Shear modulus with plastic deformation

Question

The shear modulus $G$ gives information about the linear-elastic material behavior. How does it behave in the area of plastic deformation?
With $ \tau=G\tan\gamma$, shear modulus should get smaller up to a limit value (break), right?
I am thinking of metal with plastic strains up to 5%.

Answer 1

During plastic deformation, the shear modulus $G$ is typically assumed to be zero. In other words, if deformation is already occurring, then if we overlay a elastic stress–strain experiment, zero incremental stress is required to obtain incremental positive strain, which implies an absence of stiffness, or $G=0$. Through the relations $\frac{\Delta V}{V}=1-2\nu$, where $V$ is volume, and $G=\frac{3K(1-2\nu)}{2(1+\nu)}$, where $K$ is the bulk modulus, we see that $G=0$ is equivalent to the standard assumption of constant volume and Poisson ratio $\nu=1/2$ during plastic deformation.

After plastic deformation, the shear modulus is typically about the same as it was before deformation (say, within a few percent). As an elastic modulus, the shear modulus typically depends—in metals, ceramics, and crosslinked polymers, at least—on molecular bond stretching/compression. Broadly, it doesn't really matter if the bulk material's undergone deformation slip (e.g., if dislocations have moved through the lattice) or if crack propagation has occurred elsewhere; the nature of the bonding remains unchanged. (Edge cases exist where the long-range order of the crystal has been nearly obliterated due to accumulated dislocations and grain boundaries, but this generally requires plastic strains of ≫5%.)

The relation $\tau=G\gamma$, where $\tau$ is shear stress and $\gamma$ is the engineering shear strain or decrease in corner angle, applies only to an independent elastic stress–strain test. You can't take $\gamma$ from plastic deformation and use it to calculate an elastic modulus. Does all this make sense?