Respuesta :
Solution :
Given data:
Diameter of the copper cylindrical rod = 16 mm
Yield strength = 250 MPa
Calculating the percent cold work
[tex]$\text{Percentage Cold Work} = \frac{\pi\left(\frac{d_0}{2}\right)^2-\pi\left(\frac{d_d}{2}\right)^2}{\pi\left(\frac{d_0}{2}}\right)^2} \times 100$[/tex]
[tex]$ = \frac{\pi\left(\frac{16}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{16}{2}}\right)^2} \times 100$[/tex]
= 50% CW
Therefore, at [tex]50\% \ CW[/tex], the yield strength of copper will be of the order of 330 MPa.
The ductility will be 4% elongation (EL).
Rather than performing drawing in single operation, we draw some of the fraction of total deformation, and then the anneal them to recrystallize and also finally w do cold work on the material for the second time to achieve its final diameter, ductility and yield strength.
[tex]21\% \ CW[/tex]is required for a yield strength of [tex]250 \ MPa[/tex]. Similarly, a maximum of [tex]23\% \ CW[/tex] is required for[tex]12\% \ EL[/tex].
The average of the two values is [tex]22\% \ CW[/tex]. To achieve both the [tex]\text{specified yield strength and ductility,}[/tex] the copper should be deformed to[tex]22\% \ CW[/tex]. The[tex]\text{ final diameter}[/tex] after the first drawing and the initial diameter for the second drawing is [tex]$d_0'$[/tex] , then
[tex]$22\% \ CW = \frac{\pi\left(\frac{d_0'}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{d_0'}{2}}\right)^2} \times 100$[/tex]
[tex]$d_0'=\frac{11.3}{\sqrt{1-\frac{22\% \ CW}{100}}}$[/tex]
[tex]$d_0'=12.8 \ mm$[/tex]
