Chapter 5.3, Problem 5.83P

The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density ρ_s = 1.76 × 10³ kg/m³) is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years of service until the dam becomes unsafe.

Fig. P5.82 and P5.83

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 $282\text{years}$.