4. kinematics jan 2013

Kinematics The Study of Deformation & Motion

“…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.

Definition

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

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.

Context

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.

Balance Laws and the Theory of Stress

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.

Placement of Bodies

The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as: 𝑥1, 𝑥2 = 𝑋1 + 𝑋22/2,

3𝑋2

3.5. Using

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

oafak@unilag.edu.ng 12/29/2012 Department of Systems Engineering, University of Lagos 7

Placements in 3-D Euclidean 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)

Spatial Space

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

Basic Hypotheses

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 Indestructibility

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.

Material & Spatial Vectors

The spatial vectors,

𝐱 = 𝝌 𝐗, 𝑡 ≡𝜕𝝌(𝐗, 𝑡)

𝜕𝑡

𝐱 = 𝝌 𝐗, 𝑡 ≡𝜕2𝝌(𝐗, 𝑡)

𝜕𝑡2

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.

Velocity & Acceleration

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

Further Hypotheses

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:

Convected Coordinates

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.

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

𝐅 𝐗, 𝑡 ≡𝑑𝐱

𝑑𝐗 = 𝛻𝜒 𝐗, 𝑡

is called the deformation gradient. Clearly, the deformation gradient maps infinitesimal material vectors (e.g. 𝑑𝐗)

to infinitesimal spatial vectors (e.g. 𝑑𝐱).

The Deformation Gradient

At a particular instant in time, the placement 𝜒𝑡 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.

Homogeneous Deformation

Dropping the functional dependencies, we have that, 𝑑𝐱 = 𝐅𝑑𝐗

In which the deformation gradient maps material vectors to spatial. We can also write,

𝐅−1𝑑𝐱 = 𝑑𝐗

So that the inverse of the deformation gradient maps spatial vectors to material vectors.

Material and Spatial Mapping

Consider a spatial vector 𝐬. Take its inner product with the spatial vector equation, 𝑑𝐱 = 𝐅𝑑𝐗, we obtain,

𝐬 ⋅ 𝑑𝐱 = 𝐬 ⋅ 𝐅𝑑𝐗 = 𝑑𝐗 ⋅ 𝐅T𝐬

Which clearly shows that 𝐅T𝐬 is a material vector. Clearly, 𝐅T maps spatial vectors to material vectors.

Given a material vector 𝐦, a similar consideration for the scalar equation,

𝐦 ⋅ 𝐅−1𝑑𝐱 = 𝐦 ⋅ 𝑑𝐗 = 𝑑𝐱 ⋅ 𝐅−T𝐦

Clearly shows that 𝐅−T is a map of material vectors to spatial vectors.

Transposes

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.

Polar Decomposition Theorem

The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as: 𝑥1, 𝑥2 = 𝑋1 + 𝑋2

2/2, 𝑋2 . Using Mathematica (Reddy3.15.nb) the resulting deformation is:

oafak@unilag.edu.ng 12/29/2012 25

Examples of deformation mappings

Department of Systems Engineering, University of Lagos

The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as:

𝑥1, 𝑥2 =1

418 + 4𝑋1 + 6𝑋2 ,

414 + 6𝑋2 . Using

Mathematica (Reddy3.4.3.15.nb) the resulting deformation is:

The deformation, 𝐱 = 𝜒(𝐗, 𝑡) can take the 2-D form such as:

𝑥1, 𝑥2 = 4 − 2𝑋1 − 𝑋2, 2 +3𝑋1

2. Using Mathematica

(Holzapfel72.nb) the resulting deformation is:

Lines & Circles

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

Animation

By the results of this theorem, 𝑹𝑇𝑹 = 𝑹𝑹𝑇 = 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.

oafak@unilag.edu.ng 12/29/2012

Polar Decomposition

31 Department of Systems Engineering, University of Lagos

Note that 𝐂 = 𝐅T𝐅 = 𝐔T 𝐑T𝐑 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 ≡ 𝑼𝑇𝑼 = 𝑼𝑼𝑻 = 𝑻

To prove this theorem, we must first show that 𝑭𝑇𝑭 is symmetric and positive definite. Symmetry is obvious.

