Question

a 3-m wide, 8-m high rectangular gate is located at the end of arectangular passage that is connected to a large open tankfilled with water. The gate is hinged at its bottom and heldclosed by a horizontal force, F_H , located at the centerof the gate. The maximum value for F_H is 3500kN. (a) Determine the maximum water depth, h, abovethe center of the gate that can exist without the gateopening. (b) Is the answer the same if the gate ishinged at the top?

Answer 1

Concepts and reason

The concept of hydrostatic force on a plane surface is used to calculate the resultant force acting on rectangular gate. The point of application of this resultant force from the free surface of water is also calculated .

Moment equilibrium of horizontal and hydrostatic forces about the hinge are considered resulting in maximum water depth.

The concept of hydrostatic force is used in the design of ships, dams, and storage tanks.This is used in calculation of force acting on the surface of submerged object.

Fundamentals

The magnitude of the resultant force of fluid is equal to the product of pressure acting at centroid of the area and total area.

${F_R} = \gamma {h_c}A$

Here, $\gamma$ is specific weight, ${h_c}$ is the distance from free surface of water to the centroid of area of the object, and A is area of the object.

The relation for distance of hydrostatic force from free water surface is as follows:

${y_R} = \frac{{{I_{xc}}}}{{{y_c}A}} + {y_c}$

Here, ${I_{xc}}$ is the second moment of area with respect to an axis passing through centroid and parallel to x axis, ${y_c}$ is the distance of the centroid from the free surface of water.

The distance of hydrostatic force from free surface is always below the centroid.

Magnitude of moment can be calculated by multiplying the magnitude of force with the perpendicular distance between the point where the force is applied and the point where the moment to be found.It is

$M = Fd$

Here, F is the force and d is the perpendicular distance.

A body is in equilibrium if vector sum of all the forces is equal to zero or moment of all forces about any point is equal to zero. That is,

Summation of all forces in x direction is zero, $\sum {{F_x}} = 0$

Summation of all forces in y direction is zero, $\sum {{F_y}} = 0$

Moment about point any point is zero, $\sum {{M_A}} = 0$

General sign convection for moment: The moment is considered positive in counter-clockwise direction and negative in clockwise direction.

Draw the schematic diagram of the rectangular passage.

Calculate resultant force exerted on the gate due to the pressure of water.

$\begin{array}{c}\\{F_R} = \gamma {y_c}A\\\\ = \gamma \times {y_c} \times b \times {h_g}\\\end{array}$

Here, $\gamma$ is specific weight of the water, ${y_c}$ is the distance of centroid of the gate from free surface, A is the area of the gate, b is the width of the gate and ${h_g}$ is the height of the gate.

Substitute $9.81\,{\rm{kN/}}{{\rm{m}}^{\rm{3}}}$ for $\gamma$ , $h$ for ${y_c}$ , ${\rm{3 m}}$ for b and 8 m for ${h_g}$ .

$\begin{array}{c}\\{F_R} = \left( {9.81{\rm{ }}\frac{{{\rm{kN}}}}{{{{\rm{m}}^{\rm{3}}}}}} \right) \times h \times \left( {3{\rm{ m}}} \right) \times \left( {8{\rm{ m}}} \right)\\\\ = 235.44 \times h{\rm{ kN}}\\\end{array}$

Write the expression to calculate the distance of resultant force from water surface.

$\begin{array}{c}\\{y_R} = \frac{{{I_{xc}}}}{{{y_c}A}} + {y_c}\\\\ = \frac{{\frac{{bh_g^3}}{{12}}}}{{{y_c} \times b \times {h_g}}} + {y_c}\\\end{array}$

Substitute $h$ for ${y_c}$ , ${\rm{3 m}}$ for b and 8 m for ${h_g}$ .

$\begin{array}{c}\\{y_R} = \frac{{\frac{{3 \times {8^3}}}{{12}}}}{{h \times 3 \times 8}} + h\\\\ = \frac{{5.33}}{h} + h\\\end{array}$

Calculate the distance of resultant force from hinge.

$l = h + 4 - {y_R}$

Substitute $\frac{{5.33}}{h} + h$ for ${y_R}$ .

$\begin{array}{c}\\l = h + 4 - \frac{{5.33}}{h} - h\\\\ = 4 - \frac{{5.33}}{h}\\\end{array}$

Apply moment equilibrium equation about hinge.

$\begin{array}{l}\\\sum M = 0\\\\{F_H} \times 4 - l \times {F_R} = 0\\\\{F_H} \times 4 = l \times {F_R}\\\end{array}$

Substitute $3500\,{\rm{kN}}$ for ${F_H}$ , $235.44 \times h{\rm{ kN}}$ for ${F_R}$ , and $4 - \frac{{5.33}}{h}$ for $l$ .

$\begin{array}{l}\\3500 \times 4 = \left( {4 - \frac{{5.33}}{h}} \right) \times 235.44h\\\\3500 \times 4 = 941.76h - 1254.89\\\\15254.89 = 941.76\,h\\\\h = 16.2\,{\rm{m}}\\\end{array}$

Draw the free body diagram of gate when the hinge is at top.

Calculate the distance of the resultant force from the hinge.

${l_1} = {y_R} - \left( {h - 4} \right)$

Substitute $\frac{{5.33}}{h} + h$ for ${y_R}$ .

$\begin{array}{c}\\{l_1} = \frac{{5.33}}{h} + h - \left( {h - 4} \right)\\\\ = \frac{{5.33}}{h} + 4\\\end{array}$

Apply moment equilibrium equation about the hinge.

$\begin{array}{l}\\\sum M = 0\\\\ - {F_H} \times 4 + {l_1} \times {F_R} = 0\\\\{F_H} \times 4 = {l_1} \times {F_R}\\\end{array}$

Substitute $3500\,{\rm{kN}}$ for ${F_H}$ , $235.44 \times h{\rm{ kN}}$ for ${F_R}$ , and $\frac{{5.33}}{h} + 4$ for ${l_1}$ .

$\begin{array}{l}\\3500 \times 4 = \left( {\frac{{5.33}}{h} + 4} \right) \times 235.44h\\\\3500 \times 4 = 941.76h + 1254.89\\\\941.76h = 12745.11\\\\h = 13.533\,{\rm{m}}\\\end{array}$

Therefore, the answer is not same when the hinge is at top.

Ans: Part a

The maximum water depth h above the center of the gate that can exist without gate opening is $16.2\,{\rm{m}}$ .

Part b

No, the answer is not same in both cases.