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 d 

0
     x sin 2   y sin 2  2 xy cos2  , (3.15)
1
1
1
 d    1 

 2
hence tg  2  xy . (3.16)
1
  
x y

We see that the expressions for tg 2 and tg 2 are the same.
0 1
Comparing the right sides of (3.13) and (3.15) it is easy to see that
 d 


    2 t  1   0 . (3.17)
 d   1 
The equations (3.15) - (3.17) show that extreme normal stresses
arise on the planes that are free of tangential stresses, and,
conversely, tangential stresses are zero on the planes where normal
ones take extreme values.
To determine the principal stresses we use the equation (3.9),
taking into account that the stress state is plane
   
x yx  0 . (3.18)
   
xy y
Having found the determinant, come to the quadratic equation
 2         2  0 ,
x y x y xy
hence
   1 2
  x y       4 2 . (3.19)
max 2 2 x y xy
min
By comparing the found values of principal stresses with the
value   0, rename them  ,  ,  in descending order.
z 1 2 3
To determine the position of the principal plane with the stress
   it is necessary to turn the plane with more normal stress
max 1
on the angle  (in absolute value it is not more than 45) in the
0
direction in which the vector of tangential stress acting on this
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