CONTINUUM MECHANICS for ENGINEERS

Therefore, substituting u i for P ij… in Eq 4.5-4 yields
dui
∂ui
∂ui
vi = = + vk
dt ∂t
∂x
so that by differentiating the above displacement components
v 1 = –x 1e t + v 1(1 – e t )
v 2 = x 2e –t + v 2(1 – e –t )
v 3 = –x 2(2e –2t – e –t ) + v 2(e –2t – e –t )
which results in a set of equations having the desired velocity components
on both sides of the equations. In general, this set of equations must be
solved simultaneously. In this case, the solution is quite easily obtained,
yielding
v 1 = –x 1, v 2 = x 2, v 3 = –x 2e –2t
to confirm the results of Example 4.5-1.
4.6 Deformation Gradients, Finite Strain Tensors
In deformation analysis we confine our attention to two stationary configurations
and disregard any consideration for the particular sequence by which
the final deformed configuration is reached from the initial undeformed configuration.
Accordingly, the mapping function is not dependent upon time as
a variable, so that Eq 4.2-6 takes the form
x i = χ i(X) or x = χ(X) (4.6-1)
Consider, therefore, two neighboring particles of the body situated at the
points P and Q in the undeformed configuration such that Q is located with
respect to P by the relative differential position vector
dX= dXAIˆ A
as shown in Figure 4.2. The magnitude squared of dX is
(dX) 2 = dX ⋅ dX = dX A dX A
k
(4.6-2)
(4.6-3)