Chapter 3, Problem 142P

To determine

The required force at center of gravity of gate to keep the gate closed.

Answer to Problem 142P

The required force at center of gravity of gate to keep the gate closed is $\underset{\_}{6771.61\text{N}}$.

Explanation of Solution

The radius of the semi-circular gate is $0.5\text{m}$, specific gravity of fluid in section $1$ is $0.91$, specific gravity of fluid in section $2$ is $1.26$.

The figure below shows the different type of forces acting on the gate.

  

Figure-(1)

Write the expression for diameter of semi circular gate.

  $d=2R$...... (I)

Write the expression for area of gate.

  $A=\frac{\pi }{8}{d}^2$...... (II)

Here, diameter of semicircular gate is $d$.

Write the expression for distance between centroidal axis of gate and its diametral axis.

  $h=\frac{4R}{3\pi }$...... (III)

Here, radius of semicircular gate is $R$.

Write the expression for moment of inertia of gate about its centroidal axis.

  $I=\frac{\pi }{128}{d}^4+A{h}^2$...... (IV)

Write the expression for distance of centroid of the gate to the free surface of section $1$.

  $y_{C_1}=4.74\text{m}+h$...... (V)

Write the expression for distance of centroid of the gate to the free surface of section $2$.

  $y_{C_2}=h$...... (VI)

Write the expression for center of pressure of gate from free surface of section $1$.

  $y_{P_1}=y_{C_1}+\frac{I}{A\times \left(y_{C_1}+\frac{P_1}{{\rho }_w{S}_{1}g}\right)}$...... (VII)

Here, pressure above section $1$ is $P_1$, density of water ${\rho }_w$, specific gravity of fluid in section $1$ is $S_1$ and acceleration due to gravity is $g$.

Write the expression for center of pressure of gate from free surface of section $2$.

  $y_{P_2}=y_{C_2}+\frac{I}{A\times \left(y_{C_2}+\frac{P_{atm}}{{\rho }_w{S}_{2}g}\right)}$........ (VIII)

Here, atmospheric pressure is $P_{atm}$ and specific gravity of fluid in section $2$ is $S_2$.

Write the expression for force acting on the gate due to fluid in section $1$.

  $F_{R_1}=\left(P_1+{\rho }_w{S}_{1}g{y}_{C_1}\right)A$...... (IX)

Write the expression for force acting on the gate due to fluid in section $2$.

  $F_{R_2}=\left(P_{atm}+{\rho }_w{S}_{2}g{y}_{C_2}\right)A$...... (X)

Write the expression for moment equilibrium equation about hinge point of gate.

  $\begin{array}{l}{F}_{R_2}{y}_{P_2}+F{y}_{C_2}=F_{R_1}\left(y_{P_1}-4.74\text{m}\right)\\ F=\frac{F_{R_1}\left(y_{P_1}-4.74\text{m}\right)-F_{R_2}{y}_{P_2}}{y_{C_2}}\end{array}$...... (XI)

Calculation:

Substitute $0.5\text{m}$ for $R$ in equation (I).

  $\begin{array}{c}d=2\times 0.5\text{m}\\ \text{=1}\text{m}\end{array}$

Substitute $1\text{m}$ for $d$ in equation (II).

  $\begin{array}{c}A=\frac{\pi }{8}{\left(1\text{m}\right)}^2\\ =0.39265{\text{m}}^2\end{array}$

Substitute $0.5\text{m}$ for $R$ in equation (III).

  $\begin{array}{c}h=\frac{4\times 0.5\text{m}}{3\pi }\\ =0.2122\text{m}\end{array}$

Substitute $1\text{m}$ for $d$, $0.7853{\text{m}}^2$ for $A$ and $0.2122\text{m}$ for $h$ in equation (IV).

  $\begin{array}{c}I=\frac{\pi }{128}{\left(1\text{m}\right)}^4+\left(0.39265{\text{m}}^2\right)\times {\left(0.2122\text{m}\right)}^2\\ =0.024543{\text{m}}^4+0.0833{\text{m}}^4\\ =0.10786{\text{m}}^4\end{array}$

Substitute $0.2122\text{m}$ for $h$ in equation (V).

  $\begin{array}{c}{y}_{C_1}=4.74\text{m}+0.2122\text{m}\\ =4.9522\text{m}\end{array}$

Substitute $0.2122\text{m}$ for $h$ in equation (VI).

  $y_{C_2}=0.2122\text{m}$

Substitute $4.9522\text{m}$ for $y_{C_1}$, $0.39265{\text{m}}^2$ for $A$, $80\text{kPa}$ for $P_1$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $0.91$ for $S_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.10786{\text{m}}^4$ for $I$ in equation (VII).

  $\begin{array}{c}{y}_{P_1}=4.9522\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(4.9522\text{m}+\frac{80\text{kPa}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 0.91\times 9.81\text{m}/{\text{s}}^{\text{2}}}\right)}\\ =4.9522\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(4.9522\text{m}+\frac{80\text{kPa}\times \left(\frac{\text{1000}\text{N}/{\text{m}}^{\text{2}}}{\text{1}\text{kPa}}\right)}{8927.1\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\right)}\\ =4.9522\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(4.9522\text{m}+\frac{80000\text{N}/{\text{m}}^{\text{2}}}{8927.1\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\right)}\\ =4.9522\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(4.9522\text{m}+\frac{80000\text{N}/{\text{m}}^{\text{2}}\left(\frac{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}{\text{m}}^{\text{2}}}{\text{1}\text{N}/{\text{m}}^{\text{2}}}\right)}{8927.1\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\right)}\end{array}$

  $\begin{array}{l}=4.9522\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(4.9522\text{m}+\frac{80000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}{\text{m}}^{\text{2}}}{8927.1\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\right)}\\ =4.9522\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(4.9522\text{m}+\text{8}\text{.96147}\text{m}\right)}\\ =4.9522\text{m}+\frac{0.10786{\text{m}}^4}{10.9264{\text{m}}^3}\\ =4.9522\text{m}+9.8715\times {\text{10}}^{-3}\text{m}\end{array}$

  $=4.962\text{m}$

Substitute $0.2122\text{m}$ for $y_{C_2}$, $0.39265{\text{m}}^2$ for $A$, $101325\text{N}/{\text{m}}^{\text{2}}$ for $P_{atm}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.26$ for $S_2$ and $0.10786{\text{m}}^4$ for $I$ in equation (VIII).

