Deformation Gradient

As pointed out earlier, in continuum mechanics, two different material particles always occupy two distinct places. Consider two material particles $P(X_i)$ and $Q(X_i + dX_i)$.

Here $(X_i)$ stands for the triplet $(X_1,\ X_2,\ X_3)$ used to identify particles in the reference configuration. Let points $P$ and $Q$ move to $P^\prime$ and $Q^\prime$ respectively so that the vector $\mathbf{P}\mathbf{Q}$ is deformed into the vector $\mathbf{P}^\prime \mathbf{Q}^\prime$. The vector $\mathbf{P}^\prime \mathbf{Q}^\prime$ need not, and in general will not, equal $\mathbf{P}\mathbf{Q}$. This means that the length and direction of $\mathbf{P}^\prime \mathbf{Q}^\prime$ may be different from that of $\mathbf{P}\mathbf{Q}$. Given $\mathbf{P}\mathbf{Q}$ or $\mathbf{P}^\prime \mathbf{Q}^\prime$ and the motion, our problem is to find the other vector.

Let the motion of the body be given by

$$x_i = x_i (X_1,\ X_2,\ X_3,\ t).$$ (3.7.1)

Then

$$\mathbf{P}^\prime\mathbf{Q}^\prime = [x_i (X_1 + dX_1,X_2 + dX_2,X_3 + dX_3,t) - x_i (X_1,X_2,X_3,t)]\mathbf{e}_i.$$ (3.7.2)

Using Taylor's theorem for series expansion, we get

$$\mathbf{P}^\prime\mathbf{Q}^\prime =\ \left[\frac{\partial x_i}{\partial X_1} dX_1 + \frac{\partial x_i... ...rac{\partial x_i}{\partial X_3} dX_3\right] \mathbf{e}_i + O (\vert dX\vert^2),$$ (3.7.3)

$$=\ \frac{\partial x_i}{\partial X_j} dX_j\mathbf{e}_i + O(\vert dX\vert^2).$$ (3.7.4)

If points $P$ and $Q$ are close together in the reference configuration, then higher order terms can be neglected as compared to terms linear in $dX_i$. Using this approximation, we get

$$\mathbf{P}^\prime\mathbf{Q}^\prime = \left[\frac{\partial x_i}{\partial X_j}\bigg\vert _P dX_j\right]\mathbf{e}_i.$$ (3.7.5)

Here the suffix $\big\vert _P$ reminds us that $$\frac{\partial x_i}{\partial X_j}$$ is evaluated at the point $P$. In section 2.9, it was mentioned that a comma followed by an index $i$ will be used to indicate partial differentiation with respect to $x_i$. Now if we were to use that notation, then it will not be clear whether the partial differentiation is with respect to $x_i$ or $X_i$. To clarify the situation we will, henceforth, use upper case latin indices for $X$. That is, the triplet $(X_1,\ X_2,\ X_3)$ will be denoted by $X_A$ instead of $X_i$ and eqn. (3.7.5) will be written as

$$\mathbf{P}^\prime\mathbf{Q}^\prime = \frac{\partial x_i}{\partial X_A}\bigg\vert _PdX_A \mathbf{e}_i = F_{iA}\big\vert _PdX_A\mathbf{e}_i.$$ (3.7.6)

In component form,

$$(P^\prime Q^\prime)_j =\ \mathbf{P}^\prime\mathbf{Q}^\prime \cdot\mathbf{e}_j = F_{iA\vert _P}dX_A\mathbf{e}_i \cdot\mathbf{e}_j\ ,$$

$$=\ F_{jA}\big\vert _P dX_A\ ,$$

$$=\ F_{jA}\big\vert _P(PQ)_A\ .$$ (3.7.7)

Thus $F_{jA}$ relates the components of vector $\mathbf{P}\mathbf{Q}$ in the reference configuration to the components of the vector $\mathbf{P}^\prime \mathbf{Q}^\prime$ into which $\mathbf{P}\mathbf{Q}$ is deformed. Since in Continuum Mechanics we assume that

$$J = \det \left\vert F_{iA}\right\vert > 0$$

therefore, $F_{jA}$ is an invertible matrix. This implies that once the motion (3.7.1) is known, we can find $\mathbf{P}^\prime \mathbf{Q}^\prime$ from $\mathbf{P}\mathbf{Q}$ and vice-a-versa.

