Answer
$v = 0.367$
Work Step by Step
Given:
cylindrical metal alloy 10 mm in diameter deformed in tension
force of 15,000 N reduces specimen diameter by $7 \times 10^{-3} mm$
Required:
Poisson's ratio if elastic modulus is 100 GPa
Solution:
Using a combination of Equations 6.1 and 6.5, it follows:
$ε_{z} = \frac{σ}{E} = \frac{F}{A_{0}E} = \frac{F}{π(\frac{d_{0}}{2})^{2}E} = \frac{4F}{πd_{0}^{2} E}$
Considering that traverse strain ($ε_{x} =\frac{Δd}{d_{0}}$) and Poisson's ratio can be computed using Equation 6.8, we find:
$v = \frac{ε_{x}}{ε_{z}} = -\frac{\frac{Δd}{d_{0}}}{(\frac{4F}{πd_{0}{2}E})} = -\frac{d_{0}ΔdπE}{4F} = -\frac{(10 \times 10^{-3} m)(-7 \times 10^{-6} m)(π)(100 \times 10^{9} N/m^{2})}{(4)(15,000 N)} = 0.367$
