Chapter 5.3, Problem 5.81P

(a)

To determine

The reaction forces exerted by the ground on the base of the concrete dam $AB$.

Answer to Problem 5.81P

The resultant reaction forces acts on the base of the dam is $H=44.1\text{kN}$ and $V=228\text{kN}$ acts in the upward direction.

Explanation of Solution

Given that the width of the dam section $w$ is $1\text{ft}$. The height of the dam at the point $C$ is $h=18\text{ft}$.

The free-body diagram consists of dam and the triangular section $BDE$ of water above the dam is shown in the Figure 1.

The length $x_1$ is acts on the load $W_1$, the distance from the base to the center of the dam is $x_2$ acts on the load $W_2$, the distance from the base to the point $D$ is $x_3$ acts on the load $W_3$, other forces acting on the free body are the weight of the dam represented by the weights of its components are $W_1$, $W_2$, and the weight of the water is $W_3$, and the resultant pressure forces exerted on section $BD$ by the water is $P$.

Write the equation for weight force acts on the dam.

$W=\rho ghA$ (I)

Here, the weight of the dam is $W$, density of the object is $\rho$, acceleration due to gravity is $g$, height of the dam is $h$, and the area of the cross section of the dam is $A$.

Replace $bt$ for $A$ in the equation (I).

$W=\rho gh\left(bt\right)$

Here, the thickness of the dam section is $t$ and the breadth of the dam section is $b$.

Write the equation for the weight of the dam represented by the weights of its components. $W_1={\rho }_{C}g{h}_1\left(b_{1}t\right)$

Here, the weight of the dam by the components of fist section is $W_1$, density of the concrete dam is ${\rho }_C$, acceleration due to gravity is $g$, the thickness of the dam section is $t$, the breadth of the dam section at $C$ is $b_1$, and the height of the dam from the ground to cross section of dam is $h_1$.

Substitute $2.4\times {10}^3{\text{kg/m}}^3$ for ${\rho }_C$, $9.81{\text{m/s}}^2$ for $g$, $4\text{m}$ for $h_1$, $1.5\text{m}$ for $b_1$ and $1\text{m}$ for $t$.

$\begin{array}{c}{W}_1=\left(2.4\times {10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(4\text{m}\right)\left(1.5\text{m}\right)\left(1\text{m}\right)\\ =141.264\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =141.26\text{kN}\end{array}$

Write the equation for the weight of the dam represented in the triangular section.

$W_2=\frac{1}{3}{\rho }_{C}g\left(b_2{h}_{2}t\right)$

Here, the weight of the dam by the components of second section is $W_2$, the thickness of the dam section is $t$, the breadth of the dam section at $CE$ is $b_2$, and the height of the dam from the ground section $EB$ is $h_2$.

Substitute $2.4\times {10}^3{\text{kg/m}}^3$ for ${\rho }_C$, $9.81{\text{m/s}}^2$ for $g$, $3\text{m}$ for $h_2$, $2\text{m}$ for $b_2$ and $1\text{m}$ for $t$.

$\begin{array}{c}{W}_2=\frac{1}{3}\left(2.4\times {10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(2\text{m}\right)\left(3\text{m}\right)\left(1\text{m}\right)\\ =47.088\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =47.09\text{kN}\end{array}$

Write the equation for the weight of the dam represented in the parabola section by the weights of its components.

$W_3=\frac{2}{3}{\rho }_{w}g\left(b_3{h}_{3}t\right)$

Here, the weight of the dam by the components of third section is $W_3$, density of the water is ${\rho }_w$, the breadth of the dam section at $CE$ is $b_2$, and the height of the dam from the ground section $EB$ is $h_3$.

Substitute ${10}^3{\text{kg/m}}^3$ for ${\rho }_w$, $9.81{\text{m/s}}^2$ for $g$, $3\text{m}$ for $h_3$, $2\text{m}$ for $b_3$ and $1\text{m}$ for $t$.

$\begin{array}{c}{W}_3=\frac{2}{3}\left({10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(2\text{m}\right)\left(3\text{m}\right)\left(1\text{m}\right)\\ =39.24\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =39.24\text{kN}\end{array}$

Write the equation of the force pressure exerted by the ground on the base of the dam.

$P=\frac{1}{2}Ap$ (II)

Here, the reaction force exerted on the dam is $P$, cross section area of the dam is $A$, and the pressure of the water is $p$.

Replace $\rho gh$ for $p$ and $hw$ for $A$ in equation (II).

$P=\frac{1}{2}\left(hw\right)\left({\rho }_{w}gh\right)$ (III)

Write the equilibrium equation for the section of dam acts along x axis (Refer Fig 1).

$\begin{array}{c}{\displaystyle \sum F_x}=0\\ H-P=0\end{array}$ (IV)

Here, the reaction force exerted by the ground on the base $AB$ of the dam is $H$.

Write the equilibrium equation for the section of beam acts along y axis and then calculate the reaction force (Refer Fig 1).

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ V-W_1-W_2-W_3=0\end{array}$ (V)

Here, the reaction force exerted by the ground on the base $AB$ of the dam is $V$.

Conclusion:

Substitute ${10}^3{\text{kg/m}}^3$ for ${\rho }_w$, $9.81{\text{m/s}}^2$ for $g$, $3\text{m}$ for $h$, and $1\text{m}$ for $w$. in equation (III) and solve for $P$.

$\begin{array}{c}P=\frac{1}{2}\left(3\text{m}\right)\left({10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(1\text{m}\right)\left(3\text{m}\right)\\ =44.145\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =44.145\text{kN}\end{array}$

Substitute $44.145\text{kN}$ for $P$ in equation (IV) and solve for $H$.

