Fig. Ex. 20.2 shows a free standing cantilever sheet pile with no backfill driven into homogeneous sand. The following data are available:
H = 20 ft, P = 3000 Ib/ft of wall, \gamma=115 lb / ft ^{3}, \phi=36^{\circ}.
Determine: (a) the depth of penetration, D, and (b) the maximum bending moment M_{\max }.
The equation for D is (Eq 20.11)
D^{4}+C_{1} D^{2}+C_{2} D+C_{3}=0 (20.11)
D^{4}+C_{1} D^{2}+C_{2} D+C_{3}=0
where C_{1}=-\frac{8 P}{\gamma K}=-\frac{8 \times 3000}{115 \times 3.59}=-58.133
C_{2}=-\frac{12 P H}{\gamma K}=-\frac{12 \times 3000 \times 20}{115 \times 3.59}=-1744
C_{3}=-\frac{4 P^{2}}{(\gamma K)^{2}}=-\frac{4 \times 3000^{2}}{(115 \times 3.59)^{2}}=-211.2
Substituting and simplifying, we have
D^{4}-58.133 D^{2}-1744 D-211.2=0
From the above equation D =13.5 ft.
From Eq. (20.6)
\bar{y}_{0}=\sqrt{\frac{2 P_{a}}{\gamma K}} (20.6)
\bar{y}_{0}=\sqrt{\frac{2 P_{a}}{\gamma K}}=\sqrt{\frac{2 P}{\gamma K}}=\sqrt{\frac{2 \times 3000}{115 \times 3.59}}=3.81 ft
FromEq. (20.12)
M_{\max }=P\left(H+\bar{y}_{o}\right)-\frac{\gamma \bar{y}_{o}^{3} K}{6} (20.12)
\begin{aligned}&M_{\max }=P_{a}\left(H+\bar{y}_{0}\right)-\frac{\gamma \bar{y}_{0}^{3} K}{6} \\&=3000(20+3.81)-\frac{115 \times(3.81)^{3} \times 3.59}{6} \\&=71,430-3,806=67,624 lb – ft / ft \text { of wall } \\&M_{\max }=67,624 lb – ft / ft \text { of wall }\end{aligned}