Mechanical Properties of Fluids

Question

If α is the volumetric strain and l is the longitudinal strain, then find the Poisson’s ratio in terms of a and β.

Solution

Let a wire of length l, diameter d be loaded by force F.
Let dl be the change in length and dV the change in volume, and dD be the change in the diameter.

Volume of wire before the force is applied is,
$straight V apostrophe equals πr squared straight t equals πD squared over 4 straight ell$
Volume of wire after the force is applied is,
$straight V apostrophe equals straight pi over 4 open parentheses straight D plus dD close parentheses squared open parentheses straight ell plus straight d straight ell close parentheses$
Change in volume is,
dV = V' - V
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$rightwards double arrow space space space space space space space space dV over straight V equals 2 dD over straight D plus fraction numerator straight d straight ell over denominator straight ell end fraction space rightwards double arrow space space space space space fraction numerator dVIV over denominator straight d straight ell straight I straight ell end fraction equals 2 space space fraction numerator dDID over denominator straight d straight ell straight I straight ell end fraction plus 1$
$rightwards double arrow space space straight alpha over straight beta equals negative 2 straight sigma plus 1 space space space space space space space space space space space space space space space space space space space space space space open square brackets because space straight sigma equals space space fraction numerator dDID over denominator straight d straight ell straight I straight ell end fraction close square brackets rightwards double arrow space space space straight sigma equals 1 half open parentheses 1 minus straight alpha over straight beta close parentheses$