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Substituting the deformation (6.7) in Hooke's law (6.8), we
obtain
y
  E . (6.9)
x

The analysis of the relationship (6.9) shows that the normal
stresses in cross section:
- are evenly distributed across the width of a section;
- change linearly across the height of a section;
- equal to zero if y  , that is on the axis z ;
0
- reach the largest values at the points most distant from the
neutral layer.
The geometric locus at which normal stresses in the cross
section of the beam are zero is called the neutral axis of the cross
section.
Substituting the relation (6.9) in the equation (6.6), we obtain
E E
 ydF  S  0.
z
 
F
E
Since  0 , then the cross-section static moment –

S  0 which means that the axis z – a neutral axis is the central
z
section, that is runs through the center of gravity of the section.
Substituting the formula (6.9) in the equation (6.5), we obtain
E E
 zydF  J  0 .
zy
 
F
E
Since  0 , then centrifugal moment of inertia J  ,
0
 zy
which indicates that the axes ,z y should be the main axes of
inertia.
Substituting the relation (6.9) to (6.4), we obtain

E EJ
2
M   x   y dF  z ,
z
 
F
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