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A 30 mm hole shaft assembly results in minimum and maximum clearances of 0.03 mm and 0.30 mm respectively. The hole has a unilateral tolerance with zero fundamental deviation. If the tolerance on the shaft is 0.08 mm, the maximum size of the hole is
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- 230.11
- 30.16
- 30.19
- 30.27
Correct Option: C
Min. clearance
= Holemin – Shaftmax = 0.03 .......(i)
Max clearance
= Holemin – Shaftmax = 0.30 ............(ii)
Shaftmax – Shaftmin = 0.08 ...............(iii)
Since, hole has unilateral tolerance,
Hmin = 30 mm
From eq. (i), we get
Shaftmax = 30 – 0.03 = 29.97 min
From eq. (iii), we get
Shaftmin = 29.89 mm
From eq. (ii), we get
Holemax = 30.19 mm