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

`(3)/(8)in= 19,05mm`

`(1)/(8)in=3,175mm`

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

`\sigma = (F)/(A) = (72000)/((3,175)(19,05)*10^(-6))`

`\sigma = 1190.40Mpa`

A) The elastic straint is given by,

`\epsilon_e = (\sigma)/(E)`

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

`\epsilon_e=(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