3. Kinematics

 

 

 

 

3.1 Basic Assumptions

 

Continuum mechanics is a combination of mathematics and physical laws that approximate the large-scale behavior of matter that is subjected to mechanical loading.   It is a generalization of Newtonian particle dynamics, and starts with the same physical assumptions inherent to Newtonian mechanics; and adds further assumptions that describe the structure of matter.  Specifically:

 

 The Newtonian reference frame: In classical continuum mechanics, the world is idealized as a three dimensional Euclidean space (a vector space consisting of all triads of real numbers $(x_1,x_2,x_3)$ ).  A point in space is identified by a unique set of three real numbers.   A Euclidean space is endowed with a metric, which defines the distance between points: $d=\sqrt{x_i{x}_i}$.  Vectors can be expressed as components in a basis - $\{e_1,e_2,e_3\}$ of mutually perpendicular unit vectors.  Physical quantities such as force, velocity, acceleration are expressed as vectors in this space.  A Cartesian Coordinate Frame is a fixed point O together with a basis.   A Newtonian reference frame is a particular choice of Cartesian coordinate frame in which Newton’s laws of motion hold.  

 

   The Continuum:  Matter is idealized as a continuum, which has two properties: (i) it is infinitely divisible (you can subdivide some region of the solid as many times as you wish); and (ii) it is locally homogeneous $–$ in other words if you subdivide it sufficiently many times, all sub-divisions have identical properties (eg mass density).  A continuum can be thought of as an infinite set of vanishingly small particles, connected together.

 

Both the existence of a Newtonian reference frame, and the concept of a continuum, are mathematical idealizations.   Experimental evidence suggest that the laws of motion based on these assumptions accurately approximate the behavior of most solid and fluid materials at length scales of order mm-km or so in engineering applications.  In some cases continuum models can also approximate behavior at much shorter length scales (for volumes of material containing a few 1000 atoms), but models at these length scales often require different relations between internal forces deformation measures in the solid to those used to model larger volumes.

 

 

 

3.2 Reference and deformed configuration of a solid

 

The configuration of a solid is a region of space occupied (filled) by the solid.   When we describe motion, we normally choose some convenient configuration of the solid to use as reference  - this is often the initial, undeformed solid, but it can be any convenient region that could be occupied by the solid.   The material changes its shape under the action of external loads, and at some time t occupies a new region which is called the deformed or current configuration of the solid. 

 

For some applications (fluids, problems with growth or evolving microstructures) a fixed reference configuration can’t be identified $–$ in this case we usually use the deformed material as the reference configuration.

 

Mathematically, we describe a deformation as a 1:1 mapping which transforms points from the reference configuration of a solid to the deformed configuration.  Specifically, let ${\xi }_i$ be three numbers specifying the position of some point in the undeformed solid (these could be the three components of position vector in a Cartesian coordinate system, or they could be a more general ccoordinate system, such as polar coordinates).  As the solid deforms, each the values of the coordinates change to different numbers.  We can write this in general form as ${\eta }_i=f_i({\xi }_k,t)$.  This is called a deformation mapping.

 

To be a physically admissible deformation

(i)     The coordinates must specify positions in a Newtonian reference frame.  This means that it must be possible to find some coordinate transformation $x_i({\xi }_k)$, such that $x_i$ are components in an orthogonal basis, which is taken to be ‘stationary’ in the sense of Newtonian dynamics.

(ii)   The functions $f_i({\xi }_k)$ must be 1:1 on the full set of real numbers; and $f_i$ must be invertible

(iii)  $f_i$ must be continuous and continuously differentiable (we occasionally relax these two assumptions, but this has to be dealt with on a case-by-case basis)

(iv)  The mapping must satisfy $\det (\partial {\eta }_i/\partial {\xi }_j)>0$.

 

To begin with, we will describe all motions and deformations by expressing positions of points in both undeformed and deformed solids as components in a Cartesian reference frame (which is also taken to be an inertial frame).  Thus $x_i$ will denote components of the position vector of a materal particle before deformation, and $y_i(x_k)$ will be components of its position vector after deformation.  

 

 

 

 

3.3 The Displacement and Velocity Fields

 

The displacement vector u(x,t) describes the motion of each point in the solid. To make this precise, visualize a solid deforming under external loads.  Every point in the solid moves as the load is applied: for example, a point at position x in the undeformed solid might move to a new position y at time t.  The displacement vector is defined as

$y=x+u(x,t)$

We could also express this formula using index notation, as

$y_i=x_i+u_i(x_1,x_2,x_3,t)$

Here, the subscript i  has  values 1,2, or 3, and (for example) $y_i$ represents the three Cartesian components of the vector y.

 

The displacement field completely specifies the change in shape of the solid. The velocity field would describe its motion, as

$v_i(x_k,t)=\frac{\partial y_i}{\partial t}={\frac{\partial u_i(x_k,t)}{\partial t}|}_{x_k=\text{const}}$

We also define the acceleration field

$a_i(x_k,t)=\frac{{\partial }^2{y}_i}{\partial t^2}={\frac{\partial v_i(x_k,t)}{\partial t}|}_{x_k=\text{const}}$

 

 

 

Examples of some simple deformations

 

 

 

 

 

 

3.4 Eulerian and Lagrangian descriptions of motion and deformation.

 

The displacement and velocity are vector valued functions.   In any application, we have a choice of writing the vectors as functions of the position of material particles before deformation $x_i$

$y_i=x_i+u_i(x_j,t){\frac{\partial y_i}{\partial t}|}_{x_i=\text{const}}=\frac{\partial u_i}{\partial t}=v_i(x_j,t)$

This is called the lagrangean description of motion.  It is usually the easiest way to visualize a deformation.

 

But in some applications (eg fluid flow problems, where it’s hard to identify a reference configuration) it is preferable to write the displacement, velocity and acceleration vectors as functions of the deformed position of particles.  

$y_i=x_i+u_i(y_j,t){\frac{\partial y_i}{\partial t}|}_{x_i=\text{const}}=v_i(y_j,t){\frac{{\partial }^2{y}_i}{\partial t^2}|}_{x_i=\text{const}}=a_i(y_j,t)$

These express displacement, velocity and displacement as functions of a particular point in space (visualize describing air flow, for example).  This is called the Eulerian description of motion. Of course the functions of $x_i$ and $y_i$ are not the same $–$ we just run out of symbols if we introduce different variables in the Lagrangian and Eulerian descriptions.      

 

The relationships between displacement, velocity, and acceleration are somewhat more complicated in the Eulerian description.  In the laws of motion, we normally are interested in the velocity and acceleration of a particular material particle, rather the rate of change of displacement and velocity at a particular point in space.  When computing the time derivatives, it is necessary to take into account that $y_i$ is a function of time.  Thus, displacement, velocity and acceleration are related by

${\left({\delta }_{ij}-\frac{\partial u_i}{\partial y_k}\right)\frac{\partial y_k}{\partial t}|}_{x_i=\text{const}}={\frac{\partial u_i}{\partial t}|}_{y_i=\text{const}}{\frac{{\partial }^2{y}_i}{\partial t^2}|}_{x_i=\text{const}}=a_i(y_j,t)={\frac{\partial v_i}{\partial t}|}_{y_i=\text{const}}+v_k(y_j,t)\frac{\partial v_i}{\partial y_k}$

You can derive these results by a simple application of the chain rule.

 

 

3.5 The Displacement gradient and Deformation gradient tensors

 

These quantities are defined by

 Displacement Gradient Tensor:   $\nabla u$ is a tensor with components $\frac{\partial u_i}{\partial x_k}$

 Deformation Gradient Tensor: $F=I+\nabla u\text{or in Cartesian components }{F}_{ik}={\delta }_{ik}+\frac{\partial u_i}{\partial x_k}$

where I  is the identity tensor, with components described by the Kronekor delta symbol:

${\delta }_{ik}=\{\begin{array}{l}1,i=k\\ 0,i\ne k\end{array}$

and $\nabla$ represents the gradient operator. Formally, the gradient of a vector field u(x) is defined so that

$\left[\nabla u\right]\cdot n=\underset{\alpha \to \text{0}}{\text{Lim}}\frac{u(x+\alpha n)-u(x)}{\alpha }$

 but in practice the component formula $\partial u_i/\partial x_j$ is more useful.

