Allowable bearing pressure of a 1 m square footing in sand: 16-Civ-B3 Geotechnical Design, December 2013, Q6

Question 6: Allowable bearing pressure of a 1 m square footing in sand (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	Value
Footing	square, $B \times B = 1.0 \times 1.0$ m
Embedment $D_f$ (Figure 3)	1.5 m
Unit weight $\gamma$ (question text)	20 kN/m$^3$
Friction angle $\phi$ (question text)	36°, $c = 0$ (coarse-grained)
Required factor of safety	$FS \ge 3$
Permissible settlement	25 mm
$E_s$, 0–2 m below the base	6000 kPa
$E_s$, 2–8 m below the base	12 000 kPa
$E_s$, 8–14 m below the base	10 000 kPa

Find. The maximum allowable gross bearing pressure the footing may carry when it must satisfy both a factor of safety of at least 3 against a bearing failure and a settlement of not more than 25 mm; then the reduction in ultimate capacity if the water table rises to the ground surface, and the effect of that rise on settlement.

Figure Q6 — the 1 m square footing at D_f = 1.5 m, the CPT modulus profile from Figure 3, and the Schmertmann strain-influence distribution that governs the settlement. Figure 3 measures the modulus profile from the underside of the footing, so the whole influence depth 2B = 2.0 m lies in the 6000 kPa band.

Approach. Compute the ultimate capacity from Terzaghi’s equation and divide by the factor of safety; separately, invert Schmertmann’s strain-influence method to find the net pressure that produces exactly 25 mm; the smaller of the two governs, and the water-table cases are then obtained by replacing $\gamma$ with $\gamma^{\prime}$.

