Solution Set 1
Class Allowance Hole Tol Shaft Tol
1 0.0025*cubroot(d^2) 0.0025* cubroot(d) -0.0025*cubroot(d)
2 0.0014* cubroot(d^2) 0.0013* cubroot(d) -0.0013* cubroot(d)
3 0.0009* cubroot(d^2) 0.0008* cubroot(d) -0.0008* cubroot(d)
4 0 0.0006* cubroot(d) -0.0004* cubroot(d)
5 0 0.0006* cubroot(d) 0.0004* cubroot(d)
6 -0.00025*d 0.0006* cubroot(d) 0.0004* cubroot(d)
7 -0.0005*d 0.0006* cubroot(d) 0.0004* cubroot(d)
8 -.001*d 0.0006* cubroot(d) 0.0004* cubroot(d)
d โก Nominal Size
Problem 1: Design and develop a tolerance plot of a fitting pair using the Hole Basic system.
The required fit is a Medium Fit with a nominal size of 1.125 inch (Class 3).
MMChole = Nom. Size = 1.125โ
LMChole = 1.125 + 0.0008*cubroot(1.125) = 1.125 + 0.0008*1.040 = 1.125 + .00083 = 1.1258โ
Allowance: 0.0009*cubroot(d^2) = 0.0009*1.0817 = .00097โ
MMCShaft = MMChole โ Allowance = 1.125 โ 0.00097 = 1.1240โ
LMCshaft = MMCs โ Tolerances = 1.1240 - .0008*cubroot(1.125) = 1.1240 - .00083 = 1.1232โ
Problem 2: Design and develop a tolerance plot of a fitting pair using the Shaft Basic system.
The required fit is a Free Fit with a nominal size of 0.75โ (Class 2).
MMCs = Nom. Size = 0.750โ
LMCs = 0.750 โ 0.0013*cubroot(.750) = .750 - .00118 = .7488โ
Allowance: 0.0014*cubroot(d^2) = 0.0014 * .8255 = .00116โ
MMCh = 0.750 + Allow. = 0.750 + .00116 = 0.7512โ
LMCh = MMCh + tolh = .7512 + .00118 = 0.7523โ
Problem 3: Design and develop a tolerance plot of a fitting pair using the Hole Basic system.
The required fit is a Medium Force Fit with a nominal size of 1.875โ (Class 7).
MMCh = Nom. Size = 1.875โ
LMCh = MMCh + 0.0006*cubroot(1.875) = 1.875 + 0.00074 = 1.8757โ
Interference = 0.0005*1.875 (neg. allowance!) = .00094
MMCs = MMCh + Interference + Tols = 1.875 + 0.00094 + 0.0004*cubroot(1.875) =
= 1.875 + 0.00094 + 0.00049 = 1.8764โ
LMCs = MMCh + Interference = 1.875 + 0.00094 = 1.8759โ
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