 

Note also that

$\begin{array}{c}\nabla y=\nabla \left(x+u(x)\right)=I+F\\ \text{or }\frac{\partial y_i}{\partial x_j}=\frac{\partial }{\partial x_j}\left(x_i+u_i\right)={\delta }_{ij}+\frac{\partial u_i}{\partial x_j}=F_{ij}\end{array}$

 

The concepts of displacement gradient and deformation gradient are introduced to quantify the change in shape of infinitesimal line elements in a solid body. To see this, imagine drawing a straight line on the undeformed configuration of a solid, as shown in the figure.  The line would be mapped to a smooth curve on the deformed configuration.  However, suppose we focus attention on a line segment dx, much shorter than the radius of curvature of this curve, as shown.  The segment would be straight in the undeformed configuration, and would also be (almost) straight in the deformed configuration.  Thus, no matter how complex a deformation we impose on a solid, infinitesimal line segments are merely stretched and rotated by a deformation.The infinitesimal line segments dx and dy are related by

$dy=F\cdot dx\text{or }d{y}_i=F_{ik}d{x}_k$

Written out as a matrix equation, we have

$\left[\begin{array}{c}d{y}_1\\ d{y}_2\\ d{y}_3\end{array}\right]=\left[\begin{array}{ccc}1+\frac{\partial u_1}{\partial x_1}& \frac{\partial u_1}{\partial x_2}& \frac{\partial u_1}{\partial x_3}\\ \frac{\partial u_2}{\partial x_1}& 1+\frac{\partial u_2}{\partial x_2}& \frac{\partial u_2}{\partial x_3}\\ \frac{\partial u_3}{\partial x_1}& \frac{\partial u_3}{\partial x_2}& 1+\frac{\partial u_3}{\partial x_3}\end{array}\right]\left[\begin{array}{c}d{x}_1\\ d{x}_2\\ d{x}_3\end{array}\right]$

 

To derive this result, consider an infinitesimal line element dx in a deforming solid.  When the solid is deformed, this line element is stretched and rotated to a deformed line element dy.  If we know the displacement field in the solid, we can compute dy=[x+dx+u(x+dx)]-[x+u(x)] from the position vectors of its two end points

$d{y}_i=x_i+d{x}_i+u_i(x_k+d{x}_k)-(x_i+u_i(x_k))$

Expand $u_i(x_k+d{x}_k)$ as a Taylor series

$u_i(x_k+d{x}_k)\approx u_i(x_k)+\frac{\partial u_i}{\partial x_k}d{x}_k$

so that

$d{y}_i=d{x}_i+\frac{\partial u_i}{\partial x_k}d{x}_k=\left({\delta }_{ik}+\frac{\partial u_i}{\partial x_k}\right)d{x}_k$

We identify the term in parentheses as the deformation gradient, so

$d{y}_i=F_{ik}d{x}_k$

 

The inverse of the deformation gradient $F^{-1}$ arises in many calculations.  It is defined through

$d{x}_i=F_{ik}^{-1}d{y}_k$

or alternatively

$F_{ij}^{-1}=\frac{\partial x_i}{\partial y_j}$

 

 

 

3.6 Deformation gradient resulting from two successive deformations

 

Suppose that two successive deformations are applied to a solid, as shown.  Let

$dy=F^{(1)}\cdot dxdz=F^{(2)}\cdot dy\text{or}d{y}_i=F_{ij}^{(1)}d{x}_{j}d{z}_i=F_{ij}^{(2)}d{y}_j$

map infinitesimal line elements from the original configuration to the first deformed shape, and from the first deformed shape to the second, respectively, with

$F^{(1)}=y\otimes {\nabla }_x{F}^{(2)}=z\otimes {\nabla }_y\text{or}{F}_{ij}^{(1)}=\frac{\partial y_i}{\partial x_j}{F}_{ij}^{(2)}=\frac{\partial z_i}{\partial y_j}$

The deformation gradient that maps infinitesimal line elements from the original configuration directly to the second deformed shape then follows as

$dz=F\cdot dx\text{with }F=F^{(2)}\cdot F^{(1)}\text{or}d{z}_i=F_{ij}d{x}_j{F}_{ij}=F_{ik}^{(2)}{F}_{kj}^{(1)}$

Thus, the cumulative deformation gradient due to two successive deformations follows by multiplying their individual deformation gradients.

 

To see this, write the cumulative mapping as $z_i\left(y_j(x_k)\right)$ and apply the chain rule

$d{z}_i=\frac{\partial z_i}{\partial y_j}\frac{\partial y_j}{\partial x_k}d{x}_k$

 

 

 

  

 

3.7 The Jacobian of the deformation gradient $–$ change of volume

 

The Jacobian is defined as

$J=\det \left(F\right)=\det \left({\delta }_{ij}+\frac{\partial u_i}{\partial x_j}\right)$

It is a measure of the volume change produced by a deformation.  To see this, consider the infinitessimal volume element shown with sides dx, dy, and dz in the figure above. The original volume of the element is

$d{V}_0=dz\cdot (dx\times dy)={\in }_{ijk}d{z}_{i}d{x}_{j}d{y}_k$

Here, ${\in }_{ijk}$ is the permutation symbol. The element is mapped to a paralellepiped with sides dr, dv, and dw with volume given by

$dV={\in }_{ijk}d{w}_{i}d{r}_{j}d{v}_k$

Recall that

$d{r}_i=F_{il}d{x}_l,d{v}_j=F_{jm}d{y}_m,d{w}_k=F_{kn}d{z}_n$

so that

$dV={\in }_{ijk}{F}_{il}d{x}_l{F}_{jm}d{y}_m{F}_{kn}d{z}_n={\in }_{ijk}{F}_{il}{F}_{jm}{F}_{kn}d{x}_{l}d{y}_{m}d{z}_n$

Recall that

${\in }_{ijk}{A}_{il}{A}_{jm}{A}_{kn}={\in }_{lmn}\det (A)$

so that

$dV=\det (F){\in }_{lmn}d{x}_{l}d{y}_{m}d{z}_n=\det \left(F\right)d{V}_0$

Hence

$\frac{dV}{d{V}_0}=\det \left(F\right)=J$

Observe that

 For any physically admissible deformation, the volume of the deformed element must be positive (no matter how much you deform a solid, you can’t make material disappear).  Therefore, all physically admissible displacement fields must satisfy J>0

 If a material is incompressible, its volume remains constant.  This requires J=1.

 If the mass density of the material at a point in the undeformed solid is ${\rho }_0$, its mass density in the deformed solid is $\rho ={\rho }_0/J$

 

Derivatives of J. When working with constitutive equations, it is occasionally necessary to evaluate derivatives of J with respect to the components of F.  The following result (which can be proved e.g. by expanding the Jacobian using index notation $–$ see HW1, problem 7, eg) is extremely useful

$\frac{\partial J}{\partial F_{ij}}=J{F}_{ji}^{-1}$

 

 

 

 

 

3.8 Transformation of internal surface area elements

 

 

When we deal with internal forces in a solid, we need to work with forces acting on internal surfaces in a solid.  An important question arises in this treatment: if we identify an element of area $d{A}_0$ with normal $n_0$ in the reference configuration, and then what are the area of $dA$ and normal $n$ of this area element in the deformed solid?

 

The two are related through

$dAn=J{F}^{-T}\cdot d{A}_0{n}_{0}dA{n}_i^{}=J{F}_{ki}^{-1}{n}_k^{0}d{A}_0$

 

To see this,

1.      let $d{v}_i^0,d{w}_j^0$ be two infinitesimal material fibers with different orientations at some point in the reference configuration.  These fibers bound a parallelapiped with area and normal

$d{A}_0{n}_i^0={\in }_{ijk}d{w}_j^{0}d{v}_k^0$

2.      The vectors map to $d{v}_j=F_{jq}d{v}_q^0$,  $d{w}_k=F_{kn}d{w}_n^0$ in the deformed solid

3.      In the deformed solid the area element is thus

$dA{n}_i={\in }_{ijk}{F}_{jq}d{v}_q^0{F}_{kn}d{w}_n^0=F_{ml}{F}_{li}^{-1}{\in }_{mjk}{F}_{jq}{F}_{kn}d{v}_q^{0}d{w}_n^0$

4.      Recall the identity ${\in }_{lmn}\det (A)={\in }_{ijk}{A}_{il}{A}_{jm}{A}_{kn}$ - so $dA{n}_i=F_{li}^{-1}{\in }_{lqn}Jd{v}_q^{0}d{w}_n^0=J{F}_{ki}^{-1}{n}_k^{0}d{A}_0$

 

 

3.8 The Lagrange strain tensor

 

The Lagrange strain tensor is defined as

$E=\frac{1}{2}(F^T\cdot F-I)\text{or}{E}_{ij}=\frac{1}{2}(F_{ki}{F}_{kj}-{\delta }_{ij})$

The components of Lagrange strain can also be expressed in terms of the displacement gradient as

$E_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}+\frac{\partial u_k}{\partial x_j}\frac{\partial u_k}{\partial x_i}\right)$

 

The Lagrange strain tensor quantifies the changes in length of a material fiber, and angles between pairs of fibers in a deformable solid.  It is used in calculations where large shape changes are expected.

