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Theroretical derivations of constitutive equations

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Posts: 10
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(@xbzihan)
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Joined: 15 years ago

I would like to fit a couple of constitutive equations for uniaxial strain and see which of the models performs best. I am trying to fit both the eight chain model and the mooney rivlin model.

Before I can do this, I have however a couple of questions that I am hoping you can answer. I apologize if they seem trivial or dumb

On page 91 of your Phd thesis equation 3.15 states the continuum mechanics expression for the Cauchy stress. Is it possible to elaborate the derivation between eq. 3.15 and 3.16? This would help a lot.

Regarding eq. 3.17 and 18 is it true that 3*lambda^2=I1? Which I found in another paper on polymerfem.com. Because if this is true I cant rewrite 3.17 to become 3.18 since Im stuck with lambda^2.

When differentiating eq.3.13 with respect to lambda the dependence of beta (inverse langevin) on lambda seem to be ignored.is this correct?

How is expression 3.23 for incompressible uniaxial deformation obtained from eq 3.19, and why is this eq (3.23) different from the one stated in Modeling of the dynamic mechanical response of elastomers. (No division through a langevin term in the latter)

How would one obtain the Cauchy stress from a strain energy density function like mooney rivlin? And how would uniaxial tension stress be described?

I hope I havent bothered you too much, but Im really stuck.

Kind regards

Harm Kooiker

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Posts: 10
Topic starter
(@xbzihan)
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Joined: 15 years ago

Explanation

This thread is regarding the PHD thesis of Jorgen Bergstrom.

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

(1) Eq. 3.15 -> 3.16:
This derivation is provided in different text books, for example, Holtzapfelss book.

(2) Eq. 3.17 -> 3.18:
Yes, your equation for I1 is correct. When I apply the chain rule I get Eq 3.18 as written.

(3) Eq. 3.13:
No, beta is not neglected. If you perform the calculation, a lot of terms cancel out.

(4) The Eq. 3.23 was derived using the standard approach:
* Derive expression for S22,
* Set S22=0, and solve for pressure p
* insert expression for p in expression for S11

(5) I recommend that you checkout, for example, Holzapfels book for the details.

-Jorgen

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Posts: 10
Topic starter
(@xbzihan)
Active Member
Joined: 15 years ago

Thanks a lot

I will check out Holtzapfels book. Thanks a lot for your quick and clear reply mr. Bergstrom.:)

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Posts: 10
Topic starter
(@xbzihan)
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Joined: 15 years ago

2 follow up questions

I hope I am not bothering too much.. we all have to learn right.

I have differentiated eq. 3.12 wrt gamma, and this leads to: dT12/dgamma= 3*Langevin inv(1/lambda lock)

This is not yielding the result I want, since it is a factor three too much. I am regarding the differentiation as 3 chains:

x=gamma

R=everything not dependent on gamma

K=(1+(x^2)/3)^0.5 = dependent on gamma

G=lambda lock

Lang=Langevin inv

T12=R*(x*K*Lang(K/G))

this needs a lot of chain rules

dT12/d x =1/R * [{x*d/dx(K*Lang(K/G))+1*(K*Lang(K/G))}+

{K*d/dx(x*Lang(K/G))+(dK/dx)*(x*Lang(K/G))}+

{Lang(K/G)*d/dx(x*K)+dLang(K/G)/dx*(x*K)}]

Because we need the derivative at x=0 a lot of terms can be cancelled, i think..(furthermore I can disregard a couple of further chains because of multiplication by x which is zero)

Now I am left with

dT12/d x = 1/R * [K(x=0)*Lang(1/G)+K(x=0)*Lang(1/G)+K(x=0)Lang(1/G)=

1/R*3Lang(1/G)

So this is my reasoning and my result is faulty, I am not able to track the fault down. I understand that reading (aka understanding what I am meaning 🙂 ) this kind of equation is difficult, I am hoping it will be possible.

2. Another question is the following:

With eq 3.23 is it true that B*=[lambda^2 0 0, 0 1/lambda 0, 0 0 1/lambda]?

Because now I dont understand why P isnt simply 1/3 tr(B*).

If you calculate this then P=1/3(lambda^2+2/Lambda)

If I follow your hint (first set S22 to zero for P), I get P=1/lambda. Which is correct leading to eq 3.23

I dont understand why the first method does not yield the correct answer. The hydrostatic pressure was always the trace divided by 3.

Can you help me with these follow up questions?

Kind regards Harm

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

(1) I encourage you to try the math again. If you still have problems, write a nice pdf summary of your steps and I will take a look.

(2) For imcompressible uniaxial loading your B* term is correct. Really, the only way to get the pressure P term is through the boundary condition that stress 22 = stress 33 = 0. I dont quite follow your approach.

-Jorgen

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