Forces and water pressure on a wall

[Fluidman117]

Homework Statement

I have a wall, which is basically a gate of a shipping lock. I have different water levels on both side of the wall. I made a graph to illustrate my problem and dimensions:

https://dl.dropboxusercontent.com/u/47965009/problem1.png

Homework Equations

1. Find the hydro-static pressure on left side of the wall
2. Find the total forces on this gate per unit meter width

The Attempt at a Solution

1. I find the pressure P=1000 kg/m3 * 9.81 m/s2 * 15.2 m= 149 kN/m2

2. To find the forces I divide the pressure distribution of the left side to two different geometrical objects. I do this because I figure the forces on the right side will cancel out the same amount of forces on the left side. So I will have one triangle at the top and from point h1, I have a rectangle in the bottom.

To calculate the forces on the triangle:
First I find the pressure at line h1 P(tri)=1000 kg/m3 * 9.81 m/s2 * (15.2-10.6)=45 kN/m2
Then to find the Forces on the triangle F= P * (h-h1)/2 = 45 * 4.6/2=103.5 kN/m

To calculate the forces on the rectangle:
F=P(tri) * 10.6m= 45 * 10.6= 477 kN/m


Firstly, of course I would like to ask if the above is correct?
The question I have now, is can I just add these forces together to get the total forces acting on the gate?

 

[mfb]

1. I find the pressure P=1000 kg/m3 * 9.81 m/s2 * 15.2 m= 149 kN/m2

Where do you have that pressure? Is that the answer to the question (I don't know, the problem statement is a bit unclear)?

2. To find the forces I divide the pressure distribution of the left side to two different geometrical objects. I do this because I figure the forces on the right side will cancel out the same amount of forces on the left side. So I will have one triangle at the top and from point h1, I have a rectangle in the bottom.

Okay, that is possible.

The question I have now, is can I just add these forces together to get the total forces acting on the gate?

Sure.
I did not check the numbers, but they look reasonable.
There are some units missing in the equations.

 

[Fluidman117]

Where do you have that pressure? Is that the answer to the question (I don't know, the problem statement is a bit unclear)?

Okay, that is possible.
Sure.
I did not check the numbers, but they look reasonable.
There are some units missing in the equations.

Yes, that would be the answer to the question 1.


I also have a few additional questions about the problem?

a) Since there is a hydrostatic pressure on the vertical wall also from the right side, would I calculate the total pressure on the wall in a similar fashion as I did for the total forces on the wall? (that means, that the pressure on the right side will cancel out the same amount of pressure on the left side?)

b) If I would have a safety factor of 1, what would be the design load in this case?

 

[mfb]

a) Since there is a hydrostatic pressure on the vertical wall also from the right side, would I calculate the total pressure on the wall in a similar fashion as I did for the total forces on the wall? (that means, that the pressure on the right side will cancel out the same amount of pressure on the left side?)

You don't have to do that calculation again.

You calculated the force per meter. To get the total force, you could multiply it with the length of the lock (if you would have that). To get the force for a section with a length of 1m, ...

To get the average pressure, you could divide it by the height of the lock (which is not required).

b) If I would have a safety factor of 1, what would be the design load in this case?

Just the actual load, I guess?

 

[Fluidman117]

Okay, but I am still a bit confused about the pressure part. What I'm interested is that I have pressure acting on the wall from left side and also pressure acting on the wall from the right side. Will the effective pressure on the wall be the sum of those? P(effective)=P1-P2 ?

 

[mfb]

I think "effective pressure" means that difference, yes.

 

[Fluidman117]

Thanks for the help mfb!