principal strains using eig function

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BLP
BLP el 18 de Jun. de 2015
Respondida: Eric Lin el 19 de Jun. de 2015
I am trying to use the eig function to calculate the principal strains from 3D strain tensor. Unfortunately I do get different results than from another tool (<http://www.continuummechanics.org/cm/techforms/Eigen.html>) For
A1 = [4.420000000000000 1.350000000000000 3.509000000000000;
1.350000000000000 3.832000000000000 0;
3.509000000000000 0 3.207000000000000]
Matlab eig function yields
>> [V1, D1] = eig(A1)
V1 =
-0.651153549926562 0.054428335246106 0.756991816825237
0.232116359059687 -0.935349177129471 0.266915553499784
0.722579442189515 0.349513194551449 0.596422230101738
D1 =
0.044859378106921 0 0
0 3.753442970840320 0
0 0 7.660697651052759
while the Web tool gives:
The Values look pretty the same, but their position in vector/matrix is different. As far I understand, the columns of the V1 represent the cosines of the angle between the a principal strain and the Cartesian axis. But which corresponds to which principal strain and how to calculate angles between the principal strains and the Cartesian strain axis?

Respuestas (1)

Eric Lin
Eric Lin el 19 de Jun. de 2015
What is happening is the web app is displaying the eigenvectors by row whereas MATLAB is displaying them by column. The relative order of the eigenvectors is not significant.
I am not familiar with the domain but from a quick search it seems that the vectors are related to the strains based on their eigenvalue, i.e. the max principal strain is the eigenvector with the largest eigenvalue. Again, I am not familiar with the domain but angles between vectors can be computed by dot product.

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