4. kinematics jan 2013

Kinematics
The Study of Deformation & Motion

Definition

“…the various possible types of motion in themselves,
leaving out … the causes to which the initiation of
motion may be ascribed … constitutes the science of
Kinematics.”—ET Whittaker
“Kinematics does not deal with predicting the
deformation resulting from a given loading, but rather
with the machinery for describing all possible
deformations a body can undergo” — EB Tadmore et al.

Department of Systems Engineering, University of Lagos 2 oafak@unilag.edu.ng 12/29/2012

Context
There are three major aspects that interest us of the
behavior of a continuously distributed body. The first
subject of this chapter, kinematics, is an organized
geometrical description of its displacement and motion.
We shall also look at a mathematical description of
internal forces. In the next chapter we shall look at
basic balance laws and the second law of
thermodynamics which describes the inbalance of
entropy. The emphasis here is the fact that these
principles are independent of the material considered.
While we may use the terminology of solid mechanics,
these laws are valid for any continuously distributed
material.

Balance Laws and the Theory of Stress

All materials respond to external influences by obeying
these same laws. The differences observed in their
responses are results of their constitution. Such
constitutive models distinguish between solids and
fluids, elastic and inelastic or time independent and
materials with time dependent behaviors. We shall
endeavor to engage general principles in their most
general forms.


Balance Laws and the Theory of Stress

Many books that engineering students encounter at
this point treat the three levels of relations (kinematic,
balance laws and constitutive models) differently for
different materials. The reality is that only the
constitutive models differ. The kinematics, transmission
of forces and balance laws are material independent.


Placement of Bodies

The abstract material body will be considered as a
three-dimensional manifold with boundary, consisting
of points, which we call material (in contrast to spatial
points). The body becomes observable by us when it
moves through the space. Mathematically, such a
motion is a time-dependent embedding into the
Euclidean space. We assume that at each instant, there
is a mapping of each point in the body to R 3 and that
all coordinate changes are differentiable.


Placements in 3-D Euclidean Space
2 3𝑋2
The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as: 𝑥1 , 𝑥2 = 𝑋1 + 𝑋2 /2, . Using
3.5
Mathematica (oafakDeform.nb)

Deformation of the original material can be viewed as
placements in 3-D Euclidean Space. Motion is a time dependent
sequence of placements


Spatial Space

At each instant, this embedding is called a placement of
the body B at a time 𝑡 ∈ R, and it is given by a mapping
𝜒: B → E
Or, including the time directly, we can write,
𝜒(⋅, 𝑡): B → E
So that the motion of the body is the smooth function
that assigns to each Euclidean point 𝐗 ∈ B a point,
𝐱 = 𝜒(𝐗, 𝑡)


Basic Hypotheses

1. That the mapping 𝜒(⋅, 𝑡): B → E be bijective.
Physically this one-to-one mapping guarantees that
no two material points occupy the same spatial
point at once. Physically, we are saying that the
material does not penetrate itself.
2. That the determinant of the material gradient is
never zero or,
𝐽 𝐗, 𝑡 ≡ 𝛻𝜒 𝐗, 𝑡 ≠ 0


Material Indestructibility

 The last stipulation guarantees that the deformation
must not be such that the material vanishes. For the
Jacobian to vanish, we must be able to deform a finite
material to nothingness. That situation is not
envisaged here in the hypothesis.
 Beginning from an initial state when 𝐽 𝐗, 𝑡 = 1, we
can easily conclude that for continuity, 𝐽 𝐗, 𝑡 > 0 ∀𝑡.
Otherwise, the state 𝐽 𝐗, 𝑡 = 0 shall have been
reached prior to any negative state; An impossibility!


Material & Spatial Vectors

 The “Reference Placement” of the material is defined
as an abstract state for the identification of the actual
material points. Several authors find it necessary to
use some initial or undeformed state of the material
for this purpose. For our use here, we consider it
purely imaginary and existing only for the purpose of
analysis.
 The points in the reference state are called “Material
Points”. The vector space associated with it contains
material vectors. The reference placement is time
independent.


Material & Spatial Vectors

 The body in question is seen only as it evolves
through time in the mapping we have previously
defined.
 The vectors associated with the spatial points
𝐱 = 𝜒(𝐗, 𝑡) are called spatial vectors. Vectors
associated with material points 𝐗 are material
vectors.
 Note that this separation, though necessary for
analysis is artificial and imaginary. In fact, only the
spatial placement is visible as it evolves over time.


Velocity & Acceleration

The spatial vectors,
𝜕𝝌(𝐗, 𝑡)
𝐱 = 𝝌 𝐗, 𝑡 ≡
𝜕𝑡
and
𝜕 2 𝝌(𝐗, 𝑡)
𝐱 = 𝝌 𝐗, 𝑡 ≡
𝜕𝑡 2
are the velocity and acceleration of the material point 𝐗
at time 𝑡. Let it be clear that despite the fact that 𝐱 and
𝐗 are not vectors (they are points) but 𝑑𝐱 and 𝑑𝐗 are
spatial and material vectors respectively.


Further Hypotheses

1. We further assume that material cannot cross the
boundary of a spatial region convecting with the
body.


Convected Coordinates

Imagine the coordinate system were to be fixed with
the body and deforms with it.
Such a coordinate system is said to be convected
coordinate system
Even if we started out with rectangular Cartesian, we
would end up with a curvilinear system as shown below:


The two figures above show the location near the
corner of a triangle prior to and sequel to a deformation
transformation when coordinate lines are allowed to
deform with the material. As a result of the
deformation, the coordinate locating the point of
interest did not change since we allow the coordinates
to deform with the triangle. In the deformed state,
what started as a Cartesian system has been
transformed to curvilinear coordinates. The coordinate
curves are bent and therefore the coordinate bases are
now tangents to the coordinate lines.



In the undeformed system here, the coordinate bases
and coordinate lines are one and the same. All that has
changed because of the deformation. The straight edge
of the triangle itself looks more like an arc in the
deformed state. Yet, in all this, the coordinate shift from
the highlighted point to the triangle edge remains
unchanged.
The above shows that the convected coordinates retain
the location but lose the bases.


The Deformation Gradient

At any instant, the vector differential of the mapping, 𝐱 = 𝜒(𝐗, 𝑡) in
the Gateaux sense is,
𝑑𝐱 = 𝛻𝜒 𝐗, 𝑡 𝑑𝐗
So that we can write that,
𝑑𝐱 = 𝐅 𝐗, 𝑡 𝑑𝐗
Where the Frechét derivative, the tensor
𝑑𝐱
𝐅 𝐗, 𝑡 ≡ = 𝛻𝜒 𝐗, 𝑡
𝑑𝐗
is called the deformation gradient. Clearly, the deformation
gradient maps infinitesimal material vectors (e.g. 𝑑𝐗)
to infinitesimal spatial vectors (e.g. 𝑑𝐱).

Homogeneous Deformation

At a particular instant in time, the placement
𝜒 𝑡 𝐗 ≡ 𝜒 𝐗, 𝑡
is the instantaneous displacement. 𝐅 𝐗, 𝑡 normally
varies throughout the material body. In the special case
when 𝐅 is constant through the material space, we have
“Homogeneous Deformation”.
Dropping the subscript 𝑡, we may write that, for
homogeneous deformations at a particular instant,


Homogeneous Deformation

 Dropping the subscript 𝑡, we may write that, for
homogeneous deformations at a particular instant,
for the material points 𝐗 and 𝐘,
𝜒 𝐗 − 𝜒 𝐘 = 𝐅 𝐗− 𝐘
From the above, we can see that the homogeneous
deformation gradient maps material vectors into spatial
vectors.


