Question 20.6: Fig. Example 12.6 gives a freestanding sheet pile penetratin...

FromEq. (20.13a)

\phi=0 \quad \bar{p}_{a}=\gamma H-2 c=\gamma H-q_{u} (20.13a)

From Eq. (20.28), The expression for D is

C_{1} D^{2}+C_{2} D+C_{3}=0 (20.28)

where \begin{aligned}&C_{1}=2 q_{u}=60 \\&C_{2}=-2 P=-2 \times 40=-80 \\&C_{3}=-\frac{\left(P+6 q_{u} H\right) P}{q_{u}}=-\frac{(40+6 \times 30 \times 5) 40}{30}=-1253\end{aligned}

Substituting and simplifying

or D^{2}-1.33 D-21=0

Solving D \approx 5.3 m. Increasing by 40% we have

D (design) = 1.4(5.3) = 7.42 m

From Eq. (20.29)

h=\frac{2 q_{u} D-P}{2 q_{u}} (20.29)