Page 83 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Field Balancing

(30). Solving the trigonometric functions yields 9.65925 = B + .5 C,
and 2.58819 = .866025 C.
Now the two equations can be simultaneously solved by
substituting the value of C from the first equation into the second
equation. This yields C = 2.98858 and B = 9.65925 – (.5 × 2.98858)
= 8.16495.
Thus by placing a 3-ounce weight at 30 degrees and an 8-
ounce weight at 90 degrees will produce the same result as the 10-
ounce weight at 75 degrees.
Note that although the B plane was @ 90 degrees, allowing a
simple solution for C, the same results can be achieved regardless
of the angles involved. There will always be two equations with
two unknowns; therefore, a solution is easily achieved.

TWO PLANE BALANCING

The majority of unbalance situations are not pure static in
nature; thus, utilization of single plane balancing is ineffective. In
general, the unbalance must be either compensated for in the
same plane in which it occurs, or resolved into two or more cor-
rective planes to prevent the formation of couple forces.
In Figure 5-8 a correction weight equal to the unbalance is
added in the wrong plane. The result will be a couple.

Unbalance
Center of
rotation

Center of
the shaft
Correction
weight

Figure 5-8. Locating the Correction Weight in a Different Plane than
the Unbalance