07-Str-B1 · May 2015

Question 6 of 9: Ultimate Bearing Capacity of a 1.5 m Square Footing on Sand Beneath 1.5 m of Clay

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations — May 2015 — 07-Str-B1 Geotechnical Design. Three-hour, OPEN-BOOK exam; any non-communicating calculator permitted (the candidate must record its make and model). Format: Section A carries five discussion questions of 7 marks each, of which any FOUR are to be answered; Section B carries four design problems of 24 marks each, of which any THREE are to be answered — a marked total of 100. The paper instructs candidates to state any interpretive assumptions, to identify the source of every design chart and assumed value, and to exercise sound engineering judgment where data are absent. All nine printed questions are worked below, because the set is intended as a study resource.

Reference texts: Das, B.M., Principles of Foundation Engineering (9th ed., Cengage) — general bearing-capacity equation, pile and pile-group capacity, consolidation settlement of footings, retaining walls; Das, B.M., Principles of Geotechnical Engineering (9th ed., Cengage) — lateral earth pressure, effective stress, consolidation theory; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed., 2006) — Canadian practice for site investigation, SPT/CPT interpretation, pile design and tolerable settlement; Craig, R.F. / Knappett, J.A., Craig's Soil Mechanics (8th ed., CRC Press) — shear strength and earth-pressure theory; Duncan, J.M., Wright, S.G. & Brandon, T.L., Soil Strength and Slope Stability (2nd ed., Wiley) — fully softened and residual strengths for fissured and expansive clays; Fredlund, D.G., Rahardjo, H. & Fredlund, M.D., Unsaturated Soil Mechanics in Engineering Practice (Wiley) — swelling soils and matric suction.

Note — Figure 2 is printed over a coarse halftone. The soil-property annotations inside the photograph-style Figure 2 (Question 8) are printed over a coarse dot screen. The values used below are read from the printed figure and are: upper sand $\gamma = 15\ \text{kN/m}^3$ over 1.5 m, lower sand $\gamma_{sat} = 18\ \text{kN/m}^3$ over 1.5 m, normally consolidated clay 2.5 m thick with $w = 35\%$ and $LL = 48$, over sand; groundwater table at the underside of the footing.

Assumptions declared once, applied throughout. $\gamma_w = 9.81\ \text{kN/m}^3$; reinforced concrete $\gamma_c = 24\ \text{kN/m}^3$; specific gravity of soil solids $G_s = 2.70$ where a void ratio must be back-figured from water content; loads are vertical and concentric unless stated. Every assumption that changes a numerical answer is repeated in the question where it is used.

Question 6: Ultimate Bearing Capacity of a 1.5 m Square Footing on Sand Beneath 1.5 m of 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 1.5 m square footing founded at 1.5 m depth on the top of a sand stratum, with 1.5 m of clay above it and the groundwater table 0.5 m below ground surface (Figure 1).

Given data — Question 6
Quantity	Symbol	Value
Footing width = length (square)	$B = L$	1.5 m
Founding depth (top of sand)	$D_f$	1.5 m
Depth of groundwater table below ground	$d_w$	0.5 m
Clay, moist unit weight (above GWT)	$\gamma_{total}$	17 kN/m³
Clay, saturated unit weight (below GWT)	$\gamma_{sat}$	19 kN/m³
Clay, undrained strength	$c_u,\ \phi_u$	50 kPa, 0
Sand, saturated unit weight	$\gamma_{sat}$	20 kN/m³
Sand, effective strength parameters	$c',\ \phi'$	2 kPa, 40°
Unit weight of water	$\gamma_w$	9.81 kN/m³

Find. The ultimate bearing capacity $q_u$ of the footing from the general (Meyerhof-form) bearing-capacity equation, stating all assumptions.

[Figure not reproduced: Figure 6.1 — Figure 1 redrawn to scale. The footing base sits at 1.5 m depth on the surface of the sand; the groundwater table is 1.0 m above the founding level, so the sand beneath the footing is fully submerged. See the official exam paper.]

