Modeling of a Shaft of a Non-uniform Diameter into a Shaft of Uniform Diameter:

A shaft connecting various rotors require not to be of uniform diameter but it might have segments of the various diameters along its length. One shaft of this kind is illustrated in Figure 15(i). In this shaft there are three segments however generally there might be any number of segments. The shaft of non-uniform diameter illustrated in Figure 15(i) is to be modeled into a shaft of consistent diameter illustrated in Figure 15 (ii). Let these two shafts be subjected to the similar amount of torque 'T'. The twist generated in the shaft of consistent diameter'd' is specified by

 θ= (T × l) /( J × G)                                -----------------(48)

Here J is polar moment of inertia; G is modulus of rigidity of the material.

                               (i)                                                                             (ii)

The net twist of non-uniform diameter is specified by

θ= θ₁ + θ₂  + θ₃ + .. .                     ------------------(49)

Here θ₁, θ₂, θ₃, ---------, etc. are twists of the segments.

       ----------------- (50)

If the two shafts are corresponding in action, after that, twists given by Equation (48) and Equation (50) must be equal.

If the material of both the shafts is similar, that means

                      G = G₁  = G₂  = G₃

   ----------(51)

If          d = d₁

    -------------- (52)

The diameter of the shaft of consistent diameter might be arbitrarily selected but to decrease calculations work this is better to select any one of the present diameters.