3.1 Body, configuration, motion, displacement

A body $\mathcal{B}$ is a set of particles. A representative particle is designated as Y, where Y is the name of the particle (e.g., Steve, Bob, or the color red); see Figure 3.1(a). Body $\mathcal{B}$ is seen only in its configurations. A configuration of $\mathcal{B}$ is the region of ${\mathcal{E}}^3$ (Euclidean 3-space) occupied at a particular instant by the body.

Consider the configuration of body $\mathcal{B}$ at time t, called the present configuration, in which $\mathcal{B}$ occupies an open volume $\mathcal{R}$ of ${\mathcal{E}}^3$ bounded by a closed surface $\partial \mathcal{R}$; see Figure 3.1(b). (If, for instance, the body $\mathcal{B}$ is an orange, the region occupied by the flesh constitutes an open volume, and the region occupied the peel constitutes a closed surface.) Let x be the position vector of the place occupied by the representative particle Y at time t; see Figure 3.1(b). Then

$\mathbf{x}=\overline{\mathbf{\chi }}\left(Y,t\right).$

The mapping (3.1) is called the motion of body $\mathcal{B}$. Note that we distinguish the function $\overline{\mathbf{\chi }}\left(Y,t\right)$ from its value (or output) x. The mapping $\overline{\mathbf{\chi }}$ is assumed to be differentiable as many times as necessary with respect to both Y and t. We also make the important assumption that the motion (3.1) is invertible for each t, so the function exists.

$Y={\overline{\chi }}^{-1}\left(\mathbf{x},t\right)$

One particular configuration κ of body $\mathcal{B}$ may be selected as the reference configuration, and then the body and its motion may be referred to this fixed configuration. A common practice is to choose κ to be the initial configuration, i.e., the configuration occupied by $\mathcal{B}$ at time t = 0. However, the reference configuration need not be the initial configuration; in fact, it can even be a configuration that the body $\mathcal{B}$ never actually attains. Since the selection of the reference configuration κ is taken from an infinite number of possible configurations of $\mathcal{B}$, a motion has infinitely many different referential descriptions, all equally valid.

In the reference configuration κ, the body $\mathcal{B}$ occupies an open volume ${\mathcal{R}}_{\text{R}}$ of ${\mathcal{E}}^3$ bounded by a closed surface $\partial {\mathcal{R}}_{\text{R}}$; see Figure 3.1(c). Let X be the position vector of the place occupied by the representative particle Y in this reference configuration.The mapping from Y to X is written

$\mathbf{X}=\kappa \left(Y\right).$

The mapping (3.3), called the reference map, is assumed to be differentiable as many times as desired, and invertible. The inverse mapping is

$Y={\kappa }^{-1}\left(\mathbf{X}\right).$

Using (3.4), we can rewrite the motion (3.1) as

$\mathbf{x}=\overline{\mathbf{\chi }}\left[{\kappa }^{-1}\left(\mathbf{X}\right),t\right]={}_{\kappa }\mathbf{\chi }\left(\mathbf{X},t\right).$

Equation (3.5) serves as a definition of the mapping κχ. It should be noted that κχ depends on the choice of reference configuration κ. If only one reference configuration is adopted, then the subscript κ is generally omitted, i.e.,

$\mathbf{x}=\mathbf{\chi }\left(\mathbf{X},t\right),$

and the dependence of χ on a particular reference configuration κ is understood.

The function (3.1) is called the material description of motion. In the material description, we deal directly with the particle Y of $\mathcal{B}$. The description of motion (3.6) is called the referential description of motion. In the referential description, the particle Y of $\mathcal{B}$ is labeled by the position X it occupied when $\mathcal{B}$ was in its reference configuration κ.

The displacement u of the particle Y is defined as the difference between its reference and present positions, i.e.,

$\mathbf{u}=\mathbf{x}-\mathbf{X};$

see Figure 3.2.

Consider a vector quantity f related to the motion of body $\mathcal{B}$. (Our use of a vector quantity is arbitrary; the discussion that follows is equally valid for a scalar quantity φ or a tensor quantity A.) In the material description, the quantity f is expressed as a function of particle Y and time t, i.e.,

$\mathbf{f}=\overline{\mathbf{f}}\left(Y,t\right).$

In the referential description, f is expressed as a function of the position X occupied by Y in the reference configuration κ, and time t. To see this we use (3.4) in (3.8):

$\mathbf{f}=\overline{\mathbf{f}}\left(Y,t\right)=\overline{\mathbf{f}}\left({\kappa }^{-1}\left(\mathbf{X}\right),t\right)=\widehat{\mathbf{f}}\left(\mathbf{X},t\right).$

The quantity f may also be expressed as a function of the position x occupied at time t by the particle Y, and time t. To see this we use (3.2) in (3.8):

$\mathbf{f}=\overline{\mathbf{f}}\left(Y,t\right)=\overline{\mathbf{f}}\left({\overline{\chi }}^{-1}\left(\mathbf{x},t\right),t\right)=\widehat{\mathbf{f}}\left(\mathbf{x},t\right).$

In (3.10) we have expressed f in terms of the present position x of Y, and time t. Such a description is called the spatial description. Referential and spatial descriptions are also referred to as Lagrangian and Eulerian descriptions, respectively.

