Page 216 - Machinery Component Maintenance

198 Machinery Component Maintenance and Repair

too far south, as are the feet. Therefore, net correction will be:

Move outboard feet .lo5 in. + ,017 in. = .122 in. north
Move inboard feet .017 in. north
It can be seen that our answers agree closely with those on the data
sheet, which were obtained graphically, The differences are not large
enough to cause us trouble in the actual field alignment correction.
Alternatively, this first example could be solved with another similar
“formula method.” To begin with, we draw the machine sketch, Figure
5-17. Then, we proceed by jotting down the relevant formulas:

Net Parallel
Correction at IB = f Offset of Shaft
Centerlines at
Plane A

Net Parallel
Correction at OB =
Plane A

For our example, the solutions would be as follows:

Vertical
At IB, -(.007) (!) - (F) - .0055 = -.0212

-.0157
=
say lower IB .021 in.

1\1

MACHINE TO
BE ADJUSTED
STATIONARY
MACH I NE

Figure 5-17. Machine sketch for face-and-rim alignment method.