Question 1 of 6: Shallow Foundations — square footing under biaxial moment
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, December 2019 — 07-Str-B5 Foundation Engineering (catalogued here under 16-Civ-B19). Three hours, open book (any non-communicating calculator permitted). Six questions are printed and each is of equal value (30 marks); the rubric states that five questions constitute a complete paper and that the first five appearing in the answer book will be marked. All six are solved below, because the set is intended as a study resource rather than a timed attempt.
Reference texts. B. M. Das, Principles of Foundation Engineering, 9th ed. (Cengage) — bearing capacity Ch. 3–4, elastic settlement Ch. 5, consolidation settlement Ch. 6, retaining walls Ch. 8, single piles Ch. 9, pile groups Ch. 10. R. F. Craig / J. A. Knappett, Craig's Soil Mechanics, 8th ed. — slope stability and rapid drawdown Ch. 12. D. W. Taylor, Fundamentals of Soil Mechanics (1948) — stability coefficients. Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Canadian limit-states practice, resistance factors and tolerable movements. Tomlinson & Woodward, Pile Design and Construction Practice, 6th ed. — equivalent-raft settlement.
Check — figure dimension. On the Question 4 drawing the 1.25 m dimension line terminates at the back (retained) face of the stem, not at its front face: the dimension line runs from the toe edge of the base to the stem's right face. The wall therefore has a 0.95 m toe and a 1.75 m heel, not a 1.25 m toe and a 1.45 m heel. The same drawing shows the ground line in front of the wall at the underside of the base, i.e. no fill covers the toe, so the bearing check below is taken with $D_f = 0$ and passive resistance in front of the toe is neglected. Both readings are stated again where they are used.
Question 1: Shallow Foundations — square footing under biaxial moment (30 marks)
Find. (a) the total (gross) factor of safety against bearing failure using the Brinch Hansen equation; (b) whether the elastic settlement stays within 30 mm, redesigning if it does not; and (c) whether the factored bearing resistance exceeds the factored load effect under 1.25 DL + 1.4 W.
Figure 1.1 — The biaxial eccentricity shifts the resultant off both axes, so the bearing check is made on the reduced Meyerhof area $B' \times L'$ (shaded) rather than on the full 1.8 m × 1.8 m plan.
Approach. Convert the two moments into eccentricities, reduce the footing to Meyerhof's effective area, evaluate the Brinch Hansen bearing capacity on that area with shape and depth factors, then repeat the exercise at factored loads for the ultimate limit state; settlement is checked separately with the elastic (theory-of-elasticity) formula for a rigid footing.
Part (a) — convert the moments to eccentricities and locate the resultant relative to the kern. A moment about the x-axis displaces the resultant along the L direction and a moment about the y-axis displaces it along B:
$$e_L=\frac{M_x}{V}=\frac{450}{1800}=0.250\ \text{m},\qquad e_B=\frac{M_y}{V}=\frac{360}{1800}=0.200\ \text{m}$$
Each is smaller than $B/6 = 1.8/6 = 0.300$ m on its own, but with moments about both axes the no-tension zone is the kern rhombus $e_B/B + e_L/L \le 1/6$. Here $0.200/1.8 + 0.250/1.8 = 0.250 \gt 0.167$, so the resultant lies outside it: the linear formula gives $\frac{1800}{3.24}\left(1-\frac{6(0.200)}{1.8}-\frac{6(0.250)}{1.8}\right) = -277.8$ kPa at the least-loaded corner, meaning that corner of the base lifts off. This does not invalidate the method used below — Meyerhof's effective area needs only the resultant to lie within the base ($e \lt B/2$), and it is precisely the device that handles partial contact.
Reduce the plan to the effective bearing area. Meyerhof's device replaces the actual footing carrying an eccentric load by a smaller, concentrically loaded footing whose centroid coincides with the resultant:
$$B'=B-2e_L=1.8-2(0.250)=1.300\ \text{m},\qquad L'=L-2e_B=1.8-2(0.200)=1.400\ \text{m}$$
$$A'=B'L'=1.300\times1.400=1.820\ \text{m}^2$$
Note that $B'$ is the smaller of the two reduced dimensions by definition, whichever physical direction it lies in.
