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Planar tension test = Pure shear test for rubber? Explanations

Hi,

 I know what is this test and I know under certain conditions (low deformations and specimen's wide 10 times its lenght), a planar tension test is equal to pure shear test.

But I don't know why if we are doing a tension test we have pure shear.

Let me explain:

    ^ 1-Axis
     |
-------------
 |           |    ---> 2-Axis
-------------

3-Axis is perpendicular to thickness.
Assumption 1:  Thin specimen===>  Sigma3=0
Assumption 2:  Wide specimen===>  Strain 2=0 near middle

This is a simple scheme of the test:

Scheme

We get:

Sigma2 = Sigma1*Poisson's ratio

For rubber incompressibility poisson ratio is 0.5 so we get sigma2=0,5*sigma1 and sigma3=0 

We get Mohr's circles like pure tension:

Mohr's circles of stress

 

But from the point of view of the strain we get Mohr's circles like pure shear:

Mohr's circles of strain 

 

I don't know what is my mistake. Why do we get 2 differents kind of Mohr's circles? I think the second Mohr's circles, on strain, are right but how could be Mohr's circles of stress?

Here maybe the explanation is better than mine, this guy has the same problem: http://www.eng-tips.com/viewthread.cfm?qid=126336

Thank you all and excuse me for my bad english. 

The pure shear is equivalent to the stress state σ1 = τ, 
σ2 = -τ, σ3 = 0. Where 
σi are the principal stresses.

Thans for the reply.

 

I know the state of stress of pure shear, and the Mohr's circles are similar to the second that I have posted (Mohr's circles of strain). I don't know why the state of stress of planar tension (the first Mohr's circles that I have posted) is different to the state of stress of pure shear but this test is considered the same as pure shear. 

We obtain pure shear (on strain) in plane which is parallel to x_2. Consider the plane which has the angle 450 with x_1 and x_3. The normal and tangent stresses are not zero. Consider the direction along normal stress. The normal stress lead to the strain. But in ortogonal direction the normal stresses are not zero and they lead to the strain in considered direction due to Poisson effect. It follows that the total strain is zero. Thus there is the shear only in this plane, but the the tangent and normal stresses are existed.

 P.S. I mean by the pure shear the special stress state (not strain).

Plane x_1 x_3 

If I understand correctly then the equation "stress_2 = 0.5 stress_1" is right for Hooke's law. I'm not sure that this linear model is good for rubber-like materials.

I made mistake. "We obtain pure shear (on strain) in plane which is parallel to x_2" is not right. The "pure shear (on strain)" is in plane x_1 x_3.

Matt Lewis's picture

You probably meant perpendicular to the x_2 direction.  I don't believe these equatlions apply for nonlinear elastic response or finite strains.

Matt Lewis
Los Alamos, New Mexico

Hi,

For this planar test Sigma_2 = 0.5 * Sigma_1 is only valid for incompressible materials undergoing small strain. It will not hold in general for hyperelastic materials. For the range in which it is valid however, we have a state of pure shear strain and a stress state that is the summation of a pure shear stress and a hydrostatic stress (of magnitude Sigma_2). If you plot the Mohr circle for the deviatoric stresses it will look like that of pure shear. For perfectly incompressible materials the hydrostatic stress component does not cause deformation. That’s why, under the right conditions, this test is a good approximation of pure shear (which is much harder to test experimentally).

Here is a figure I just created showing the Mises stress vs. strain response for an Ogden material loaded in uniaxial tension, simple shear and planar tension. You can see that for small strains the planar tension is a good approximation of simple shear.

Nagi Elabbasi

Veryst Engineering

Stress strain curves for Ogden material in uniaxial tension, planar tension and simple shear

wvmars's picture

Here is an analysis of the simple shear vs pure shear cases, and here is an explanation of the nomenclature.  Hope this clarifies. 

Hi

Could you please sent the simple shear and pure shear cases again?  the links are empty . thnx alot

 

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wvmars's picture

Here is an analysis of the simple shear vs pure shear cases, and here is an explanation of the nomenclature.  Hope this clarifies. 

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