Continuum Mechanics - Deformation gradient


by tricha122
Tags: continuum, deformation, gradient, mechanics
tricha122
tricha122 is offline
#1
Dec27-12, 10:59 AM
P: 14
Hi all,

I am trying to self-learn continuum mechanics, and I have a question regarding the development of the deformation gradient (which ultimately leads to green's deformation tensor).

I have attached the specifics of the question in a attached photo.

Ultimately, there comes a point when determining the deformation using the change in magnitude of the square of dX and dx:

dx^2 - dX^2 = dxidxi-dXadXa

However, somehow using a previous equation (dxi = xi,adXa) and the susbtitution property of the kronecker delta, they come up with:

dx^2 - dX^2 = xi,adXa*xi,bdXb - delta(ab)*dXa*dXb

My question is - how was the kronecker delta substituted in? There is no direction associated with magnitudes. Further - where did the subscript "B" come from and what does it represent physically?

Any help would be greatly appreciated.
Attached Thumbnails
Continuum Mechanics - Deformation Gradient.png  
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Studiot
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#2
Dec27-12, 05:16 PM
P: 5,462
First the subscript B is introduced as part of the kronecker delta, δij or δAB here.

It is only a dummy and does not really matter since if B ≠ a then δAB = 0

Which brings us to why the kronecker?

Well this is a way of writing the dot product of the vectors dXi and dxi etc.

Don't forget that the quantities in this diagram are vectors so additions and multiplicasions are vector additions and multiplications.
tricha122
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#3
Dec30-12, 08:06 PM
P: 14
Quote Quote by Studiot View Post
First the subscript B is introduced as part of the kronecker delta, δij or δAB here.

It is only a dummy and does not really matter since if B ≠ a then δAB = 0

Which brings us to why the kronecker?

Well this is a way of writing the dot product of the vectors dXi and dxi etc.

Don't forget that the quantities in this diagram are vectors so additions and multiplicasions are vector additions and multiplications.

Thank you for your response. However, I still am trying to wrap my head around the introduction of a new variable.

After substiuting out dXa,

(dx)^2 - (dX)^2 = (xi,a*xi,a - 1)*dXa*dXa

how does the kronecker delta substitute (dijej = ei) in here to get the next equation?

(dx^2) - (dX)^2 = (xi,a*xi,b - dab)*dXa*dXb

if a = b, then you get the previous equation.

If a =/ b, then you get

(dx^2) - (dX)^2 = (xi,a*xi,b)*dXa*dXb

which im not sure what that means physically.

tricha122
tricha122 is offline
#4
Dec30-12, 08:14 PM
P: 14

Continuum Mechanics - Deformation gradient


Quote Quote by tricha122 View Post
Thank you for your response. However, I still am trying to wrap my head around the introduction of a new variable.

After substiuting out dXa,

(dx)^2 - (dX)^2 = (xi,a*xi,a - 1)*dXa*dXa

how does the kronecker delta substitute (dijej = ei) in here to get the next equation?

(dx^2) - (dX)^2 = (xi,a*xi,b - dab)*dXa*dXb

if a = b, then you get the previous equation.

If a =/ b, then you get

(dx^2) - (dX)^2 = (xi,a*xi,b)*dXa*dXb

which im not sure what that means physically.
Further, Cab = xi,a*xi,b = F^t * F confusing me even further suggesting that xi,a xi,b are the transpose of each other when a=b.
Chestermiller
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#5
Jan4-13, 08:05 PM
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PF Gold
P: 4,446
You realize, of course, that the Einstein summation convention is being used here. Thus,

[tex]dX_adX_a=\delta_{a,b}dX_adX_b=(dX_1)^2 +(dX_2)^2 +(dX_3)^2[/tex]

In tensorial notation, δi,j are the components of the identity tensor (aka, unit tensor or metric tensor) I.

It is often more revealing to write these equations in dyadic tensor notation. Let dx represent a differential position vector between two material points in the deformed configuration of the body, and let dX represent the differential position vector between these same two material points prior to the deformation. The vectors dx and dX can be related to one another in terms of the deformation gradient tensor F:

[tex]\mathbf{dx}=\mathbf{F}\cdot\mathbf{dX}[/tex]

The squared length of the vector dx in the deformed configuration of the body can be determined by taking the dot product of dx with itself:

[tex]\mathbf{dx}\cdot\mathbf{dx}=\mathbf{dX}\cdot (\mathbf {F^T}\cdot\mathbf{F})\cdot\mathbf{dX}[/tex]

The change in the squared length of the differential position vector between the two material points in the deformed and undeformed configurations of the body is given by:

[tex]\mathbf{dx}\cdot\mathbf{dx}-\mathbf{dX}\cdot\mathbf{dX} =\mathbf{dX}\cdot (\mathbf {F^T}\cdot\mathbf{F}-\mathbf{I})\cdot\mathbf{dX}[/tex]

The Cauchy-Green tensor is defined by:

[tex]\mathbf{C}=\mathbf {F^T}\cdot\mathbf{F}[/tex]

So [tex]\mathbf{dx}\cdot\mathbf{dx}-\mathbf{dX}\cdot\mathbf{dX} =\mathbf{dX}\cdot (\mathbf {C-I})\cdot\mathbf{dX}[/tex]

It is also possible to define the finite strain tensor E as:

[tex]2\mathbf{E}=\mathbf{C-I}[/tex]

I hope this helps.

Chet


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