Analysis of Deformation

We will now study the deformation of the body at fixed instant of time, say the current time instant $t$. In particular, we will see that all the information relevant to the change in shape of the body from the initial instant of time to the current instant of time is contained in a quantity called the deformation gradient.

Deformation

Let us consider, as before, a motion $(\chi_t)_{t \in I}$ of a body $\mathcal{B}$ as it moves and deforms in the three dimensional space $\mathbb{R}^3$. Since it is much easier to deal with maps between open subsets of Euclidean spaces, our attention henceforth will almost exclusively be on the deformation map $\varphi_t:B_0 \to B_t$, where $B_0 = \chi_0(\mathcal{B})$ and $B_t = \chi_t(\mathcal{B})$ are the initial and current placements of the body, respectively.

Remark

Before we proceed further, we will introduce a useful notational convention that will turn out to quite useful in the sequel. We will, for the most part, use uppercase symbols for quantities associated with the initial configuration, and lowercase symbols for quantities associated with the current configuration. This also includes the indices for components of vectors and tensors with respect to suitably chosen bases.

Let us now introduce a basis for the three dimensional Euclidean space $\mathbb{R}^3$. For the sake of conceptual clarity, let us introduce two sets of bases $(\mathsf{E}_I)_{I=1}^3$ and $(\mathsf{e}_i)_{i=1}^3$; note that these are the bases chosen by a single observer. We will use the basis $(\mathsf{E}_I)$ to denote points in $B_0$ and the basis $(\mathsf{e}_i)$ to denote points in $B_t$. Notice how we have used uppercase and lowercase symbols for the initial and current placements, according the convention just outlined.

Remark

For all practical purposes, we can identify the bases $(\mathsf{E}_I)$ and $(\mathsf{e}_i)$ of $\mathbb{R}^3$. A formal way to express this equivalence is using the equation $\mathsf{e}_i = \sum \delta_{iI} \mathsf{E}_I$. More generally, the bases could be related via an orthogonal matrix as follows: $\mathsf{e}_i = \sum \hat{Q}_{iJ} \mathsf{E}_J$, with the matrix $[\hat{\mathsf{Q}}]$ whose $(i,J)^{\text{th}}$ entry is $\hat{Q}_{iJ}$ being orthogonal. Even though it is sufficient to identify the two bases for the following development, we will retain the upper and lower case notations for the purpose of conceptual clarity.

Note

In a more general development of continuum mechanics, the background space is itself modeled as a manifold. Since a manifold is, in general, not a linear space, the choice of a basis for space doesn't make sense in the general development. Rather, coordinates are assigned to every point on the manifold by the choice of an appropriate coordinate system. The choice of the two bases for $\mathbb{R}^3$ as we have done here can be understood as the choice of two Cartesian coordinate systems, one for $B_0$, and another for $B_t$.

With this background, let us work out the coordinate representations of the deformation map. If $(x_i)$ and $(X_I)$ are the Cartesian coordinates of the points $\mathsf{X} \in B_0$ and $\mathsf{x} \in B_t$, respectively, the deformation map $\mathsf{x} = \varphi(\mathsf{X}, t)$ can be written in coordinate form as follows: $$x_i = \varphi_i(X_1, X_2, X_3, t), \quad i = 1,2,3.$$ Here $\varphi_i(\cdot, t):B_0 \to \mathbb{R}$ are the component maps of $\varphi(\cdot,t):B_0 \to \mathbb{R}^3$. We will often write this equation as $x_i = \varphi_i(\mathsf{X},t)$.

Example

Consider the following deformation: $$x_1 = \lambda_1 X_1, \quad x_2 = \lambda_2 X_2, \quad x_3 = \lambda_3 X_3,$$ with $\lambda_1, \lambda_2, \lambda_3 > 0$. This corresponds to a volumetric strain. To understand why this is the case, note that a cube of side length $1$ gets transformed to a cuboid of side lengths $\lambda_1, \lambda_2, \lambda_3$ as a consequence of this deformation.

Example

As another example of an elementary deformation, let us consider the following: $$x_1 = X_1 + X_2 \tan \theta, \quad x_2 = X_2, \quad x_3 = X_3.$$ This is an example of a pure shear.

Deformation gradient

We now turn to the critical notion of the deformation gradient. As remarked earlier, this quantity carries all the information that is required to characterize the deformation of the body. But in its essence, it is a very simple idea. Note that the deformation map $$\varphi_t:B_0 \to B_t$$ is a nonlinear function of $\mathsf{X} \in B_0$. Recall that the preferred means to study general nonlinear maps is to construct their local linearized version, and study this linearized map locally. This, in particular, involves the computation of the Frechet derivative of the nonlinear map. The Frechet derivative of the deformation map is called the deformation gradient.

We will use the fact that when the Frechet derivative exists, it is identical to the Gateaux derivative, and use the latter to derive the expression for the deformation gradient. At any $\mathsf{X} \in B_0$, choose an arbitrary vector $\Delta \mathsf{X} \in T_{\mathsf{X}}B_0$ in the tangent space to $B_0$ at $\mathsf{X}$. Consider the line $\bar{\mathsf{X}}:[0,\delta] \to B_0$ defined as follows: for any $s \in [0,\delta]$, $$\bar{\mathsf{X}}(s) = \mathsf{X} + s\Delta\mathsf{X}.$$ Since $B_0$ is open, it is always possible to choose an appropriate value of $\delta > 0$ so as to ensure that the image of this line lies entirely in $B_0$. In the limit when $\delta \to 0$, the material curve $\chi_0^{-1} \circ \bar{\mathsf{X}}:[0,\delta] \to \mathcal{B}$ is called an infinitesimal material line segment. It follows from a simple calculation that $$\frac{d}{ds}\bigg\vert_{s=0}\bar{\mathsf{X}}(s) = \Delta \mathsf{X}.$$ This just expresses the fact that the tangent vector to the curve $\bar{\mathsf{X}}$ at $\mathsf{X}$ is $\Delta\mathsf{X}$.

