Question

QUESTION 2 The height of a 50-mm Cylindrical part with 20 mm OD is reduced to 25 mm in an open die forging operation Calculate the forging force if the flow stress of the material is 300 MPa. Assume homogeneous deformation conditions is 1). (careful with the units!) 1) The final area of the part is x104m2 (round to two decimal places) 2) The forging force is KN

Answer 1

summary: In forging, the material is conserved and there is no wastage. Due to which the volume remains constant and the volume before and after forging will be equal. The forging force is calculated by multiplying flow stress and final area.

- - Open die Eliging Given, Height of Cylindrical part is hi = 50mm Cinitial height) Outer diameter di= 20 mm di Final height hg= 25 mm (a Before Folging cbs After Forging Final diameter is df. Flow stress of the material is Ğ = 300 Mpa Assume homogeneous deformation Conditions. Since no material is wasted, Volume remains Constant. Volume = 7 d; ² he = I det he [volume = Area xheight] 7 x 20 x 50 = 77 x dp x 25 = x 20'x 2 = Af [.: Apa de] - Ag = 628.318 mm² → Ag = 628.318 x 106 m² AF = 6.28 x 10-4 m2 final Area of the part is Af = 6.28 x 10 m2 The - ② The Forging face is given by F = Ğ Af [.: impa = 10% w/m) - 300 x 6.28 x 10-Yx106 = 300x6.28x102 F = 188.4 KN Forging Force is F=188, 4 KN . The The Force Forging - x x