Chapter 5.3, Problem 5.81P

The cross section of a concrete dam is as shown. For a 1-m-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

Fig. P5.81

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.

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 $A$.

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 $41.6°$.