What is a hardening function? What does it represent?

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A hardening function represents how a material deforms under stress, specifically in terms of strain. Common types include linear and kinematic hardening functions, which are essential for modeling material behavior in simulations like those in COMSOL multiphysics software. Choosing the appropriate hardening function for specific materials, such as Earth's heated crust, can be challenging due to a lack of available literature on parameters like isotropic and kinematic tangent moduli. An elastic perfectly plastic model may be suitable, where tangent moduli can be set to zero, indicating constant shear stress post-yielding. Understanding the material's elastic constants and yield strength is crucial for accurate modeling, especially when considering pressure and temperature effects.
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So, I was introduced tangentially to plastic deformation and hardening functions. I see we have three basic function ... and the linear and kinematic seem to be the most common defaults ... I see it is a measure of the strain of a material under stress.

But how would you know which version to use for some material? Is it some tabled value like thermal conductivity or specific heat?

For instance, I need to choose a "hardening function" for Earth's heated crust as it undergoes the pressure of an inflating magma reservoir. I cannot find such a function in the literature so how would I go about choosing some reasonable function?

As an added note, this has to be applied in COMSOL multiphysics software which requires a "Isotropic tangent modulus" and a "kinematic tangent modulus" ... I cannot find ANY value for those parameters. I'm not even sure WHAT they are. Are these parts of the hardening function to describe the curvature of the line?

Since mechanics and materials is not the mainstay of what I do, this is all new to me.

Thanks for any help!
 
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Without knowing the details of the COMSOL models, I would pick an elastic perfectly plastic model. Since they are using tangent moduli, setting them equal to zero would probably do this. This means that after the material yields, the shear stress will remain constant. This will require 2 elastic constants and a yield strength which should be findable. Depending on whether you are near the surface, to improve the results i would either take into account the pressure/temperature sensitivity or the fracture behavior next if needed.
 
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