Determine the depth of embedment for the sheet pile given in Fig. Example 20.7 using the design charts given in Section 20.7.

Given: $H=5 m , h_{1}=2 m , h_{2}=3 m , h_{a}=1 m , h_{3}=4 m , \phi=30^{\circ}$

Question

Determine the depth of embedment for the sheet pile given in Fig. Example 20.7 using the design charts given in Section 20.7.

Given: [latex]H=5 m , h_{1}=2 m , h_{2}=3 m , h_{a}=1 m , h_{3}=4 m , \phi=30^{\circ}[/latex]

Accepted Answer

[latex]\frac{h_{a}}{H}=\frac{1}{5}=0.2[/latex]

From Fig. 20.18

For [latex]h_{a} / H=0.2, ext { and } \phi=30^{\circ}[/latex] we have

[latex]G_{d}=0.26, G_{t}=0.084, G_{m}=0.024[/latex]

From Fig. 20.19 for [latex]\phi=30, h_{a} / H=0.2 ext { and } h_{1} / H=0.4[/latex] we have

[latex]C_{d}=1.173, C_{t}=1.073, C_{m}=1.036[/latex]

Now from Eq. (20.46 a)

[latex]D=G_{d} C_{d} H[/latex] (20.46a)

[latex]D=G_{d} C_{d} H=0.26 imes 1.173 imes 5=1.52 m[/latex]

[latex]D(\operatorname{design})=1.4 imes 1.52=2.13 m[/latex].

[latex]T_{a}=G_{t} C_{t} Y_{a} H^{2}[/latex]

where [latex]\gamma_{a}=\frac{\left(\gamma_{m} h_{1}^{2}+\gamma_{b} h_{2}^{2}+2 \gamma_{m} h_{1} h_{2} ight)}{H^{2}}[/latex]

[latex]=\frac{13 imes 2^{2}+8.19 imes 3^{2}+2 imes 13 imes 2 imes 3}{5^{2}}=11.27 kN / m ^{2}[/latex]

Substituting and simplifying

[latex]T_{a}=0.084 imes 1.073 imes 11.27 imes 5^{2}=25.4 kN[/latex]

[latex]M_{\max }=G_{m} C_{m} \gamma_{a} H^{3}[/latex]

[latex]=0.024 imes 1.036 imes 11.27 imes 5^{3}=35.03[/latex] kN-m/m of wall

The values of D (design) and [latex]T_{a}[/latex] from Ex. 20.7 are

[latex]D(\operatorname{design})=3.18 m[/latex]

[latex]T_{a}=28.9 kN[/latex]

The design chart gives less by 33% in the value of D (design) and 12% in the value of [latex]T_{a}[/latex].

Question 20.10: Determine the depth of embedment for the sheet pile given in...