Page 272 - Machinery Component Maintenance

254 Machinery Component Maintenance and Repair

principal inertia axis from the bearing axis ( and the eccentricity e of CG)
in the rotor is therefore:

20 oz in.
e= = 0.01 in.
2000 oz

If the weight distribution is not equal between the two bearings but is,
say, 60 percent on the left bearing and 40 percent on the right bearing,
then the unbalance in the left plane must be divided by 60 percent of the
rotor weight to arrive at the approximate displacement in the left bearing
plane, whereas the unbalance in the right plane must be divided by 40
percent of the rotor weight.
An assumed unbalance of 10 oz - in. in the left plane (close to the bear-
ing) will thus cause an approximate eccentricity in the left bearing of

10 oz in.
e= = 0.00833 in.
2000 oz 0.6

and in the right bearing of

10 oz in.
e= = 0.0125 in.
2000 oz 0.4

Quite often the reverse calculation is of interest. In other words, the
unbalance is to be computed that results from a known displacement.
Again the assumption is madc that the resulting unbalance is static.
For example, assume an armature and fan assembly weighing 2000 lbs
and having a bearing load distribution of 70 percent at the armature (left)
end and 30 percent at the fan end. See Figure 6-13. Assume further that
the assembly has been balanced on its journals and that the rolling ele-
ment bearings added afterwards have a total indicated runout of 0.001
in., causing an eccentricity of the shaft axis of 112 of the TIR or 0.0005 in.

Question: How much unbalance does the bearing runout cause in each
side of the rotor?
Answer: In the armature end

U = 1400 lb * 16 oz/lb * 0.0005 in. = 11.2 oz in.
In the fan end

U = 600 lb 16 ozllb 0.0005 in. = 4.8 oz in.