To show positive definiteness, For an arbitrary real vector 𝒖 consider the expression, 𝒖 ⋅ 𝑭𝑇𝑭𝒖. Let the vector 𝒃 = 𝑭𝒖. Then we can write,

𝒖 ⋅ 𝑭𝑇𝑭𝒖 = 𝒃 ⋅ 𝒃 = 𝒃 2 > 0

as the magnitude of any real vector must be positive. Hence 𝑪 = 𝑭𝑇𝑭 is positive definite.

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

𝑪 = 𝜔𝑖𝒖𝑖 ⊗ 𝒖𝑖

𝑖=1

Given that 𝜔𝑖 = 𝜆𝑖2 is the eigenvalue corresponding

to the normalized eigenvector 𝒖𝑖. Every quadratic form with this spectral representation must be greater than zero.

Uniqueness of the Root

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

𝒖𝑗 ⋅ 𝑪𝒖𝑗 = 𝜔𝑖𝒖𝑗 ⋅ 𝒖𝑖 ⊗ 𝒖𝑖 𝒖𝑗

𝑖=1

= 𝜔𝑗 > 0.

(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 = 𝜔𝑖 > 03𝑖=1 . And

since the determinant of a matrix is an invariant, this holds true even in non spectral forms of 𝑪.

Now, let

𝑼 = 𝜆𝑖𝒖𝑖 ⊗ 𝒖𝑖

𝑖=1

Clearly,

𝑼𝟐 = 𝜆𝑖𝒖𝑖 ⊗ 𝒖𝑖

𝑖=1

𝜆1𝒖1 ⊗ 𝒖1 + 𝜆2𝒖2 ⊗ 𝒖2 + 𝜆3𝒖3 ⊗ 𝒖3

= 𝜆𝑖2𝒖𝑖 ⊗ 𝒖𝑖 =

𝑖=1

𝜔𝑖𝒖𝑖 ⊗ 𝒖𝑖 = 𝑪

𝑖=1

And this square root is unique, for were it not so, there would be another positive definite tensor 𝑼 such that,

𝑼𝟐 = 𝑼 2 = 𝑪.

The eigenvalue equation, 𝑪 − 𝜆2𝑰 𝒖 = 𝟎

is satisfied by each eigenvalue/vector pair for 𝑪. From the above, we may write,

𝑪 − 𝜆2𝑰 𝒖 = 𝑼𝟐 − 𝜆2𝑰 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 − 𝜆2𝑰 𝒖 = 𝑼 𝟐 − 𝜆2𝑰 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

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 𝑹𝑻𝑹 = 𝑼−𝑻𝑭𝑻𝑭𝑼−𝟏 = 𝑼−𝟏𝑼2𝑼−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−𝟏is also positive definite. Hence we can see that, det 𝑹 = +𝟏. Which, when added to the fact that 𝑹𝑻𝑹 = 𝟏 means that 𝑹 is a rotation.

The Rotation

It is an easy matter now to find the tensor 𝑽 such that 𝑭 = 𝑹𝑼 = 𝑽𝑹

It is obvious that 𝑽 = 𝑹𝑼𝑹𝑇 is symmetric and is the square root of the Finger tensor,

𝑩−𝟏 = 𝑭𝑭𝑻 = 𝑽𝑹𝑹𝑇𝑽𝑇 = 𝑽2

𝑼 is the Right Stretch tensor while 𝑽 is called the Left Stretch Tensor

The Stretch Tensors

The tensor,

𝑬 =1

2𝑪 − 𝟏 =

2𝑼2 − 𝟏

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:

The Strain Tensor

In fact any tensor satisfying,

𝑚𝑼𝑚 − 𝟏 𝑚 ≠ 0

log 𝑼 when 𝑚 = 0

𝑂𝑅

𝑚𝑽𝑚 − 𝟏 𝑚 ≠ 0

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.

