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While answering this question on DIY.SE, I wanted to calculate the amount of pressure (in psi) that would be exerted on the walls of the container. However, I don't know much about how to calculate such pressures.

Given:

The container:
Hot Tub Frame with Banding and Bracing Assume the interior of the container is solid (e.g. covered in plywood and a rubber liner)
Container outside dimensions: 8' x 8' x 3'
Container inside dimensions: 89" x 89" x 36"
Container volume: 165.020917824 cubic feet ((89x89x36) x 0.000578704)
Amount of water held by container : 1234.521486241344 gallons (165.020917824 x 7.481)
Weight of water held by container : 10,297.3052722176 pounds (165.020917824 x 62.4)

How would I calculate the amount of pressure on the walls in pounds per square inch?

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closed as off-topic by Chris White, Qmechanic Sep 20 '13 at 21:42

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2 Answers 2

up vote 5 down vote accepted

You first need to clarify what you are really asking about. You ask about force, but then mention units of PSI (pounds per square inch). PSI is a unit of pressure, not force.

Your container is a tank that will be filled with water. The pressure at any point of the container wall is very simple, as it's directly proportional to the water depth at that point. The pressure will be .434 PSI per foot of depth. Since your container is 3 feet deep, the pressure at the bottom will be 1.3 psi relative to outside. Similarly, half way down, the pressure will be half that, or .65 psi.

To compute the force, simply integrate the pressure over the area. Since the pressure varies linearly with depth, the average pressure is half the bottom pressure by inspection. One side is 89 x 36 inches, so 3204 square inches. This times the average pressure of .65 psi yields 2084 pounds, or about one ton pushing outward on each side.

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Perfect. Thank you. –  Tester101 Sep 20 '13 at 13:44
    
Could you elaborate on how you reach the conclusion of .434 PSI per foot? –  The Evil Greebo Sep 20 '13 at 14:21
1  
That's a constant -- He looked it up. –  Chris Cudmore Sep 20 '13 at 14:31
3  
It's the weight of a column of water at max density 1 x 1 x 12, or 62.43 lbs/ 144 ft^2 –  Chris Cudmore Sep 20 '13 at 14:38
2  
@David: I agree that homework shouldn't be answered outright, but this isn't homework. I believe this is a genuine question from someone that didn't know how to go about solving a real-life problem. It is always good to show the method of solution, as I did, but for genuine real-life problems it is perfectly fine to give the solution, in fact we should. Some folks here are being over-sensitive to "homework" problems. –  Olin Lathrop Sep 20 '13 at 16:20

Water pressure over a horizontal surface is simply the weight of the column of water over the wet surface divided by the area of the wet surface.

If you're working in PSI, then take a column of water, 1" x 1" x depth, and figure out how much it weighs. Since our column is one square inch, this weight in pounds will be numerically equal to the pressure in psi.

This is a common calculation, working out to .434 psi per foot of depth.

We have the formula now p = .434 d (p in psi, d in feet) which is linear.

There's one more fact we need. Pressure works in all directions. So if we had an ceiling at depth (an undercut), the upwards pressure on that ceiling would be the same as the downward pressure.

So, the outward pressure is the same as the downward pressure is the same as the upward pressure.

This will be 0 at the water surface and .434 d at the bottom.

What might be of more interest is the outward force. That is, what is the force your structure needs to resist in that direction.

It's simply the average pressure times the area of the wet surface, acting perpendicular to the surface.

If you need the bending moments (torque), then you locate the force at the centroid of the pressure curve, which is triangular. This means the point of action will be 1/3 of the way up from the base of the triangle, or 1/3 of the way up the wall.

And that is how you can calculate the pressure and force against the sides of a one-side submerged rectangular wall.

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