Question 12.5: A 2 m × 3 m spread foundation placed at a depth of 2 m carri...
A 2 m × 3 m spread foundation placed at a depth of 2 m carries a vertical load of 3000 kN and moment of 300 kN⋅m, as shown in Figure 12.8, Determine the factor of safety.
From Eq. (12.17), eccentricity e =\frac{M}{Q}=\frac{300}{3000} = 0.10 m. Therefore, B′ = B – 2e = 2.00 – 2 × 0.1 = 1.80 m. Also L′ = L = 3 m.
For φ′ = 32° 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 = 35.49, N_q = 23.18, and N_\gamma = 30.22.
Shape Factors
F_{cs} = 1 +\frac{B^\prime}{L} \frac{N_q}{N_c}=1+\frac{1.8}{3}\times \frac{23.18}{35.49}=1.39
F_{qs} = 1 +\frac{B^\prime}{L}\tan \phi^\prime=1+\frac{1.8}{3}\tan 32=1.37
F_{\gamma s} = 1 – 0.4\frac{B^\prime}{L}=1-0.4\times \frac{1.8}{3}=0.76
Depth Factors
F_{cd} = 1 + 0.4\frac{D_f}{B}=1+0.4\times \frac{2.0}{2.0}=1.40
F_{qd}=1+2\tan\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B}=1+2\tan 32(1-\sin 32)^2\times \frac{2}{2} = 1.28
F_{qi} = 1.00
Inclination factors are F_{ci}=1, F_{qi}=1, and F_{\gamma i}=1.
From Eq. (12.20), the ultimate bearing capacity is given by
= (5)(35.49)(1.39) (1.40)(1) + (2\times 18.5)(23.18)(1.37)(1.28)(1) + 0.5(18.5\times 1.8)(30.22)(0.76)(1.0)(1)
= 2231.17 kN/m²
From Eq. (12.21), Q_{ult} = q_u^\prime \times B^\prime L = 2231.7\times 1.8\times 3 = 12951.2 kN
FS = 12951.9/3000 = 4.01