Rate of Deformation

Let us consider a material element $\mathbf{P}^\prime \mathbf{Q}^\prime$ emanating from a material point located at $P^\prime (x_1,x_2,x_3)$ in the present configuration at time $t$. We wish to compute $$\frac{D(\mathbf{P}^\prime\mathbf{Q}^\prime)}{Dt}$$, the rate of change of length and direction of $\mathbf{P}^\prime \mathbf{Q}^\prime$. Let $P^\prime$ be the deformed position of the material point located at $P(X_1,X_2,X_3)$ in the reference configuration and $Q^\prime$ that of $Q(X_A + dX_A)$. Then

$$(P^\prime Q^\prime )_i = x_i (X_A + dX_A,t) - x_i (X_A,t).$$

Thus,

$$\begin{split}\frac{D}{Dt}(P^\prime Q^\prime)_i =\ & v_i(X_A +... ...partial v_i}{\partial X_A}dX_A,\\ =\ & v_{i,A}dX_A. \end{split}$$ (3.12.1)

Equation (3.12.1) expresses $$\frac{D(\mathbf{P}^\prime\mathbf{Q}^\prime)}{Dt}$$ in a material description. To obtain $$\frac{D(\mathbf{P}^\prime\mathbf{Q}^\prime)}{Dt}$$ in a spatial description, we note that since $v_i(X_A,t)$ is the velocity of a material point $P$ presently at the position $x_i$, therefore, if a spatial description of velocity is employed, this velocity is given by $v_i = v_i(x_j,t)$. Note that $v_i(x_j,t)$ and $v_i(X_A,t)$ are, in general, different functions. Thus,

$$\frac{D(\mathbf{P}^\prime\mathbf{Q}^\prime)_i}{Dt} =\ v_i(x_j + dx_j,t) - v_i(x_j,t),$$

$$\simeq\ \frac{\partial v_i}{\partial x_j}dx_j = v_{i,j}dx_j,$$ (3.12.2)

$$=\ [1/2(v_{i,j}+v_{j,i}) + 1/2 (v_{i,j} - v_{j,i})]dx_j,$$

$$=\ (D_{ij} + W_{ij})dx_j,$$ (3.12.3)

where we have set

$$D_{ij} =\ 1/2 (v_{i,j} + v_{j,i}),$$ (3.12.4)

$$W_{ij} = \ 1/2 (v_{i,j} - v_{j,i}).$$

$D_{ij}$, the symmetric part of the velocity gradient $v_{i,j}$, is known as the rate of deformation tensor or the strain-rate tensor, and $W_{ij}$, the antisymmetric part of the velocity gradient $v_{i,j}$, is known as the spin tensor.

In the following, we give a geometric interpretation of the elements of $D_{ij}$. Let

$$\mathbf{P}^\prime\mathbf{Q}^\prime = ds\mathbf{n}$$

where $\mathbf{n}$ is a unit vector in the direction of $\mathbf{P}^\prime\mathbf{Q}^\prime = d\mathbf{x}$. Then

$$\frac{D}{Dt} (ds^2) =\ \frac{D}{Dt} (dx_idx_i) = 2dx_i\frac{D}{Dt}(dx_i)\ ,$$

$$\hspace{.2in} ds\frac{D}{Dt}(ds) =\ dx_i(D_{ij} + W_{ij})dx_j\ ,$$

where we have substituted for $$\frac{D}{Dt}(dx_i)$$ from (3.12.3). Since $W_{ij} = -W_{ji}$, therefore

$$dx_iW_{ij}dx_j = 0\ ,$$

and we get

$$ds\frac{D}{Dt}(ds) = dx_iD_{ij}dx_j\ ,$$

or

$$\frac{1}{ds}\frac{D(ds)}{Dt} = n_iD_{ij}n_j\ .$$ (3.12.6)

