Question

Determine P necessary to just start opening the 2 m wide gate.

Determine P necessary to just start opening the 2 m wide gate.

Answer 1

Concepts and reason

This question is based on the concepts hydrostatic force and center of pressure.

Hydrostatic Force: A fluid always exerts a pressure on the submerged bodies. This pressure causes a force on the submerged body known as hydrostatic Force.

Center of pressure: On a submerged body pressure force acts through a point, this particular point is known as the center of pressure.

First, draw a labeled diagram of the gate, calculate the distance of the centroid of the gate from the free surface. Now, calculate the hydrostatic force working on the gate and obtain the location of the center of pressure for the gate. Consider the moment of all forces about the hinge to obtain the value of the force P.

Fundamentals

Hydrostatic force is calculated as,

$F = \rho gAh$

Here, $\rho$ is the density of the water, $A$ is the projected area and $h$ is the projected height.

Center of pressure is calculated as,

${h_c} = h + \frac{{{I_{GG}}{{\sin }^2}\theta }}{{Ah}}$

Here, $h$ is the projected height from the top to the point where hydrostatic force works, ${I_{GG}}$ is the moment of inertia about the gravitational center and $A$ is the projected area where the center of pressure is determined.

The moment of inertia about the center of gravity of rectangular gate is calculated as,

${I_{GG}} = \frac{{b{d^3}}}{{12}}$

Here, $b$ is the width and $d$ is the depth of the gate.

Draw the free body diagram of the gate.

Consider the inclination of the gate with the horizontal.

$\begin{array}{l}\\\tan \theta = \frac{4}{3}\\\\\theta = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)\\\\ = 53.13^\circ \\\end{array}$

Here, $\theta$ is the inclination of the gate with the horizontal.

‎

Calculate the length AD of the gate.

$\begin{array}{c}\\AD = \sqrt {\left( {{4^2} + {3^2}} \right)} \\\\ = 5{\rm{ m}}\\\end{array}$

Calculate the vertical distance of centroid of the gate from the free surface.

$\begin{array}{c}\\h = 1{\rm{ m}} + \left( {2.5{\rm{ m}}} \right)\sin 53.13^\circ \\\\ = {\rm{3 m}}\\\end{array}$

Here, $h$ is the vertical distance of centroid of the gate from the free surface.

Calculate the area of the gate.

$A = AD\left( {2{\rm{ m}}} \right)$

Here, $A$ is area of the gate.

Substitute $5{\rm{ m}}$ for AD.

$\begin{array}{c}\\A = 5{\rm{ m}}\left( {2{\rm{ m}}} \right)\\\\ = 10{\rm{ }}{{\rm{m}}^2}\\\end{array}$

Write the expression for the hydrostatic force on the gate.

$F = \rho gAh$

Here, $\rho$ is density of the water and g is acceleration due to gravity.

Substitute $1{\rm{000 kg/}}{{\rm{m}}^{\rm{3}}}$ for $\rho$ , ${\rm{9}}{\rm{.81 m/}}{{\rm{s}}^2}$ for $g$ , ${\rm{10 }}{{\rm{m}}^2}$ for $A$ and ${\rm{3 m}}$ for $h$ in above expression.

$\begin{array}{c}\\F = \left( {1{\rm{000 kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {{\rm{9}}{\rm{.81 m/}}{{\rm{s}}^2}} \right)\left( {10{\rm{ }}{{\rm{m}}^2}} \right)\left( {{\rm{3 m}}} \right)\\\\ = 29{\rm{4}}{\rm{.3 kN}}\\\end{array}$

Calculate the area moment of inertia of the gate about the axis passing through the centroid.

${I_{GG}} = \frac{{2{\rm{ m}}{{\left( {AD} \right)}^3}}}{{12}}$

Here, ${I_{GG}}$ is area moment of inertia of gate about the axis passing the centroid.

Substitute $5{\rm{ m}}$ for AD.

$\begin{array}{c}\\{I_{GG}} = \frac{{\left( {2{\rm{ m}}} \right){{\left( {{\rm{5 m}}} \right)}^3}}}{{12}}\\\\ = 20.8{\rm{33 }}{{\rm{m}}^{\rm{4}}}\\\end{array}$

Write the expression for the center of pressure of gate.

${h_c} = h + \frac{{{I_{GG}}{{\sin }^2}\theta }}{{Ah}}$

Here, ${h_c}$ is distance of the vertical height of the gate from the free surface.

Substitute ${\rm{3 m}}$ for $h$ , $20.8{\rm{33 }}{{\rm{m}}^{\rm{4}}}$ for ${I_{GG}}$ , $53.13^\circ$ for $\theta$ and ${\rm{10 }}{{\rm{m}}^2}$ for $A$ .

$\begin{array}{c}\\{h_c} = {\rm{3 m}} + \frac{{\left( {20.8{\rm{33 }}{{\rm{m}}^{\rm{4}}}} \right){{\sin }^2}\left( {53.13^\circ } \right)}}{{10{\rm{ }}{{\rm{m}}^2}\left( {{\rm{3 m}}} \right)}}\\\\ = {\rm{3}}{\rm{.4 m}}\\\end{array}$

Consider the moment of all the forces about the hinge and apply equilibrium of moment.

$P\cos \theta \left( {AD} \right) - F\left( {{h_c}\sin \theta } \right) = 0$

Substitute ${\rm{3}}{\rm{.4 m}}$ for ${h_c}$ , $29{\rm{4}}{\rm{.3 kN}}$ for $F$ , $5{\rm{ m}}$ for AD and $53.13^\circ$ for $\theta$ .

$\begin{array}{l}\\P\left( {\cos 53.13^\circ } \right)\left( {5{\rm{ m}}} \right) - \left( {29{\rm{4}}{\rm{.3 kN}}} \right)\left( {{\rm{3}}{\rm{.4 m}}} \right)\left( {\sin 53.13^\circ } \right) = 0\\\\P = {\rm{266}}{\rm{.832 kN}}\\\\ \sim {\rm{267 kN}}\\\end{array}$

Ans:

The necessary value of P to just start opening the $2$ meter-wide gate ${\rm{267 kN}}$ .