Moment about shaft from hydrostatic forces

AI Thread Summary
The discussion focuses on calculating the net moment about a horizontal shaft due to hydrostatic forces acting on a gate composed of a square and a semicircular section. The hydrostatic forces are derived using the equations for centroid height, force, and moment. The user attempts to find the location of the center of pressure for the square section but struggles with the formula, leading to confusion about the correct approach. Ultimately, the net moment is determined to be 750 kNm by considering the clockwise and anticlockwise moments produced by the forces on both sections of the gate. The thread emphasizes the importance of applying the parallel axis theorem and understanding the geometry involved in the calculations.
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Homework Statement



A gate with the shape as shown is mounted on a horizontal shaft. Determine the net moment about the shaft from the hydrostatic forces acting on the gate.

Answer: 750kNm

Homework Equations



h_c: height of centroid = y_c * sin(theta)
y_r: location of force or centre of pressure= y_c + [I_xc/(y_c * A)]
F_r= density * gravity * h_c * Area
Moment= F_r * [y_r - y_c]

The Attempt at a Solution



I split the gate into 2 parts- square and semicircle.
hc_semicircle= 7.27m
F_semicircle= 1000 * 9.8 * 7.27 * (pi * 0.5 * 3 * 3) = 1007.2kN
y_semicirlce= 7.36m

hc_square= 6/2 = 3m
F_square = 1058.4kN
y_square =?

How do i get y_r for square? if i use the formula above, then i would not get the correct answer for moment.

I was told to use h/3 to find y_square in which y_square= 2m
But i do no understand why can't i use the formula above to find y_square.

Can anyone help me with this?
 

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you are already half way there.

If you know the parallel axis theorem which you already put in. y_r: location of force or centre of pressure= y_c + [I_xc/(y_c * A)]

use it. y_square is y_c + [I_xc/(y_c * A)]
for square or rectangle I_xc is (1/12)(b)(a)^3 doesn't matter what b and a are because its a square and both are 6.
y_c is 3 as we know the centroid of square is just the middle of the square since it is the middle it is just 3 m below.
hence after working out we will find y_square to be 4.

Now we take a moment. Not literally. A moment about the shaft.
now working we need to find the net moment about the shaft. so we can look at it this way we know the Force on the square will cause an clockwise moment about the shaft. F_square is 4m down from the water surface. with geometry it is 2m above the shaft. the product of F_square and 2m will be clockwise moment about the shaft.

F_semicircle is 7.359m below the surface of water. minus away the length of the square it is just 1.359m below the shaft. the product of F_semicircle and 1.359m will be an anticlockwise moment about the shaft. the Nett moment is just the difference between clockwise and anticlockwise moments about the shaft. you will get 750kNm.

It's 5 years late you probably graduated. but I just want to help others.
 
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