00:01
Hi, let's see stress tensor sigma equal to sigma xx, sigma x, sigma x, cyg, and sigma y x, sigma y, syg, sigma y, z, and sigma z, x, sigma z, x, sigma z, y, and sigma z, z.
00:19
Let's substitute the value directly from the equation.
00:22
So we have 60 minus 40, 30, minus 40, minus 40, 30 20 20 20 let's take this is equation number 1 now we'll calculate the principal stresses now from the characteristic equation that is mod of sigma minus lambda equal to 0 let's simplify it so mod of 60 minus lambda minus 40 30 minus 40 minus 40 minus lambda 20 30 20 20 minus lambda equal to 0 let's take this is equation number 2 now let's determine of the matrix that is minus lambda cube plus 40 lambda square plus 4 ,900 lambda minus 1 1 1600 0 .00 equal to 0.
01:34
Let's take this is equation 3.
01:37
Now roots of equation 3 is lambda 1 comma 2 .3 is equal to 82 .385, 21 .90 and minus 64 .288.
01:54
And their cosines direction first calculate eigenvectors corresponding to eigenvalues.
02:00
So put lambda 1 equal to.
02:03
82 .38 5 in equation 2.
02:08
So we get minus 22 .38 minus 40 30 minus 40 minus 122 .38 20 .30.
02:22
20 .30.
02:24
20.
02:24
20.
02:24
20 minus 62.
02:26
38.
02:27
38.
02:28
And n1.
02:34
N2.
02:34
N2...