Strain Tensors

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

ii. 𝐼2 𝑪 =1

2tr2 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

6tr3 𝑺 − 3tr 𝑺 tr 𝑺2 + 2tr(𝑺3) to show that

𝐼2 𝑪 = 8𝐼3 𝑬 + 4𝐼2 𝑬 + 2𝐼1 𝑬 + 1

Homework

Consider two infinitesimal fibers 𝐟R and 𝐠R 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 ⋅ 𝐅𝐠𝑅 = 𝐠R ⋅ 𝐅T𝐟 = 𝐠R ⋅ 𝐅T𝐅𝐟𝑅 = 𝐠R ⋅ 𝐔𝐑T𝐑𝐔𝐟R = 𝐠R ⋅ 𝐔𝐔𝐟𝑅 = 𝐔𝐟R ⋅ 𝐔𝐠R

Infinitesimal Fibers

Setting 𝐟 = 𝐠, we immediately obtain, 𝐟2 = 𝐔𝐟R

So that 𝐟 = 𝐔𝐟R

The deformed length of an infinitesimal fiber is characterized by the Right Stretch Tensor.

Infinitesimal Fibers

The cosine of the angle between two deformed infinitesimal fibers can be obtained from,

𝐟 ⋅ 𝐠

𝐟 𝐠=

𝐔𝐟𝑅 ⋅ 𝐔𝐠𝑅

𝐔𝐟𝑅 𝐔𝐠𝑅

Which is the same as the cosine of the angle between the vectors 𝐔𝐟𝑅 and 𝐔𝐠𝑅.

Clearly, 𝐔 also characterizes the angles between infinitesimal fibers.

Contained Angle

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,

lim𝐿→0

Δ𝐱

𝐿= 𝐅 𝐗 𝐞

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 ⋅ 𝐅T 𝐗 𝐳 = 𝐞 ⋅ 𝐅T 𝐗 𝐅 𝐗 𝐞 = 𝐞 ⋅ 𝐂(𝐗)𝐞

The Stretch Vector

The Right Stretch Tensor is symmetric and positive definite. It can therefore be written in its spectral form as:

𝐔 = Σ𝑖=13 𝜆𝑖𝐮𝑖 ⊗ 𝐮𝑖

We are in a position to write the spectral forms of other important tensors as follows:

𝐕 = 𝐑𝐔𝐑T = Σ𝑖=1

3 𝜆𝑖𝐑 𝐮𝑖 ⊗ 𝐮𝑖 𝐑T

= Σ𝑖=13 𝜆𝑖𝐑𝐮𝑖 ⊗ 𝐑𝐮𝑖 = Σ𝑖=1

3 𝜆𝑖𝐯𝑖 ⊗ 𝐯𝑖

Where 𝐯𝑖 = 𝐑𝐮𝑖

Showing that the Left Stretch Tensor has the same eigenvalues but a rotated eigenvector from its corresponding Right Stretch Tensor.

Principal Stretches & Directions

Moreover, 𝐂 = 𝐔2 = 𝐔𝐔

= Σ𝑖=13 𝜆𝑖

2𝐮𝑖 ⊗ 𝐮𝑖

A result that follows immediately we realize that 𝐮𝑖 , 𝑖 = 1,2,3 is an orthonormal set.

𝐅 = 𝐑𝐔 = Σ𝑖=13 𝜆𝑖𝐑 𝐮𝑖 ⊗ 𝐮𝑖

= Σ𝑖=13 𝜆𝑖𝐯𝑖 ⊗ 𝐮𝑖

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,

𝑬 =1

2𝑪 − 𝟏 =

2Σ𝑖=1

3 𝜆𝑖2 − 1 𝐮𝑖 ⊗ 𝐮𝑖

Spectral Forms

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.

Volume & Area Changes

Deformation of Volume

𝑑𝐗

𝑑𝐙

𝑑𝐘 𝑑𝐱

𝑑𝐲

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

𝑑𝑉=

𝐅𝑑𝐗, 𝐅𝑑𝐘,𝐅𝑑𝐙

𝑑𝐗, 𝑑𝐘, 𝑑𝐙= 𝐼3 𝐅 = det 𝐅 ≡ 𝐽 > 0

We can also write, 𝑑𝑣 = 𝐽𝑑𝑉

The Volume change

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 = 𝑭𝑇𝑑𝒂 − 𝐽𝑑𝑨 ⋅ 𝑑𝐗

where 𝒐 is the zero vector.

Area Changes

For an arbitrary vector 𝑑𝑿, we have: 𝑭𝑇𝑑𝒂 − 𝐽𝑑𝑨 = 𝒐

so that, 𝑑𝒂 = 𝐽𝑭−𝑇𝑑𝑨 = 𝑭𝐜𝑑𝑨

where 𝑭𝐜 is the cofactor tensor of the deformation gradient.