To review: A vector quantity f (or a scalar quantity φ or a tensor quantity A) related to the motion of body $\mathcal{B}$ may be expressed in any of three different descriptions:

$\begin{array}{lll}\mathbf{f}\hfill & =\overline{\mathbf{f}}\left(Y,t\right)\hfill & \left(\text{material}\right)\hfill \\ \hfill & =\widehat{\mathbf{f}}\left(\mathbf{X},t\right)\hfill & \left(\text{referential}\right)\hfill \\ \hfill & =\tilde{\mathbf{f}}\left(\mathbf{x},t\right)\hfill & \left(\text{spatial}\right).\hfill \end{array}$

This freedom is due to our assumptions that the mappings (3.1) and (3.3) are invertible. In the referential description, each particle is labeled by its reference position; in the spatial description, each particle is labeled by its present position. We note that the material description is seldom used in continuum mechanics since it becomes intractable to find a different name for each particle in the body; the other two representations, however, are useful and will be employed.

Problem 3.1

In a particular motion of body $\mathcal{B}$, the reference configuration is chosen to be the configuration at time t = 0, i.e., the initial configuration. At t = 0, the red particle is at (1, 1, 1), the blue particle is at (0, 0, 0), and the white particle is at (−1, −1, −1). At t = 5 (the present configuration), the red particle is at (2, 2, 2), the blue particle is at (1, 1, 1), and the white particle is at (3, 3, 3), and their temperatures are 5°, 0°, and 10°, respectively; see Figure 3.3.

Solution

(a) The equation $\mathbf{x}=\overline{\mathbf{\chi }}\left(Y,t\right)$ answers the question: “Where is particle Y at time t?” That is, the input to the function $\overline{\mathbf{\chi }}\left(Y,t\right)$ is the particle name Y and the time t, and the output is the present position x of the particle at time t. For instance, $\overline{\mathbf{\chi }}\left(\text{red},5\right)=\left(2,2,2\right)$ and $\overline{\mathbf{\chi }}\left(\text{blue},5\right)=\left(1,1,1\right)$.

(b) The equation $Y={\overline{\chi }}^{-1}\left(\mathbf{x},t\right)$ answers the question: “Which particle is at x at time t?” That is, the input to the function ${\overline{\chi }}^{-1}\left(\mathbf{x},t\right)$ is the time t and the present position x of the particle at time t, and the output is the name Y of the particle. For instance, ${\overline{\chi }}^{-1}\left(\left(1,1,1\right),5\right)=\text{blue}$.

(c) The equation X = κ (Y) answers the question: “Where is particle Y in the reference configuration?” That is, the input to the function κ (Y) is the particle name Y, and the output is the reference position X of the particle. For instance, κ (red) = (1,1,1).

(d) The equation Y = κ⁻¹(X) answers the question: “Which particle is at X in the reference configuration?” That is, the input to the function κ⁻¹(X) is the reference position X of the particle, and the output is the particle name Y. For instance, κ⁻¹(−1, −1, −1) = white.

(e) The equation x = χ (X, t) answers the question: “Where is the particle that was at X in the reference configuration now located at time t?” That is, the input to the function χ (X, t) is the reference position X of the particle and the time t, and the output is the present position x of the particle at time t. For instance, χ ((1,1,1),5) = (2,2,2). To see this, note that

$\mathbf{\chi }\left(\left(1,1,1\right),5\right)=\mathbf{\chi }\left(\kappa \left(\text{red}\right),5\right)=\overline{\mathbf{\chi }}\left(\text{red,}5\right)=\left(2,2,2\right).$

(f) Let Θ be the temperature. The equation $\text{Θ}=\overline{\text{Θ}}\left(Y,t\right)$ answers the question: “What is the temperature of particle Y at time t?” The input to the function $\overline{\text{Θ}}\left(Y,t\right)$ is the particle name Y and the time t, and the output is the temperature Θ of the particle at time t. For instance, $\overline{\text{Θ}}\left(\text{white},5\right)=10°$.

