Question

A stainless-steel specimen from the same material characterized up above, was formed into a rectangular cross-section of dimensions (3/4 in. x 1/8 in). The specimen is subject to a tensile force of 72,000 N.
(a) Determine the elastic and plastic strain values when the specimen is subject to the 72,000 N tensile load.
(b) If the original length is 24 in., what will be its final length after the 72,000 N tensile load is released

Answer 1

Our dates are in different Units, we will do a unique system. I will use International System Units.

F=72000N

$\frac{3}{8}in= 19,05mm$

$\frac{1}{8}in=3,175mm$

We can now calculate the stress with the force applied, that is

$\sigma = \frac{F}{A} = \frac{72000}{(3,175)(19,05)*10^{-6}}$

$\sigma = 1190.40Mpa$

A) The elastic straint is given by,

$\epsilon_e = \frac{\sigma}{E}$

Where E is the Elastic coefficient. This coefficient for Steel is 201Gpa or $207*10^3Mpa$

$\epsilon_e=\frac{1190.40}{207*10^3}$

$\epsilon_e =5.7507*10^{-3}$

We know for the tables of Strain Steel that the Total Strain is $\epsilon_f=0.01$

Then the Plastic Strain is given by,

$\epsilon_p=\epsilon_f-\epsilon_e$

$\epsilon_p= 0.01-5.7507*10^{-3}$

$\epsilon_p = 4.2492*10^{-3}$

B) Now with the original length of 24in, i.e 609,6mm we have that

$l_i = l_0 (1+\epsilon_p)$

$l_i = (609.6mm)(1+4.294*10^{-3})$

$l_i = 612.2176mm$

Therefore the final length after the 72000N will be 612.2176mm