Voltage drop testing is being discussed.
1 answer:
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This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
= - ( εˣ / v )
= - ( -0.00088 / 0.34 )
= - ( - 0.002588 )
= 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
