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Deformation gradient for biaxial test

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Posts: 6
Topic starter
(@AliTamaddon)
Active Member
Joined: 15 years ago

Hi Jorgen,

First, thank you very much for providing such a professional platform for material modeling discussions. This is almost the only one that can provide useful information. Your kindness and expertise are greatly appreciated.

I have a few questions about biaxial test on soft tissue. Generally, for a biaxial test we can assume homogeneous deformation mode in the central region of the specimen as following

x1=lambda1*X1+kappa1*X2
x2=lambda2*X2+kappa2*X1
x3=lambda3*X3

So the deformation gradient F can be written as F=:

lambda1, kappa1, 0
kappa2, lambda1,0
0,0,lambda3

The Green-Lagrange strain can be calculated as:

E11=(lambda1^2+kappa2^2-1)/2
E22=(lambda2^2+kappa1^2-1)/2
E12=(lambda1*kappa1+lambda2*kappa2)/2

In biaxial test, if the specimen is stretched simultaneously and symmetrically along the two orthogonal directions, then it is reasonable to assume that the in-plane shearing is negligible.

[B]Now my question is about the exact meaning of negligible in-plane shearing. Specifically, it refers to
(1) E12=0 ?
or
(2) kappa1=0 and kappa2=0?[/B]

If it were the case of (1), then kappa1 and kappa2 are not necessarily be zero. Therefore even though the Lagrange stress P12=0, the true stress (sigma12) and 2nd P-K stress (S12) will not be zero. Is this correct mathematical description for symmetric biaxial testing?

If it were case (2), i.e., kappa1=kappa2=0, then E12=0, sigma12=S12=0. Is this correct description for biaxial testing?

[COLOR=Red]In summary, my first question is: between (1) and (2), which is the correct description of biaxial stretching test?[/COLOR]

6 Replies
Posts: 6
Topic starter
(@AliTamaddon)
Active Member
Joined: 15 years ago

Now comes my second question related to 1st question:

Suppose I want to use a hyperelastic model (for example, Mooney-Rivlin) to fit biaxial testing data. Also, suppose I use non-contact optical method to measure the motion of 4 markers on the tissue surface, and extracted the (lambda1, lambda2, kappa1, kappa2) using interpolate function. Since it is impossible to implement exact symmetric stretching, I believe kappa1 and kappa2 will not be zero.

Now my test data are in the form of:

(lambda1, lambda2, kappa1, kappa2) --> (P11, P22) or (T11, T22) or (S11, S22)

Now, I have two choices to fit my biaxial data.

[B]Choice 1:[/B]
using (lambda1, lambda2, kappa1, kappa2) --> deformation gradient F=

lambda1, kappa1, 0
kappa2, lambda1,0
0,0,lambda3

(here lambda3 can be determined from det(F)=1.)

From F we can obtain C or B matrix. Then, we can use the formula (6-63) or (6-64) of Holzapfels book to derive expressions for stress tensor. Letting S33=0 or T33=0 the Lagrange multiplier p can be solved. Substituting p back and we can obtain curve-fitting formula for T11 and T22.

So in this approach, the original data of kappa1 and kappa2 are involved.

[B]Choice 2[/B]
Here we assume kappa1 and kappa2 are zero, and the measured lambda1 and lambda2 are principal stretches. Therefore, the deformation gradient is

lambda1, 0, 0
0, lambda1,0
0,0,1/(lambda1*lambda2)

Then Invariants I1 and I2 expressed in lambda1 and lambda2 are substitute into Mooney-Rivlin function W=C1(I1-3)+C2(I2-3)
Then the Lagrange stress can be obtained by directly differentiate W with respect to lambda1 and lambda2 --> obtain curve-fitting formula.

In this 2nd choice, the original data of kappa1 and kappa2 are discarded, the curve fitting involves just lambda1 and lambda2, very simple.

[COLOR=Red]So, my question is: which choice shall I choose to fit test data. I prefer the second choice, but I am not sure the assumption of biaxial stretches are principal stretches is correct. In other words, is it correct to discard kappa1 and kappa2 in fitting data?[/COLOR]

Thanks a lot for your instruction.

6 Replies
Posts: 3998
(@jorgen)
Member
Joined: 4 years ago

Biaxial loading to me is a case where kappa1 = kappa2 = 0. Which as you said gives E12 = 0.

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Posts: 3998
(@jorgen)
Member
Joined: 4 years ago

I would use option (1) when fitting the experimental data since that should give the most accurate results.

You might be able to determine, however, if the influence of kappa1 and kappa2 is small enough to be neglected. If so, then you can use option 2.

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Posts: 6
Topic starter
(@AliTamaddon)
Active Member
Joined: 15 years ago

Thanks a lot, Jorgen !....but...I feel your two answers are a little contradicting..

The (1) approach of fitting data is meaningful only when kappa1 and kappa2 are not zero. Since you think kappa(s) are zero for biaxial test, why you still stick with (1) instead of (2) data fitting? I mean, since you assume kappa=0, that means the biaxial stretching is principal stretching, then (2) is as accurate as (1) for fitting, right?

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Posts: 3998
(@jorgen)
Member
Joined: 4 years ago

Oh, good point. Lets see if I can explain myself. My reasoning was that experimentally you might get finite values for kappa1 and kappa2, that is, the experiment might not be 100% pure biaxial.

If the experimental kappas are quite small, then I would just use option 2.

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Posts: 6
Topic starter
(@AliTamaddon)
Active Member
Joined: 15 years ago

Cool, your new explanation confirmed what I guessed...... Then I know the way of next steps.....Thank you soooo much!

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