Since the motion (3.7.1) gives how the body deforms and $F_{iA}$ is the gradient of the motion, $F_{iA}$ is called deformation gradient. $F_{iA}$ relates a vector $\mathbf{P}\mathbf{Q}$ in the reference configuration to the vector $\mathbf{P}^\prime \mathbf{Q}^\prime$ (in the present configuration) into which $\mathbf{P}\mathbf{Q}$ is deformed. In terms of the displacement $\mathbf{u}$, $F_{iA}$ can be written as follows.

$$\begin{split}u_i =\ & x_i - X_A\delta_{iA}\ ,\\ \frac{\partia... ...\ ,\\ F_{iA} =\ & x_{i,A}= \delta_{iA} + u_{i,A}\ . \end{split}$$ (3.7.8)

The gradient $u_{i,A}$ of the displacement $\mathbf{u}$ is known as the displacement gradient.

Example: The deformation of a body is given by

$$u_1 = (3X^2_1 + X_2),\ u_2 = (2X^2_2 + X_3),\ u_3 = (4X^2_3 + X_1).$$

Compute the vector into which the vector $10^{-2}\left(\displaystyle\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3}\right)$ passing through the point $(1,1,1)$ in the reference configuration is deformed.

Solution

Here the point $P$ is $(1,1,1)$ and the vector $\mathbf{P}\mathbf{Q}$ has components $$\left(\frac{10^{-2}}{3},\frac{10^{-2}}{3},\frac{10^{-2}}{3}\right )$$. The deformation gradient at any point is given by

$$\left[F_{iA}\right] =\left[\begin{array}{ccc}1 + 6X_1 & 1 & 0\\ 0 & 1 + 4X_2 & 1\\ 1 & 0 & 1 + 8X_3\end{array}\right]\ .$$

The deformation gradient evaluated at point $P$ is

$$\left[F_{iA}\right]\big\vert _P = \left[\begin{array}{ccc} 7 & 1 & 0\\ 0 & 5 & 1\\ 1 & 0 & 9\end{array}\right]\ .$$

Therefore, the components of the vector $\mathbf{P}^\prime \mathbf{Q}^\prime$ are given by

$$\left[\begin{array}{c} (P^\prime Q^\prime )_1\\ \noalign{\vski... ...ign{\vskip .1in} 6\\ \noalign{\vskip .1in} 10\end{array}\right].$$

Note that, in the preceding example, the vector $\mathbf{P}^\prime \mathbf{Q}^\prime$ is neither parallel to $\mathbf{P}\mathbf{Q}$ nor is it equal to $\mathbf{P}\mathbf{Q}$ in magnitude. The ratio of the length of $\mathbf{P}^\prime \mathbf{Q}^\prime$ to that of $\mathbf{P}\mathbf{Q}$ is called stretch in the direction of $\mathbf{P}\mathbf{Q}$.

It is emphasized that in going from (3.7.4) to (3.7.5) we assumed that the length of the vector $\mathbf{P}\mathbf{Q}$ is infinitesimal. However, no assumption was made as to the magnitude¹ of $F_{iA}$. Thus (3.7.5) is valid no matter how large or small the components of deformation gradient are. Said differently (3.7.5) is valid both for small and large deformations so long as we study the deformation of infinitesimal vectors passing through the point $P$. Thus $F_{iA}$ describes the deformation of material particles in an infinitesimal neighborhood of $P$.

A special case in which (3.7.5) follows exactly from (3.7.4) is that when $F_{iA}$ is constant. That is, each of the nine quantities $\partial x_i/\partial X_A$ is a constant. A deformation for which $F_{iA}$ is a constant is called a homogeneous deformation.

Example: Given the following displacement components

$$u_1 = 0\cdot 1 X^2_2,\, u_2 = u_3 = 0.$$

a)

Is this deformation possible in a continuously deformable body? Prove your answer.

b)

Find the deformed vectors of the material vectors $0.01\mathbf{e}$, and $0.015\mathbf{e}_2$ which pass through the point $P\ (1,1,0)$ in the reference configuration.

c)

Determine the stretches at the point $(1,1,0)$ in the $X_1$ and $X_2$ directions.

d)

Determine the change in angle between lines through the point $P\ (1,1,0)$ that were parallel to $X_1$ and $X_2$ axes.

