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 Â with normal Â in a deformable solid.Â  Suppose the solid is deformed, and let Â denote the components of the deformation gradient tensor. The area element deforms with the solid, and has a new area Â and normal n.Â  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 , Â in the deformed configuration.Â  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.Â  Let Â 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). LetÂ  Â denote a closed region within the undeformed solid, and let V be the same region of the solid in the deformed configuration.Â  Suppose that

denotes the velocity field within the body.Â  We shall show that

and also that

At first glance, this looks obvious Â just take the derivative under the integral sign.Â  You canâ€™t do this, however, because the volume V changes with time, as the solid is deforming.Â  In addition, the mass density varies with time, because of the deformation, so even if we could take the time derivative under the integral, weâ€™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.Â  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.Â  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.

Â Let Â denote the position vector of a material particle on the axis of the undeformed rod;

Â Let Â denote the arc-length coordinate of this particle after deformation.

Â Define basis vectors Â attached to the deformed rod, following the convention described in Section 10.2.

Â  Define an angular velocity vector Â and curvature vector Â through

Â We shall show that the gradient of the angular velocity vector characterizing the rotation of the rodâ€™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.

(c) A.F. Bower, 2008
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