Nanson Formula

For the uniform biaxial deformation, given that {𝑥1, 𝑥2, 𝑥3} = {𝜆1𝑋1, 𝜆2𝑋2, 𝑋3}. Compute the Deformation Gradient tensor, the Lagrangian Strain Tensor as well as the Eulerian Strain Tensor components.

𝐹 =𝜕 𝜆1𝑋1, 𝜆2𝑋2, 𝑋3

𝜕 𝑋1, 𝑋2, 𝑋3=

𝜆1 0 00 𝜆2 00 0 1

Clearly in this case,

𝑪 = 𝑭𝑇𝑭 = 𝑩−1 = 𝑭𝑭𝑇 =𝜆1

0 𝜆22 0

Examples

And the Piola

Tensor 𝑩 = 𝑭−𝑇𝑭−1 =𝜆1

−2 0 0

0 𝜆2−2 0

Now, the Lagrangian Strain Tensor

𝑬 =1

2𝑪 − 𝑰 =

𝜆12 − 1 0 0

0 𝜆22 − 1 0

And the Eulerian Strain Tensor

𝒆 =1

2𝑰 − 𝑩 =

1 − 𝜆1−2 0 0

0 1 − 𝜆2−2 0

Show that the tensor C 163.24 34.6 4.234.6 19. −30.4.2 −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: = 𝝀𝟏𝑿𝟏 + 𝒌𝟏𝑿𝟐 𝐠𝟏 + 𝒌𝟐𝑿𝟏 + 𝝀𝟐𝑿𝟐 𝐠𝟐 +𝝀𝟑𝑿𝟑𝐠𝟑 Compute the tensors 𝑭, 𝑪, 𝑬, 𝑼 and 𝑹. Show that 𝑹𝑻𝑹 = 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.

𝐹 =

𝜆1 𝑘1 0𝑘2 𝜆2 00 0 𝜆3

𝐶 =

𝑘22 + 𝜆1

2 𝑘1𝜆1 + 𝑘2𝜆2 0

𝑘1𝜆1 + 𝑘2𝜆2 𝑘12 + 𝜆2

0 0 𝜆32

Full code @ Taber02.nb

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.

Given the Deformation Gradient Tensor 1

0 1 00 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,

𝑭𝑝 =𝜆1cos𝜃 𝜆2sin𝜃 0−𝜆1sin𝜃 𝜆2cos𝜃 0

. Show that 𝜆1 = 𝜆2−1

Homework

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.

Material & Spatial Derivatives

We assume that the Frechét derivative,𝑑𝐱

𝑑𝐗 has a non-

vanishing determinant 𝐽 =𝑑𝐱

𝑑𝐗 so that the inverse,

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) Ξ𝑅(𝐗, 𝑡) over the reference placement, a simple change of variables gives,

Ξ𝑅 𝐗, 𝑡 = Ξ𝑅 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 ≡ Ξ𝑅 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)

Reference Map

Material or substantial derivative of a field defined over the reference placement can be written as,

𝜕Ξ𝑅 𝐗, 𝑡

𝜕𝑡 𝐗

= Ξ 𝑅

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 ≡ Ξ𝑅 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

𝑑Ξ 𝐱, 𝑡 =𝜕Ξ 𝐱, 𝑡

𝜕𝐱𝑑𝐱 +

𝜕Ξ 𝐱, 𝑡

𝜕𝑡𝑑𝑡

So that the material time derivative can be computed directly:

𝜕Ξ 𝐗, 𝑡

𝜕𝑡 𝐗

=𝜕Ξ 𝐱, 𝑡

𝜕𝐱

𝜕𝑡+

𝜕Ξ 𝐱, 𝑡

𝜕𝑡 𝐱

= grad Ξ 𝐱, 𝑡 𝐯 +𝜕Ξ 𝐱, 𝑡

𝜕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 .

The second term, 𝜕Ξ 𝐱,𝑡

𝜕𝑡 𝐱 depending on fixing the

spatial coordinate is the local derivative.

