10. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
10
Products
• Shapes
– I-beams, railroad tracks
• Sections
– door frames, gutters
• Flat plates
• Rings
• Screws
11. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
11
Products
• A greater volume of metal is rolled
than processed by any other
means.
12. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
12
Rolling Analysis
• Objectives
– Find distribution of roll pressure
– Calculate roll separation force (“rolling
force”) and torque
– Processing Limits
– Calculate rolling power
13. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
13
Flat Rolling Analysis
• Consider rolling of a flat plate in a 2-high
rolling mill
Width of plate w is large Æ plane strain
hb hf
R
V0 Vf (> V0)
θ
14. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
14
Flat Rolling Analysis
• Friction plays a critical role in enabling rolling Æ
cannot roll without friction; for rolling to occur
• Reversal of frictional forces at neutral plane (NN)
hb hf
V0 Vf (> V0)
α
L
Entry Zone Exit Zone
N
N
tan
µ α
≥
15. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
15
Flat Rolling Analysis
hb hf
φ
N
N
dx
p
p
µp
µp
σx
σx + dσx
Stresses on Slab in Entry Zone
Stresses on Slab in Exit Zone
p
p
µp
µp
σx
σx + dσx
16. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
16
Equilibrium
Simplifying and ignoring HOTs
( ) ( )
φ
µ
φ
φ
σ
cos
sin
2 m
⋅
= pR
d
h
d x
( ) ( ) 0
cos
2
sin
2 =
−
⋅
⋅
±
⋅
⋅
−
+
⋅
+ h
d
pR
d
pR
dh
h
d x
x
x σ
φ
φ
µ
φ
φ
σ
σ
• Appling equilibrium in x (top entry, bottom exit)
17. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
17
Simplifying
• Since α << 1, then sinφ = φ, cosφ = 1
• Plane strain, von Mises
( ) ( )
µ
φ
φ
σ
m
⋅
= pR
d
h
d x 2
flow
flow
x Y
Y
p ′
≡
⋅
=
− 15
.
1
σ
18. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
18
Differentiating
• Substituting
• or
( )
[ ] ( )
µ
φ
φ
m
⋅
=
⋅
′
−
pR
d
h
Y
p
d flow
2
( )
µ
φ
φ
m
⋅
=
⋅
−
′
⋅
′ pR
h
Y
p
Y
d
d
flow
flow 2
1
19. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
19
Differentiating
( ) ( )
µ
φ
φ
φ
m
⋅
=
⋅
′
⋅
−
′
+
′
⋅
⋅
′ pR
h
Y
d
d
Y
p
Y
p
d
d
h
Y flow
flow
flow
flow 2
1
Rearranging, the variation Y’flow.h with respect to φ is small compared to
the variation p/ Y’flow with respect to φ so the second term is ignored
( )
µ
φ
φ
m
h
R
Y
p
Y
p
d
d
flow
flow 2
=
′
′
20. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
20
Thickness
( )
φ
cos
1
2 −
⋅
+
= R
h
h f
or, after using a Taylor’s series expansion, for small φ
2
φ
⋅
+
= R
h
h f
L
!
4
!
2
1
cos
4
2
φ
φ
φ +
−
=
0
from the definition
of a circular segment
R
R
φ
2
f
b h
h −
L
2
b
h
2
f
h
21. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
21
Substituting and integrating
C
h
R
h
R
R
h
Y
p
f
f
f
ln
tan
2
ln
ln 1
+
=
′
−
φ
µ
m
( ) φ
µ
φ
φ
d
R
h
R
Y
p
Y
p
d
f
flow
flow
∫
∫ ⋅
⋅
+
=
′
′
m
2
2
In[1]:= ‡
2 R Hφ − µL
hf + R φ2
φ
Out[1]= 2 R −
µ ArcTanB
R φ
hf
F
hf R
+
LogAhf + R φ2E
2 R
22. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
22
Eliminating ln()
( )
H
R
h
Y
C
p flow µ
m
exp
⋅
′
⋅
=
= −
f
f h
R
h
R
H φ
1
tan
2
23. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
23
Entry region
• at φ = α, H = Hb ,
( )
H
R
h
Y
C
p flow µ
−
⋅
′
⋅
= exp
( )
b
b
H
h
R
C µ
exp
=
( ) [ ]
( )
H
H
h
h
Y
p b
b
xb
flow −
−
′
= µ
σ exp
= −
f
f h
R
h
R
H φ
1
tan
2
= −
f
f
b
h
R
h
R
H α
1
tan
2
[ ]
( )
H
H
h
h
Y
p b
b
flow −
′
= µ
exp
With back tension=(Y’flow – σxb)
24. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
24
Exit region
at φ = 0, H = Hf =0,
f
h
R
C =
( ) ( )
H
h
h
Y
p
f
xf
flow µ
σ exp
−
′
=
= −
f
f h
R
h
R
H φ
1
tan
2
( ) ( )
H
h
h
Y
p
f
flow µ
exp
′
=
With forward tension
25. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
25
Effect of back and front tension
Y
xb
Y σ
− xf
Y σ
−
Y
maximum pressure
pressure
distance
26. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
26
Flat Rolling Analysis Results –
without front and back tension
p
p
µp
µp
σx
σx + dσx
Stresses on Slab in Entry Zone Stresses on Slab in Exit Zone
p
p
µp
µp
σx
σx + dσx
Using slab analysis we can derive roll pressure distributions for the entry
and exit zones as: h0 and hb are the same thing
1
2 tan
f f
R R
H
h h
φ
−
=
( )
0
0
2
3
H H
f
h
p Y e
h
µ −
=
2
3
H
f
f
h
p Y e
h
µ
=
Entry Zone Exit Zone
α
φ =
= @
0 H
H
27. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
27
Average rolling pressure – per unit
width
∫
∫ =
−
−
=
n
n
Rd
p
R
p
Rd
p
R
p exit
n
exit
ave
entry
n
entry
ave
φ
φ
α
φ
φ
φ
φ
α 0
,
,
1
;
)
(
1
28. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
28
Rolling force
• F = pave,entry x Areaentry + pave,exit x Areaexit
29. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
29
Force
• An alternative method
• again, very difficult to do.
