivers a gross factor of safety of 2.5 against bearing failure in the short term, the width that delivers the same factor of safety in the long term, and the dimension to be recommended for construction.

Approach. Apply Meyerhofʹs general bearing-capacity equation twice on the same geometry — once with the undrained parameters and once with the drained parameters — and in each case solve for the width at which the ultimate capacity is 2.5 times the applied gross pressure, noting that B appears on both sides of the equation so the solution is iterative.

1. Write the governing equation and confirm which terms survive. Meyerhofʹs general expression, with shape, depth and inclination factors on each term, is $$q_u = c\,N_c F_{cs}F_{cd}F_{ci} + q\,N_q F_{qs}F_{qd}F_{qi} + \tfrac{1}{2}\gamma B\,N_{\gamma}F_{\gamma s}F_{\gamma d}F_{\gamma i}$$ The column load is vertical and concentric, so all three inclination factors are unity throughout. The surcharge at founding level is $$q = \gamma D_f = 19(2.0) = 38.0\ \text{kPa}$$ and because the water table sits at 8.0 m while the zone that governs bearing capacity extends only from D_f to about D_f + B ≈ 3 m, no unit-weight correction for groundwater is required — the 8.0 m datum is given precisely so that this can be checked and dismissed.
2. Assemble the short-term (undrained) case. With φ_u = 0 the factors are N_c = 5.14, N_q = 1 and N_γ = 0, so the width term vanishes altogether and only the cohesion and surcharge terms remain. For a square footing F_cs = 1 + (B/L)(N_q/N_c) = 1 + 1/5.14 = 1.195 and F_qs = 1 + (B/L)tanφ = 1, while the depth factors for φ = 0 are F_qd = 1 and $$F_{cd} = 1 + 0.4\tan^{-1}\!\left(\frac{D_f}{B}\right)\ \text{(radians, since } D_f/B > 1)$$ so that $$q_u = 150(5.14)(1.195)F_{cd} + 38.0 = 921.1\,F_{cd} + 38.0$$ Note that B now enters only through F_cd, which is why the undrained capacity is almost independent of width and the whole variation comes from the applied pressure.
3. Solve the short-term case for B. The gross applied pressure at founding level is q_applied = Q/B² + γD_f, so the requirement is $$\frac{q_u(B)}{\dfrac{300}{B^{2}} + 38.0} = 2.5$$ Iterating gives F_cd = 1.482 and q_u = 1403 kPa at $$\boxed{B_{\text{short-term}} = 0.757\ \text{m}}$$ a strikingly small footing, which is simply a consequence of a very stiff clay: at c_u = 150 kPa the cohesion term alone supplies over 900 kPa of capacity.
4. Assemble the long-term (drained) case. Now cʹ = 0.5 kPa and φʹ = 28°, giving $$\begin{aligned} N_q &= e^{\pi\tan\varphi^{\prime}}\tan^{2}\!\left(45^{\circ} + \frac{\varphi^{\prime}}{2}\right) = 14.72 \\ N_c &= (N_q - 1)\cot\varphi^{\prime} = 25.80 \\ N_{\gamma} &= 2(N_q + 1)\tan\varphi^{\prime} = 16.72 \end{aligned}$$ with shape factors F_cs = 1 + 14.72/25.80 = 1.570, F_qs = 1 + tan28° = 1.532 and F_γs = 1 − 0.4 = 0.6, and depth factors $$\begin{aligned} F_{qd} &= 1 + 2\tan\varphi^{\prime}(1 - \sin\varphi^{\prime})^{2}\tan^{-1}\!\left(\frac{D_f}{B}\right) \\ F_{cd} &= F_{qd} - \frac{1 - F_{qd}}{N_c\tan\varphi^{\prime}} \\ F_{\gamma d} &= 1 \end{aligned}$$ Because cʹ is essentially zero, the drained capacity comes almost entirely from the surcharge term — which is exactly why embedment matters so much more than width in this case.
5. Solve the long-term case, and identify which governs. Imposing the same condition on the drained expression gives $$\boxed{B_{\text{long-term}} = 0.800\ \text{m}}$$ The long-term requirement is the larger of the two, so the drained condition governs the design. That ordering is worth pausing over, because it is the opposite of the usual expectation for a clay: a normally consolidated clay ordinarily gains strength as it consolidates, and the short term is critical. Here it is not, because c_u = 150 kPa is very high relative to cʹ = 0.5 kPa — the given parameter set describes a clay whose drained cohesion is negligible, so the long-term capacity has to be built almost entirely from friction on a modest surcharge.
6. Check the validity of the shallow-foundation idealisation. At B = 0.800 m the embedment ratio is D_f/B = 2.0/0.800 = 2.50. Meyerhofʹs general equation is a shallow-foundation formulation, and at an embedment ratio of 2.5 it is at the edge of its calibration — the failure surface is becoming a punching mechanism rather than a general shear surface, and the depth factors are doing most of the work. This is a reason to adopt a wider footing than the calculation strictly demands.
7. Recommend a practical dimension. A footing 0.80 m square is not buildable for a 300 kN column: the column itself, its reinforcement cage, cover and the construction tolerance on a 2 m deep excavation all argue for a minimum plan dimension of about 1 m. Adopting B = L = 1.0 m gives an applied gross pressure of q = 300/1.0² + 38.0 = 338 kPa against $$q_{u,\text{short}} = 1367\ \text{kPa}, \qquad q_{u,\text{long}} = 1263\ \text{kPa}$$ and therefore $$\boxed{FS_{\text{short}} = 4.04, \qquad FS_{\text{long}} = 3.74}$$ Both comfortably exceed 2.5, the embedment ratio falls to a sounder 2.0, and the extra 0.2 m of width costs almost nothing.

The recommendation is therefore a 1.0 m × 1.0 m square footing at 2.0 m depth. It is worth adding that at these pressures on a stiff clay the bearing check is unlikely to be the binding constraint: settlement should be verified, and a footing this small on a clay with c_u = 150 kPa will settle only a few millimetres, so bearing capacity genuinely does govern here — the reverse of the usual situation in this subject, and a point worth stating rather than assuming.