 

To visualize the physical significance of E, suppose we mark out an imaginary tensile specimen with (very short) length $l_0$ on our deforming solid, as shown in the picture. The orientation of the specimen is arbitrary, and is specified by a unit vector m, with components $m_i$.  Upon deformation, the specimen increases in length to $l=l_0+\delta l$. Define the strain of the specimen as

${\epsilon }_L(m_i)=\frac{l^2-l_0{}^2}{2l_0{}^2}=\frac{\delta l}{l_0}+\frac{{\left(\delta l\right)}^2}{2l_0^2}$

Note that this definition of strain is similar to the definition $\epsilon =\delta l/l_0$ you are familiar with, but contains an additional term.  The additional term is negligible for small $\delta l$. Given the Lagrange strain components $E_{ij}$, the strain of the specimen may be computed from

${\epsilon }_L(m)=m\cdot E\cdot m\text{or }{\epsilon }_L(m_i)=E_{ij}{m}_i{m}_j$

We proceed to derive this result. Note that

$d{x}_i=l_0{m}_i$

is an infinitesimal vector with length and orientation of our undeformed specimen.  From the preceding section, this vector is stretched and rotated to

$d{y}_k=\left({\delta }_{kj}+\frac{\partial u_k}{\partial x_j}\right)d{x}_j=\left({\delta }_{kj}+\frac{\partial u_k}{\partial x_j}\right)l_0{m}_j$

The length of the deformed specimen is equal to the length of  dy, so we see that

$\begin{array}{l}{l}^2=d{y}_{k}d{y}_k=\left({\delta }_{kj}+\frac{\partial u_k}{\partial x_j}\right)l_0{m}_j\left({\delta }_{ki}+\frac{\partial u_k}{\partial x_i}\right)l_0{m}_i\\ =\left({\delta }_{ij}+\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}+\frac{\partial u_k}{\partial x_i}\frac{\partial u_k}{\partial x_j}\right)l_0{}^2{m}_j{m}_i\end{array}$

Hence,  the strain for our line element is

${\epsilon }_L(m_i)=\frac{l^2-l_0{}^2}{2l_0{}^2}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}+\frac{\partial u_k}{\partial x_j}\frac{\partial u_k}{\partial x_i}\right)m_i{m}_j$

giving the results stated.

 

 

 

3.9 The Eulerian strain tensor

 

The Eulerian strain tensor is defined as

$E^{*}=\frac{1}{2}(I-F^{-T}\cdot F^{-1})\text{or}{E}_{ij}^{*}=\frac{1}{2}({\delta }_{ij}-F_{ki}^{-1}{F}_{kj}^{-1})$

Its physical significance is similar to the Lagrange strain tensor, except that it enables you to compute the strain of an infinitesimal line element from its orientation after deformation.

 

Specifically, suppose that n denotes a unit vector parallel to the deformed material fiber, as shown in the picture.  Then

${\epsilon }_E(n)=\frac{l^2-l_0{}^2}{2l^2}=n\cdot E^{*}\cdot n\text{or }{\epsilon }_E(n_i)=E_{ij}^{*}{n}_i{n}_j$

The proof is left as an exercise.

 

 

 

3.10 The Infinitesimal strain tensor

 

The infinitesimal strain tensor is defined as

$\epsilon =\frac{1}{2}\left(u\nabla +{\left(u\nabla \right)}^T\right)\text{or }{\epsilon }_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$

where u is the displacement vector.  Written out in full

${\epsilon }_{ij}\equiv \left[\begin{array}{ccc}\frac{\partial u_1}{\partial x_1}& \frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right)& \frac{1}{2}\left(\frac{\partial u_1}{\partial x_3}+\frac{\partial u_3}{\partial x_1}\right)\\ \frac{1}{2}\left(\frac{\partial u_2}{\partial x_1}+\frac{\partial u_1}{\partial x_2}\right)& \frac{\partial u_2}{\partial x_2}& \frac{1}{2}\left(\frac{\partial u_2}{\partial x_3}+\frac{\partial u_3}{\partial x_2}\right)\\ \frac{1}{2}\left(\frac{\partial u_3}{\partial x_1}+\frac{\partial u_1}{\partial x_3}\right)& \frac{1}{2}\left(\frac{\partial u_3}{\partial x_2}+\frac{\partial u_2}{\partial x_3}\right)& \frac{\partial u_3}{\partial x_3}\end{array}\right]$

 

The infinitesimal strain tensor is an approximate deformation measure, which is only valid for small shape changes.  It is more convenient than the Lagrange or Eulerian strain, because it is linear.

 

Specifically, suppose the deformation gradients are small, so that all $\partial u_i/\partial x_j<<1$. Then the Lagrange strain tensor is

$E_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}+\frac{\partial u_k}{\partial x_j}\frac{\partial u_k}{\partial x_i}\right)\approx \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)\approx {\epsilon }_{ij}$

so the infinitesimal strain approximates the Lagrange strain.  You can show that it also approximates the Eulerian strain with the same accuracy.

 

Properties of the infinitesimal strain tensor

 

 For small strains, the engineering strain of an infinitesimal fiber aligned with a unit vector m can be estimated as

${\epsilon }_e(m)=\frac{l-l_0}{l_0}\approx {\epsilon }_{ij}{m}_i{m}_j$

 Note that

$\text{trace}\left(\epsilon \right)\equiv {\epsilon }_{kk}=\frac{\partial u_1}{\partial x_1}+\frac{\partial u_2}{\partial x_2}+\frac{\partial u_3}{\partial x_3}=\frac{dV-d{V}_0}{d{V}_0}$

 (see below for more details)

 The infinitesimal strain tensor is closely related to the strain matrix introduced in elementary strength of materials courses.  For example, the physical significance of the (2 dimensional) strain matrix

$\left[\begin{array}{cc}{\epsilon }_{xx}& {\gamma }_{xy}\\ {\gamma }_{yx}& {\epsilon }_{yy}\end{array}\right]$

is illustrated in the figure.

 

To relate this to the infintesimal strain tensor, let $\{e_1,e_2,e_3\}$ be a Cartesian basis, with $e_1$ parallel to x and $e_2$ parallel to y as shown.  Let ${\epsilon }_{ij}$ denote the components of the infinitesimal strain tensor in this basis.  Then

$\begin{array}{l}{\epsilon }_{11}={\epsilon }_{xx}\\ {\epsilon }_{22}={\epsilon }_{yy}\\ {\epsilon }_{12}={\epsilon }_{21}={\gamma }_{xy}/2={\gamma }_{yx}/2\end{array}$

 

 

3.11 Engineering shear strains

 

For a general strain tensor (which could be any of $E$, $E^{*}$ or $\epsilon$, among others), the diagonal strain components ${\epsilon }_{11},{\epsilon }_{22},{\epsilon }_{33}$ are known as `direct’ strains, while the off diagonal terms  ${\epsilon }_{12}={\epsilon }_{21}{\epsilon }_{13}={\epsilon }_{31}{\epsilon }_{23}={\epsilon }_{32}$ are known as ‘shear strains’

 

The shear strains are sometimes reported as ‘Engineering Shear Strains’ which are related to the formal definition by a factor of 2 i.e.

${\gamma }_{12}=2{\epsilon }_{12}{\gamma }_{13}=2{\epsilon }_{13}{\gamma }_{23}=2{\epsilon }_{23}$

 

This factor of 2 is an endless source of confusion.  Whenever someone reports shear strain to you, be sure to check which definition they are using.  In particular, many commercial finite element codes output engineering shear strains.

