# Kinematic Description

For geometrically nonlinear problems, that is, problems in which rigid body rotations and deformation are large, a large number of measures of deformation are possible but most theoretical work and computer software employ the following three measures:
• The velocity strain (or rate of deformation)

${D}_{ij}=\frac{1}{2}\left(\frac{\partial {v}_{i}}{\partial {x}_{j}}+\frac{\partial {v}_{j}}{\partial {x}_{i}}\right)$

• The Green strain tensor (Lagrangian strain tensor) measured with respect to initial configuration

${E}_{ij}=\frac{1}{2}\left(\frac{\partial {u}_{i}}{\partial {X}_{j}}+\frac{\partial {u}_{j}}{\partial {X}_{i}}+\frac{\partial {u}_{k}}{\partial {X}_{i}}\frac{\partial {u}_{k}}{\partial {X}_{j}}\right)$

• The Almansi strain tensor (Eulerian strain tensor) measured with respect to deformed configuration

${E}^{A}{}_{ij}=\frac{1}{2}\left(\frac{\partial {u}_{i}}{\partial {x}_{j}}+\frac{\partial {u}_{j}}{\partial {x}_{i}}-\frac{\partial {u}_{k}}{\partial {x}_{i}}\frac{\partial {u}_{k}}{\partial {x}_{j}}\right)$

All three measures of strains can be related to each other and can be used with any type of mesh.

## Velocity Strain or Deformation Rate

The strain rate is derived from the spatial velocity derivative:

${\stackrel{˙}{\epsilon }}_{ij}=\frac{d{\epsilon }_{ij}}{dt}={D}_{ij}=\frac{1}{2}\left(\frac{\partial {v}_{i}}{\partial {x}_{j}}+\frac{\partial {v}_{j}}{\partial {x}_{i}}\right)$

or in matrix form:

$\stackrel{˙}{\epsilon }=D=\frac{1}{2}\left(L+{L}^{T}\right)$

Where, the velocity gradient in the current configuration is:

${L}_{ij}=\frac{\partial {v}_{i}}{\partial {x}_{j}}$

The velocity of a material particle is:

${v}_{i}=\frac{\partial {x}_{i}}{\partial t}$

Where, the partial differentiation with respect to time $t$ means the rate of change of the spatial position $x$ of a given particle. The velocity difference between two particles in the current configuration is given by:

$d{v}_{i}=\frac{\partial {v}_{i}}{\partial {x}_{j}}d{x}_{j}={L}_{ij}d{x}_{j}={L}_{ij}{F}_{jk}d{X}_{k}$

In matrix form:

$dv=Ldx=LFdX$

On the other hand, it is possible to write the velocity difference directly as:

$dv=\frac{\partial }{\partial t}\left(FdX\right)=\stackrel{˙}{F}dX$

Where,

$\stackrel{˙}{F}=\frac{\partial F}{\partial t}$

One has as a result:

$L=\stackrel{˙}{F}{F}^{-1}$

Now, $L$ is composed of a rate of deformation and a rate of rotation or spin:

$L=D+\Omega$

Since these are rate quantities, the spin can be treated as a vector. It is thus possible to decompose $L$ into a symmetric strain rate matrix and an anti symmetric rotation rate matrix just as in the small motion theory the infinitesimal displacement gradient is decomposed into an infinitesimal strain and an infinitesimal rotation. The symmetric part of the decomposition is the strain rate or the rate of deformation and is:

$\stackrel{˙}{\epsilon }=D=\frac{1}{2}\left(\stackrel{˙}{F}{F}^{-1}+{F}^{-T}{\stackrel{˙}{F}}^{T}\right)$

The anti symmetric part of the decomposition is the spin matrix:

$\Omega =\frac{1}{2}\left(\stackrel{˙}{F}{F}^{-1}-{F}^{-T}{\stackrel{˙}{F}}^{T}\right)$

The velocity-strain measures the current rate of deformation, but provides no information about the total deformation of the continuum. In general, Equation 13 is not integral analytically; except in the unidimensional case, where one obtains the true strain:

$\epsilon =In\left(l/L\right)$

$l$ and $L$ are respectively the dimensions in the deformed and initial configurations. Furthermore, the integral in time for a material point does not yield a well-defined, path-independent tensor so that information about phenomena such as total stretching is not available in an algorithm that employs only the strain velocity. Therefore, to obtain a measure of total deformation, the strain velocity has to be transformed to some other strain rate.