(g) The equation $\text{Θ}=\widehat{\text{Θ}}\left(\mathbf{X},t\right)$ answers the question: “What is the temperature at time t of the particle Y that occupied reference position X?” The input to the function $\widehat{\text{Θ}}\left(\mathbf{X},t\right)$ is the reference position X of the particle and the time t, and the output is the temperature Θ of the particle at time t. For instance, $\widehat{\text{Θ}}\left(\left(0,0,0\right),5\right)=0°$.

(h) The equation $\text{Θ}=\widetilde{\text{Θ}}\left(\mathbf{x},t\right)$ answers the question: “What is the temperature at time t of the particle Y that occupies present position x at time t?” The input to the function $\widetilde{\text{Θ}}\left(\mathbf{x},t\right)$ is the time t and the present position x of the particle at time t, and the output is the temperature Θ of the particle at time t. For instance, $\widetilde{\text{Θ}}\left(\left(2,2,2\right),5\right)=5°$.

(i) Note that the material form $\overline{\text{Θ}}\left(Y,t\right)$, the referential form $\widehat{\text{Θ}}\left(\mathbf{X},t\right)$, and the spatial form $\widetilde{\text{Θ}}\left(\mathbf{x},t\right)$ are different functions. To illustrate this, note that $\widehat{\text{Θ}}\left(\left(1,1,1\right),5\right)=5°$ but $\widetilde{\text{Θ}}\left(\left(1,1,1\right),5\right)=0°$, i.e., different functions generally provide different output when given identical input.

3.2 Material derivative, velocity, acceleration

Consider once again a vector quantity f related to the motion of body $\mathcal{B}$. The material derivative of f, denoted by $\stackrel{.}{\mathbf{f}}$, is defined as the partial derivative of the material description (3.8) of f with respect to time t, holding the label Y fixed:

$\stackrel{.}{\mathbf{f}}=\frac{\partial }{\partial t}\overline{\mathbf{f}}\left(Y,t\right).$

The quantity $\stackrel{.}{\mathbf{f}}$ can be described as the time rate change of f following a particle. It follows that the velocity v and acceleration a of the particle Y are defined by

$\mathbf{v}=\stackrel{.}{\mathbf{x}}=\frac{\partial }{\partial t}\overline{\mathbf{\chi }}\left(Y,t\right),\mathbf{a}=\stackrel{.}{\mathbf{v}}=\frac{\partial }{\partial t}\overline{\mathbf{v}}\left(Y,t\right)=\frac{{\partial }^2}{\partial t^2}\overline{\mathbf{\chi }}\left(Y,t\right).$

We now determine the referential and spatial descriptions of the material derivative $\stackrel{.}{\mathbf{f}}$, which was defined in its impractical material description in (3.11). Using (3.3), (3.9), (3.11), and the chain rule (2.95), we have

$\begin{array}{ll}\stackrel{.}{\mathbf{f}}\hfill & =\frac{\partial }{\partial t}\overline{\mathbf{f}}\left(Y,t\right)\hfill \\ \hfill & =\frac{\partial }{\partial t}\widehat{\mathbf{f}}\left(\mathbf{X},t\right)+\frac{\partial }{\partial \mathbf{X}}\widehat{\mathbf{f}}\left(\mathbf{X},t\right)\frac{\partial \mathbf{X}}{\partial t}\hfill \\ \hfill & =\frac{\partial }{\partial t}\widehat{\mathbf{f}}\left(\mathbf{X},t\right)+\frac{\partial }{\partial \mathbf{X}}\widehat{\mathbf{f}}\left(\mathbf{X},t\right)\frac{\partial }{\partial t}\kappa \left(Y\right)\hfill \\ \hfill & =\frac{\partial }{\partial t}\widehat{\mathbf{f}}\left(\mathbf{X},t\right).\hfill \end{array}$

Therefore, in the referential description, $\stackrel{.}{\mathbf{f}}$ is

$\stackrel{.}{\mathbf{f}}=\frac{\partial }{\partial t}\widehat{\mathbf{f}}\left(\mathbf{X},t\right),$

so the referential forms of the velocity v and acceleration a are

$\mathbf{v}=\stackrel{.}{\mathbf{x}}=\frac{\partial }{\partial t}\mathbf{\chi }\left(\mathbf{X},t\right),\mathbf{a}=\stackrel{.}{\mathbf{v}}=\frac{\partial }{\partial t}\widehat{\mathbf{v}}\left(\mathbf{X},t\right)=\frac{{\partial }^2}{\partial t^2}\mathbf{\chi }\left(\mathbf{X},t\right).$

These forms give the velocity and acceleration at time t of a particle Y that occupied position X in the reference configuration.

