16-Civ-B3 Geotechnical Design · May 2015

Question 6 of 9: Ultimate Bearing Capacity of a Square Footing on Sand beneath a Clay Cover

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada National Examinations, May 2015 — 98-Civ-B3 Geotechnical Design. Three hours, OPEN BOOK, any non-communicating calculator. Section A carries five discussion questions of 7 marks each (answer any four); Section B carries four design questions of 24 marks each (answer any three); the examinable total is 4 × 7 + 3 × 24 = 100 marks. All nine questions are worked below, because the set is a study resource rather than a timed attempt.

Reference texts (98-Civ-B3 / 16-Civ-B3 Geotechnical Design).

B. M. Das, Principles of Foundation Engineering, 9th ed. — bearing capacity (Ch. 3), stress increase in a soil mass (Ch. 6), retaining walls (Ch. 8), pile foundations (Ch. 11), subsurface exploration (Ch. 2).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — consolidation (Ch. 11), shear strength (Ch. 12), lateral earth pressure (Ch. 13).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for site investigation, bearing resistance, deep foundations and earth-retaining structures.
R. F. Craig, Craig's Soil Mechanics, 9th ed. — earth pressure theory, effective stress and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 3rd ed. — in-situ testing, settlement of shallow and deep foundations.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay, pile-group behaviour, load testing.

Sources of design charts and assumed values (page-1 Note 6). Note 6 of this paper requires the candidate to identify the source of every design chart used and of every value assumed in the absence of data. They are named where used and collected here:

Q6 — bearing capacity factors $N_c$, $N_q$, $N_{\gamma}$ from Das, Principles of Foundation Engineering, Table 3.3 (Prandtl–Reissner $N_q$, Vesic $N_{\gamma}=2(N_q+1)\tan\phi'$); shape factors after De Beer (1970) and depth factors after Hansen (1970), Das Table 3.4. Assumed: unit weight of water $\gamma_w = 9.81$ kN/m$^3$; general shear failure; the sand extends at least $2B$ below the base.
Q7 — adhesion factor $\alpha = 1.0$ for soft clay ($c_u \le 50$ kPa) from NAVFAC DM-7.2 Fig. 1 and Tomlinson & Woodward Table 4.6; $\lambda = 0.24$ at an embedded length of 12 m from Vijayvergiya & Focht (1972) as tabulated by Das, Table 11.7; bearing factor $N_c^{*}=9$ for $L/D \ge 4$ (Skempton). Assumed: pile spacing $s = 3d = 1.5$ m centre to centre, driven closed-end concrete piles, clay $\gamma_{sat} = 17$ kN/m$^3$ with the water table at ground level.
Q8 — compression index from the Skempton correlation $C_c = 0.009\,(LL-10)$, Das Principles of Geotechnical Engineering Eq. (11.42); $2{:}1$ stress distribution, Das Eq. (6.31); Boussinesq rectangular influence factor (Das Table 6.6) used as the cross-check.
Q9 — Coulomb active pressure coefficient, Das Principles of Foundation Engineering Eq. (8.13) with the wall friction angle prescribed by the question, $\delta = 0.6\phi' = 18^{\circ}$. Assumed: unit weight of reinforced concrete $\gamma_c = 24$ kN/m$^3$ (CSA A23.3 nominal); passive resistance in front of the toe neglected; the backfill surface is horizontal and carries no surcharge.

Question 6: Ultimate Bearing Capacity of a Square Footing on Sand beneath a Clay Cover (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.

Quantity	Symbol	Value
Footing plan dimensions (square)	$B = L$	1.5 m
Depth of footing base below ground	$D_f$	1.5 m
Depth of groundwater table below ground	$d_w$	0.5 m
Clay, 0 to 0.5 m (above water table)	$\gamma_{total}$	17 kN/m$^3$
Clay, 0.5 to 1.5 m (below water table)	$\gamma_{sat}$	19 kN/m$^3$
Clay undrained parameters	$c_u,\ \phi_u$	50 kPa, $0^{\circ}$
Sand (bearing stratum)	$\gamma_{sat}$	20 kN/m$^3$
Sand effective parameters	$c',\ \phi'$	2 kPa, $40^{\circ}$
Unit weight of water (assumed)	$\gamma_w$	9.81 kN/m$^3$

Find. The ultimate bearing capacity $q_u$ of the footing from the general (Meyerhof–Vesic) bearing capacity equation, and the corresponding net allowable bearing pressure at a factor of safety of 3.

