- #1
peter2108
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Safe to remove inner wall of pond or will it burst??
1.Problem
I have a fishpond. It is raised 0.4m above the ground, the wall being interlocking wooden blocks. The blocks are all locked together all the way round so they could be picked up and put down somewhere else. The pond is 2.4m long by 1.6m. In fact the bounding wall is double skinned, the inner wall is connected to outer by a buttreses.I don't think the inner wall is needed. I would like to remove it. But is the force on the walls enough to make this imprudent.The guy selling the blocks originally said you needed a double skin. ... but he would wouldn't he?
Can I remove the inner wall safely? Might the wall burst?
After removing inner wall total force on long side is
[itex]\int ^{h}_{0} \rho \; g \; w \; y \; dy[/itex]
where [itex]\rho[/itex] is density of water, [itex]g[/itex] is acceleration due to gravity, [itex]h[/itex] is height and [itex]w[/itex] is length of wall.
Well the total force on the long side is [itex]1/2 \times 1000 \times 9.81 \times 2.4 \times 0.4^{2} \ = \ 1883[/itex]. Sounds quite a lot. But the force on the North side of the wall is exactly balanced by the force in the opposite direction on the South side. So it can be ignored? Surely not. Again the length of the wall affects the total force so that a side of 24m instead of 2.4m would be under 10 times the force. But my intuition is that the length of the sides does not matter. It's all about the dam bursting and that seems to depend only the pressure which is [itex]1883/2.4 \ = \ 784[/itex]. But how to interpret this? I think if I divide by [itex]g[/itex] I get the "weight" of the water pressing on the wall which would be [itex]784/9.81 \ = \ 79[/itex]. I think that is far too high. So now I am puzzled.
1.Problem
I have a fishpond. It is raised 0.4m above the ground, the wall being interlocking wooden blocks. The blocks are all locked together all the way round so they could be picked up and put down somewhere else. The pond is 2.4m long by 1.6m. In fact the bounding wall is double skinned, the inner wall is connected to outer by a buttreses.I don't think the inner wall is needed. I would like to remove it. But is the force on the walls enough to make this imprudent.The guy selling the blocks originally said you needed a double skin. ... but he would wouldn't he?
Can I remove the inner wall safely? Might the wall burst?
Homework Equations
After removing inner wall total force on long side is
[itex]\int ^{h}_{0} \rho \; g \; w \; y \; dy[/itex]
where [itex]\rho[/itex] is density of water, [itex]g[/itex] is acceleration due to gravity, [itex]h[/itex] is height and [itex]w[/itex] is length of wall.
The Attempt at a Solution
Well the total force on the long side is [itex]1/2 \times 1000 \times 9.81 \times 2.4 \times 0.4^{2} \ = \ 1883[/itex]. Sounds quite a lot. But the force on the North side of the wall is exactly balanced by the force in the opposite direction on the South side. So it can be ignored? Surely not. Again the length of the wall affects the total force so that a side of 24m instead of 2.4m would be under 10 times the force. But my intuition is that the length of the sides does not matter. It's all about the dam bursting and that seems to depend only the pressure which is [itex]1883/2.4 \ = \ 784[/itex]. But how to interpret this? I think if I divide by [itex]g[/itex] I get the "weight" of the water pressing on the wall which would be [itex]784/9.81 \ = \ 79[/itex]. I think that is far too high. So now I am puzzled.