Question

5. Bar stock of initial diameter = 90 mm is drawn with a draft = 10 mm. The draw die has an entrance angle = 18, and the coefficient of friction at the work die interface = 0.08. The metal behaves as a perfectly plastic material with yield stress = 105 MPa. Determine (a) area reduction, (b) draw stress, (c) draw force required for the operation, and (d) power to perform the operation if exit velocity = 1.0 m/min.

Answer 1

Answer:

a)  0.2099

b)  46.5 MPa

c)  233765 N

d)  3896 W

Explanation:

a)

r = (A'' - A') / A'', where

A'' = 1/4 * π * D''²

A'' = 1/4 * 3.142 * 90²

A'' = 6362.55 mm²

D' = D'' - d = 90 - 10 = 80 mm

A' = 1/4 * π * D'²

A' = 1/4 * 3.142 * 80²

A' = 5027.2 mm²

r = (A'' - A') / A'

r = (6362.55 - 5027.2) / 6362.55

r = 1335 / 6362.55

r = 0.2099

b)

Draw stress = σd

Y' = k = 105 MPa

Φ = 0.88 + 0.12(D/Lc), where

D = 0.5 (90 + 80) = 85 mm

Lc = 0.5 [(90 - 80)/sin 18] = 16.18 mm

Φ = 0.88 + 0.12(85/16.18) = 1.51

σd = Y' * (1 + μ/tan α) * Φ * In(A''/A')

σd = 105 * (1 + 0.08/tan18) * 1.51 * In(6362.55/5027.2)

σd = 105 * 1.246 * 1.51 * 0.2355

σd = 46.5 MPa

c)

F = A' * σd

F = 5027.2 * 46.5

F = 233764.8 N

d)  

P = 233764.8 (1 m/min)

P = 233764.8 Nm/min

P = 3896.08 Nm/s

P = 3896.08 W