Answer
$l_{0} = 105.69\ mm$
Work Step by Step
Given:
cylindrical specimen of metal alloy stressed in compression
original diameter- 30.00 mm
final diameter- 30.04 mm
final length - 105.20 mm
elastic modulus- 65.5 GPa
shear modulus - 25.4 GPa
Required:
original length
Solution:
Computing the lateral strain $ε_{x}$, we find:
$ε_{x} = \frac{Δd}{d_{0}} = \frac{(30.04 mm - 30.00 mm)}{30.00 mm} = 1.33 \times 10^{-3}$
Using Equation 6.9, compute Poisson's ratio:
$v = \frac{E}{2G} - 1= \frac{65.5 \times 10^{3} MPa}{(2)(25.4 \times 10^{3} MPa)} - 1= 0.289$
Using Equation 6.8:
$ε_{z} = -\frac{ε_{x}}{v} = -\frac{1.33 \times 10^{-3}}{0.289} = -4.60 \times 10^{-3}$
Using Equation 6.2 to solve $l_{0}$:
$l_{0} = \frac{l_{f}}{1 + ε_{z}} = \frac{105.20 mm}{1 - 4.60 \times 10^{-3} } = 105.69 mm$