∫
∫ ⋅
⋅
⋅
+
⋅
⋅
⋅
=
n
n
d
R
p
w
d
R
p
w
F exit
entry
φ
α
φ
φ
φ
0
30. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
30
Force - approximation
F / roller = L w pave
∆h = hb - hf
h
R
L ∆
≈
=
L
h
f
p ave
ave
31. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
31
Derivation of “L”
( )
φ
cos
1
2 −
⋅
+
= R
h
h f
L
!
4
!
2
1
cos
4
2
φ
φ
φ +
−
=
0
2
φ
⋅
+
= R
h
h f
L
R =
⋅φ
circular segment
Taylor’s expansion
R
R
φ
2
f
b h
h −
L
2
b
h
2
f
h
32. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
32
Derivation of “L”
setting h = hb at φ = α, substituting, and rearranging
2
⋅
=
∆
=
−
R
L
R
h
h
h f
b
or
h
R
L ∆
⋅
=
33. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
33
Approximation based on forging
plane strain – von Mises
+
⋅
=
ave
flow
ave
h
L
Y
p
2
1
15
.
1
µ
average flow stress:
due to shape of element
34. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
34
Small rolls or small reductions
• friction is not significant (µ -> 0)
1
>>
=
∆
L
have
flow
ave Y
p ⋅
= 15
.
1
+
⋅
=
ave
flow
ave
h
L
Y
p
2
1
15
.
1
µ
0
35. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
35
Large rolls or large reductions
• Friction is significant (forging
approximation)
1
<<
≡
∆
L
have
+
⋅
=
ave
flow
ave
h
L
Y
p
2
1
15
.
1
µ
36. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
36
Force approximation: low
friction
1
>>
≡
∆
L
have
flow
Y
Lw
roller
F ⋅
= 15
.
1
37. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
37
Force approximation: high
friction
1
<<
≡
∆
L
have
+
⋅
=
ave
flow
h
L
Y
Lw
roller
F
2
1
15
.
1
µ
38. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. 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
material
pull-in
vb
vf
vR
vR
vR
material
pull-back
39. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. 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.
( )
( )
( )
( )
n
b
n
b
f
b H
H
H
H
h
h
2
exp
2
exp
exp
−
=
= µ
µ
µ
40. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
40
Neutral point
−
=
f
b
b
n
h
h
H
H ln
1
2
1
µ
=
R
h
H
R
h f
n
f
n
2
tan
φ
41. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
41
Torque
R
Froller
L/2
paveA
h
R
L ∆
≈
∆h = hb - hf
A
p
F
F
ave
roller
y
=
∴
=
∑ 0
2
2
/
L
F
F
L
F
r
roller
Torque roller
roller
roller =
⋅
=
⋅
=
φ
µ
φ
µ
φ
α
φ
d
pR
w
d
pR
w
T
n
n
∫
∫ −
=
0
2
2
entry exit
42. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
42
Power
Power / roller = Tω = FrollerLω / 2
ω = 2πN
N = [rev/min]
43. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. 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).
