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P Pe Pe
  x  y y  z z , (1.16)
F J J
z y
or

P  e y e 
   1 y  z z  , (1.17)

2
F   i z 2 i y 
J J y
where i  z , i  – radii of inertia of the cross section
z y
F F
of the rod.
Determine the position of the neutral line. For this equated
to zero right-hand side of the expression (1.17):
P  e y e z 
 1 y  z  0 .

F   i z 2 i 2 y 

P
As  0 , then
F
e y e
1 y  z z  0 . (1.18)
i 2 i 2
z y
Expression (1.18) is the equation of the neutral axis. It can
be represented as the equation of the line in segments
y z
  1,
a b
where
2
i 
a  y   z ,
н 
e y 
 , (1.19)
2
i н 
b  z  
н 
e z 
this segments that cuts off the neutral line on the coordinate axes
(fig. 1.8).
Analyzing the formula (1.19), we see:

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