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

– I-beams, railroad tracks

• 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 )

dφ

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 )

dφ

• 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

dφ

• 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]

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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

Kε

= 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

= 0.089 rad = 5.12o

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

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60

ME 6222: Manufacturing Processes and Systems

Prof. J.S. Colton

61

Widening of material

φ

Side view

Top view

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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

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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 deflect under load

Rolls can be crowned

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Prof. J.S. Colton

65

Roll deflection

Rolls can be stacked for stiffness

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66

Method to reduce roll deflection

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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

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69

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

– I-beams, railroad tracks

• 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 )

dφ

ME 6222: Manufacturing Processes and Systems

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14

Simplifying

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

d (σ x h )

= 2 pR ⋅ (φ m μ )

dφ

• 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

dφ

• 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

Kε

= 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

= 0.089 rad = 5.12o

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 deflect under load

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