Chapter 2, Problem 2.7CP

To determine

To estimate: the angle at which levee floats.

Answer to Problem 2.7CP

$\theta \approx 26.3°$

Explanation of Solution

Given information:

Height of levee is $30m$

Width of levee is $5m$

Depth is $20m$

Density is equal to $300kg/m^3$

Find the summation of moments with respect to hinge to find the value of angle

The forces acting on the levee are shown below,

Assume,

Density of seawater as,

${\rho }_{seawater}=1025kg/m^3$

Calculation:

The submerged length $L_{AS}$ of levee on adiabatic side is equal to,

$L_{AS}=25/\cos \theta$

Therefore, calculate the force $F_{AS}$ on levee for unit width,

$\begin{array}{l}{F}_{AS}=\rho g{h}_{AS}{L}_{AS}\\ =\left(1025kg/m^3\right)\left(9.81m/s^2\right)\left(\frac{25}{2}\right)\left(\frac{25}{\cos \theta }\right)\\ =\frac{3142265.625}{\cos \theta }\end{array}$

The submerged length $L_{LS}$ of levee on lagoon side is equal to,

$L_{LS}=24/\cos \theta$

Therefore, calculate the force $F_{LS}$ on levee for unit width,

$\begin{array}{l}{F}_{LS}=\rho g{h}_{LS}{L}_{LS}\\ =\left(1025kg/m^3\right)\left(9.81m/s^2\right)\left(\frac{24}{2}\right)\left(\frac{24}{\cos \theta }\right)\\ =\frac{2895912}{\cos \theta }\end{array}$

Find the weight of levee,

$\begin{array}{l}W=A{\rho }_{levee}g\\ =\left(30m\right)\left(5m\right)\left(300kg/m^3\right)\left(9.81m/s^2\right)\\ =441450N\end{array}$

Find the buoyancy force,

The average submerged length is equal to,

$L_{average}=\left(\frac{24m+25m}{2}\right)=24.5m$

Therefore,

$\begin{array}{l}{F}_{Bouyancy}=\rho g{L}_{average}{\left(width\right)}_{leeve}\\ =\left(1025kg/m^3\right)\left(9.81m/s^2\right)\left(24.5m\right)\left(5m\right)\\ =1231768.125N\end{array}$

Calculate the moments about hinge,

${\displaystyle \sum M=}{F}_{AS}\left(\frac{25/\cos \theta }{3}\right)+W\left(15\sin \theta \right)-F_{LS}\left(\frac{24/\cos \theta }{3}\right)-F_{Bouyancy}\left(\frac{24.5}{2}\sin \theta \right)=0$

Substitute for forces,

$\begin{array}{l}{\displaystyle \sum M=}\left(\frac{3142265.625}{\cos \theta }\right)\left(\frac{25/\cos \theta }{3}\right)+\left(441450N\right)\left(15\sin \theta \right)\\ -\left(\frac{2895912}{\cos \theta }\right)\left(\frac{24/\cos \theta }{3}\right)-\left(1231768.125N\right)\left(\frac{24.5}{2}\sin \theta \right)=0\end{array}$

Solve to find angle $\theta$,

$\theta \approx 26.3°$

Conclusion:

The levee floats at an angle of $\theta \approx 26.3°$.