Deformation of Areas and Volumes

Consider two different infinitesimal line elements $\mathbf{P}\mathbf{Q}$ and $\mathbf{P}\mathbf{R}$ emanating from a point $P$ in the reference configuation. During the deformation lines $\mathbf{P}\mathbf{Q}$ and $\mathbf{P}\mathbf{R}$ are deformed into

$\mathbf{P}^\prime \mathbf{Q}^\prime$ and $\mathbf{P}^\prime\mathbf{R}^\prime$ respectively. Hence the parallelogram whose adjacent sides are $\mathbf{P}\mathbf{Q}$ and $\mathbf{P}\mathbf{R}$ in the reference configuration is deformed into the one with adjacent sides as $\mathbf{P}^\prime \mathbf{Q}^\prime$ and $\mathbf{P}^\prime\mathbf{R}^\prime$. Let us denote the areas of these by $d\mathbf{A}$ and $d\mathbf{a}$ respectively. Then

$$d\mathbf{A} =\ \mathbf{P}\mathbf{Q}\times \mathbf{P}\mathbf{R}$$

$$\therefore\hspace{.2in} dA_B =\ \varepsilon_{BCD}(PQ)_C(PR)_D\ .$$

Also

$$\mathbf{d}\mathbf{a}=\ \mathbf{P}^\prime\mathbf{Q}^\prime\times \mathbf{P}^\prime\mathbf{R}^\prime\ ,$$

$$da_i =\ \varepsilon_{ijk}(P^\prime Q^\prime )_j(P^\prime R^\prime )_k\ .$$

Recalling that $(P^\prime Q^\prime)_j = F_{jC}(PQ)_C$, we obtain

$$da_i =\ \varepsilon_{ijk}F_{jC}(PQ)_CF_{kD}(PR)_D\ ,$$

$$=\ \varepsilon_{pjk}F_{jC}F_{kD}(F_{pB}(F^{-1}_{Bi}))(PQ)_C (PR)_D\ ,$$

$$=\ J\varepsilon_{BCD}(F^{-1})_{Bi}(PQ)_C(PR)_D\ ,$$

$$=\ J(F^{-1})_{Bi}dA_B.$$

Hence

$$\mathbf{d}\mathbf{a} = J(\mathbf{F}^{-1})^T\mathbf{d}\mathbf{A}.$$ (3.10.1)

Now consider the parallelepiped formed by three nonplanar infinitesimal line elements $\mathbf{P}\mathbf{Q}$, $\mathbf{P}\mathbf{R}$ and $\mathbf{P}\mathbf{S}$ passing through a point $P$ in the reference configuration. Because of the deformation, the parallelepiped is deformed into the one whose three concurrent sides are $\mathbf{P}^\prime \mathbf{Q}^\prime$, $\mathbf{P}^\prime\mathbf{R}^\prime$ and $\mathbf{P}^\prime\mathbf{S}^\prime$. If $dV$ and $dv$ denote the volumes of these in the reference and the current configurations respectively, then

$$dV = \mathbf{P}\mathbf{Q}\times \mathbf{P}\mathbf{R}\cdot\mathbf{P}\mathbf{S} = \varepsilon_{BCD}(PQ)_B(PR)_C(PS)_D.$$

Similarly

$$dv = \mathbf{P}^\prime\mathbf{Q}^\prime\times \mathbf{P}^\prime\m... ...psilon_{ijk}(P^\prime Q^\prime )_i(P^\prime R^\prime )_j(P^\prime S^\prime )_k.$$

Substituting

$$(P^\prime Q^\prime)_i =\ F_{iB}(PQ)_B\ etc.,\$$

$$dv =\ \varepsilon_{ijk}F_{iB}(PQ)_BF_{jC}(PR)_CF _{kD}(PS)_D\ ,$$

$$=\ J\varepsilon_{BCD}(PQ)_B(PR)_C(PS)_D\ ,$$

$$=\ J\, dV\ .$$

Hence

$$dv = J\, dV\ .$$ (3.10.2)

A deformation such that

$$dv = dV$$

at every material point in the body is called an isochoric (volume preserving) deformation. Thus for an isochoric deformation, $J = 1$ at each material point of the body. Examples of isochoric deformations are the simple shearing deformation given in the example problem on page 3-2 and the one given in the exercise on page 3-31. Note that the latter is not a homogeneous deformation even though it is isochoric.

Exercise: Given the following deformation

$$u_1 = 2X^2_1 + X_1X_2,\hspace{.2in} u_2 = X^2_2,\hspace{.2in} u_3 = 0.$$

At the material point $(1/2,0,0)$ in the reference configuration, consider an infinitesimal plane formed by the vectors $10^{-2}(1,0,0)$ and $10^{-2}(1,1,0)$. Find the (vector) area of the element into which this plane is deformed.

Exercise: The displacement components for a body are $u_1 = 2X_1 + X_2,\ u_2 = X_3,\ u_3 = X_3-X_2$. At the material point $(1,0,0)$ on the surface of the body in the reference configuration, an element of area has components $10^{-2}(1,1,1)$. Find the components of the area into which this is deformed.