Question 14.2: Figure 14.11 shows a gravity retaining wall for a granular (...

The retaining wall and the soil regions of interest are divided into the rectangles and triangles shown in Figure 14.12 for case of computation of the weights and moment arms.
For φ′ = 35° and α = 15°, we find from Table 11.2 that K_a = 0.2968. Thus,
P_a=\frac{1}{2}\gamma H^{\prime 2}K_a=\frac{1}{2}(18.5)(6.54)^2(0.2968)= 117.4 kN/m
P_h=P_a \cos \alpha = 113.4 kN/m
P_v=P_a \sin \alpha= 30.4 kN/m

Stability with Respect to Overturning
The overturning moment is
M_O=P_h\frac{H^\prime}{3}=113.4\times 2.18= 247.2 kN/m
The resisting moment is
\sum M_R = 1143.0 kN/m
Therefore,
FS_{(overturning)} = \frac{\sum M_g}{M_O}=\frac{1143.0}{247.2} = 4.62
Stability with Respect to Sliding
For φ′ = 35°, from Eq. 11.28,
K_p=\tan^2\left(45+\frac{\phi^\prime}{2}\right)= 3.690
P_p=\frac{1}{2}\gamma D^2 K_p=\frac{1}{2}(18.5)(1.0)^2(3.690)= 34.1 kN/m
Take δ′ as {}^2/3 φ′. So δ′ = 23.3°, we have
FS_{(sliding)} =\frac{ (\sum V) \tan \delta^\prime+ P_p}{P_h}=\frac{475.4\times  \tan 23.3 + 34.1}{113.4} = 2.11
Stability with Respect to Bearing Capacity Failure
e = \frac{B}{2}-\frac{ \sum M_R – \sum M_O}{\sum V} =\frac{ 4.0}{2} -\frac{ 1143.0 – 247.2}{475.4} = 0.116 m \lt \frac{B}{6} \left(=\frac{4}{6}=0.67 m\right)
From Eq. (14.11),
q_{max} = q_{toe} = \frac{\sum V}{B}\left(1+\frac{6e}{B}\right)=\frac{475.4}{4.0} \left(1+\frac{(6)(0.116)}{4.0}\right)= 139.5 kN/m²
In granular soil, Eq. 14.13 becomes
q_u^\prime = q N_qF_{qd} F_{qi} +0.5 \gamma B^\prime N_\gamma F_{\gamma d}F_{\gamma i}
where, B′ = B – 2e = 3.768 m and q = γD = (18.5)(1.0) = 18.5 kN/m².
For φ′ = 35°, from Table 12.1

N_q = 33.30 and N_\gamma = 48.03. Hence,
F_{qd}=1+2\tan\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B}=1+2 \tan 35(1- \sin 35)^2 \frac{1}{4}= 1.07
F_{\gamma d} =1
\Psi=\tan^{-1}\left(\frac{P_a \cos \alpha}{\sum V}\right)=\tan^{-1}\left(\frac{113.4}{475.4}\right)=13.4°
F_{qi}=\left(1-\frac{\Psi}{90}\right)^2=\left(1-\frac{13.4}{90}\right)= 0.72
F_{\gamma i}=\left(1+\frac{\Psi}{\phi^\prime}\right)^2=\left(1-\frac{15}{35}\right)^2= 0.33
q_u = (18.5)(33.30)(1.07)(0.72) + (0.5)(18.5)(3.768)(48.03)(1.0)(0.33)= 1027.0 kN/m²
FS_{(bearing  capacity)} =\frac{q_u^\prime}{q_{max}}=\frac{1027.0}{139.5} = 7.4