2 A shaft and a hole assembly is exprssed as 60 H8n5 fit A E

2. A shaft and a hole assembly is exprssed as 60 H8/n5 fit. A) Establish the limits on the hole dimensions B) Establish the limits on th shaft dimensions C) Esimate the maximum interfernce/clearance D) Esimate the minimum interfernce/clearance E) Classify the fit as clearance, interference or transition.

Solution

solution:

1)here for given kole shaft system by calculation dimensions are as given below

basic size=60 mm

2)here new basic size lies between 50 to 65

hence new basic size is geometric mean of them,

D=(D1*D2)^.5=(50*65)^.5=57 mm

2)here basic tolerance unit is given by

i=.45*(D)^(1/3)+.001*D

i=1.788

3)here hole tolerance is

H8=IT8=25*i=44.72 micron=.04472 mm

4)for shaft tolerance is

n5=IT5=7i=12.516 micron=.012516 mm

5)here hole limit are

Lower limit of hole=LLH=basic size+fundamental deviation=60+0=60 mm

ULH=basic size+F.D.+tolerance=60+0+.04472=60.04472 mm

hole tolerance=ULH-LLH=.04472 mm

6)for given shaft limit

lower deviation is

ei=5*D^.34=19.7682 micron=.01976 mm

upper deviation=es=ei+IT5=19.7682+12.516=32.2842 micron

here shaft limit are

ULS=basic size+es=60+.03228=60.03228 mm

LLS=basic size+ei=60+.01976=60.01976 mm

shaft tolerance=ULS-LLS=.01252 mm

5)here maximum and minimum interference

max interference=ULH-LLS=60.04472-60.01976=.02496 mm

min. interference=ULH-ULS=.01219 mm

6)here maximum and minimum clearance

max clearance=ULS-LLH=.03228 mm

min .clearance=LLS-LLH=.01976 mm

7)as here LLH<LLS,hence here interference fit exist between shaft and hole