Question

A hole shaft pair has been designated as 64 mm \(H_8f_7\).The diameter steps are is 50 mm and 80 mm. For shaft designation f, upper deviation is assumed as \(-5.5 D^{0.41}\)

\(\begin{array}{|l|l|l|}\hline
\text{For}&\text{Tolerance}\\\hline
H_8&25 i\\\hline
f_7& 16 i\\\hline
\end{array}\)

The allowance for the above fit is

Answer 1

64 mm diameter lies in the diameter step of 50-80 mm. Therefore,

Geometric mean diameter,

\(D =\sqrt{D_{min}\times D_{max}}\)

\(=\sqrt{50\times 80}\)

= 63.246 mm

Fundamental tolerance unit(i)

\( =(045 D^{\dfrac{1}{3}} +0.001 D)\mu m\)

\( =[0.45(63.246)^{\dfrac{1}{3}} +0.001(63.24
6)]\)

\(=1.859 \mu m\)

= 0.00186 mm

\(25 i = 25\times 0.00186 = 0.04646~ mm\)

Fundamental deviation of 'f' shaft

\( =-5.5 D^{0.41}\)

\( =-5.5[63.246]^{0.41}\)

= -0.30115 mm

\(16 i = 16\times0.00186 = 0.029\)

\(\text{Shaft size is} ~~~ 64{\begin{matrix}
-0.030\\
-0.060
\end{matrix}}\)

\(\text{Hole size is} ~~~ 64{\begin{matrix}
+0.046\\
0.000
\end{matrix}}\)

Allowance = 64.000 -63.970

\( = 0.030 mm = 30\mu m\)