Safe bearing capacity against allowable bearing capacity in dense sand: 07-Str-B1, Undated paper, Q4

Question 4: Safe bearing capacity against 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 — and for dense sand it is true by a wide margin, because the two quantities answer two different questions and in dense sand the settlement question is far more demanding than the shear question.

The definitions do the work. The net safe bearing capacity is the net ultimate bearing capacity divided by a factor of safety against shear failure, $q_{s(net)} = q_{u(net)}/FS$, and the safe bearing capacity $q_s$ adds back the overburden that was there before, $q_s = q_{u(net)}/FS + \gamma D_f$. It is a strength criterion and nothing else. The safe bearing pressure is a separate quantity: the net pressure that produces exactly the tolerable settlement. The allowable bearing capacity $q_a$ is defined as the smaller of the two. It follows directly from that definition that $q_a \le q_s$ always, with equality only when shear failure happens to govern.

Why the inequality is strict, and large, in dense sand. Terzaghi's bearing capacity factors grow roughly exponentially with $\phi'$. At $\phi' = 40^\circ$, $N_q = 81.3$ and $N_\gamma = 100.4$; at $\phi' = 30^\circ$ they are only 22.5 and 19.1. A dense sand therefore has an enormous ultimate bearing capacity. But dense sand is also stiff-but-not-rigid, its settlement is essentially immediate, and the permissible settlement is set by the structure, not the soil — typically 25 mm total for an isolated footing. That limit is reached at a pressure that has nothing to do with $\phi'$ and everything to do with the modulus, and it caps the design pressure long before shear failure is approached. As a footing gets wider the gap widens further, because the ultimate capacity grows with $B$ through the $N_\gamma$ term while the settlement-controlled pressure falls with $B$ through the $[(B+0.3)/B]^2$ factor.

A numerical illustration. Take a 4.0 m square footing at $D_f = 1.5\ \text{m}$ in a dry dense sand with $\gamma = 18\ \text{kN/m}^3$, $\phi' = 40^\circ$ and $N_{60} = 45$, with a tolerable settlement of 25 mm. Terzaghi gives

$$q_u = qN_q + 0.4\gamma BN_\gamma = (27)(81.27) + 0.4(18)(4.0)(100.39) = 5086\ \text{kPa}$$

so the net safe bearing capacity at a factor of safety of 3 is $q_{s(net)} = (5086 - 27)/3 = 1686\ \text{kPa}$. The settlement criterion, from Meyerhof's relation as modified by Bowles with the depth factor $F_d = 1 + 0.33(1.5/4.0) = 1.124$, gives

$$q_{net} = \frac{N_{60}}{0.08}\left(\frac{B+0.3}{B}\right)^2 F_d \frac{S_e}{25} = \frac{45}{0.08}(1.156)(1.124)(1) = 731\ \text{kPa}$$

and therefore

$$\boxed{q_a = \min(1686,\ 731) = 731\ \text{kPa} \ \text{(settlement governs)} \ \lt \ q_{s(net)} = 1686\ \text{kPa}}$$

The allowable pressure is less than half the safe bearing capacity, and the design is a settlement design. Two riders complete the answer. If the two terms are used loosely as synonyms, as some texts do, the statement would read as false; the question is only meaningful under the standard definitions set out above, and it is those definitions that make it true. And the inequality can reverse in direction of governance — not of magnitude — for a narrow footing on a loose sand, where the ultimate capacity is small, the settlement-controlled pressure is comparatively large, and shear governs so that $q_a = q_s$.

Topic. The distinction between the shear-failure criterion and the settlement criterion for a shallow foundation, and which one governs in dense sand.

Key relations. $q_{u(net)} = q_u - \gamma D_f$; net safe bearing capacity $q_{u(net)}/FS$; allowable bearing capacity $q_a = \min(q_{s},\ q_{settlement})$; Terzaghi $N_q = e^{2(3\pi/4-\phi/2)\tan\phi}/[2\cos^2(45^\circ+\phi/2)]$; Meyerhof settlement relation as above.

Why this works. A foundation has two independent limit states, and a design is only adequate if it satisfies both. Increasing $\phi'$ improves the strength limit state dramatically because the bearing capacity factors are exponential in $\phi'$, but it improves the serviceability limit state only weakly, because settlement is governed by stiffness and by the geometry of the stressed zone. The two therefore diverge as the sand gets denser, and the smaller of the two — the settlement limit — is what the engineer may actually apply to the ground.

Common pitfall. Computing a very large $q_u$ for a dense sand, dividing by three, and issuing that as the allowable bearing pressure without ever performing a settlement check. The second pitfall is mixing net and gross pressures: the factor of safety is applied to the net ultimate capacity, and the overburden term is added back afterwards, not before.

Source. Das, Principles of Foundation Engineering (9th ed.), Ch. 3 (Ultimate Bearing Capacity) and Ch. 5 (Allowable Bearing Capacity and Settlement); CFEM (4th ed.) Ch. 8 and Ch. 11 on tolerable movement.

Question 4 of 9: Safe bearing capacity against allowable bearing capacity in dense sand

Question 4: Safe bearing capacity against allowable bearing capacity in dense sand (7 marks)