Question 12.17: Refer to Example 12.16. For the clay, cu = 270 lb/ft². If th...
Refer to Example 12.16. For the clay, c_u = 270 lb/ft². If the required factor of safety against bearing capacity failure is 3, determine the depth of the foundation.
From Eq. (12.58), we have
FS=\frac{5.14 c_u\left(1+\frac{0.195 B}{L}\right)\left(1+0.4\frac{D_f}{B}\right)}{\frac{Q}{A}-\gamma D_f}
Here, FS = 3, c_u = 270 lb/ft², B/L = 90/120 = 0.75, and Q/A = (45\times 10^6)/(90\times 120)= 4166.7 lb/ft². Substituting these values into Eq. (12.58) yields
3=\frac{(5.14)(270)\left[1+(0.195)(0.75)\right] \left[1+0.4\left(\frac{D_f}{90}\right) \right] }{4166.7- (120)D_f}12,500.1 – 360 D_f = 1590.77 + 7.07 D_f
10,909.33 = 367.07 D_f
or
D_f≈29.72 ft
Related Answered Questions
Question: 12.6
Verified Answer:
With c′ = 0, Eq. 12.20 becomes
q_u^\prime =...
Question: 12.5
Verified Answer:
From Eq. (12.17), eccentricity e =\frac{M}{...
Question: 12.4
Verified Answer:
With c′ = 0, Eq. (12.20) gives
q_u^\prime =...
Question: 12.1
Verified Answer:
From Eq. (12.7), noting that F_{ci} = F_{qi...
Question: 12.3
Verified Answer:
With c′ = 0, the ultimate bearing capacity [Eq. (1...
Question: 12.2
Verified Answer:
For φ′ = 23° from Table 12.1,
φ′
N_c[/l...
Question: 12.7
Verified Answer:
Given B = 3 ft and L = 6 ft. Note that \bar...
Question: 12.16
Verified Answer:
From Eq. (12.56a), we have
D_f=\frac{Q}{A\g...
Question: 12.15
Verified Answer:
From Eq. (12.43),
F_d =1+0.33(D_f/B)=1+0.33...
Question: 12.14
Verified Answer:
From Eq. (12.54),
q_{all(net)}=1.713 c_u\le...