To obtain the spatial description of $\stackrel{.}{\mathbf{f}}$, we use (3.1), (3.10), (3.11), (3.12), and the chain rule (2.95):

$\begin{array}{ll}\stackrel{.}{\mathbf{f}}\hfill & =\frac{\partial }{\partial t}\overline{\mathbf{f}}\left(Y,t\right)\hfill \\ \hfill & =\frac{\partial }{\partial t}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)\frac{\partial \mathbf{x}}{\partial t}\hfill \\ \hfill & =\frac{\partial }{\partial t}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)\frac{\partial }{\partial t}\overline{\mathbf{\chi }}\left(Y,t\right)\hfill \\ \hfill & =\frac{\partial }{\partial t}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)\mathbf{v}.\hfill \end{array}$

Therefore, the spatial description of $\stackrel{.}{\mathbf{f}}$ is

$\stackrel{.}{\mathbf{f}}=\frac{\partial }{\partial t}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{f}}\left(\mathbf{x},t\right)\mathbf{v},$

so the spatial form of the acceleration a is

$\mathbf{a}=\stackrel{.}{\mathbf{v}}=\frac{\partial }{\partial t}\tilde{\mathbf{v}}\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{v}}\left(\mathbf{x},t\right)\mathbf{v}.$

This form gives the acceleration at time t of a particle Y that occupies position x at time t.

Similarly, for a scalar quantity φ, we have material, referential, and spatial descriptions of the material derivative:

$\stackrel{.}{\phi }=\frac{\partial }{\partial t}\overline{\phi }\left(Y,t\right),\stackrel{.}{\phi }=\frac{\partial }{\partial t}\widehat{\phi }\left(\mathbf{X},t\right),\stackrel{.}{\phi }=\frac{\partial }{\partial t}\tilde{\phi }\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\phi }\left(\mathbf{x},t\right)·\mathbf{v},$

and, for a tensor quantity A,

$\stackrel{.}{\mathbf{A}}=\frac{\partial }{\partial t}\overline{\mathbf{A}}\left(Y,t\right),\stackrel{.}{\mathbf{A}}=\frac{\partial }{\partial t}\widehat{\mathbf{A}}\left(\mathbf{X},t\right),\stackrel{.}{\mathbf{A}}=\frac{\partial }{\partial t}\tilde{\mathbf{A}}\left(\mathbf{x},t\right)+\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{A}}\left(\mathbf{x},t\right)\mathbf{v}.$

For a scalar-, vector-, or tensor-valued function of position and time associated with the motion of body $\mathcal{B}$, there are four possible partial derivatives, depending on if the function is considered in its referential form or its spatial form:

$\begin{array}{llll}\stackrel{.}{\phi }=\frac{\partial }{\partial t}\widehat{\phi }\left(\mathbf{X},t\right),\hfill & {\phi }^{\prime }=\frac{\partial }{\partial t}\tilde{\phi }\left(\mathbf{x},t\right),\hfill & \text{Grad}\phi =\frac{\partial }{\partial \mathbf{X}}\widehat{\phi }\left(\mathbf{X},t\right),\hfill & \text{grad}\phi =\frac{\partial }{\partial \mathbf{x}}\tilde{\phi }\left(\mathbf{x},t\right),\hfill \\ \stackrel{.}{\mathbf{f}}=\frac{\partial }{\partial t}\widehat{\mathbf{f}}\left(\mathbf{X},t\right),\hfill & {\mathbf{f}}^{\prime }=\frac{\partial }{\partial t}\tilde{\mathbf{f}}\left(\mathbf{x},t\right),\hfill & \text{Grad}\mathbf{f}=\frac{\partial }{\partial \mathbf{X}}\widehat{\mathbf{f}}\left(\mathbf{X},t\right),\hfill & \text{grad}\mathbf{f}=\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{f}}\left(\mathbf{x},t\right),\hfill \\ \stackrel{.}{\mathbf{A}}=\frac{\partial }{\partial t}\widehat{\mathbf{A}}\left(\mathbf{X},t\right),\hfill & {\mathbf{A}}^{\prime }=\frac{\partial }{\partial t}\tilde{\mathbf{A}}\left(\mathbf{x},t\right),\hfill & \text{Grad}\mathbf{A}=\frac{\partial }{\partial \mathbf{X}}\widehat{\mathbf{A}}\left(\mathbf{X},t\right),\hfill & \text{grad}\mathbf{A}=\frac{\partial }{\partial \mathbf{x}}\tilde{\mathbf{A}}\left(\mathbf{x},t\right).\hfill \end{array}$

