Chapter 18, Problem 18.1P

Refer to Figure 18.9. A cantilever sheet pile is driven into a granular soil where the water table is 2 m (L₁) below the top of the sand. The properties of the sand are ${\varphi }^{\prime }=40°$, $\gamma =17.5\text{kN}/{\text{m}}^3$, and ${\gamma }_{\text{sat}}=19\text{kN}/{\text{m}}^3$. It is proposed to excavate to a depth of 6 m (L) below the ground level. Determine the actual depth to which the sheet pile must be driven (L + D), using the net lateral pressure diagram. Note:

$D_{\text{actual}}=1.3{\left(L_3+L_4\right)}_{\text{theory}}$

To determine

Find the required actual depth of the sheet pile.

Answer to Problem 18.1P

The required actual depth of the sheet pile is $\underset{\_}{12.25\text{m}}$.

Explanation of Solution

Given information:

The depth of water level below the sand $\left(L_1\right)$ is 2 m.

The angle of internal friction for sand $\left({\varphi }^{\prime }\right)$ is $40°$.

The unit weight of the sand $\left(\gamma \right)$ is $\text{17}\text{.5}{\text{kN/m}}^{\text{3}}$.

The saturated unit weight of the sand $\left({\gamma }_{sat}\right)$ is $\text{19}{\text{kN/m}}^{\text{3}}$.

The depth below the ground level (L) is 6 m.

Calculation:

Show the cross section of cantilever sheet pile with dimensions as in Figure (1).

Refer Figure (1),

Find the depth of water table level $\left(L_2\right)$:

$\begin{array}{c}{L}_2=L-L_1\\ =6-2\\ =4\text{m}\end{array}$

Find the rankine active pressure coefficient $\left(K_a\right)$ using the relation:

$\begin{array}{c}\left(K_a\right)={\tan }^2\left(45-\frac{{\varphi }^{\prime }}{2}\right)\\ ={\tan }^2\left(45-\frac{40°}{2}\right)\\ =0.217\end{array}$

Find the rankine passive pressure coefficient $\left(K_p\right)$ using the relation:

$\begin{array}{c}{K}_p={\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ ={\tan }^2\left(45+\frac{40°}{2}\right)\\ =4.599\end{array}$

Find the difference between the rankine active pressure coefficient and rankine passive pressure coefficient:

$\begin{array}{c}{K}_p-K_a=4.599-0.217\\ =4.382\end{array}$

Find the effective unit weight of sand $\left({\gamma }^{\prime }\right)$:

$\begin{array}{c}{\gamma }^{\prime }={\gamma }_{sat}-{\gamma }_w\\ =19-9.81\\ =9.19{\text{kN/m}}^{\text{3}}\end{array}$

At the water table level:

Find the intensity of the active pressure to the right of the pile $\left({{\sigma }^{\prime }}_1\right)$:

$\begin{array}{c}\left({{\sigma }^{\prime }}_1\right)=K_a{L}_1\gamma \\ =0.217\times 2\times 17.5\\ =7.60{\text{kN/m}}^{\text{2}}\end{array}$

At the excavation level:

Find the intensity of the active pressure to the right of the pile $\left({{\sigma }^{\prime }}_2\right)$:

$\begin{array}{c}\left({{\sigma }^{\prime }}_2\right)=K_a\left(L_1\gamma +L_2\times {\gamma }^{\prime }\right)\\ =0.217\times \left(2\times 17.5+4\times 9.19\right)\\ =19.45{\text{kN/m}}^{\text{2}}\end{array}$

Find the depth below the dredge line $\left(L_3\right)$:

$\begin{array}{c}{L}_3=\frac{{{\sigma }^{\prime }}_2}{\left(K_p-K_a\right){\gamma }^{\prime }}\\ =\frac{19.45}{4.382\times 9.19}\\ =0.48\text{m}\end{array}$

Find the area of the pressure diagram (P):

$\begin{array}{c}P=0.5\times {{\sigma }^{\prime }}_1\times L_1+{{\sigma }^{\prime }}_1\times L_2+0.5\times \left({{\sigma }^{\prime }}_2-{{\sigma }^{\prime }}_1\right)\times L_2+0.5\times {{\sigma }^{\prime }}_2\times L_3\\ =0.5\times 7.60\times 2+7.60\times 4+0.5\times \left(19.45-7.60\right)\times 4+0.5\times 19.45\times 0.48\\ =66.37\text{kN/m}\end{array}$

Find the area of the pressure diagram into center of pressure $\left(P\overline{z}\right)$:

$\begin{array}{c}P\overline{z}=\left\{\begin{array}{l}0.5\times {{\sigma }^{\prime }}_1\times L_1\left(\frac{1}{3}{L}_1+L_2+L_3\right)+{{\sigma }^{\prime }}_1\times L_2\left(L_1+L_3\right)+0.5\times \left({{\sigma }^{\prime }}_2-{{\sigma }^{\prime }}_1\right)\times L_2\left(\frac{1}{3}{L}_2+L_3\right)\\ +0.5\times {{\sigma }^{\prime }}_2\times L_3\left(\frac{2}{3}{L}_3\right)\end{array}\right\}\\ =\left\{\begin{array}{l}0.5\times 7.60\times 2\left(\frac{1}{3}\times 2+4+0.48\right)+7.60\times 4\left(2+0.48\right)\\ +0.5\times \left(19.45-7.60\right)\times 4\left(\frac{1}{3}\times 4+0.48\right)\\ +0.5\times 19.45\times 0.48\left(\frac{2}{3}\times 0.48\right)\end{array}\right\}\\ =158.92\text{kN-m/m}\end{array}$

Find the center of pressure to the area $\overline{z}$:

$\begin{array}{c}\overline{z}=\frac{P\overline{z}}{P}\\ =\frac{158.92}{66.37}\\ =2.39\text{m}\end{array}$

Find the intensity of the passive pressure $\left({{\sigma }^{\prime }}_5\right)$:

$\begin{array}{c}\left({{\sigma }^{\prime }}_5\right)=K_p\left(L_1\gamma +L_2\times {\gamma }^{\prime }\right)+\left(K_p-K_a\right){\gamma }^{\prime }\times L_3\\ =4.599\times \left(2\times 17.5+4\times 9.19\right)+4.382\times 9.19\times 0.48\\ =349.35{\text{kN/m}}^{\text{2}}\end{array}$

Find the area $\left(A_1\right)$:

$\begin{array}{c}{A}_1=\frac{{{\sigma }^{\prime }}_5}{{\gamma }^{\prime }\left(K_p-K_a\right)}\\ =\frac{349.35}{9.19\times 4.382}\\ =8.68\text{m}\end{array}$

Find the area $\left(A_2\right)$:

$\begin{array}{c}{A}_2=\frac{8P}{{\gamma }^{\prime }\left(K_p-K_a\right)}\\ =\frac{8\times 66.37}{9.19\times 4.382}\\ =13.18{\text{m}}^2\end{array}$

Find the area $\left(A_3\right)$:

$\begin{array}{c}{A}_3=\frac{6P\left(2\overline{z}{\gamma }^{\prime }\left(K_p-K_a\right)+{{\sigma }^{\prime }}_5\right)}{{{\gamma }^{\prime }}^2{\left(K_p-K_a\right)}^2}\\ =\frac{6\times 66.37\left[2\times 2.39\times 9.19\left(4.382\right)+349.35\right]}{{9.19}^2\times {4.382}^2}\\ =133.05{\text{m}}^3\end{array}$

Find the area $\left(A_4\right)$:

$\begin{array}{c}{A}_4=\frac{P\left(6\overline{z}{{\sigma }^{\prime }}_5+4P\right)}{{{\gamma }^{\prime }}^2{\left(K_p-K_a\right)}^2}\\ =\frac{66.37\left[6\times 2.39\times 349.35+4\times 66.37\right]}{{9.19}^2\times {4.382}^2}\\ =215.89{\text{m}}^4\end{array}$

Find the depth below point E $\left(L_4\right)$:

$\begin{array}{l}{\left(L_4\right)}^4+A_1{\left(L_4\right)}^3-A_2{\left(L_4\right)}^2-A_3\left(L_4\right)-A_4=0\\ {\left(L_4\right)}^4+8.68{\left(L_4\right)}^3-13.18{\left(L_4\right)}^2-133.05\left(L_4\right)-215.89=0\end{array}$ (1)

Use trial and error method to calculate depth below point E $\left(L_4\right)$.

Try $\left(L_4\right)=4.30\text{m}$:

Substitute 4.30 m for $L_4$ in Equation (4).

$\begin{array}{l}{\left(4.30\right)}^4+8.68{\left(4.30\right)}^3-13.18{\left(4.30\right)}^2-133.05\left(4.30\right)-215.89=0\\ 341.88+690.12-243.70-572.115-215.89=0\\ 0=0\end{array}$

Hence, the assumption is correct.

The depth below point E is 4.30 m.

Find the depth below the dredge line to bottom of the pile (D):

$\begin{array}{c}D=L_4+L_3\\ =4.30+0.48\\ =4.78\text{m}\end{array}$

Increase the depth below the dredge line to bottom of the pile by 30%. The depth below the dredge line to bottom of the pile (D) is 6.214 m.

Find required actual depth of the sheet pile $\left(L+D\right)$:

$\begin{array}{c}\left(L+D\right)=6+6.214\\ =12.214\text{m}\\ \approx 12.25\text{m}\end{array}$

Thus, the required actual depth of the sheet pile is $\underset{\_}{12.25\text{m}}$.