As a consequence of the deformation, the line $\bar{\mathsf{X}}$ gets deformed to the line $\bar{\mathsf{x}}:[0,\delta] \to B_t$, where, for any $s \in [0,\delta]$, $$\bar{\mathsf{x}}(s) = \varphi_t(\bar{\mathsf{X}}(s)).$$ We will use the notation $\Delta \mathsf{x} \in T_{\mathsf{x}}B_t$ to denote the tangent vector to the deformed curve $\bar{\mathsf{x}}$ at $\mathsf{x} = \varphi_t(\mathsf{X})$.

We will now define the deformation gradient at $\mathsf{X}$ as the map $$\mathsf{F}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$$ that maps the tangent vector $\Delta\mathsf{X} \in T_{\mathsf{X}}B_0$ to the tangent vector $\Delta\mathsf{x} \in T_{\mathsf{X}}B_t$: $$\Delta \mathsf{x} = \mathsf{F}(\mathsf{X},t)\Delta\mathsf{X}.$$ To compute the expression for the deformation gradient $\mathsf{F}(\mathsf{X},t)$ in terms of the deformation map, note that $$\begin{split} \Delta \mathsf{x} &= \frac{d}{ds}\bigg\vert_{s=0} \bar{\mathsf{x}}(s)\\ &= \frac{d}{ds}\bigg\vert_{s=0} \varphi_t(\bar{\mathsf{X}}(s))\\ &= \nabla \varphi_t(\mathsf{X}) \cdot \frac{d}{ds}\bigg\vert_{s=0} \bar{\mathsf{X}}(s)\\ &= \nabla \varphi_t(\mathsf{X}) \cdot \Delta \mathsf{X}. \end{split}$$ For reasons that will become clear later on, we will introduce a new notation: $$\nabla \varphi_t(\mathsf{X}) \cdot \Delta \mathsf{X} = \text{GRAD } \varphi_t(\mathsf{X}) \Delta \mathsf{X}.$$

Remark

Notice that we dropped the generalized dot product in $\nabla\varphi_t(\mathsf{X})\cdot\Delta\mathsf{X}$ in writing it as $\text{GRAD }\varphi_t(\mathsf{X})\Delta\mathsf{X}$. This is to due to the fact that the deformation gradient is actually the Frechet derivative of the deformation map. As long as we are working in Euclidean spaces, this distinction isn't really important.

We will therefore write the action of the deformation map $\mathsf{F}(\mathsf{X},t)$ on $\Delta\mathsf{X} \in T_{\mathsf{X}}B_0$ as follows: $$\Delta \mathsf{x} = \mathsf{F}(\mathsf{X},t)\Delta\mathsf{X} = \text{GRAD } \varphi_t(\mathsf{X}) \Delta\mathsf{X}.$$ Since the deformationg gradient maps tangent vectors to curves in the initial placement to tangent vectors to curves in the current placement, it is also called as the tangent map of the deformation.

Remark

The symbol $\text{GRAD }\cdot$ stands for the gradient operation with respect to the coordinate system chosen on $B_0$. The uppercase notation is chosen to be consistent with the notational convention we chose earlier.

Let us work out the coordinate representation of the deformation gradient. Since we are working within a Cartesian coordinate setting, we can choose the same basis $(\mathsf{E}_I)$ for all the tangent spaces of $B_0$, and the basis $(\mathsf{e}_i)$ for all the tangent spaces of $B_t$. If $$\Delta \mathsf{x} = \sum \Delta x_i \mathsf{e}_i, \quad\text{and}\quad \Delta \mathsf{X} = \sum \Delta X_I \mathsf{E}_I,$$ then it follows from a straightforward calculation that $$\Delta x_i = \sum F_{iI}\Delta X_I = \sum \frac{\partial \varphi_i(\mathsf{X},t)}{\partial X_I} \Delta X_I.$$ The coordinate representation of the deformation gradient follows immediately as $$\mathsf{F}(\mathsf{X},t) = \sum F_{iJ}(\mathsf{X},t) \mathsf{e}_i \otimes \mathsf{E}_J,$$ where $$F_{iJ}(\mathsf{X},t) = \frac{\partial \varphi_i(\mathsf{X},t)}{\partial X_J}.$$ We will at times use the shorthand notation $F_{iJ}(\mathsf{X},t) = \varphi_{i,J}(\mathsf{X},t)$ to write this equation. Note that the components of $\mathsf{F}(\mathsf{X},t)$ can be arranged into a matrix $[\mathsf{F}]$ whose $(i,J)^{\text{th}}$ entry is $F_{iJ}(\mathsf{X},t)$. In terms of this matrix, we see that $$\begin{bmatrix} \Delta x_1\\ \Delta x_2\\ \Delta x_3\end{bmatrix} = \begin{bmatrix} F_{11} & F_{12} & F_{13}\\ F_{21} & F_{22} & F_{23}\\ F_{31} & F_{32} & F_{33} \end{bmatrix} \begin{bmatrix} \Delta X_1\\ \Delta X_2\\ \Delta X_3\end{bmatrix},$$ or, in a much more concise form, $[\Delta \mathsf{x}] = [\mathsf{F}][\Delta\mathsf{X}]$.