Approach. Because the footing bears on the sand, the bearing-capacity factors and shape/depth factors are computed from the sand's $\phi' = 40^\circ$; the overlying clay contributes only as an effective surcharge at founding level, and the groundwater table, being above the base, requires the buoyant unit weight in the $N_\gamma$ term.

Assumptions (as the question requires). (i) The sand stratum extends at least $2B = 3$ m below the base, so a general shear failure develops entirely within the sand and no punching through to a weaker layer occurs. (ii) The sand is dense ($\phi' = 40^\circ$), so general shear rather than local shear applies and no $\phi'$ reduction is made. (iii) The load is vertical and concentric, so all inclination factors $F_{ci} = F_{qi} = F_{\gamma i} = 1$. (iv) The clay above founding level acts only as surcharge; its undrained strength $c_u = 50$ kPa is not mobilised in the bearing failure and is not used. (v) The base of the footing is at 1.5 m and the footing thickness is neglected in computing surcharge. (vi) $N_\gamma$ is taken in the Meyerhof form $2(N_q+1)\tan\phi'$, as tabulated by Das. (vii) The groundwater table is at its stated position for the design case; a fall would increase capacity, so this is the governing assumption.

Effective surcharge at founding level. Above the water table the clay acts at its moist unit weight, below it at its buoyant weight: $$q = d_w\,\gamma_{total} + (D_f - d_w)(\gamma_{sat} - \gamma_w) = 0.5(17) + 1.0(19 - 9.81)$$ $$q = 8.50 + 9.19 = \boxed{17.69\ \text{kPa}}$$
Bearing-capacity factors for the sand, $\phi' = 40^\circ$. Using the standard closed forms, $$N_q = e^{\pi\tan\phi'}\tan^2\!\left(45^\circ + \frac{\phi'}{2}\right) = e^{\pi(0.8391)}\tan^2(65^\circ) = 64.20$$ $$N_c = (N_q - 1)\cot\phi' = \frac{64.20 - 1}{0.8391} = 75.31 \qquad N_\gamma = 2(N_q + 1)\tan\phi' = 2(65.20)(0.8391) = 109.41$$ These agree with the tabulated values in Das for $\phi' = 40^\circ$.
Shape factors (De Beer), for a square footing $B/L = 1$. $$F_{cs} = 1 + \frac{B}{L}\cdot\frac{N_q}{N_c} = 1 + \frac{64.20}{75.31} = 1.852$$ $$F_{qs} = 1 + \frac{B}{L}\tan\phi' = 1 + 0.8391 = 1.839 \qquad F_{\gamma s} = 1 - 0.4\frac{B}{L} = 0.600$$
Depth factors (Hansen), noting $D_f/B = 1.5/1.5 = 1.0 \le 1$ so the direct (non-arctangent) forms apply: $$F_{qd} = 1 + 2\tan\phi'(1-\sin\phi')^2\frac{D_f}{B} = 1 + 2(0.8391)(1-0.6428)^2(1.0) = 1.214$$ $$F_{cd} = F_{qd} - \frac{1 - F_{qd}}{N_c\tan\phi'} = 1.214 + \frac{0.214}{75.31(0.8391)} = 1.218 \qquad F_{\gamma d} = 1.000$$
Unit weight for the $N_\gamma$ term. The water table lies 1.0 m above the base, so the entire failure wedge beneath the footing is submerged and the buoyant unit weight of the sand governs: $$\gamma' = \gamma_{sat} - \gamma_w = 20 - 9.81 = 10.19\ \text{kN/m}^3$$
Assemble the general bearing-capacity equation. With all inclination factors unity, $$q_u = c'N_cF_{cs}F_{cd} + q\,N_qF_{qs}F_{qd} + \tfrac{1}{2}\gamma' B\,N_\gamma F_{\gamma s}F_{\gamma d}$$ Cohesion term: $2(75.31)(1.852)(1.218) = 339.7$ kPa.
Surcharge term: $17.69(64.20)(1.839)(1.214) = 2535.6$ kPa.
Self-weight term: $0.5(10.19)(1.5)(109.41)(0.600)(1.000) = 501.7$ kPa.
Ultimate bearing capacity. Adding the three contributions, $$q_u = 339.7 + 2535.6 + 501.7 = \boxed{3377\ \text{kPa}}$$ The surcharge term supplies three-quarters of the capacity, which is the expected result for a footing embedded a full width into ground with a high $\phi'$.
Net and allowable values for design. Removing the surcharge that was there before construction gives the net ultimate capacity, and a factor of safety of 3 applied to the net value gives the gross allowable pressure: $$q_{u(net)} = q_u - q = 3377 - 17.7 = 3359\ \text{kPa} \qquad q_{all} = \frac{3359}{3} + 17.7 = 1137\ \text{kPa}$$ The corresponding ultimate column load is $Q_u = q_u B L = 3377(1.5)(1.5) = 7.60$ MN.

