A cylindrical specimen of copper having an original diameter of 15.2mm is pulled in tension with 9,500 N force, producing only elastic deformation.what is the resulting engineering stress in the specimen?
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The stress in the specimen will be 52.4 x 10⁶
Explanation :
Given diameter of the cylinder = 15.2 mm
=> Radius of the cylinder = 15.2/2 = 7.6 mm = 7.6 x 10⁻³ m
Area of the cylinder = πr² = 3.14 x (7.6 x 10⁻³)²
=> A= 181.3 x 10⁻⁶ m²
Given force applied, F = 9500 N
we know that stress is given as the ratio of force and area,
S = F/A
= 9500/(181.3 x 10⁻⁶)
= 52.4 x 10⁶ N/m²
Hence the stress in the specimen will be 52.4 x 10⁶
I hope this answer helps you, please feel free to ask any doubt regarding this in the comment section.
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