renasis
Mechanical
- Dec 29, 2002
- 56
Does anybody know how the recommended holes sizes for screws were established in the ASME B18.3-1986 & B18.2.8 tables? For example, the following recommendations are given for a 1/4-20 screw.
ASME B18.3 - Close Fit .266" (17/64 drill)
ASME B18.3 - Normal Fit .281" (9/32 drill)
ASME B18.2.8 - Close Fit .266"/.272" (17/64 drill)
ASME B18.2.8 - Normal Fit .281"/.290" (9/32 drill)
ASME B18.2.8 - Loose Fit .297"/.311" (19/64 drill)
Military standard AND10387 gives a +.006/-.001 tolerance on drilled holes between .251-.500. Given this, each of the hole sizes would give a min position clearance of the following:
Min hole - Max screw = Possible position error
ASME B18.3 - Close Fit .265-.250 =.015"
ASME B18.3 - Normal Fit .280-.250 =.030"
ASME B18.2.8 - Close Fit .265-.250 =.015"
ASME B18.2.8 - Normal Fit .280-.250 =.030"
ASME B18.2.8 - Loose Fit .296-.250 =.046"
In general, most clearance and threaded holes have ±.005 tolerances in the x and y. Position error for each hole would be .014", for both would be .028". So where does the .030" come from on a normal fit?
I obtained these values from references to the standards. I don't own them. Maybe there is a note in them about how clearance was determined for each of the fits?
ASME B18.3 - Close Fit .266" (17/64 drill)
ASME B18.3 - Normal Fit .281" (9/32 drill)
ASME B18.2.8 - Close Fit .266"/.272" (17/64 drill)
ASME B18.2.8 - Normal Fit .281"/.290" (9/32 drill)
ASME B18.2.8 - Loose Fit .297"/.311" (19/64 drill)
Military standard AND10387 gives a +.006/-.001 tolerance on drilled holes between .251-.500. Given this, each of the hole sizes would give a min position clearance of the following:
Min hole - Max screw = Possible position error
ASME B18.3 - Close Fit .265-.250 =.015"
ASME B18.3 - Normal Fit .280-.250 =.030"
ASME B18.2.8 - Close Fit .265-.250 =.015"
ASME B18.2.8 - Normal Fit .280-.250 =.030"
ASME B18.2.8 - Loose Fit .296-.250 =.046"
In general, most clearance and threaded holes have ±.005 tolerances in the x and y. Position error for each hole would be .014", for both would be .028". So where does the .030" come from on a normal fit?
I obtained these values from references to the standards. I don't own them. Maybe there is a note in them about how clearance was determined for each of the fits?