Material and Spatial Mapping

Dropping the functional dependencies, we have that,
𝑑𝐱 = 𝐅𝑑𝐗
In which the deformation gradient maps material
vectors to spatial. We can also write,
𝐅 −1 𝑑𝐱 = 𝑑𝐗
So that the inverse of the deformation gradient maps
spatial vectors to material vectors.


Transposes
Consider a spatial vector 𝐬. Take its inner product with
the spatial vector equation, 𝑑𝐱 = 𝐅𝑑𝐗, we obtain,
𝐬 ⋅ 𝑑𝐱 = 𝐬 ⋅ 𝐅𝑑𝐗 = 𝑑𝐗 ⋅ 𝐅 T 𝐬
Which clearly shows that 𝐅 T 𝐬 is a material vector.
Clearly, 𝐅 T maps spatial vectors to material vectors.
Given a material vector 𝐦, a similar consideration for
the scalar equation,
𝐦 ⋅ 𝐅 −1 𝑑𝐱 = 𝐦 ⋅ 𝑑𝐗 = 𝑑𝐱 ⋅ 𝐅 −T 𝐦
Clearly shows that 𝐅 −T is a map of material vectors to
spatial vectors.


Polar Decomposition Theorem

For a given deformation gradient 𝐅, there is a unique
rotation tensor 𝐑, and unique, positive definite
symmetric tensors 𝐔 and 𝐕 for which,
𝐅 = 𝐑𝐔 = 𝐕𝐑
This is a fundamental theorem in continuum mechanics
called the Polar decomposition theorem.


Examples of deformation mappings

The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such
2
as: 𝑥1 , 𝑥2 = 𝑋1 + 𝑋2 /2, 𝑋2 . Using Mathematica
(Reddy3.15.nb) the resulting deformation is:


The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as:
1 1
𝑥1 , 𝑥2 = 18 + 4𝑋1 + 6𝑋2 , 14 + 6𝑋2 . Using
4 4
Mathematica (Reddy3.4.3.15.nb) the resulting deformation
is:


The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as:
3𝑋1 𝑋2
𝑥1 , 𝑥2 = 4 − 2𝑋1 − 𝑋2 , 2 + − . Using Mathematica
2 2
(Holzapfel72.nb) the resulting deformation is:


Lines & Circles


Animation

An example of the function 𝐱 = 𝜒(𝐗, 𝑡) evolving temporally
and spatially.
This Mathematica animation demonstrates all the issues
discussed previously including:
1. Reference Placement
2. Motion Function
3. Spatial Placements
4. Time dependency
File presently at OAFAKAnimate.nb


Polar Decomposition

By the results of this theorem,
𝑹 𝑇 𝑹 = 𝑹𝑹 𝑇 = 𝑰
𝑹 is called the rotation tensor while 𝑼 and 𝑽 are the
right (or material) stretch tensor and the left (spatial)
stretch tensors respectively. Being a rotation tensor, 𝑹
must be proper orthogonal. In addition to being an
orthogonal matrix, the matrix representation of 𝑹 must
have a determinant that is positive:
det 𝑹 = +1.


Note that
𝐂 = 𝐅 T 𝐅 = 𝐔T 𝐑T 𝐑 𝐔 = 𝐔T 𝐈 𝐔 = 𝐔2 .
Definition: Positive Definite. A tensor 𝑻 is positive
definite if for every real vector 𝒖, the quadratic form
𝒖 ⋅ 𝑻𝒖 > 𝟎. If 𝒖 ⋅ 𝑻𝒖 ≥ 𝟎 Then 𝑻 is said to be positive
semi-definite.
Now every positive definite tensor 𝑻 has a square root
𝑼 such that,
𝑼2 ≡ 𝑼 𝑇 𝑼 = 𝑼𝑼 𝑻 = 𝑻


Proof

To prove this theorem, we must first show that 𝑭 𝑇 𝑭 is
symmetric and positive definite. Symmetry is obvious.
To show positive definiteness, For an arbitrary real
vector 𝒖 consider the expression, 𝒖 ⋅ 𝑭 𝑇 𝑭𝒖. Let the
vector 𝒃 = 𝑭𝒖. Then we can write,
𝒖 ⋅ 𝑭 𝑇 𝑭𝒖 = 𝒃 ⋅ 𝒃 = 𝒃 2 > 0
as the magnitude of any real vector must be positive.
Hence 𝑪 = 𝑭 𝑇 𝑭 is positive definite.


Uniqueness of the Root

A spectral decomposition of the symmetric, positive
definite tensor 𝑪 can be written as,
3

𝑪= 𝜔𝑖 𝒖𝑖 ⊗ 𝒖𝑖
𝑖=1
Given that 𝜔 𝑖 = 𝜆 𝑖 2 is the eigenvalue corresponding
to the normalized eigenvector 𝒖 𝑖 . Every quadratic form
with this spectral representation must be greater than
zero.



 It follows easily that each eigenvalue is positive because
contracting with each eigenvector from the left and right,
we have,
3

𝒖 𝑗 ⋅ 𝑪𝒖 𝑗 = 𝜔 𝑖 𝒖 𝑗 ⋅ 𝒖 𝑖 ⊗ 𝒖 𝑖 𝒖 𝑗 = 𝜔 𝑗 > 0.
𝑖=1
(note very carefully the suppression of the summation convention here)
 Above proves that each eigenvalue is greater than zero
and in the spectral form, 𝐽 = det 𝐶 = 3 𝜔 𝑖 > 0 . And
𝑖=1
since the determinant of a matrix is an invariant, this holds
true even in non spectral forms of 𝑪.


Now, let
3

𝑼= 𝜆𝑖 𝒖𝑖 ⊗ 𝒖𝑖
𝑖=1
Clearly,
3

𝑼𝟐 = 𝜆𝑖 𝒖𝑖 ⊗ 𝒖𝑖 𝜆1 𝒖1 ⊗ 𝒖1 + 𝜆2 𝒖2 ⊗ 𝒖2 + 𝜆3 𝒖3 ⊗ 𝒖3
𝑖=1
3 3
2
= 𝜆𝑖 𝒖𝑖 ⊗ 𝒖𝑖 = 𝜔𝑖 𝒖𝑖 ⊗ 𝒖𝑖 = 𝑪.
𝑖=1 𝑖=1
And this square root is unique, for were it not so, there would
be another positive definite tensor 𝑼 such that,
𝑼 𝟐 = 𝑼2 = 𝑪.


