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E  1  

2  C 1 1   C 2 2    p 1 ,
1   r 1 

E  1  

2  C 1 1   C 2 2    p 2 ,
1   r 2 
from which we find

2 2
2
2
1  r p  r p 1  r r  p  p 
C   1 1 2 2 , C   1 2 1 2 . (9.14)
1 2 2 2 2 2
E r  r E r  r
2 1 2 1
Thus, the formula for the stresses and radial displacement finally
takes the form
2 2
2
2
  r p  r p r r  p  p  1
t 1 1 2 2 1 2 1 2
    , (9.15)
2
2
 r  r  r 1 2 r  r 1 2 r 2
2
2
2 2
2
2
1  r p  r p 1  r r 2  p  p 2  1
1
1
u   1 1 2 2 r     . (9.16)
2
2
E r  r 2 E r  r 2 r
2 1 2 1
With the formula for  and  see that their sum - a constant,
t r
i.e.
    const
t r
Therefore, the relative linear deformation of the ring in the
direction of the axis of the cylinder is a constant

       const .
x t r
E
If the ends of the cylinder are present and the longitudinal
strength N is parallel to axis, in its cross-section stress appears
N
  ,
x 2 2 (9.17)
 r  r 
 2 1
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