Respuesta :
Explanation:
The cross-sectional area of the specimen is calculated as follows.
[tex]A_{o} = \frac{pi}{4} d^{2}[/tex]
= [tex]\frac{3.14}{4} \times (\frac{4.4}{1000})^{2}[/tex]
= [tex]1.5197 \times 10^{-5} m^{2}[/tex]
Equation of stress is as follows.
[tex]\sigma = \frac{F}{A_{o}}[/tex]
And, the equation of strain is as follows.
[tex]\epsilon = \frac{\Delta l}{l_{o}}[/tex]
Hence, the Hook's law is as follows.
E = [tex]\frac{\sigma}{\epsilon}[/tex]
E = [tex]\frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}[/tex]
= [tex]\frac{F \times l_{o}}{A_{o} \times \Delta l}[/tex]
or, [tex]l_{o} = \frac{E \times \Delta l \times A_{o}}{F}[/tex]
= [tex]\frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}[/tex]
= 0.6554 m
or, [tex]l_{o}[/tex] = 655.4 mm
Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.