The eigenvalue equation,
𝑪 − 𝜆2 𝑰 𝒖 = 𝟎
is satisfied by each eigenvalue/vector pair for 𝑪. From the above,
we may write,
𝑪 − 𝜆2 𝑰 𝒖 = 𝑼 𝟐 − 𝜆2 𝑰 𝒖
= 𝑼 + 𝜆𝑰 𝑼 − 𝜆𝑰 𝑵
= 𝟎.
In the last expression, 𝑼 − 𝜆𝑰 𝒖 must be equal to zero. If not, we
then have the fact that
𝑼 + 𝜆𝑰 𝒖 = 𝟎
This would mean that – 𝜆 is an eigenvalue of 𝑼. An impossibility
because 𝑼 is positive definite and can only have positive
eigenvalues. If we had started with,
𝑪 − 𝜆2 𝑰 𝒖 = 𝑼 𝟐 − 𝜆2 𝑰 𝒖 = 𝑼 + 𝜆𝑰 𝑼 − 𝜆𝑰 𝒖 = 𝟎
we would equally reach the conclusion that 𝑼 − 𝜆𝑰 𝒖 = 𝟎.
And this will remain true as we use each eigenvalue of 𝑼. is also an eigenvalue/vector for 𝑼. That
proves that they are the same tensor. Hence the square root of the tensor 𝑪 is unique

Polar Decomposition: Physical
Meaning

Photo from wikipedia


The Rotation
To complete the Polar Decomposition Theorem, we
now need to show that the 𝑹 in
𝑭 = 𝑹𝑼
is a rotation. Now, from the above equation, we have
that,
𝑭𝑼−𝟏 = 𝑹
so that
𝑹 𝑻 𝑹 = 𝑼−𝑻 𝑭 𝑻 𝑭𝑼−𝟏 = 𝑼−𝟏 𝑼2 𝑼−𝟏 = 𝟏
Which shows 𝑹 to be an orthogonal tensor. But
det 𝑹 = det 𝑭𝑼−𝟏 = det 𝑭 × det 𝑼−𝟏 > 0.
From physical considerations, we know that determinant of the deformation gradient is necessarily
positive and that of the inverse of 𝑼 is positive because 𝑼−𝟏 is also positive definite. Hence we can see
that, det 𝑹 = +𝟏. Which, when added to the fact that 𝑹 𝑻 𝑹 = 𝟏 means that 𝑹 is a rotation.


The Stretch Tensors

It is an easy matter now to find the tensor 𝑽 such that
𝑭 = 𝑹𝑼 = 𝑽𝑹
It is obvious that 𝑽 = 𝑹𝑼𝑹 𝑇 is symmetric and is the
square root of the Finger tensor,
𝑩−𝟏 = 𝑭𝑭 𝑻 = 𝑽𝑹𝑹 𝑇 𝑽 𝑇 = 𝑽2
𝑼 is the Right Stretch tensor while 𝑽 is called the Left
Stretch Tensor


The Strain Tensor
The tensor,
1 1 2
𝑬= 𝑪− 𝟏 = 𝑼 − 𝟏
2 2
is called the Green-St Venant or Lagrange Strain Tensor.
Note immediately that this tensor vanishes if the
deformation gradient is a rotation or the identity
tensor. This is a general property of all strain tensors.
This is a general property of all strain tensors. Guided by
this fact, other strain tensors can be defined:


Strain Tensors
In fact any tensor satisfying,
1
𝑼𝑚− 𝟏 𝑚≠0
𝑚
log 𝑼 when 𝑚=0
ℰ = 𝑂𝑅
1
𝑽𝑚− 𝟏 𝑚≠0
𝑚
log 𝑽 when 𝑚=0
is a strain tensor. Clearly, 𝑚 = 2 in the first case gives
the Lagrange strain tensor while 𝑚 = 2 in the second
gives the Eulerian strain tensor.

Homework
1. Starting with the mapping properties of the deformation
gradient, show that
i. 𝑼, 𝑪 and 𝑬 map material vectors to material vectors
ii. 𝑽 and 𝑩 map spatial vectors to spatial vectors
iii. 𝑹 maps material vectors to spatial vectors
2. Using the definition of the principal invariants of a
tensor, show
i. 𝐼1 𝑪 = 2𝐼1 𝑬 + 3
1
ii. 𝐼2 𝑪 = 2 tr 2 𝑪 − tr 𝑪2
= 4𝐼2 𝑬 + 4tr 𝑬 + 3
= 4𝐼2 𝑬 + 4𝐼1 𝑬 + 3
3. And use the fact that for any tensor 𝑺, 𝐼3 𝑺 =
1
tr 3 𝑺 − 3tr 𝑺 tr 𝑺2 + 2tr(𝑺3 ) to show that
6
𝐼2 𝑪 = 8𝐼3 𝑬 + 4𝐼2 𝑬 + 2𝐼1 𝑬 + 1


Infinitesimal Fibers

 Consider two infinitesimal fibers 𝐟R and 𝐠 R in the
undeformed state. These can be represented by the
two material vectors. The equivalent spatial fibers are
spatial vectors 𝐟 and 𝐠. A dot product of these has a
physical meaning:
𝐟 ⋅ 𝐠 = 𝐟 ⋅ 𝐅𝐠 𝑅
= 𝐠 R ⋅ 𝐅 T 𝐟 = 𝐠 R ⋅ 𝐅 T 𝐅𝐟 𝑅
= 𝐠 R ⋅ 𝐔𝐑T 𝐑𝐔𝐟R
= 𝐠 R ⋅ 𝐔𝐔𝐟 𝑅 = 𝐔𝐟R ⋅ 𝐔𝐠 R


Infinitesimal Fibers

Setting 𝐟 = 𝐠, we immediately obtain,
𝐟 2 = 𝐔𝐟R 2
So that
𝐟 = 𝐔𝐟R
The deformed length of an infinitesimal fiber is
characterized by the Right Stretch Tensor.


Contained Angle

The cosine of the angle between two deformed
infinitesimal fibers can be obtained from,
𝐟⋅ 𝐠 𝐔𝐟 𝑅 ⋅ 𝐔𝐠 𝑅
=
𝐟 𝐠 𝐔𝐟 𝑅 𝐔𝐠 𝑅
Which is the same as the cosine of the angle between
the vectors 𝐔𝐟 𝑅 and 𝐔𝐠 𝑅 .
Clearly, 𝐔 also characterizes the angles between
infinitesimal fibers.


The Stretch Vector

Consider the material vector
Δ𝐗 = 𝐿𝐞
Where we have chosen the unit vector 𝐞 coinciding with
the particular material fibre. The corresponding spatial
fibre at a given time (suppressing the dependency on
time), is given by,
Δ𝐱 = 𝐅 𝐗 Δ𝐗 = 𝐿𝐅 𝐗 𝐞
Plus some terms that will vanish as we make Δ𝐗 small.


The Stretch Vector
In the limit as 𝐿 approaches zero,
Δ𝐱
lim = 𝐅 𝐗 𝐞
𝐿→0 𝐿
The magnitude of this quantity is defined as the
material stretch
𝜆= 𝐅 𝐗 𝐞
Clearly,
𝜆2 = 𝐅 𝐗 𝐞 ⋅ 𝐅 𝐗 𝐞 = 𝐳 ⋅ 𝐅 𝐗 𝐞
If we write 𝐳 = 𝐅 𝐗 𝐞. By the definition of the
transpose, we can see that,
𝜆2 = 𝐞 ⋅ 𝐅 T 𝐗 𝐳 = 𝐞 ⋅ 𝐅 T 𝐗 𝐅 𝐗 𝐞 = 𝐞 ⋅ 𝐂(𝐗)𝐞

Principal Stretches & Directions
The Right Stretch Tensor is symmetric and positive definite. It
can therefore be written in its spectral form as:
𝐔 = Σ3 𝜆 𝑖 𝐮 𝑖 ⊗ 𝐮 𝑖
𝑖=1
We are in a position to write the spectral forms of other
important tensors as follows:
𝐕 = 𝐑𝐔𝐑T
= Σ 3 𝜆 𝑖 𝐑 𝐮 𝑖 ⊗ 𝐮 𝑖 𝐑T
𝑖=1
= Σ 3 𝜆 𝑖 𝐑𝐮 𝑖 ⊗ 𝐑𝐮 𝑖 = Σ 3 𝜆 𝑖 𝐯 𝑖 ⊗ 𝐯 𝑖
𝑖=1 𝑖=1
Where 𝐯 𝑖 = 𝐑𝐮 𝑖
Showing that the Left Stretch Tensor has the same
eigenvalues but a rotated eigenvector from its corresponding
Right Stretch Tensor.


