Page 161 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Advanced Machine Alignment

The moment of inertia (I) for a solid shaft is:

4
I = πd /64 (8.8)

If the modulus of elasticity (E) for steel is assumed to be 30
6
× 10 , and

I = 63.62. The total deflection at the center is dx =
6
3
5(1131)(144) /384(30 × 10 )(63.62) = .02304 inches.
Step 2. Since the catenary for shafts deflecting due to their own
weight is very flat, the tangent at the bearing location is closely
approximated by a simple triangle. That is, the curvature of the
catenary approaches a straight line from its center to the bearing
location. This slight error will have no effect on the final answer.
The slope of the curve at the bearing is dy/dx or .02304/72 =
.00032. Since the tangent is also defined as the slope at any point,
the arctangent function provides the angle the face of the shaft
makes with the vertical plane. Thus arctangent of .00032 is .0183
degrees.

Step 3. The offset in the face reading due to the deflection of the
shaft is then expressed as dF = d sin(A). Since the diameter (d) is
6 inches, dF = 6 × sin(.0183) = .00192 inches or 1.9 mils. This would
be rounded up to 2 mil offset for the face reading. From this ex-

▼
A
▼

Figure 8-28. The Angle of the Face Due to Deflection of a Shaft