Question 12.3: A square column foundation to be constructed on a sandy soil...
A square column foundation to be constructed on a sandy soil has to carry a gross allowable inclined load of 59.6 kip. The depth of the foundation will be 2.3 ft. The load will be inclined at an angle of 20° to the vertical (Figure 12.5).
Assume that the unit weight of the soil is 114.5 lb/ft³. Determine the width of the foundation, B. Use a factor of safety of 3
With c′ = 0, the ultimate bearing capacity [Eq. (12.7)] becomes
q_u = q N_qF_{qs}F_{qd} F_{qi} +\frac{1}{2} \gamma BN_\gamma F_{\gamma s}F_{\gamma d}F_{\gamma i}q = (2.3)(114.5) = 263.35 lb/ft²
γ = 114.5 lb/ft³
From Table 12.1,
| φ′ | N_c | N_q | N_\gamma | φ′ | N_c | N_q | N_\gamma |
| 0 | 5.14 | 1.00 | 0.00 | 23 | 18.05 | 8.66 | 8.20 |
| 1 | 5.38 | 1.09 | 0.07 | 24 | 19.32 | 9.60 | 9.44 |
| 2 | 5.63 | 1.20 | 0.15 | 25 | 20.72 | 10.66 | 10.88 |
| 3 | 5.90 | 1.31 | 0.24 | 26 | 22.25 | 11.85 | 12.54 |
| 4 | 6.19 | 1.43 | 0.34 | 27 | 23.94 | 13.20 | 14.47 |
| 5 | 6.49 | 1.57 | 0.45 | 28 | 25.80 | 14.72 | 16.72 |
| 6 | 6.81 | 1.72 | 0.57 | 29 | 27.86 | 16.44 | 19.34 |
| 7 | 7.16 | 1.88 | 0.71 | 30 | 30.14 | 18.40 | 22.40 |
| 8 | 7.53 | 2.06 | 0.86 | 31 | 32.67 | 20.63 | 25.99 |
| 9 | 7.92 | 2.25 | 1.03 | 32 | 35.49 | 23.18 | 30.22 |
| 10 | 8.35 | 2.47 | 1.22 | 33 | 38.64 | 26.09 | 35.19 |
| 11 | 8.80 | 2.71 | 1.44 | 34 | 42.16 | 29.44 | 41.06 |
| 12 | 9.28 | 2.97 | 1.69 | 35 | 46.12 | 33.30 | 48.03 |
| 13 | 9.81 | 3.26 | 1.97 | 36 | 50.59 | 37.75 | 56.31 |
| 14 | 10.37 | 3.59 | 2.29 | 37 | 55.63 | 42.92 | 66.19 |
| 15 | 10.98 | 3.94 | 2.65 | 38 | 61.35 | 48.93 | 78.03 |
| 16 | 11.63 | 4.34 | 3.06 | 39 | 67.87 | 55.96 | 92.25 |
| 17 | 12.34 | 4.77 | 3.53 | 40 | 75.31 | 64.20 | 109.41 |
| 18 | 13.10 | 5.26 | 4.07 | 41 | 83.86 | 73.90 | 130.22 |
| 19 | 13.93 | 5.80 | 4.68 | 42 | 93.71 | 85.38 | 155.55 |
| 20 | 14.83 | 6.40 | 5.39 | 43 | 105.11 | 99.02 | 186.54 |
| 21 | 15.82 | 7.07 | 6.20 | 44 | 118.37 | 115.31 | 224.64 |
| 22 | 16.88 | 7.82 | 7.13 | 45 | 133.88 | 134.88 | 271.76 |
for φ′ = 30°, we find
N_q = 18.4
N_\gamma = 22.4
From Table 12.2,
| Factor | Relationship | Source |
| Shape* | F_{cs} = 1 +\frac{B}{L}\frac{ N_q}{ N_c} | De Beer (1970) |
| F_{qs} = 1 +\frac{B}{L}\tan \phi^\prime | ||
| F_{\gamma s} = 1 – 0.4\frac{B}{L} | ||
| where L = length of the foundation (L > B) | ||
| Depth{}^† | Condition (a): Df/B \leq 1 | Hansen (1970) |
| F_{cd} = 1 + 0.4\frac{D_f}{B} | ||
| F_{qd}=1+2\tan\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B} | ||
| F_{\gamma d} = 1 | ||
| Condition (b): D_f /B \gt 1 | ||
| F_{cd} = 1 +( 0.4)\tan^{-1}\left(\frac{D_f}{B}\right) | ||
| F_{qd}=1+2\tan\phi^\prime(1-\sin \phi^\prime)^2\tan^{-1}\left(\frac{D_f}{B}\right) | ||
| F_{\gamma d} = 1 | ||
| Inclination | F_{ci}=F_{qi}=\left(1-\frac{\beta^\circ}{90^\circ}\right)^2 | Meyerhof (1963); Hanna and Meyerhof (1981) |
| F_{\gamma i}=\left(1-\frac{\beta}{\phi^\prime}\right)^2 | ||
| where β = inclination of the load on the foundation with respect to the vertical |
F_{qs}=1+\left(\frac{B}{L}\right)\tan \phi^\prime=1+0.577=1.577
F_{\gamma s}=1-0.4\left(\frac{B}{L}\right)=0.6
F_{qd}=1+2\tan \phi^\prime(1-\sin \phi^\prime)^2 \frac{D_f}{B}=1+\frac{(0.289)(2.3)}{B}=1+\frac{0.665}{B}
F_{\gamma d}=1
F_{qi}=\left(1-\frac{\beta^\circ}{90^\circ}\right)^2=\left(1-\frac{20}{90}\right)^2=0.605
F_{\gamma i}=\left(1-\frac{\beta^\circ}{\phi^\prime}\right)^2=\left(1-\frac{20}{30}\right)^2=0.11Hence,
q_u = (263.35)(18.4)(1.577)\left(1+\frac{0.665}{B}\right)(0.605)+ (0.5)(114.5)(B)(22.4)(0.6)(1)(0.11)
= 4623.15 +\frac{3074.4}{B}+ 84.6B (a)
Thus,
q_{all}= \frac{q_u}{3} = 1541.05 +\frac{1024.8}{B}+28.2B (b)
For Q = total vertical allowable load = q_{all}\times B^2 or
q_{all} = \frac{Q^\prime \cos 20}{B^2}=\frac{ (59,600)(\cos 20)}{B^2} (c)
Equating the right-hand sides of Eqs. (b) and (c) gives
\frac{56,006}{B^2}= 1541.05 +\frac{1024.8}{B}+28.2 B
By trial and error, we find B ≈ 5.5 ft