Question 20.5: Refer to Fig. Ex. 20.5. Solve the problem in Ex. 20.4 if the...
\begin{aligned}&h_{1}=2.5 m , h_{2}=6-2.5=3.5 m , \gamma_{ b }=17-9.81=7.19 kN / m ^{3} \\&p_{1}=\gamma h_{1} K_{A}=17 \times 2.5 \times 1 / 3=14.17 kN / m ^{2} \\&\bar{p}_{a}=p_{1}+\gamma_{ b } h_{2} K_{A}=14.17+7.19 \times 3.5 \times 1 / 3=22.56 kN / m ^{2} \\&P_{a}=\frac{1}{2} p_{1} h_{1}+p_{1} h_{2}+\frac{1}{2}\left(\bar{p}_{a}-p_{1}\right) h_{2} \\&=\frac{1}{2} \times 14.17 \times 2.5+14.17 \times 3.5+\frac{1}{2}(22.56-14.17) \times 3.5 \\&=17.71+49.6+14.7=82 kN / m\end{aligned}
Determination of \bar{y} (Refer to Fig. 20.12)
Taking moments of all the forces above dredge line about C we have
\begin{aligned}&82 \bar{y}=17.71\left(3.5+\frac{2.5}{3}\right)+49.6 \times \frac{3.5}{2}+14.7 \times \frac{3.5}{3} \\&=76.74+86.80+17.15=180.69 \\&\bar{y}=\frac{180.69}{82}=2.20 m\end{aligned}
From Eq. (20.27), the equation for D is
C_{1} D^{2}+C_{2} D+C_{3}=0 (20.27)
C_{1} D^{2}+C_{2} D+C_{3}=0
where \begin{aligned}C_{1} &=\left[2 q_{u}-\left(\gamma h_{1}+\gamma_{b} h_{2}\right)\right] \\&=[140-(17 \times 2.5+7.19 \times 3.5)]=72.3\end{aligned}
\begin{aligned}&C_{2}=-2 P_{a}=-2 \times 82=-164 \\&C_{3}=-\frac{\left(P_{a}+6 q_{u} \bar{y}\right) P_{a}}{q_{u}+\left(\gamma h_{1}+\gamma_{b} h_{2}\right)}=-\frac{(82+6 \times 70 \times 2.2) \times 82}{70+17 \times 2.5+7.19 \times 3.5)}=-599\end{aligned}
Substituting we have,
72.3 D^{2}-164 D-599=0
or D^{2}-2.27 D-8.285=0
solving we have D = 4.23 m
Increasing D by 40%; the design value is
D (design) = 1.4(4.23) = 5.92 m
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