Answer:
Engineering strain =2499
True strain = 7.82
Explanation:
Given that
Initial diameter D= 25 mm
Final diameter = 0.5 mm
As we know that ,the volume of material will remain constant
[tex]A_iL_i=A_fL_f[/tex]
[tex]\dfrac{\pi }{4}D^2L_i=\dfrac{\pi }{4}d^2L_f[/tex]
So now by putting the values
[tex]\dfrac{\pi }{4}\times 25^2L_i=\dfrac{\pi }{4}\times 0.5^2L_f[/tex]
[tex]\dfrac{L_f}{L_i}=2500[/tex]
We know that
[tex]Engineering\ strain=\dfrac{L_f-L_i}{L_i}[/tex]
[tex]Engineering\ strain\ e=\dfrac{2500L_i-L_i}{L_i}[/tex]
e=2499
We also know that
[tex]True\ strain=\ln\left ( 1+e \right )[/tex]
[tex]True\ strain=\ln\left ( 1+2499 \right )[/tex]
True strain = 7.82
