Civil Engineering - Hydraulics - Discussion

Discussion Forum : Hydraulics - Section 5 (Q.No. 44)
44.
The depth of the centre of pressure on a vertical rectangular gate (4 m wide, 3 m high) with water upto top surface, is
1.0 m
1.5 m
2.0 m
2.5 m.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 1 of 2.

Jero M said:   2 years ago
We know that, the centre of pressure for a plane uniform vertical surface lies at the depth of two-thirds of the height of immersed portion.
So, here we get 2/3 of 3m which is equal to 2m.

The given answer is correct.
(1)

Abhay said:   9 years ago
The center of pressure is located at the centroid of the rectangular shaped pressure field 2/3 from the top of the water line.

b = 4 m, d = 3 m.

Center of pressure = 2/3 d = (2/3)*3 = 2.0 m.

M.Atiq said:   6 years ago
Center of pressure is always below the centre of gravity and CG is located at point H/2 = 1.5m so simply we can make an idea that it will be at 2m.

Muhammad Hashim said:   4 years ago
CP = (Ig/Ax)+x.
CP = ((bd3/12)/bd*d/2)+1.5,
CP = (d/6)+1.5,
CP = 1/2+1.5,
CP = 2.
C is the right one.
(1)

Salahuddin said:   6 years ago
CG equals CP when water surface coincides the body surface. Therefore cg=cp=3/2=1.5.

Am I right?

Aadarsh Devkate said:   7 years ago
@All.

It's very simple
Gate is vertical.
Means H = 4m.
D/2 = 4/2 = 2.

Imran said:   5 years ago
Area = b*d.
CG (y)= d/2.
Ig = bd3/12.
Center of pressure = y + Ig/Ay.

Dinesh prasad sao said:   8 years ago
2/3 height if water is full, if not full then it's 5 of water.

Santhosh said:   6 years ago
Depth of center of gravity,

H = 2h/3
= 2*3 / 3
= 2 m.

Niranjan said:   2 years ago
The depth of the center of pressure= 2H/3.
(2*3)/3 =2m.
(1)


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