Answer
$E= 99.8 GPa $
Work Step by Step
Given:
cylindrical specimen with diameter of 10.0 mm
tensile force of 1500 N that produces an elastic reduction in diameter of $6.7 \times 10^{-4} mm$
Required:
elastic modulus if Poisson's ratio is 0.35
Solution:
Using a combination of Equations 6.1 and 6.5:
$E = \frac{σ}{ε_{z}} = \frac{F}{A_{0}ε_{z}} = \frac{F}{ε_{z}π(\frac{d_{0}}{2})^{2}} = \frac{4F}{ε_{z}πd_{0}^{2}}$
Using Equation 6.8 and considering that $ε_{x} = \frac{Δd}{d_{0}}$, we find:
$ε_{z} = - \frac{ε_{x}}{v} = - \frac{Δd}{d_{0}v}$
Substituting this expression for $ε_{z}$:
$E= \frac{4F}{ε_{z}πd_{0}^{2}} = \frac{4Fv}{πd_{0}Δd} = \frac{(4)(1500 N)(0.35)}{(π)(10 \times 10^{-3} m)(6.7 \times 10^{-7} m)} = 9.98 \times 10^{10} Pa = 99.8 GPa $
