Chapter 2, Problem 2.83P

To determine

The force $F$ sufficient to keep the gate from opening.

Answer to Problem 2.83P

The force $F$ sufficient to keep the gate from opening is $7481\text{lbf}$.

Explanation of Solution

Write the expression for horizontal component of hydrostatic force.

$F_H=\gamma A{h}_{CG}$ …… (I)

Here, the specific weight of water is $\gamma$, the projected area of the curve is $A$ and the height of centre of gravity is $h_{CG}$.

Write the area of projected plane.

$A=rw$

Substitute $rw$ for $A$ in Equation (I).

$F_H=\gamma rw{h}_{CG}$ …… (II)

Write the moment of inertia of projected plane.

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

Here, the radius of the curve is $r$ and the width of the curve is $w$.

Write the position of the horizontal force.

$y=\frac{I}{A{h}_{CG}}+h_{CG}$ …… (III)

Write the height of centre of gravity.

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

Substitute $rw$ for $A$, $\frac{r}{2}$ for $h_{CG}$ and $\frac{r^{3}w}{12}$ for $I$ in Equation (III).

$\begin{array}{c}y=\frac{\left(\frac{r^{3}w}{12}\right)}{\left(rw\right)\frac{r}{2}}+\frac{r}{2}\\ =\frac{r}{6}+\frac{r}{2}\\ =\frac{2}{3}r\end{array}$ …… (IV)

The vertical force is the difference of the hydrostatic force of the square part and hydrostatic force of quarter circle part.

Write the expression for vertical force.

$\begin{array}{c}{F}_v=\gamma V_{sq}-\gamma V_q\\ =\gamma \left(V_{sq}-V_q\right)\end{array}$ …… (V)

Here, the volume of the square part is $V_{sq}$ and the volume of the quarter circle part is $V_q$.

Write the expression for volume of the square part.

$V_{sq}=w{r}^2$

Write the expression for volume of the quarter circle part.

$V_q=w\left(\frac{\pi }{4}{r}^2\right)$

Substitute $w\left(\frac{\pi }{4}{r}^2\right)$ for $V_q$ and $w{r}^2$ for $V_{sq}$ in Equation (V).

$\begin{array}{c}{F}_v=\gamma \left[w{r}^2-w\left(\frac{\pi }{4}{r}^2\right)\right]\\ =\gamma w{r}^2\left[1-\frac{\pi }{4}\right]\end{array}$ …… (VI)

Write the moment about $B$ for the vertical forces.

$\begin{array}{l}{F}_v\cdot k=F_{sq}\times \frac{r}{2}-F_q\left(r-\frac{4r}{3\pi }\right)\\ k=\frac{\gamma w\left(r^2\right)\times \frac{r}{2}-\gamma w\left(\frac{\pi }{4}{r}^2\right)\left(r-\frac{4r}{3\pi }\right)}{F_v}\end{array}$ …… (VII)

The weight of the gate acts at its centre of gravity.

Write the center of gravity of the quarter circle from $B$

$c_q=r-\frac{2r}{\pi }$ …… (VIII)

The following diagram shows the forces on the gate.

Figure-(1)

In Figure-(1), the weight of the gate is $W$.

Write the moment about $B$.

$F\times r+W\times c_q-F_v\times k-F_H\left(r-y\right)=0$ (IX)

Conclusion:

Substitute $62.4\text{lbf}/{\text{ft}}^3$ for $\gamma$, $8\text{ft}$ for $r$, $10\text{ft}$ for $w$ and $\frac{8\text{ft}}{2}$ for $h_{CG}$ in Equation (II).

$\begin{array}{c}{F}_H=\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(8\text{ft}\right)\left(10\text{ft}\right)\frac{8\text{ft}}{2}\\ =\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(80{\text{ft}}^2\right)\left(4\text{ft}\right)\\ =19968\text{lbf}\end{array}$

Substitute $8\text{ft}$ for $r$ in Equation (IV).

$\begin{array}{c}y=\frac{2}{3}\left( 8\text{ft}\right)\\ =\frac{16}{3}\text{ft}\\ =5.33\text{ft}\end{array}$

Substitute $62.4\text{lbf}/{\text{ft}}^3$ for $\gamma$, $10\text{ft}$ for $w$ and $8\text{ft}$ for $r$ in Equation (VI).

$\begin{array}{c}{F}_v=\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(10\text{ft}\right){\left(8\text{ft}\right)}^2\left[1-\frac{\pi }{4}\right]\\ =39936\text{lbf}\times \left[1-\frac{\pi }{4}\right]\\ =8570.4\text{lbf}\end{array}$

Substitute $62.4\text{lbf}/{\text{ft}}^3$ for $\gamma$, $10\text{ft}$ for $w$, $8570.4\text{lbf}$ for $F_v$, and $8\text{ft}$ for $r$ in Equation (VII).

$\begin{array}{c}k=\frac{\left[\begin{array}{l}\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(10\text{ft}\right){\left(8\text{ft}\right)}^2\times \frac{\left(8\text{ft}\right)}{2}\\ -\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(10\text{ft}\right)\left(\frac{\pi }{4}{\left(8\text{ft}\right)}^2\right)\left(\left(8\text{ft}\right)-\frac{4\left(8\text{ft}\right)}{3\pi }\right)\end{array}\right]}{8570.4\text{lbf}}\\ =1.78\text{ft}\end{array}$

Substitute $8\text{ft}$ for $r$ in Equation (VIII).

$\begin{array}{c}{c}_q=\left(8\text{ft}\right)-\frac{2\left(8\text{ft}\right)}{\pi }\\ =2.9\text{ft}\end{array}$

Substitute $8\text{ft}$ for $r$, $2.9\text{ft}$ for $c_q$, $1.78\text{ft}$ for $k$, $8570.4\text{lbf}$ for $F_v$, for $W$, $19968\text{lbf}$ for $F_H$ and $5.33\text{ft}$ for $y$ in Equation (IX).

$\begin{array}{l}\left[\begin{array}{l}F\times \left(8\text{ft}\right)+\left(3000\text{lbf}\right)\times \left(2.9\text{ft}\right)-\left(8570.4\text{lbf}\right)\times \left(1.78\text{ft}\right)\\ -\left(19968\text{lbf}\right)\left(\left(8\text{ft}\right)-\left(5.33\text{ft}\right)\right)\end{array}\right]=0\\ F=\frac{59848\text{lbf}\cdot \text{ft}}{8\text{ft}}\\ F=7481\text{lbf}\end{array}$

Thus, the force $F$ sufficient to keep the gate from opening is $7481\text{lbf}$.