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Hyperelastic material characterization: why they call it the pure shear specimen

wvmars's picture

Many people puzzle over the nomenclature of the pure shear test. They rightly point out that 1) the Pure Shear test piece is loaded in tension by extending the specimen in the axial direction, and 2) a shearing deformation, by definition, involves the lateral motion of parallel planes. They wonder where is the "shear"? and what does it mean to say that the shear is "pure"?  Here, we review the origins of the terminology. 

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Lixiang Yang's picture

Hi Sir,

    Maybe you know more than me. This is my comments.

   In hyperelastic materials,  the first order shear stress will cause the second order compression and tension stress for both incompressible and compressible hyperelastic materials.

 

 

wvmars's picture

You are correct that direct stresses are coupled to shear stresses for finite straining.  Note that the arguments we present here refer only to the kinematics of the Pure Shear testpiece, not to its state of loading.  Another interesting challenge would be to compute the associated stress state.  I believe you will find that it does not give a pure shear stress, even though the rate of deformation calculation does. 

Lixiang Yang's picture

I calculated the rate of deformation tensor of simple shear stress state. I found the tensor is not symmetric. I think 3-d mohr circle can only be used to plot the symmetric matrix of L, that is D.  And if W vanishes, the simple shear is equal to pure shear, isn't it?  

 I would like to know what is the relationship between stress tensor and rate of deformation tensor.  

Even the rate of deformation tensor of the simple shear strain becomes pure shear strain tensor,   stress states of them between them are still different. Why?

wvmars's picture

When we first wrote about this topic, we posed the question of how pure shear compares to simple shear.  Here is our analysis of the two cases.  Enjoy!

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