In the assembly shown below, the part dimensions are: L1

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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
2. 2.0^{± 0.012}
3. 2.0^±0.016
4. 2.0^±0.020

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