Appendix E
Miscellaneous
derivations
E.1. Relation between the
areas of the faces of a tetrahedron
For the tetrahedron
shown, we prove that:

where
 Â is the area of the face with normal n, and  Â 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  Â bounding the area element in the undeformed
configuration

Note that the infinitesimal vectors map to  ,
  in the deformed configuration. Therefore

Let  Â 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  Â denote the mass density in the deformed
configuration. (Both  Â and  Â 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.
E.4. Time Derivatives of the Curvature Vector for
Deforming Rods
Consider
a deforming rod, as shown in the figure.
 Let   denote the position vector of a material
particle on the axis of the undeformed rod;
 Let   denote the arc-length coordinate of this
particle after deformation.
 Define basis
vectors  Â attached to the deformed rod, following the
convention described in Section 10.2.
 Define an angular velocity vector   and curvature vector   through

 We shall show that the gradient of the angular
velocity vector characterizing the rotation of the rod’s cross-section is
related to the time derivative of the curvature vector by

To see this, start by
differentiating the definitions of the angular velocity vector and the
curvature vector

and, similarly,

where  Â with  Â held fixed. The preceding two results show
that

Next, note that we can
expand the triple cross-products (see Appendix A) as

Hence, we conclude that

This result must hold for
all three vectors  ,
and therefore

as stated.
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