Terzaghi bearing-capacity factors for $\phi = 36^{\circ}$. Terzaghi’s factors follow from $N_q = \dfrac{e^{2\pi(0.75-\phi/360)\tan\phi}}{2\cos^{2}(45^{\circ}+\phi/2)}$, with $N_{\gamma}$ taken from the tabulated Kumbhojkar (1993) values:$$\begin{aligned}N_q &= 47.16 \\ N_{\gamma} &= 54.36 \quad (c = 0, \text{ so } N_c \text{ is not needed}).\end{aligned}$$
Surcharge at founding level. With the water table more than 14 m down, the sand above the base is moist and the full unit weight applies:$$q = \gamma D_f = 20 \times 1.5 = 30.0\ \text{kPa}.$$
Ultimate bearing capacity, square footing. Terzaghi’s square -footing form is $q_u = 1.3\,c\,N_c + q\,N_q + 0.4\,\gamma\,B\,N_{\gamma}$; the cohesion term vanishes:$$q_u = 30.0(47.16) + 0.4(20)(1.0)(54.36) = 1414.7 + 434.9$$$$\boxed{q_u = 1849.6\ \text{kPa}}$$
Allowable pressure from the bearing-capacity criterion. Applying the factor of safety to the net capacity, which is the correct form because the overburden $q$ was there before the footing:$$q_{all(net)} = \frac{q_u - q}{FS} = \frac{1849.6 - 30.0}{3} = 606.5\ \text{kPa}, \quad \text{i.e. a gross pressure of } 636.5\ \text{kPa}.$$This is very large, as it always is for a small footing in dense sand, so settlement is expected to govern — which is exactly why the question sets a 25 mm limit.
Set up Schmertmann’s strain-influence method. Settlement is$$S_e = C_1 C_2\,\Delta q \sum \frac{I_z}{E_s}\,\Delta z ,$$where for a square (axisymmetric) footing $I_z = 0.1$ at the base, rises linearly to $I_{zp}$ at $z = B/2$ and falls linearly to zero at $z = 2B$. Here $B/2 = 0.5$ m and $2B = 2.0$ m below the base, i.e. between 1.5 m and 3.5 m below ground level.
Peak influence factor. The peak depends on the net pressure through $I_{zp} = 0.5 + 0.1\sqrt{\Delta q/\sigma^{\prime}_{zp}}$, in which $\sigma^{\prime}_{zp}$ is the effective overburden at the depth of the peak:$$\sigma^{\prime}_{zp} = \gamma\left(D_f + \tfrac{B}{2}\right) = 20(1.5 + 0.5) = 40.0\ \text{kPa}.$$Because $I_{zp}$ contains $\Delta q$, the calculation is iterative; converging on $\Delta q = 213.4$ kPa gives$$I_{zp} = 0.5 + 0.1\sqrt{213.4/40.0} = 0.731 .$$
Sum $I_z\Delta z/E_s$ over the influence depth. Figure 3 draws the $E_s$ axis along the underside of the footing and measures its depth axis from there, so the whole influence zone (0 to $2B$ = 2.0 m below the base) lies in the 6000 kPa band; the 12 000 kPa band begins only 2.0 m below the base. Splitting the distribution at its peak leaves two linear segments whose mean values may be used:$$\begin{aligned}0 \le z \le 0.5\ \text{m}: & \quad \bar{I}_z = \tfrac{0.1+0.731}{2} = 0.415, \quad \frac{\bar{I}_z \Delta z}{E_s} = \frac{0.415(0.5)}{6000} = 3.462\times10^{-5} \\0.5 \le z \le 2.0\ \text{m}: & \quad \bar{I}_z = \tfrac{0.731+0}{2} = 0.365, \quad \frac{\bar{I}_z \Delta z}{E_s} = \frac{0.365(1.5)}{6000} = 9.137\times10^{-5}\end{aligned}$$$$\sum \frac{I_z}{E_s}\Delta z = 1.260\times10^{-4}\ \text{m/kPa}.$$
Correction factors. $C_1$ removes the part of the applied pressure that merely replaces the excavated overburden, and $C_2$ allows for creep in the sand:$$\begin{aligned}C_1 &= 1 - 0.5\frac{q}{\Delta q} = 1 - 0.5\frac{30.0}{213.4} = 0.930 \\ C_2 &= 1 + 0.2\log_{10}\!\frac{t}{0.1}.\end{aligned}$$Taking $t = 0.1$ year gives the immediate settlement, $C_2 = 1.00$.
Net pressure that produces 25 mm. Substituting and solving $S_e = 0.025$ m:$$0.025 = 0.930(1.00)\,\Delta q \,(1.260\times10^{-4}) \;\Rightarrow\; \boxed{\Delta q = 213.4\ \text{kPa}}$$which is a gross pressure of $213.4 + 30.0 = 243.4$ kPa.
Allow for creep over a design life of 10 years. With $t = 10$ yr, $C_2 = 1 + 0.2\log_{10}(100) = 1.40$; repeating the solution gives $I_{zp} = 0.702$, $C_1 = 0.908$, and$$\Delta q = 162.5\ \text{kPa} \quad (\text{gross } 192.5\ \text{kPa}).$$
Governing allowable pressure. Settlement controls by a wide margin (213.4 kPa against 606.5 kPa from bearing capacity), so$$\boxed{q_{all(net)} \approx 213\ \text{kPa immediately, reducing to } 162\ \text{kPa over 10 years}}$$A design value of 160 kPa net is recommended; at that pressure the factor of safety against a bearing failure is $(1849.6-30.0)/160 = 11.4$.
Water table risen to the ground surface. Both the surcharge and the self-weight terms now act at the buoyant unit weight $\gamma^{\prime} = 20 - 9.81 = 10.19$ kN/m$^3$:$$q_{u(sat)} = \gamma^{\prime} D_f N_q + 0.4\gamma^{\prime} B N_{\gamma} = 10.19(1.5)(47.16) + 0.4(10.19)(1.0)(54.36) = 720.8 + 221.6$$$$\boxed{q_{u(sat)} = 942.4\ \text{kPa}, \quad \text{a reduction of } 49.1\ \%}$$The capacity is very nearly halved, which is the familiar result that submergence roughly halves the bearing capacity of a granular soil because $\gamma^{\prime} \approx \gamma/2$. The corresponding net allowable pressure falls to $(942.4 - 15.3)/3 = 309.0$ kPa.
Effect of submergence on settlement. Two effects act in the same direction. First, the overburden at the depth of peak strain halves, $\sigma^{\prime}_{zp} = 10.19(2.0) = 20.4$ kPa, so at the same net pressure $I_{zp} = 0.5 + 0.1\sqrt{213.4/20.4} = 0.823$ instead of 0.731 — a 12 % larger strain-influence area and therefore a 12 % larger settlement. Second, and more important, the modulus itself would be lower: $E_s$ was inferred from a CPT run above the water table, and the cone resistance of the same sand at half the effective confining stress is markedly smaller, so the profile in Figure 3 would have to be re-measured rather than re-used. Taken together, the settlement at a given pressure would rise by well over half, and the settlement-limited allowable pressure would fall below about 140 kPa immediately and about 105 kPa over 10 years.

