Chapter 8, Problem 8.118P

A concrete seawater dam is shown in cross section. Is the dam safe against tipping about edge A? The mass densities are $24000{\text{kg/m}}^{\text{3}}$ for concrete and $1030{\text{kg/m}}^{\text{3}}$ for seawater.

To determine

Whether the dam safe against tipping about edge A

Answer to Problem 8.118P

The distance $x$ is equal to $2.035m$ which is less than $2.5m$, therefore safe against tipping about edge A.

Explanation of Solution

Given information:

Densities of concrete and sea water are $2400kg/m^3,1030kg/m^3$ respectively.

The resultant force P is defined as:

  $P=pA$

In above equation

  $p$ - Pressure

  $A$ - Area

The weight 'W' of water is defined as:

  $W=\rho g\times V$

  $V$ - Volume

Calculation:

Assume $R$ as the resultant force and $N$ as the normal force.

Assume $W_1,W_2,W_3$ as the weight of water acting on the dam.

Assume $x,y^1$ as the line of action.

The resultant force $R$ of 1m length of dam

  $\begin{array}{l}R=\frac{1}{2}ph\\ =\frac{1}{2}\rho g{h}^2\\ =\frac{1}{2}\left(1030kg/m^3\right)\left(9.81m/s^2\right){\left(4m\right)}^2\\ =80834.4N\end{array}$

The distance $y^1$

  $y^1=\frac{4}{3}=1.333m$

The weight $W_1$ of water

  $\begin{array}{l}{W}_1=\frac{2}{3}\rho gah\\ =\frac{2}{3}\left(1030kg/m^3\right)\left(9.81m/s^2\right)\left(1.5m\right)\left(4m\right)\\ =40417.2N\end{array}$

The distance $x_1$

  $x_1=\frac{2}{5}a=\frac{2}{5}\left(1.5\right)=0.6m$

The weight $W_2$ of concrete

  $\begin{array}{l}{W}_2=g{\rho }_{c}bh\\ =\left(2400kg/m^3\right)\left(9.81m/s^2\right)\left(1m\right)\left(4m\right)\\ =94176N\end{array}$

The distance $x_2$

  $\begin{array}{l}{x}_2=a+\frac{b}{2}\\ =2m\end{array}$

The weight $W_3$ of concrete

  $\begin{array}{l}{W}_3=\frac{1}{3}{\rho }_{c}gah\\ =\frac{1}{3}\left(2400kg/m^3\right)\left(9.81m/s^2\right)\left(1.5m\right)\left(4m\right)\\ =47088N\end{array}$

The distance $x_3$

  $x_3=\frac{7}{10}a=\frac{7}{10}\left(1.5m\right)=1.05m$

Write equilibrium equation in y direction

  $\uparrow {\displaystyle \sum F_y=0}$

  $N-W_1-W_2-W_3=0$

Substitute and solve

  $\begin{array}{l}N=40417.2N+94176N+47088N\\ =181681.2N\end{array}$

For the equilibrium of the dam, the moment about point O is equal to zero.

  ${\displaystyle \sum M_O=0}$

  $\begin{array}{l}Nx=\left(80834.4N\right)\left(1.333m\right)+\left(40417.2N\right)\left(0.6m\right)+\left(94176N\right)\left(2m\right)+\left(47088N\right)\left(1.05m\right)\\ =369.8\times {10}^{3}Nm\end{array}$

Therefore

  $x=\frac{369796.975Nm}{181681.2N}=2.035m$

The distance $x$ is less than $2.5m$, therefore safe against tipping about edge A.

Conclusion:

The distance $x$ is equal to $2.035m$ which is less than $2.5m$, therefore safe against tipping about edge A.