Chapter 5.3, Problem 5.83P

To determine

How many years does the dam becomes unsafe?

Answer to Problem 5.83P

The dam becomes unsafe in is service until $282\text{years}$.

Explanation of Solution

Sketch the free body diagram of the dam as shown in the Figure 1.

Write the equation for the pressure.

$P=\frac{F}{A}$ (I)

Here, the pressure is $P$, force exerted on the dam is $F$, and the area of the dam is $A$

Write the equation for force exerted on the dam without the silt.

$P_w=\frac{1}{2}A{p}_w$ (II)

Here, the force exerted on the dam is $P_w$, area of the dam is $A$, and the pressure of the horizontal force of the water is $p_w$.

Write the equation for gage pressure in a liquid.

$p_w=\rho gh$

Here, the density of the water is $\rho$, aceeleration due to gravity is $g$, and the height of the dam is $h$.

Replace $lb$ for $A$ and $\rho gh$ for $p_w$ in equation (II).

$P_w=\frac{1}{2}\left(lb\right)\rho gh$ (III)

Here, the length of the dam is $l$, breadth of the dam is $b$, aceeleration due to gravity is $g$, and the height of the dam is $h$.

Substitute $6.6\text{m}$ for $l$, $1\text{m}$ for $b$, $9.81{\text{m/s}}^2$ for $g$, $1000{\text{kg/m}}^3$ for $\rho$, and $6.6\text{m}$ for $h$ in equation (III).

$\begin{array}{c}{P}_w=\frac{1}{2}\left[\left(6.6\text{m}\right)\left(1\text{m}\right)\right]\left(1000{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(6.6\text{m}\right)\\ =213661.8\text{N}\end{array}$

Write the equation for 120 percentage of resisting force exerted on the dam.

$P_{\text{r}}=1.2P_w$

Here, the resisting force exerted on the dam is $P_{\text{r}}$.

Substitute $213661.8\text{N}$ for $P_w$ in above equation to solve for $P_{\text{r}}$.

$\begin{array}{c}{P}_{\text{r}}=1.2\left(213661.8\text{N}\right)\\ =256394.16\text{N}\end{array}$

Write the equation for force exerted on the dam after a depth that the silt has settled.

${P^{\prime }}_w=\frac{1}{2}\left(l-d\right)\left(b\right)\rho g\left(h-d\right)$ (IV)

Here, force exerted on the dam after silt is settled is ${P^{\prime }}_w$ and distance of the silt settled in the dam is $d$.

Substitute $6.6\text{m}$ for $l$, $1\text{m}$ for $b$, $9.81{\text{m/s}}^2$ for $g$, $1000{\text{kg/m}}^3$ for $\rho$, and $6.6\text{m}$ for $h$ in equation (IV).

$\begin{array}{c}{P^{\prime }}_w=\frac{1}{2}\left[\left(6.6\text{m}-d\right)\left(1\text{m}\right)\right]\left[\left(1000{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(6.6-d\right)\text{m}\right]\\ =4905{\left(6.6-d\right)}^2\text{N}\end{array}$

Write the equation for pressure force exerted on the dam above the silt at region I

(Refer fig 1).

$P_{\text{I}}=db\rho g\left(h-d\right)$ (V)

Here, force exerted on the dam above the silt at region I is $P_{\text{I}}$ and width of the dam is $b$.

Substitute $1\text{m}$ for $b$, $9.81{\text{m/s}}^2$ for $g$, $1000{\text{kg/m}}^3$ for $\rho$, and $6.6\text{m}$ for $h$ in equation (V).

$\begin{array}{c}{P}_{\text{I}}=\left[d\left(1\text{m}\right)\right]\left[\left(1000{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(6.6-d\right)\text{m}\right]\\ =9810\left(6.6d-d^2\right)\text{N}\end{array}$

Write the equation for pressure force exerted on the dam surface of the silt at region II

(Refer fig 1).

$P_{\text{II}}=\frac{1}{2}\left(db\right){\rho }_{s}g\left(d\right)$ (VI)

Here, force exerted on the surface of the silt at region II is $P_{\text{II}}$ and density of the silt is ${\rho }_s$.

Substitute $1\text{m}$ for $b$, $9.81{\text{m/s}}^2$ for $g$, and $1.76\times {10}^3{\text{kg/m}}^3$ for ${\rho }_s$ in equation (VI).

$\begin{array}{c}{P}_{\text{II}}=\frac{1}{2}\left[d\left(1\text{m}\right)\right]\left[\left(1.76\times {10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(d\right)\text{m}\right]\\ =\left(8632.8d^2\right)\text{N}\end{array}$

The net force exerted on the dam on both the regions is,

$P={P^{\prime }}_w+P_{\text{I}}+P_{\text{II}}$ (VII)

Here, the net force exerted on the dam is $P$.

Conclusion:

Substitute $4905{\left(6.6-d\right)}^2\text{N}$ for ${P^{\prime }}_w$, $9810\left(6.6d-d^2\right)\text{N}$ for $P_{\text{I}}$, and $\left(8632.8d^2\right)\text{N}$ for $P_{\text{II}}$ in Equation (VII).

$\begin{array}{c}P=4905{\left(6.6-d\right)}^2\text{N}+9810\left(6.6d-d^2\right)\text{N}+\left(8632.8d^2\right)\text{N}\\ =\left[4905\left(43.56-13.2d+d^2\right)+9810\left(6.6d-d^2\right)+\left(8632.8d^2\right)\right]\text{N}\\ =\left[213661.8-64746d+4905d^2+64746d-9810d^2+8632.8d^2\right]\text{N}\\ =\left[3727.8d^2+213661.8\right]\text{N}\end{array}$

The net force exerted on the dam is equal to the resisting force exerted on the dam.

$P=P_{\text{r}}$

Substitute $\left[3727.8d^2+213661.8\right]\text{N}$ for $P$ and $256394.16\text{N}$ for $P_{\text{r}}$ in above Equation.

$\begin{array}{c}\left[3727.8d^2+213661.8\right]\text{N}=256394.16\text{N}\\ 3727.8d^2=256394.16\text{N}-213661.8\text{N}\\ 3727.8d^2=42732.2\end{array}$

Solve the above equation for $d$

$\begin{array}{c}{d}^2=11.4631{\text{m}}^2\\ d=3.386\text{m}\end{array}$

Write the equation for number of years dam becomes unsafe.

$d=NR$

Here, the number of years dam becomes unsafe is represented as $N$ and rate of the dam to be unsafe is $R$.

Substitute $3.386\text{m}$ for $d$ and $12\text{mm/year}$ for $R$ to solve for $N$.

$\begin{array}{c}3.386\text{m}=N\left(12\text{mm/year}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ 3.386\text{m}=\left(12\times {10}^{-3}\text{m/year}\right)N\\ N=\frac{3.386\text{m}}{12\times {10}^{-3}\text{m/year}}\\ =282\text{years}\end{array}$

The dam becomes unsafe in is service until $282\text{years}$.