Answer:
DRAWING LOAD IS [tex]3.67 A_{O}\sigma[/tex]
Explanation:
wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire
drawing load for wire drawing is given as P = [tex]A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})[/tex]
Where A f is initial area, Ao is original area, σ is yield stress
as given in question sectional area reduce 60%, therefore
[tex]A_{f} = A_{O}- 0.6A_{O}[/tex]
= [tex]0.4 A_{O}[/tex]
Due to change in area ,drawing load p is
p = [tex] 0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})[/tex]
p = [tex]3.67 A_{O}\sigma[/tex]
