Answer:
strains for the respective cases are
0.287
0.318
0.127
and for the entire process 0.733
Step-by-step explanation:
The formula for the true strain is given as:

Where
True strain
l= length of the member after deformation
original length of the member
Now for the first case we have
l= 1.6m

thus,


similarly for the second case we have
l= 2.2m
(as the length is changing from 1.6m in this case)
thus,


Now for the third case
l= 2.5m

thus,


Now the true strain for the entire process
l=2.5m

thus,
