# Rolling

of 69

## Content

Deformation Processing Rolling

ver. 1

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

1

Overview

Process
Equipment
Products
Mechanical Analysis
Defects

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

2

Process

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

3

Process

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

4

Process

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

5

Ring Rolling

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

6

Equipment

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

7

Equipment

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

8

Products

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

9

Products
• Shapes

• Sections
– door frames, gutters

• Flat plates
• Rings
• Screws

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

10

Products
• A greater volume of metal is
rolled than processed by any
other means.

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

11

Model – flat sheet
zero
slip
point

φ
α
hb

σxb

vb

R
vb < vR < vf

vf

hf

σxf

Exit
Entrance

vR
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

12

Plane strain analysis
Entrance
φ

μp

Exit
φ

p

μp

p

h + dh

h + dh
h
σx + dσx

dx

μp

p

h
σx σx + dσx

dx

μp
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

σx

p
13

Equilibrium
• (top) = entry, (bottom) = exit

(σ x + dσ x ) ⋅ (h + dh ) − 2 pR ⋅ dφ ⋅ sin φ ± 2μpR ⋅ dφ ⋅ cos φ = 0
Simplifying and ignoring HOTs

d (σ x h )
= 2 pR ⋅ (sin φ m μ cos φ )

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

14

Simplifying
• Since α << 1, then sinφ = φ, cosφ = 1

d (σ x h )
= 2 pR ⋅ (φ m μ )

• Plane strain, von Mises

p − σ x = 1.15 ⋅ Y flow ≡ Y flow
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

15

Differentiating
• Substituting

[(

) ] = 2 pR⋅ (φ m μ )

′ ⋅h
d p − Yflow

• or

⎞ ⎤
⎛ p
d ⎡
⎢Y flow
′ ⋅⎜
− 1⎟ ⋅ h ⎥ = 2 pR ⋅ (φ m μ )
⎟ ⎥
⎜ Y flow

dφ ⎢
⎠ ⎦

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

16

Differentiating
⎞ d
d ⎛⎜ p ⎞⎟ ⎛⎜ p
′ ⋅h⋅
′ ⋅ h = 2 pR ⋅ (φ m μ )
Y flow
Y flow
+
− 1⎟ ⋅
⎟ dφ
′ ⎟ ⎜ Y flow

dφ ⎜⎝ Y flow
⎠ ⎝

(

)

rearranging

d ⎛⎜ p ⎞⎟
′ ⎟
dφ ⎜⎝ Y flow

p

Y flow

2R
(φ m μ )
=
h

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

17

Thickness

R

h = h f + 2 R ⋅ (1 − cos φ )
from the definition
of a circular segment

φ R
hb − h f

L

hb
2

2

hf

2
or, after using a Taylor’s series expansion, for small φ

cos φ = 1 −

h = h f + R ⋅φ

2

φ2
2!

+

φ4
4!

L

0

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

18

Substituting and integrating

⎛ p ⎞

d⎜
⎜ Y flow
′ ⎟

⎠=
p

Y flow

∫h

2R
f

+ R ⋅φ

(
)
m
φ
μ
⋅dφ
2

p
h
R
R
−1⎜
⎟ + ln C
ln
tan φ
= ln m 2 μ
⎜ hf ⎟
Y f′
R
hf

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

19

Eliminating ln()

h
′ ⋅ exp(m μH )
p = C ⋅ Y flow
R

R
R
−1⎜

tan φ
H =2
⎜ hf ⎟
hf

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

20

Entry region
• at φ = α, H = Hb , and p =(Y’flow – σxb)

R
C=
exp(μH b )
hf

(

′ − σ xb
p = Y flow

)

h
exp(μ [H b − H ])
hb

R
R ⎞⎟
−1 ⎜
tan α
Hb = 2

hf
hf ⎟

R
R ⎞⎟
−1⎜
tan φ
H =2
⎜ hf ⎟
hf

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

21

Exit region
at φ = 0, H = Hf =0, and p =(Y’flow – σxf)
R
C=
hf
h
′ − σ xf ) exp(μH )
p = (Y flow
hf

