Refer to Example 20.7. Determine for the pile (a) the bending moment Mmax, and (b) the reduced moment by Rowe's method.

Question

Refer to Example 20.7. Determine for the pile (a) the bending moment [latex]M_{\max }[/latex], and (b) the reduced moment by Rowe's method.

Accepted Answer

Refer to Fig. Ex. 20.7. The following data are available

[latex]H=5 m , h_{1}=2 m , h_{2}=3 m , h_{3}=4 m[/latex]

[latex]\gamma_{d}=13 kN / m ^{3}, \gamma_{b}=8.19 kN / m ^{3} ext { and } \phi=30^{\circ}[/latex]

The maximum bending moment occurs at a depth [latex]h_{m}[/latex] from ground level where the shear is zero. The equation which gives the value of [latex]h_{m}[/latex] is

[latex]\frac{1}{2} \bar{p}_{1} h_{1}-T_{a}+\bar{p}_{1}\left(h_{m}-h_{1} ight)+\frac{1}{2} \gamma_{b} K_{A}\left(h_{m}-h_{1} ight)^{2}=0[/latex]

where [latex]\bar{p}_{1}=8.67 kN / m ^{2}[/latex]

[latex]h_{1}=2 m , T_{a}=28.9 kN / m[/latex]

Substituting

[latex]\frac{1}{2} imes 8.67 imes 2-28.9+8.67\left(h_{m}-2 ight)+\frac{1}{2} imes 8.19 imes \frac{1}{3}\left(h_{m}-2 ight)^{2}=0[/latex]

Simplifying, we have

[latex]h_{m}^{2}+2.35 h_{m}-23.5=0[/latex]

or [latex]h_{m} \approx 3.81 m[/latex]

Taking moments about the point of zero shear,

[latex]\begin{aligned}M_{\max } &=-\frac{1}{2} \bar{p}_{1} h_{1} h_{m}-\frac{2}{3} h_{1}+T_{a}\left(h_{m}-h_{a} ight)-\bar{p}_{1} \frac{\left(h_{m}-h_{1} ight)^{2}}{2}-\frac{1}{6} \gamma_{b} K_{A}\left(h_{m}-h_{1} ight)^{3} \&=-\frac{1}{2} imes 8.67 imes 23.81-\frac{2}{3} imes 2+28.9(3.81-1)-8.67 \frac{(3.81-2)^{2}}{2}-\frac{1}{6} 8.19 imes \frac{1}{3}(3.81-2)^{3} \&=-21.41+81.2-14.2-2.70 \approx 42.8 kN - m / m\end{aligned}[/latex]

From Ex. 20.7

D (design) = 3.18m, H = 5m

Therefore [latex]\bar{H}=8.18 m[/latex]

From Eq. (20.47a) [latex] ho=109 imes 10^{-6} \frac{\bar{H}^{4}}{E I}[/latex]

[latex] ho=109 imes 10^{-6} \frac{\bar{H}^{4}}{E I}[/latex] (20.47a)

Assume [latex]E=20.7 imes 10^{4} MN / m ^{2}[/latex]

For a section [latex]P_{z}-27, \quad I=25.2 imes 10^{-5} m ^{4} / m[/latex]

Substituting [latex] ho=\frac{10.9 imes 10^{-5} imes(8.18)^{4}}{2.07 imes 10^{5} imes 25.2 imes 10^{-5}}=9.356 imes 10^{-3}[/latex]

[latex]\log ho=\log \frac{9.36}{10^{3}}=-2.0287 ext { or say } 2.00[/latex]

Assuming the sand backfill is loose, we have from Fig. 20.20 (a)

[latex]\frac{M_{d}}{M_{\max }}=0.32 ext { for } \log ho=-2.00[/latex]

Therefore [latex]M_{d}(\operatorname{design})=0.32 imes 42.8=13.7 kN - m / m[/latex]

Question 20.11: Refer to Example 20.7. Determine for the pile (a) the bendin...

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