We employ this notation throughout the remainder of the book. In words, $\stackrel{.}{\mathbf{f}}$ denotes the partial derivative of the referential description of f with respect to time (this is a restatement of (3.13)); f′ denotes the partial derivative of the spatial description of f with respect to time; Grad f denotes the gradient of the referential description of f; and grad f denotes the gradient of the spatial description of f. Also, Div f and div f are the divergence of f when considered in its referential and spatial descriptions, respectively. With use of the notation introduced in (3.19), the spatial forms of $\stackrel{.}{\phi }$, $\stackrel{.}{\mathbf{f}}$, and $\stackrel{.}{\mathbf{A}}$ (refer to (3.15), (3.17)₃, and (3.18)₃) become

$\stackrel{.}{\phi }={\phi }^{\prime }+\text{grad}\phi ·\mathbf{v},\stackrel{.}{\mathbf{f}}={\mathbf{f}}^{\prime }+\left(\text{grad}\mathbf{f}\right)\mathbf{v},\stackrel{.}{\mathbf{A}}={\mathbf{A}}^{\prime }+\left(\text{grad}\mathbf{A}\right)\mathbf{v}.$

Note that to relate the operations Grad and grad, we have, for instance,

$\text{Grad}\phi ={\mathbf{F}}^{\text{T}}\text{grad}\phi ,\text{Grad}\mathbf{f}=\left(\text{grad}\mathbf{f}\right)\mathbf{F},$

where F = Grad x = ∂x/∂X is called the deformation gradient (there is more on this in Section 3.3).

The following forms of the product rule can be verified:

$\begin{array}{lll}\stackrel{·}{\overline{\phi \mathbf{f}}}=\stackrel{.}{\phi }\mathbf{f}+\phi \stackrel{.}{\mathbf{f}},\hfill & \stackrel{·}{\overline{\mathbf{f}·\mathbf{g}}}=\stackrel{.}{\mathbf{f}}·\mathbf{g}+\mathbf{f}·\stackrel{.}{\mathbf{g}},\hfill & \stackrel{·}{\overline{\mathbf{f}\times \mathbf{g}}}=\stackrel{.}{\mathbf{f}}\times \mathbf{g}+\mathbf{f}\times \stackrel{.}{\mathbf{g}},\hfill \\ \stackrel{·}{\overline{\mathbf{Af}}}=\stackrel{.}{\mathbf{A}}\mathbf{f}+\mathbf{A}\stackrel{.}{\mathbf{f}},\hfill & \stackrel{·}{\overline{\mathbf{AB}}}=\stackrel{.}{\mathbf{A}}\mathbf{B}+\mathbf{A}\stackrel{.}{\mathbf{B}},\hfill & \stackrel{·}{\overline{\mathbf{A}·\mathbf{B}}}=\stackrel{.}{\mathbf{A}}·\mathbf{B}+\mathbf{A}·\stackrel{.}{\mathbf{B}},\hfill \end{array}$

where φ is a scalar, f and g are vectors, and A and B are tensors. We also have

$\stackrel{·}{\overline{\mathbf{A}+\mathbf{B}}}=\stackrel{.}{\mathbf{A}}+\stackrel{.}{\mathbf{B}},\stackrel{·}{\overline{{\mathbf{A}}^{\text{T}}}}={\left(\stackrel{.}{\mathbf{A}}\right)}^{\text{T}}.$

Exercises

1. In Cartesian component notation, verify the following forms of the product rule:

(a) $\stackrel{·}{\overline{\phi \mathbf{f}}}=\stackrel{.}{\phi }\mathbf{f}+\phi \stackrel{.}{\mathbf{f}}$.

(b) $\stackrel{·}{\overline{\mathbf{f}·\mathbf{g}}}=\stackrel{.}{\mathbf{f}}·\mathbf{g}+\mathbf{f}·\stackrel{.}{\mathbf{g}}$.

(c) $\stackrel{·}{\overline{\mathbf{f}\times \mathbf{g}}}=\stackrel{.}{\mathbf{f}}\times \mathbf{g}+\mathbf{f}\times \stackrel{.}{\mathbf{g}}$.

(d) $\stackrel{·}{\overline{\mathbf{Af}}}=\stackrel{.}{\mathbf{A}}\mathbf{f}+\mathbf{A}\stackrel{.}{\mathbf{f}}$.

(e) $\stackrel{·}{\overline{\mathbf{AB}}}=\stackrel{.}{\mathbf{A}}\mathbf{B}+\mathbf{A}\stackrel{.}{\mathbf{B}}$.

(f) $\stackrel{·}{\overline{\mathbf{A}·\mathbf{B}}}=\stackrel{.}{\mathbf{A}}·\mathbf{B}+\mathbf{A}·\stackrel{.}{\mathbf{B}}$.

