CONTINUUM MECHANICS for ENGINEERS

which upon substitution into the above velocity expressions yields the spatial
components,
v 1 = –x 1, v 2 = x 2, v 3 = –x 2e –2t
Therefore, we may now calculate dθ/dt in spatial form using Eq 4.5-4,
dθ
dt
= –e –t (x 1 – 2x 2 + 3x 3) – x 1e –t – 2x 2e –t – 3x 3e –t
which may be converted to its material form using the original motion
equations, resulting in
dθ
dt
= –2X 1e –2t – 3X 2(2e –2t – e –t ) – 3X 3e –t
An interesting and rather unique situation arises when we wish to determine
the velocity field in spatial form by a direct application of Eq 4.5-4 to
the displacement field in its spatial form. The following example illustrates
the point.
Example 4.5-2
Verify the spatial velocity components determined in Example 4.5-1 by
applying Eq 4.5-4 directly to the displacement components in spatial form
for the motion in that example.
Solution
We may determine the displacement components in material form directly
from the motion equations given in Example 4.5-1,
u 1 = x 1 – X 1 = X 1(e –t – 1)
u 2 = x 2 – X 2 = X 2(e t – 1)
u 3 = x 3 – X 3 = X 2(e –t – 1)
and, using the inverse equations X = χ –1 (x,t) computed in Example 4.5-1, we
obtain the spatial displacements
u 1 = x 1(1 – e t )
u 2 = x 2(1 – e –t )
u 3 = x 2 (e –2t – e –t )