0
$\begingroup$

What causes us to think that the stress is distributed uniformly over the cross sections A and B? Force A applied to a stationary bar. Figure: Force F applied to a stationary bar

$\endgroup$
2
  • $\begingroup$ Do you mean on an experimental point of view or a theoretical derivation? $\endgroup$
    – LPZ
    Commented Aug 22, 2022 at 16:30
  • $\begingroup$ @lpz Ideally I'd like to know both, but in the OP I meant a theoretical derivation. $\endgroup$
    – Naghi
    Commented Aug 22, 2022 at 19:37

1 Answer 1

1
$\begingroup$

What causes us to think that the stress is distributed uniformly over the cross sections A and B?

In real situations, the stress is never exactly uniformly distributed. However, it's convenient and often reasonably accurate to use such a model.

This treatment is justified by Saint-Venant's principle, which holds that sufficiently far from the left end, the stress state is nearly indistinguishable from that produced by a distributed load over the left end under the condition that the left end is free to laterally contract from Poisson effects. (One is encouraged to investigate discussions of this principle in the literature through Google Search, for instance.)

We'll also assume an isotropic and homogeneous beam, of course.

Under these conditions, there's no source of lateral stresses and no reason to privilege any part of the cross section as having any different stress from any other part. Put another way, model parsimony allows and encourages an assumption of uniform stress across the cross section.

This scenario lies in sharp contrast to bending, for example, in which the axial stress has to be positive (negative) in the top half of the cross section and negative (positive) in the bottom half to transmit a bending moment through the material. Here, the stress is required to vary in some way across the cross section.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.