Solution:
For the cylindrical specimen
Diameter, d = 15 mm
Length, L = 214 mm
Change in diameter, d = 0.00788 mm
Poission's ratio, = 0.35
Lateral strain,
ela = (d/d) = (0.00788 mm) /(15 mm) = 0.0005253333
longitudinal strain, elo = ela/
elo = 0.0005253333/0.35
elo = 0.001501
At elongation strain elo =0.001501
Stress = 128 MPa
Therefore, required force, F
= (128)d2/4
= (128)(15)2/4
= 22619.5 N
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