Time Derivatives

Let Ξ 𝐱, 𝑡 = 𝜙 𝐱, 𝑡 , a scalar spatial field. Then the substantial derivative becomes,

𝜕𝜙 𝐗, 𝑡

𝜕𝑡 𝐗

=𝜕𝜙 𝐱, 𝑡

𝜕𝐱⋅𝜕𝐱

𝜕𝑡+

𝜕𝜙 𝐱, 𝑡

𝜕𝑡 𝐱

= grad𝜙 𝐱, 𝑡 ⋅ 𝐯 +𝜕𝜙 𝐱, 𝑡

𝜕𝑡 𝐱

The product now being a dot product on account of the fact that grad𝜙 𝐱, 𝑡 is a vector.

Scalar Function

Let Ξ 𝐱, 𝑡 = 𝐠 𝐱, 𝑡 , a vector spatial field. Then the substantial derivative becomes,

𝜕𝐠𝐑 𝐗, 𝑡

𝜕𝑡 𝐗

=𝜕𝐠 𝐱, 𝑡

𝜕𝐱⋅𝜕𝐱

𝜕𝑡+

𝜕𝐠 𝐱, 𝑡

𝜕𝑡 𝐱

= grad 𝐠 𝐱, 𝑡 𝐯 +𝜕𝐠 𝐱, 𝑡

𝜕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 = grad 𝐯 𝐯 +𝜕𝐯

𝜕𝑡 𝐱

Vector Function

Let Ξ 𝐱, 𝑡 = 𝐓 𝐱, 𝑡 , a tensor spatial field. Then the substantial derivative becomes,

𝜕𝐓𝐑 𝐗, 𝑡

𝜕𝑡 𝐗

=𝜕𝐓 𝐱, 𝑡

𝜕𝐱⋅𝜕𝐱

𝜕𝑡+

𝜕𝐓 𝐱, 𝑡

𝜕𝑡 𝐱

= grad 𝐓 𝐱, 𝑡 𝐯 +𝜕𝐓 𝐱, 𝑡

𝜕𝑡 𝐱

The product now being a contraction operation on account of the fact that grad 𝐓 𝐱, 𝑡 is third order tensor.

Tensor Function

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.

Spatial Derivatives

Consider a scalar field 𝜙. Differentiating in spatial

coordinates gives us, 𝜕𝜙

𝜕𝑋𝑖. Applying the chain rule, we

𝜕𝜙

𝜕𝑋𝑖=

𝜕𝜙

𝜕𝑥𝑗

𝜕𝑋𝑖

Which in full vector form, Grad𝜙 = 𝛻𝜙 = 𝐅Tgrad𝜙

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.

Material & Spatial Gradients

We similarly apply the gradient operator to a vector 𝐠 defined over the material placement:

𝜕𝑔𝑖

𝜕𝑋𝑗=

𝜕𝑔𝑖

𝜕𝑥𝑘

𝜕𝑋𝑗

Again the above components show that Grad𝐠 = 𝛻𝐠 = grad𝐠 𝐅

Vector Gradients

It follows immediately that Div 𝐠 ≡ tr 𝛻𝐠 = tr grad 𝐠 𝐅

= grad 𝐠 T: 𝐅 = grad 𝐠 : 𝐅T = 𝐅T: 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.

Vector Gradients, contd

Given a motion 𝐱 = 𝜒(𝐗, 𝑡) in the explicit form,

x1, x2, x3 = X1,X2

1 + 𝑡−

1 + 𝑡, X3

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

Example

In motion 1 + 𝑡 𝑋1, 1 + 𝑡 2𝑋2, 1 + 𝑡2 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.

Exercise

find the tensor as well as physical components of the deformation gradient if the material and spatial frames are referred to spherical polar coordinates

