Page 269 - Machinery Component Maintenance

Balancing oj Machinery Components 251

Weight of disc = 999 02
Unbalance mass m = 1 02
Total rotor weight W - 1000 oz
Journal Axis
Displacement
Center of of C G
Gravity eccentricity")

Unbalance U - m - r
- 1 oz 10 in
s 10 oz-in

Figure 6-10. Disc-shaped rotor with displaced center of gravity due to unbalance.

Assume a perfectly balanced disc, as shown in Figure 6-10, rotating
about its shaft axis and weighing 999 ounces. An unbalance mass m of
one ounce is added at a ten in. radius, bringing the total rotor weight W
up to lo00 ounces and introducing an unbalance equivalent to 10 ounce
in. This unbalance causes the CG of the disc to be displaced by a distance
e in the direction of the unbalance mass.
Since the entire mass of the disc can be thought to be concentrated in its
center-of-gravity, it (the CG) now revolves at a distance e about the shaft
axis, constituting an unbalance of U = We. Substituting into this for-
mula the known values for the rotor weight, we get:

10 oz * in. = 1ooOoz *e

Solving for e we find
10 oz . in.
e= = 0.01 in.
lo00 02

In other words, we can find the displacement e by the following for-
mula:
u (oz in.)
e (in.) =
w (oz)

For example, if a fan is first balanced on a tightly fitting arbor, and
subsequently installed on a shaft having a diameter 0.002 in. smaller than