2. In Cartesian component notation, verify that $\stackrel{·}{\overline{\mathbf{A}+\mathbf{B}}}=\stackrel{.}{\mathbf{A}}+\stackrel{.}{\mathbf{B}}$.

3. Prove in direct notation that $\stackrel{·}{\overline{{\mathbf{A}}^{\text{T}}}}={\left(\stackrel{.}{\mathbf{A}}\right)}^{\text{T}}$.

3.3 Deformation and strain

A fundamental postulate of classical continuum mechanics is that the response at a particle of the continuum to a deformation at a particular time depends only on the deformation of a small neighborhood of the particle up to that time. We will therefore devote a sizeable amount of space to (1) defining what is meant by the small neighborhood of a particle, (2) defining what is meat by deformation, and (3) describing the kinematical quantities that characterize deformation.

3.3.1 Deformation gradient

The small neighborhood of a particle Y consists of particle Y and all of its nearest neighbors. Suppose Y occupies position X in the reference configuration ${\mathcal{R}}_{\text{R}}$ and position x in the present configuration $\mathcal{R}$; refer to Figure 3.1. A vector dX is a filament of infinitesimal length dS in the reference configuration, radiating from X in the direction along unit vector N:

$\text{d}\mathbf{X}=\mathbf{N}\text{d}S.$

  (3.24)

The line element dX connects particle Y to one of its nearest neighbors in the reference configuration; see Figure 3.4. The small neighborhood of particle Y in the reference configuration ${\mathcal{R}}_{\text{R}}$ is the set of all dXs radiating from position X.

The motion x = χ (X, t) is defined for all particles in the body, i.e., for all X in ${\mathcal{R}}_{\text{R}}$. During this motion, the line element dX will be deformed—in general stretched (or compressed) and rotated, but not bent (dX is too short to be bent)—into a line element dx of length ds, radiating from the new position x of particle Y in the direction along unit vector n:

$\text{d}\mathbf{x}=\mathbf{n}\text{d}s;$

  (3.25)

see Figure 3.4. Recall from Section 3.1 that the motion x = χ (X, t) of the body maps each particle Y from its reference position X to its present position x. Since the motion is defined for all particles in the body, the deformation, or change in local geometry, of the small neighborhood of particle Y can be found by evaluating the function χ (X, t) on the particle and all of its nearest neighbors. However, a more tractable approach to quantifying the deformation of this small neighborhood involves defining a deformation measure that can be evaluated at Y alone, so the need to track the motion of all of the nearest neighbors of Y is obviated. This is accomplished through the use of a Taylor series.

A Taylor series is a mathematical tool that expresses the value of a function at one point in terms of the value of the function and its derivatives at another point. Suppose two particles that occupy points $\stackrel{1}{\mathbf{X}}$ and $\stackrel{2}{\mathbf{X}}$ in the reference configuration are mapped to points $\stackrel{1}{\mathbf{x}}$ and $\stackrel{2}{\mathbf{x}}$ in the present configuration (see Figure 3.5). A Taylor series expansion at $\stackrel{1}{\mathbf{X}}$ gives

$\stackrel{2}{\mathbf{x}}=\mathbf{\chi }\left(\stackrel{2}{\mathbf{X}},t\right)=\mathbf{\chi }\left(\stackrel{1}{\mathbf{X}},t\right)+\left[\frac{\partial \mathbf{\chi }}{\partial \mathbf{X}}\left(\stackrel{1}{\mathbf{X}},t\right)\right]\left(\stackrel{2}{\mathbf{X}}-\stackrel{1}{\mathbf{X}}\right)+\frac{1}{2}\left[\frac{{\partial }^2\mathbf{\chi }}{\partial {\mathbf{X}}^2}\left(\stackrel{1}{\mathbf{X}},t\right)\right]\left(\stackrel{2}{\mathbf{X}}-\stackrel{1}{\mathbf{X}}\right)\otimes \left(\stackrel{2}{\mathbf{X}}-\stackrel{1}{\mathbf{X}}\right)+\dots$

  (3.26)

If $\stackrel{1}{\mathbf{X}}$ and $\stackrel{2}{\mathbf{X}}$ are nearest neighbors, then the higher-order terms are negligible, leaving

$\stackrel{2}{\mathbf{x}}=\stackrel{1}{\mathbf{x}}+\left[\frac{\partial \mathbf{\chi }}{\partial \mathbf{X}}\left(\stackrel{1}{\mathbf{X}},t\right)\right]\left(\stackrel{2}{\mathbf{X}}-\stackrel{1}{\mathbf{X}}\right).$

Upon replacing $\stackrel{1}{\mathbf{X}}$ with X, $\stackrel{2}{\mathbf{X}}-\stackrel{1}{\mathbf{X}}$ with dX, and $\stackrel{2}{\mathbf{x}}-\stackrel{1}{\mathbf{x}}$ with dx, we have