Two design comments belong with this number. First, an allowable pressure over 1000 kPa will never be reached in practice: settlement of a 1.5 m footing on sand will limit the working pressure to a few hundred kPa, so the serviceability limit state governs. Second, if the sand layer is thinner than about $2B$ and is underlain by the soft clay, a punching-shear check through the sand into the clay must be made and will give a much lower capacity.

Final results — Question 6
Quantity	Symbol	Value
Effective surcharge at founding level	$q$	17.69 kPa
Bearing-capacity factors ($\phi' = 40^\circ$)	$N_c,\ N_q,\ N_\gamma$	75.31, 64.20, 109.41
Shape factors	$F_{cs},\ F_{qs},\ F_{\gamma s}$	1.852, 1.839, 0.600
Depth factors	$F_{cd},\ F_{qd},\ F_{\gamma d}$	1.218, 1.214, 1.000
Cohesion / surcharge / self-weight terms	—	339.7 / 2535.6 / 501.7 kPa
Ultimate bearing capacity	$q_u$	3377 kPa
Net ultimate bearing capacity	$q_{u(net)}$	3359 kPa
Gross allowable bearing pressure (FS = 3 on net)	$q_{all}$	1137 kPa
Ultimate column load	$Q_u$	7.60 MN

Topic. The general bearing-capacity equation with shape, depth and groundwater corrections for a shallow square footing.

Key relations. $q_u = c'N_cF_{cs}F_{cd}F_{ci} + qN_qF_{qs}F_{qd}F_{qi} + \tfrac{1}{2}\gamma B N_\gamma F_{\gamma s}F_{\gamma d}F_{\gamma i}$; $N_q = e^{\pi\tan\phi'}\tan^2(45^\circ+\phi'/2)$; $N_c = (N_q-1)\cot\phi'$; $N_\gamma = 2(N_q+1)\tan\phi'$; $\gamma' = \gamma_{sat} - \gamma_w$ when the water table is at or above the base.

Why this works. Terzaghi's superposition splits the failure mechanism into three independent contributions — the cohesion on the slip surface, the surcharge that must be lifted as the wedges rotate, and the self-weight of the failing soil — each scaled by a factor derived from the same log-spiral geometry. Shape factors convert the plane-strain solution to a three-dimensional one; depth factors credit the shear resistance mobilised in the soil above founding level, which the plane mechanism ignores. Water enters only through effective stress: the surcharge is computed as an effective vertical stress, and the wedge weight is buoyant wherever it is below the water table.

Common pitfall. Using the clay's $c_u = 50$ kPa in the cohesion term because it is the larger cohesion on the page. The failure surface forms in the bearing stratum, so the sand's $c' = 2$ kPa is correct and the clay appears only as surcharge. The second classic slip is using the total unit weight of the sand in the $N_\gamma$ term, which here would overstate that term by 96 % and $q_u$ by about 14 %.

Source. Das, Principles of Foundation Engineering, 9th ed., §3.6–§3.9 (general bearing-capacity equation, shape, depth and inclination factors; effect of the water table); CFEM 4th ed., §10.

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