CONTINUUM MECHANICS for ENGINEERS

The difference (dx) 2 – (dX) 2 may also be developed in terms of the spatial
variables in a similar way as
where the symmetric tensor
(dx) 2 – (dX) 2 = δ ij dx idx j – (X A,i dx i)(X A,j dx j)
= (δ ij – X A,i X A,j)dx idx j
= (δ ij – c ij)dx i dx j
(4.6-15)
c ij = X A,i X A,j or c = (F –1 ) T ⋅ (F –1 ) (4.6-16)
is called the Cauchy deformation tensor. From it we define the Eulerian finite
strain tensor e as
so that now
2e ij = (δ ij – c ij) or 2e = (I – c) (4.6-17)
(dx) 2 – (dX) 2 = 2 e ij dx i dx j = dx ⋅ 2e ⋅ dx (4.6-18)
Both E AB and e ij are, of course, symmetric second-order tensors, as can be
observed from their definitions.
For any two arbitrary differential vectors dX (1) and dX (2) which deform into
dx (1) and dx (2) , respectively, we have from Eq 4.6-9 together with Eqs 4.6-12
and 4.6-13,
dx (1) ⋅ dx (2) = F ⋅ dX (1) ⋅ F ⋅ dX (2) = dX (1) ⋅ F T ⋅ F ⋅ dX (2)
= dX (1) ⋅ C ⋅ dX (2) = dX (1) ⋅ (I + 2E) ⋅ dX (2)
= dX (1) ⋅ dX (2) + dX (1) ⋅ 2E ⋅ dX (2) (4.6-19)
If E is identically zero (no strain), Eq 4.6-19 asserts that the lengths of all line
elements are unchanged [we may choose dX (1) = dX (2) = dX so that (dx) 2 = (dX) 2 ],
and in view of the definition dx (1) ⋅ dx (2) = dx (1) dx (2) cosθ, the angle between
any two elements will also be unchanged. Thus in the absence of strain, only
a rigid body displacement can occur.
The Lagrangian and Eulerian finite strain tensors expressed by Eqs 4.6-13
and 4.6-17, respectively, are given in terms of the appropriate deformation
gradients. These same tensors may also be developed in terms of displacement
gradients. For this purpose we begin by writing Eq 4.4-3 in its time-independent