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Engineering
Mechanical Engineering
Body forces in static equilibrium (FEM issue)
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[QUOTE="skrat, post: 6596327, member: 455703"] Yes, you are correct, my definition of ##\sigma## is a wrong by factor 2 on all off-diagonal terms. What if, excuse me for brainstorming, I wanted to take the last term from the weak formulation of the hookean law $$ \int _\Omega \nabla\cdot \sigma v \mathrm{d} \Omega $$ and decided to play with a little. For example use the calculus identity $$\nabla \cdot (\sigma v) = (\nabla \cdot \sigma)v + \sigma \nabla v.$$ This would allow me to rewrite the term to $$ \int _\Omega \nabla\cdot \sigma v \mathrm{d} \Omega = \int _\Omega \sigma \nabla v \mathrm{d} \Omega + \int _\Omega \nabla\cdot (\sigma v) \mathrm{d} \Omega $$ and using the Divergence theorem the last integral can be over ##\partial \Omega## finally resulting in $$ \int _\Omega \nabla\cdot \sigma v \mathrm{d} \Omega = \int _\Omega \sigma \nabla v \mathrm{d} \Omega+ \int _{\partial \Omega} \hat{n} \cdot \sigma v \mathrm{d} \partial \Omega.$$ I may be shooting blanks with this but the second interval could be zero as ##v## can always be set to 0 on the boundary ##\partial \Omega##. If that is true than computing the first integral is all that remains. Does that make any sense? Would this help at all? If nothing else it should improve the numerical stability where the cube is fixed as computation of derivatives of stress tensor components there is no longer required. [/QUOTE]
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Body forces in static equilibrium (FEM issue)
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