True or false — safe versus allowable bearing capacity in dense sand: 16-Civ-B3 Geotechnical Design, Undated paper, Q4

Question 4: True or false — safe versus allowable bearing capacity in dense sand (7 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.

Answer: TRUE. It is true by definition for every soil, and in dense sand the gap between the two is very wide.

The definitions do the first half of the work. The three quantities are distinct and are often blurred. The net ultimate bearing capacity $q_{nu}=q_u-\gamma D_f$ is the pressure at which the soil shears. The safe bearing capacity applies a factor of safety to that shear failure alone, $q_s=q_{nu}/\mathrm{FS}+\gamma D_f$. Separately, the safe bearing pressure $q_{np}$ is the pressure at which the footing settles by the tolerable amount. The allowable bearing capacity is the smaller of the two, $q_a=\min(q_s,\;q_{np})$, because a foundation must satisfy both limit states. It follows immediately that $q_a\le q_s$ always, and $q_a \lt q_s$ whenever settlement governs.

Given. A representative dense sand: $\phi'=41^{\circ}$, $\gamma=19\ \text{kN}/\text{m}^3$, corrected $N_{60}=45$, a 2.0 m square pad at $D_f=1.5\ \text{m}$, tolerable settlement 25 mm, $\mathrm{FS}=3$.

Find. The safe bearing capacity from shear and the settlement-limited pressure, to show which one becomes the allowable value.

The two criteria as functions of footing width for a dense sand. The shear-based curve rises with B because the $N_\gamma$ term is proportional to B; the settlement-based curve falls. In dense sand the settlement curve is the lower one over the whole practical range, so it is the one that becomes $q_a$.

Approach. Compute the shear-based capacity with Meyerhof’s closed-form factors, compute the settlement-limited pressure from the modified Meyerhof SPT equation, and compare.

Bearing-capacity factors at 41 degrees. Meyerhof’s factors are closed-form, so no chart is needed and the source is unambiguous: $N_q=e^{\pi\tan\phi'}\tan^{2}(45^{\circ}+\phi'/2)$ and $N_\gamma=(N_q-1)\tan(1.4\phi')$. With $\phi'=41^{\circ}$, $$\begin{aligned}N_q&=e^{\pi\tan 41^{\circ}}\tan^{2}(65.5^{\circ})=73.90 \\ N_\gamma&=(73.90-1)\tan(57.4^{\circ})=113.99 .\end{aligned}$$ The size of these numbers is the whole point: they roughly triple between 30 and 41 degrees.
Net ultimate bearing capacity of the 2 m pad. For a square footing on sand, with $c=0$, $$q_{nu}=\gamma D_f (N_q-1)+0.4\,\gamma B N_\gamma =19(1.5)(72.90)+0.4(19)(2.0)(113.99)$$ $$q_{nu}=2077.6+1732.5=\boxed{3810\ \text{kPa}}$$ and the safe bearing capacity at a factor of safety of 3 is $q_s=3810/3=1270\ \text{kPa}$.
Settlement-limited pressure for the same pad. Using the modified Meyerhof equation for $B \gt 1.22\ \text{m}$, with $F_d=1+0.33D_f/B=1.248$ and $S_e=25\ \text{mm}$, $$\begin{aligned}q_{np}&=11.98\,N_{60}\left(\frac{3.28B+1}{3.28B}\right)^{2}F_d\,\frac{S_e}{25.4} \\ &=11.98(45)(1.328)(1.248)(0.984) \\ &=\boxed{879\ \text{kPa}}\end{aligned}$$
Compare and conclude. $q_s=1270\ \text{kPa}$ against $q_{np}=879\ \text{kPa}$, so $q_a=879\ \text{kPa}$ and $q_s/q_a=1.45$. The statement is true, and settlement is the governing limit state.

The reason the gap is so wide in dense sand is structural rather than numerical. $N_q$ and $N_\gamma$ grow roughly exponentially with $\phi'$, so once $\phi'$ passes about 38 degrees the shear-based capacity climbs into the thousands of kilopascals, far beyond any pressure a real footing applies. Stiffness, by contrast, grows only moderately with density, so the settlement-limited pressure stays in the hundreds. In loose sand and in soft clay the ordering can reverse and shear can govern, which is why the definition, not the arithmetic, is the safe way to answer this question.

Question 4 — the two limit states compared for a 2 m square pad on dense sand
Quantity	Symbol	Value
Meyerhof factor	$N_q$	73.90
Meyerhof factor	$N_\gamma$	113.99
Net ultimate bearing capacity	$q_{nu}$	3810 kPa
Safe bearing capacity, FS = 3	$q_s$	1270 kPa
Settlement-limited pressure, 25 mm	$q_{np}$	879 kPa
Allowable bearing capacity	$q_a=\min(q_s,q_{np})$	879 kPa
Verdict on the statement	—	TRUE

Topic. Safe bearing capacity, safe bearing pressure and allowable bearing capacity.

Key relations. $q_{nu}=q_u-\gamma D_f$; $q_s=q_{nu}/\mathrm{FS}+\gamma D_f$; $q_a=\min(q_s,q_{np})$; Meyerhof $N_q=e^{\pi\tan\phi'}\tan^{2}(45^{\circ}+\phi'/2)$.

Why this works. Bearing capacity and settlement are two independent limit states with independent governing mechanisms. Defining the allowable value as the minimum of the two is what makes a single number safe to hand to a structural engineer.

Common pitfall. Applying the factor of safety to the gross rather than the net pressure. The overburden $\gamma D_f$ was already there before the footing was built, so factoring it inflates the answer; on a deep footing the error is large.

Source. Das, Principles of Foundation Engineering, 8th ed., ch. 3 and ch. 5; CFEM 4th ed., ch. 10 (limit states for shallow foundations).

Question 4 of 9: True or false — safe versus allowable bearing capacity in dense sand

Question 4: True or false — safe versus allowable bearing capacity in dense sand (7 marks)