[Figure not reproduced: Figure 1 (redrawn) — square footing bearing on the surface of the sand stratum at 1.5 m depth, with the water table 0.5 m below ground level. See the official exam paper.]

Approach. The footing bears on sand, so the analysis is a drained effective-stress one using $c'$ and $\phi'$ of the sand: the clay contributes only as surcharge, the effective overburden at base level supplies $q$, and because the water table stands above the base the submerged unit weight governs the $N_{\gamma}$ term.

Identify the bearing stratum and the governing drainage condition. The base of the footing sits on the top of the sand, so the failure surface develops entirely within the sand. Sand is free-draining, so no excess pore pressure is generated and the correct analysis is a drained one with $c' = 2$ kPa and $\phi' = 40^{\circ}$. The overlying clay is not part of the bearing stratum; its only role is to apply surcharge to the failure mechanism, and its undrained strength $c_u = 50$ kPa is not used in the bearing capacity calculation at all.
Effective surcharge at foundation level. The water table is 0.5 m below ground, so the top 0.5 m of clay is at $\gamma_{total}$ and the next 1.0 m is submerged: $$q = 0.5\,\gamma_{total} + 1.0\,(\gamma_{sat}-\gamma_w) = 0.5(17) + 1.0(19-9.81)$$ which gives $q = 8.50 + 9.19 = 17.69$ kPa. Note that this is the effective surcharge; using the total overburden of $0.5(17)+1.0(19) = 27.5$ kPa would overstate the surcharge term by more than 50 per cent.
Effective unit weight beneath the base. The water table is above the foundation level, so the whole of the failure wedge below the base is submerged and the unit weight entering the $N_{\gamma}$ term is $\gamma' = \gamma_{sat} - \gamma_w = 20 - 9.81 = 10.19$ kN/m$^3$.
Bearing capacity factors for the sand. With $\phi' = 40^{\circ}$, $$N_q = e^{\pi\tan\phi'}\tan^2\!\left(45+\tfrac{\phi'}{2}\right) = e^{2.6360}(2.1445)^2 = 64.20$$ and then $N_c = (N_q-1)\cot\phi' = 63.20/0.8391 = 75.31$ and $N_{\gamma} = 2(N_q+1)\tan\phi' = 2(65.20)(0.8391) = 109.41$. These are the values tabulated by Das, Principles of Foundation Engineering Table 3.3.
Shape factors (De Beer). For a square footing $B/L = 1$: $$F_{cs} = 1 + \frac{B}{L}\cdot\frac{N_q}{N_c} = 1 + \frac{64.20}{75.31} = 1.852$$ $$\begin{aligned} F_{qs} &= 1 + \frac{B}{L}\tan\phi' = 1 + 0.8391 = 1.839 \\ F_{\gamma s} &= 1 - 0.4\frac{B}{L} = 0.600 \end{aligned}$$ The shape factors raise the cohesion and surcharge terms substantially and cut the self-weight term, as they should for a square rather than a strip footing.
Depth factors (Hansen). Here $D_f/B = 1.5/1.5 = 1.0 \le 1$, so the simple form applies: $$F_{qd} = 1 + 2\tan\phi'(1-\sin\phi')^2\frac{D_f}{B} = 1 + 2(0.8391)(0.3572)^2(1.0) = 1.214$$ $$\begin{aligned} F_{cd} &= F_{qd} - \frac{1-F_{qd}}{N_c\tan\phi'} = 1.214 + \frac{0.214}{63.19} = 1.217 \\ F_{\gamma d} &= 1.000 \end{aligned}$$ The load is vertical and concentric, so all three inclination factors are unity.
Assemble the general bearing capacity equation. Substituting term by term, $$q_u = c'N_cF_{cs}F_{cd} + q\,N_qF_{qs}F_{qd} + \tfrac{1}{2}\gamma' B\,N_{\gamma}F_{\gamma s}F_{\gamma d}$$ gives a cohesion term of $2(75.31)(1.852)(1.217) = 339.7$ kPa, a surcharge term of $17.69(64.20)(1.839)(1.214) = 2535.7$ kPa, and a self-weight term of $0.5(10.19)(1.5)(109.41)(0.600)(1.000) = 501.7$ kPa. Hence $$\boxed{q_u = 339.7 + 2535.7 + 501.7 = 3377\ \text{kPa}}$$
Net ultimate and allowable bearing pressures. The net ultimate value is $q_{u(net)} = q_u - q = 3377 - 17.69 = 3359$ kPa, so at the usual factor of safety of 3 on net pressure the net allowable bearing pressure is $q_{all(net)} = 3359/3 = 1120$ kPa and the gross allowable pressure is $1120 + 17.69 \approx 1137$ kPa. A capacity of this order will never be reached in practice: a 1.5 m square footing on sand at anything approaching 1100 kPa would settle far beyond tolerance, so the design will be governed by settlement, not by bearing capacity.