∆h/2
Fn
Ff
R
α
α
α
α
α sin
cos n
f F
F ≥
µ
α =
tan
µ = friction coefficient
n
f F
F ⋅
= µ
44. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
44
Processing limits
R
h
R
h
R
2
1
2
cos
∆
−
=
∆
−
=
α
R
h <<
∆ α
α 2
cos
1
sin −
=
Also
and
2
2
2
1
1
sin
∆
−
∆
+
−
=
R
h
R
h
α
0
R
h
∆
≈
α
sin
R
h
h
R
h
R
h
R
h
R
h
∆
≈
∆
−
∆
≅
∆
+
∆
−
∆
= 2
2
1
tanα
45. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
45
Processing limits
So, approximately
( )
R
h
∆
=
= 2
2
tan µ
α
Hence, maximum draft
R
h 2
max µ
=
∆
Maximum angle of acceptance
µ
α
φ 1
max tan−
=
=
46. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
46
Processing Limits
( ) 2
max
h R
µ
∆ =
Max. angle of acceptance
h0 hf
α
R
µ
α
φ 1
max tan−
=
=
Max. reduction in thickness
47. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
47
Cold rolling
(below recrystallization point)
strain hardening, plane strain – von
Mises
1
15
.
1
15
.
1
2
+
⋅
=
⋅
=
n
K
Y
n
flow
flow
ε
τ
average flow stress:
due to shape of element
48. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
48
Hot rolling –
(above recrystallization point)
strain rate effect, plane strain - von
Mises
• Average strain rate
=
=
f
b
R
h
h
L
V
t
ln
ε
ε
&
m
flow
flow C
Y ε
τ &
⋅
⋅
=
⋅
= 15
.
1
15
.
1
2
average flow stress:
due to shape of element
49. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
49
Example 1.1
• Cold roll a 5% Sn-bronze
• Calculate force on roller
• Calculate power
• Plot pressure in nip (no back or forward
tension)
50. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
50
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
51. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
51
Example 1.3
• Maximum draft:
∆hmax = µ2R
= (0.1)2 • 75 = 0.75 mm
∆hactual = hb - hf = 2 - 1.4
= 0.6 mm
52. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
52
Example 1.4
• Maximum angle of acceptance
φmax = tan-1 µ = tan-1(0.1) = 0.1 radian
( ) ( )
o
12
.
5
089
.
0
75
4
.
1
2
=
=
−
=
−
=
rad
R
h
h f
b
α
53. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
53
Example 1.5
• Roller force: F = L w pave
• L = (R∆h)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
+
⋅
=
ave
flow
h
L
Y
Lw
roller
F
2
1
15
.
1
µ
54. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
54
Example 1.6
36
.
0
2
4
.
1
ln
ln =
=
=
b
f
f
h
h
ε
( ) MPa
n
K
Y
n
flow
f
354
46
.
1
36
.
0
720
15
.
1
1
15
.
1
15
.
1
2
46
.
0
=
⋅
⋅
=
+
⋅
=
⋅
=
ε
τ
55. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
55
Example 1.7
tons
N
h
L
Y
Lw
roller
F
ave
flow
2
.
3
392
,
28
7
.
1
2
7
.
6
1
.
0
1
10
354
10
10
10
7
.
6
2
1
15
.
1
6
3
3
=
=
×
×
+
×
×
⋅
×
⋅
×
=
+
⋅
=
−
−
µ
56. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
56
Example 1.8
( )
R
V
L
F
T
roller
kW
Power R
⋅
⋅
⋅
=
×
=
2
ω
( )
hp
roll
kW
roll
kW
Power
35
.
1
/
01
.
1
075
.
0
2
8
.
0
10
7
.
6
392
,
28
/
3
=
=
⋅
⋅
×
⋅
=
−
57. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
57
Example 1.9
• Entrance
• Exit
( ) ( )
( )
H
H
h
h
Y
p b
b
xb
flow −
−
= µ
σ exp
'
( ) ( )
( )
H
h
h
Y
p
f
xf
flow µ
σ exp
'
−
=
58. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
58
Example 1.10
( )
R
h
h f
−
=
φ
= −
f
f h
R
h
R
H φ
1
tan
2
59. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
59
Example 1.11
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
0.00 0.20 0.40 0.60 0.80 1.00
φ / φ max
pressure
/
2Tflow
Exit Entrance
φ
Friction hill
60. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
60
Rolling
Shear stress
Normal Stress
62. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
62
Widening of material
φ
Side view
Top view
63. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
63
Residual stresses -
due to frictional constraints
a) small rolls or small reduction (ignore friction)
b) large rolls or large reduction (include friction)
64. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
64
Defects
• a) wavy edges
– roll deflection
• b) zipper cracks
– low ductility
• c) edge cracks
– barreling
• d) alligatoring
– piping, inhomogeniety
65. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
65
Roll deflection
Rolls can deflect under load
Rolls can be crowned
66. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
66
Roll deflection
Rolls can be stacked for stiffness
67. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
67
Method to reduce roll deflection
68. Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
68
Summary
• Process
• Equipment
• Products
• Mechanical Analysis
• Defects
