Question 12.1: Consider a shallow foundation with the following: ● Foundati...
Consider a shallow foundation with the following:
● Foundation: B = 3 ft
L = 5 ft
D_f = 3 ft
● Soil: γ = 110 lb/ft³
φ′ = 25°
c′ = 400 lb/ft²
Determine the gross allowable load the foundation can carry. Use FS = 4.
From Eq. (12.7), noting that F_{ci} = F_{qi} = F_{\gamma i} = 1
q_u = c^\prime N_cF_{cs}F_{cd} + qN_qF_{qs}F_{qd} +\frac{1}{2} \gamma BN_\gamma F_{\gamma s}F_{\gamma q}
For φ′ = 25°, 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 |
N_c = 20.72
N_q = 10.66
N_\gamma = 10.88
q = \gamma D_f = (110)(3) = 330 lb/ft²
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_{cs} = 1 +\frac{B N_q}{L N_c}= 1 + \left(\frac{3}{5}\right)\left(\frac{10.66}{20.72}\right) = 1.309
F_{qs} = 1 +\frac{B}{L}\tan \phi^\prime=1+\left(\frac{3}{5}\right)\tan 25 = 1.28
F_{\gamma s} = 1 – 0.4\frac{B}{L}=1-(0.4)\left(\frac{3}{5}\right)= 0.76
F_{cd} = 1 + 0.4\frac{D_f}{B}= 1 + (0.4)\left(\frac{3}{3}\right)= 1.4
F_{qd}=1+2\tan\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B}=1+2\tan 25(1-\sin 25)^2\left(\frac{3}{3}\right) = 1.311
F_{\gamma d} = 1
q_u = (400)(20.72)(1.309)(1.4) + (330)(10.66)(1.28)(1.311) +\frac{1}{2}(110)(3)(10.88)(0.76)(1)
= 15,188.6 + 5903.1 + 1364.4 = 22,456.1 lb/ft²
q_{all} =\frac{q_u}{FS} = \frac{22,456.1}{4} ≈ 5614 lb
Q_{all} = (q_{all})(BL) = (5614)(3\times 5) = 84,210 lb≈ 84.2 kip