R
R ⎞⎟
−1 ⎜
H =2
tan φ
⎜ hf ⎟
hf

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

22

Effect of back and front tension
maximum pressure
pressure

Y

Y

Y − σ xf

Y − σ xb

distance
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

23

Average rolling pressure – per unit width

pave,entry

1
Hb − H n

1
=
Hb − Hn

Hb

Hn

1
Hn − H f

Hb

∫p

entry

dH ; pave ,exit

Hn

1
pentry ⋅ dH =
Hb − H n
Hn

Hf

pexit ⋅ dH =

1
=
Hn − H f

Hb

∫(

′ − σ xb
Y flow

Hn

1
Hn − H f

)

Hf

∫p

exit

⋅ dH

Hn

h
exp(μ [H b − H ]) ⋅ dH
hb

Hn

h
′ − σ xf ) exp(μH ) ⋅ dH
(
Y flow

h

Hf

f

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

24

Average rolling pressure - entry
1
pave (entry ) =
Hb − H n

Hb

′ − σ xb )
(
Y flow

Hn

h
exp(μ [H b − H ]) ⋅ dH
hb

Taylor’s series expansion
0
2

0
3

n

x
x
x
+
+L+
=
exp(x ) = 1 + x +
2! 3!
n!

n

k =0

xk
k!

′ − σ xb )
(
Y flow
pave (entry ) =
h ⋅ (1 + μ [H b − H ]) ⋅ dH
hb ⋅ (H b − H n ) ∫
Hb

Hn

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

25

Average rolling pressure - entry
′ − σ xb )
(
Y flow
pave (entry ) =
h ⋅ (1 + μ [H b − H ]) ⋅ dH
hb ⋅ (H b − H n ) ∫
Hb

Hn

R
R ⎞⎟
−1 ⎜
tan φ
H =2
⎜ hf ⎟
hf

h = h f + R ⋅φ 2
solve numerically
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

26

Average rolling pressure - exit
1
pave (exit ) =
Hn − H f

Hn

∫(

′ − σ xf
Y flow

Hf

)

h
exp(μH ) ⋅ dH
hf

Taylor’s series expansion
0
2

0
n

3

x
x
x
+
+L+
=
exp(x ) = 1 + x +
2! 3!
n!

n

k =0

xk
k!

′ − σ xf )
(
Y flow
pave (exit ) =
h ⋅ (1 + μH ) ⋅ dH

h f ⋅ (H n − H f )
H
Hn

f

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

27

Average rolling pressure - exit
′ − σ xf )
(
Y flow
pave (exit ) =
h ⋅ (1 + μH ) ⋅ dH

h f ⋅ (H n − H f )
H
Hn

f

R
R ⎞⎟
−1 ⎜
tan φ
H =2
⎜ hf ⎟
hf

h = h f + R ⋅φ 2
solve numerically
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

28

Rolling force
• F = pave,entry x Areaentry + pave,exit x Areaexit

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

29

Force
• An alternative method
α

F=

∫φ

n

φn

w ⋅ pentry ⋅ R ⋅ dφ + w ⋅ pexit ⋅ R ⋅ dφ
0

• again, very difficult to do.
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

30

Force - approximation
F / roller = L w pave

L ≈ RΔh
Δh = hb - hf

pave

⎛ have ⎞
= f⎜

⎝ L ⎠
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

31

Derivation of “L”
circular segment

h = h f + 2 R ⋅ (1 − cos φ )
0

Taylor’s expansion

cos φ = 1 −

h = h f + R ⋅φ

φ

2

2!

+

φ

hb
2

L

φ R
hb − h f

hf

2

2

4

4!

