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2
  p sin  p cos  cos sin   sin  
t  x  y x yx

2
 cos   sin cos . 
xy y
From this we obtain:
- normal stress on an arbitrary plane
   cos 2    sin 2    sin 2  ; (3.12)
 x y xy
- tangential stress on an arbitrary plane
  
  x y sin 2    cos 2  . (3.13)
t yx
2

3.5.2 Finding the position of principal planes and the
value of principal stresses

Find the position of principal planes on condition that on these
planes tangential stresses equal to zero:
 
x y
sin 2 0  yx cos 2 0  0,
2
hence
 2 yx
tg  2  . (3.14)
0
  
x y
From the equation (3.14) we find two values of the angle  ,
0
that differ from each other at 90. These angles determine the
position of principal planes. One value of the angle corresponds to
the plane of action of the maximum principal stress, and the
second - the minimum.
To make sure whether really the angles  correspond to plane
0
actions of the maximum and the minimum normal stresses we
should study the extreme expression (3.12):

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