Question

In the assembly shown below, the part dimensions are :


${\text{L}}_1={22.0}^{\pm 0.01}\text{mm},{\text{L}}_2={\text{L}}_3={10.0}^{\pm 0.005}\text{mm}$,
Assuming the normal distributions of part dimensions, the dimension ${\text{L}}_4$ in mm for assembly condition would be,
Diagram 1

• A. ${2.0}^{\pm 0.008}$
  
• B. ${2.0}^{\pm 0.020}$
  
• C. ${2.0}^{\pm 0.016}$
  
• D. ${2.0}^{\pm 0.012}$
  

Answer 1

The correct option is B ${2.0}^{\pm 0.020}$


${\text{L}}_{4\text{max}}={\text{L}}_{1\text{max}}-({\text{L}}_{2\text{min}}+{\text{L}}_{3\text{min}})$

$=22.01-(9.995+9.995)$

$=2.02$

${\text{L}}_{4\text{min}}={\text{L}}_{1\text{min}}-({\text{L}}_{2\text{max}}+{\text{L}}_{3\text{max}})$

$=21.99-(10.005+10.005)$

$=1.98$

${\text{L}}_4=2^{\pm 0.02}$