Chapter 2, Problem 2.82P

To determine

The horizontal component of hydrostatic force against the dam.

The vertical component of hydrostatic force against the dam.

The point $CP$ where resultant strikes the dam.

Answer to Problem 2.82P

The horizontal component of hydrostatic force against the dam is $97.9\text{MN}$.

The vertical component of hydrostatic force against the dam is $153.8\text{MN}$.

The point $CP$ where resultant strikes the dam is $\left(3.1,10.7\right)$.

Explanation of Solution

Write the expression for horizontal hydrostatic force.

$F_H=P_{1}A$ …… (I)

Here, the pressure is $P_1$ and the area of projected plane is $A$.

Write the expression for the pressure acting on dam.

$P_1={\gamma }_w{h}_{CG}$

Here, the specific weight of water is ${\gamma }_w$ and the height of center of gravity from projected plane is $h_{CG}$.

Substitute ${\gamma }_w{h}_{CG}$ for $P_1$ in Equation (I).

$F_H={\gamma }_w{h}_{CG}A$ …… (II)

Write the area of the projected plane.

$A=w\times h$

Here, the width of dam is $w$ and the height of the dam is $h$.

Write the height of center of gravity from projected plane.

$h_{CG}=\frac{h}{2}$

Substitute $\frac{h}{2}$ for $h_{CG}$ and $w\times h$ for $A$ in Equation (II).

$F_H=\frac{h}{2}w\times h\times {\gamma }_w$ …… (III)

Write the position of center of pressure on vertical plane.

$y_{CP}=\frac{I}{A{h}_{CG}}$ …… (IV)

Here, the moment of inertia is $I$.

Write the moment of inertia for dam.

$I=\frac{w{h}^3}{12}$

Substitute $\frac{w{h}^3}{12}$ for $I$, $\frac{h}{2}$ for $h_{CG}$ and $w\times h$ for $A$ in Equation (IV).

$\begin{array}{c}{y}_{CP}=\frac{\left(\frac{w{h}^3}{12}\right)}{\left(w\times h\right)\left(\frac{h}{2}\right)}\\ =\frac{h}{6}\end{array}$

Write the line of action of horizontal force.

$y=10+y_{CP}$ …… (V)

Substitute $\frac{h}{6}$ for $y_{CP}$ in Equation (V).

$y=10+\frac{h}{6}$ …… (VI)

Write the vertical component of hydrostatic force.

$F_v={\gamma }_w{V}_{\text{curved}\text{part}}$ …… (VII)

Here, the volume of the curved part is $V_{\text{curved}\text{part}}$.

Write the volume of the curved part.

$V_{\text{curved}\text{part}}=\left(\frac{\pi h^2}{4}\right)w$

Substitute $\left(\frac{\pi h^2}{4}\right)w$ for $V_{\text{curved}\text{part}}$ in Equation (VII).

$F_v={\gamma }_w\left(\frac{\pi h^2}{4}\right)w$ …… (VIII)

Write the line of action of the vertical component of hydrostatic force.

$x=\frac{4h}{3\pi }$ …. (IX)

Write the resultant force at $CP$.

$F=\sqrt{F_H^2+F_V^2}$ …… (X)

Write the angle of the forces.

$\tan \theta =\frac{F_v}{F_H}$ …… (XI)

The below diagram shows the force $F$ at $CP$.

Figure-(1)

In Figure-(1), the height of $CP$ from base is $a$ and the horizontal distance is $b$.

Write the height of $CP$ from base.

$a=h-h\cdot \sin \theta$ …… (XII)

Write the horizontal distance of $CP$.

$b=h\cos \theta$ …… (XIII)

Conclusion:

Substitute $20\text{m}$ for $h$, $50\text{m}$ for $w$, $9790\text{N}/{\text{m}}^3$ for ${\gamma }_w$ in Equation (III).

$\begin{array}{c}{F}_H=\frac{\left(20\text{m}\right)}{2}\left(50\text{m}\right)\times \left(20\text{m}\right)\times \left(9790\text{N}/{\text{m}}^3\right)\\ =10000{\text{m}}^3\times \left(9790\text{N}/{\text{m}}^3\right)\\ =97.9\text{MN}\end{array}$

Thus, the horizontal component of hydrostatic force against the dam is $97.9\text{MN}$.

Substitute $9790\text{N}/{\text{m}}^3$ for ${\gamma }_w$, $20\text{m}$ for $h$ and $50\text{m}$ for $w$ in Equation (VIII).

$\begin{array}{c}{F}_v=\left(9790\text{N}/{\text{m}}^3\right)\left(\frac{\pi {\left(20\text{m}\right)}^2}{4}\right)\left(50\text{m}\right)\\ =\left(9790\text{N}/{\text{m}}^3\right)\times \left(15707.96{\text{m}}^3\right)\\ =153.8\text{MN}\end{array}$

Thus, the vertical component of hydrostatic force against the dam is $153.8\text{MN}$.

Substitute $153.8\text{MN}$ for $F_v$ and $97.9\text{MN}$ for $F_H$ in Equation (XI).

$\begin{array}{l}\tan \theta =\frac{153.8\text{MN}}{97.9\text{MN}}\\ \tan \theta =1.58\\ \theta ={\tan }^{-1}\left(1.58\right)\\ \theta =57.68°\end{array}$

Substitute $20\text{m}$ for $h$ and $57.68°$ for $\theta$ in Equation (XII).

$\begin{array}{c}a=20\text{m}-20\text{m}\left(\sin 57.68°\right)\\ =20\text{m}-16.9\text{m}\\ =\text{3}\text{.1}\text{m}\end{array}$

Substitute $20\text{m}$ for $h$ and $57.68°$ for $\theta$ in Equation (XIII).

$\begin{array}{c}b=\left(20\text{m}\right)\cos \left(57.68°\right)\\ =\left(20\text{m}\right)\left(0.5346\right)\\ \approx 10.7\text{m}\end{array}$

Thus, the point $CP$ where resultant strikes the dam is $\left(3.1,10.7\right)$.