Spectral Forms

Moreover,
𝐂 = 𝐔2 = 𝐔𝐔
= Σ 3 𝜆2 𝐮 𝑖 ⊗ 𝐮 𝑖
𝑖=1 𝑖
A result that follows immediately we realize that
𝐮 𝑖 , 𝑖 = 1,2,3 is an orthonormal set.
𝐅 = 𝐑𝐔 = Σ 3 𝜆 𝑖 𝐑 𝐮 𝑖 ⊗ 𝐮 𝑖
𝑖=1
3
= Σ 𝑖=1 𝜆 𝑖 𝐯 𝑖 ⊗ 𝐮 𝑖
Remember that 𝐅 is not symmetric. Its product bases is
made up of eigenvectors from the left and right stretch
tensors.

Spectral Forms

The Lagrangian Strain Tensor,

1 1 3
𝑬= 𝑪 − 𝟏 = Σ 𝑖=1 𝜆2 − 1 𝐮 𝑖 ⊗ 𝐮 𝑖
𝑖
2 2


Volume & Area Changes
Consider an elemental volume in the reference state in
the form of a parallelepiped with dimensions 𝑑𝐗, 𝑑𝐘 and
𝑑𝐙. Let this deform to the paralellepiped bounded by
𝑑𝐱, 𝑑𝐲 and 𝑑𝐳 in the current placement caused by a
deformation gradient 𝐅.
 We require that this parallelepiped be of a non-trivial
size, ie 𝑑𝐗, 𝑑𝐘, 𝑑𝐙 ≠ 0
 This means the material vectors 𝑑𝐗, 𝑑𝐘 and 𝑑𝐙 are
linearly independent.
 Clearly, we must have that 𝑑𝐱 = 𝐅 𝑑𝐗, 𝑑𝐲 = 𝐅 𝑑𝐘 and
𝑑𝐳 = 𝐅 𝑑𝐙.


Deformation of Volume

𝑑z
𝑑𝐙 𝑑𝐲
𝑑𝐱
𝑑𝐘
𝑑𝐗


The Volume change

The undeformed volume is given by,
𝑑𝑉 = 𝑑𝐗, 𝑑𝐘, 𝑑𝐙
and the deformed volume
𝑑𝑣 = 𝑑𝐱, 𝑑𝐲, 𝑑𝐳 = 𝐅𝑑𝐗, 𝐅𝑑𝐘,𝐅𝑑𝐙
Clearly seeing that 𝑑𝐗, 𝑑𝐘 and 𝑑𝐙 are independent
vectors,
𝑑𝑣 𝐅𝑑𝐗, 𝐅𝑑𝐘,𝐅𝑑𝐙
= = 𝐼3 𝐅 = det 𝐅 ≡ 𝐽 > 0
𝑑𝑉 𝑑𝐗, 𝑑𝐘, 𝑑𝐙
We can also write, 𝑑𝑣 = 𝐽𝑑𝑉

Area Changes

For an element of area 𝑑𝒂 in the deformed body with a
vector 𝑑𝒙 projecting out of its plane (does not have to be
normal to it) we have the following relationship:
𝑑𝒗 = 𝐽𝑑𝑽 = 𝑑𝒂 ⋅ 𝑑𝒙 = 𝐽𝑑𝑨 ⋅ 𝑑𝑿
where 𝑑𝑨 is the element of area that transformed to 𝑑𝒂 and
𝑑𝑿 is the image of 𝑑𝒙 in the undeformed material. Noting
that, 𝑑𝒙 = 𝑭𝑑𝑿 we have,
𝑑𝒂 ⋅ 𝑭𝑑𝑿 − 𝐽𝑑𝑨 ⋅ 𝑑𝑿 = 𝒐
= 𝑭 𝑇 𝑑𝒂 − 𝐽𝑑𝑨 ⋅ 𝑑𝐗
where 𝒐 is the zero vector.


Nanson Formula

For an arbitrary vector 𝑑𝑿, we have:
𝑭 𝑇 𝑑𝒂 − 𝐽𝑑𝑨 = 𝒐
so that,
𝑑𝒂 = 𝐽𝑭−𝑇 𝑑𝑨 = 𝑭 𝐜 𝑑𝑨
where 𝑭 𝐜 is the cofactor tensor of the deformation
gradient.


Examples
For the uniform biaxial deformation, given that
{𝑥1 , 𝑥2 , 𝑥3 } = {𝜆1 𝑋1 , 𝜆2 𝑋2 , 𝑋3 }. Compute the
Deformation Gradient tensor, the Lagrangian Strain
Tensor as well as the Eulerian Strain Tensor
components.
𝜕 𝜆1 𝑋1 , 𝜆2 𝑋2 , 𝑋3 𝜆1 0 0
𝐹= = 0 𝜆2 0
𝜕 𝑋1 , 𝑋2 , 𝑋3
0 0 1
Clearly in this case,
𝜆1 2 0 0
𝑪 = 𝑭 𝑇 𝑭 = 𝑩−1 = 𝑭𝑭 𝑇 = 0 𝜆2 2 0
0 0 1


And the Piola
𝜆1 −2 0 0
Tensor 𝑩 = 𝑭−𝑇 𝑭−1 = 0 𝜆2 −2 0
0 0 1
Now, the Lagrangian Strain Tensor
1 1 𝜆1 2 − 1 0 0
𝑬= 𝑪− 𝑰 = 0 𝜆2 2 − 1 0
2 2
0 0 0
And the Eulerian Strain Tensor
1 1 1 − 𝜆1 −2 0 0
𝒆= 𝑰− 𝑩 = 0 1 − 𝜆2 −2 0
2 2
0 0 0


163.24 34.6 4.2
 Show that the tensor C 34.6 19. −30. is
4.2 −30. 178.
positive definite. (a) Find the square root of the C by
finding its spectral decomposition from its
eigenvalues and eigenvectors. (b) Use the
Mathematica function MatrixPower[C, ½] to
compare your result.