 

 

 

 

 

 

3.12 Decomposition of infinitesimal strain into volumetric and deviatoric parts

 

The volumetric infinitesimal strain is defined as $\text{trace(}\epsilon )\equiv {\epsilon }_{kk}$

The deviatoric infinitesimal strain is defined as $e\text{=}\epsilon -\frac{1}{3}I\text{trace(}\epsilon )\equiv e_{ij}={\epsilon }_{ij}-\frac{1}{3}{\delta }_{ij}{\epsilon }_{kk}$

 

The volumetric strain is a measure of volume changes, and for small strains is related to the Jacobian of the deformation gradient by ${\epsilon }_{kk}\approx J-1$.  To see this, recall that

$J=\det \left[\begin{array}{ccc}1+\frac{\partial u_1}{\partial x_1}& \frac{\partial u_1}{\partial x_2}& \frac{\partial u_1}{\partial x_3}\\ \frac{\partial u_2}{\partial x_1}& 1+\frac{\partial u_2}{\partial x_2}& \frac{\partial u_2}{\partial x_3}\\ \frac{\partial u_3}{\partial x_1}& \frac{\partial u_3}{\partial x_2}& 1+\frac{\partial u_3}{\partial x_3}\end{array}\right]\approx \left(1+\frac{\partial u_1}{\partial x_1}\right)\left(1+\frac{\partial u_2}{\partial x_2}\right)\left(1+\frac{\partial u_3}{\partial x_3}\right)\approx 1+\frac{\partial u_1}{\partial x_1}+\frac{\partial u_2}{\partial x_2}+\frac{\partial u_3}{\partial x_3}$

The deviatioric strain is a measure of shear deformation (shear deformation involves no volume change).

 

 

 

 

 

3.13 The Infinitesimal rotation tensor

 

The infinitesimal rotation tensor is defined as

$w=\frac{1}{2}\left(u\nabla -{\left(u\nabla \right)}^T\right)\text{or }{w}_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i}\right)$

Written out as a matrix, the components of $w_{ij}$ are

$w_{ij}\equiv \left[\begin{array}{ccc}0& \frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right)& \frac{1}{2}\left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)\\ \frac{1}{2}\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)& 0& \frac{1}{2}\left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right)\\ \frac{1}{2}\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right)& \frac{1}{2}\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)& 0\end{array}\right]$

Observe that $w_{ij}$ is skew symmetric: $w_{ij}=-w_{ji}$. 

 

A skew tensor represents a rotation through a small angle.  Specifically, the operation $d{y}_i=\left({\delta }_{ij}+w_{ij}\right)d{x}_j$ rotates the infinitesimal line element $d{x}_j$ through a small angle $\theta =\sqrt{w_{ij}{w}_{ij}/2}$ about an axis parallel to the unit vector $n_i={\in }_{ijk}{w}_{kj}/(2\theta )$.  (A skew tensor also sometimes represents an angular velocity). 

To visualize the significance of  $w_{ij}$, consider the behavior of an imaginary, infinitesimal, tensile specimen embedded in a deforming solid.  The specimen is stretched, and then rotated through an angle $\varphi$ about some axis q.  If the displacement gradients are small, then $\varphi <<1$.

 

The rotation of the specimen depends on its original orientation, represented by the unit vector m.  One can show (although one would rather not do all the algebra) that $w_{ij}$ represents the average rotation, over all possible orientations of m, of material fibers passing through a point.

 

As a final remark, we note that a general deformation can always be decomposed into an infinitesimal strain and rotation

$\frac{\partial u_i}{\partial x_j}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)+\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i}\right)={\epsilon }_{ij}+w_{ij}$

Physically, this sum of ${\epsilon }_{ij}$ and $w_{ij}$ can be regarded as representing two successive deformations $–$ a small strain, followed by a rotation, in the sense that

$d{y}_i=\left({\delta }_{ik}+w_{ik}\right)\left({\delta }_{kj}+{\epsilon }_{kj}\right)d{x}_j\approx d{x}_i+\left({\epsilon }_{ij}+w_{ij}\right)d{x}_j$

first stretches the infinitesimal line element, then rotates it.

 

 

 

3.14 Principal values and directions of the infinitesimal strain tensor

The three principal values $e_i$ and directions $n^{(i)}$ of the infinitesimal strain tensor satisfy

$\begin{array}{c}\epsilon \cdot n^{(i)}=e_i{n}^{(i)}\\ \text{or }{\epsilon }_{kl}{n}_l^{(i)}=e_i{n}_l^{(i)}\end{array}$

Clearly, $e_i$ and $n^{(i)}$ are the eigenvalues and eigenvectors of $\epsilon$.  There are three principal strains and three principal directions, which are always mutually perpendicular. 

 

Their significance can be visualized as follows.

1.      Note that the decomposition $\frac{\partial u_i}{\partial x_j}={\epsilon }_{ij}+w_{ij}$ can be visualized as a small strain, followed by a small rigid rotation, as shown in the picture.

2.      The formula $\epsilon \cdot n^{(i)}=e_i{n}^{(i)}$ indicates that a vector n is mapped to another, parallel vector by the strain.

3.      Thus, if you draw a small cube with its faces perpendicular to $n^{(i)}$ on the undeformed solid, this cube will be stretched perpendicular to each face, with a fractional increase in length $e_i=\delta l_i/l_0$.  The faces remain perpendicular to $n^{(i)}$ after deformation.

4.      Finally, w rotates the small cube through a small angle onto its configuration in the deformed solid.

 

 

 

3.15  Strain Equations of Compatibility for infinitesimal strains

 

It is sometimes necessary to invert the relations between strain and displacement $–$ that is to say, given the strain field, to compute the displacements. In this section, we outline how this is done, for the special case of infinitesimal deformations.

 

For infinitesimal motions the relation between strain and displacement is

${\epsilon }_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$

Given the six strain components ${\epsilon }_{ij}$ (six, since ${\epsilon }_{ij}={\epsilon }_{ji}$ ) we wish to determine the three displacement components $u_i$. First, note that you can never completely recover the displacement field that gives rise to a particular strain field.  Any rigid motion produces no strain, so the displacements can only be completely determined if there is some additional information (besides the strain) that will tell you how much the solid has rotated and translated.  However, integrating the strain field can tell you the displacement field to within an arbitrary rigid motion.

 

Second, we need to be sure that the strain-displacement relations can be integrated at all. The strain is a symmetric second order tensor field, but not all symmetric second order tensor fields can be strain fields. The strain-displacement relations amount to a system of six scalar differential equations for the three displacement components u_i.

 

To be integrable, the strains must satisfy the compatibility conditions, which may be expressed as

${\in }_{ipm}{\in }_{jqn}\frac{{\partial }^2{\epsilon }_{mn}}{\partial x_p\partial x_q}=0$

Or, equivalently

$\frac{{\partial }^2{\epsilon }_{ij}}{\partial x_k\partial x_l}+\frac{{\partial }^2{\epsilon }_{kl}}{\partial x_i\partial x_j}-\frac{{\partial }^2{\epsilon }_{il}}{\partial x_j\partial x_k}-\frac{{\partial }^2{\epsilon }_{jk}}{\partial x_i\partial x_l}=0$

Or, once more equivalently

$\begin{array}{l}\frac{{\partial }^2{\epsilon }_{11}}{\partial x_2^2}+\frac{{\partial }^2{\epsilon }_{22}}{\partial x_1^2}-2\frac{{\partial }^2{\epsilon }_{12}}{\partial x_1\partial x_2}=0\\ \frac{{\partial }^2{\epsilon }_{11}}{\partial x_3^2}+\frac{{\partial }^2{\epsilon }_{33}}{\partial x_1^2}-2\frac{{\partial }^2{\epsilon }_{13}}{\partial x_1\partial x_3}=0\\ \frac{{\partial }^2{\epsilon }_{22}}{\partial x_3^2}+\frac{{\partial }^2{\epsilon }_{33}}{\partial x_2^2}-2\frac{{\partial }^2{\epsilon }_{23}}{\partial x_2\partial x_3}=0\end{array}$      $\begin{array}{l}\frac{{\partial }^2{\epsilon }_{11}}{\partial x_2^{}\partial x_3^{}}-\frac{\partial }{\partial x_1}\left(-\frac{\partial {\epsilon }_{23}}{\partial x_1^{}}+\frac{\partial {\epsilon }_{31}}{\partial x_2^{}}+\frac{\partial {\epsilon }_{12}}{\partial x_3^{}}\right)=0\\ \frac{{\partial }^2{\epsilon }_{22}}{\partial x_3^{}\partial x_1^{}}-\frac{\partial }{\partial x_2}\left(-\frac{\partial {\epsilon }_{31}}{\partial x_2^{}}+\frac{\partial {\epsilon }_{12}}{\partial x_3^{}}+\frac{\partial {\epsilon }_{23}}{\partial x_1^{}}\right)=0\\ \frac{{\partial }^2{\epsilon }_{33}}{\partial x_1^{}\partial x_2^{}}-\frac{\partial }{\partial x_3}\left(-\frac{\partial {\epsilon }_{12}}{\partial x_3^{}}+\frac{\partial {\epsilon }_{23}}{\partial x_1^{}}+\frac{\partial {\epsilon }_{31}}{\partial x_2^{}}\right)=0\end{array}$

It is easy to show that all strain fields must satisfy these conditions -  you simply need to substitute for the strains in terms of displacements and show that the appropriate equation is satisfied.  For example,

$\frac{{\partial }^2{\epsilon }_{11}}{\partial x_2^2}+\frac{{\partial }^2{\epsilon }_{22}}{\partial x_1^2}-2\frac{{\partial }^2{\epsilon }_{12}}{\partial x_1\partial x_2}=\frac{{\partial }^4{u}_1}{\partial x_1^{}\partial x_2^2}+\frac{{\partial }^4{u}_2}{\partial x_2^{}\partial x_1^2}-2\frac{{\partial }^2}{\partial x_1\partial x_2}\frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right)=0$

and similarly for the other expressions.

