How is stress related to force and area in a solid body under equilibrium?

AI Thread Summary
The discussion centers on the relationship between stress, force, and area in a solid body under equilibrium. It emphasizes that internal stresses must sum to zero to maintain equilibrium; otherwise, the body would fail. The integral of the pressure difference (Delta P) must equal the sum of the pressures at the surfaces removed (P1 and P2) when bisecting the body. The participants clarify that calculating stress as force divided by area is essential, and integrating stress over the area can yield the resultant force. Overall, understanding these principles is crucial for analyzing solid bodies in equilibrium.
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Why is it the integral of delta P that must equal P1 and P2 in the second diagram (half of the original body)? I thought it is simply that Delta P = (P4 + P3).
 

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I don't really get what your asking? I think the diagram is basically just stating that internal stresses in a solid 3D body must sum to zero, if they do not then the body is exploding.
 
It says that if you bisect the body, that is already in equilibrium, the resultant surface integral has to equal P1+P2, which were removed, for the body to stay in equilibrium. It would also be correct to say what you are saying, but I think the main idea was that the surface integral has to be equal to what was removed.
 
So stress=force/area. Does it mean that if I find the stress over the area (Delta A) and integrate it with respect to Delta A, then I get the new force which is equal to the sum of P1 and P2?
 
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