$\text{d}\mathbf{x}=\left[\frac{\partial \mathbf{\chi }}{\partial \mathbf{X}}\left(\mathbf{X},t\right)\right]\text{d}\mathbf{X}.$

We label

$\mathbf{F}\equiv \frac{\partial \mathbf{\chi }\left(\mathbf{X},t\right)}{\partial \mathbf{X}}=\text{Grad}\mathbf{\chi }=\text{Grad}\mathbf{x},$

  (3.27)

so

$\text{d}\mathbf{x}=\mathbf{F}\text{d}\mathbf{X},$

  (3.28)

where F is called the deformation gradient.

Using definition (3.28), one can verify that F satisfies requirements (2.7) and is thus a tensor. Hence, F linearly maps each line element dX radiating from X in the reference configuration into a line element dx radiating from x in the present configuration. (dX is not bent when it is mapped by F to dx.) Therefore F describes the mapping of the small neighborhood of X to the small neighborhood of x, and thus quantifies the changes in the relative positions of nearest neighbors in the body; see Figure 3.6. In other words, F, a function evaluated only at the point X, is a measure of the deformation, or change of local geometry, in the neighborhood of X as it moves to x.

Since F is a linear map, if the small neighborhood of X in the reference configuration is a sphere, then the deformation maps it into an ellipsoid in the present configuration; see Figure 3.6. (If a spherical volume in the reference configuration deforms into a shape other than an ellipsoid, then it is too large to qualify as a small neighborhood, thereby invalidating our truncation of the series (3.26).) Note that although F is a tensor, which is by definition a linear map, the deformation is not restricted to be “linear,” in the sense of infinitesimal. The stretch and rotation of the map from dX to dx can be large.

The tensor F, in general, is a function of position and time, i.e.,

$\mathbf{F}=\widehat{\mathbf{F}}\left(\mathbf{X},t\right)=\tilde{\mathbf{F}}\left(\mathbf{x},t\right).$

  (3.29)

If F is independent of position, i.e., F = F(t), the deformation is said to be homogeneous. If F is independent of time, i.e., $\mathbf{F}=\widehat{\mathbf{F}}\left(\mathbf{X}\right)=\tilde{\mathbf{F}}\left(\mathbf{x}\right)$, the deformation is said to be static.

Exercises

1. Prove that the deformation gradient F is a tensor.

Exercises

1. Prove in direct notation that $\frac{1}{{\lambda }^2}=\mathbf{n}·\mathbf{c}\mathbf{n}=\mathbf{c}·\left(\mathbf{n}\otimes \mathbf{n}\right)$.

2. Prove in direct notation that the Cauchy deformation tensor c is symmetric and positive definite.

3.3.3 Polar decomposition, stretch tensors, rotation tensor

We return to our discussion of the deformation gradient F. Since F is invertible (see (3.37)), it can be decomposed using the polar decomposition theorem (refer to Section 2.3) into the form

$\mathbf{F}=\mathbf{RU}$

  (3.41a)

or

$\mathbf{F}=\mathbf{VR},$

  (3.41b)

where R is an orthogonal tensor (it will be shown in Sections 3.6 and 4.12 that det F > 0, so R is in fact proper orthogonal), and U and V are symmetric, positive-definite tensors. For physical reasons to be revealed shortly, the tensors R, U, and V are called the rotation tensor, right stretch tensor, and left stretch tensor, respectively.

The effect of the multiplicative decompositions (3.41a) and (3.41b) is to replace the linear transformation (3.28) by either of two pairs of linear transformations:

$\text{d}\mathbf{x}=\mathbf{F}\text{d}\mathbf{X}=\left(\mathbf{RU}\right)\text{d}\mathbf{X}=\mathbf{R}\left(\mathbf{U}\text{d}\mathbf{X}\right),$

i.e.,

$\text{d}{\mathbf{X}}^{*}=\mathbf{U}\text{d}\mathbf{X}\text{followed by}\text{d}\mathbf{x}=\mathbf{R}\text{d}{\mathbf{X}}^{*},$

  (3.42a)

or

$\text{d}\mathbf{x}=\mathbf{F}\text{d}\mathbf{X}=\left(\mathbf{VR}\right)\text{d}\mathbf{X}=\mathbf{V}\left(\mathbf{R}\text{d}\mathbf{X}\right),$

i.e.,

$\text{d}{\mathbf{X}}^{*}=\mathbf{R}\text{d}\mathbf{X}\text{followed by}\text{d}\mathbf{x}=\mathbf{V}\text{d}{\mathbf{x}}^{*};$

  (3.42b)

see Figure 3.7. We emphasize that (3.42a) and (3.42b) involve sequential (or serial) mappings, a distinguishing feature of multiplicative decompositions. Physically, it can be shown that (3.42a) represents stretch followed by pure rotation (refer to Problem 3.6), while (3.42b) represents pure rotation followed by stretch (refer to Problem 3.7).

