5
$\begingroup$

What is meant by yield of isotropic plastic material? How is yield of Isotropic plastic material different from the term normal yield which is referred as transition from elastic behaviour to plastic. I am bit confused of term Yield of Isotropic plastic material.

Edit: yes I am talking about Isotropic material and requesting for answers based on what graph of yield criteria explains

Please help, thanks!

$\endgroup$
2
  • $\begingroup$ I think you are taking about plasticity of isotropic material. If we see yield criteria for isotropic material then for an isotropic material the yield criteria will be a function of the invariants of the stress deviator. So it's bit different from normal yield I guess. $\endgroup$ – Rajakr Nov 3 '20 at 16:42
  • $\begingroup$ If you mean polymers, their yield is time sensitive . So strain rate matters and when you measure it matters . I expect ASTM will have written standard procedures . $\endgroup$ – blacksmith37 Apr 7 at 0:55
0
$\begingroup$

In terms of yield surfaces, the 2D yield surface of an isotropic material will be mirrored over the line sigma_1 = sigma_2. In 3D, an isotropic yield surface will have rotational symmetry about the line sigma_1 = sigma_2 = sigma_3 (the hydrostatic axis). These symmetry constraints imply that the yield surface will intersect each principal stress axis at the same value. Additionally, if the material is isotropically hardened, the yield surface will expand outward concentrically to the original yield surface.

In contrast, a the yield surface for an anisotropic yield surface does not have these symmetry requirements, and can intersect the different principal stress axes at different values. For a 2D yield surface, the degree of anisotropy, R, is given by R = sigma_2 / sigma_1, where R=1 corresponds to an isotropic material.

$\endgroup$

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.