Evaluate the Brinch Hansen bearing-capacity factors at $\phi'=36^\circ$. Hansen retains Reissner's $N_q$ and Prandtl's $N_c$ and adopts his own $N_\gamma$:
$$N_q=e^{\pi\tan\phi'}\tan^2\!\left(45+\tfrac{\phi'}{2}\right)=e^{\pi(0.7265)}\tan^2 63^\circ=9.804\times3.852=37.75$$
$$N_c=(N_q-1)\cot\phi'=\frac{36.75}{0.7265}=50.59,\qquad N_\gamma=1.5\,(N_q-1)\tan\phi'=1.5(36.75)(0.7265)=40.05$$
The Hansen $N_\gamma$ is deliberately the most conservative of the common expressions; Vesic's $2(N_q+1)\tan\phi'=56.3$ would raise the answer, so the formula named in the question must be the one used.
Compute the shape and depth factors. Hansen's shape factors take the effective aspect ratio $B'/L'=1.300/1.400=0.9286$, whereas the depth factors take the real width $B$ because the embedment is a physical property of the excavation:
$$s_c=1+\frac{N_q}{N_c}\frac{B'}{L'}=1+\frac{37.75}{50.59}(0.9286)=1.693,\qquad s_q=1+\frac{B'}{L'}\tan\phi'=1.675$$
$$s_\gamma=1-0.4\frac{B'}{L'}=0.629$$
With $D_f/B=1.8/1.8=1.0 \le 1$,
$$d_c=1+0.4\frac{D_f}{B}=1.400,\qquad d_q=1+2\tan\phi'(1-\sin\phi')^2\frac{D_f}{B}=1+1.453(0.1699)=1.247,\qquad d_\gamma=1.0$$
The load is vertical — the moments are carried by the eccentricity, not by a horizontal thrust — so every inclination factor is unity.
Assemble the ultimate bearing capacity. The surcharge at founding level is $q=\gamma D_f=18.0(1.8)=32.4$ kPa. The water table at 6.5 m lies below the depth $D_f+B'=1.8+1.3=3.1$ m over which the failure wedge develops, so no submergence correction is applied to either the $q$ term or the $\gamma$ term:
$$q_u=c'N_cs_cd_c+qN_qs_qd_q+\tfrac{1}{2}\gamma B'N_\gamma s_\gamma d_\gamma$$
$$q_u=(20)(50.59)(1.693)(1.400)+(32.4)(37.75)(1.675)(1.247)+\tfrac{1}{2}(18.0)(1.300)(40.05)(0.629)$$
$$q_u=2398+2554+295=\boxed{5247\ \text{kPa}}$$
The cohesion and surcharge terms each supply about 46 % of the capacity while the self-weight term supplies only 6 % — a direct consequence of the footing being narrow relative to its embedment.
Form the total factor of safety. The applied gross pressure on the effective area and the resulting capacity are
$$q_{app}=\frac{V}{A'}=\frac{1800}{1.820}=989.0\ \text{kPa},\qquad Q_{ult}=q_uA'=5247\times1.820=9549\ \text{kN}$$
$$FS=\frac{q_u}{q_{app}}=\frac{Q_{ult}}{V}=\frac{9549}{1800}=\boxed{5.30}$$
A factor of safety of 5.3 is generously above the conventional minimum of 3, which already hints that this footing will be governed by deformation rather than by collapse.