Let us quickly look into the physical meaning of the deformation gradient $\mathsf{F}(\mathsf{X},t)$. Suppose that $\mathsf{X},\mathsf{X}' \in B_0$ are two neighboring points, and let $\Delta \mathsf{X} = \mathsf{X}' - \mathsf{X}$. Let the corresponding points in the current placement be $\mathsf{x}, \mathsf{x}' \in B_t$, and let $\Delta \mathsf{x} = \mathsf{x} - \mathsf{x}'$. In the limit when the line joining $\mathsf{X}$ and $\mathsf{X}'$ in $B_0$ corresponds to an infinitesimal material line segment in $\mathcal{B}$, it follows from a simple application of Taylor's theorem that $$\begin{split} x'_i &= \varphi_i(X'_1, X'_2, X'_3, t)\\ &= \varphi_i(X_1 + \Delta X_1, X_2 + \Delta X_2, X_3 + \Delta X_3, t)\\ &= \varphi_i(X_1,X_2,X_3,t) + \sum \frac{\partial \varphi_i(X_1,X_2,X_3,t)}{\partial X_I} \Delta X_I + o(\lVert \Delta \mathsf{X}\rVert^2). \end{split}$$ Rearranging this, we see that $$\Delta x_i \simeq \sum F_{iI}(\mathsf{X},t)\Delta X_I.$$ The deformation gradient $\mathsf{F}(\mathsf{X},t)$ thus tells us how two neighboring points in the initial placement get relatively deformed as a consequence of the deformation.

Remark

In the derivation above, $o(\cdot)$ stands for the little 'o' symbol. This is defined as follows: given functions $f, g:\mathbb{R} \to \mathbb{R}$, we say that $f = o(g)$ at $x_* \in \mathbb{R}$ if it is true that $$\lim_{x \to x_*} \frac{f(x)}{g(x)} = 0.$$ The extension to higher dimensional functions is similarly defined.

Let us know revisit the examples we studied in the previous section and compute their deformation gradient

Example

For the case of a volumetric strain deformation with $x_1 = \lambda_1 X_1$, $x_2 = \lambda_2 X_2$ and $x_3 = \lambda_3 X_3$, it follows from a simple computation that $$F_{iJ}(\mathsf{X},t) = \sum \lambda_i \delta_{iJ} \mathsf{e}_i \otimes \mathsf{E}_J.$$ In matrix form, $$[\mathsf{F}(\mathsf{X},t)] = \begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0 & \lambda_3 \end{bmatrix}.$$ It is a good idea to relate this deformation gradient to the effect the volumetric strain deformation has on a unit cube.

Example

Let us consider the example of pure shear, with $x_1 = X_1 + \tan \theta X_2$, $x_2 = X_2$ and $x_3 = X_3$. It follows from a straightforward calculation that $$[\mathsf{F}(\mathsf{X},t)] = \begin{bmatrix} 1 & s & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix},$$ where $s = \tan \theta$ denotes the amount of shear. Note that the deformation gradient can be equivalenlty written as $$\mathsf{F}(\mathsf{X},t) = \mathsf{I} + s \mathsf{e}_1 \otimes \mathsf{E}_2.$$ Here, $\mathsf{I}:T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$ is the identity map that maps the basis $(E_I)$ of $T_{\mathsf{X}}B_0$ to the basis $(\mathsf{e}_i)$ of $T_{\mathsf{x}}B_t$.

In practice, we will often have a few other restrictions on the deformation gradient $\mathsf{F}(\mathsf{X},t)$. In particular, we will always assume that the deformation gradient is invertible. This implies, in particular that the determinant of $\mathsf{F}(\mathsf{X},t)$ is non-zero. We will, in fact, show shortly that for physically admissible deformations, $\text{det}(\mathsf{F}(\mathsf{X},t)) > 0$. We will therefore assume in the sequel the existence of the inverse deformation map $\mathsf{F}^{-1}(\mathsf{x},t):T_{\mathsf{x}}B_t \to T_{\mathsf{X}}B_0$.

Length transformation

Let us now see how the length of an infinitesimal line element changes as a consequence of deformation. In what follows, we will identify, as in the preceeding discussions, an infinitesimal line element with an element of the appropriate tangent space. Thus, suppose that the vector $\Delta\mathsf{X} \in T_{\mathsf{X}}B_0$ is mapped to $\Delta\mathsf{x} \in T_{\mathsf{x}}B_t$ by the deformation gradient $\mathsf{F}(\mathsf{X},t)$. It follows from a simple calculation that $$\begin{split} \lVert \Delta \mathsf{x}\rVert^2 &= \Delta \mathsf{x} \cdot \Delta \mathsf{x}\\ &= (\mathsf{F}(\mathsf{X},t)\Delta\mathsf{X}) \cdot (\mathsf{F}(\mathsf{X},t)\Delta\mathsf{X})\\ &= \Delta \mathsf{X} \cdot \mathsf{F}^T(\mathsf{x},t)\mathsf{F}(\mathsf{X},t)\Delta\mathsf{X}, \end{split}$$ where $\mathsf{x} = \varphi(\mathsf{X},t)$. Note that $\mathsf{F}^T:T_{\mathsf{x}}B_t \to T_{\mathsf{X}}B_0$. We now introduce the right Cauchy-Green tensor $\mathsf{C}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$ as follows: $$\mathsf{C}(\mathsf{X},t) = \mathsf{F}^T(\mathsf{x},t)\mathsf{F}(\mathsf{X},t).$$

Remark

Many engineering textbooks write the foregoing equation as $\mathsf{C} = \mathsf{F}^T\mathsf{F}$. We will adopt such a simplified notation when it suits our needs. But it should always be kept in mind what such shorthand notations actually mean.