𝑭 =

𝐹11 𝐹2

1 𝐹31

𝐹12 𝐹2

2 𝐹32

𝐹13 𝐹2

3 𝐹33

𝜕𝜚

𝜕𝜌

𝜕𝜚

𝜕𝜃

𝜕𝜚

𝜕𝜙𝜕𝜗

𝜕𝜌

𝜕𝜗

𝜕𝜃

𝜕𝜗

𝜕𝜙𝜕𝜑

𝜕𝜌

𝜕𝜑

𝜕𝜃

𝜕𝜑

𝜕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 𝜂𝑖and that of the

spatial are ℎ𝑖then, the physical component, 𝐹 𝑖𝑗 =𝐹𝑗

𝑖𝜂𝑖

ℎ𝑗. The vector ℎ𝑖 = 1, 𝜌, 𝜌 sin 𝜃 ,

and 𝜂𝑖 = {1, 𝜚, 𝜚 sin 𝜗}. Accordingly,

𝐹 𝑖𝑗 =𝐹𝑗

𝑖𝜂𝑖

ℎ𝑗=

𝐹𝜚𝜌 𝐹𝜚𝜃 𝐹𝜚𝜙

𝐹𝜗𝜌 𝐹𝜗𝜃 𝐹𝜗𝜙

𝐹𝜑𝜌 𝐹𝜑𝜃 𝐹𝜑𝜙

𝜕𝜚

𝜕𝜌

𝜕𝜚

𝜕𝜃

𝜌 sin 𝜃

𝜕𝜚

𝜕𝜙

𝜚𝜕𝜗

𝜕𝜌

𝜕𝜗

𝜕𝜃

𝜌 sin 𝜃

𝜕𝜗

𝜕𝜙

𝜚 sin 𝜗𝜕𝜑

𝜕𝜌

𝜚 sin 𝜗

𝜕𝜑

𝜕𝜃

𝜚 sin 𝜗

𝜌 sin 𝜃

𝜕𝜑

𝜕𝜙

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 = 𝜕

𝜕𝑡Grad 𝝌(𝐗, 𝑡)

= Grad 𝝌 𝐗, 𝑡 = grad 𝐯(𝐱, 𝑡)𝐅

So that 𝐅 = 𝐋𝐅. Therefore, 𝐋 = 𝐅 𝐅−1

Velocity Gradient

Beginning with 𝐅 = 𝐋𝐅, the transpose yields,

𝐅 T

= 𝐅T𝐋T

Differentiating 𝐅𝐅−1 = 𝟏, we can see that

𝐅 𝐅−1 = −𝐅 𝐅−1

So that,

𝐅−1 = −𝐅−1𝐅 𝐅−1 = −𝐅−1𝐋

Velocity Gradient

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:

𝐃 ≡1

2𝐋 + 𝐋T =

2grad[𝐯 𝐱, 𝑡 ] + gradT[𝐯 𝐱, 𝑡 ]

is defined as the rate of deformation or stretching tensor. And the anti-symmetric part,

𝐖 ≡1

2𝐋 − 𝐋T =

2grad 𝐯 𝐱, 𝑡 − gradT[𝐯 𝐱, 𝑡 ]

is the spin tensor.

Deformation Rates & Spins

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−1𝑑𝐱 = 𝐋𝐅𝐅−1𝑑𝐱 = 𝐋𝑑𝐱

Using the fact that for an element of spatial length 𝑑𝑠, 𝑑𝑠2 = 𝑑𝐱 ⋅ 𝑑𝐱

We can differentiate the latter, 𝑑

𝑑𝑡𝑑𝑠2 = 𝑑𝐱 ⋅ 𝑑𝐱 + 𝑑𝐱 ⋅ 𝑑𝐱

= 2𝑑𝐱 ⋅ 𝑑𝐱 = 2𝑑𝐱 ⋅ 𝐋𝑑𝐱 = 2𝑑𝐱 ⋅ 𝐃 + 𝐖 𝑑𝐱 = 2𝑑𝐱 ⋅ 𝐃𝑑𝐱 + 2𝑑𝐱 ⋅ 𝐖𝑑𝐱 = 2𝑑𝐱 ⋅ 𝐃𝑑𝐱

For any tensor 𝐓 dependent on a parameter 𝛼, Liouville Formula (previously established) says that

𝑑𝛼det 𝐓 = tr

𝑑𝐓

𝑑𝛼𝐓−1 det 𝐓

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

where 𝐽 = det 𝐓 and 𝐽 =𝑑𝐽

𝑑𝑡.

We now consider rigid, irrotational and isochoric motions.

Special 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 𝛿2 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 Motions

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 ∈ B𝒕

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 𝐱, 𝑡 = 𝐯 𝐲, 𝑡 − gradT 𝐯 𝐱, 𝑡 𝐱 − 𝐲 = 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.

