I want to make a circular floating pond on the ground with an area of 1000m square and a height of 1.4m to raise fish. My solution is to use sheet iron with a width of 1.4m to form a circle with a radius of about 17.8m, then line it with HDPE tarpaulin and pump water in. Please help me calculate the required thickness of iron plate to make the pond? Thank you!
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$\begingroup$ What forces on the wall have you considered? The water force due to depth is easy and there are so many references for that. BUT will you need to have ladders? other equipment? holes for filtering ? $\endgroup$– Solar MikeMar 9 at 7:49
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$\begingroup$ Also will you make sure that area is flat? $\endgroup$– Solar MikeMar 9 at 8:33
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$\begingroup$ yes, the ground is flat! $\endgroup$– Ghsjs KhaMar 10 at 14:11
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1 Answer
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The hydrostatic pressure of water is dependent on the height only. Assuming the water density at 1kg/l:
For simplicity, we measure the maximum hydrostatic pressure at the depth of 1.4m and ignore the additional strength from the plate mechanism working above this level.
$$P_w=1.4m*1kgfcm^2/10m=0.14kgf/cm^2$$
The tensile force exerted on a centimeter-wide strip of the cylindrical tank is:
$$T=2*17.8*100cm/m*0.14=498.00kg/cm \ width$$ Assuming we use mild steel with a yield stress of 250mPa and a working strength of $$ \ 150 mPa=\pm 1500kg/cm^2$$ the total thickness, T, of the sheet is,
$$T=498/1500=0.33cm=3.3mm$$
Edit
This divided by two will be the thickness of each side. $$3.3/2=1.65mm$$ So, we pick a round thickness that is better capable to be bolted or welded, say 2mm thick.
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$\begingroup$ Shouldn't it be just a half of that? With pressure $P=\rho\cdot g \cdot h = .0137 MPa$, allowable stress $f=150 MPa$ and radius $R=17800 mm$, minimum thickness should be about $\frac{P\cdot R}{f} = 1.63 mm$. $\endgroup$ Mar 9 at 21:23
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$\begingroup$ the force is what you'd get if you cut the circle in half exposed to the projection of pressure perpendicular to your cut. So it's 2pr/f $\endgroup$– kamranMar 9 at 23:45
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$\begingroup$ In a cylinder with a length $L$, the hoop force equilibrium with thickness $t$ would be $R\cdot L\cdot P = t\cdot L\cdot f$. The 2 term is there if you use diameter instead of radius. Here is a reference to Barlow formula (probably the oldest one in the field), which uses external diameter for some additional safety. $\endgroup$ Mar 10 at 13:41
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$\begingroup$ Thanks for everyone's answers! I have another option, replacing the iron plate with 1.4m long wooden bars that are joined together to form a circle with the same size. The outside uses round iron, welded to form a belt to keep it fixed. (construction iron used for concrete pillars and beams) Assume 4 belts are 0 apart; 0.5; 1.0; 1.4m from top to bottom respectively. So what is the proper diameter of each belt in MM? (Assuming that the wooden bar has good bending force, it will definitely not break. The foot of the column is left underground, not buried in the ground) $\endgroup$ Mar 10 at 14:49
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1$\begingroup$ @TomášLétal, my apologies. you are right. the force I got was the total pressure on the hoop, it needs to be divided by 2. to get the share of each side. I edit my answer. $\endgroup$– kamranMar 10 at 18:31