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M
  x     , (5.19)
GJ

where    – allowable angle of twist per unit length, which is
selected for the different designs and different types of loads from
a range 0,15 2 of 1m length.
In the design calculating from the condition (5.19), we find the
rod diameter:
- for a circular cross-section
32M
D  4 x ; (5.20)
G   
- for a ring cross-section

32M
D  4 x . (5.21)
G  1 c  4 

In the formulas (5.19) and (5.20) the value    should be
substituted in rad / m.
From the two rod diameters derived from the strength
conditions (5.10) and the condition of hardness (5.19), we take a
large value.

5.6 Potential energy of deformation

We will show the relationship between the twisting moment and
the angle of twist of a rod graphically (fig. 5.7). Within the
Hooke’s law this dependence is represented by a straight line OA.
Under the static load the twisting moment increases from zero to
some value M . Accordingly, the angle of twist increases from
x
zero to some value  .

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