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

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SUMMARY

The discussion centers on the relationship between stress, force, and area in a solid body under equilibrium conditions. It establishes that internal stresses in a solid 3D body must sum to zero to prevent failure, with the surface integral of stresses equating to the forces removed (P1 and P2) when the body is bisected. The equation stress = force/area is emphasized, indicating that integrating stress over an area (Delta A) yields the resultant force, confirming equilibrium in the system.

PREREQUISITES
  • Understanding of solid mechanics principles
  • Familiarity with stress and strain concepts
  • Knowledge of equilibrium conditions in physics
  • Basic calculus for integration of stress over area
NEXT STEPS
  • Study the concept of internal stress distribution in solid mechanics
  • Learn about equilibrium equations in three-dimensional bodies
  • Explore the application of integration in calculating resultant forces
  • Investigate the implications of stress concentration in materials
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Students and professionals in engineering, particularly those specializing in solid mechanics, structural analysis, and materials science, will benefit from this discussion.

v_pino
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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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