Determine the depth of embedment and the force in the tie rod of the anchored bulkhead shown in Fig. Ex. 20.7(a). The backfill above and below the dredge line is sand, having the following properties
G_{s}=2.67, \gamma_{s a t}=18 kN / m ^{3}, \gamma_{d}=13 kN / m ^{3} \text { and } \phi=30^{\circ}Solve the problem by the free-earth support method. Assume the backfill above the water table remains dry.
Assume the soil above the water table is dry
For \phi=30^{\circ}, \quad K_{A}=\frac{1}{3}, \quad K_{P}=3.0
and K=K_{P}-K_{A}=3-\frac{1}{3}=2.67
\gamma_{b}=\gamma_{\text {sat }}-\gamma_{w}=18-9.81=8.19 kN / m ^{3}.
where \gamma_{w}=9.81 kN / m ^{3}.
The pressure distribution along the bulkhead is as shown in Fig. Ex. 20.7(b)
\begin{aligned}&\bar{p}_{1}=\gamma_{d} h_{1} K_{A}=13 \times 2 \times \frac{1}{3}=8.67 kN / m ^{2} \text { at } GW \text { level } \\&\bar{p}_{a}=\bar{p}_{1}+\gamma_{b} h_{2} K_{A}=8.67+8.19 \times 3 \times \frac{1}{3}=16.86 kN / m ^{2} \text { at dredge line level }\end{aligned}
\begin{aligned}y_{0}=& \frac{\bar{p}_{a}}{\gamma_{b} \times K}=\frac{16.86}{8.19 \times 2.67}=0.77 m \\P_{a}=& \frac{1}{2} \times \bar{p}_{1} \times h_{1}+\bar{p}_{1} \times h_{2}+\frac{1}{2}\left(\bar{p}_{a}-\bar{p}_{1}\right) h_{2}+\frac{1}{2} \bar{p}_{a} y_{0} \\=& \frac{1}{2} \times 8.67 \times 2+8.67 \times 3+\frac{1}{2}(16.86-8.67) 3 \\&+\frac{1}{2} \times 16.86 \times 0.77=53.5 kN / m \text { of wall }\end{aligned}
To find \bar{y}, taking moments of areas about 0, we have
53.5 \times \bar{y}=\frac{1}{2} \times 8.67 \times 2 \frac{2}{3}+3+0.77+8.67 \times 3(3 / 2+0.77)
+\frac{1}{2}(16.86-8.67) \times 3(3 / 3+0.77)+\frac{1}{2} \times 16.86 \times \frac{2}{3} \times 0.77^{2}=122.6
We have \bar{y}=\frac{122.6}{53.5}=2.3 m , \quad \bar{y}_{a}=4+0.77-2.3=2.47 m
Now P_{p}=\frac{1}{2} \times \gamma_{b} \times K \times D_{0}^{2}=\frac{1}{2} \times 8.19 \times 2.67 D_{0}^{2}=10.93 D_{0}^{2}
and its distance from the anchor rod is
h_{4}=h_{3}+y_{0}+2 / 3 D_{0}=4+0.77+2 / 3 D_{0}=4.77+0.67 D_{0}
Now, taking the moments of the forces about the tie rod, we have
\begin{aligned}&P_{a} \times \bar{y}_{a}=P_{p} \times h_{4} \\&53.5 \times 2.47=10.93 D_{0}^{2} \times\left(4.77+0.67 D_{0}\right)\end{aligned}
Simplifying, we have
\begin{aligned}&D_{0} \approx 1.5 m , D=y_{0}+D_{0}=0.77+1.5=2.27 m \\&D(\operatorname{design})=1.4 \times 2.27=3.18 m\end{aligned}
For finding the tension in the anchor rod, we have
P_{a}-P_{p}-T_{a}=0
Therefore, T_{a}=P_{a}-P_{p}=53.5-10.93(1.5)^{2}=28.9 kN/m of wall for the calculated depth D_{o}.