In Cartesian Coordinates, the deformation of a rectangular
sheet is given by: = 𝝀 𝟏 𝑿 𝟏 + 𝒌 𝟏 𝑿 𝟐 𝐠 𝟏 + 𝒌 𝟐 𝑿 𝟏 + 𝝀 𝟐 𝑿 𝟐 𝐠 𝟐 +
𝝀 𝟑 𝑿 𝟑 𝐠 𝟑 Compute the tensors 𝑭, 𝑪, 𝑬, 𝑼 and 𝑹. Show that
𝑹 𝑻 𝑹 = 𝟏. For 𝜆1 = 1.1, 𝜆2 = 1.25, 𝑘1 = 0.15, 𝑘2 = −0.2,
determine the principal values and directions of 𝑬. Verify
that the principal directions are mutually orthogonal.
Compute the strain invariants and show that they are
consistent with the characteristic equation.
𝜆1 𝑘1 0
𝐹 = 𝑘2 𝜆2 0
0 0 𝜆3
2 2
𝑘2 + 𝜆1 𝑘1 𝜆1 + 𝑘2 𝜆2 0
𝐶 = 𝑘1 𝜆1 + 𝑘2 𝜆2 𝑘1 + 𝜆2
2
2 0
0 0 𝜆2
3
Full code @ Taber02.nb


Homework
 A body undergoes a deformation defined by,
𝑦1 = 𝛼𝑥1 , 𝑦2 = − 𝛽𝑥2 + 𝛾𝑥3 , 𝑎𝑛𝑑 𝑦3 = 𝛾𝑥2 − 𝛽𝑥3
where 𝛼, 𝛽 𝑎𝑛𝑑 𝛾 are constants. Determine 𝑭, 𝑪, 𝑬, 𝑼
and 𝑹.
3 4
1
2 3
 Given the Deformation Gradient Tensor 0 1 0
0 0 1
Find the rotation tensor, the right stretch tensor and
the left stretch tensor. Demonstrate that the
Rotation tensor is true orthogonal.


Homework
 In the isochoric deformation gradient,
𝜆1 cos𝜃 𝜆2 sin𝜃 0
𝑭 𝑝 = −𝜆1 sin𝜃 𝜆2 cos𝜃 0 . Show that 𝜆1 = 𝜆−1
2
0 0 1


Material & Spatial Derivatives

Our main concern in this section is with scalars, vectors
and tensors of different orders defined over the
Euclidean Point Space. We call them Tensor Fields or
Tensor Point Functions.
By motion, we mean the mapping,
𝜒: ℰ × ℛ → ℰ
Which is a smooth function that assigns to each
material point 𝐗 ∈ ℰ and time 𝑡 ∈ ℛ a point
𝐱 = 𝜒(𝐗, 𝑡)
In the Euclidean point space occupied by the reference
particle at 𝐗.


Reference Map

𝑑𝐱
We assume that the Frechét derivative, has a non-
𝑑𝐗
𝑑𝐱
vanishing determinant 𝐽 = so that the inverse,
𝑑𝐗
𝐗 = 𝜒 −1 (𝐱, 𝑡)
exists. It is called the Reference Map. A field description
of any tensor with respect to 𝐗 and 𝑡 is a material
description while a description with respect to 𝐱 and 𝑡 is
a spatial description. The motion and the reference
maps provide a way to obtain a spatial description from
a material description and vice versa.


Reference Map

For a given arbitrary-order tensor field (scalar, vector, or
higher-order tensor) Ξ 𝑅 (𝐗, 𝑡) over the reference
placement, a simple change of variables gives,
Ξ 𝑅 𝐗, 𝑡 = Ξ 𝑅 𝜒 −1 (𝐱, 𝑡), 𝑡 ≡ Ξ 𝐱, 𝑡
By a simple application of the reference map. The
reverse operation for a field over a spatial placement,
Ξ 𝐱, 𝑡 = Ξ 𝜒(𝐗, 𝑡), 𝑡 ≡ Ξ 𝑅 𝐗, 𝑡
results from the motion description directly.
(Not distinguishing between the functions, subscript 𝑅 or free, can cause a lot of confusion. Some
writers try to avoid this by using uppercase variables for the material functions while using lower case
for spatial)


Time Derivatives

Material or substantial derivative of a field defined over
the reference placement can be written as,
𝜕Ξ 𝑅 𝐗, 𝑡
=Ξ𝑅
𝜕𝑡 𝐗
To compute this derivative for a tensor Ξ 𝐱, 𝑡 over a
spatial placement requires that we perform the change
of variables with the motion function, 𝐱 = 𝜒(𝐗, 𝑡) to
first obtain, Ξ 𝜒(𝐗, 𝑡), 𝑡 ≡ Ξ 𝑅 𝐗, 𝑡 and then perform
the material time derivative.


Time Derivatives
We know from calculus that the total differential of a
composite function Ξ 𝐱, 𝑡
𝜕Ξ 𝐱, 𝑡 𝜕Ξ 𝐱, 𝑡
𝑑Ξ 𝐱, 𝑡 = 𝑑𝐱 + 𝑑𝑡
𝜕𝐱 𝜕𝑡
So that the material time derivative can be computed
directly:

𝜕Ξ 𝐗, 𝑡 𝜕Ξ 𝐱, 𝑡 𝜕𝐱 𝜕Ξ 𝐱, 𝑡
= +
𝜕𝑡 𝐗
𝜕𝐱 𝜕𝑡 𝜕𝑡 𝐱
𝜕Ξ 𝐱, 𝑡
= grad Ξ 𝐱, 𝑡 𝐯 +
𝜕𝑡 𝐱


Time Derivatives

 On the RHS, the first term, grad Ξ 𝐱, 𝑡 is the
convective term and the product with the velocity
depends on the size of the object Ξ 𝐱, 𝑡 .
𝜕Ξ 𝐱,𝑡
 The second term, depending on fixing the
𝜕𝑡 𝐱
spatial coordinate is the local derivative.


Scalar Function

Let Ξ 𝐱, 𝑡 = 𝜙 𝐱, 𝑡 , a scalar spatial field. Then the
substantial derivative becomes,
𝜕𝜙 𝐗, 𝑡 𝜕𝜙 𝐱, 𝑡 𝜕𝐱 𝜕𝜙 𝐱, 𝑡
= ⋅ +
𝜕𝑡 𝐗
𝜕𝜙 𝐱, 𝑡
= grad𝜙 𝐱, 𝑡 ⋅ 𝐯 +
𝜕𝑡 𝐱
The product now being a dot product on account of the
fact that grad𝜙 𝐱, 𝑡 is a vector.

Vector Function
Let Ξ 𝐱, 𝑡 = 𝐠 𝐱, 𝑡 , a vector spatial field. Then the
𝜕𝐠 𝐑 𝐗, 𝑡 𝜕𝐠 𝐱, 𝑡 𝜕𝐱 𝜕𝐠 𝐱, 𝑡
= ⋅ +
𝜕𝑡 𝐗
𝜕𝐠 𝐱, 𝑡
= grad 𝐠 𝐱, 𝑡 𝐯 +
𝜕𝑡 𝐱
 The product now being a contraction operation on
account of the fact that grad 𝐠 𝐱, 𝑡 is second order
tensor. In particular, the acceleration is given by
𝜕𝐯
𝐯 = grad 𝐯 𝐯 +
𝜕𝑡 𝐱


Tensor Function

Let Ξ 𝐱, 𝑡 = 𝐓 𝐱, 𝑡 , a tensor spatial field. Then the
𝜕𝐓 𝐑 𝐗, 𝑡 𝜕𝐓 𝐱, 𝑡 𝜕𝐱 𝜕𝐓 𝐱, 𝑡
= ⋅ +
𝜕𝑡 𝐗
𝜕𝐓 𝐱, 𝑡
= grad 𝐓 𝐱, 𝑡 𝐯 +
𝜕𝑡 𝐱
The product now being a contraction operation on
account of the fact that grad 𝐓 𝐱, 𝑡 is third order
tensor.

Spatial Derivatives

 We have seen that the material evolves in space in a
continuous sequence of spatial placements. This is
time dependent. We also have a reference, time
independent placement. It is always necessary to
distinguish between these two.
 Accordingly we have referred to the tensor fields in
one as spatial tensors and those in the other as
material tensors.


