Question

What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?

Answer 1

DRAWING LOAD IS
`3.67 A_(O)\sigma`

Step-by-step 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 =
`A_(F)*\sigma*ln((A_(O))/(A_(F)))`

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

`A_(f) = A_(O)- 0.6A_(O)`

=
`0.4 A_(O)`

Due to change in area ,drawing load p is

p =
`0.4A_(O)*\sigma*ln((A_(O))/(0.4A_(O)))`

p =
`3.67 A_(O)\sigma`