Solve Example 20.7 by applying Fs = 2 to the passive earth pressure.

Question

Solve Example 20.7 by applying [latex]F_{s}=2[/latex] to the passive earth pressure.

Accepted Answer

Refer to Fig. Ex. 20.8

The following equations may be written

[latex]P_{p}^{\prime}=\frac{1}{2} \gamma_{b} K_{P} D^{2} \cdot \frac{1}{F_{s}}=\frac{1}{2} imes 8.19 imes 3 D^{2} imes \frac{1}{2}=6.14 D^{2}[/latex]

[latex]p_{p}=\gamma_{b} K_{p} D=8.19 imes 3 D \approx 24.6 D[/latex]

[latex]\frac{F G}{B C}=\frac{\alpha p_{p}}{p_{p}}=\frac{D-h}{D} ext { or } h=D(1-\alpha)[/latex]

Area [latex]A B E F=\frac{D+h}{2} \quad \alpha p_{p}=\frac{D+h}{2} \quad \alpha imes 24.6 D[/latex]

or [latex]6.14 D^{2}=\alpha(D+h) imes 12.3 D[/latex]

Substituting for [latex]h=D(1-\alpha)[/latex] and simplifying

we have

[latex]2 \alpha^{2}-4 \alpha+1=0[/latex]

Solving the equation, we get [latex]\alpha=0.3[/latex].

Now [latex]h=D(1-0.3)=0.7 D ext { and } A G=D-0.7 D=0.3 D[/latex]

Taking moments of the area ABEF about the base of the pile, and assuming [latex]\alpha p_{p}=1[/latex] in Fig. Ex. 20.8 we have

[latex]\frac{1}{2}(1) imes 0.3 D \frac{1}{3} imes 0.3 D+0.7 D+(1) imes 0.7 D imes \frac{0.7 D}{2}[/latex]

[latex]=\frac{1}{2}(1) 0.3 D+(1) imes 0.7 D \bar{y}_{p}[/latex]

simplifying we have [latex]\bar{y}_{p}=0.44 D[/latex]

[latex]\bar{y}_{p}=0.44 D[/latex]

Now [latex]h_{4}=h_{3}+\left( D -\overline{y_{p}} ight)=4+(D-0.44 D)=4+0.56 D[/latex]

From the active earth pressure diagram (Fig. Ex. 20.8) we have

[latex]\begin{aligned}\bar{p}_{1} &=\gamma_{d} h_{1} K_{A}=13 imes 2 imes \frac{1}{3}=8.67 kN / m ^{2} \\bar{p}_{a} &=p_{1}+\gamma_{b}\left(h_{2}+D ight) K_{A}=8.67+\frac{8.19(3+D)}{3}=16.86+2.73 D \P_{a} &=\frac{1}{2} p_{1} h_{1}+\frac{\left(\bar{p}_{1}+\bar{p}_{a} ight)}{2}\left(h_{2}+D ight) \&=\frac{1}{2} imes 8.67 imes 2+\frac{8.67+16.86+2.73 D}{2}(3+D)=1.36 D^{2}+16.86 D+47\end{aligned}[/latex]

Taking moments of active and passive forces about the tie rod, and simplifying , we have

(a) for moments due to active forces [latex]=0.89 D^{3}+13.7 D^{2}+66.7 D+104[/latex]

(b) for moments due to passive forces [latex]=6.14 D^{2}(4+0.56 D)=24.56 D^{2}+3.44 D^{3}[/latex]

Since the sum of the moments about the anchor rod should be zero, we have

[latex]0.89 D^{3}+13.7 D^{2}+66.7 D+104=24.56 D^{2}+3.44 D^{3}[/latex]

Simplifying we have

[latex]D^{3}+4.26 D^{2}-26.16 D-40.8=0[/latex]

By solving the equation we obtain D = 4.22 m with [latex]F_{s}=2.0[/latex]

Force in the anchor rod

[latex]T_{a}=P_{a}-P_{p}^{\prime}[/latex]

where [latex]P_{a}=1.36 D^{2}+16.86 D+47=1.36 imes(4.22)^{2}+16.86 imes 4.25+47=143[/latex] kN

[latex]P_{p}^{\prime}=6.14 D^{2}=6.14 imes(4.22)^{2}=109[/latex] kN

Therefore [latex]T_{a}=143-109=34[/latex] kN/m length of wall.

Question 20.8: Solve Example 20.7 by applying Fs = 2 to the passive earth p...

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