in this presentation i have discussed about Forming Process of Mechanical Engineering. it can be beneficial for mechanical engineering competitive exams. watch it on video form https://youtu.be/zo6q_7teNGI

1. 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

4. Mechanism of Rolling

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.

7. Rolling 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)