$\begin{array}{c}H-44.145\text{kN}=0\\ H=44.145\text{kN}\\ =44.1\text{kN}\end{array}$

Substitute $141.26\text{kN}$ for $W_1$, $47.09\text{kN}$ for $W_2$, and $39.24\text{kN}$ for $W_3$, in equation (V) and solve for $V$.

$\begin{array}{c}V-141.26\text{kN}-47.09\text{kN}-39.24\text{kN}=0\\ V-227.59\text{kN}=0\\ V=227.6\text{kN}\\ =228\text{kN}\end{array}$

Therefore, the resultant reaction forces acts on the base of the dam is $H=44.1\text{kN}$ and $V=228\text{kN}$ acts in the upward direction.

(b)

To determine

The point of forces acts on the base $AB$ of the dam from the resultant of part a.

Answer to Problem 5.81P

The point in which the forces acts on the base $AB$ of the dam is $x=1.167\text{m}$ acting toward right side of $A$.

Explanation of Solution

The distance from the base of the dam to the point $A$ acts on the load $W_1$ is

$\begin{array}{c}{x}_1=\frac{1}{2}\left(1.5\text{m}\right)\\ =0.75\text{m}\end{array}$

The distance from the base of the dam to the side $BD$ acts on the load $W_2$ is.

$\begin{array}{c}{x}_2=1.5\text{m}+\frac{1}{4}\left(2\text{m}\right)\\ =2\text{m}\end{array}$

The distance from the base of the dam to the point $DE$ acts on the load $W_3$ (Refer fig 1) is.

$\begin{array}{c}{x}_3=1.5\text{m}+\frac{5}{8}\left(2\text{m}\right)\\ =2.75\text{m}\end{array}$

Write the equilibrium equation for the section on the base $AB$ of the dam and calculate the moment labeled at point $A$ (Refer Fig 1).

$\begin{array}{c}{\displaystyle \sum M_A}=0\\ xV-W_1{x}_1-W_2{x}_2-W_3{x}_3+Pw=0\end{array}$ (VI)

Here, the different section of the dam is represented as $x_1$, $x_2$, and $x_3$.

Conclusion:

Substitute $144.26\text{kN}$ for $W_1$, $47.09\text{kN}$ for $W_2$, $39.24\text{kN}$ for $W_3$, $228\text{kN}$ for $V$, $0.75\text{m}$ for $x_1$, $2\text{m}$ for $x_2$, $2.75\text{m}$ for $x_3$, $44.145\text{kN}$ for $P$, and $1\text{m}$ for $w$ in equation (VI) and to solve $x$.

$\begin{array}{c}x\left(228\text{kN}\right)-\left(144.26\text{kN}\right)\left(0.75\text{m}\right)-\left(47.09\text{kN}\right)\left(2\text{m}\right)-\left(39.24\text{kN}\right)\left(2.75\text{m}\right)+\left(44.145\text{kN}\right)\left(1\text{m}\right)=0\\ x\left(228\text{kN}\right)-\left(108.195\text{kN}\cdot \text{m}\right)-\left(94.18\text{kN}\cdot \text{m}\right)-\left(107.91\text{kN}\cdot \text{m}\right)+\left(44.145\text{kN}\cdot \text{m}\right)=0\\ x\left(228\text{kN}\right)-\left(310.285\text{kN}\text{.m}\right)+\left(44.145\text{kN}\cdot \text{m}\right)=0\end{array}$

Solve the above equation for $x$.

$\begin{array}{c}x\left(228\text{kN}\right)-\left(310.285\text{kN}\text{.m}\right)+\left(44.145\text{kN}\cdot \text{m}\right)=0\\ x\left(228\text{kN}\right)-\left(266.14\text{kN}\cdot \text{m}\right)=0\\ x\left(228\text{kN}\right)=\left(266.14\text{kN}\cdot \text{m}\right)\\ x=1.167\text{m}\end{array}$

Therefore, the point in which the forces acts on the base $AB$ of the dam is $x=1.167\text{m}$ acting toward right side of $A$.

(c)

To determine

The resultant pressure force exerted by the water on the face $BC$ of the dam.

Answer to Problem 5.81P

The resultant pressure force exerted by the water on the face $BC$ of the dam is $R=59.1\text{kN}$ at the angle of $41.6°$.

Explanation of Solution

The free body diagram of the water section $BDE$ in the dam as shown in figure 2:

Write the equilibrium equation for the s resultant pressure force exerted by the water on the face $BC$ of the dam (Refer Fig 2).

$\begin{array}{c}{\displaystyle \sum F}=0\\ R=\sqrt{P^2+W_3^2}\end{array}$ (VII)

Here, the resultant pressure force exerted by the water on the dam is $R$.

Solve for the angle of resultant force exerted by the water on the dam by using trigonometric relation (Refer fig 3).

$\begin{array}{c}\tan \theta =\frac{opp}{adj}\\ \theta ={\tan }^{-1}\left(\frac{W_3}{P}\right)\end{array}$ (VIII)

Conclusion:

Substitute $39.24\text{kN}$ for $W_3$ and $44.145\text{kN}$ for $P$ in equation (VII) and to solve for $R$.

$\begin{array}{c}R=\sqrt{{\left(44.145\text{kN}\right)}^2+{\left(39.24\text{kN}\right)}^2}1948.781025\\ =59.06\text{kN}\\ =59.1\text{kN}\end{array}$

Substitute $39.24\text{kN}$ for $W_3$ and $44.145\text{kN}$ for $P$ in equation (VIII) and to solve for $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{39.24\text{kN}}{44.145\text{kN}}\right)\\ ={\tan }^{-1}\left(0.8889\right)\\ =41.6°\end{array}$

Therefore, the resultant pressure force exerted by the water on the face $BC$ of the dam is $R=59.1\text{kN}$ at the angle of $41.6°$.