16-Civ-B3 Geotechnical Design · December 2014

Question 9 of 10: Width of a square footing for a short-term factor of safety of 2

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada National Examinations, December 2014 — 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 five design questions of 24 marks each (answer any three); the examinable total is 4 × 7 + 3 × 24 = 100 marks. All ten 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).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — lateral earth pressure (Ch. 13), shear strength (Ch. 12), slope stability (Ch. 15), subsurface exploration (Ch. 17).
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, 8th ed. — earth pressure theory and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 2nd ed. — in-situ testing and SPT correlations.
J. E. Bowles, Foundation Analysis and Design, 5th ed. — bearing capacity factors and retaining-wall stability tables.

Sources of charts and assumed values (page-1 Note 6). Note 6 of this paper requires the candidate to identify the source of every design chart and every assumed value. Each chart reading and each assumption below is therefore named where it is used, and the values assumed in the absence of data are collected here:

Q6 — adhesion factor α from Das, Principles of Foundation Engineering, Table 11.6 (Terzaghi, Peck & Mesri form, α against $c_u/p_a$); $\lambda$ from Vijayvergiya & Focht (1972) as tabulated by Das, Table 11.7.
Q7 — overburden correction $C_N$ from Liao & Whitman (1986); $\phi'$ from Wolff (1989) and from Hatanaka & Uchida (1996), both reproduced in Das, Ch. 2; settlement-controlled bearing pressure from Meyerhof (1965) as given by Das, Ch. 5, used only as a serviceability check because the question forbids direct correlations of bearing capacity to penetration index. Table I prints the blow counts as field values $N_f$; with no hammer data they are converted as $N_{60} = N_f$, i.e. a safety hammer at the reference 60 per cent energy ratio with borehole, sampler and rod-length factors of 1 (Das, Ch. 2, hammer-efficiency and correction-factor tables).
Q8 — embankment influence factor from Osterberg (1957), reproduced as Das Fig. 6.24; the closed form of that chart is used so the reading carries no chart-scaling error.
Q9 — Meyerhof general bearing-capacity equation with the shape factors of De Beer (1970) and the depth factors of Hansen (1970), as set out in Das, Ch. 3.
Q10 — Coulomb active earth-pressure coefficient, Das Eq. 13.31; unit weight of the mass-concrete wall assumed $\gamma_c = 24\ \text{kN/m}^3$ (CFEM 4th ed., normal-density concrete), the only value the figure does not supply.

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 square footing at 2.0 m depth in a normally consolidated clay, carrying a 600 kN column load, with the water table at 7.0 to 8.0 m and therefore well below the failure zone.

Given data — Question 9
Quantity	Symbol	Value
Column load	$Q$	600 kN
Founding depth	$D_f$	2.0 m
Footing shape	—	square, $B = L$
Unit weight	$\gamma$	19 kN/m³
Undrained parameters	$c_u$, $\phi_u$	150 kPa, 0
Drained parameters	$c'$, $\phi'$	0.5 kPa, 36°
Groundwater table	—	7.0 to 8.0 m depth
Required short-term factor of safety	$FS$	2

Find. The footing width $B$ such that the short-term (undrained) factor of safety on net bearing capacity is exactly 2, together with a confirmation that the long-term drained case does not govern.

Figure 9.1 — Square footing at 2.0 m depth in normally consolidated clay. The water table at 7.0 to 8.0 m lies more than three footing widths below founding level and therefore does not enter the bearing-capacity calculation.

Approach. Short-term means undrained, so the analysis uses $c = c_u = 150\ \text{kPa}$ with $\phi = 0$. Meyerhof's general equation then reduces to the cohesion term plus the surcharge term, with $N_c = 5.14$ and $N_q = 1$; because the depth factor depends on $D_f/B$ the equation is implicit in $B$ and is solved by iteration.

