Question 7 of 9: Friction angle from SPT data, and design of a 3.0 m square footing

Question 7: Friction angle from SPT data, and design of a 3.0 m square footing (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.

Footing 3.0 m × 3.0 m, founding depth D_f = 2.0 m; water table at 18 m depth, that is 10 m below the base and far below the 2B = 6 m zone of influence, so no buoyancy correction applies anywhere in this design.

Find. A defensible design value of φ' from the SPT profile, and then, using that φ' and not any direct N-to-bearing-capacity rule, the allowable bearing pressure and the corresponding allowable column load for the 3.0 m square footing, with settlement checked explicitly.

Approach. Normalise the field blow counts to (N₁)₆₀, read φ' from a published (N₁)₆₀-to-φ' relation over the depth interval that actually carries the footing, compute the ultimate bearing capacity from that φ' with Vesic's factors and De Beer's shape and depth factors, and then check settlement by Schmertmann's strain-influence method. Whichever of the two criteria gives the lower allowable pressure is the design value.

1. Build the effective overburden profile. With the water table at 18 m the whole tested depth is above it, so total and effective vertical stresses coincide. Accumulating the tabulated unit weights over 2 m increments, $$\sigma'_v(2) = 19.0 \times 2 = 38.0\ \text{kPa},\qquad \sigma'_v(8) = 19.0 \times 6 + 21.5 \times 2 = 157.0\ \text{kPa}$$ and so on, giving the fourth column of the table above.
2. Correct the blow counts for overburden. The field values are taken as already energy-corrected, $N_{60} \approx N_f$ — the paper gives no hammer or energy information, and this is the standard reading of a bare tabulated N_f. Liao and Whitman's overburden correction is $$C_N = \sqrt{\frac{p_a}{\sigma'_v}}, \qquad p_a = 100\ \text{kPa},\qquad C_N \le 1.7$$ Applying it gives $(N_1)_{60}$ in the last column. Notice what the correction reveals: the raw counts climb steadily from 6 to 26 and look like a soil getting denser with depth, whereas the normalised values sit between 9.7 and 15.4 throughout. The deposit is of essentially uniform relative density; the rise in N_f is almost entirely a confinement effect.
3. Choose the design (N₁)₆₀. The footing stresses the soil from its base at 2 m down to roughly 2B = 6 m below the base, that is to 8 m depth. Averaging the four tests inside that interval, $$(N_1)_{60,\,\text{design}} = \frac{9.73 + 11.47 + 13.11 + 14.37}{4} = 12.17$$ The average over the whole 16 m profile is 13.42, so the choice of interval is worth about 10 per cent and the answer is not sensitive to it.
4. Estimate φ' from the normalised blow count. Using the Peck, Hanson and Thornburn chart as fitted by Wolff (1989), $$\phi' = 27.1 + 0.3\,(N_1)_{60} - 0.00054\,(N_1)_{60}^{2} = 27.1 + 3.651 - 0.080 = 30.67^\circ$$ Kulhawy and Mayne's alternative, $\phi' = \tan^{-1}\!\left[\dfrac{N_{60}}{12.2 + 20.3\,(\sigma'_v/p_a)}\right]^{0.34}$, gives 33.6° at 2 m rising to 36.4° at 8 m. The two families bracket the answer; taking the lower, more conservative estimate and rounding, $$\boxed{\phi' = 31^\circ}$$ is adopted for design. This is a medium-dense sand, consistent with (N₁)₆₀ of about 12 and a relative density near 45 per cent.
5. Compute the bearing-capacity factors from φ'. As the note in the question requires, everything from here is derived from φ', not from N. Vesic's expressions give $$N_q = e^{\pi\tan\phi'}\tan^2\!\left(45^\circ + \frac{\phi'}{2}\right) = 20.63,\qquad N_c = (N_q - 1)\cot\phi' = 32.67,$$ $$N_{\gamma} = 2(N_q + 1)\tan\phi' = 25.99$$ The sand is cohesionless, c' = 0, so N_c plays no part.
