## difference of 1st, 2nd ,3rd principal stretch

Dear all,

I have problem on defining first principal stretch, second principal stretch and third principal stretch.

Does it means in x axis we definite it as first, y axis is second, third is z axis?

what if my load is applied on a cubic from the -y direction in a uniaxial testing tension, so my first principal stretch remains as x axis, or will it change?

I hope someone can understand my statement, if not, do post it. i try to make myself clear.

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 Quote by fruitkiwi Dear all, I have problem on defining first principal stretch, second principal stretch and third principal stretch. Does it means in x axis we definite it as first, y axis is second, third is z axis? what if my load is applied on a cubic from the -y direction in a uniaxial testing tension, so my first principal stretch remains as x axis, or will it change? I hope someone can understand my statement, if not, do post it. i try to make myself clear.
Are you talking about 1st 2nd and 3rd principal stresses in Engineering? If so, I can move this thread to the Engineering forums. If not, it can stay here.

 Hello Mike, i think FK is referring to principal strains not stresses. Edit: Fruitkiwi, Continuing with this guess, I am further guessing that you are referring to the tensor shear strain εij = 0.5(εij + εji) i,j = 1,2,3 and not the engineering strains γ which are twice this. Please confirm which system we are working in. Hopefully you realise that the axes of principal strain do not, in general, coincide with the normal x,y,z axes?

## difference of 1st, 2nd ,3rd principal stretch

Hi, Berkeman,

I think it is related to continuum mechanics so i choose to post it here.
I have not problem on engineering application, however, in terms of physic definition, erm, I have problem.

Hi, Studiot,

Thanks for save my post :)
The system you refer to is Lagrangian or Eulerian system, am i right?
I am in Eulerian system,
"since axes of principal strain do not, in general, coincide with the normal x,y,z axes, "
May i know how can i determine the axes of principal strain? as looks like it is determined by the stretch you prescribed on the material.

 Dear Studiot, Thanks for your detail evaluations. I think I can differentiate stress and strain, they have clear distinction, :) The one I make mistake is strain and stretch. Currently read about Holzapfel "A continuum Approach for engineering" , that's why i use the term stretch. stretch =1+strain, I overlook it. My first language is chinese...now I see the reason why we need the coordinate system to describe, ad indicated in the book which is full of equation. I will derive it to strengthen the understanding.thanks again.

 My first language is chinese...
Sorry I know nothing of chinese texts, I was hoping that you were going to say you were from a French influenced part of the East, I would then have recommended

Mecanique de Materiaux Solides by Lemaitre and Chaboche
Which has an english translation from Cambrideg University Press.

I don't know Holzapfel and can't find direct reference to 'stretch', however from you comments I think it corresponds to the deformation of a line segment.

In continuum mechanics the analysis is pointwise - that is strains refer to the limiting displacement and deformation shrunk to a point.
Note strain and dispalcement are different tensors.

If, however we consider a short line segment ds between P (x,y,z) and Q (x+dx,y+dy,z+dz) in the undeformed body

Then P moves to P*(x*,y*,z*) and Q moves to Q*(x*+dx*,y*+dy*,z*+dz*) and ds becomes ds*, so the components of the change are u,v,w

x* = x+u, y*=y+v, z*=z+w

and define the quantity εE as

$${\varepsilon _E} = \frac{{d{s^*} - ds}}{{ds}}$$

Then some manipulation of differential coefficients leads to

$$\begin{array}{l} \left( {1 + {\varepsilon _E}} \right){l^*} = \left( {1 + \frac{{\partial u}}{{\partial x}}} \right)l + \frac{{\partial u}}{{\partial y}}m + \frac{{\partial u}}{{\partial z}}n \\ \left( {1 + {\varepsilon _E}} \right){m^*} = \frac{{\partial v}}{{\partial x}}l + \left( {1 + \frac{{\partial v}}{{\partial y}}} \right)m + \frac{{\partial v}}{{\partial z}}n \\ \left( {1 + {\varepsilon _E}} \right){n^*} = \frac{{\partial w}}{{\partial x}}l + \frac{{\partial w}}{{\partial y}}m + \left( {1 + \frac{{\partial w}}{{\partial z}}} \right)n \\ \end{array}$$

Where l, m and n are direction cosines of the principal strain axes.

I wonder if your stretch quantity is (1+εE)

εE is the engineering strain.

Edit
I have found some definitions in my continuum mechanics books defining 'stretch' as εE so there is some difference between authorities about this one.

 Dear Studiot, Thank for agree with me on the stretch term i used. I think the strech is 1+εE. Finally I can start to read the books. "In continuum mechanics the analysis is pointwise - that is strains refer to the limiting displacement and deformation shrunk to a point. Note strain and dispalcement are different tensors." Thank you.
 Now we are getting somewhere would you like to refine you question? I note that your wiki reference is to stress. Does that mean you wish to approach principal axes via stress analysis? That would make things much easier.
 Thanks for points out my mistake. I edited the post. My mistake, use the wrong image. Ya, my intention is to understand the stretch, not stress. You answered my question.

 Quote by fruitkiwi Dear all, I have problem on defining first principal stretch, second principal stretch and third principal stretch.

Hi fruitkiwi --

If you were to do a tension test on your material, for example, and no rigid body rotation is present, then the "principal stretch" in the longitudinal direction would be "1" plus the quantity of engineering strain that is measured, as you know.

Strictly speaking, recall (from "polar decomposition") that V (the left stretch tensor) and U (the right stretch tensor) have the same eigenvalues and note that these scalar values (the eigenvalues, λ) are the principal stretches. This is standard terminology in continuum mechanics, as you know.

