Answer:
Explanation:
From the information given:
original diameter [tex]d_o[/tex] = 10 mm
final diameter [tex]d_f =[/tex] 7.5 mm
Cold work tensile strength of brass = 380 MPa
Recall that;
[tex]\text {The percentage CW }= \dfrac{\pi (\dfrac{d_o}{2})^2 - \pi(\dfrac{d_f}{2})^2 }{\pi(\dfrac{d_o}{2})^2} \times 100[/tex]
[tex]\implies \dfrac{\pi (\dfrac{10}{2})^2 - \pi(\dfrac{7.5}{2})^2 }{\pi(\dfrac{10}{2})^2} \times 100[/tex]
[tex]\implies43.87\% \ CW[/tex]
→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.
→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.
To achieve 15% EL, 28% CW is allowed at most
i.e
The lower bound cold work = 15%
The upper cold work = 28%
The average = [tex]\dfrac{15+28}{2}[/tex] = 21.5 CW
Now, after the first drawing, let the final diameter be [tex]d_o^'[/tex]; Then:
[tex]4.5\% \ CW = \dfrac{\pi (\dfrac{d_o^'}{2})^2 - (\dfrac{7.5}{2})^2}{\pi (\dfrac{d_o^'}{2})^2}\times 100[/tex]
By solving:
[tex]d_o^'} = 8.46 mm[/tex]
To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.