In terms of the bases $(\mathsf{E}_I)$ and $(\mathsf{e}_i)$ of the respective tangent spaces, it follows from a simple calculation that $$\begin{split} \mathsf{F}^T(\mathsf{x},t) &= \sum F_{iJ}(\mathsf{X},t) \mathsf{E}_J \otimes \mathsf{e}_i,\\ \mathsf{C}(\mathsf{X},t) &= \sum F_{iI}(\mathsf{X},t)F_{iJ}(\mathsf{X},t) \mathsf{E}_I \otimes \mathsf{E}_J. \end{split}$$

Remark

It is instructive to pause here and reflect on the mathematical difference between the quantities $\mathsf{F}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$ and $\mathsf{C}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$. Both these are linear maps, and can be converted into tensors on the product spaces $T_{\mathsf{X}}B_0 \otimes T_{\mathsf{x}}B_t$ and $T_{\mathsf{X}}B_0 \otimes T_{\mathsf{X}}B_0$, respectively. Notice how the definition of $\mathsf{F}(\mathsf{X},t)$ involves two distinct vector spaces, while $\mathsf{C}(\mathsf{X},t)$ depends only on one vector space. For this reason, $\mathsf{F}(\mathsf{X},t)$ is called a two-point tensor, while $\mathsf{C}(\mathsf{X},t)$ is a tensor in the usual sense of the term as we studied earlier.

Returning to our discussion of the change in the length of an infinitesimal line segment as a consequence of deformation, it follows from the developments so far that $$\lVert \Delta\mathsf{x} \rVert^2 - \lVert \Delta\mathsf{X} \rVert^2 = \Delta\mathsf{X} \cdot \left(\mathsf{F}^T(\mathsf{x},t)\mathsf{F}(\mathsf{X},t) - \mathsf{I}\right)\Delta\mathsf{X},$$ where $\mathsf{I}$ is the identity map on $T_{\mathsf{X}}B_0$. Introducing the Green-Lagrange strain tensor, also called the Lagrangian strain tensor, $\mathsf{E}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ as $$\mathsf{E}(\mathsf{X},t) = \frac{1}{2}\left(\mathsf{F}^T(\mathsf{x},t)\mathsf{F}(\mathsf{X},t) - \mathsf{I}\right),$$ we see that the change in length of an infinitesimal line segment can be expressed in terms of the Lagrangian strain tensor as $$\lVert \Delta\mathsf{x} \rVert^2 - \lVert \Delta\mathsf{X} \rVert^2 = 2\Delta\mathsf{X} \cdot \mathsf{E}(\mathsf{X},t)\Delta\mathsf{X}.$$ We thus see that the change in length of an infinitesimal line segment is dictated by the Lagrangian strain tensor.

In terms of the basis $(\mathsf{E}_I)$ of $T_{\mathsf{X}}B_0$, it is straightfoward to verify that $$\mathsf{E}(\mathsf{X},t) = \sum E_{IJ}(\mathsf{X},t) \mathsf{E}_I \otimes \mathsf{E}_J, \qquad E_{IJ}(\mathsf{X},t) = \frac{1}{2}\left(C_{IJ}(\mathsf{X},t) - \delta_{IJ}\right).$$ Here, $C_{IJ}(\mathsf{X},t) = \sum F_{iI}(\mathsf{X},t)F_{iJ}(\mathsf{X},t)$ are the components of the right Cauchy-Green tensor, and $\delta_{IJ}$ is the Kronecker delta symbol. The diagonal component fields $E_{II}(\mathsf{X},t)$ are called axial, or tensile strains, while the off-diagonal component fields $E_{IJ}(\mathsf{X},t)$, with $I \neq J$ are called shear strains.

The preceeding discussion allowed us to study the change in length of an infinitesimal line segment in the Lagrangian description. Alternatively, we can describe the same phenomena from the Eulerian viewpoint. Following the same line of thought as above, we see that $$\Delta\mathsf{X}\cdot\Delta\mathsf{X} = \mathsf{F}^{-1}(\mathsf{x},t)\Delta\mathsf{x} \cdot \mathsf{F}^{-1}(\mathsf{x},t)\Delta\mathsf{x}.$$ Defining the left Cauchy-Green tensor $\mathsf{b}(\mathsf{x},t):T_{\mathsf{x}}B_t \to T_{\mathsf{x}}B_t$ as $$\mathsf{b}(\mathsf{x},t) = \mathsf{F}(\mathsf{X},t)\mathsf{F}^T(\mathsf{x},t),$$ where $\mathsf{X} = \varphi_t^{-1}(\mathsf{x})$, it follows that $$\begin{split} \Delta \mathsf{X} \cdot \Delta \mathsf{X} &= \Delta \mathsf{x} \cdot \mathsf{F}^{-T}(\mathsf{X},t)\mathsf{F}^{-1}(\mathsf{x},t)\Delta\mathsf{x}\\ &= \Delta\mathsf{x} \cdot \mathsf{b}^{-1}(\mathsf{x},t)\Delta\mathsf{x}. \end{split}$$ The Eulerian strain tensor, also called the Almansi strain tensor, is the map $\mathsf{e}:T_{\mathsf{x}}B_t \to T_{\mathsf{x}}B_t$ defined as follows: $$\mathsf{e}(\mathsf{x},t) = \frac{1}{2}\left(\mathsf{F}^{-T}(\mathsf{X},t)\mathsf{F}^{-1}(\mathsf{x},t) - \mathsf{I}\right).$$ Here, $\mathsf{I}$ is the identity map on $T_{\mathsf{x}}B_t$.

