Kinematic Deformation: Trajectories, Acceleration & Tensors Explained

In summary, the conversation discusses a motion of a continuum defined by a given velocity field and orthonormal vectors. The lagrangian representation of motion is determined, along with the trajectories, streamlines, and streaklines. The acceleration field is also calculated. The deformation gradient tensor, expansion tensor, deformation tensor, and strain rate tensor are all determined, along with the eigenvalues and principle directions. It is also mentioned that if the value of the given constant is small, there will be little to no deformation. The conversation ends with a request for an explanation of the original equation and its relation to mechanics.
  • #1
mullzer
11
0

Homework Statement



Let a and b be two given orthonormal vectors around a fixed point O. The motion of a continuum is defined by the following velocity field:

[tex] v(M) = \alpha \vec{a} (\vec{b} . \vec{OM}) \\ [\tex]

where [tex] \alpha [\tex] is a known positive constant.

1. Choosing a marker, determine the lagrangian representation of motion, given that at time t= 0, the particle occupies the position X= ([X][/1], [X][/2], [X][/3]).

Determine the trajectories, streamlines and streaklines of the motion. Calculate the acceleration field.

2. Determine the composants of the deformation gradient tensor F, the expansion tensor C, the derformation tensor X, the strain rate tensor D and the tensors G = [F][/-1] and B = [G][/T] G

Calculate the eigenvalues and principle directions of C, X and B. What would happen to these if the value of [tex]\alpha t [\tex] was small? Interperate the case.



Homework Equations


The Lagrangian stament of motion: [tex] \vec {x_{i}} (t) = \vec{\phi_{i}} (\vec {X}, i = 1, 2, 3\\

and \vec{v} (\vec {u}, t) = \frac{d \vec{x}}{dt} \\[\tex]


The Attempt at a Solution



First of all, is it alright to substitute [tex]\vec{a} [\tex] and \vec{b} [\tex] with basis vectors [tex]\vec{e_4{1}} [\tex] and \vec{e_{2}} [\tex] which would correspond with [tex]\vec{x_{i}} [\tex] ?

1. use the second relevant eq:

[tex] \vec{v} (\vec {u}, t) = \frac{d \vec{x}}{dt} \\[\tex]

with respect to [tex] x_{1}, x_{2} and x_{3}. According to some notes i have jotted down, this yields:

[tex] \frac{d \vec{x_{1}}}{dt} = \alpha x_{1} \\

\frac{d \vec{x_{2}}}{dt} = -\alpha x_{2} \\

\frac{d \vec{x_{3}}}{dt} = 0 \\ [\tex]

which is the main source of my confusion. How come it is different for each of these? Is it do with the wedge product of a and b?

After that I rearrange these eq's and solve for [tex] x_{i} \ [\tex] using the intial condition of X= ([x=[X][/1], [X][/2], [X][/3]) at t = 0:

[tex] x_{1} = C_{1}e^(\alphat) \ [\tex] etc; where [tex]C_{1} \ [\tex]becomes [tex] X_{1}\\ [\tex]

I believe that these eq's are also the trajectories of the motion.

Deriving these further with respect to t gives the lagrangian speed and then the lagrangian acceleration:

[tex] \Gamma_{1} = \alpha^2 X_{1}e^(\alphat) \ [\tex] etc.

2. To obtain F, I derived the eq. of motion [tex] x_{1} = C_{1}e^(\alphat) \ [\tex] with respect to [tex] x_{1}, x_{2} and x_{3} [\tex] and then the other eq's of motion with respect to [tex] x_{1}, x_{2} and x_{3} [\tex]. These nine values were used to create a diagonal tensor with trace [tex] e^(\alpha t) + e^(-\alpha t) + 1. \\ [\tex]

C is then equal to [F][/T]F

X = [tex] \frac {1}{2} (C-I). \\ [tex]

I know how to get eigenvaues by solving C - [tex]\lambda[\tex]I.

Mainly, I would just like an explanation of how the original equation came about (providing the rest of what I have done is correct). If you could explain it as simply as possible as I am now to this field of mechanics. Thanks in advance!
 
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  • #2
This time with better notation (hopefully):

1. Homework Statement

Let a and b be two given orthonormal vectors around a fixed point O. The motion of a continuum is defined by the following velocity field:

[tex] v(M) = \alpha \vec{a} (\vec{b} . \vec{OM}) [/tex]

where [tex] \alpha [/tex] is a known positive constant.

1. Choosing a marker, determine the lagrangian representation of motion, given that at time t= 0, the particle occupies the position X= ([tex]X_{1}[/tex],[tex]X_{2}[/tex],[tex]X_{3}[/tex]).

