Page 61 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Vibration Due to Unbalance

(.00005") from its balanced position to its field assembly position.
Thus the inch-ounces of unbalance was 320 × .00005 = .016. Using
2
formula (4.1) yields F = 1.774 × (23200/1000) × .016 = 15.2 pounds
of force. That is over 75% of its weight!
Although most of the equipment does not rotate at that high
an rpm, much of the more common electric motor driven equip-
ment does operate at 3,450 or 1,750 rpm.

Example 4.2
Steel weighs approximately 485 pounds per cubic foot or
about 4.49 ounces per cubic inch. If an electric motor had a 3-inch
shaft and rotated at 3,450 rpm, and had a 1/2" × 1/2" × 3" keyway
that was not filled with key material, how much force would be
generated?
3,450 RPM

Missing Key
Material

3” Dia.

Figure 4-8. Example of Missing Key Material
Step 1. The missing key material is .75 cubic inches which weighs
approximately 3.37 ounces. The center of mass for the key material
would be 1-1/4 inches from the motor centerline. Thus 4.21 inch-
ounces of unbalance would be present. Using the formula below
to determine the force,
2
F = 1.774 × (3450/1000) × 4.21 = 88.89 pounds of force.
Although most machines’ bearings can withstand loads of up
to eight times the normal static load, for a short period, it can be
easily seen that the life of these parts will be greatly reduced.