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APPENDIX C. HYPERELASTIC DERIVATIONS
F∗ = F · Q T
(C.2)
Eq. (C.2) is what we wanted to find, and it should be expected. Recall that V·R is physically understood to be a rotation, R, followed by a deformation (axial strains and shear), V. Thus, F∗ = F · QT is an explicit rigid body rotation, QT , followed by the total deformation + rotation, F. In other words, the rotation, QT , is applied to the initial configuration dX. Our new definition of F∗ is difficult to physically interpret from a Lagrangian point-of-view, but we will use it in order to show that σ = f (V), as follows. We know that σ ∗ = f (F∗ ), where our new definition of “∗” requires that σ ∗ = σ and F∗ = F · QT So, σ = f (F · QT ) We know that F = V · R Substituting −→ σ = f (V · R · QT ) Again, since we are simply applying our “∗” operator to both σ and f (F), we can take Q to be anything we want, as doing so is analogous to operating on both sides of any ordinary equation. Here, we can take Q to be R. This allows us to directly arrive at the desired result: σ = f (V)
(C.3)
The result (eq. (C.3)) is expected since it was shown in Chapter 5 that σ and V are work-conjugate. Recall also that V2 = B.