Solution:

a)

$$\left[F_{iA}\right] =\ \left[\begin{array}{ccc}1 & 0.2X_2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]$$

$$\det\left[F_{iA}\right] =\ 1 > 0.$$

Therefore, the given deformation is possible in a continuously deformable body.

b)

The material vector $0\cdot 01\mathbf{e}_1$ through the material point $(1,1,0)$ is deformed into the vector given by

$$\left[\begin{array}{ccc} 1 & 0.2(1) & 0\\ 0 & 1 & 0\\ 0 & 0 & 1... ...\end{array}\right] = \left[\begin{array}{c} 0.01\\ 0\\ 0\end{array}\right]\ ,$$

and the material vector $0.015\mathbf{e}_2$ through the material point $(1,1,0)$ is deformed into

$$\left[\begin{array}{ccc} 1 & 0.2(1) & 0\\ 0 & 1 & 0\\ 0 & 0 & 1... ...array}\right] = \left[\begin{array}{c} 0.003\\ 0.015\\ 0\end{array}\right]\ .$$

c)

Stretch at $P\ (1,1,0)$ in $X_1$-direction$= \displaystyle\frac{0.01}{0.01} = 1$.
Stretch at $P\ (1,1,0)$ in $X_2$-direction$= \displaystyle\frac{\sqrt{0.003^2 + .015^2}}{0.015} = 1.02$.

d)

Angle between the vectors into which vectors $0.01\mathbf{e}_1$ and $0.015\mathbf{e}_2$ through the point $(1,1,0)$ are deformed $= \cos^{-1}\displaystyle\frac{(0.01)(0.003 + 0 + 0)}{(0.01)(0.003^2 + 0.015^2)^{1/2}}= 78.7^\circ$.
Change in angle $= 11.3^\circ$.

Example: Given the following displacement components

$$u_1 = 2X^2_1 + X_1X_2,\ u_2 = X^2_2,\ u_3 = 0,$$

and that for points in the body, $X_1\ge 0,\ X_2\ge 0$.

a)

Find the vector in the reference configuration that ends up parallel to $x_1$-axis through the point $(1,0,0)$ in the current configuration.

b)

Find stretch of a line element that ends up parallel to $x_1$-axis through the point $(1,0,0)$ in the present configuration.

Solution: For the given displacement components,

$$x_1 = X_1 + 2X^2_1 + X_1X_2,\ x_2 = X_2 + X^2_2,\ x_3 = X_3.$$

Therefore the undeformed position of point $(1,0,0)$ is obtained by solving

$$1 =\ X_1 + 2X^2_1 + X_1X_2,$$

$$0 =\ X_2 + X^2_2,$$

$$0 =\ X_3.$$

The solution of these equations which satisfies $X_1\ge 0$ is $(1/2,0,0)$. The material particle which currently is at $P^\prime (1,0,0)$ was at $P(1/2,0,0)$ in the reference configuration.

$$\left[F_{iA}\right] =\ \left[\begin{array}{ccc} 1+4X_1+X_2 & X_1 & 0\\ 0 & 1 + 2X_2 & 0\\ 0 & 0 & 1\end{array}\right]$$

$$\left[F_{iA}\big\vert _P\right] =\ \left[\begin{array}{ccc} 3 & 0.5 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]$$

$$\therefore \hspace*{.2in} \left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] =\ \left[\begin{array}{ccc} 3 & 0.5 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{c} dX_1 \\ dX_2\\ dX_3\end{array}\right]$$

where $(dX_1,dX_2,dX_3)$ is the vector in the reference configuration that ends up into a vector $(1,0,0)$ through the point $(1,0,0)$ in the current configuration.

$$\left[\begin{array}{c} dX_1\\ dX_2\\ dX_3\end{array}\right] = 1... ... 0\end{array}\right] = \left[\begin{array}{c} 1/3\\ 0 \\ 0\end{array}\right].$$

Thus a vector parallel to $x_1$-axis through $(1/2,0,0)$ in the reference configuration ends up into a vector parallel to $x_1$-axis through $(1,0,0)$ in the current configuration.

b)

Stretch along the desired line $=\displaystyle\frac{\sqrt{ 1^2+0+0}}{\sqrt{(1/3)^2 + 0+0}} = 3$.

Exercise: Given the displacement field

$$u_1 = 10^{-2}(2X_1 + X^2_2),\ u_2 = 10^{-2}(X^2_1 - X^2_2), \ u_3 = 0.$$

Find the stretches and the change of angle for the material lines $(0.1,0,0)$ and $(0,0.1,0)$ that emanate from the material particle $(1,-1,0)$.

Exercise: The displacement components for a body are

$$u_1 = 2X_1 + X_2,\ u_2 = X_3,\ u_3 = X_3 - X_2\ .$$

a)

Verify that this displacement vector is possible for a continuously deformed body.

b)

Is this deformation homogeneous?

c)

Determine the stretch in the direction $(1/3,\ 1/3,\ 1/3)$ through the point $(1,1,1)$ in the reference configuration.

d)

Determine the direction cosines of the line element in the reference configuration that ends up in the $x_3$-direction at the place $(1,1,0)$.

e)

Determine the change in angle between the lines through the point $(1,1,1)$ (in the reference configuration) whose directions in the reference configuration are $1,0,0$ and $1/3,\ 1/3,\ 1/3$.