Symmetry of Infinitesimal (linear) strain tensor

Jayadeep,

The strain tensor (infinitesimal or not) in a classical continuum is symmetric due to invariance requirements.  A superposed rigid body rotation applied to the continuum should not affect the strain energy or the deformation power.  This is why only the symmetric part of the displacement gradient is retained in the definition, as the skew symmetric part captures such rotations.  The same reasons apply to the case of finite deformation as well, and is the reason we use E = 1/2(F^t*F-I) instead of F (F is the deformation gradient and ^t signifies transpose) as a strain measure.

Jason

Dear Jayadeep:

"1. It is defined to be symmetric so that it behaves like a tensor. " --> A tensor does not have to be symmetric. Symmetry may not even be meaningful for, e.g. a two-point tensor (a multilinear map between two separate manifolds) like deformation gradient in continuum mechanics.

"2. The stress tensor, which is its energy conjugate, is symmetric, and hence the skew-symmetric part has no contribution towards strain energy." --> You need to be more specific about stress. Cauchy stress is symmetric (as a consequence of balance of angular momentum) and so is the second Piola-Kirchhoff stress tensor but the first Piola-Kirchhoff stress is not "symmetric"; as a matter of fact symmetry is not meaningful for this two-point tensor.
    This is all valid when you have a "simple" or "classical" continuum where you ignore microstructure. In the presence of couple stresses (or any generalized stress), Cauchy stress tensor may lose its symmetry.

Classical linear elasticity has traditionally been formulated ab-initio, e.g. with no reference to nonlinear elasticity. This has also been the case in the physics literature. In my opinion, this has been the source of much confusion in the literature. The natural approach would be to linearize nonlinear elasticity with respect to a given reference motion (in this sense linearized elasticity is no unique).

There are different measures of deformation, "deformation gradient" is one example. Another example is the right Cauchy-Green strain tensor (C_AB), which is symmetric (and geometrically is the pull back of the spatial metric by deformation mapping). If you linearize deformation gradient (F), you would not find a symmetric strain tensor. What is understood as "linear strain" is linearization of "material strain tensor", which is one half C minus the material metric (E=1/2(C-G)). Because E is symmetric, its linearization will be symmetric too. It turns out that geometrically this is nothing but one half the Lie derivative of spatial metric tensor with respect to the displacement field. Metric is symmetric and so is its Lie derivative with respect to any vector field.

In summary, there are different measures of deformation (strain). Symmetric or "non-symmetric" tensors can be measures of strain. If one chooses a symmetric measure of strain, then its linearization will be symmetric as well (symmetry is preserved under linearization).

Regards,
Arash

Dear Jayadeep,

I also have wondered extensively about this question, and eventually concluded that the symmetry of the infinitesimal strain tensor is only a supposition. You may wish to find a memoire by Cauchy from 1850 whereby he appears to state this himself.

 As it turns out, symmetry is not required for invariance in terms of Euclid objectivity or satisfaction of the principle of material frame-indifference. Both can be obtained with a skewed strain tensor. This is evident from the existence of Cosserat theory.

However, I have posited that symmetry of the infinitesimal strain tensor is in fact valid and rational provided that the material undergoing the strain is isotropic and free of any micropolar moment.

The corollary to this is that should the material be orthotropic or anisotropic, there is a rational and general description that suggest the infinitesimal strain tensor may be asymmetric. This reasoning is provided in a paper authored by myself and colleagues (doi.wiley.com/10.1002/nme.2379). 

Arash,

I'm very interested in your comment (copied below) regarding linear strain E being defined in terms of G & Lie derivative. Is there a source that discusses this in greater detail ?  I'm trying to understand linear elastic deformation from a Lie Theory perspective (groups, algebras, derivative). Thanks, John.

 " What is understood as "linear strain" is linearization of "material strain tensor", which is one half C minus the material metric (E=1/2(C-G)). Because E is symmetric, its linearization will be symmetric too. It turns out that geometrically this is nothing but one half the Lie derivative of spatial metric tensor with respect to the displacement field. Metric is symmetric and so is its Lie derivative with respect to any vector field."

Dear John:
 
You may want to look at the book by Marsden and Hughes (Mathematical Foundations of Elasticity). There is a typo where they discuss linearization of the Lagrangian strain. We can discuss this after you read the material.

Regards,
Arash