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

Answer

given,

diameter of work piece = 50 mm

height = 40 mm

$\epsilon = ln(\dfrac{40}{30}) = 0.2876$

$Y_f = 600(0.2876)^{0.12} = 516.67 MPa$

$V = \dfrac{\pid^2L}{4}=\dfrac{\pi\times 50^2\times 40}{4}$

V = 78,739 mm³

at h = 30 mm $A = \dfrac{V}{h}=\dfrac{78739}{30}$ = 2,618 mm²

$A = \dfrac{\pi d^2}{4}$

$d = \sqrt{\dfrac{2,618 \times 4 }{\pi}}$

d = 57.73 mm

$k_f = 1+\dfrac{0.4(0.2)(57.73)}{30}$ = 1.154

F = 1.154 × 516.67×  2618

F = 1560.92 kN

At h = 20 mm

$\epsilon = ln(\dfrac{40}{20}) = 0.693$

$Y_f = 600(0.693)^{0.12} = 574.18 MPa$

$V = \dfrac{\pid^2L}{4}=\dfrac{\pi\times 50^2\times 40}{4}$

V = 78,739 mm³

at h =20 mm $A = \dfrac{V}{h}=\dfrac{78739}{20}$ = 3936.95 mm²

$A = \dfrac{\pi d^2}{4}$

$d = \sqrt{\dfrac{3936.95 \times 4 }{\pi}}$

d = 70.8 mm

$k_f = 1+\dfrac{0.4(0.2)(70.8)}{20}$ = 1.283

F = 1.154 × 574.18× 3936.95

F = 2900 kN