Determine the trajectories, streamlines and streaklines of the motion. Calculate the acceleration field.

2. Determine the composants of the deformation gradient tensor F, the expansion tensor C, the derformation tensor X, the strain rate tensor D and the tensors G =[tex]F^{-1}[/tex] and B = [tex]G^{T}[/tex] G

Calculate the eigenvalues and principle directions of C, X and B. What would happen to these if the value of [tex]\alpha t [/tex] was small? Interperate the case.



2. Homework Equations
The Lagrangian stament of motion: [tex] \vec {x_{i}} (t) = \vec{\phi_{i}} (\vec {X},t) [/tex], i = 1, 2, 3

and [tex] \vec{v} (\vec {u}, t) = \frac{d \vec{x}}{dt} [/tex]


3. The Attempt at a Solution

First of all, is it alright to substitute [tex]\vec{a} [/tex] and [tex] \vec{b} [/tex] with basis vectors [tex]\vec{ e_1} [/tex] and \vec{e_2} [/tex] which would correspond with [tex]\vec{x_{i}} [/tex] ?

1. use the second relevant eq:

[tex] \vec{v} (\vec {u}, t) = \frac{d \vec{x}}{dt} [/tex]

with respect to [tex] x_{1}, x_{2} [/tex] and [tex]x_{3}[/tex]. According to some notes i have jotted down, this yields:

[tex] \frac{d \vec{x_{1}}}{dt} = \alpha x_{1} [/tex]

[tex]\frac{d \vec{x_{2}}}{dt} = -\alpha x_{2} [/tex]

[tex]\frac{d \vec{x_{3}}}{dt} = 0 [/tex]

which is the main source of my confusion. I don't know how to get this solution :grumpy: How come it is different for each of these? Is it do with the wedge product of a and b?

After that I rearrange these eq's and solve for [tex] x_{i} \ [/tex] using the intial condition of

x= ([tex]X_{1}[/tex],[tex]X_{2}[/tex],[tex]X_{3}[/tex]) at t = 0:

[tex] x_1 = C_{1}e^{\alpha t} [/tex] etc;

where [tex]C_{1} [/tex] becomes [tex] X_{1} [/tex]

I believe that these eq's are also the trajectories of the motion.

Deriving these further with respect to t gives the lagrangian speed and then the lagrangian acceleration:

[tex] \Gamma_{1} = \alpha^2 X_{1}e^{\alpha t} \ [/tex] etc.

2. To obtain F, I derived the eq. of motion [tex] x_{1} = C_{1}e^(\alpha t) \ [/tex] with respect to [tex] x_{1}, x_{2} and x_{3} [/tex] and then the other eq's of motion with respect to [tex] x_{1}, x_{2} and x_{3} [/tex]. These nine values were used to create a diagonal tensor with trace [tex] e^{\alpha t} + e^{-\alpha t} + 1. [/tex]

C is then equal to [F][/T]F

X = [tex] \frac {1}{2} (C-I). [/tex]

I know how to get eigenvaues by solving C - [tex]\lambda[/tex]I.

Am I right in saying that if the value of [tex]\alpha t [/tex] was small, there would be little or no deformation because these are directly proportional to it?

Mainly, I would just like an explanation of how the original equation came about (providing the rest of what I have done is correct). If you could explain it as simply as possible as I am now to this field of mechanics. I am also wondering how to interperate the trafectories etc into rough sketches.

Thanks in advance!
 
Last edited:

1. What is kinematic deformation?

Kinematic deformation refers to the change in shape or position of a body or object over time. It is a fundamental concept in mechanics and is often used to describe the motion of objects in engineering and physics.

2. What are trajectories in kinematic deformation?

Trajectories in kinematic deformation refer to the path or curve that an object takes as it moves through space. It is often described in terms of position, velocity, and acceleration.

3. How is acceleration related to kinematic deformation?

Acceleration is a measure of the rate at which the velocity of an object changes over time. In kinematic deformation, acceleration is used to describe how the shape or position of an object changes over time.

4. What is the role of tensors in kinematic deformation?

Tensors are mathematical objects that are used to describe the deformation of a body or object. In kinematic deformation, tensors are used to represent the change in shape or position of an object over time.

5. How is kinematic deformation used in real-world applications?

Kinematic deformation has many practical applications, such as in engineering, robotics, and biomechanics. It is used to study the motion of objects and can be applied to design and analyze structures, predict the movement of objects, and understand the biomechanics of human movement.

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