A couple of questions about the behavior of normal stress.

For both of these, assume that the axial force applied is ideal, meaning distributed along the whole face of the member on both sides, and perfectly perpendicular to the face of the member, so that average normal stress is guaranteed to be equivalent to the exact stress on any point on the cross section. (This is opposed to having the axial force targeted at part of the face on both sides, so certain places on the cross-sectional area would be experiencing more stress.)

For simplicity, also assume we are dealing with an ordinary cylindrical member.

  1. Is the normal stress identical no matter where along the beam you take a cross-section? Or is it larger the closer you get to the sides of the member, where the axial force is applied?

  2. If it is, how would the member start deforming once the stress became too large? Where in the member would the material start to give.

I understand that these scenarios are ideal and never possible in the real world but hypothetically, if they were to happen, the member would still undergo deformation so I was curious what that would look like. Thank you!


1 Answer 1


The normal stress is identical at any section you consider as long as the beam is in static equilibrium. Consider any section of some finite thickness. The force on each face must be equal, otherwise the section would be accelerating. Therefore, the forces at all cross-section are equal.

This might be easier to understand if you consider a rope with a person pulling on each end. If the rope isn't moving, then the force on each person is the same. It's impossible for one person to be pulling harder than the other if the system is stationary.

The beam will deform all along the length when the load is applied. If the stress exceeds the yield point, then plastic deformation will occur. No material or shape is perfect, so there will be some section(s) that are weaker than others and that is where the permanent deformation/fracture will occur.

  • $\begingroup$ Right, I understand that the magnitude of the force must be the same on both sides to avoid acceleration. I think the confusion about deformation extended to confusion about pressure at each cross-section. Thank you! $\endgroup$ Commented Jun 9, 2022 at 20:09

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