6. Evaluate the shape and depth factors. For a square footing (B/L = 1) with D_f/B = 2/3 = 0.667, De Beer's shape factors and Hansen's depth factors are $$F_{qs} = 1 + \frac{B}{L}\tan\phi' = 1.601,\qquad F_{\gamma s} = 1 - 0.4\frac{B}{L} = 0.600$$ $$F_{qd} = 1 + 2\tan\phi'\,(1-\sin\phi')^2\frac{D_f}{B} = 1.188,\qquad F_{\gamma d} = 1.000$$
7. Compute the ultimate bearing capacity. The surcharge at founding level is $q = \gamma D_f = 19.0 \times 2.0 = 38.0$ kPa, and the soil for one footing width below the base has $\gamma = 19.0$ kN/m³. Then $$q_u = q N_q F_{qs} F_{qd} + \tfrac{1}{2}\gamma B N_{\gamma} F_{\gamma s} F_{\gamma d}$$ $$q_u = 38.0(20.63)(1.601)(1.188) + \tfrac{1}{2}(19.0)(3.0)(25.99)(0.600)(1.000) = 1491.5 + 444.5$$ $$\boxed{q_u = 1936\ \text{kPa}}$$ Subtracting the surcharge already there gives the net ultimate value $q_{u,\text{net}} = 1936.0 - 38.0 = 1898.0$ kPa, and with a conventional FS = 3 on net bearing capacity the allowable net pressure would be 632.7 kPa. Adopting Meyerhof's $N_{\gamma}$ instead of Vesic's lowers $q_u$ to 1809 kPa and the allowable net pressure to 590 kPa — a 7 per cent difference that changes nothing, because bearing capacity is not what governs here.
8. Set up the settlement check. A pressure of 633 kPa on a 3 m footing on a medium-dense sand would be absurd in service, so the design must be completed by a settlement calculation. Schmertmann's strain-influence method for a square (axisymmetric) footing places the peak influence factor at z = B/2 below the base and zero at z = 2B. The effective stress at the peak, 3.5 m below ground, is $\sigma'_{zp} = 19.0 \times 3.5 = 66.5$ kPa, and Bowles' correlation $E_s = 500(N_{60}+15)$ kPa gives layer moduli of 10 500, 12 500, 14 500 and 16 500 kPa for the bands centred on the tests at 2, 4, 6 and 8 m. This is a settlement correlation, not a bearing-capacity correlation, so it is fully consistent with the note in the question.
9. Compute the settlement at a trial pressure. Taking a net applied pressure $\Delta q = 170$ kPa, the peak influence factor is $$I_{zp} = 0.5 + 0.1\sqrt{\frac{\Delta q}{\sigma'_{zp}}} = 0.5 + 0.1\sqrt{\frac{170}{66.5}} = 0.660$$ the embedment factor is $C_1 = 1 - 0.5(\sigma'_D/\Delta q) = 1 - 0.5(38.0/170) = 0.888$, and with $C_2 = 1$ for the immediate condition, $$S_e = C_1 C_2 \Delta q \sum \frac{I_z}{E_s}\Delta z = 0.888 \times 170 \times 1.6085\times10^{-4} = 0.0243\ \text{m}$$ that is 24.3 mm. Repeating the calculation at other pressures, the net pressure that produces exactly 25 mm is 174 kPa, while 200 kPa would give 29.7 mm.
10. Choose the design pressure and load. Twenty-five millimetres is the conventional tolerable total settlement for an isolated footing on sand, because it limits differential settlement between adjacent footings to about 20 mm and the angular distortion to well within 1/500. Taking the round figure below the computed limit, $$\boxed{q_{\text{all,net}} = 170\ \text{kPa}\quad\Longrightarrow\quad Q_{\text{all}} = 170 \times 3.0 \times 3.0 = 1530\ \text{kN}}$$ Against the bearing-capacity ceiling of 632.7 kPa this represents a true factor of safety on bearing of $1898.0/170 = 11.2$, so the footing is nowhere near a bearing failure. As a completely independent check, Meyerhof's direct SPT rule for 25 mm of settlement, $q_{\text{all}} = 11.98 N_{60}\left[(3.28B+1)/(3.28B)\right]^2$ with $N_{60} = 12$, gives 174.5 kPa — within 3 per cent of the Schmertmann result. (It is quoted only as a check; the design itself is built on φ', as the question directs.)