Question 12.7: A rigid shallow foundation 3 ft × 6 ft is shown in Figure 12...
A rigid shallow foundation 3 ft × 6 ft is shown in Figure 12.15. Calculate the elastic settlement at the center of the foundation.
Given B = 3 ft and L = 6 ft. Note that \bar{\zeta} = 15 ft = 5B. From Eq. (12.32)
E_s=\frac{\sum{E_{s(i)}}\Delta \zeta}{\bar{\zeta}}
=\frac{(1450)(6)+(1160)(3)+(1750)(6)}{15}=1512 lb/in² = 217,728 lb/ft²
For the center of the foundation,
α = 4
\frac{L}{B}=\frac{6}{3}=2
n^\prime=\frac{H}{\left(\frac{B}{2}\right)}=\frac{15}{\left(\frac{3}{2}\right)}=10
From Tables 12.3 and 12.4, F_1 = 0.641 and F_2 = 0.031. From Eq. (12.24),
I_s= F_1+\frac{2-\mu s}{1-\mu s}F_2
= 0.641+\frac{2 – 0.3}{1 – 0.3}(0.031)= 0.716
Again \frac{D_f}{B}=\frac{3}{3}=1,\frac{L}{B}=2,\mu_s = 0.3. From Figure 12.12, I_f = 0.71. Hence,
S_{e(flexible)}=q_o(\alpha B^\prime)\frac{1-\mu_s^2}{E_s}I_sI_f
=(3150)\left(4\times \frac{3}{2}\right)\left(\frac{1-{0.3}^2}{217,728}\right)(0.716)(0.71)= 0.0402 ft = 0.48 in.
Since the foundation is rigid, from Eq. (12.31),
S_{e(rigid)} =(0.93)(0.48)\approx 0.45 in.
