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  How is it possible to develop the strain energy function around a stressed configuration ?

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Strain elastic energies are generally defined around a stress-free configuration.

Is it possible in some case to define it around a stressed configuration ?

Meaning in general we have \(W=W(F)\).

For example for an hyper-elastic material : \(W=\frac{μ}{2}(α^2_x+α^2_y+α^2_z−3)\) , where the α are the deformation from a stress-free configuration. But could we also expand the energy around a stressed configuration so that : \(W=W_0+\frac{μ}{2}(β^2_x+β^2_y+β^2_z−3)\) with β the deformation from the stressed configuration and \(W_0\) the strain energy of this stressed configuration ?

EDIT: after thinking a little bit, I thought that it's obvious that if you go from a configuration 1 to 2 to 3 and 1 is stress-free, then \(\mathbf{F}=\mathbf{F}_{12}\mathbf{F}_{23}\). And if \(W_{23}=W_{13}-W_{12}\) is right, then the answer is no because you can't developp it like the expression I wrote before. But is it right that  \(W_{23}=W_{13}-W_{12}\) ?

PS: I also asked this question on Physics SE.

asked Dec 17, 2019 in Theoretical Physics by JA (20 points) [ revision history ]
edited Dec 18, 2019 by JA

Strictly, you cannot, W is not linear. You must refer to the alpha xyz of Wo..

Edit: for your second question, the answer is trivially yes.

use the tex button in the editor instead of $

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