Chapter 3, Problem 90P

A 4-m-long quarter-circular gate of radius 3 m and of negligible weight is hinged about its upper edge A, as shown in Fig. P3-90. The gate controls the flow of water over the ledge at B, where the gate is pressed by a spring. Determine the minimum spring force required to keep the gate closed when the water level rises to A at the upper edge of the gate.

To determine

The minimum spring force required to keep the gate closed when the water level rises to A at the upper edge of the gate.

Answer to Problem 90P

The minimum spring force required to keep the gate closed when the water level rises to $A$at the upper edge of the gate is $\underset{\_}{150\text{kN}}$

Explanation of Solution

Given information:

The radius of quarter circle is $3\text{m}$, length of quarter circle is $4\text{m,}$ and weigh of quarter circle is negligible about its upper edge $A$

The Figure below shows the free body diagram of gate with all forces acting on it.

Figure-(1)

Due to hydrostatic force there is horizontal and vertical force on the gate.

Write the expression for the horizontal force on gate.

  $F_H=\rho g{h}_{cg}A$....... (I)

Here, the horizontal force acting on the gate is $F_H$, the density of water is $\rho$, the height of center of gravity of the projected portion of the gate from its base is $h_{cg}$, acceleration due to gravity is $g,$ and the area of projected portion of the gate $A$.

The area of projected portion of the gate is in the shape of rectangle whose length is $4\text{m}$ and width is $3\text{m}$

Write the expression for area of gate

  $A=\text{length×width}$....... (II)

Write the expression the height of center of gravity of the projected portion of the gate from its base.

  $h_{cg}=\frac{\text{Width}}{2}$....... (III)

Write the expression of the vertical force on gate.

  $F_y=\rho g{h}_{bt}A$....... (IV)

Here, the vertical force acting on the gate is $F_y$, the density of water is $\rho$, the height of liquid column from free surface to the bottom of the gate is $h_{bt}$, acceleration due to gravity $g,$ and the area of projected portion of the gate. $A$.

Write the expression for weight of fluid.

  $W=\rho gV$....... (V)

Here, the weight of fluid is $W$, the density of water is $\rho$, acceleration due to gravity is $g$ is and the volume of fluid block $V$.

Write the expression for volume of fluid block.

  $V=w\text{×}\frac{\pi }{4}{r}^2$....... (VI)

Substitute $w\text{×}\frac{\pi }{4}{r}^2$ for $V$ in Equation (V).

  $W=\rho g\left(w\text{×}\frac{\pi }{4}{r}^2\right)$..... (VII)

Write the expression to resolve the force in vertical direction.

  $\begin{array}{l}\sum F_y=0\\ A_{vt}+F_y-W=0\\ A_{vt}=-F_y+W\end{array}$....... (IX)

Here, the vertical reaction at $A$ is $A_{vt}$.

Write the expression for centre of gravity of fluid element.

  $CG=\frac{4r}{3\pi }$........ (X)

Write the expression for moments of all the forces about hinge $A$.

  $\begin{array}{l}\sum M_A=0\\ \left(F_H\text{×}{d}_1\right)+\left(F_y\text{×}{d}_2\right)-\left(W\text{×}{d}_2\right)-\left(F_s\text{×}r\right)=0\\ \left(F_H\text{×}{d}_1\right)+\left(F_y\text{-W}\right)\text{×}{d}_2-\left(F_s\text{×}r\right)=0\end{array}$....... (XI)

Here, the perpendicular distance from the force $F_H$ is $d_1$, the perpendicular distance from the force $F_y$ is $d_2$ and spring force is $F_s$

Calculation:

Substitute, $3\text{m}$ for width and $4\text{m}$ for length in equation (II).

  $\begin{array}{c}A=\text{4}\text{m×3}\text{m}\\ =\text{12}{\text{m}}^2\end{array}$

Substitute, $3\text{m}$ for width in equation (III).

  $\begin{array}{c}{h}_{cg}=\frac{3\text{m}}{2}\\ =1.5\text{m}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1.5\text{m}$ for $h_{cg}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $12{\text{m}}^{\text{2}}$ for $A$ in equation (I).

  $\begin{array}{c}{F}_H=1000\text{kg}/{\text{m}}^{\text{3}}\text{×}9.81\text{m}/{\text{s}}^{\text{2}}\text{×}1.5\text{m×}12{\text{m}}^{\text{2}}\\ =176580\text{N}\\ \text{=}176580\text{N×}\frac{1\text{kN}}{1000\text{N}}\\ =176.580\text{kN}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $3\text{m}$ for $h_{bt}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $12{\text{m}}^{\text{2}}$ for $A$ in Equation (IV).

  $\begin{array}{c}{F}_y=1000\text{kg}/{\text{m}}^{\text{3}}\text{×}9.81\text{m}/{\text{s}}^{\text{2}}\text{×3}\text{m×}12{\text{m}}^{\text{2}}\\ =353200\text{N}\\ =353200\text{N}\frac{1\text{kN}}{1000\text{N}}\\ =353.2\text{kN}\end{array}$

Substitute, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $3\text{m}$ for $r$ and $4\text{m}$ for $w$ in Equation (VII).

  $\begin{array}{c}W=1000\text{kg}/{\text{m}}^{\text{3}}\text{×}9.81\text{m}/{\text{s}}^{\text{2}}\text{×}\left(\text{4}\text{m×}\frac{\pi }{4}{\left(3\text{m}\right)}^2\right)\\ =277370\text{N}\\ \text{=}277370\text{N×}\frac{1\text{kN}}{1000\text{N}}\\ =277.37\text{kN}\end{array}$

Substitute $353.2\text{kN}$ for $F_y$ and $277.37\text{kN}$ for $W$ in above equation (IX).

  $\begin{array}{c}{A}_{vt}=-353.2\text{kN}+277.37\text{kN}\\ =75.83\text{kN}\end{array}$

Substitute $3\text{m}$ for $r$ in equation (X).

  $\begin{array}{c}CG=\frac{4\text{×}3\text{m}}{3\pi }\\ =\frac{12\text{m}}{3\pi }\\ =1.273\text{m}\end{array}$

Substitute $176.58\text{kN}$ for $F_H$, $353.2\text{kN}$ for $F_y$, $277.37\text{kN}$ for $W$, $2m$ for $d_1$ and $1.27\text{m}$ for $d_2$ in equation (XI).

  $\begin{array}{l}\left(176.58\text{kN×}2\text{m}\right)+\left(353.2\text{kN}-277.37\text{kN}\right)\text{×}1.27\text{m}-\left(F_s\text{×3}\text{m}\right)=0\\ 353.7+96.304-\left(F_s\text{×3}\text{m}\right)=0\\ F_s=150\text{kN}\end{array}$

Conclusion:

The minimum spring force required is $\underset{\_}{150\text{kN}}$.