Page 301 - Machinery Component Maintenance

Balancing of Machinery Components 283

Example:

W = 1OOOgrams
e = V2 TIR (mounting surface to shaft axis), say 1/2 of 0.004 in.
= 0.002 in. + 1/2 the total clearance between in. rotor and arbor,
say 1/2 of 0.002 = 0.001 in. = 0.003 in.

1. U, = lOOOg - 0.003 in. = 3 gram - inches
To this may have to be added:
2. Unbalance caused by eccentricity and thread clearance of the
clamping nut, assume:
W = 100 grams
e = 0.001 in.
U2 = l00g 0.001 in. = 0.1 g in.
(For simplification, the residual unbalance of the nut is ig-
nored)

3. Residual unbalance Us of the arbor, assume 0.1 g * in.
4. Eccentricity plus V2 fit clearance in mounting surfaces of the final
rotor installation. Assuming that similar tolerances prevail as were
used in making the arbor, the same unbalance will result, or:

U4 = UI = 3 gmin.

Total unbalance caused by arbor eccentricity and fit clearance UI , nut ec-
centricity U2, arbor residual unbalance, and installation error therefore
may add up to a maximum of

u, + uz + u3 + u4 = 3 + 0.1 + 0.1 + 3
= 6.2 (g in.)

Statistical Evaluation of Errors

One can readily see that if the rotor balance tolerance is, say, 10
g in., 62 percent of it (6.2 g * in.) is already used up by tooling and
mounting errors. Thus, the balancing machine operator is forced to bal-
ance each part to 10 - 6.2 = 3.8 (g in.) or better, to be sure that the
maximum permissible residual unbalance of 10 g in. will be attained in
the final assembly. This may be rather time consuming and, therefore,
costly. To allow a larger working tolerance, the various tooling errors
could be reduced by a more precisely machined arbor and final shaft.
However, this too may be costly or impractical.