Question

A Circular shaft having diameter ${65.00}_{-0.05}^{+0.01}mm$ is manufactured by a turning process. A 50 $\mu m$ thick coating of TiN is deposited on the shaft. Allowed variation in TiN film thickness is $\pm 5\mu m$. The minimum hole diameter (in mm) to just provide clearance fit is

• A. 65.01
• B. 65.12
• C. 65.10
• D. 64.95

Answer 1

The correct option is B 65.12

$\text{Shaft :}{65}_{-0.05}^{+0.01}mm$

$\text{Coating thickness =}50\pm 5\mu m$

$\text{Clearance fit}$

This is a question if clearance fit, for just clearance fit Min.C will be zero.

$\text{UL of shaft = LL of hole.}$

After coating, the largest shaft will be product if we start with the largest shaft and add maximum coating.


$t_{max}=55\mu m=0.055mm$

$\text{Largest shaft after plating}$

$=0.055+65.01+0.055mm$

$=65.12mm$