Finding the inverse of 4th rank elasticity tensor

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The discussion focuses on the challenge of finding the inverse of a fourth-order elasticity tensor, specifically transitioning from equations 2.7.11 to 2.7.16. The user seeks guidance or references on this topic, indicating a lack of available online resources. They express a need for assistance in tensor algebra to demonstrate that the product of the tensor and its inverse yields the identity tensor. The conversation highlights the complexity of tensor operations in elasticity theory. Overall, the user is looking for clear explanations or examples to aid their understanding.
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Homework Statement
I need to find how to obtain the expression of the inverse of a rank four elasticity tensor.
Relevant Equations
C = k 1x1 + 2µ[I-1/3*1x1] where C in the foutrth order tensor
C^-1 = k^(-1)/9 1x1 + 2µ^(-1)[I-1/3*1x1]
I'm desperately trying to understand how to get from 2.7.11 to 2.7.16 and cannot find any reference online on how to find the inverse of an elastic tangent modulus (fourth_order tensor). Can someone help me or give me a reference I can check where they do a similar thing? I would really appreciate it !

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If you know how to do tensor algebra, you should be able to show that ##C/otimes C^{-1}=\text{ the identity tensor.}## (I don't know how to do it.)
 

Thread 'Help needed with Elliptic Integrals'

I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW
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