The volumetric strain is calculated from density. For one dimensional deformation:

$\mu =\frac{\rho }{{\rho }_{0}}-1=\frac{-\delta \Omega }{\Omega }=\frac{-\delta l}{l}$

## Green Strain Tensor

The square of the distance which separates two points in the final configuration is given in matrix form by:

$d{x}^{T}dx=d{X}^{T}{F}^{T}FdX$

Subtracting the square or the initial distance, you have:

$d{x}^{T}dx-d{X}^{T}dX=d{X}^{T}\left({F}^{T}F-\text{I}\right)dX=2Ed{X}^{T}dX$
$E=\frac{1}{2}\left({F}^{T}{F}^{T}-\text{I}\right)$

$C={F}^{T}F$ and $B=F{F}^{T}$ are called respectively right and left Cauchy-Green tensor.

$E=\frac{1}{2}\left(A+{A}^{T}+{A}^{T}A\right)$
${E}_{ij}=\frac{1}{2}\left(\frac{\partial {u}_{i}}{\partial {X}_{j}}+\frac{\partial {u}_{j}}{\partial {X}_{i}}+\frac{\partial {u}_{k}}{\partial {X}_{i}}\frac{\partial {u}_{k}}{\partial {X}_{j}}\right)$

In the unidimensional case, the value of the strain is:

$E=\left({l}^{2}-{L}^{2}\right)/\left(2{L}^{2}\right)$

Where, $l$ and $L$ are respectively the dimensions in the deformed and initial configurations.

It can be shown that any motion $F$ can always be represented as a pure rigid body rotation followed by a pure stretch of three orthogonal directions:

$F=RU=R\left(I+H\right)$

with the rotation matrix $R$ satisfying the orthogonality condition:

${R}^{T}R=I$

and $H$ symmetric.

The polar decomposition theorem is important because it will enable to distinguish the straining part of the motion from the rigid body rotation.

From Equation 20 and Equation 24:

$E=H+\frac{{H}^{2}}{2}=\frac{1}{2}\left({F}^{T}F-I\right)$
$R=F{\left(I+H\right)}^{-1}$

Equation 26 allows the computation of $H$ , and Equation 27 of $R$ .

As the decomposition of the Jacobian matrix $F$ exists and is unique, $H$ is a new measure of strain which is sometimes called the Jaumann strain. Jaumann strain requires the calculation of principal directions.

If rotations are small,

$R=I+\Omega$
${R}^{T}R={\left(I+\Omega \right)}^{T}\left(I+\Omega \right)$
${\Omega }^{T}+\Omega =0$

if second order terms are neglected.

As a result, for the Jacobian matrix:

$F=I+A=\left(I+\Omega \right)\left(I+H\right)$

leading, if the second order terms are neglected, to the classical linear relationships:

$A=\Omega +H$
$H=\frac{1}{2}\left({A}^{T}+A\right)$
$\Omega =\frac{1}{2}\left(A-{A}^{T}\right)$

So for Equation 32 and Equation 33, when rigid body rotations are large, the linear strain tensor becomes non-zero even in the absence of deformation.

The preceding developments show that the linear strain measure is appropriate if rotations can be neglected; that means if they are of the same magnitude as the strains and if these are of the order of 10-2 or less. It is also worth noticing that linear strains can be used for moderately large strains of the order of 10-1 provided that the rotations are small. On the other hand, for slender structures which are quite in extensible, nonlinear kinematics must be used even when the rotations are order of 10-2 because, if you are interested in strains of 10-3 - 10-4, using linear strain the error due to the rotations would be greater than the error due to the strains.

Large deformation problems in which nonlinear kinematics is necessary, are those in which rigid body rotation and deformation are large.