# Decomposition of deformation into bend, stretch and twist?

I'm wondering is there any way to decompose the deformation of an object into different components? For example, into stretching, bending and twisting part respectively? The decomposition could be applied to any description of deformation, either to deformation gradient tensor, strain tensor or stress tensor.

To my knowledge, there is some literature using the decomposition of deformation gradient tensor into rotation and deformation part. But what I want is to further decompose the deformation, leave away the rigid transformation like translating and rotating.

Thanks!

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## 1 Answer

There is no unique way to "further decompose" the deformation into the "rigid transformation" and "others" because whatever "rigid part" you choose, you may always calculate "others" as a simple difference (that's because there's really no global constraint on the "other" part). So the "rigid part" may be anything you want.

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Thanks for replying. I'm not going to further get a rigid transformation from deformation because by deformation I mean pure deformation. I just want to decompose the deformation into different part for further analysis. Is it that there isn't any way to decompose it because deformation components like bending, twisting are coupled together in one metric? – Fei Zhu Sep 21 '12 at 14:30
It's exactly the same question, @FeiZhu, and I think I have already answered it. There is no way to define a "pure deformation". For example, if you declare that a shear deformation sliding horizontal planes on others is a "pure deformation", then you should allow the sliding vertical planes to be a "pure deformation" as well - but these two shear deformations differ by a simple rotation. There is no way to say "how much rotation" is included in a general deformation. – Luboš Motl Jun 13 '13 at 11:02
In other words, one may say whether a linear transformation is a rotation or not. But one can't say whether a linear transformation is a "pure deformation" because no deformations are "purer" than others. The condition doesn't exist. At most, you may define a "pure deformation" as anything that isn't a rotation-and-translation but almost every deformation would be "pure" by this criterion, it's a negative, not a positive, condition. Do you understand that my answer is No? It sounds like you don't like it and only want to hear the answer yes but this answer is wrong. – Luboš Motl Jun 13 '13 at 11:05
Thank you @Luboš Motl, I think I get it. – Fei Zhu Jun 14 '13 at 4:32