For the tetrahedron shown, we prove that:
where is the area of the face with normal n, and is the face with normal
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:
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
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
where , giving the required result.
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
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.
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 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
where with 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
(c) A.F. Bower, 2008