Calculate lateral force/pressure of a body water

In summary, the hydrostatic pressure is negligible and a flexible liner should be there to avoid small cracks. The force acting on the longest wall comes to 4905 Newtons at the center of pressure.
  • #1
Vetmora
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So I'm planning to build an above ground concrete fish pond. The size of the inside of the pond will be (L)2000mm x (W)500mm x (D)500mm. What I want to know before I go any further is will the poured concrete walls hold the pressure of the water inside. So if I know the pressure of water I'll hopefully be able to know how thick to make the walls etc.

I know there are formulas for this but I'm pretty green horned when it comes to physics: https://en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_pressure.

Is there someone who would be able to help me out? I'd like to use metric units.
 
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  • #2
Vetmora said:
The size of the inside of the pond will be (L)2000mm x (W)500mm x (D)500mm.

Cant help directly with the calc., those measurements are not overly large, I had a tropical fish tank that size made out of 4mm thick glass
 
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  • #3
Ah, good to know. Didn't think it would be a problem but thought it couldn't hurt to double check.
 
  • #4
Vetmora said:
So I'm planning to build an above ground concrete fish pond. The size of the inside of the pond will be (L)2000mm x (W)500mm x (D)500mm. What I want to know before I go any further is will the poured concrete walls hold the pressure of the water inside. So if I know the pressure of water I'll hopefully be able to know how thick to make the walls etc.

I know there are formulas for this but I'm pretty green horned when it comes to physics: https://en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_pressure.

Is there someone who would be able to help me out? I'd like to use metric units.
I would say that the hydrostatic forces are negligible compared with the practical possibilities of frost damage or plant roots or ground moving due to drought. A flexible liner should be there as that will avoid problems resulting from small cracks (old carpet underneath it or expensive equivalent from the pond shop). I think you are more likely to lose water through cracks than with a tsunami across your garden. The advantage you have here is that the loading is very steady and you have no shock loads to deal with. Your pond will not be approaching the Hoover Dam in height and that you would, in any case, be considering a wall thickness of more than just one vertical skin of brick or blocks; a radius between floor and walls and a few degrees of slope in towards the top would 'look right'. Also, if you are planning to cap the wall, it could look good with a reasonable thickness so that you can lean on it and get your face close enough to see the inhabitants.. Look at images of ponds on Google and that will give you an idea of what works structurally - also you can choose a design that's pleasing to look at. I have never heard of Pond Wall Failure so I don't think there's anything to worry about.

PS It's essential that you have some way in and out of it for visitors (unless you just want ornamental fish which I, personally find a bit boring on their own.) But an emergency exit for small mammals to get themselves out would be good for them and good for you to avoid small bodies floating about.

PPS P = ρgh will give you the hydrostatic pressure at depth h. Bearing in mind that 1Atm corresponds to about 10m of water, your depth of 0.5m gives you just 1/20Atm excess pressure on the floor (and outwards at the bottom too!)
 
  • #5
If I'm not completely rusty then your pressure comes to about:
P=1000x9.81x0.5 = 4905 Pascal
The force acting on the longest wall comes to F=4905 x 2.0x0.5 = 4905 Newtons at the centre of pressure.
I'm not even sure if it will really be that bad because the pressure will be a worst case scenario right at the bottom of the tank. I have to admit that I'm very rusty with these calcs, we did this way back in our second semester (about 10 years ago). I cannot remember what assumptions you're supposed to make.
 

What is lateral force/pressure of a body water?

The lateral force/pressure of a body water is the amount of force or pressure exerted by a body of water on a surface in a direction perpendicular to the surface. This force/pressure is caused by the weight of the water and can vary depending on the depth and density of the water.

How is lateral force/pressure of a body water calculated?

The lateral force/pressure of a body water can be calculated using the formula F = ρgh, where F is the force in Newtons, ρ is the density of the water in kilograms per cubic meter, g is the gravitational acceleration in meters per second squared, and h is the depth of the water in meters.

What factors can affect the lateral force/pressure of a body water?

The lateral force/pressure of a body water can be affected by the depth and density of the water, as well as the shape and orientation of the surface it is exerting force/pressure on. Other factors such as wind, waves, and currents can also influence the force/pressure.

Why is it important to calculate lateral force/pressure of a body water?

Calculating the lateral force/pressure of a body water is important in various fields such as engineering, architecture, and marine science. It helps in designing structures that can withstand the force/pressure of water, predicting erosion and sedimentation patterns, and understanding the impact of water on marine ecosystems.

Are there any safety precautions to consider when dealing with lateral force/pressure of a body water?

Yes, it is important to consider safety precautions when dealing with lateral force/pressure of a body water. This can include using proper equipment and techniques when conducting measurements, understanding the potential hazards of working in or near water, and following safety protocols to prevent accidents or injuries.

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