# Thread: Hencky strain versus Stretch tensor

1. Junior Member
Join Date
2006-07
Posts
23

## Hencky strain versus Stretch tensor

Hello everybody,

I need some help with computation of any Strain tensor (or def. grad.) from logarithmic (natural) Hencky strain tensor. I have found the following relation in many references:

En=ln(U)

where En is Hencky strain tensor and U is right stretch tensor. The situation is that I have En and need to calculate U (or any other deformation tensor: V, C, E, e, F, ...). I though that I can do that this way:

U[i,j]=exp(En[i,j])

but this does not work even for undeformed configuration, where En shoud be ZERO matrix and U should be IDENTITY matrix (i.e. zero non-diagonal components).

Does anyone know how to calculate U[i,j] correctly?
Thank you for any responces,
Paul

2. Hi Paul,

The tensor U can indeed be obtained from: U = exp[ En ]
where exp[.] is the matrix exponential.

You can check out the matrix exponential (and other matrix functions) using Matlab (or the free alternative octave).

There are also many good books on these kinds of computations, for example: "Matrix Computations" by Golub and Van Loan.

What I often do to calculate the matrix exponential is to first find the eigenvalues and eigenvectors of En, then take the exponential of the eigenvalues, and then rotate the results back to the original coordinate system.

- Jorgen

3. Junior Member
Join Date
2006-07
Posts
23
Thanks a lot Jorgen for your responce,

I am just trying to account for rigid body rotation in my USERMAT subroutine in ANSYS (hyperelastic model). Since ansys provide me only STRAIN (Hencky) in rotated coord. system and DEFGRAD in global coord. system I consider about three following ways how to calculate the "output" STRESS tensor (in rotated coord. system):

1/ to use STRAIN tensor as input, transform it into C (right C-G) tensor and calculate STRESS as output

2/ to use DEFGRAD tensor as input, transform in into F in rotated coord. system (via rotation matrix R calculated separately from input DEFGRAD) and calculate STRESS as output

3/ to use DEFGRAD tensor as input, simply calculate Cauchy stress in global coord. system and then transform it into STRESS in rotated coord. system as output (via rotation matrix R calculated separately from DEFGRAD). May be, Jacobian would has to be transformed as well into rotated system.

Could anybody, please, tell me what way is the best?

Paul

4. Hi Paul,

All three of your options seem OK to me, and in fact should give the same answer. I suggest that you pick your personal favorite and try it out!

- Jorgen

5. Junior Member
Join Date
2006-07
Posts
23
Hello Jorgen (and others),

finally I choosed the way 3 ... to calculate stress and jacobian (DDSDDE) in global coordinates based on def. gradient and want to rotate stress and elasticity matrix DDSDDE into corotated coordinate system.

I have calculated rotation matrix R from polar decomposition of def. gradient (F=R*U) and would like to rotate DDSDDE ... but I don't know how :-( . Could anybody please advise me how to rotate (4th order) elasticity tensor into other coordinate system based on rotation matrix R?

Paul

6. The following rules apply in index notation.

A first order tensor (i.e. vector) v is rotated as follows:
R_ij v_j

A second order tensor A is rotated as follows:
R_ik R_jm A_km

A forth order tensor D is rotated as follows:
R_im R_jn R_kp R_lq D_mnpq

- Jorgen

7. Junior Member
Join Date
2006-07
Posts
23
Thank you Jorgen for your attention,

but (probably stupid question) ... what does Q tensor mean?

Paul

8. Oops, that was a typing error. It should have been R.
I have corrected the error.

- Jorgen

9. Junior Member
Join Date
2006-07
Posts
23
Thanks Jorgen for the correction.

I have one more question about the rotation of second order tensor through:
R_ik R_jm A_km

Is this applicable also to non-symmetric tensor such as deformation gradient tensor F? If I do the transformation of F based on R tensor (calculated from polar decomposition of F) through the above relation, I do not obtain symmetrical tensor. However I expect symmetrical tensor should be obtained in rotated coordinate system.

If I perform transformation through the alternative relation: R(-1).F, everything seems to be OK. (R(-1) means the inverse of R).

Does anybody please know why "R_ik R_jm A_km" relation does not work with F tensor? Or there is something wrong in the considered approach?

Paul
Last edited by Paul_S; 2007-11-19 at 02:12.

10. Hi Paul,

The R_ik R_jm A_km equation is true in general (both for symmetric and unsymmetric 2nd order tensors A). But keep in mind that it simply tells you the components of a tensor in a different coordinate system that is rotated by R relative to the original.

The Polar decomposition: F = RU = VR tells you how a deformation gradient can be decomposed into a symmetric stretch tensor and a rotation tensor.

The "trick" is to use the right tensor operations at the right time, which will depend on what you are trying to do.

- Jorgen

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