Using these definitions, we can now compute the change in length of an infinitesimal line segment in the Eulerian description as follows: $$\begin{split} \lVert \Delta \mathsf{x} \rVert^2 - \lVert \Delta \mathsf{X} \rVert^2 &= \Delta\mathsf{x}\cdot\Delta\mathsf{x} - \Delta\mathsf{x} \cdot \mathsf{b}^{-1}(\mathsf{x},t)\Delta\mathsf{x}\\ &= 2\Delta\mathsf{x}\cdot\mathsf{e}(\mathsf{x},t)\Delta\mathsf{x}. \end{split}$$ It is important to note the following identity: $$\frac{1}{2}\left(\lVert \Delta \mathsf{x} \rVert^2 - \lVert \Delta \mathsf{X} \rVert^2\right) = \Delta\mathsf{X} \cdot \mathsf{E}(\mathsf{X},t)\Delta\mathsf{X} = \Delta\mathsf{x}\cdot\mathsf{e}(\mathsf{x},t)\Delta\mathsf{x}.$$ Thus, the same physical phenomena can be expressed in two complementary viewpoints. It is however important to develop concrete relations like the one above that express the mutual compatibility of such distinct but equivalent descriptions.

Example

We would intuitively expect a body undergoing a pure translation or a pure rotation to have zero strain since the relative positions of the body do not change under these kinds of deformations. Let us verify this. Suppose that we are given a deformation $\varphi_t:B_0 \to B_t$ of the following form: for any $\mathsf{X} \in B_0$ $$\varphi_t(\mathsf{X}) = \mathsf{c} + \mathsf{Q}(\mathsf{X} - \mathsf{O}),$$ where $\mathsf{c} \in \mathbb{R}^3$ is a constant vector, and $\mathsf{O} \in \mathbb{R}^3$ is some fixed point in $\mathbb{R}^3$. Further, we will assume that the constant map $\mathsf{Q}:\mathbb{R}^3 \to \mathbb{R}^3$ is orthogonal. With these assumptions, it is clear that this denotes a deformation consisting of a rotation according to the rotation map $\mathsf{Q}$ about the point $\mathsf{O}$ followed by a uniform translation $\mathsf{c}$. It is left as a straightforward exercise to compute that $$\mathsf{F}(\mathsf{X},t) = \mathsf{Q},$$ and, consequently, that $\mathsf{E}(\mathsf{X},t) = \mathsf{0}$, and $\mathsf{e}(\mathsf{x},t) = \mathsf{0}$. We thus see that the definition of strain that we have developed is consistent with our intuitive sense of how a strain ought to behave.

Once we know how an infinitesimal line element changes in length as a consequence of deformation, the change in length of an arbitrary curve in the initial placement can be computed. Specifically, suppose that $\gamma:I \subseteq \mathbb{R} \to B_0$ is a curve in $B_0$. The length of this curve, $L_\gamma$ is computed as follows: $$L_{\gamma} = \int_I \lVert D\gamma(t)\rVert dt.$$

Remark

It is not difficult to show that the length thus defined is independent of the parametrization of the curve. Indeed, suppose that there exists a smooth, monotonically increasing and surjective function $\tau:I \to J \subseteq \mathbb{R}$ such that we can formulate an alternative parametrization of the line $\gamma(I)$ as the curve $\tilde{\gamma}:J \to B_0$ as follows: $\tilde{\gamma} = \gamma \circ \tau^{-1}$. Let $\tilde{L}_\gamma$ denote the length of the curve $\tilde\gamma$; we expect, based on the fact that $\gamma(I) = \tilde{\gamma}(J)$ that $\tilde{L}_\gamma = L_\gamma$. Let us verify this: $$\begin{split} L_\gamma &= \int_I \lVert D\gamma(t) \rVert dt\\ &= \int_I \lVert D(\tilde\gamma \circ \tau)(t) \rVert dt\\ &= \int_I \lVert D\tilde\gamma(\tau(t))\rVert D\tau(t) dt\\ &= \int_J \lVert D\tilde\gamma(s) \rVert ds\\ &= \tilde{L}_\gamma. \end{split}$$ In deriving this equation, use has been made of the fact that $D\tau(t) > 0$ for every $t \in I$.

We now wish to find the length $l_{\gamma}$ of the deformed curve $\varphi_t \circ \gamma: I \to B_t$. This follows at once from the following calculation: $$\begin{split} l_\gamma &= \int_I \lVert D(\varphi_t \circ \gamma)(t) \rVert dt\\ &= \int_I \lVert D\varphi(\gamma(t), t) D\gamma(t)\rVert dt\\ &= \int_I \sqrt{\mathsf{F}(\gamma(t),t)D\gamma(t) \cdot \mathsf{F}(\gamma(t),t)D\gamma(t)} \,dt\\ &= \int_I \sqrt{D\gamma(t) \cdot \mathsf{C}(\gamma(t),t)D\gamma(t)} \,dt. \end{split}$$ We thus see that a knowledge of the deformation gradient allows us to compute the lengths of any curve in the initial placement.

Volume transformation

Let us now study how a volume of space occupied by the body at the initial time changes as a consequence of the deformation. Towards that end, we will first find out how an infinitesimal volume transforms as a consequence of deformation, and use this information to compute the change in volume of an arbitray initial volume.

