Our dates are in different Units, we will do a unique system. I will use International System Units.
F=72000N
[tex]\frac{3}{8}in= 19,05mm[/tex]
[tex]\frac{1}{8}in=3,175mm[/tex]
We can now calculate the stress with the force applied, that is
[tex]\sigma = \frac{F}{A} = \frac{72000}{(3,175)(19,05)*10^{-6}}[/tex]
[tex]\sigma = 1190.40Mpa[/tex]
A) The elastic straint is given by,
[tex]\epsilon_e = \frac{\sigma}{E}[/tex]
Where E is the Elastic coefficient. This coefficient for Steel is 201Gpa or [tex]207*10^3Mpa[/tex]
[tex]\epsilon_e=\frac{1190.40}{207*10^3}[/tex]
[tex]\epsilon_e =5.7507*10^{-3}[/tex]
We know for the tables of Strain Steel that the Total Strain is [tex]\epsilon_f=0.01[/tex]
Then the Plastic Strain is given by,
[tex]\epsilon_p=\epsilon_f-\epsilon_e[/tex]
[tex]\epsilon_p= 0.01-5.7507*10^{-3}[/tex]
[tex]\epsilon_p = 4.2492*10^{-3}[/tex]
B) Now with the original length of 24in, i.e 609,6mm we have that
[tex]l_i = l_0 (1+\epsilon_p)[/tex]
[tex]l_i = (609.6mm)(1+4.294*10^{-3})[/tex]
[tex]l_i = 612.2176mm[/tex]
Therefore the final length after the 72000N will be 612.2176mm
