Answer:
The maximum length of the specimen before the deformation was 358 mm or 0.358 m.
Explanation:
The specific deformation ε for the material is:
[tex] \epsilon = \deltaL /L[/tex] (1)
Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:
[tex] \sigma = E/ \epsilon[/tex] (2)
Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:
[tex] L= \delta L/ \epsilon[/tex]
[tex] \epsilon =\sigma /E[/tex] Â
So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:
[tex] \sigma=F/A[/tex]
[tex]\sigma=\frac{F} {\pi*d^2/4}[/tex]
[tex] \sigma=\frac{2170 N}{\pi*4.5 mm^2/4}[/tex]
[tex] \sigma= 136000000 Pa= 136 Mpa[/tex] Â
Then de specific defotmation:
[tex] \epsilon =136 MPa / 108 GPa = 1.26*10^{-3}[/tex]
Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:
[tex]L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m[/tex]
