I came across a problem where I was asked to find out the direction of the maximum principal stress in a given stress tensor , with the help of eigen vectors I was able to construct an equation which has three roots ,the three roots are the three principal stresses , but I am confused about how i will be able to find out the direction of the maximum principal stress , kindly help me.


Assuming the following stress tensor:

$$A = \begin{bmatrix}100,20,0\\20,0,20\\0,20,50\end{bmatrix}$$

you can find the eigenvalues and eigenvectors in octave/matlab

A = [100, 20 ,0 ;20 ,0,20; 0,20 ,50]
[k,l] = eig(A)

The printout is

k =

   0.169653  -0.132430  -0.976565
  -0.935341   0.290482  -0.201883
   0.310410   0.947672  -0.074586

l =

Diagonal Matrix

   -10.265         0         0
         0    56.130         0
         0         0   104.135

The principal stress are ordered from the smallest to the largest(algebraically)


principal Stress principal Stress $\sigma_i$ principal direction $\vec{v}_i$
$\sigma_1$ 104.135 $ \begin{bmatrix} 0.169653 \\ -0.935341 \\0.310410\end{bmatrix}$
$\sigma_2$ 56.130 $ \begin{bmatrix} -0.132430\\ 0.290482 \\0.947672\end{bmatrix}$
$\sigma_3$ -10.265 $ \begin{bmatrix}-0.976565 \\ -0.201883\\ -0.074586 \end{bmatrix}$

The maximum shear stress exists on the plane at 45 degrees to the planes of the largest and the smallest principal stress.

The value of the max shear stress is $\frac{104.135 -( -10.265)}{2} $

The max shear stress direction is:

$$\frac{\vec{v}_1+\vec{v}_3}{\sqrt{2}}= \begin{bmatrix} -0.4035 \\ -0.5686\\ 0.1179 \end{bmatrix}$$

NOTE: I am a bit rusty, so if you find an error please point it out.

  • $\begingroup$ just to be sure about the solution , are you finding out the direction cosines for each of the principal stresses as the principal directions $\endgroup$ Sep 23 at 11:30
  • $\begingroup$ I am not using the direction cosines at all. That would require to define for each vector three values, which I find confusing. If you want to do it that way you can though. $\endgroup$
    – NMech
    Sep 23 at 11:32
  • $\begingroup$ i appreciate your attempt , but at this moment (I) need to basically pen down the solution for assessment , so direction cosines would be more comfortable . Thank you for the help. $\endgroup$ Sep 23 at 11:42

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