0
$\begingroup$

The same geometry and the same free body diagram. However, if we replace the material from steel to glass, for example. How does that influence the analysis?

Two things I think are 1. steel is ductile and glass is brittle. This means the cross section when it fails will be different. 2. glass has lower yield stress, which means it is much easier to fail.

Is there anything else that I omit? Thank you very much.

$\endgroup$
3
  • $\begingroup$ What would you have to change if you replaced steel with aluminium or titanium? $\endgroup$ – Solar Mike Sep 22 '18 at 21:13
  • $\begingroup$ One: Steel is not always ductile, so the statement 'steel is ductile' is false. Two: The distribution of stresses would remain the same, hint think about photoelasticity, however the results would not be same. $\endgroup$ – Sam Farjamirad Sep 22 '18 at 21:39
  • $\begingroup$ in addition, if you replace glass, then you'll see three different regions after failure, one is the origine, it is spherical, two is mist or hackle region if remember correctly , and then a very smooth region, mirror region, not all are visible with naked eye, but a magnifier is enough. $\endgroup$ – Sam Farjamirad Sep 22 '18 at 21:46
1
$\begingroup$

If you are dealing with a simple Back-of-the-envelope calculation, then it should not matter, from stress perspective, what type of material you chose. All you have to do is to compare the results to the material allowable strength.

However, if you are talking about a FEA problem, you should take care closely of the material you choose. Tools like ANSYS calculate the object deformation (which is totally affected by the material Young's modulus and Poisson's ratio) and then derive the stresses out of it.

$\endgroup$
4
  • 1
    $\begingroup$ Well, sure, FEA uses deformation to obtain stress, but the end result (assuming no failure and reasonably elastic material properties) will be identical: if material A has a Young's modulus which is twice as much as B's, then A's deformation will be half of B's; the stresses in both materials will be the same. $\endgroup$ – Wasabi Sep 22 '18 at 21:34
  • $\begingroup$ You are right. But, this is true as long שד the problem is simple. For instance, in the case there are several materials involving, each one will carry a load in proportion to its relative stiffness, which is directly related the its material properties. I guess other local phenomena may also be influenced by the material, so being cautious is commanded. $\endgroup$ – Yaniv Ben David Sep 22 '18 at 21:44
  • 1
    $\begingroup$ @Wasabi The stress will only be the same if (1) you are doing a linear analysis assuming small displacements and strains, and the external loads are independent of the displacements (i.e. no "follower forces"), and (2) Poisson's ratio is the same for both materials. The sensitivity to Poisson's ratio can be large, if one material is almost incompressible and the other is not (e.g. a metal compared with rubber). Note, some glasses have Poisson's ratio > 0.4 compared with typical metals about 0.3.. $\endgroup$ – alephzero Sep 22 '18 at 23:48
  • $\begingroup$ … as a well known analogous situation to the sensitivity to Poisson's ratio, consider the difference between plane stress and plane strain boundary conditions in 2D analysis. The effective value of Youngs modulus changes by a factor of $(1-\nu)$. $\endgroup$ – alephzero Sep 22 '18 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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