Problem 3.5

Prove in direct notation that C = U².

Solution

From definition (3.34) and the polar decomposition (3.41a), we have

$\mathbf{C}={\mathbf{F}}^{\text{T}}\mathbf{F}={\left(\mathbf{RU}\right)}^{\mathbf{T}}\left(\mathbf{RU}\right).$

It follows from result (2.14)₂ that

$\mathbf{C}={\mathbf{U}}^{\text{T}}{\mathbf{R}}^{\text{T}}\mathbf{RU}.$

Recall that the rotation tensor R is orthogonal, so R^TR = I (refer to (2.51)). It follows that

$\mathbf{C}={\mathbf{U}}^{\text{T}}\mathbf{IU}={\mathbf{U}}^{\text{T}}\mathbf{U}=\mathbf{UU}\equiv {\mathbf{U}}^2$

since U is symmetric.

Problem 3.6

Prove in direct notation that decomposition (3.42a) represents stretch followed by pure rotation. That is, show that the deformation (3.42a)₁ of dX into dX* involves stretching (and perhaps rotation), and the deformation (3.42a)₂ of dX* into dx is rotation without stretch (i.e., pure rotation).

Solution

To prove this, we demonstrate that the intermediate element dX* has the same length ds as the deformed element dx; this, in turn, implies that the part of the deformation (3.42a) described by the map U contains all of the stretch of the total deformation from dX to dx, so the part of the deformation (3.42a) described by the map R is pure rotation. To begin, we have

${\left|\text{d}{\mathbf{X}}^{*}\right|}^2=\text{d}{\mathbf{X}}^{*}·\text{d}{\mathbf{X}}^{*}=\mathbf{U}\text{d}\mathbf{X}·\mathbf{U}\text{d}\mathbf{X}.$

Using definitions (2.11) and (2.13), and noting that U is symmetric, we find that

${\left|\text{d}{\mathbf{X}}^{*}\right|}^2=\text{d}\mathbf{X}·{\mathbf{U}}^{\text{T}}\left(\mathbf{U}\text{d}\mathbf{X}\right)=\text{d}\mathbf{X}·\left({\mathbf{U}}^{\text{T}}\mathbf{U}\right)\text{d}\mathbf{X}=\text{d}\mathbf{X}·{\mathbf{U}}^2\text{d}\mathbf{X}.$

It follows that

${\left|\text{d}{\mathbf{X}}^{*}\right|}^2=\mathbf{N}·{\mathbf{U}}^2\mathbf{N}\text{d}{S}^2=\mathbf{N}·\mathbf{CN}\text{d}{S}^2={\text{λ}}^2\text{d}{S}^2={\left(\text{λd}S\right)}^2=\text{d}{s}^2,$

where we have used (3.24), (3.30), (3.33), and C = U² (refer to Problem 3.5). We then conclude that

$\left|\text{d}{\mathbf{X}}^{*}\right|=\left|\text{d}\mathbf{x}\right|=\text{d}s.$

Problem 3.7

Prove in direct notation that decomposition (3.42b) represents pure rotation followed by stretch. That is, show that the deformation (3.42b)₁ of dX into dx* is pure rotation (rotation without stretch), and the deformation (3.42b)₂ of dx* into dx involves stretching (and perhaps rotation).

Solution

To prove this, we demonstrate that the intermediate element dx* has the same length dS as the undeformed element dX; this, in turn, implies that the part of the deformation (3.42b) described by the map R is only rotation, and all of the stretch of the total deformation from dX to dx is contained in the map V. To begin, we have

${\left|\text{d}{\mathbf{x}}^{*}\right|}^2=\text{d}{\mathbf{x}}^{*}·\text{d}{\mathbf{x}}^{*}=\mathbf{R}\text{d}\mathbf{X}·\mathbf{R}\text{d}\mathbf{X}.$

Since R is orthogonal, it follows from definition (2.50) that

${\left|\text{d}{\mathbf{x}}^{*}\right|}^2=\mathbf{R}\text{d}\mathbf{X}·\mathbf{R}\text{d}\mathbf{X}=\text{d}\mathbf{X}·\text{d}\mathbf{X}={\left|\text{d}\mathbf{X}\right|}^2,$

so

$\left|\text{d}{\mathbf{x}}^{*}\right|=\left|\text{d}\mathbf{X}\right|=\text{d}S.$