The eigenvalues of V and the eigenvalues of U would also be in the same order if no rigid body rotation is present.

 Quote by fruitkiwi Does it means in x axis we definite it as first, y axis is second, third is z axis?
Your question relates to the direction of these principal stretches.
The eigenvectors of V and U differ when rigid body rotation is present. The direction of principal stretch depends on whether you are working in "spatial" (V) or "material" (U) coordinates.

In other words, the values of λ are in a certain order and this order will differ, if rigid body rotation is present, depending on how you obtain your values of λ.

 Quote by fruitkiwi what if my load is applied on a cubic from the -y direction in a uniaxial testing tension, so my first principal stretch remains as x axis, or will it change?
It will be in the y direction.

This isn't really the important question though, in my opinion.

My question to you: what direction do the principal stretches act if you perform that tension test while simultaneously performing a rigid body rotation on the specimen?

The principal stretch magnitude along the length of the specimen is always "1" plus the measured engineering strain. This is one of your eigenvalues.

However, the direction of principal stretch (the order of your eigenvalues) will depend on how you go about calculating them (due to the rigid body rotation).

I recommend you solve such a problem (make one up). For example, start with x1=-1/2X1, x2=1/2X2, x3=2X2. Draw what this looks like -- should look like a tension test with a 90degree rigid body motion. Find F, C, B, their eigenvalues and eigenvectors -- etc. This is the great thing about strains (or "stretches") - you can see them!

I think that the comments regarding stress and isotropy, etc., are important too, but I won't hijack. I have a feeling you already understand which strain measure and stress measures form appropriate pairs anyway.

 Thank you anyway afreiden, Quality contributions are always welcome.
 Was a mistake in my little exercise - plus I wanted to write it more clearly: Fruitkiwi -- start with: $x_1=-\frac{1}{2}X_1$ $x_2=\frac{1}{2}X_3$ $x_3=2X_2$ Also - Studiot: I guess what I would mention just to address everything in the thread is that in large strain isotropy some people pair up the Cauchy stress tensor, σ, with the strain tensor B since they behave the same under rigid body motion. Similarly, in spatial principal stress space (the main topic of this thread), these folks (or FEM codes) may then choose to pair up the principal Cauchy stresses, $\sigma_i$, with the principal stretches, $\lambda_i$. In this case, those $\lambda_i$ values are the eigenvalues of V (recall V2=B), which would put the eigenvalues, $\lambda_1, \lambda_2, \lambda_3$, in the order that corresponds to the principal Cauchy stress values, $\sigma_1, \sigma_2, \sigma_3$. In linear infinitesimal isotropy, the stress used by the FEM codes is analogous to the Second-Piola-Kirchhoff stress tensor, $\hat{\sigma}$ and the corresponding strain is analogous to C (or U -- since U2=C). I think I talked more about this in another thread awhile back.

 Similarly, in spatial principal stress space (the main topic of this thread),
Hello afreiden,

However as far as I can see fruitkiwi wants to attack strain directly, without necessarily relating it to stress.

This is more difficult (although in real life we actually measure strain or load not stress) and why I have been feeling my way.

Further, in the uniaxial tension situation outlined by both afreiden and fruitkiwi there is more to it than meets the eye. Some authors use a magnification factor for the uniaxial stress depending upon the restraint conditions on the other two axes.

 Quote by Studiot Further, in the uniaxial tension situation outlined by both afreiden and fruitkiwi there is more to it than meets the eye. Some authors use a magnification factor for the uniaxial stress depending upon the restraint conditions on the other two axes.
Too much reality for the classical physics forum :-) :-p

 Too much reality for the classical physics forum
Sometimes it is good to understand the difference between theory and reality.

Tensor stress analysis describes stress at point. You can't shrink (or expand) a point.

Consider this.

Let a bar, of modulus E and poissons ratio 1/m, be stretched by stress s1 in the first principal direction in such a manner as to prevent all lateral strain (e2 and e3).

Then

$$\begin{array}{l} {e_1} = \frac{{{s_1}}}{E} - \frac{{{s_2}}}{{mE}} - \frac{{{s_3}}}{{mE}} \\ {e_2} = \frac{{{s_2}}}{E} - \frac{{{s_1}}}{{mE}} - \frac{{{s_3}}}{{mE}} = 0 \\ {e_3} = \frac{{{s_3}}}{E} - \frac{{{s_2}}}{{mE}} - \frac{{{s_1}}}{{mE}} = 0 \\ \end{array}$$

$${s_1} = {e_1}E\frac{{m\left( {m - 1} \right)}}{{\left( {m - 2} \right)\left( {m + 1} \right)}}$$

Now if the bar is steel with a typical poissons ration of 1/3, then m=3

This means that the effective modulus in this situation along the first principal axis is modified by a factor of

$$\frac{{3\left( {3 - 1} \right)}}{{\left( {3 - 2} \right)\left( {3 + 1} \right)}} = \frac{6}{4} = 1.5$$

So to calculate the strain from the stress we need to use an effective modulus of

Eeffective = 1.5E

Quite a difference.

 Hi, Afreiden, Thanks for physic of classical mechanics. now progressing in this thought "what direction do the principal stretches act if you perform that tension test while simultaneously performing a rigid body rotation on the specimen?" Hi, Studiot, Thanks for the application of continuum mechanics. i thought it always a black hole in continuum mechanics :P update your guys after i finish the thought. Thanks a lot.