Moment about shaft from hydrostatic forces

Click For Summary
SUMMARY

The net moment about the shaft from hydrostatic forces acting on a gate is calculated to be 750 kNm. The gate is divided into two parts: a square and a semicircle. The forces acting on each part are determined using the formulas for hydrostatic pressure and the parallel axis theorem. The final calculation involves determining the moments caused by each force about the shaft, leading to the net moment result.

PREREQUISITES
  • Understanding of hydrostatic pressure calculations
  • Familiarity with the parallel axis theorem
  • Knowledge of centroid and center of pressure concepts
  • Proficiency in basic mechanics and moment calculations
NEXT STEPS
  • Study hydrostatic force calculations in fluid mechanics
  • Learn about the parallel axis theorem in detail
  • Explore centroid and center of pressure for various shapes
  • Practice calculating moments in mechanical systems
USEFUL FOR

Engineering students, mechanical engineers, and professionals involved in fluid mechanics and structural analysis will benefit from this discussion.

zerogoal
Messages
21
Reaction score
0

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?
 

Attachments

  • shaft.jpg
    shaft.jpg
    6.8 KB · Views: 1,093
Physics news on Phys.org
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.
 
Last edited by a moderator:

Similar threads

Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
10K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
1
Views
3K
Replies
2
Views
5K
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
6K