
#1
Feb2314, 01:03 AM

P: 47

Say for example I have a hollow cube with equal length/width on all sides. Let's also assume that the outside walls of this cube is unbreakable / perfect insulator / unable to deform in anyway.
Now let's assume that we put a divider in the middle of the hollow cube of a certain thickness. The property of the divider material has a yield strength of lets say 30ksi. And now somehow one side of the cube is pressurized to 5ksi and the other side of the divider is 14psi (atmospheric). So my question is, if the thickness of that divider is 1mm, by intuition even though the pressure is less than the yielding point the divider will break because it's so thin. So than what is the governing point or equation that introduces the thickness of a divider as a variable? 



#2
Feb2314, 01:29 AM

P: 1,260

How large is your cube?




#3
Feb2314, 02:51 AM

P: 47

inside lets assume....... 1ftx1ftx1ft?? How would that matter? 



#4
Feb2314, 03:49 AM

P: 1,260

I'm thinking of a conceptual problem.. how does this work?
Rupture will depend upon the shear stress on the membrane, and the tensile stress on the membrane.
with regards to shear only, For the 12inch cubic, the divider has 144 square inches of area. The pressure differential is 4986 psi giving a total load on the divider ( 144 in^2 ) of 717,984 pounds. The membrane has a shear area of 1.92 inches so it would shear off if it it did not rupture from bulging and tensile stress. Compare that to a 1inch cube where the load on the divider is now only 4986 pounds. Shear area ( perimeter) is 0.16, so the shear would be 4986/0.16 = 31 ksi. That is getting close to the yield strength, but the membrane would probably shear, and the bulging would be less pronounced. or a 0.1 cube, area of membrane = 0.001 sq inches, load = 4.986 pounds , shear area = 0.016 sq inch, shear stress = 311 pounds, in this case shear should not occur. Less bulging also. Area of the membrane does count. 


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