Question

A cylindrical part is warm upset forged in an open die. The starting work piece is 50 mm in diameter, and 40 mm in hieght. The final height=20mm. Coefficent of friction at the die-work interface =0.2. The work material has a flow curve defined by K=600 MPa, and n=0.12. Determine the force in the operation at process begins, at intermediate height of 30 mm, and at the final height of 20 mm.

Answer 1

given,

diameter of work piece = 50 mm

height = 40 mm

`\epsilon = ln((40)/(30)) = 0.2876`

`Y_f = 600(0.2876)^(0.12) = 516.67 MPa`

`V = (\pid^2L)/(4)=(\pi* 50^2* 40)/(4)`

V = 78,739 mm³

at h = 30 mm
`A = (V)/(h)=(78739)/(30)` = 2,618 mm²

`A = (\pi d^2)/(4)`

`d = \sqrt{(2,618 * 4 )/(\pi)}`

d = 57.73 mm

`k_f = 1+(0.4(0.2)(57.73))/(30)` = 1.154

F = 1.154 × 516.67× 2618

F = 1560.92 kN

At h = 20 mm

`\epsilon = ln((40)/(20)) = 0.693`

`Y_f = 600(0.693)^(0.12) = 574.18 MPa`

`V = (\pid^2L)/(4)=(\pi* 50^2* 40)/(4)`

V = 78,739 mm³

at h =20 mm
`A = (V)/(h)=(78739)/(20)` = 3936.95 mm²

`A = (\pi d^2)/(4)`

`d = \sqrt{(3936.95 * 4 )/(\pi)}`

d = 70.8 mm

`k_f = 1+(0.4(0.2)(70.8))/(20)` = 1.283

F = 1.154 × 574.18× 3936.95

F = 2900 kN