Not that for planar problems for which ${\epsilon }_{13}={\epsilon }_{23}=0\text{ and }\frac{d{\epsilon }_{ij}}{d{x}_3}=0,$ all of these compatibility equations are satisfied trivially, with the exception of the first: $\frac{{\partial }^2{\epsilon }_{11}}{\partial x_2^2}+\frac{{\partial }^2{\epsilon }_{22}}{\partial x_1^2}-2\frac{{\partial }^2{\epsilon }_{12}}{\partial x_1\partial x_2}=0$

 

It can be shown that

 (i) If the strains do not satisfy the equations of compatibility, then a displacement vector can not be integrated from the strains.

(ii) If the strains satisfy the compatibility equations, and the solid simply connected (i.e. it contains no holes that go all the way through its thickness), then a displacement vector can be integrated from the strains. 

(iii) If the solid is not simply connected, a displacement vector can be calculated, but it may not be single valued $–$ i.e. you may get different solutions depending on how the path of integration encircles the holes.

 

Now, let us return to the question posed at the beginning of this section.  Given the strains, how do we compute the displacements?

2D strain fields

For 2D (plane stress or plane strain) the procedure is quite simple and is best illustrated by working through a specific case

 

As a representative example, we will use the strain field in a 2D (plane stress) cantilever beam with Young’s modulus E and Poisson’s ratio $\nu$ loaded at one end by a force P. The beam has a rectangular cross-section with height 2a and out-of-plane width b.  We will show later (Sect 5.2.4) that the strain field in the beam is

${\epsilon }_{11}=2C{x}_1{x}_2{\epsilon }_{22}=-2\nu C{x}_1{x}_2{\epsilon }_{12}=\left(1+\nu \right)C\left(a^2-x_2^2\right),C=\frac{3P}{4E{a}^{3}b}$

 

We first check that the strain is compatible.  For 2D problems this requires

$\frac{{\partial }^2{\epsilon }_{11}}{\partial x_2^2}+\frac{{\partial }^2{\epsilon }_{22}}{\partial x_1^2}-2\frac{{\partial }^2{\epsilon }_{12}}{\partial x_1\partial x_2}=0$

which is clearly satisfied in this case.

 

For a 2D problem we only need to determine $u_1(x_1,x_2)$ and $u_2(x_1,x_2)$ such that

$\frac{\partial u_1}{\partial x_1^{}}={\epsilon }_{11},\frac{\partial u_2}{\partial x_2^{}}={\epsilon }_{22}\text{and }\frac{\partial u_1}{\partial x_2^{}}+\frac{\partial u_2}{\partial x_1^{}}=2{\epsilon }_{12}$.

The first two of these give

${\epsilon }_{11}=\frac{\partial u_1}{\partial x_1}=2C{x}_1{x}_2{\epsilon }_{22}=\frac{\partial u_2}{\partial x_2}=-2\nu C{x}_1{x}_2$

We can integrate the first equation with respect to $x_1$ and the second equation with respect to $x_2$ to get

$u_1=C{x}_1^2{x}_2+f_1(x_2)u_2=-\nu C{x}_1{x}_2^2+f_2(x_1)$

where $f_1(x_2)$ and $f_2(x_1)$ are two functions of $x_2$ and $x_1$, respectively, which are yet to be determined.  We can find these functions by substituting the formulas for $u_1(x_1,x_2)$ and $u_2(x_1,x_2)$ into the expression for shear strain

$\begin{array}{l}{\epsilon }_{12}=\frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right)=\left(1+\nu \right)C\left(a^2-x_2^2\right)\\ \frac{1}{2}\left(C{x}_1^2-\nu C{x}_2^2+\frac{d{f}_1}{d{x}_2}+\frac{d{f}_2}{d{x}_1}\right)=\left(1+\nu \right)C\left(a^2-x_2^2\right)\end{array}$

We can re-write this as

$\left(\frac{d{f}_2}{d{x}_1}+C{x}_1^2\right)+\left(\frac{d{f}_1}{d{x}_2}-\nu C{x}_2^2-2\left(1+\nu \right)C\left(a^2-x_2^2\right)\right)=0$

The two terms in parentheses are functions of $x_1$ and $x_2$, respectively.  Since the left hand side must vanish for all values of  $x_1$ and $x_2$, this means that

$\begin{array}{l}\left(\frac{d{f}_2}{d{x}_1}+C{x}_1^2\right)=\omega \\ \left(\frac{d{f}_1}{d{x}_2}-\nu C{x}_2^2-2\left(1+\nu \right)C\left(a^2-x_2^2\right)\right)=-\omega \end{array}$

where $\omega$ is an arbitrary constant.  We can now integrate these expressions to see that

$\begin{array}{l}{f}_1=\left(2(1+\nu )C{a}^2-\omega \right)x_2-\frac{C}{3}(2+\nu )x_2^3+c\\ f_2=\omega x_1-\frac{C}{3}{x}_1^3+d\end{array}$

where c and d are two more arbitrary constants. Finally, the displacement field follows as

$\begin{array}{l}{u}_1=C{x}_1^2{x}_2-\frac{C}{3}(2+\nu )x_2^3+2(1+\nu )C{a}^2{x}_2-\omega x_2+c\\ u_2=-\nu C{x}_1{x}_2^2-\frac{C}{3}{x}_1^3+\omega x_1+d\end{array}$

The three arbitrary constants $\omega$, c and d can be seen to represent a small rigid rotation through angle $\omega$ about the $x_3$ axis, together with a displacement (c,d) parallel to $(x_1,x_2)$ axes, respectively.

 

 

3D strain fields

 

For a general, three dimensional field a more formal procedure is required. Since the strains are the derivatives of the displacement field, so you might guess that we compute the displacements by integrating the strains.  This is more or less correct.  The general procedure is outlined below.

 

We first pick a point $x_0$ in the solid, and arbitrarily say that the displacement at $x_0$ is zero, and also take the rotation of the solid at  $x_0$ to be zero. Then, we can compute the displacements at any other point x in the solid, by integrating the strains along any convenient path.  In a simply connected solid, it doesn’t matter what path you pick.

Actually, you don’t exactly integrate the strains $–$ instead, you must evaluate the following integral

$u_i(x)={\displaystyle \underset{x0}{\overset{x}{\int }}{U}_{ij}(x,\text{ξ})d{\xi }_j}$

where

$U_{ij}(x,\text{ξ})={\epsilon }_{ij}(\text{ξ})+(x_k-{\xi }_k)\left[\frac{\partial {\epsilon }_{ij}(\text{ξ})}{\partial {\xi }_k}-\frac{\partial {\epsilon }_{kj}(\text{ξ})}{\partial {\xi }_i}\right]$

Here, $x_k$ are the components of the position vector at the point where we are computing the displacements, and ${\xi }_j$ are the components of the position vector $\text{ξ}$ of a point somewhere along the path of integration. The fact that the integral is path-independent (in a simply connected solid) is guaranteed by the compatibility condition.  Evaluating this integral in practice can be quite painful, but fortunately almost all cases where we need to integrate strains to get displacement turn out to be two-dimensional.