If $\mathbf{n} = (1,0,0)$, the right-hand side of (3.12.6) equals $D_{11}$. Thus $D_{11}$ gives the rate of change of length per unit length known as stretching or rate of extension for a material line presently parallel to $x_1$-axis. Similarly $D_{22}$ and $D_{33}$ give, respectively, the stretching of a material line presently in the $x_2$- and $x_3$-direction. To obtain a physical interpretation of the off-diagonal elements of $D_{ij}$, let

$$\mathbf{P}^\prime\mathbf{Q}^\prime = ds_1\mathbf{n} \quad \mathbf{P}^\prime\mathbf{R}^\prime = ds_2\mathbf{m}\ .$$

Then

$$\frac{1}{ds_1ds_2}\frac{D}{Dt}(\mathbf{P}^\prime\mathbf{Q}^\prime... ...rime\mathbf{R}^\prime) = m_i (D_{ij} + W_{ij})n_j + n_i (D_{ij} + W_{ij})m_j\ ,$$

$$\frac{1}{ds_1ds_2}\frac{D(ds_1ds_2)}{Dt}\cos\theta_{(m,n)} -\sin\theta_{(m,n)}\dot\theta_{(m,n)} = 2m_iD_{ij}n_j\ .$$ (3.12.7)

Here $\theta_{(m,n)}$ is the angle between directions $\mathbf{m}$ and $\mathbf{n}$. Thus for $\mathbf{m} = (1,0,0)$ and $\mathbf{n} = (0,1,0)$, the right-hand side of (3.12.7) equals $2D_{12}$ and the left-hand side equals $-\dot\theta_{(m,n)}$. Hence $2D_{12}$ equals the rate of decrease of angle from $$\frac{\pi}{2}$$ of two line elements presently parallel to $x_1$ and $x_2$-axes, known as shearing or rate of shear. A similar interpretation holds for $D_{23}$ and $D_{31}$.

Since $D_{ij}$ is symmetric, we also have the result that there always exist three mutually perpendicular directions (eigenvectors of $D_{ij}$) along which the stretching (an eigenvalue of $D_{ij}$) is either maximum or minimum among stretchings for all differential elements extending from the material particle which currently is at $P^\prime$.

Note that the strain-rate tensor does not, in general, equal the time rate of change of the strain tensor. To prove this, we first conclude from eqns. (3.12.2) and (3.7.7) that

$$\dot F_{iA} dX_A = v_{i,j} F_{jA} dX_A\ ,$$

which must hold for all choices of $dX_A$. Thus

$$v_{i,j} = \dot F_{iA}(F^{-1})_{Aj},\ \ \mathbf{L} = \dot\mathbf{F}\mathbf{F}^{-1}\ ,$$ (3.12.8)

where $\mathbf{L}$ is the spatial velocity gradient. Differentiation of both sides of (3.8.2) with respect to time and the definition (3.8.6)$_1$ of the strain tensor $\mathbf{E}$ gives

$$\dot\mathbf{E} =\ 2\dot\mathbf{C} = \dot\mathbf{F}^T\mathbf{F} + \mathbf{F}^T\dot\mathbf{F} = \mathbf{F}^T(\mathbf{L}^T + \mathbf{L})\mathbf{F}\ ,$$

$$=\ 2\mathbf{F}^T\mathbf{D}\mathbf{F}\ .$$ (3.12.9)

Hence $\dot\mathbf{E} \ne \mathbf{D}$, and the strain-rate tensor $\mathbf{D}$ should not be confused with the time rate of change of the strain tensor $\mathbf{E}$.