Part (b) — estimate the elastic settlement of the loaded area. A spread footing on a silty sand settles essentially immediately, so the drained modulus is used with the theory-of-elasticity expression. Treating the footing as rigid (its thickness and the reinforced-concrete pedestal make it far stiffer than the soil) the influence factor is $I_s\approx0.82$, and the settlement is computed on the same effective area used for bearing so that the two checks describe the same loading:
$$S_e=\frac{q_{app}\,B'\,(1-\nu^2)}{E_s}\,I_s=\frac{(989.0)(1.300)(1-0.3^2)}{40\,000}(0.82)$$
$$S_e=\boxed{24.0\ \text{mm}}\ \lt\ 30\ \text{mm}\quad\checkmark$$
Conclude on the settlement tolerance and on redesign. The predicted 24.0 mm is about 80 % of the 30 mm allowance, so the footing does satisfy the serviceability requirement and no redesign is required. Two conservatisms are worth naming because they widen the margin rather than narrow it: the embedment (Fox) correction $I_f$ for $D_f/B=1.0$ is of order 0.7–0.8 and has been omitted, and using the full 1.8 m × 1.8 m plan with the average pressure $q_{avg}=1800/3.24=555.6$ kPa gives only 18.7 mm. The design carried into part (c) is therefore the original 1.8 m × 1.8 m footing at $D_f=1.8$ m.
Part (c) — factor the loads and recompute the eccentricities. The axial load is dead and the moments are wind, so each is factored at its own value:
$$V_f=1.25(1800)=2250\ \text{kN},\qquad M_{x,f}=1.4(450)=630\ \text{kN}\cdot\text{m},\qquad M_{y,f}=1.4(360)=504\ \text{kN}\cdot\text{m}$$
Because the moments are factored more heavily than the load, the factored eccentricities grow:
$$e_L=\frac{630}{2250}=0.280\ \text{m},\qquad e_B=\frac{504}{2250}=0.224\ \text{m}$$
$$B'=1.8-2(0.280)=1.240\ \text{m},\qquad L'=1.8-2(0.224)=1.352\ \text{m},\qquad A'=1.677\ \text{m}^2$$
This is the essential difference between the working-stress check of part (a) and the limit-states check here: applying separate factors changes the shape of the loading, not merely its magnitude.
Recompute the bearing capacity on the smaller factored effective area. With $B'/L'=0.9172$ the shape factors become $s_c=1.685$, $s_q=1.666$, $s_\gamma=0.633$; the depth factors are unchanged because $D_f$ and $B$ are unchanged. Substituting,
$$q_u=2386+2542+283=5210\ \text{kPa}$$
The reduction from 5247 kPa is slight because the loss of area is partly offset by the larger shape factor on the dominant cohesion term.
Compare the factored resistance with the factored load effect. Applying the bearing resistance factor to the ultimate capacity and the factored load to the factored effective area,
$$q_r=f_{bc}\,q_u=0.5(5210)=2605\ \text{kPa},\qquad q_f=\frac{V_f}{B'L'}=\frac{2250}{1.677}=1342\ \text{kPa}$$
$$q_r=2605\ \text{kPa}\ \gt\ q_f=1342\ \text{kPa}\quad\Longrightarrow\quad \boxed{\text{ULS satisfied, } q_r/q_f=1.94}$$
Comment on the results. The footing carries roughly twice the factored bearing demand and settles to 80 % of its allowance, so the serviceability limit state, not the ultimate limit state, is the binding constraint — the classic signature of a small, deeply embedded footing on a competent granular soil. The two calculations are also mutually consistent: a working-stress factor of safety of 5.3 corresponds, once the ratio of factored to unfactored load (2250/1800 = 1.25) and the resistance factor of 0.5 are accounted for, to a limit-states ratio close to $5.30 \times 0.5 / 1.25 \approx 2.1$, against the 1.94 obtained — the small shortfall being the extra eccentricity produced by factoring wind more heavily than dead load. Practically, the resultant already lies outside the biaxial kern rhombus (0.250 at service load and 0.280 at factored load, against 1/6), so a corner of the base loses contact under the design wind. That is generally acceptable for a transient wind load, because the effective-area checks above already account for it, but any revision that enlarges the sign face should come with a larger footing rather than a re-check of the same one.