Rigid Motion

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,

𝐯 = 𝐯′ + grad 𝐯 𝐯 = 𝐯′ +1

2grad 𝐯2

Irrotational Motions

𝟐𝐖 = 𝐋 − 𝐋T

so that, 2𝐖𝐯 = 𝐋𝐯 − 𝐋T𝐯 = 𝐋𝐯 −1

2grad 𝐯2

𝐯 = 𝐯′ + 𝐋𝐯 = 𝐯′ +1

2grad 𝐯2 + 2𝐖𝐯

When flow is irrotational, 𝐖(x,t)=0

Hence,

𝐯 = 𝐯′ +1

2grad 𝐯2

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

𝑑𝑉= 𝐽

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.

Isochoric Motion

Differentiation of spatial integrals. Consider the time derivative of the spatial integral,

𝑑𝑡 𝜑 𝐱, 𝑡 𝑑𝑣

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,

𝑑𝑡 𝜑 𝐱, 𝑡 𝐽𝑑𝑣𝑅

converting the domain to a fixed referential placement.

Reynolds’ Transport Theorem

It is now possible to differentiate under the integral and write,

𝑑𝑡 𝜑 𝐱, 𝑡 𝐽𝑑𝑣𝑅

𝑑𝑡𝜑 𝐱, 𝑡 𝐽𝑑𝑣𝑅

= 𝜑 𝐱, 𝑡 𝐽 + 𝜑 𝐱, 𝑡 𝐽 𝑑𝑣𝑅B

= 𝜑 𝐱, 𝑡 𝐽 + 𝜑 𝐱, 𝑡 𝐽div 𝐯 𝑑𝑣𝑅B

= 𝜑 𝐱, 𝑡 + 𝜑 𝐱, 𝑡 div 𝐯 𝐽𝑑𝑣𝑅B

= 𝜑 𝐱, 𝑡 + 𝜑 𝐱, 𝑡 div 𝐯 𝑑𝑣B𝑡

We can therefore write,

= 𝜑 𝐱, 𝑡 + 𝜑 𝐱, 𝑡 div 𝐯 𝑑𝑣B𝑡

Setting the function 𝜑 𝐱, 𝑡 = 1, we can calculate the material derivative of the spatial volume:

𝑑𝑡 𝑑𝑣

= div 𝐯𝑑𝑣B𝑡

so that if the volume does not change over time, div 𝐯 = 0.

Reynolds’ Transport Theorem

Motion is said to be steady when the local acceleration 𝐯′ 𝐱, t ∀ 𝐱 ∈ B𝑡 (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, B𝑡 is independent of time. Hence,

B𝑡 = 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,

𝐱 = 𝐱0 + 𝐕 𝑿, 𝑡 𝑑𝑡𝑡

or, equivalently the solution to the differential equation, 𝑑𝐱

𝑑𝑡= 𝐯(𝐱, 𝑡)

Given a steady motion, solutions to the differential equation, 𝑑𝐬(𝑡)

𝑑𝑡= 𝐯(𝒔 𝑡 )

are called streamlines. For steady motion, these two equations coincide and the path lines become streamlines.

Steady Motion

Romano 4.71 Given the motion 𝐱 = 1 +𝑡

2𝑋1, 𝑋2, 𝑋3

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 𝑣𝑖 ,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 𝐱, 𝑡 = 𝐯 𝐲, 𝑡 − gradT 𝐯 𝐱, 𝑡 𝐱 − 𝐲 = 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

Exercises

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

original material

material gradient

material indestructibility

material spatial vectors

university of lagos7

time r

placement ofthe body

distributed body

Education

jan 2013 newsletter

robot kinematics: forward and inverse kinematics -...

bookshoppe jan 2013

chap. 3: kinematics (2d) - physics and astronomy at...

kinematics jan 27

kinematics of fluid flow · 2013. 4. 23. · kinematics of...

kinematics -...

benchmark jan 2013

kinematics – graphing motion unit #2 kinematics

architecture jan 2013

3 jan 2013

jurutera jan 2013

university of groningen galactic kinematics and dynamics...

sat jan 2013

robotics 2013 03 kinematics chain

introduction and kinematics lecture 1 - spotlights | web...

grade boundaries jan 2013 [edexcel]jan 2013 [edexcel

kinematics review key 2013

kinematics & dynamics - 18- kinematics . · pdf filesummary...

jan - march 2013