Spatial Derivatives

 Apart from time derivatives, we need spatial
derivatives to compute gradients, divergences, curls,
etc. of field variables. For this purpose it is necessary
to distinguish between the derivatives of variables in
the Material and in the Spatial description.


Material & Spatial Gradients
 Consider a scalar field 𝜙. Differentiating in spatial
𝜕𝜙
coordinates gives us, 𝑖. Applying the chain rule, we
𝜕𝑋
have,
𝜕𝜙 𝜕𝜙 𝜕𝑥 𝑗
𝑖
=
𝜕𝑋 𝜕𝑥 𝑗 𝜕𝑋 𝑖
Which in full vector form,
Grad𝜙 = 𝛻𝜙 = 𝐅 T grad𝜙
Here we have referred to the material gradient in upper
case and the spatial in lower case using the nabla sign
only for the material. Such notations are not consistent
in the Literature.


Vector Gradients

We similarly apply the gradient operator to a vector
𝐠 defined over the material placement:
𝜕𝑔 𝑖 𝜕𝑔 𝑖 𝜕𝑥 𝑘
=
𝜕𝑋 𝑗 𝜕𝑥 𝑘 𝜕𝑋 𝑗
Again the above components show that
Grad𝐠 = 𝛻𝐠 = grad𝐠 𝐅


Vector Gradients, contd
It follows immediately that
Div 𝐠 ≡ tr 𝛻𝐠 = tr grad 𝐠 𝐅
= grad 𝐠 T : 𝐅 = grad 𝐠 : 𝐅 T
= 𝐅 T : grad 𝐠
by the definition of the inner product.
We also note that,
𝛻𝐠 𝐅 −1 = grad 𝐠
Again remember here that the definition of divergence
is the trace of grad so that,
div 𝐠 = tr grad 𝐠 = tr 𝛻𝐠 𝐅 −1 = 𝛻𝐠: 𝐅 −T = 𝐅 −T : 𝛻𝐠
as required to be shown.

Example

Given a motion 𝐱 = 𝜒(𝐗, 𝑡) in the explicit form,
X2 2X3
x1, x2, x3 = X1, − , X3
1+ 𝑡 1+ 𝑡
Calculate the acceleration by differentiating twice.
Find the same acceleration by expressing velocity in
spatial terms and taking the material derivative.
Full dialog in Kinematics.nb


Exercise

In motion 1 + 𝑡 𝑋1 , 1 + 𝑡 2 𝑋2 , 1 + 𝑡 2 𝑋3 , Find the
velocity and acceleration by using a material
description. Show that the same result can be obtained
from a spatial description using the Substantive
derivative of the spatial velocity. Comment on the
practical implications of your results (Tadmore 3.9)
Use Mathematica to illustrate this motion, Find the
deformation gradient and the stretch tensors of the
motion.


find the tensor as well as physical components of the deformation gradient if the
material and spatial frames are referred to spherical polar coordinates
𝜕𝜚 𝜕𝜚 𝜕𝜚
1 1 1 𝜕𝜌 𝜕𝜃 𝜕𝜙
𝐹1 𝐹2 𝐹3
2 2 2
𝜕𝜗 𝜕𝜗 𝜕𝜗
𝑭 = 𝐹1 𝐹2 𝐹3 =
3 3 3 𝜕𝜌 𝜕𝜃 𝜕𝜙
𝐹1 𝐹2 𝐹3 𝜕𝜑 𝜕𝜑 𝜕𝜑
𝜕𝜌 𝜕𝜃 𝜕𝜙
To obtain physical components we note that the contravariant component is spatial while
the covariant is material. If the magnitudes of the material vectors are 𝜂 𝑖 and that of the
𝑖
𝐹𝑗 𝜂 𝑖
spatial are ℎ 𝑖 then, the physical component, 𝐹 𝑖𝑗 = . The vector ℎ 𝑖 = 1, 𝜌, 𝜌 sin 𝜃 ,
ℎ𝑗
and 𝜂 𝑖 = {1, 𝜚, 𝜚 sin 𝜗}. Accordingly,
𝜕𝜚 1 𝜕𝜚 1 𝜕𝜚
𝜕𝜌 𝜌 𝜕𝜃 𝜌 sin 𝜃 𝜕𝜙
𝑖 𝐹𝜚𝜌 𝐹 𝜚𝜃 𝐹 𝜚𝜙
𝐹𝑗 𝜂 𝑖 𝜕𝜗 𝜚 𝜕𝜗 𝜚 𝜕𝜗
𝐹 𝑖𝑗 = = 𝐹 𝜗𝜌 𝐹 𝜗𝜃 𝐹 𝜗𝜙 = 𝜚
ℎ𝑗 𝜕𝜌 𝜌 𝜕𝜃 𝜌 sin 𝜃 𝜕𝜙
𝐹 𝜑𝜌 𝐹 𝜑𝜃 𝐹 𝜑𝜙
𝜕𝜑 𝜚 sin 𝜗 𝜕𝜑 𝜚 sin 𝜗 𝜕𝜑
𝜚 sin 𝜗
𝜕𝜌 𝜌 𝜕𝜃 𝜌 sin 𝜃 𝜕𝜙


Velocity Gradient

The spatial tensor field, 𝐋 = grad[𝐯 𝐱, 𝑡 ]is defined as
the velocity gradient. Recall that the deformation
gradient,
𝐅 = Grad 𝝌(𝐗, 𝑡)
The material derivative of this equation,
𝜕
𝐅= Grad 𝝌(𝐗, 𝑡) = Grad 𝝌 𝐗, 𝑡 = grad 𝐯(𝐱, 𝑡)𝐅
𝜕𝑡 𝐗
So that 𝐅 = 𝐋𝐅. Therefore,
𝐋 = 𝐅 𝐅 −1


Velocity Gradient

Beginning with 𝐅 = 𝐋𝐅, the transpose yields,
T
𝐅 = 𝐅 T 𝐋T
Differentiating 𝐅𝐅 −1 = 𝟏, we can see that
𝐅 𝐅 −1 = −𝐅 𝐅 −1
So that,
𝐅 −1 = −𝐅 −1 𝐅 𝐅 −1 = −𝐅 −1 𝐋


Deformation Rates & Spins

We are now able to define tensors that quantify the
deformation and spin rates. Recall that we are always
able to break a second-order tensor into its symmetric
and antisymmetric parts. The symmetric part:
1 T =
1
𝐃≡ 𝐋+ 𝐋 grad[𝐯 𝐱, 𝑡 ] + gradT [𝐯 𝐱, 𝑡 ]
2 2
is defined as the rate of deformation or stretching
tensor. And the anti-symmetric part,
1 T =
1
𝐖≡ 𝐋− 𝐋 grad 𝐯 𝐱, 𝑡 − gradT [𝐯 𝐱, 𝑡 ]
2 2
is the spin tensor.


