# Forming process

Student at vit-east
2. Jul 2021
1 von 16

### Forming process

• 2. Bulk Forming ◦Rolling Process ◦Extrusion Process ◦Forging Process ◦Wire Drawing ◦Squeezing Sheet metal Forming Bending Deep Drawing Shearing Powder Metal Forming Powder Forging Powder Injection Powder Extrusion Moulding
• 3. Rolling The main purpose of this rolling process is to reduce the thickness of the metal plates. In the forming processes, metal is not removed. Production of flat plate, sheet, and foil in long length at high speeds and with good surface finish
• 5. Velocity = ∏DN/60 Reduction in flow h= h1-h2 From figure, L2 = (2R – h/2) (h/2) Since h2 is very small L2 = R*h Bite angle tanΦ = L/ (R-h/2) tanΦ = (h1-h2 / R)1/2 Roll Force F= LW*avg Shear Stress L= (R*h)1/2
• 6. Planetary Mill  It consists of a heavy backing rolls surrounded by a large number of planetary mill.
• 8. Ring Rolling Application Large rings for rockets and turbines, gearwheels rims, ball bearing races, flanges and reinforcing ring for pipes.
• 9. Thread Rolling  It is cold forming process by which straight or tapered thread are formed round rods by passing them between dies.
• 10. Rolling Defects  Wavy Edges  Spread  Crocodile Crack
• 11. Q 1 A Copper strip of 200 mm width and 30 mm thickness is to be rolled to a thickness of 25 mm. The roll of radius 300 mm rotates at 100 rpm. The average shear strength of the work 180 Mpa. The roll strip contact length and roll forces are Solution :- width 200 mm H1 = 30 mm H2 = 25 mm R = 300 mm N = 100 rpm Shear Stress = 180 Mpa we know L = (R*h)1/2 = (5*300)1/2 = 38.73 mm shear stress = Force / area Force = shear stress * Area F= 180*200*38.73 = 1.39 MN
• 12. Q 2 An annealing copper strip 228 mm wide and 25 mm thick is rolled to a thickness of 20 mm in one pass. The roll radius is 300 mm and the roll rotate at 100 rpm. Calculate the Roll force and Power required in the operation? Take shear stress 26000N/mm2. Answer:- W = 228 mm h1 = 25 mm h2 = 20mm R= 300 mm N = 100 rpm F= L*W*Shear stress L = (R*h)1/2 = (300*5)1/2 = 38.72 mm True strain of strip = ln(h1/h2)= ln(25/20)=0.223 F = 38.72*228*26000=229.53 MN Power = 2*3.14 *N*T/60 = 2*3.14*100*229.53*38.72/60 Power= 46.53kw
• 13. Q 3 Rolling very thin strips of mild steel requires a) Large diameter Rolls b) Small diameter rolls c) High speed rolling d) Rolling without a lubricant (b) Q 4 In Rolling process separating force can be decreased by a) Reducing roll diameter b) Increasing roll diameter c) Providing back up Roll d) Increasing the friction between roll and metals (a)
• 14. Q 5 The thickness of a plate is reduced from 30 mm to 10 mm by successive cold rolling passes using identical rolls of diameter 600 mm. Assume that there is no changes in width. If the coefficient of friction between the rolls and the work piece is 0.1, the minimum number of passes required is Sol:- h1=30 mm h2 = 10 mm h= h2-h1 = 20 mm D = 600 mm R = 300 mm coefficient of friction(µ) = 0.1 hmax = µ2 R = 0.1*0.1*300 = 3 mm Number of Passes = htotal / hmax = 20/3 = 7 pass
• 15. Q 6 which of the following assumption are correct for Rolling 1. The material is plastic 2. The arc of contact is circular with radius greater than radius of roll 3. Coefficient of friction is constant over the arc of contact and act in one direction through out the arc of contact. Select the correct answer a) 1 and 2 b) 1 and 3 c) 2 and 3 d) 1,2 and 3 (a)
• 16. Q 7 In one setting of rolls in a 3 high rolling mill one gets a) One reduction in thickness b) Two reduction in thickness c) There reduction in thickness d) Two or three reaction in thickness depending upon setting (b)