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Stretching and Spin Tensors Relations
Can someone help me to understand this please:
The stretching tensor (D) is the symmetric part of the velocity gradient (L), and if I derive D from L with using the polar decomposition of the deformation gradient (F=RU) I get the relation:
D=0.5 R (Udot Uinv + Uinv Udot) R' ----(Eq.1) where Udot is the time derivative of U, Uinv is the inverse of U, and the prime indicates the tranpose of the matrix.
I derive D also from the Lagrangian Green strain tensor E=0.5(FTF-I)=0.5(U2-I), and i get the relation:
D=R Udot Uinv R'-----(Eq.2)
I am not sure if Eq.1 and Eq.2 are always equivalent, if yes is that mean the spin tensor (skew-symmetric part of L) is always equal to W= Rdot R' (because in some book, I found W=Rdot R' + 0.5 R (Udot Uinv - Uinv Udot) R')
Thanks alot
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Re: Stretching and spin tensors
U^2 = U.U' = U.U
d/dt(U^2) = Udot . U + U . Udot which is not = 2 U. Udot
A.B not eq B. A
See also http://en.wikiversity.org/wiki/Continuum_mechanics/Time_derivatives_and_rate s
-- Biswajit
Thanks a lot Biswajit. It is
Thanks a lot Biswajit. It is clear now.