Question

A shaft of 80 mm dia is made to rotate in a bush. The tolerances for both the shaft and bush are 0.06 mm. Give a maximum clearance of 0.13 mm.
The limit size of the shaft with the hole based system is

• A. $80\begin{array}{c}+0.06\\ +0.00\end{array}$
• B. $80\pm 0.03$
• C. $80\begin{array}{c}+0.01\\ -0.07\end{array}$
• D. $80\begin{array}{c}-0.01\\ -0.07\end{array}$

Answer 1

The correct option is D $80\begin{array}{c}-0.01\\ -0.07\end{array}$

U.L of hole - L.L of hole = 0.06 mm

U.L of shaft - L.L of shaft = 0.06 mm

Max clearance = U.L of hole - L.L of shaft

= 0.13 mm

As it is a hole based system,

Hole size will be $80\begin{array}{c}+0.06\\ +0.00\end{array}$

$\therefore$U.L of hole - 0.13 mm = L.L of shaft

80.06 - 0.13 mm = L.L of shaft

L.L of shaft = 79.93 mm

$\therefore$U.L of shaft = L.L of shaft +0.06mm

=79.99 mm

Allowance of assembly

= L.L of hole - U.L of shaft

= 80.00- 79.99 = 0.01 mm