The work metal at this elevated temperature yields at 90 MPa

The work metal at this elevated temperature yields at 90 MPa. Coefficient of friction at the die-work interface = 0.40. Determine (a) the final height of the part and (b) the maximum force in the operation. A cup-shaped part is backward extruded from an aluminum slug that is 50 mm in diameter. The final dimensions of the cup are: outside diameter = 50 mm, inside diameter = 40 mm, height = 100 mm, and thickness of base = 5 mm. Determine (a) extrusion ratio, (b) shape factor, and (c) height of starting slug required to achieve the final dimensions, (d) If the metal has flow curve parameters K = 400 MPa and n = 0.25, and the constants in the Johnson extrusion strain equation are:o = 0.8 and b = 1.5, determine the extrusion force.

Solution

Solution:

(a) r_x = A_o/A_f

A_o = 0.25p(50)² = 1963.75 mm²

A_f = 0.25p(50² - 40²) = 706.86 mm²

r_x = 1963.75/706.86 = 2.778

(b) To determine the die shape factor we need to determine the perimeter of a circle whose area is equal to that of the extruded cross section, A = 706.86 mm². The radius of the circle is R = (706.86/p)^0.5 = 15 mm, C_c = 2p(15) = 94.25 mm.

The perimeter of the extruded cross section C_x = p(50 + 40) = 90p = 282.74 mm.

K_x = 0.98 + 0.02(282.74/94.25)^2.25 = 1.217

(c) Volume of final cup consists of two geometric elements: (1) base and (2) ring.

(1) Base t = 5 mm and D = 50 mm. V₁ = 0.25p(50)²(5) = 9817.5 mm³

(2) Ring OD = 50 mm, ID = 40 mm, and h = 95 mm.

V₂ = 0.25p(50² - 40²)(95) = 0.25p(2500 - 1600)(95) = 67,151.5 mm³

Total V = V₁ + V₂ = 9817.5 + 67,151.5 = 76,969 mm³

Volume of starting slug must be equal to this value V = 76,969 mm³

V = 0.25p(50)²(h) = 1963.5h = 76,969 mm³

h = 39.2 mm

(d) e = ln 2.778 = 1.0218

e_x = 0.8 + 1.5(1.0218) = 2.33

Y_f = 400 ( 1.0218)^0.25 / 1.25 = 321.73 MPa

p = K_x Y_f e_x = 1.217 (321.73) (2.33) = 912.3 MPa

A_o = 0.25p(40)² = 1256.6 mm²

F = 912.3(1256.6) = 1,146,430 N