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Assumptions:

  1. Cross terms in strain tensor are defined as equal $\varepsilon_{xy} = \varepsilon_{yx}$.
  2. pure mode I crack.
  3. Far from crack tip, material is purely elastic and we are way below yield stress => $\varepsilon_{ij} \propto \sigma_{ij}$.
  4. Cross term $\sigma_{yx} $in linear elastic fracture mechanics (LEFM) contains factor $\sin{\frac{\theta}{2}}$.

Now, cross term is not symmetric respect to $\theta$ (contains sin). The material however should be symmetric and thus $\varepsilon_{xy} = -\varepsilon_{yx}$. What am I missing?

What is the difference between positive and negative shear stress?

LEFM formulas can be found e.g. in page 6 in:

http://www.public.iastate.edu/~gkstarns/ME417/LEFM.pdf

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  • $\begingroup$ Link now dead . $\endgroup$ – Qmechanic May 13 '19 at 4:48
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You are confusing the global coordinate $\theta$ with the local direction. Both the strain and the stress tensors are symmetric locally, that is, index-wise:$\epsilon_{ij}=\epsilon_{ji}$ and likewise for the stress. The symmetry of the strain is by definition and that of the stress is due to torque balance. This is true for (almost) all systems.

However, the stress and strain, as fields, exhibit a global symmetry, which results from the symmetry of the loading, and is system-specific: $$ \sigma_{xx}(r,\theta)= \sigma_{xx}(r,-\theta) \qquad \sigma_{yy}(r,\theta)= \sigma_{yy}(r,-\theta) \qquad \sigma_{xy}(r,\theta)=- \sigma_{xy}(r,-\theta)$$

For mode I cracks (for mode II it's the opposite).

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