Question 12.6: A square foundation is shown in Figure 12.9. Assume that the...
A square foundation is shown in Figure 12.9. Assume that the load eccentricity e = 0.5 ft. Determine the ultimate load, Q_{ult}.

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With c′ = 0, Eq. 12.20 becomes
q_u^\prime = q N_qF_{qs}F_{qd} F_{qi} +\frac{1}{2} \gamma B^\prime N_\gamma F_{\gamma s}F_{\gamma a}F_{\gamma i}
q = (3)(115) = 345 lb/ft²
For φ′ = 30°, 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_q = 18.4 and N_\gamma = 22.4.
B′=5 – 2(0.5) = 4 ft
L′= 5 ft
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+\frac{B^\prime}{L^\prime}\tan \phi^\prime=1+\left(\frac{4}{5}\right) \tan 30^\circ =1.462
F_{qd}=1+2\tan \phi^\prime(1-\sin \phi^\prime)^2 \frac{D_f}{B}=1+\frac{(0.289)(3)}{5}= 1.173
F_{\gamma s}=1-0.4\left(\frac{B^\prime}{L^\prime}\right)=1-0.4\left(\frac{4}{5}\right)=0.68
F_{\gamma d}=1
So
q_u^\prime = (345)(18.4)(1.462)(1.173)+ \frac{1}{2}(115)(4)(22.4)(0.68)(1)
= 10,886.4 + 3503.4 ≈ 14,390 lb/ft²
Hence,
Q_{ult}=B^\prime L^\prime (q_u^\prime)=(4)(5)(14,390)= 287,800 lb
≈ 287.8 kip