We now attempt to provide a physical interpretation of the spin tensor. Let $\mathbf{n}$ be a unit eigenvector of $\mathbf{D}$, i.e., $\mathbf{D}\mathbf{n} = \lambda\mathbf{n}$ where $\lambda$ is the eigenvalue corresponding to $\mathbf{n}$. From (3.12.3), we have

$$(dsn_i)^\cdot =\ \dot{\overline{ds}}n_i + ds\dot n_i = (D_{ij} + W_{ij})dsn_j\ ,$$

$$\dot n_i =\ D_{ij} n_j + W_{ij} n_j - (n_kD_{kl}n_l)n_i\ ,$$

$$=\ W_{ij} n_j\ ,$$ (3.12.10)

where we have used (3.12.6). It states that the spin tensor operating on a unit eigenvector of $\mathbf{D}$ gives the rate of change of that unit vector. Thus the spin equals the angular velocity of the principal axes of stretching.

The axial vector

$$w_i = -\epsilon_{ijk}W_{jk}\ \ \ \mathbf{w} = \mathbf{v}\ ,$$

is the vorticity vector; its direction is the axis of spin, and its magnitude is the vorticity magnitude, $w$.

$$w = \sqrt{w_iw_i} = \sqrt{sW_{jk}W_{jk}}\ .$$

We now show that the spin tensor defined by (3.12.4)$_2$ does not equal the rate of change of the rotation matrix $\mathbf{R}$. Equations (3.12.9) and (3.13.1)$_1$ give

$$\mathbf{L} = (\dot\mathbf{R}\mathbf{U} + \mathbf{R}\dot\mathbf{U})(\mathbf{U}^{-1}\mathbf{R}^T)\ ,$$ (3.13.15)

or

$$\mathbf{D} + \mathbf{W} =\ \dot\mathbf{R}\mathbf{R}^T + \mathbf{R}\dot\mathbf{U}\mathbf{U}^{-1}\mathbf{R}^T\ ,$$

$$=\ \dot\mathbf{R}\mathbf{R}^T + \frac{1}{2}\mathbf{R}(\dot\mathbf{U}... ... (\dot\mathbf{U}\mathbf{U}^{-1} - \mathbf{U}^{-1}\dot\mathbf{U})\mathbf{R}^T\ ,$$ (3.13.16)

where we have added and subtracted $$\frac{1}{2} \mathbf{R}\mathbf{U}^{-1}\dot\mathbf{U}\mathbf{R}^T$$ to the right-hand side. Equating symmetric and skew-symmetric tensors on both sides, we obtain

$$\begin{split}\mathbf{D} =\ & \frac{1}{2}\mathbf{R}(\dot\mathb... ...{-1} - \mathbf{U}^{-1}\dot\mathbf{U})\mathbf{R}^T\ .\end{split}$$ (3.13.17)

These equations clearly evince that $\mathbf{D}\ne \dot\mathbf{U}$ and $\mathbf{W}\ne \dot\mathbf{R}$.

Taking the material derivative of both sides of eqn. (3.10.2) we obtain

$$\frac{D}{Dt} (dv) =\ \dot J\, dV,$$ (3.12.11)

$$=\ J\left(\frac{\partial v_i}{\partial x_i}\right)\, dV,$$ (3.12.12)

$$=\ \frac{\partial v_i}{\partial x_i}\, dv,$$

and, therefore,

$$\frac{1}{dv}\frac{\cdot}{dv} = \frac{\partial v_i}{\partial x_i} = D_{ii} = I_D.$$ (3.12.13)

In going from (3.12.11) to (3.12.12) we used

$$\dot J = J\frac{\partial v_i}{\partial x_i} = JI_D$$

which can be obtained from eqns. (3.11.4), (3.11.5) and (3.11.6). Thus the first principal invariant $I_D$ of the rate of deformation tensor $D_{ij}$ gives the rate of change of volume per unit volume.

