Question 20.7: Determine the depth of embedment and the force in the tie ro...

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)

To find $\bar{y}$, taking moments of areas about 0, we have

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

Now, taking the moments of the forces about the tie rod, we have

Simplifying, we have

For finding the tension in the anchor rod, we have

Therefore, $T_{a}=P_{a}-P_{p}=53.5-10.93(1.5)^{2}=28.9$ kN/m of wall for the calculated depth $D_{o}$.