 

 

 

 

3.16 Cauchy-Green Deformation Tensors

 

There are two Cauchy-Green deformation tensors $–$ defined through

 The Right Cauchy Green Deformation Tensor   $C=F^T\cdot F{C}_{ij}=F_{ki}{F}_{kj}$

 The Left Cauchy Green Deformation Tensor     $B=F\cdot F^T{B}_{ij}=F_{ik}{F}_{jk}$

They are called `left’ and `right’ tensors because of their relation to the `left’ and ‘right’ stretch tensors defined below.  They can be regarded as quantifying the squared length of infinitesimal fibers in the deformed configuration, by noting that if a material fiber $dx=l_{0}m$ in the undeformed solid is stretched and rotated to $dy=ln$ in the deformed solid, then

$\frac{l^2}{l_0^2}=m\cdot C\cdot m\frac{l_0^2}{l^2}=n\cdot B^{-1}\cdot n$

 

 

 

3.17 Rotation tensor, and Left and Right Stretch Tensors

 

The definitions of these quantities are

 The Right Stretch Tensor $U=C^{1/2}={\left(F^T\cdot F\right)}^{1/2}{U}_{ij}=C_{ij}^{1/2}$

 The Left Stretch Tensor   $V=B^{1/2}{V}_{ij}=B_{ij}^{1/2}$

 The Rotation Tensor         $R=F\cdot U^{-1}=V^{-1}\cdot F{R}_{ij}=F_{ik}{U}_{kj}^{-1}=V_{ik}^{-1}{F}_{kj}$

 

To calculate these quantities you need to remember how to calculate the square root of a matrix.  For example, to calculate the square root of C, you must

1.      Calculate the eigenvalues of C $–$ we will call these ${\lambda }_n^2$, with n=1,2,3.  Since C and B are both symmetric and positive definite, the eigenvalues ${\lambda }_n^2$ are all positive real numbers, and therefore their square roots ${\lambda }_n$ are also positive real numbers.

2.      Calculate the eigenvectors of C and normalize them so they have unit magnitude.  We will denote the eigenvectors by $c^{(n)}$.  They must be normalized to satisfy $c^{(n)}\cdot c^{(n)}=1$

3.      Finally, calculate $C^{1/2}={\displaystyle {\sum }_{n=1}^3{\lambda }_n{c}^{(n)}\otimes c^{(n)}}$, where $c$ denotes a dyadic product (See Appendix B). In components, this can be written $C_{ij}^{1/2}={\displaystyle {\sum }_{n=1}^3{\lambda }_n{c}_i^{(n)}{c}_j^{(n)}}$

4.      As an additional bonus, you can quickly compute the inverse square root (which is needed to find R) as

$C^{-1/2}={\displaystyle {\sum }_{n=1}^3\frac{1}{{\lambda }_n}{c}^{(n)}\otimes c^{(n)}}\text{or }{C}_{ij}^{-1/2}={\displaystyle {\sum }_{n=1}^3\frac{1}{{\lambda }_n}{c}_i^{(n)}{c}_j^{(n)}}$

 

To see the physical significance of these tensors, observe that

1.      The definition of the rotation tensor shows that

$\begin{array}{l}R=F\cdot U^{-1}\iff F=R\cdot U\\ R=V^{-1}\cdot F\iff F=V\cdot R\end{array}$

2.      The multiplicative decomposition of a constant tensor $F=R\cdot U$ can be regarded as a sequence of two homogeneous deformations $–$ U, followed by R.  Similarly, $F=V\cdot R$ is R followed by V.

3.      R is proper orthogonal (it satisfies $R\cdot R^T=R^T\cdot R=I$ and det(R)=1), and therefore represents a rotation.  To see this, note that U is symmetric, and therefore satisfies $U^{-T}=U^{-1}$, so that

$\begin{array}{l}{R}^{T}R={\left(F\cdot U^{-1}\right)}^T\cdot \left(F\cdot U^{-1}\right)\\ =U^{-T}\cdot F^T\cdot F\cdot U^{-1}\\ =U^{-1}\cdot U^2\cdot U^{-1}=I\end{array}$

and det(R)=det(F)det(U^-1)=1

4. U can be expressed in the form

$U={\lambda }_1{u}^{(1)}\otimes u^{(1)}+{\lambda }_2{u}^{(2)}\otimes u^{(2)}+{\lambda }_3{u}^{(3)}\otimes u^{(3)}$

where $u^{(i)}$ are the three (mutually perpendicular) eigenvectors of U. (By construction, these are identical to the eigenvectors of C).  If we interpret $u^{(i)}$ as basis vectors, we see that U is diagonal in this basis, and so corresponds to stretching parallel to each basis vector, as shown in the figure below.

The decompositions

$F=R\cdot U$ and $F=V\cdot R$

are known as the right and left polar decomposition of F. (The right and left refer to the positions of U and V).  They show that every homogeneous deformation can be decomposed into a stretch followed by a rigid rotation, or equivalently into a rigid rotation followed by a stretch. The decomposition is discussed in more detail in the next section.

 

3.18 Principal stretches

 

The principal stretches can be calculated from any one of the following (they all give the same answer)

1. The eigenvalues of the right stretch tensor U
2. The eigenvalues of the left stretch tensor V
3. The square root of the eigenvalues of the right Cauchy-Green tensor C
4. The square root of the eigenvalues of the left Cauchy-Green tensor B

The principal stretches are also related to the eigenvalues of the Lagrange and Eulerian strains.  The details are left as an exercise.

 

There are two sets of principal stretch directions, associated with the undeformed and deformed solids.

1. The principal stretch directions in the undeformed solid are the (normalized) eigenvectors of U or C.  Denote these by $u^{(i)}$.
2. The principal stretch directions in the deformed solid are the (normalized) eigenvectors of V or B. Denote these by $v^{(i)}$.

 

To visualize the physical significance of principal stretches and their directions, note that a deformation can be decomposed as $F=R\cdot U$ into a sequence of a stretch followed by a rotation. 

 

Note also that

1. The principal directions $u^{(i)}$ are mutually perpendicular.  You could draw a little cube on the undeformed solid with faces perpendicular to these directions, as shown above.
2. The stretch U will stretch the cube by an amount ${\lambda }_i$ parallel to each $u^{(i)}$.  The faces of the stretched cube remain perpendicular to $u^{(i)}$.
3. The rotation R will rotate the stretched cube so that the directions $u^{(i)}$ rotate to line up with $v^{(i)}$.
4. The faces of the deformed cube are perpendicular to $v^{(i)}$

 

The decomposition $F=V\cdot R$ can be visualized in much the same way.  In this case, the directions $u^{(i)}$ are first rotated to coincide with $v^{(i)}$.  The cube is then stretched parallel to each $v^{(i)}$ to produce the same shape change.

 

We could compare the undeformed and deformed cubes by placing them side by side, with the vectors $v^{(i)}$ and $u^{(i)}$ parallel, as shown in the figure. 

 

 

 

3.19 Generalized strain measures

 

The polar decompositions $F=R\cdot U$ and $F=V\cdot R$ provide a way to define additional strain measures.  Let ${\lambda }_i$ denote the principal stretches, and let $u^{(i)}$ and $v^{(i)}$ denote the normalized eigenvectors of U and V.  Then one could define strain tensors through

$\begin{array}{l}\text{Lagrangian Nominal strain: }{\displaystyle \sum _{\text{i=1}}^{\text{3}}({\lambda }_i-1)u^{(i)}\otimes u^{(i)}}\\ \text{Lagrangian Logarithmic strain: }{\displaystyle \sum _{\text{i=1}}^{\text{3}}\log ({\lambda }_i)u^{(i)}\otimes u^{(i)}}\end{array}$

The correspoinding Eulerian strain measures are

$\begin{array}{l}\text{Eulerian Nominal strain: }{\displaystyle \sum _{\text{i=1}}^{\text{3}}({\lambda }_i-1)v^{(i)}\otimes v^{(i)}}\\ \text{Eulerian Logarithmic strain: }{\displaystyle \sum _{\text{i=1}}^{\text{3}}\log ({\lambda }_i)v^{(i)}\otimes v^{(i)}}\end{array}$

Another strain measure can be defined as

$\text{Green's strain: }{E}_G\text{= }{\displaystyle \sum _{\text{i=1}}^{\text{3}}\frac{1}{2}({\lambda }_i^2-1)v^{(i)}\otimes v^{(i)}}$

This can be computed directly from the deformation gradient as

$E_G\text{= }\frac{\text{1}}{\text{2}}\left(F\cdot F^T-I\right)$

and is very similar to the Lagrangean strain tensor, except that its principal directions are rotated through the rigid rotation R.