Quantity	Value
Effective surcharge at base, $q$	17.69 kPa
Submerged unit weight of sand, $\gamma'$	10.19 kN/m$^3$
Bearing capacity factors $N_c$ / $N_q$ / $N_{\gamma}$	75.31 / 64.20 / 109.41
Cohesion term	339.7 kPa
Surcharge term	2535.7 kPa
Self-weight term	501.7 kPa
Ultimate bearing capacity, $q_u$	3377 kPa
Net ultimate, $q_{u(net)}$	3359 kPa
Net allowable at FS = 3	1120 kPa

Check — assumptions made, as the question requires.

General shear failure is assumed, appropriate to a dense sand with $\phi' = 40^{\circ}$; a loose sand would fail by local or punching shear and the reduced parameters $\phi'_{red} = \tan^{-1}(0.67\tan\phi')$ would be used instead.
The sand is assumed to extend at least $2B = 3.0$ m below the base, so the failure wedge is contained within it. If a soft layer lay within that depth the two-layer punching analysis of Das §3.11 would govern instead.
The load is vertical, concentric and static; there is no adjacent excavation, slope or neighbouring footing.
$\gamma_w = 9.81$ kN/m$^3$; the water table is taken as stable at 0.5 m depth. If it fell below a depth $B$ beneath the base the same calculation with $q = 25.5$ kPa and $\gamma = 20$ kN/m$^3$ gives $q_u = 4980$ kPa, so the submerged condition analysed here is the conservative and correct design case.
The clay cover is treated as surcharge only; no adhesion on the sides of the footing and no contribution from the clay's undrained strength has been credited.
Bearing capacity, shape and depth factors are from Das, Principles of Foundation Engineering 9th ed. Tables 3.3 and 3.4 (Prandtl–Reissner, Vesic, De Beer, Hansen), as required by page-1 Note 6.

Topic. The general (Meyerhof–Vesic) bearing capacity equation with shape and depth factors, and the effect of groundwater.

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

Why this works. The three terms are the superposition of three separate limit-equilibrium problems: a weightless soil with cohesion, a weightless soil with surcharge, and a cohesionless soil with self-weight. Superposition is not exact but is slightly conservative, which is why it has survived since Terzaghi. The water table enters in two independent places — through the effective surcharge above the base, and through the effective unit weight of the wedge below it — and here it does both, because it stands above the foundation level. Notice that the surcharge term dominates at 75 per cent of the total: on a high-friction sand, embedment buys far more capacity than width does.

Common pitfall. Using the total unit weight of the clay to compute $q$ when part of it is below the water table, and forgetting to submerge the sand in the $N_{\gamma}$ term. A second, subtler trap is to use $c_u = 50$ kPa somewhere in the calculation because it appears in the question: the clay's undrained strength is given only so that the candidate can demonstrate it is not relevant to a footing bearing on the sand beneath it.

Source. Das, Principles of Foundation Engineering 9th ed. §3.6–3.9 and Tables 3.3, 3.4; CFEM 4th ed. Ch. 10.

← Question 5 Next: Question 7 →