Page 85 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Field Balancing

1. The sum of the forces from the correction weight(s) and the
unbalance, must equal zero.

2. The sum of the moments about the unbalance must equal
zero.

Example 5-3
A rotating element has a 10-pound unbalance. A 5-pound
weight is placed 5 feet to the right of the unbalance, 180 degrees
opposite it. Two additional weights are added to the left of the
unbalance, also 180 degrees opposite the unbalance, a 2-pound
weight at 8 feet and a 3-pound weight at 3 feet. Show that the
corrections made will balance the shaft.

▲ 10#

8’ 3’ 5’

▼
2# ▼
3#
▼
5#

Figure 5-11. Example 5-3

Step 1. The two basic criteria must be met for the shaft to be in
balance. Since the correction weights are all placed 180 degrees
opposite the unbalance, the sum of their weights must equal the
unbalance to satisfy the first criteria. 2 + 3 + 5 = 10, thus the
amount of added weight is correct.

Step 2. The second criteria states the sum of the moments must
equal zero. Selecting the point of unbalance as the reference point,
the unbalance has no moment, since the distance is zero. Assum-
ing measurements to the right of the reference are positive and
measurements to the left are negative, the moments are 2 × –8 =
–16, 3 × –3 = –9, and 5 × 5 = 25. Adding these moments yields –
16 – 9 + 25 = 0, and thus the second criteria is also met.