 

 

3.20  Measure of rate of deformation - the velocity gradient

 

We now list several measures of the rate of deformation. The velocity gradient is the basic measure of deformation rate, and is defined as

$L={\nabla }_{y}v\equiv L_{ij}=\frac{\partial v_i}{\partial y_j}$

It quantifies the relative velocities of two material particles at positions y and y+dy in the deformed solid, in the sense that

$d{v}_i=v_i(y+dy)-v_i(y)=\frac{\partial v_i}{\partial y_j}d{y}_j$

The velocity gradient can be expressed in terms of the deformation gradient and its time derivative as

${\nabla }_{y}v=\dot{F}\cdot F^{-1}\frac{\partial v_i}{\partial y_j}={\dot{F}}_{ik}{F}_{kj}^{-1}$

To see this, note that

$d{v}_i=\frac{d}{dt}d{y}_i=\frac{d}{dt}\left(F_{ij}d{x}_j\right)={\dot{F}}_{ij}d{x}_j$

and recall that $d{y}_j=F_{ji}d{x}_i\Rightarrow d{x}_j=F_{jk}^{-1}d{y}_k$, so that

$d{v}_i={\dot{F}}_{ij}{F}_{jk}^{-1}d{y}_k$

 

 

 

3.21 Stretch rate and spin (vorticity) tensors

 

The stretch rate tensor is defined as $D=\left(L+L^T\right)/2D_{ij}=\left(L_{ij}+L_{ji}\right)/2$

The spin tensor or Vorticity tensor is defined as $W=\left(L-L^T\right)/2W_{ij}=\left(L_{ij}-L_{ji}\right)/2$

 

A general velocity gradient can be decomposed into the sum of stretch rate and spin, as

$L=D+W{L}_{ij}=D_{ij}+W_{ij}$

The stretch rate quantifies the rate of stretching of material fibers in the deformed solid, in the sense that

$\frac{1}{l}\frac{dl}{dt}=n\cdot D\cdot n=n_i{D}_{ij}{n}_j$

is the rate of stretching of a material fiber with length l and orientation n in the deformed solid.  To see this, let $dy=ln$, so that

$\frac{d}{dt}dy=\frac{dl}{dt}n+l\frac{dn}{dt}$

By definition,

$\frac{d}{dt}dy=\frac{d}{dt}\left(F\cdot dx\right)=\dot{F}\cdot dx=\dot{F}\cdot \left(F^{-1}dy\right)=\dot{F}{F}^{-1}\cdot dy=L\cdot dy=(D+W)\cdot ln$

Hence

$(D+W)\cdot ln=\frac{dl}{dt}n+l\frac{dn}{dt}$

Finally, take the dot product of both sides with n, note that since n is a unit vector $dn/dt$ must be perpendicular to n and therefore $n\cdot dn/dt=0$.  Note also that $n\cdot W\cdot n=0$, since W is skew-symmetric.  It is easiest to show this using index notation: $n_i{W}_{ij}{n}_j=n_i(L_{ij}-L_{ji})n_j/2=0$.  Therefore

$n\cdot (D+W)\cdot ln=\frac{dl}{dt}n\cdot n+ln\cdot \frac{dn}{dt}\Rightarrow n\cdot D\cdot ln=\frac{dl}{dt}$

 

The spin tensor W can be shown to provide a measure of the average angular velocity of all material fibers passing through a material point. 

 

The vorticity vector is another measure of the angular velocity.  It is defined as

$w=curl(v)w_i={\in }_{ijk}\frac{\partial v_k}{\partial y_j}$

 

It is related to the spin tensor as

$\omega =2dual(W){\omega }_i=-{\in }_{ijk}{W}_{jk}$

Where dual (W) denotes the dual vector of the skew tensor W.

 

The vorticity vector has the property that, for any vector g, $Wg=\frac{1}{2}\omega \times g{W}_{ji}{g}_i=\frac{1}{2}{\in }_{jki}{\omega }_k{g}_i$.

 

A motion satisfying W=curl(v)= 0 is said to be irrotational $–$ such motions are of interest in fluid mechanics.

 

 

 

 

3.22 Spatial (Eulerian) description of acceleration

 

The acceleration of a material particle is, by definition

$a={\frac{\partial v}{\partial t}|}_{x=const}$

 

In fluid mechanics, it is often convenient to use a spatial description of velocity and acceleration $–$ that is to say the velocity field is expressed as a function of position y in the deformed solid as $v(y,t)$.   The acceleration of the material particle with instantaneous position y in the deformed solid can be expressed as

$\begin{array}{l}{a}_i=\frac{\partial v_i}{\partial y_k}\frac{\partial y_k}{\partial t}+{\frac{\partial v_i}{\partial t}|}_{y_i=const}=L_{ik}{v}_k+{\frac{\partial v_i}{\partial t}|}_{y_i=const}\\ =\left(D_{ik}+W_{ik}\right)v_k+{\frac{\partial v_i}{\partial t}|}_{y_i=const}=\left(D_{ik}+W_{ik}\right)v_k+{\frac{\partial v_i}{\partial t}|}_{y_i=const}\end{array}$

 

 

3.23 Acceleration - spin $–$ vorticity relations

 

In fluid mechanics, equations relating the acceleration to the spatial velocity field are useful.  In particular, it can be shown that

  $a_i={\frac{\partial v_i}{\partial t}|}_{x_k=const}={\frac{\partial v_i}{\partial t}|}_{y_k=const}+\frac{1}{2}\frac{\partial }{\partial y_i}(v_k{v}_k)+2W_{ik}{v}_k$

 $a_i={\frac{\partial v_i}{\partial t}|}_{x_k=const}={\frac{\partial v_i}{\partial t}|}_{y_k=const}+\frac{1}{2}\frac{\partial }{\partial y_i}(v_k{v}_k)+{\in }_{ijk}{\omega }_j{v}_k$

 ${\in }_{ijk}\frac{\partial a_k}{\partial y_j}={\frac{\partial {\omega }_i}{\partial t}|}_{x=const}-D_{ij}{\omega }_j+\frac{\partial v_k}{\partial y_k}{\omega }_i$

 

Deriving these relations is left as an exercise.

 

 

 

3.24 Rate of change of volume

 

We have seen that

$J=\det (F)$

quantifies the volume change associated with a deformation, in that

$JdV=d{V}_0$

 

In rate form:

$\frac{dJ}{dt}=\frac{dJ}{d{F}_{ij}}\frac{d{F}_{ij}}{dt}=J{F}_{ji}^{-1}{\dot{F}}_{ij}=J{L}_{ii}=J\frac{\partial v_i}{\partial y_i}=J{D}_{ii}$.

 

The trace of D, trace of L or the trace of grad(v)  are therefore measures of rate of change of volume.

 

 

 

 

3.25 Infinitesimal strain rate and rotation rate

 

For small strains the rate of deformation tensor is approximately equal to the infinitesimal strain rate, while the spin can be approximated by the time derivative of the infinitesimal rotation tensor

$\begin{array}{l}\frac{d}{dt}\epsilon =\frac{d}{dt}\frac{1}{2}\left(u\otimes \nabla +{\left(u\otimes \nabla \right)}^T\right)\approx D\text{or }{\dot{\epsilon }}_{ij}\approx D_{ij}\\ \frac{d}{dt}w=\frac{d}{dt}\frac{1}{2}\left(u\otimes \nabla -{\left(u\otimes \nabla \right)}^T\right)\approx W\text{or }{\dot{w}}_{ij}\approx W_{ij}\end{array}$

The approximation is because the infinitesimal strain and rotation involve derivatives with respect to position in the reference configuration, while the stretch rate and spin tensors are defined in terms of spatial derivatives.   Similarly, you can show that

                                                    $\frac{d}{dt}\frac{\partial u_i}{\partial x_j}={\dot{F}}_{ij}={\dot{\epsilon }}_{ij}+{\dot{w}}_{ij}\approx L_{ij}$

 

 

 

3.26 Other deformation rate measures

 

The rate of deformation tensor can be related to time derivatives of other strain measures.  For example the time derivative of the Lagrange strain tensor can be shown to be

$\frac{dE}{dt}=F^T\cdot D\cdot F{\dot{E}}_{ij}=F_{ki}{D}_{kl}{F}_{lj}$

Other useful results are

 For a pure rotation $\dot{R}\cdot R^T+R\cdot {\dot{R}}^T=0$, or equivalently $\dot{R}\cdot R^T=-{\left(\dot{R}\cdot R^T\right)}^T$.  To see this, recall that $R{R}^T=I$ and evaluate the time derivative. 