We now derive an expression for the rate of change of an area element. Rewriting eqn. (3.10.1) as $\mathbf{F}^Td\mathbf{a} = Jd\mathbf{A}$ and taking the time derivative of both sides, we obtain

$$\dot\mathbf{F}^Td\mathbf{a} + \mathbf{F}^T\dot{\overline{d\mathbf{a}}} = \dot Jd\mathbf{A}\ ,$$

$$\mathbf{F}^T\mathbf{L}^Td\mathbf{a} + \mathbf{F}^T\dot{\overline{d\mathbf{a}}} = JI_Dd\mathbf{A}\ ,$$

$$\noalign{\mbox{or}} \dot{\overline{d\mathbf{a}}} = (I_D - \mathbf{L}^T)d\mathbf{a},\ ... ...da_i}} = \left(I_D\delta_{ij} - \frac{\partial v_j}{\partial x_i}\right)da_j\ .$$ (3.12.14)

In an isochoric deformation, $\dot{\overline{dv}} = 0$ but $\dot{\overline{d\mathbf{a}}}\ne \mathbf{0}$ in general.

Example: Given the velocity field

$$v_i = 2x_2\delta_{1i},$$

find

(a)

the rate of deformation and the spin tensors,

(b)

the rate of extension per unit length of the line element $\mathbf{P}^\prime\mathbf{Q}^\prime = 10^{-2}(1,2,0,)$,

(c)

the maximum and the minimum rates of extension.

Solution

(a)

The matrix of the velocity gradient is

$$[v_{i,j}] = \left[\begin{array}{ccc} 0 & 2 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{array}\right]\ .$$

Therefore,

$$D_{ij} =\ 1/2[v_{i,j} + v_{j,i}] = \left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{array}\right],$$

$$W_{ij} = \ 1/2[v_{i,j} - v_{j,i}] = \left[\begin{array}{rcc} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{array}\right].$$

(b)

Given $\mathbf{P}^\prime\mathbf{Q}^\prime = 10^{-2}(1,2,0) = 10^{-2}(\sqrt 5 ) \left(\displaystyle\frac{1}{\sqrt 5},\frac{2}{\sqrt 5},0\right)$. Thus $\mathbf{n} = (1/\sqrt 5,\ 2/\sqrt 5,\ 0)$. From eqn. (3.12.6)

$$\frac{1}{ds}\frac{D(ds)}{Dt} =\ n_iD_{ij}n_j = \{\mathbf{n}\}^T [\mathbf{D}]\{\mathbf{n}\}$$

$$=\ (1/\sqrt 5\ \ 2/\sqrt 5\ \ 0)\left[\begin{array}{ccc}0 & 1 & 0\\ ... ...ght]\left[\begin{array}{c} 1/\sqrt 5\\ 2/\sqrt 5\\ 0\end{array}\right] = 4/5\ .$$

(d)

From the characteristic equation

$$\det [D_{ij} - \lambda\delta_{ij}] = 0,$$

we determine the eigenvalues of the tensor $D_{ij}$ as $\lambda = 0,\ \pm 1$. Therefore, 1 is the maximum and $-1$ the minimum rate of extension. The eigenvector $\mathbf{n}^{(1)}$ (for $\lambda_1 = 1$) determined from

$$D_{ij}n^{(1)}_j - \lambda_1\delta_{ij}n^{(1)}_j = 0$$

$$\hspace{.2in} n^{(1)}_jn^{(1)}_j = 1$$

is $\mathbf{n}^{(1)} = (1/\sqrt 2) (1,1,0)$. Similarly $\mathbf{n}^{(2)}$ corresponding to $\lambda_2 = -1$ is $(1/\sqrt 2)(1,-1,0)$. The third eigenvector is $\mathbf{n}^{(3)} = (0,0,1)$.

Exercise: For the velocity field

$$v_i = 2x^2_2\delta_{i1},$$

find

a)

the rate of extension per unit length of a material line element which in the present configuration is given by $10^{-2}(1,1,0)$ through the point $(5,3,0)$,

b)

the deformation corresponding to the given velocity field if $x_i = X_i$ at time $t = 0$,

c)

the principal stretches and the deformed position of the axis of the maximum principal stretch for the material particle which at $t = 1$ is at $(0,1/2,0)$,

d)

principal stretchings and their axes for the material particle which currently is at $(0,1/2,0)$.