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}.

12.9
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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

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