Respuesta :
Answer:
The elastic modulus is 99.7 GPa
Solution:
As per the question:
Diameter of the metal alloy, D = 10.0 mm = 0.01 m
Tensile force, F = 1500 N
Elastic reduction in the diameter of the metal alloy, [tex]\Delat D = 6.7\times 10^{- 4} mm = 6.7\times 10^{- 7} m[/tex]
Poisson's ratio, [tex]\mu = 0.35[/tex]
Now,
Lateral strain is given by:
[tex]\varepsilon_{D} = \frac{\Delta D}{D} = \frac{6.7\times 10^{- 4}}{10} = 6.7\times 10^{- 5}[/tex]
Longitudinal strain, [tex]\varepsilon_{l}[/tex]:
[tex]\varepsilon_{l} = \frac{\varepsilon_{D}}{\mu} = \frac{6.7\times 10^{- 5}}{0.35} = 1.914\times 10^{- 4}[/tex]
Now, stress is given by:
[tex]\sigma = \frac{F}{Area,\ A}[/tex]
where
[tex]A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.01}{2})^{2} = 7.854\times 10^{- 5} m^{2}[/tex]
Now,
[tex]\sigma = \frac{1500}{7.854\times 10^{- 5}} = 19.098\times 10^{6} = 19.098 MPa[/tex]
Now, the elastic modulus is given by:
[tex]E = \frac{\sigma}{\varepsilon_{l}}[/tex]
[tex]E = \frac{19.098\times 10^{6}}{1.914\times 10^{- 4}} = 99.784\times 10^{9}Pa = 99.7 GPa[/tex]
