Respuesta :
Explanation:
The given data is as follows.
Cross-section area of the rectangular is as follows.
Area = [tex]10 mm \times 12.7 mm[/tex]
= [tex]127 \times 10^{-6} m^{2}[/tex]
= [tex]1.27 \times 10^{-4} m^{2}[/tex]
Tension applied on the specimen is 35,500 N
Now, formula to calculate the modulus of elasticity for a material of aluminium is as follows.
E = 70 GPa
= [tex]70 \times 10^{9} Pa[/tex]
or, = [tex]70 \times 10^{9} N/m^{2}[/tex]
Now, stress on the specimen is as follows.
[tex]\sigma = \frac{P}{A}[/tex]
= [tex]\frac{35000}{1.27 \times 10^{-4}}[/tex]
= [tex]275.59 \times 10^{6} N/m^{2}[/tex]
According to Hook's law, we will calculate the resulting stain as follows.
[tex]\sigma = E \epsilon[/tex]
[tex]\epsilon = \frac{\sigma}{E}[/tex]
= [tex]\frac{275.59 \times 10^{6}}{70 \times 10^{9}}[/tex]
= [tex]3.937 \times 10^{-3} mm/mm[/tex]
Thus, we can conclude that the resulting strain is [tex]3.937 \times 10^{-3} mm/mm[/tex].
