(a) The stability number N_{s} from Eq. (20.57a) is
N_{s}=\frac{\gamma H}{c} \leq 4 (20.57a)
N_{s}=\frac{\gamma H}{c}=\frac{115 \times 25}{800}=3.6<4
The soil is stiff fissured clay. As such the pressure envelope shown in Fig. 20.28(c) is applicable. Assume p_{a}=0.3 \gamma H
p_{a}=0.3 \times 115 \times 25=863 lb / ft ^{2}
The pressure envelope is drawn as shown in Fig. Ex. 20.18(b).
(b) Strut loads
Taking moments about the strut head B_{1}(B)
\begin{aligned}&R_{A} \times 7.5=\frac{1}{2} 863 \times 6.25\left(\frac{6.25}{3}+6.25\right)+863 \times \frac{(6.25)^{2}}{2} \\&=22.47 \times 10^{3}+16.85 \times 10^{3}=39.32 \times 10^{3} \\&R_{A}=5243 lb / ft \\&R_{ B 1}=\frac{1}{2} \times 863 \times 6.25+863 \times 6.25-5243=2848 lb / ft\end{aligned}
Due to symmetry
\begin{aligned}&R_{A}=R_{C}=5243 lb / ft \\&R_{B 2}=R_{B 1}=2848 lb / ft\end{aligned}
Strut loads are:
\begin{aligned}&P_{A}=5243 \times 12=62,916 lb =62.92 kips \\&P_{B}=2 \times 2848 \times 12=68,352 lb =68.35 kips \\&P_{C}=62.92 kips\end{aligned}
(c) Moments
The shear force diagram is shown in Fig. 20.18c for sections D B_{1} \text { and } B_{2} E
Moment at A=\frac{1}{2} \times 5 \times 690 \times \frac{5}{3}=2,875 Ib-ft/ft of wall
Moment at m=2848 \times 3.3-863 \times 3.3 \times \frac{3.3}{2}=4699 Ib-ft/ft
Because of symmetrical loading
Moment at A = Moment at C = 2875 Ib-ft/ft of wall
Moment at m = Moment at n = 4699 Ib-ft/ft of wall
Hence, the maximum moment = 4699 Ib-ft/ft of wall.
The section modulus and the required sheet pile section can be determined in the usual way.