Page 270 - Machinery Component Maintenance

252 Machinery Component Maintenance and Repair

the arbor, the total play resulting from the loose fit may be taken up in
one direction by a set screw. Thus the entire fan is displaced by one half
of the play or 0.001 in. from the axis about which it was originally bal-
anced. If we assume that the fan weighs 100 pounds, the resulting unbal-
ance will be:

U = 100 lb 16 oz/lb 0.001 in. = 1.6 oz * in.

The same balance error would result if arbor and shaft had the same
diameter, but the arbor (or the shaft) had a total indicated runout (TIR) of
0.002 in. In other words, the displacement is always only one half of the
total play or TIR.
The CG displacement e discussed above equals the shaft displacement
only if there is no influence from other sources, a case seldom encoun-
tered. Nevertheless, for balancing purposes, the theoretical shaft respec-
tively CG displacement is used as a guiding parameter.
On rotors having a greater length than a disc, the formula e = U/W
for finding the correlation between unbalance and displacement still
holds true if the unbalance happens to be static only. However, if the un-
balance is anything other than static, a somewhat more complicated situa-
tion arises.
Assume a balanced roll weighing 2000 oz, as shown in Figure 6-1 1,
having an unbalance mass m of 1 oz near one end at a radius r of 10 in.
Under these conditions the displacement of the center-of-gravity (e) no
longer equals the displacement of the shaft axis (d) in the plane of the
bearing. Since shaft displacement at the journals is usually of primary

m 1 02 (total weight Including
unbalance mass)

L-
1

Figure 6-1 1. Roll with unbalance.