  $\begin{array}{c}{y}_{P_2}=0.2122\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(0.2122\text{m}+\frac{101325\text{N}/{\text{m}}^{\text{2}}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 1.26\times 9.81\text{m}/{\text{s}}^{\text{2}}}\right)}\\ =0.2122\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(0.2122\text{m}+\frac{101325\text{N}/{\text{m}}^{\text{2}}\left(\frac{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}{\text{m}}^{\text{2}}}{\text{1}\text{N}/{\text{m}}^{\text{2}}}\right)}{12360.6\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\right)}\\ =0.2122\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(0.2122\text{m}+\frac{101325\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}{\text{m}}^{\text{2}}}{12360.6\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\right)}\\ =0.2122\text{m}+\frac{0.10786{\text{m}}^4}{0.39265{\text{m}}^2\times \left(0.2122\text{m}+8.19741\text{m}\right)}\end{array}$

  $\begin{array}{l}=0.2122\text{m}+\frac{0.10786{\text{m}}^4}{3.3{\text{m}}^3}\\ =0.2122\text{m}+.03268\text{m}\\ \text{=0}\text{.24488}\text{m}\end{array}$

Substitute $80\text{kPa}$ for $P_1$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $0.91$ for $S_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $4.9522\text{m}$ for $y_{C_1}$ and $0.39265{\text{m}}^2$ for $A$ in equation (IX).

  $\begin{array}{l}{F}_{R_1}=\left(80\text{kPa}+1000\text{kg}/{\text{m}}^{\text{3}}\times 0.91\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 4.9522\text{m}\right)0.39265{\text{m}}^2\\ =\left(80\text{kPa}\left(\frac{\text{1000}\text{N}/{\text{m}}^{\text{2}}}{\text{1}\text{kPa}}\right)+44208.784\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)0.39265{\text{m}}^2\\ =\left(80000\text{N}/{\text{m}}^{\text{2}}\left(\frac{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}{\text{m}}^{\text{2}}}{\text{1}\text{N}/{\text{m}}^{\text{2}}}\right)+44208.784\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)0.39265{\text{m}}^2\\ =\left(80000\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}+44208.784\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)0.39265{\text{m}}^2\end{array}$

  $\begin{array}{l}=48770.579\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =48770.579\text{N}\end{array}$

Substitute $101325\text{N}/{\text{m}}^{\text{2}}$ for $P_{atm}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.26$ for $S_2$, $0.2122\text{m}$ for $y_{C_2}$ and $0.39265{\text{m}}^2$ for $A$ in equation (X).

  $\begin{array}{c}{F}_{R_2}=\left(101325\text{N}/{\text{m}}^{\text{2}}+1000\text{kg}/{\text{m}}^{\text{3}}\times 1.26\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.2122\text{m}\right)0.39265{\text{m}}^2\\ =\left(101325\text{N}/{\text{m}}^{\text{2}}+26222.919\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)0.39265{\text{m}}^2\\ =\left(101325\text{N}/{\text{m}}^{\text{2}}+26222.919\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\right)0.39265{\text{m}}^2\\ =\left(101325\text{N}/{\text{m}}^{\text{2}}+26222.919\text{N}/{\text{m}}^{\text{2}}\right)0.39265{\text{m}}^2\end{array}$

  $\begin{array}{c}{F}_{R_2}=\left(127547.919\text{N}/{\text{m}}^{\text{2}}\right)\times \left(0.39265{\text{m}}^2\right)\\ =50081.69\text{N}\end{array}$

Substitute $48770.579\text{N}$ for $F_{R_1}$, $4.962\text{m}$ for $y_{P_1}$, $50081.69\text{N}$ for $F_{R_2}$, $\text{0}\text{.24488}\text{m}$ for $y_{P_2}$ and $0.2122\text{m}$ for $y_{C_2}$ in equation (XI).

  $\begin{array}{c}F=\frac{48770.579\text{N}\left(4.962\text{m}-4.74\text{m}\right)-50081.69\text{N}\times \text{0}\text{.24488}\text{m}}{0.2122\text{m}}\\ =\frac{48770.579\text{N}\times \text{0}\text{.222}\text{m}-50081.69\text{N}\times \text{0}\text{.24488}\text{m}}{0.2122\text{m}}\\ =\frac{10827.068\text{N}\cdot \text{m}-12264.004\text{N}\cdot \text{m}}{0.2122\text{m}}\\ =\frac{-1436.936\text{N}\cdot \text{m}}{0.2122\text{m}}\end{array}$

  $=-6771.61\text{N}$

Conclusion:

The values are calculated by putting values in the equations : $F_{R_2}=\left(P_{atm}+{\rho }_w{S}_{2}g{y}_{C_2}\right)A$

  $y_{P_2}=y_{C_2}+\frac{I}{A\times \left(y_{C_2}+\frac{P_{atm}}{{\rho }_w{S}_{2}g}\right)}$