General Engineering · Applied Mechanics · 2 marks · Numerical answer
A homogeneous isotropic linear-elastic cylindrical component of diameter 10 mm is under uniform uniaxial tension. A 14 kN force reduces its diameter uniformly by $3\times10^{-3}$ mm. With E = 150 GPa and π = 3.14, find Poisson’s ratio (two decimals).
Area is $3.14(10)^2/4=78.5$ mm², so axial stress is $14000/78.5=178.344$ MPa. Axial strain is $178.344/150000=0.00118896$; transverse strain is $-0.003/10=-0.0003$. Hence $\nu=-\epsilon_t/\epsilon_a=0.25232$, rounding to **0.25**.
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