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Why do we need to define the strain tensor if we already have a tensor that describes the deformation between relative particles in the continuum which is the deformation gradient tensor?

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    $\begingroup$ You don't need to. The deformation gradient is enough to calculate stresses etc. However, the deformation gradient is a two-point tensor with one foot in the reference configuration and the other in the current configuration. It's easier to measure quantities either completely in the reference or completely in the current configuration. Hence the popularity of "strain" measures. $\endgroup$ – Biswajit Banerjee Mar 19 at 5:04
  • $\begingroup$ can you please give an example? $\endgroup$ – user134613 Mar 19 at 9:34
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Biswajit Banerjee Mar 19 at 17:59
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For a rigid body rotation, the deformation tensor is non-zero but the strain tensor is zero, and so is the stress in the body.

The strain tensor can be derived mathematically from the deformation tensor, but it does not represent the same physical concept.

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  • $\begingroup$ In the mathematical equation of deformation tensor there is no description of rotation it just say that dx = F dX, so it just do the transformation of the vector segment from the reference to current configuration. So how come it is non-zero in case of rigid body rotation? Is it because dx and dX are vectors hence the direction of them changes, and so F assumes that there has been a deformation but in reality the whole object has rotated bu not changed shape (as a rigid body). So why do we care about deformation tensor if its not so clear and might give us a wrong interpretation? $\endgroup$ – user134613 Mar 19 at 9:13
  • $\begingroup$ In mechanics, the term "deformation" includes both stretches and rotations. See the figure at the top of en.wikipedia.org/wiki/Deformation_(physics). $\endgroup$ – Biswajit Banerjee Mar 19 at 18:02

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