Consider three curves $\hat{\mathsf{X}}_1, \hat{\mathsf{X}}_2, \hat{\mathsf{X}}_3:[-\epsilon,\epsilon] \to B_0$ such that $\hat{\mathsf{X}}_1(0) = \hat{\mathsf{X}}_2(0) = \hat{\mathsf{X}}_3(0) = \mathsf{X} \in B_0$. Let us consider the scalar triple product $$\Delta V = \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_1(s) \cdot \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_2(s) \times \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_3(s)$$ of the tangent vectors to these curves at $\mathsf{X}$. We will implicitly assume that the tangent vectors to the chosen curves at $\mathsf{X}$ are non-collinear. The scalar triple product $\Delta V$ is called an infinitesimal volume element. Note that that $\Delta V \in \mathbb{R}$. We wish to study how the scalar triple product, $\Delta v \in \mathbb{R}$, of the tangents corresponding deformed curves, $\hat{x}_i = \varphi_t \circ \hat{X}_i:[-\epsilon,\epsilon] \to B_t$, $i = 1,2,3$, at $\mathsf{x} = \varphi_t(\mathsf{X}) \in B_t$ is related to $\Delta V$. This relationship is easily computed: $$\begin{split} \Delta v &= \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{x}}_1(s) \cdot \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{x}}_2(s) \times \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{x}}_3(s)\\ &= \frac{d}{ds}\bigg\vert_{s=0}\varphi_t \circ \hat{\mathsf{X}}_1(s) \cdot \frac{d}{ds}\bigg\vert_{s=0}\varphi_t \circ \hat{\mathsf{X}}_2(s) \times \frac{d}{ds}\bigg\vert_{s=0}\varphi_t \circ \hat{\mathsf{X}}_3(s)\\ &= \mathsf{F}(\mathsf{X},t)\frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_1(s) \cdot \mathsf{F}(\mathsf{X},t)\frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_2(s) \times \mathsf{F}(\mathsf{X},t)\frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_3(s)\\ &= \text{det}(\mathsf{F}(\mathsf{X},t))\,\frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_1(s) \cdot \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_2(s) \times \frac{d}{ds}\bigg\vert_{s=0}\hat{\mathsf{X}}_3(s)\\ &= \text{det}(\mathsf{F}(\mathsf{X},t))\,\Delta V. \end{split}$$ In deriving this expression, we have made use of the chain rule of differentiation, and the definition of the determinant of a linear map in terms of volume forms. Introducing the Jacobian $$J(\mathsf{X},t) = \text{det}(\mathsf{F}(\mathsf{X},t)),$$ we can succinctly write the transformation rule for infinitesimal volume elements as follows: $$\Delta v = J(\mathsf{X},t)\Delta V.$$ Again, notice how a knowledge of the deformation gradient is sufficient to compute how volumes transform.

If $\Omega \subseteq B_0$ is some sub-region of $B_0$, we can compute the volume $V_\Omega$ occupied by the region $\Omega$ using the integral $$V_\Omega = \int_\Omega dV.$$ Here the integral is to be understood as the standard triple integral used to compute volumes in $\mathbb{R}^3$. As a consequence of deformation, the region $\Omega$ gets transformed to $\varphi_t(\Omega) \subseteq B_t$. The volume $v_\Omega$ of this deformed region can be computed as $$v_\Omega = \int_{\varphi_t(\Omega)} dv.$$ Employing the standard change of variables rules for integration, we can transform this volume integral over $\varphi_t(\Omega)$ to an integral over $\Omega$: $$v_\Omega = \int_\Omega J(\mathsf{X},t) \, dV.$$ Notice the similarity between this and the transformation rule for infinitesimal volumes derived eariler.

Area transformation

Finally, let us consider how infinitesimal area elements transform as a consequence of deformation. The starting point is the volume transformation rule that we studied earlier.

Suppose that we have three curves $\hat{\mathsf{X}}_1, \hat{\mathsf{X}}_2, \hat{\mathsf{X}}_3:[-\epsilon,\epsilon] \to B_0$ such that $\hat{\mathsf{X}}_1(0) = \hat{\mathsf{X}}_2(0) = \hat{\mathsf{X}}_3(0) = \mathsf{X} \in B_0$, such that their tangents are non-collinear at $\mathsf{X}$, as before. Let us now consider an infinitesimal area element $\Delta A \in T_{\mathsf{X}}B_0$, defined as follows: $$\Delta A = \frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{X}}_2(s) \times \frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{X}}_3(s).$$ Without loss of generality, we will suppose that the curves $\hat{\mathsf{X}}_2,\hat{\mathsf{X}}_3$ are fixed, but $\hat{\mathsf{X}}_1$ is arbitrary, as long as the tangents of these curves at $\mathsf{X}$ are not collinear. With this definition, we can write the infinitesimal volume element $\Delta V$ at $\mathsf{X}$ defined by the curves $\hat{\mathsf{X}}_1, \hat{\mathsf{X}}_2, \hat{\mathsf{X}}_3$ as $$\Delta V = \frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{X}}_1(s) \cdot \Delta A.$$ As a consequence of deformation, the infinitesimal volume element $\Delta V$ gets transformed to the volume element $\Delta v$, and the infinitesimal area element $\Delta A$ gets transformed to $\Delta a \in T_{\mathsf{\varphi_t(\mathsf{X})}}B_t$. Using the transformation rule $\Delta v = J(\mathsf{X},t)\Delta V$ for volume elements, we see that $$\begin{split} J(\mathsf{X},t) \frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{X}}_1(s) \cdot \Delta A &= \frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{x}}_1(s) \cdot \Delta a\\ &= \frac{d}{ds}\bigg\vert_{s=0} \varphi_t \circ \hat{\mathsf{X}}_1(s) \cdot \Delta a\\ &= \mathsf{F}(\mathsf{X},t)\frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{X}}_1(s) \cdot \Delta a\\ &= \frac{d}{ds}\bigg\vert_{s=0} \hat{\mathsf{X}}_1(s) \cdot \mathsf{F}^T(\varphi_t(\mathsf{X}), t)\Delta a. \end{split}$$ Since the curve $\hat{\mathsf{X}}_1$ is arbitrary, we see at once that $$J(\mathsf{X},t) \Delta A = \mathsf{F}^T(\varphi_t(\mathsf{X}), t) \Delta a.$$ We obtain the transformation rule for areas after a simple rearrangement of this equation as $$\Delta a = J(\mathsf{X},t) \mathsf{F}^{-T}(\mathsf{X},t)\Delta A.$$ It is worth keeping in mind that $\Delta a \in T_{\mathsf{x}}B_t$ and $\Delta A \in T_{\mathsf{X}}B_0$, where $\mathsf{x} = \varphi_t(\mathsf{X})$.

