Answer:
strains for the respective cases are
0.287
0.318
0.127
and for the entire process 0.733
Explanation:
The formula for the true strain is given as:
[tex]\epsilon =\ln \frac{l}{l_{o}}[/tex]
Where
[tex]\epsilon =[/tex] True strain
l= length of the member after deformation
[tex]l_{o} = [/tex] original length of the member
Now for the first case we have
l= 1.6m
[tex]l_{o} = 1.2m[/tex]
thus,
[tex]\epsilon =\ln \frac{1.6}{1.2}[/tex]
[tex]\epsilon =0.287[/tex]
similarly for the second case we have
l= 2.2m
[tex]l_{o} = 1.6m[/tex] (as the length is changing from 1.6m in this case)
thus,
[tex]\epsilon =\ln \frac{2.2}{1.6}[/tex]
[tex]\epsilon =0.318[/tex]
Now for the third case
l= 2.5m
[tex]l_{o} = 2.2m[/tex]
thus,
[tex]\epsilon =\ln \frac{2.5}{2.2}[/tex]
[tex]\epsilon =0.127[/tex]
Now the true strain for the entire process
l=2.5m
[tex]l_{o} = 1.2m[/tex]
thus,
[tex]\epsilon =\ln \frac{2.5}{1.2}[/tex]
[tex]\epsilon =0.733[/tex]
