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Appendix E
Miscellaneous
derivations
E.1. Relation between the
areas of the faces of a tetrahedron
For the tetrahedron
shown, we prove that:
where
is the area of the face with normal n, and is the face with normal
Note that
Note also that we can compute the area of the face with
normal n by taking cross products
of the vectors defining the sides of the face:
so that
as
required.
E.2. Relation between area elements before and after
deformation
Consider an element of area with normal in a deformable solid. Suppose the solid is deformed, and let denote the components of the deformation
gradient tensor. The area element deforms with the solid, and has a new area and normal n. We plan to prove that
the deformed area element is related to its undeformed area through
where
Start by noting that the area before deformation can be
computed by taking the cross product of two infinitesimal vectors bounding the area element in the undeformed
configuration
Note that the infinitesimal vectors map to ,
in the deformed configuration. Therefore
Let denote the inverse of the deformation
gradient tensor, i.e. .
Then, we could write
Now,
recall the identity
so that
where ,
giving the required result.
E.3. Time derivatives of integrals over volumes
within a deforming solid
Consider a deformable solid. Let denote the mass density of the solid in the
original configuration, and let denote the mass density in the deformed
configuration. (Both and can vary with position in the solid).
Let denote a closed region within the undeformed
solid, and let V be the same region of the solid in the deformed
configuration. Suppose that
denotes the velocity
field within the body. We shall show
that
and also that
At
first glance, this looks obvious just take the derivative under the integral
sign. You can’t do this, however,
because the volume V changes with
time, as the solid is deforming. In
addition, the mass density varies with time, because of the deformation, so
even if we could take the time derivative under the integral, we’d end up
with an additional term. To do the derivative properly, we first need to
change variables so the integral is evaluated over the undeformed volume (which is independent of time). Thus
where
and we have recalled a
result from the Kinematics section
Now,
we can happily differentiate. The mass
density in the undeformed configuration does not vary with time, so that
The
last expression was obtained by changing variables in the integral back to
the deformed configuration. This is
the first result we wanted.
To show the second
result, follow exactly the same procedure, until you obtain
Now, observe that
(the
cross product of two parallel vectors is zero) so substituting this into the
preceding equation and changing variables in the integral as before gives the
required result.
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