To conclude this section, let us look at how the area transformation rule is typically applied in practice. Suppose that we are give a vector field $\mathsf{w}(\mathsf{x},t)$ on $B_t$. The flux $F_{\mathsf{w}}$ associated with this vector field is typically computed as $$F_{\mathsf{w}} = \int_{\partial B_t} \mathsf{w}(\mathsf{x},t) \cdot \mathsf{n}(\mathsf{x},t) \, da,$$ where $\mathsf{n}(\mathsf{x},t)$ is the outward normal to $B_t$ at $\mathsf{x} \in \partial B_t$. We can compute the flux associated with this vector field equivalently by evaluating the integral over the initial placement $B_0$: $$F_{\mathsf{w}} = \int_{\partial B_0} J(\mathsf{X},t) \, \mathsf{w}(\varphi(\mathsf{X},t),t) \cdot \mathsf{F}^{-T}(\mathsf{X},t)\mathsf{N}(\mathsf{X},t) \, dA.$$ Notice how the area transformation rule derived earlier for infinitesimal volumes, "$\mathsf{n}da = J\mathsf{F}^{-T}\mathsf{N}dA$", is used in writing the foregoing equation. Defining the Piola transform of the vector field $\mathsf{w}$ on $B_t$ as the vector field $\mathsf{W}$ on $B_0$, where, for any $\mathsf{X} \in B_0$, $$\mathsf{W}(\mathsf{X},t) = J(\mathsf{X},t)\mathsf{F}^{-1}(\varphi(\mathsf{X},t),t)\mathsf{w}(\varphi(\mathsf{X},t),t),$$ we see that $$\int_{\partial B_0} \mathsf{w}(\mathsf{x},t)\cdot\mathsf{n}(\mathsf{x},t)\,da = \int_{\partial B_0} \mathsf{W}(\mathsf{X},t)\cdot\mathsf{N}(\mathsf{X},t)\,dA.$$ This formula will prove to be quite useful to us later on.

Remark

Note that all the area integrals encountered here are to be understood in the sense of the integrals of parametrized surfaces, as was studied earlier.

Polar decomposition theorem

The previous sections illustrate the fact that all the relevant information regarding the change in shape of a body as a consequence of a deformation is captured by the deformation gradient. We will now study a very important decomposition of the deformation gradient that will prove to be very useful later on. The basic idea is very simple: a deformation characterized by a pure rotation does not produce any strain; hence, if we are interested in the strain caused by a general deformation, can we extract out its purely rotational components so that we are left with only that part of the deformation that is responsible for an actual shape change?

The polar decomposition theorem provides a means to effect this decomposition in a precise manner. Specifically, this theorem states that any deformation graident can be decomposed uniquely as either a pure stretch followed by a pure rotation, or a pure rotation followed by a pure stretch. Specifically, given a deformation $\varphi_t:B_0 \to B_t$, the deformation gradient $\mathsf{F}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$, where $\mathsf{X} \in B_0$, and $\mathsf{x} = \varphi_t(\mathsf{X}) \in B_t$, can be decomposed uniquely as follows: $$\mathsf{F}(\mathsf{X},t) = \mathsf{R}(\mathsf{X},t)\mathsf{U}(\mathsf{X},t) = \mathsf{V}(\mathsf{x},t)\mathsf{R}(\mathsf{X},t),$$ where $\mathbf{R}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$ is an orthogonal map, called the rotation map, that quantifies the purely rotational part of the deformation gradient, and $\mathsf{U}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ and $\mathsf{V}(\mathsf{x},t):T_{\mathsf{x}}B_t \to T_{\mathsf{x}}B_t$ are symmetric and positive definite tensors, called the right stretch tensor and the left stretch tensor, respectively. The decomposition $\mathsf{F}(\mathsf{X},t) = \mathsf{R}(\mathsf{X},t)\mathsf{U}(\mathsf{X},t)$ is called the right polar decomposition of $\mathsf{F}(\mathsf{X},t)$, and $\mathsf{F}(\mathsf{X},t) = \mathsf{V}(\mathsf{x},t)\mathsf{R}(\mathsf{X},t)$ is called the left polar decomposition of $\mathsf{F}(\mathsf{X},t)$.