 If the deformation gradient is decomposed into a stretch followed by a rotation as $F=R\cdot U$ then $D=R\cdot \left(\dot{U}\cdot U^{-1}+U^{-1}\cdot \dot{U}\right)\cdot R^T/2$ and $W=\dot{R}\cdot R^T+R\cdot \left(\dot{U}\cdot U^{-1}-U^{-1}\cdot \dot{U}\right)\cdot R^T/2$

 

For small strains the rate of change of  Lagrangian strain E  is approximately equal to the rate of change of infinitesimal strain $\text{ε}$:

$\frac{dE}{dt}\approx \frac{d}{dt}\epsilon {\dot{E}}_{ij}\approx \frac{d}{dt}{\epsilon }_{ij}$

 

3.27 Path lines, streamlines, and vortex lines

 

Path lines, streamlines, and vortex lines are useful concepts in fluid mechanics.

 

 A path line is the curve traced by a material particle as it moves through space.   If the curve is described in parametric form by $y_i(\lambda )$, with $\lambda$ a scalar, then the curve satisfies

$\frac{d{y}_i(t)}{dt}=v_i(X,t)$

 

 A stream line is a curve that is everywhere tangent to the spatial velocity vector.   In general, streamlines may be functions of time.  If $y_i(\lambda ,t)$ is the parametric representation of the curve, at time t , then $y_i(\lambda ,t)$ is a member of the family of solutions to the differential equation

$\frac{d{y}_i(\lambda ,t)}{d\lambda }=v_i\left(y(\lambda ),t\right)$

For the particular case of a steady flow, the spatial velocity field is (by definition) independent of time, and therefore the curves are fixed in space.

 

 A vortex line is a curve that is everywhere tangent to the vorticity vector.   These curves satisfy the differential equation

$\frac{d{y}_i(\lambda ,t)}{d\lambda }={\omega }_i\left(y(\lambda ),t\right)$

Again, for the special case of a steady flow the vortex lines are independent of time.

 

 

 

3.27 Reynolds Transport Relation

 

The Reynolds transport theorem is a useful way to calculate the rate of change of a quantity inside a volume that deforms with a solid (e.g. the total mass of a volume).  Let $\varphi (y,t)$ be any scalar valued property of a material particle at position y in the deformed solid.   The Reynolds transport relation states that rate of change of the total value of this property within a volume V of a deformed solid can be calculated as

$\frac{d}{dt}{\displaystyle \underset{V}{\int }\varphi dV}={\displaystyle \underset{V}{\int }\left({\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi \frac{\partial v_i}{\partial y_i}\right)}dV={\displaystyle \underset{V}{\int }\left({\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi D_{kk}\right)}dV={\displaystyle \underset{V}{\int }\left({\frac{\partial \varphi }{\partial t}|}_{y=const}\right)}dV+{\displaystyle \underset{S}{\int }\left(\varphi v_k{n}_k\right)}dA$

 

Note that the material volume V and surface S convect with the deforming solid $–$ they are not control volumes.

 

To see this, note that we can’t take the time derivative inside the integral because the volume changes with time as the solid deforms.   But we can map the integral back to the reference configuration, which is time independent $–$ the derivative can then be taken inside the integral.

$\begin{array}{l}\frac{d}{dt}{\displaystyle \underset{V}{\int }\varphi dV}=\frac{d}{dt}{\displaystyle \underset{V_0}{\int }\varphi JdV}={\displaystyle \underset{V_0}{\int }\left({J\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi \frac{\partial J}{\partial t}\right)}dV\\ ={\displaystyle \underset{V_0}{\int }\left({\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi D_{kk}\right)}JdV={\displaystyle \underset{V}{\int }\left({\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi D_{kk}\right)}dV\end{array}$

 

The last result follows by noting that ${\frac{\partial \varphi }{\partial t}|}_{x=const}={\frac{\partial \varphi }{\partial t}|}_{y=const}+\frac{\partial \varphi }{\partial y_i}{v}_i$.  Then note that $\frac{\partial \varphi }{\partial y_i}{v}_i+\varphi \frac{\partial v_i}{\partial y_i}=\frac{\partial (\varphi v_i)}{\partial y_i}$ and apply the divergence theorem to this term.

 

3.28 Transport Relations for material curves and surfaces

 

Similar transport relations can be derived for material curves and surfaces which convect with a deformable solid or fluid.

 

Let C be a material curve in a deformable solid; and let S be an interior surface with normal vector n.  Let $\varphi (y,t)$ be any scalar valued property of a material particle at position y in the deformed solid.   Then

1. $\frac{d}{dt}{\displaystyle \underset{C}{\int }\varphi {\tau }_{i}ds}={\displaystyle \underset{C}{\int }\left({{\delta }_{ij}\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi \frac{\partial v_i}{\partial y_j}\right){\tau }_j}ds$
2. $\frac{d}{dt}{\displaystyle \underset{S}{\int }\varphi n_{i}dA}={\displaystyle \underset{S}{\int }\left({{\delta }_{ij}\frac{\partial \varphi }{\partial t}|}_{x=const}+{\delta }_{ij}\varphi \frac{\partial v_k}{\partial y_k}-\varphi \frac{\partial v_j}{\partial y_i}\right)n_j}dA$

 

To show the first result, start by mapping the integral to the reference configuration, then take the time derivative, and map back to the current configuration, as follows

$\begin{array}{l}\frac{d}{dt}{\displaystyle \underset{C}{\int }\varphi {\tau }_{i}ds}=\frac{d}{dt}{\displaystyle \underset{C_0}{\int }\varphi F_{ij}{\tau }_j^{0}d{s}_0}={\displaystyle \underset{C_0}{\int }\left({\frac{d\varphi }{dt}|}_x{F}_{ij}+\varphi \frac{d{F}_{ij}}{dt}\right){\tau }_j^{0}d{s}_0}\\ ={\displaystyle \underset{C}{\int }\left({\frac{d\varphi }{dt}|}_x{\delta }_{ik}+\varphi \frac{d{F}_{ij}}{dt}{F}_{jk}^{-1}\right){\tau }_{k}ds}={\displaystyle \underset{C}{\int }\left({{\delta }_{ij}\frac{\partial \varphi }{\partial t}|}_{x=const}+\varphi \frac{\partial v_i}{\partial y_j}\right){\tau }_j}ds\end{array}$

To show the second, apply the same process to the surface integral.  The details are left as an exercise…

 

 

3.28 Circulation and the circulation transport relation

The circulation of the velocity field around a closed curve C is defined as

$I_C={\displaystyle \underset{C}{\int }v\cdot \tau ds}$,

where $\tau$ is a unit vector tangent to the curve.  If C is a reducible curve (i.e. if there is a regular, open surface S bounded by C  that lies within the configuration) then Stokes theorem shows that

$I_C={\displaystyle \underset{C}{\int }v\cdot \tau ds}={\displaystyle \underset{S}{\int }curl(v)\cdot mdA}$

The circulation transport relation states that

${\frac{\partial I_c}{\partial t}|}_{x=const}={\displaystyle \underset{C}{\int }{\frac{\partial v}{\partial t}|}_{x=const}\cdot \tau ds}$

for any material curve (i.e. a curve that convects with material particles within a body).  To see this recall the transport relation for a material curve, and set $\varphi =v_i$

$\frac{d}{dt}{\displaystyle \underset{C}{\int }{v}_i{\tau }_{i}ds}={\displaystyle \underset{C}{\int }\left({\frac{\partial v_j}{\partial t}|}_{x=const}+v_i\frac{\partial v_i}{\partial y_j}\right){\tau }_j}ds$

Note that

$v_i\frac{\partial v_i}{\partial y_j}=\frac{1}{2}\frac{\partial (v_i{v}_i)}{\partial y_j}$

and hence

${\displaystyle \underset{C}{\int }\left(v_i\frac{\partial v_i}{\partial y_j}\right){\tau }_j}ds={\displaystyle \underset{C}{\int }\frac{d(v_i{v}_i)}{ds}}ds=0$

because C is a closed curve.

 

Kelvin’s circulation theorem is a direct consequence of this result.   The theorem states that if the acceleration is the gradient of a potential, then the circulation around any closed material curve remains constant.   To see this, let

$\begin{array}{l}{\frac{\partial v_i}{\partial t}|}_{x=const}=\frac{\partial \varphi }{\partial y_i}\\ {\frac{\partial I_c}{\partial t}|}_{x=const}={\displaystyle \underset{C}{\int }\frac{\partial \varphi }{\partial y_i}\cdot {\tau }_{i}ds}={\displaystyle \underset{C}{\int }\frac{\partial \varphi }{\partial s}ds}=0\end{array}$