Collecting the results:

Quantity	Value
Ultimate bearing capacity, water table deep	1849.6 kPa
Net allowable pressure from $FS = 3$	606.5 kPa
Net pressure giving 25 mm immediate settlement	213.4 kPa
Net pressure giving 25 mm after 10 years ($C_2 = 1.40$)	162.5 kPa
Governing (recommended) net allowable pressure	160 kPa — settlement controls
Ultimate capacity with the water table at the surface	942.4 kPa
Reduction in ultimate capacity on submergence	49.1 %
Peak influence factor $I_{zp}$, dry / submerged	0.731 / 0.823

Check — the alternative reading of Figure 3. If the figure’s own values are used instead of the question text ($B = 2.5$ m, $\gamma = 18$ kN/m$^3$, $\phi^{\prime} = 38^{\circ}$), the method is identical but every number changes; the text has been followed here because it is the instruction to the candidate and because it supplies the parameters the figure does not. A candidate should state which set has been adopted, as paper Note 1 invites. Separately, Figure 3 draws the $E_s$ axis along the underside of the footing, so the modulus bands are measured from the base; reading them from ground level instead (6000 kPa over only the top 0.5 m of the influence zone) would give 308.5 kPa immediately and 234.2 kPa at 10 years — an unconservative reading the drawing does not support.

Topic. Bearing capacity and settlement of a shallow footing in sand.

Key relations. $q_u = 1.3cN_c + qN_q + 0.4\gamma B N_{\gamma}$ (Terzaghi, square); $S_e = C_1C_2\Delta q\sum (I_z/E_s)\Delta z$; $I_{zp} = 0.5+0.1\sqrt{\Delta q/\sigma^{\prime}_{zp}}$; $C_1 = 1-0.5(q/\Delta q)$, $C_2 = 1+0.2\log_{10}(t/0.1)$; $\gamma^{\prime} = \gamma - \gamma_w$.

Why this works. Terzaghi’s equation describes a rupture surface that has to develop through the soil, so it is controlled by strength; Schmertmann’s method integrates the vertical strain through the stress bulb, so it is controlled by stiffness. For small footings in dense sand the strength term is enormous but the stiffness is finite, so the two criteria give very different answers and settlement almost always governs. Submergence attacks both, because effective stress appears directly in the bearing equation and indirectly in every correlation for $E_s$.

Common pitfall. Applying the factor of safety to the gross rather than the net capacity (which is unconservative for a deep footing), and re-using a modulus profile measured above the water table for a submerged condition. A third pitfall is forgetting that $I_{zp}$ itself depends on $\Delta q$, so the settlement calculation must be iterated.

Source. Das, Principles of Foundation Engineering (9th ed.) §3.6 (Terzaghi’s equation and Table 3.1), §3.8 (water-table correction) and §5.6 (strain-influence method, after Schmertmann, Hartman & Brown 1978); CFEM (4th ed.) §10.

Question 6 of 9: Allowable bearing pressure of a 1 m square footing in sand

Question 6: Allowable bearing pressure of a 1 m square footing in sand (24 marks)