While these two resemble the definition for the small
strain tensors and rotation as they relate to the
displacement gradient, the quantities here are not
approximations but apply even in large deformation
and spin rates. Using the two equations above, we are
able to write,
𝐋= 𝐃+ 𝐖



Now we note that,
𝑑𝐯 = 𝑑𝐱 = 𝐅 𝑑𝐗 = 𝐅 𝐅 −1 𝑑𝐱 = 𝐋𝐅𝐅 −1 𝑑𝐱 = 𝐋𝑑𝐱
Using the fact that for an element of spatial length 𝑑𝑠,
𝑑𝑠 2 = 𝑑𝐱 ⋅ 𝑑𝐱
We can differentiate the latter,
𝑑
𝑑𝑠 2 = 𝑑𝐱 ⋅ 𝑑𝐱 + 𝑑𝐱 ⋅ 𝑑𝐱
𝑑𝑡
= 2𝑑𝐱 ⋅ 𝑑𝐱 = 2𝑑𝐱 ⋅ 𝐋𝑑𝐱
= 2𝑑𝐱 ⋅ 𝐃 + 𝐖 𝑑𝐱 = 2𝑑𝐱 ⋅ 𝐃𝑑𝐱 + 2𝑑𝐱 ⋅ 𝐖𝑑𝐱
= 2𝑑𝐱 ⋅ 𝐃𝑑𝐱


Special Motions

For any tensor 𝐓 dependent on a parameter 𝛼, Liouville
Formula (previously established) says that
𝑑 𝑑𝐓 −1
det 𝐓 = tr 𝐓 det 𝐓
𝑑𝛼 𝑑𝛼
Now substitute the deformation gradient 𝐅 for 𝐓 and let the
parameter 𝛼 be elapsed time 𝑡. It follows easily that the above
equation becomes,
𝐽 = 𝐽 div 𝐯 𝐱, 𝑡
𝑑𝐽
where 𝐽 = det 𝐓 and 𝐽 = .
𝑑𝑡
We now consider rigid, irrotational and isochoric motions.


Rigid Motions

Whenever motion evolves in such a way as to keep the
distances between two spatial points unchanged in
time, we have rigid motion. Consider a small material
fibre lying between the points 𝐗 and Y. As the motion
evolves, the length 𝛿(𝑡) of the fibre is,
𝛿 𝑡 = 𝐱 − 𝐲 = 𝜒 𝐗, 𝑡 − 𝜒 𝐘, 𝑡
or 𝛿 2 𝑡 = 𝜒 𝐗, 𝑡 − 𝜒 𝐘, 𝑡 ⋅ 𝜒 𝐗, 𝑡 − 𝜒 𝐘, 𝑡
differentiating,
𝛿 𝑡 𝛿 𝑡 = 𝐱 − 𝐲 ⋅ 𝜒 𝐗, 𝑡 − 𝜒 𝐘, 𝑡


Rigid Motion
We now proceed to show that in such a motion, the
stretching rate, 𝐃 𝐱, 𝑡 = 𝟎.
We note that, for a rigid motion, the time rate of
change, 𝛿 𝑡 = 0. This clearly means that,
𝛿 𝑡 𝛿 𝑡 = 𝐱 − 𝐲 ⋅ 𝜒 𝐗, 𝑡 − 𝜒 𝐘, 𝑡 = 0
Differentiating the above, bearing in mind that
grad 𝐮 ⋅ 𝐯 = gradT 𝐮 𝐯 + gradT 𝐯 𝐮, we have,
gradT 𝐯 𝐱, 𝑡 𝐱 − 𝐲 + 𝐯 𝐱, 𝑡 − 𝐯 𝐲, 𝑡 = 0
so that
𝐯 𝐱, 𝑡 = 𝐯 𝐲, 𝑡 − gradT 𝐯 𝐱, 𝑡 𝐱− 𝐲


Rigid Motion
Differentiating wrt 𝐲
grad 𝐯 𝐲, 𝑡 = −gradT 𝐯 𝐱, 𝑡
which shows in particular, when we allow 𝐱 = 𝐲, that
grad𝐯 𝐱, 𝑡 = −gradT 𝐯 𝐱, 𝑡
or that the velocity gradient is skew. This immediately
implies that the symmetric part, 𝐃 𝐱, 𝑡 = 𝟎.
Furthermore, the above equations, taken together
implies that, ∀ 𝐱, 𝐲 ∈ B 𝒕
grad𝐯 𝐱, 𝑡 = grad𝐯 𝐲, 𝑡
so that grad𝐯 𝐱, 𝑡 = 𝐋 𝐱, 𝑡 = 𝐖(𝑡) where 𝐖 is a
spatially constant skew tensor.


Rigid Motion

The velocity of a rigid motion can therefore be
expressed as,
= 𝐯 𝐲, 𝑡 + 𝐖(𝑡) 𝐱 − 𝐲
= 𝛂 𝑡 + 𝛌 𝑡 × 𝐱− 𝒐
where 𝛂 is the velocity of the origin and the axial vector
𝛌 is the vector cross of 𝐖.


Irrotational Motions
Define vorticity; the spatial vector field,
𝛚(𝐱, 𝑡) = curl 𝐯
But for any two vectors 𝐮 and 𝐯, 𝐮 × : 𝐯 × = 2𝐮 ⋅ 𝐯
and, 𝐮 × : grad 𝐯 = 𝐮 ⋅ curl 𝐯,
Given any tensor ,
𝒂 ⋅ 𝛚 = 𝒂 ⋅ curl 𝐯 = 𝒂 × : grad 𝐯
= 𝒂× : 𝐋= 𝒂× : 𝐃+ 𝐖
= 𝒂× : 𝐖= 𝒂× : 𝒘×
= 𝟐𝒂 ⋅ 𝒘
Clearly, the vorticity 𝛚 is twice the axial spin vector 𝒘


Irrotational Motions

 Motion is irrotational if 𝐖(x,t)=0 or, equivalently,
curl 𝐯(𝐱, 𝑡) = 𝒐
 This implies that ∃𝜑(𝐱, 𝑡)such that 𝐯 𝐱, 𝑡 = grad𝜑.
The velocity in an irrotational flow is the gradient of a
potential field.
In irrotational motion, the material substantial
acceleration takes the form,
1
′ + grad 𝐯 𝐯 = 𝐯 ′ + grad 𝐯 2
𝐯= 𝐯
2


Proof

𝟐𝐖 = 𝐋 − 𝐋T
1
so that, 2𝐖𝐯 = 𝐋𝐯 − 𝐋T 𝐯 = 𝐋𝐯 − grad 𝐯 2
2
1
𝐯 = 𝐯 + 𝐋𝐯 = 𝐯 + grad 𝐯 2 + 2𝐖𝐯
′ ′
2
When flow is irrotational, 𝐖(x,t)=0
Hence,
1
′ + grad 𝐯 2
𝐯= 𝐯
2


Isochoric Motion

If during the motion, the volume of any arbitrary
material region does not change, the motion is called
isochoric or isovolumic.
Recall that the volume ratio
𝑑𝑣
= 𝐽
𝑑𝑉
Furthermore, 𝐽 = 𝐽 div 𝐯 𝐱, 𝑡 . Consequently, isochoric
motion results when 𝐽 = 0 or div 𝐯 𝐱, 𝑡 = 0.
The last condition derives from the Reynold’s transport
theorem that we next discuss.

Reynolds’ Transport Theorem
Differentiation of spatial integrals. Consider the time
derivative of the spatial integral,
𝑑
𝜑 𝐱, 𝑡 𝑑𝑣
𝑑𝑡 B 𝑡
The domain of integration is varying with time, hence
we cannot simply convert this to a differentiation under
the integral sign. By Liouville’s formula, we can write,
𝑑 𝑑
𝜑 𝐱, 𝑡 𝑑𝑣 = 𝜑 𝐱, 𝑡 𝐽𝑑𝑣 𝑅
𝑑𝑡 B 𝑡 𝑑𝑡 B
converting the domain to a fixed referential placement.