Set up the undrained bearing-capacity factors. For $\phi_u = 0$ the Prandtl solution gives $$N_c = \pi + 2 = 5.14, \qquad N_q = 1, \qquad N_\gamma = 0$$ so the general equation reduces to $$q_u = c_u N_c F_{cs} F_{cd} + q N_q F_{qs} F_{qd}$$ with $q = \gamma D_f = 19(2.0) = 38\ \text{kPa}$ and, for $\phi = 0$, $F_{qs} = F_{qd} = 1$.
Shape factor. For a square footing $B/L = 1$ and the De Beer shape factor for the cohesion term is $$F_{cs} = 1 + \frac{B}{L}\cdot\frac{N_q}{N_c} = 1 + \frac{1}{5.14} = 1.1946$$ This value does not change during the iteration because the footing stays square.
Depth factor. Anticipating a footing narrower than it is deep, the Hansen depth factor for $\phi = 0$ and $D_f/B > 1$ is $$F_{cd} = 1 + 0.4\tan^{-1}\!\left(\frac{D_f}{B}\right) \quad \text{(radians)}$$ and it is this term that makes the equation implicit: a narrower footing sits relatively deeper and gains capacity.
Write the design condition. The applied net pressure is the column load spread over the base, and the net ultimate resistance is the gross value less the surcharge that was there before construction: $$q_{u(net)} = c_u N_c F_{cs} F_{cd}, \qquad q_{net} = \frac{Q}{B^2}, \qquad FS = \frac{q_{u(net)}}{q_{net}} = 2$$ so that the required width satisfies $$B^2 = \frac{2 Q}{c_u N_c F_{cs} F_{cd}}$$
Iterate. Starting from a trial $B = 1.0\ \text{m}$: $D_f/B = 2.0$, so $F_{cd} = 1 + 0.4\tan^{-1}(2.0) = 1 + 0.4(1.1071) = 1.4429$ and $$q_{u(net)} = 150(5.14)(1.1946)(1.4429) = 1329\ \text{kPa}$$ $$B = \sqrt{\frac{2(600)}{1329}} = 0.950\ \text{m}$$ Repeating with $B = 0.950\ \text{m}$: $D_f/B = 2.105$, $F_{cd} = 1.4513$, $q_{u(net)} = 1337\ \text{kPa}$ and $B = 0.947\ \text{m}$. A third cycle does not move the answer, so $$\boxed{B_{required} = 0.95\ \text{m}}$$
Round to a constructible size and confirm. Adopting a square footing $B = 1.0\ \text{m}$, the applied net pressure is $q_{net} = 600/1.0^2 = 600\ \text{kPa}$ against a net ultimate resistance of 1329 kPa, so $$\boxed{FS = \frac{1329}{600} = 2.21 > 2 \quad \text{(adopt 1.0 m square)}}$$ The gross ultimate bearing capacity at that width is $q_u = 1329 + 38 = 1367\ \text{kPa}$.
Check the long-term drained case. A normally consolidated clay is weakest in the short term, but the check must be made because the question supplies drained parameters. With $c' = 0.5\ \text{kPa}$ and $\phi' = 36^\circ$, $N_c = 50.59$, $N_q = 37.75$ and $N_\gamma = 56.31$; the shape factors become $F_{cs} = 1.746$, $F_{qs} = 1.727$, $F_{\gamma s} = 0.6$, and with $D_f/B = 2.0$ the depth factors are $F_{qd} = 1.273$ and $F_{cd} = 1.281$. Substituting at $B = 1.0\ \text{m}$ gives $q_u = 56.6 + 3154.0 + 321.0 = 3531\ \text{kPa}$, hence $q_{u(net)} = 3493\ \text{kPa}$ and a long-term factor of safety of $3493/600 = 5.8$. The undrained case governs, as expected.
Note what the answer does not cover. A 1.0 m square footing at 2.0 m depth is unusually small for a 600 kN load; the reason is the very high undrained strength of 150 kPa, which is that of a hard rather than a soft clay. In practice such a footing would be sized by the structural requirements of the column base and by the settlement check, and the 600 kPa contact pressure would be reviewed against the preconsolidation pressure of the deposit before being accepted — a normally consolidated clay with $c_u = 150\ \text{kPa}$ and the usual ratio $c_u/\sigma'_v \approx 0.25$ implies an effective overburden of the order of 600 kPa at that level, which is irreconcilable with a founding depth of 2 m and is flagged below.

Final results — Question 9
Quantity	Symbol	Value
Undrained bearing-capacity factor	$N_c$	5.14
Shape factor (square)	$F_{cs}$	1.1946
Depth factor at the converged width	$F_{cd}$	1.451
Net ultimate bearing capacity	$q_{u(net)}$	1337 kPa
Required footing width for FS = 2	$B$	0.95 m square
Adopted (constructible) width	$B$	1.0 m square
Factor of safety at the adopted width	$FS$	2.21
Long-term drained factor of safety at $B = 1.0$ m	$FS_{LT}$	5.8 (does not govern)

Check: the stated soil is described as normally consolidated yet is given $c_u = 150\ \text{kPa}$ and $\phi' = 36^\circ$, a combination that belongs to a hard, heavily overconsolidated clay or a clay till rather than to a normally consolidated deposit at 2 m depth. The calculation above answers the question exactly as set, using the undrained parameters supplied. If the deposit is genuinely normally consolidated the governing check would instead be consolidation settlement under a 600 kPa contact pressure, which would demand a very much larger footing; page-1 Note 1 of the paper invites the candidate to record precisely this kind of assumption. The soil weight above the base and the self-weight of the footing have been taken as equal to the excavated soil, which is why the net formulation is used.

Topic: undrained (short-term) bearing capacity of a square footing on clay by Meyerhof's general equation.

Key relations: $q_u = c N_c F_{cs} F_{cd} + q N_q F_{qs} F_{qd} + \tfrac{1}{2}\gamma B N_\gamma F_{\gamma s} F_{\gamma d}$; for $\phi = 0$: $N_c = 5.14$, $N_q = 1$, $N_\gamma = 0$; $F_{cs} = 1 + (B/L)(N_q/N_c)$; $F_{cd} = 1 + 0.4\tan^{-1}(D_f/B)$ for $D_f/B > 1$; $FS = q_{u(net)}/(Q/B^2)$.

Why this works: immediately after loading, a saturated clay cannot change volume, so the strength available is the undrained strength and the friction angle is zero. That collapses the general equation to a cohesion term and a surcharge term, and because $N_\gamma = 0$ the footing width enters only through the depth factor — which is why the solution converges in two iterations rather than needing a full numerical root find.

Common pitfall: mixing gross and net quantities. If the gross ultimate capacity is divided by the gross applied pressure the factor of safety is inflated, because the surcharge appears on both sides. The net formulation compares only what the construction has actually added; here it moves the required width from 0.93 m to 0.95 m, a small difference on this problem but a large one on a shallow, lightly loaded footing.

Source: B. M. Das, Principles of Foundation Engineering, 9th ed., Ch. 3 — general bearing-capacity equation, shape factors after De Beer (1970), depth factors after Hansen (1970); CFEM 4th ed., Ch. 10.

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