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Consider two bars one rigid and the other deformable, acted upon by two equal and opposite point loads P as shown. In either of the cases, if we cut the beam from an imaginary section, then, to bring (say) the left part of the beams (obtained after cut) in equilibrium an equal and opposite force, equal to P, must be developed in that part of the beam.

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In case (b), the intensity of these internal resistive forces developed on the section is called stress, and is equal to the internal resistive forces divided by the area.

In case (a), since the bar is rigid it doesn't undergo any deformation. Hence, there must be no stresses, since stresses appear to resist deformations. However, we see that even in (a) internal resistive forces are developed, and hence we can define their intensity via stress.

So, my question is, is there any stress in the rigid bar? If no, how am I to conceive the internal force developed in case (a)?

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All real objects deform under stress. If you create an imaginary object that does not deform, you can decide how this imaginary object reacts to a stress.

Stress transfer in an object is limited by the speed of sound, you may start there. Or if you want to ignore that too, you can account for limit of the speed of light for the information propagation.

But right now you just took a bad model, and now ask why is it bad. Well, you made it this way for whatever reason, you answer why you made it this way.

If you just want a simple model for ease of calculation, just assume that stress propagates instantly and doesnt depend on deformation, only on forces and areas.

Anyway, my main point: what you are asking is not physical. It is more related to your model for some simplified calculation. That you did not mention, so I cant know what properties of this simplified model are important for your calculation. Describe that, and then it will be easier to answer your questions.

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  • $\begingroup$ Thanks for the insight. What if I consider in case (a)not a rigid body, but a body with very high Young's Modulus, E. In that case, the body in (a) will deform (slightly), and the stress in case a and b should be the same, since P/A is same in each case. Am I right on this? $\endgroup$ Mar 22 at 13:54
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    $\begingroup$ @HarshitRajput yes. But if you also consider transient effects, objects may behave differently. Vibrate with a different frequency, like a split rubber and glass would. $\endgroup$ Mar 22 at 13:57

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