A bar of 20 mm dia is tested in tension. It is observed that when a load of 38 kN is applied the extension measured over a gauge length of 200 mm is 0.12 mm and
contraction in diameter is 0.0036 mm. Find the Poisson's ratio
0.2
0.3
0.25
0.33
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0
Answer:
0.33 please brainliest me
Answered by
0
Answer:
0.3
Explanation:
linear strain=dL/L=0.12/200=6×10^-4mm
lateral strain=dD/D=0.0036/20=1.8×10^-4mm
Poisson's ratio=lateral strain/linear strain
=1.8×10^-4/6×10^-4
=0.3
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