# Help: calculate elasticity tensor for hyperelastic material with plane stress configuration

Dear all,

I am confused with the calculation of elasticity tensor of hyperelastic material in plane stress configuration. Your helps are very much appreciated. Below is my question:

Incompressible strain energy function is expressed as, saying, U=***+p(J-1).

Therefore, the second piola-kirchhoff stress is: S=***+pC^-1 (the Lagrange muliplier p is treated as an arbitrary constant here, right?).

Using the plane stress condition, S<33>=0 so that p is updated as a function of deformation.

Now, to calculate the elasticity tensor, H=2*dS/dC. Should I continue to treat p as a constant, independent of deformation, and get:

(1): H=***+2*p*d(C^-1)/dC    otherwise, should I treat it here as a function of deformation and get:

(2): H=***+2*(C^-1*(dp/dC)+p*d(C^-1)/dC)

I am not sure which one is correct (or none of them?). The elasticity tensor H possesses both major and minor symmetry (is this always correct?), if H takes the form (1), major symmetry holds;

but if H takes the form (2), it lost its major symmetry. Any comment about this? Thank you all very much indeed!

Best regards

Haofei Liu

Hi Haofei,

I am new to this field. Your question is very interested.

First, you have a full 3-d incompressible hyperelastic model with p undetermined.

Then you applied boundary condition to it and find p. The model then reduced from 3-d to 2-d.

So 2rd order tensor reduce to 2 by 2 matrix. Then all tensor requirement may not apply to new 2 dimension model.

It is because that  2 dimension model is not a tensor. It is also because that this 2 dimension model is not strict constitutive relationship of materials because of boundary condition was applied. The boundary condition could change in other cases.

About your question about major and minor symmetry, I don't why 2nd one don't satisfy major while the first one does? Can you explain a little bit? I am also not sure that major and minor symmetry is a validating point if stress and strain are not tensor.

Best,

Lixiang Yang

### 'p' is a Lagragne Multiplier in incompressible material

Hi Liu,

I am learning Non-linear Solid Mechanics. Let me make an attempt to answer your questions.

In case of incompressible material, 'p' is a Lagrange multiplier. It is a constant. When we write the strain energy density, we add the constraint, J=1 as p(J-1) to the strain energy function. To compute the stress (say 2nd PK = S), we take the derivative of strain energy density wrt 'C' (Right Cauchy-Green tensor). When derivative of p(J-1) is taken wrt 'C', it becomes pdJ/dC, where 'p' is a constant. Here, 'p' is estimated from boundary conditions. As you rightly said about the plane strain condition => S33 = 0. This gives the value of pressure.

When elasticity tensor, H, is computed, we take the derivative of 'S' wrt 'C'. 'S' has two parts - isochoric (Siso) and volumetric (Svol). Svol = pC^-1. When Svol is differentiated wrt 'C' to get Hvol, 'p' again remains constant. Since 'p' (constant) is a Lagrange Multiplier.

Hence your first equation is correct and H' is symmetric

If the material is compressible, then, strain energy has two parts = Isochoric (Uiso) + volumetric (Uvol). Note that that Uvol = U(J).  In this case, 'p' is obtained by taking the derivative of U(J) wrt 'J'. Here, 'p' is a variable. Hence the second equation is correct for compressible material. In the second equation, 'J' also will appear along with 'p'. Then 'H' becomes symmetric.

When there is no body couple and rotary inertia is neglected, then shear stresses are complementary. This gives the minor symmetry. This exists for all the materials.

Major symmetry is material specific and exists for hyperelastic materials. In hyperelastic materials, potential exists. Stresses can be computed by taking the derivative of potential (strain energy density) wrt strains.

With best regards,