Example 2.7: A disc of mass 13.6 kg and the polar mass moment of inertia 0.02 kg-m2, is mounted at the mid-span of a shaft with a span length of 0.4064 m. Assume the shaft to be simply supported at bearings. The rotor is known to have an unbalance of 0.2879 kg-cm. Determine forces exerted on bearings at the spin speed of 6000 rpm. The diameter of the steel shaft is 2.54 cm with E = 200 GNm-2.
Solution: The following data are available:
Bearing forces are obtained by considering firstly, shaft as rigid and then by considering shaft as flexible. In both cases, bearings are considered as rigid in transverse directions.
(i). For the rigid shaft & rigid bearings (Fig. 2.26):
The component of forces in the horizontal & vertical directions are given as, respectively,
(ii) For the flexible shaft and rigid bearings (Method 1) (Fig. 2.27):
Since fy = ky, bearing reaction forces can be written as (Figure 2.27c)
The stiffness of the rotor system as shown in Figure 2.26 is given as
EOM of the disc, from the free body diagram of the disc (Figure 2.27b), is given as
For the simple harmonic motion, we have 
, hence the above equation can be written as
Hence the bearing reaction force can be obtained as
Components of bearing force in the horizontal and vertical directions can be obtained as, respectively,
