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I am a little confused about the expression of the strain in virtual work equations. The small strain tensor can be defined as followed: $\boldsymbol{\epsilon} = \frac{1}{2}(\nabla \textbf{u}+(\nabla \textbf{u})^T)$. herefore, a virtual strain field $\delta\boldsymbol{\epsilon}$ should be: $$ \delta \epsilon_{ij} = \frac{1}{2}(\frac{\partial \delta u_{i}}{\partial x_j} + \frac{\partial \delta u_{j}}{\partial x_i}) $$ Unfortunately, in the context of equilibrium equations and virtual work we generally find $\delta \boldsymbol{\epsilon} = \nabla \delta \textbf{u}$ instead of $\delta\boldsymbol{\epsilon} = \frac{1}{2}(\nabla \delta \textbf{u}+(\nabla \delta \textbf{u})^T)$.

Can someone help me understand that?

Thanks!

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The first definition is written that way to force symmetry and coordinate system independence:

enter image description here enter image description here enter image description here

All three animations show the same shear strain with different (rotated) coordinate systems. Since Nature doesn’t care which way we draw our axes, we’d like to describe all three cases in an invariant manner. We do this by averaging the relative displacements in orthogonal directions.

The second, simpler definition must carry an associated promise that we’re going to write a symmetric strain matrix; then, there’s no need to perform the averaging.

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  • $\begingroup$ Yes, the rotation would have to be zero (or already separated into a distinct term in the analysis). $\endgroup$ Commented Feb 13, 2022 at 15:32

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