What should be the allowance for press fitting?

[DaveH]

Unal,

This will give you some idea, this considers the Hole is dead to size and the Shaft dia. is altered.

It also assumes the Shaft is softer or equal to the hardness of the material that contains the hole.

From ¼” to 2” dia. Shaft/Hole (Sorry it is in inches)


All these numbers are in Thou’s. ie 0.5 = 0.0005”

For a Force fit the Shaft dia will be + 0.5 + (0.75 thousands of an inch per inch of dia.)

So put another way it is Half a Thou PLUS Three Quarters of a Thou for every inch in dia.


For a 1” hole the shaft will need to be .0005” + (.00075 / inch) = .00125” So the shaft needs to be (near enough) 1.0013”


For a ½” hole the shaft will need to be .0005” + (.00075 / inch) = .0005 + (.00075 x ½”) = .0005” + .000375” = .000875” So the shaft needs to be .5009 (near enough)



This gives you a “run of the mill” force fit for mild steel, brass, phosphor bronze, aluminium.

The finish is important and assumes a “reamer” type finish, to the hole and shaft.

It is also very important to take notice of what Ken has said.

Some times it is much better to “Loctite” the parts together, doesn’t stress the parts.


DaveH