R

L

2

R ⋅φ = L
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

32

Derivation of “L”
setting h = hb at φ = α, substituting, and rearranging

⎛L⎞
hb − h f = Δh = R ⋅ ⎜ ⎟
⎝R⎠

2

or

L = R ⋅ Δh

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

33

Approximation based on forging
plane strain – von Mises

pave

μL ⎞
⎟⎟
= 1.15 ⋅ Y flow ⎜⎜1 +
⎝ 2have ⎠
average flow stress:
due to shape of element

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

34

Small rolls or small reductions
have
Δ=
>> 1
L
• friction is not significant (μ -> 0)

pave

μL ⎞
⎟⎟
= 1.15 ⋅ Y flow ⎜⎜1 +
⎝ 2have ⎠
0

pave = 1.15 ⋅ Y flow
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

35

Large rolls or large reductions
have
Δ≡
<< 1
L
• Friction is significant (forging
approximation)
pave

μL ⎞
⎟⎟
= 1.15 ⋅ Y flow ⎜⎜1 +
⎝ 2have ⎠

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

36

Force approximation: low friction
have
Δ≡
>> 1
L

F

roller

= 1.15 ⋅ LwY flow

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

37

Force approximation: high friction
have
Δ≡
<< 1
L
F

μL ⎞
⎟⎟
= 1.15 ⋅ LwY flow ⎜⎜1 +
roller
⎝ 2have ⎠

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

38

Zero slip (neutral) point
• Entrance: material is pulled into the nip
– roller is moving faster than material

• Exit: material is pulled back into nip
– roller is moving slower than material
vf

vb
vR
material
pull-in

vR

vR
material
pull-back

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

39

System equilibrium
• Frictional forces between roller and
material must be in balance.
– or material will be torn apart

• Hence, the zero point must be where
the two pressure equations are equal.
hb
exp(μH b )
=
= exp(μ (H b − 2 H n ))
hf
exp(2 μH n )

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

40

Neutral point
1 ⎛⎜
1 hb
Hn =
H b − ln
2 ⎜⎝
μ hf
φn =

⎛H
tan ⎜ n
⎜ 2
R

hf

h f ⎞⎟
R ⎟

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

41

Torque
paveA

L ≈ RΔh

L/2
R

Δh = hb - hf

∑F

y

=0

∴ Froller = pave A

Froller
φn

α

T=

∫φ

n

Torque / roller = r ⋅ Froller

wμpR 2 dφ −
entry

wμpR 2 dφ

0

exit

Froller L
L
= ⋅ Froller =
2
2

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

42

Power
Power / roller = Tω = FrollerLω / 2
ω = 2πN
N = [rev/min]

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

43

Processing limits
• The material will be drawn into the nip if the
horizontal component of the friction force (Ff) is
larger, or at least equal to the opposing horizontal
component of the normal force (Fn).

F f cos α ≥ Fn sin α
α
α
Ff

R

Δh/2

α

F f = μ ⋅ Fn

tan α = μ
Fn

μ = friction coefficient

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

44

Processing limits
Also
cos α =

Δh
2 = 1 − Δh
R
2R

R−

and
Δh << R

sin α = 1 − cos 2 α

0
Δh ⎛ Δh ⎞
sin α = 1 − 1 +
−⎜ ⎟
2R ⎝ 2R ⎠
Δh
R
tan α =

2
Δh ⎛ Δh ⎞
1−
+⎜

R ⎝ 2R ⎠

2

sin α ≈

Δh
R

Δh
Δh

R − Δh
R

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

45

Processing limits
So, approximately

(tan α )

2

Δh
=μ =
R
2

Hence, maximum draft

Δhmax = μ 2 R
Maximum angle of acceptance

φmax = α = tan −1 μ
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

46

Cold rolling
(below recrystallization point)
strain hardening, plane strain – von Mises

2τ flow = 1.15 ⋅ Y flow

= 1.15 ⋅
n +1
n

average flow stress:
due to shape of element

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

47

Hot rolling –
(above recrystallization point)
strain rate effect, plane strain - von Mises

• Average strain rate

V
h
R
b
&

ε = = ln⎜
t
L ⎝ hf

ε

2τ flow = 1.15 ⋅ Y flow

m
&
= 1.15 ⋅ C ⋅ ε

average flow stress:
due to shape of element
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

