# Converting stress-strain curve to shear stress-shear strain

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1. Apr 2, 2015

### turpy

1. The problem statement, all variables and given/known data
For a crystalline metal material
- Elastic modulus E
- Poisson ratio v
- A table with test data of stresses vs. total strains, from a monotonic uniaxial tension test, which generates a stress-strain curve.

How would you use this data to find the corresponding pure shear stress-strain curve?

2. Relevant equations
ε_elastic = (σ/E)
γ_elastic = (τ/G)
G=E/[2*(1+v)]

3. The attempt at a solution
Using the crystal structure of metal, the normal stresses from the table could be converted to shear stresses via the τ_crss equation (http://virtualexplorer.com.au/special/meansvolume/contribs/wilson/Critical.html [Broken])

Then, the elastic shear strains can be obtained from τ/G. But what about the plastic shear strains? This is where I'm stuck. Hints/help would be highly appreciated!

Last edited by a moderator: May 7, 2017
2. Apr 3, 2015

### Staff: Mentor

You use the tensile test to determine the Young's modulus and the poisson ratio. Then you use your equation to calculate the shear modulus G from these. Then you plot shear stress vs shear strain with a slope of G.

Chet

Last edited by a moderator: May 7, 2017
3. Apr 3, 2015

### turpy

Hi Chet,
Thanks for the response. That covers the linear elastic region of the shear stress-shear strain curve, but what about the plastic region?

4. Apr 3, 2015

### Staff: Mentor

I don't have the answer to this immediately up my sleeve. I want to spend a little time playing with the equations.

Chet

5. Apr 3, 2015

### Staff: Mentor

If there is a way of doing it in the plastic region, I have not been able to figure out how. It certainly can't be done directly from the experimental measurements because, for all possible plane orientations within the sample, with this kind of uniaxial loading, there is no orientation in which there is a pure shear stress on the plane. There is always a normal component of the stress (except, of course, at 90 degrees to the load, where the shear stress is zero).

Chet

6. Apr 18, 2015

### afreiden

In the same way that 3D elasticity tells you $G$, based on $E$ and $\nu$, plasticity (von Mises) tells you that the material will "yield" in pure shear at a value of $\tau_y$, which is known, based on your known uniaxial yield stress, $\sigma_y$. This value is:
$\tau=\frac{\sigma_y}{\sqrt{3}}$

Again, the assumption there is von Mises plasticity.

Hope that helps

Last edited: Apr 18, 2015
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