It is now possible to differentiate under the integral and
write,
𝑑 𝑑
𝜑 𝐱, 𝑡 𝐽𝑑𝑣 𝑅 = 𝜑 𝐱, 𝑡 𝐽𝑑𝑣 𝑅
𝑑𝑡 B B 𝑑𝑡
= 𝜑 𝐱, 𝑡 𝐽 + 𝜑 𝐱, 𝑡 𝐽 𝑑𝑣 𝑅
B
= 𝜑 𝐱, 𝑡 𝐽 + 𝜑 𝐱, 𝑡 𝐽div 𝐯 𝑑𝑣 𝑅
B
= 𝜑 𝐱, 𝑡 + 𝜑 𝐱, 𝑡 div 𝐯 𝐽𝑑𝑣 𝑅
B
= 𝜑 𝐱, 𝑡 + 𝜑 𝐱, 𝑡 div 𝐯 𝑑𝑣
B𝑡


Reynolds’ Transport Theorem
We can therefore write,
𝑑
𝜑 𝐱, 𝑡 𝑑𝑣 = 𝜑 𝐱, 𝑡 + 𝜑 𝐱, 𝑡 div 𝐯 𝑑𝑣
𝑑𝑡 B 𝑡 B𝑡
Setting the function 𝜑 𝐱, 𝑡 = 1, we can calculate the
material derivative of the spatial volume:
𝑑
𝑑𝑣 = div 𝐯𝑑𝑣
𝑑𝑡 B 𝑡 B𝑡
so that if the volume does not change over time,
div 𝐯 = 0.


Steady Motion
Motion is said to be steady when the local acceleration
𝐯 ′ 𝐱, t ∀ 𝐱 ∈ B 𝑡 (at every point) is zero. In this case,
the substantial acceleration,
𝐯 = 𝐯 ′ + grad 𝐯 𝐯 = grad 𝐯 𝐯
In this case, the deformed body, B 𝑡 is independent of
time. Hence,
B 𝑡 = B ∀𝑡
In steady motion, all the particles that pass through a
particular spatial point (coincides here with material
point) does so at the same velocity.


Steady Motion
A particle path is the trajectory of an individual particle as the
flow evolves. This path is,
𝑡
𝐱 = 𝐱0 + 𝐕 𝑿, 𝑡 𝑑𝑡
0
or, equivalently the solution to the differential equation,
𝑑𝐱
= 𝐯(𝐱, 𝑡)
𝑑𝑡
Given a steady motion, solutions to the differential equation,
𝑑𝐬(𝑡)
= 𝐯(𝒔 𝑡 )
𝑑𝑡
are called streamlines. For steady motion, these two
equations coincide and the path lines become streamlines.


Exercises

𝑡 2
Romano 4.71 Given the motion 𝐱 = 1+ 𝑋1 , 𝑋2 , 𝑋3
𝑇
Find the Material and spatial representation of the
velocity and acceleration.
Romano 4.73 Explain why the following Mathematica
code shows that the kinetic field v is rigid:
𝑣1 : = 2𝑥3 − 5𝑥2 ; 𝑣2 : = 5𝑥1 − 3𝑥3 ; 𝑣3 : = 3𝑥2 − 2𝑥1 ;
xx: = {𝑥1 , 𝑥2 , 𝑥3 }; vv:={v1,v2,v3};𝜕{xx} vv


Exercises

Romano 4.74. Show that a rigid motion is also isochoric.
In a rigid motion, 𝐃 = Sym grad 𝐯 𝐱, 𝑡 = 𝟎. Because
grad 𝐯 𝐱, 𝑡 is skew and there must be a vector 𝐰 such
that grad 𝐯 𝐱, 𝑡 = 𝐰 ×. The trace of this must vanish.
This trace is the div 𝐯 𝐱, 𝑡 = 0.
Now for isochoric motion, 𝐽 = 𝐽div 𝐯 = 0. A rigid motion
is therefore necessarily isochoric.


Exercises
For a vector field 𝐯 𝐱, 𝑡 , if sym grad 𝐯 = 𝟎, Show that
div 𝐯 = 0. Is the converse true?
div 𝐯 = tr grad 𝐯
= tr sym grad 𝐯 + skw grad 𝐯
= tr sym grad 𝐯 + tr skw grad 𝐯
=0+0
We have used the fact that trace operation is linear and that
the trace of any skew tensor is zero. The converse is NOT
true. For any tensor 𝐓
tr 𝐓 = 0 ⟽ 𝐓 = 0
The implication is one directional because there are non-zero
tensors with zero traces.
Question: Correlate this with Slide 100


Exercises

Romano 4.75 Find a class of isochoric, non-rigid
motions. In isochoric motion, div 𝐯 = 0. Is it possible to
find 𝐷 ≠ 0?
div 𝐯 = tr grad 𝐯 = 𝟎
For the stretching tensor still to remain nonzero, we
must have that grad 𝐯 is not skew. This is possible if
𝑣 𝑖 , 𝑖 = 𝟎 but 𝑣1 ,1 ≠ 𝑣2 ,2 ≠ 𝑣3 ,3 ≠ 0.


Exercises

Gurtin 10.1 Show that a motion whose velocity field is
rigid is itself rigid.
If velocity is constant, then
gradT 𝐯 𝐱, 𝑡 = 𝟎.
In this case, we have a rigid motion with,
= 𝐯 𝐲, 𝑡 + 𝐖(𝑡) 𝐱 − 𝐲
= 𝛂 𝑡 + 𝛌 𝑡 × 𝐱− 𝒐
So that 𝛌 𝑡 = 𝒐.


Exercises

oafak 3.21. When a blood vessel is under pressure, the
following deformation transformations were observed,
𝑟 = 𝑟 𝑅 , 𝜙 = Φ + 𝜓𝑍 , 𝑧 = 𝜆𝑍 Compute the
deformation gradient, Cauchy-Green Tensor,
Lagrangian. and Eulerian strain tensors for this
deformation. Do this manually as well as with
Mathematica


Exercises

Taber 141 oafak 3.22 A cylindrical tube undergoes the
deformation given by 𝑟 = 𝑅, 𝜙 = Θ + 𝜗 𝑅 , 𝑧 = 𝑍 +
𝑤(𝑅) where 𝑅, Φ, 𝑍 and 𝑟, 𝜙, 𝑧 , are polar
coordinates of a point in the tube before and after
deformation respectively, 𝜗 and 𝑤 are scalar functions
of 𝑅. (a) Explain the meaning of the situation where (i)
𝜗 = 0, (ii) 𝑤 = 0. (b) Compute 𝑭, 𝑪 and 𝑬, (c) Find the
Lagrangian and Eulerian strain components


oafak3.23 A body is in the state of plane strain relative to the
𝑥 − 𝑦 plane. Assume all the components of the strain are
known relative to Cartesian axes 𝑥, 𝑦, 𝑧 . Find the stress
components relative to another axes rotated along the 𝑧-axis
by an angle 𝜃
oafak 3.25 A velocity field has components of the form,
𝑣1 = 𝛼𝑦1 − 𝛽𝑦2 𝑡, 𝑣2 = 𝛽𝑦1 − 𝛼𝑦2 and 𝑣3 = 0 where 𝛼 and
𝛽 are positive constants. Assume that the spatial mass
density is independent of the current position so that
grad 𝜚 = 𝒐, 𝑎 express 𝜚 so that the conservation of mass is
satisfied. (𝑏) Find a condition for which the motion is
isochoric.


4. kinematics jan 2013

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