Finding the inverse of 4th rank elasticity tensor

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SUMMARY

The discussion focuses on finding the inverse of a fourth rank elasticity tensor, specifically in the context of transitioning from version 2.7.11 to 2.7.16 of a software tool. The user seeks guidance on tensor algebra to demonstrate that the product of the elasticity tensor and its inverse yields the identity tensor. This indicates a need for a deeper understanding of tensor operations and their applications in material science or engineering contexts.

PREREQUISITES
  • Tensor algebra fundamentals
  • Understanding of elasticity theory
  • Familiarity with fourth order tensors
  • Basic knowledge of software versioning and updates
NEXT STEPS
  • Study the properties of fourth order tensors in elasticity
  • Learn how to compute the inverse of a tensor
  • Explore software documentation for version 2.7.16 updates
  • Research applications of tensor algebra in material science
USEFUL FOR

Material scientists, mechanical engineers, and students studying elasticity and tensor algebra who need to understand the mathematical foundations of elasticity tensors and their inverses.

Pilou115
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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.)
 

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