The proof of the polar decomposition theorem is quite simple. To begin with, notice that the right Cauchy-Green tensor $\mathsf{C}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ is both symmetric and positive definite. The symmetric of $\mathsf{C}(\mathsf{X},t)$ follows from the fact that $$\mathsf{C}^T(\mathsf{X},t) = (\mathsf{F}^T(\mathsf{x},t)\mathsf{F}(\mathsf{X},t))^T = \mathsf{F}^T(\mathsf{x},t)\mathsf{F}(\mathsf{X},t) = \mathsf{C}(\mathsf{X},t).$$ The positive definiteness of $\mathsf{C}(\mathsf{X},t)$ follows from the fact that for any $\Delta \mathsf{X} \in T_{\mathsf{X}}B_0$, $$\Delta \mathsf{X} \cdot \mathsf{C}(\mathsf{X},t)\Delta \mathsf{X} = \mathsf{F}(\mathsf{X},t)\Delta \mathsf{X} \cdot \mathsf{F}(\mathsf{X},t)\Delta \mathsf{X} = \lVert \mathsf{F}(\mathsf{X},t)\Delta \mathsf{X}\rVert^2 \ge 0.$$ Further, it follows from the same argument that $\Delta \mathsf{X} \cdot \mathsf{C}(\mathsf{X},t)\Delta \mathsf{X} = 0$ iff $\Delta X = \mathsf{0}$. Recall from our discussion of eigenvalues and eigenvectors that since $\mathsf{C}(\mathsf{X},t)$ is symmetric and positive definite, we can compute its square root. Define the tensor $\mathsf{U}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ as follows: $$\mathsf{U}(\mathsf{X},t) = \sqrt{\mathsf{C}(\mathsf{X},t)}.$$ It follows from the definition that $\mathsf{U}(\mathsf{X},t)$ is both symmetric and positive definite. Define the map $\mathsf{R}:T_{\mathsf{X}}B_0\to T_{\mathsf{x}}B_t$ as follows: $$\mathsf{R}(\mathsf{X},t) = \mathsf{F}(\mathsf{X},t)\mathsf{U}^{-1}(\mathsf{X},t).$$ It is easy to check that the map $\mathsf{R}(\mathsf{X},t)$ is an orthogonal map. Indeed, note that $$\mathsf{R}^T\mathsf{R} = \mathsf{U}^{-T}\mathsf{F}^T\mathsf{F}\mathsf{U} = \mathsf{U}^{-1}\mathsf{U}^2\mathsf{U} = \mathsf{I},$$ where $\mathsf{I}:T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ is the identity tensor on $T_{\mathsf{X}}B_0$. Note that we have suppressed the arguments of the various maps to keep the notation simple. A similar argument shows that $\mathsf{R}\mathsf{R}^T = \mathsf{I}$, where $\mathsf{I}$ is the identity map on $T_{\mathsf{x}}B_t$. This shows that $\mathsf{R}(\mathsf{X},t)$ is an orthogonal map. We have thus shown that the deformation gradient $\mathsf{F}(\mathsf{X},t)$ can be written as the action of a symmetric tensor followed by the action of a rotation map.

We still need to show that this decomposition is unique. Towards this end, suppose that $\mathsf{F}(\mathsf{X},t)$ also admits another right polar decomposition of the form $\mathsf{F} = \tilde{\mathsf{R}}\tilde{\mathsf{U}}$, where, in the interest of notational simplicity, we have suppressed the arguments of the various maps. It follows from the following argument, $$\mathsf{U}^2 = \mathsf{F}^T\mathsf{F} = \tilde{\mathsf{U}}^T\tilde{\mathsf{R}}\mathsf{R}\mathsf{U} = \tilde{\mathsf{U}}^2,$$ that $\mathsf{U} = \tilde{\mathsf{U}}$. This, along with the equation $\mathsf{R}\mathsf{U} = \tilde{\mathsf{R}}\tilde{\mathsf{U}}$ shows that $\mathsf{R} = \tilde{\mathsf{R}}$, thereby establishing the uniqueness of the right polar decomposition.

To prove the left decomposition, define the tensor $\mathsf{V}(\mathsf{x},t):T_{\mathsf{x}}B_t \to T_{\mathsf{x}}B_t$ as follows: $$V(\mathsf{x},t) = \sqrt{\mathsf{b}(\mathsf{x},t)},$$ where $\mathsf{b}(\mathsf{x},t)$ is the left Cauchy-Green tensor. It is easily checked that $\mathsf{V}(\mathsf{x},t)$ is both symmetric and positive definite. It follows from an argument analogous to the previous one that the map $\mathsf{S}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{x}}B_t$ defined as $$S(\mathsf{X},t) = \mathsf{V}^{-1}(\mathsf{x},t)\mathsf{F}(\mathsf{X},t),$$ is orthogonal. We thus see that the the deformation gradient can be written in the left polar decomposition form as $\mathsf{F}(\mathsf{X},t) = \mathsf{V}(\mathsf{x},t)\mathsf{S}(\mathsf{S},t)$. We will now show that $\mathsf{S} \equiv \mathsf{R}$. To see this, note that we can use the orthogonality of $\mathsf{S}(\mathsf{X},t)$ in the following way: $$\mathsf{F} = \mathsf{V}\mathsf{S} = (\mathsf{S}\mathsf{S}^T)\mathsf{V}\mathsf{S} = \mathsf{S}(\mathsf{S}^T\mathsf{V}\mathsf{S}).$$ Notice now that since $\mathsf{S}$ is orthogonal, and $\mathsf{S}^T\mathsf{V}\mathsf{S}$ is symmetric and positive definite, it follows from the uniqueness of the right polar decomposition that $$\begin{split} \mathsf{R}(\mathsf{X},t) &= \mathsf{S}(\mathsf{X},t),\\ \mathsf{U}(\mathsf{X},t) &= \mathsf{R}^T(\mathsf{x},t)\mathsf{V}(\mathsf{x},t)\mathsf{R}(\mathsf{X},t). \end{split}$$ We have thus shown that $\mathsf{F}(\mathsf{X},t) = \mathsf{V}(\mathsf{x},t)\mathsf{R}(\mathsf{X},t)$, which this the left polar decomposition. The uniqueness of the left polar decomposition follows from an argument analogous to that of the uniqueness of the right polar decomposition. This completes the proof of the polar decomposition theorem.

As a quick consistency check, let us see compute the Lagrangian strain tensor using the polar decomposition theorem: $$\mathsf{E} = \frac{1}{2}(\mathsf{F}^T\mathsf{F} - \mathsf{I}) = \frac{1}{2}(\mathsf{U}^2 - \mathsf{I}).$$ We have suppressed the arguments for notational convenience. We thus see that the rotation map $\mathsf{R}(\mathsf{X},t)$ does not contribute to the strain, as expected. A similar caculation shows that this is true for the Eulerian strain tensor too. The strain is entirely captured by the right and left stretch tensors $\mathsf{U}(\mathsf{X},t)$ and $\mathsf{V}(\mathsf{x},t)$, respectively. In this context, it is useful to mention that the eigenvalues of the right and left stretch vectors are called the right principal stretches and left principal stretches, respectively.