Materials Science and Manufacturing Engineering Miscellaneous Easy Questions and Answers | Page - 96

Materials Science and Manufacturing Engineering Miscellaneous

A cylindrical pin of 25^{+0.010^+0.020} mm diameter is electroplated, plating thickness is 2.0^±0.005mm. Neglecting the gauge tolerance, the diameter (in mm, up to 3 decimal points accuracy) of the GO ring gauge to inspect the plated pin is ____.

1. 9.030 mm.
1. 2.030 mm.
1. 29.030 mm.
1. 11.030 mm.

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D = 25.02 + 2 × 2.005 = 29.030 mm.

Correct Option: C

D = 25.02 + 2 × 2.005 = 29.030 mm.

Which one of the following statements is TRUE?

1. The 'GO' gauge controls the upper limit of a hole
1. The 'NO GO' gauge controls the lower limit of a shaft
1. The 'GO' gauge controls the lower limit of a hole
1. The 'NO GO' gauge control the lower limit of a hole

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Flatness of machine bed can be measured from Auto collimator.

Correct Option: C

Flatness of machine bed can be measured from Auto collimator.

In the assembly shown below, the part dimensions are:
L₁ = 22.0^±0.01 mm
L₂ = L₃ = 10.0^±0.005 mm
Assuming the normal distribution of part dimensions, the dimension L₄ in mm for assembly condition would be:

1. 2.0^±0.008
1. 2.0^{± 0.012}
1. 2.0^±0.016
1. 2.0^±0.020

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Since all dimensions have bilateral tolerances
L₄ = L₁ – L₂ – L₃
= 22 – 10 – 10
L₄ = 2 mm
Tolerance = 0.01 + 0.0015 + 0.0005 = 0.012
&there; L₄ = 2.0^±0.02mm T
olerance is calculated assuming L₄ to be sink and tolerance of sink will be cumulative sum of all tolerances

Correct Option: D

Since all dimensions have bilateral tolerances
L₄ = L₁ – L₂ – L₃
= 22 – 10 – 10
L₄ = 2 mm
Tolerance = 0.01 + 0.0015 + 0.0005 = 0.012
&there; L₄ = 2.0^±0.02mm T
olerance is calculated assuming L₄ to be sink and tolerance of sink will be cumulative sum of all tolerances

Holes of diameter 25^{+0.020^+0.040} mm are assembled interchangeably with the pins of diameter 25^{−0.008^+0.005} mm. The minimum clearance in the assembly will be

1. 0.048 mm
1. 0.015 mm
1. 0.005 mm
1. 0.008 mm

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Minimum clearance ⇒ minimum hole – maximum shaft
= 25 + .020 – 25 + .005
= 0.015 mm

Correct Option: B

Minimum clearance ⇒ minimum hole – maximum shaft
= 25 + .020 – 25 + .005
= 0.015 mm

A Go-NO Go plug gauge is to be designed for measuring a hole of nominal diameter 25 mm with a hole tolerance of ± 0.015 mm. considering 10% of work tolerance to be the gauge tolerance and no wear condition, the dimension (in mm) of the GO plug gauge as per the unilateral tolerance system is

1. 24.985^{−0.003^+0.003}
1. 25.015^{−0.006^+0.000}
1. 24.985^{−0.03^+0.03}
1. 24.985^{−0.003^+0.000}

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25 ± 0.015
GO-Gauge
U.L = 24.985
L.L = 24.985
24.985^{+0.003^+0.000}

Correct Option: D

25 ± 0.015
GO-Gauge
U.L = 24.985
L.L = 24.985
24.985^{+0.003^+0.000}