48

Example 1.1
• Cold roll a 5% Sn-bronze
• Calculate force on roller
• Calculate power
• Plot pressure in nip (no back or forward
tension)

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

49

Example 1.2
• w = 10 mm
• hb = 2 mm
• height reduction = 30% (hf = 0.7 hb)
– hf = 1.4 mm

R = 75 mm
vR = 0.8 m/s
mineral oil lubricant (μ = 0.1)
K = 720 MPa, n = 0.46
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

50

Example 1.3
• Maximum draft:
Δhmax = μ2R
= (0.1)2 • 75 = 0.75 mm
Δhactual = hb - hf = 2 - 1.4
= 0.6 mm

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

51

Example 1.4
• Maximum angle of acceptance
φmax = tan-1 μ = tan-1(0.1) = 0.1 radian

α=

(hb − h f )
R

=

(2 − 1.4)

75

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

52

Example 1.5
• Roller force: F = L w pave
• L = (RΔh)0.5 = [75 x (2-1.4)]0.5
= 6.7 mm
• w = 10 mm
• have = (hb+hf) / 2 = 1.7 mm
have / L = 1.7 / 6.7 = 0.25 < 1
∴ friction is important
F

μL ⎞

⎟⎟
= 1.15 ⋅ LwY flow ⎜1 +
roller
⎝ 2have ⎠

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

53

Example 1.6
⎛ hf
ε f = ln⎜⎜
⎝ hb

1.4 ⎞
⎟ = ln⎛⎜
⎟ = 0.36

⎝ 2 ⎠

2τ flow = 1.15 ⋅ Y = 1.15 ⋅
720 ⋅ (0.36)
= 1.15 ⋅
1.46

Kε nf
n +1

0.46

= 354 MPa

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

54

Example 1.7

μL ⎞
⎟⎟
= 1.15 ⋅ LwY flow ⎜⎜1 +
roller
⎝ 2have ⎠
−3
−3
6
= 6.7 × 10 ⋅10 × 10 ⋅ 354 × 10
F

⎛ 0 .1 × 6 .7 ⎞
× ⎜1 +

2 × 1 .7 ⎠

= 28,392 N = 3.2 tons
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

55

Example 1.8
Power (kW )

F ⋅ L ⋅ VR
= T ×ω =
roller
2⋅R
−3

28,392 ⋅ 6.7 × 10 ⋅ 0.8
Power (kW ) / roll =
2 ⋅ 0.075
= 1.01 kW / roll = 1.35 hp
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

56

Example 1.9
• Entrance

p=

(

'
Y flow

− σ xb

)

− σ xf

)

• Exit

p=

(

'
Y flow

h
exp(μ (H b − H ))
hb

h
exp(μ (H ))
hf

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

57

Example 1.10

φ=

(h − h f )
R

R
R

tan −1⎜ φ
H =2
⎜ hf ⎟
hf

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

58

Example 1.11
Friction hill
pressure / 2Tflow

1.35
1.3
1.25

Exit

Entrance
φ

1.2
1.15
1.1
1.05
1
0.95
0.9
0.00

0.20

0.40

0.60

0.80

1.00

φ / φ max

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

59

Rolling
Normal Stress

Shear stress

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

60

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

61

Widening of material
φ
Side view

Top view

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

62

Residual stresses due to frictional constraints
a) small rolls or small reduction (ignore friction)
b) large rolls or large reduction (include friction)

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

63

Defects
• a) wavy edges

– barreling

– roll deflection

• b) zipper cracks
– low ductility

• c) edge cracks
• d) alligatoring
– piping, inhomogeniety

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

64

Roll deflection

Rolls can be crowned

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

65

Roll deflection
Rolls can be stacked for stiffness

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

66

Method to reduce roll deflection

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

67

Summary

Process
Equipment
Products
Mechanical Analysis
Defects

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

68

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton

69

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