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Chapter 3

Rolling of metals
Subjects of interest
• Introduction/objectives • Rolling mills • Classification of rolling processes • Hot rolling • Cold rolling • Forces and geometry relationships in rolling • Simplified analysis of rolling load: Rolling variables • Problems and defects in rolled products • Rolling-mill control • Theories of cold rolling • Theories of hot rolling • Torque and power
Suranaree University of Technology
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Jan-Mar 2007

Objectives
• This chapter provides information on different types of metal rolling processes which can also be divided in to hot and cold rolling process. • Mathematical approaches are introduced for the understanding of load calculation in rolling processes. • Finally identification of defects occurring during and its solutions are included.

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Introduction- Definition of rolling process
• Definition of Rolling : The process of plastically deforming metal by passing it between rolls. • Rolling is the most widely used forming process, which provides high production and close control of final product. • The metal is subjected to high compressive stresses as a result of the friction between the rolls and the metal surface.

www.world-aluminium.org

Rolling process

Note: rolling processes can be mainly divided into 1) hot rolling and 2) cold rolling.
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Introduction- Hot and cold rolling processes www.vatech.at Hot rolling
• The initial breakdown of ingots into blooms and billets is generally done by hot-rolling. This is followed by further hotrolling into plate, sheet, rod, bar, pipe, rail.

Cold rolling
• The cold-rolling of metals has played a major role in industry by providing sheet, strip, foil with good surface finishes and increased mechanical strength with close control of product dimensions.
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Sheet rolling machines
Rolled strips

Rollforming machine

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Bloom

Billet

Slap

Terminology
• Bloom is the product of first breakdown of ingot (cross sectional area > 230 cm2). Semifinished products • Billet is the product obtained from a further reduction by hot rolling (cross sectional area > 40x40 mm2). • Slab Further rolling steps is the hot rolled ingot (cross sectional area > 100 cm2 and with a width ≥ 2 x thickness). Plate • Plate is the product with a thickness > 6 mm. Mill products • Sheet is the product with a thickness < 6 mm and width > 600 mm. • Strip is the product with a thickness < 6 mm and width < 600 mm. Sheet Strip

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Mill rolls

Rolls
Mill rolls

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Ring rolls
• Ring rolls are used for tube rolling, ring rolling. • Ring rolls are made of spheroidized graphite bainitic and pearlitic matrix or alloy cast steel base.

Cantilever mill roll

Tube mill roll

Universal roll

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Typical arrangement of rollers for rolling mills
Two-high mill, pullover
The stock is returned to the entrance for further reduction.

Four-high mill
Small-diameter rolls (less strength & rigidity) are supported by larger-diameter backup rolls

Two-high mill, reversing
The work can be passed back and forth through the rolls by reversing their direction of rotation.

Cluster mill or Sendzimir mill
Each of the work rolls is supported by two backing rolls.

Three-high mill
Consist of upper and lower driven rolls and a middle roll, which rotates by friction.

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Typical arrangement of rollers for rolling mills Continuous rolling σb σf
• Use a series of rolling mill and each set is called a stand. • The strip will be moving at different velocities at each stage in the mill.
A four stand continuous mill or tandem mil.

• The speed of each set of rolls is synchronised so that the input speed of each stand is equal to the output speed of preceding stand. • The uncoiler and windup reel not only feed the stock into the rolls and coiling up the final product but also provide back tension and front tension to the strip.

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Typical arrangement of rollers for rolling mills Planetary mill
• Consist of a pair of heavy backing rolls surrounded by a large number of planetary rolls. • Each planetary roll gives an almost constant reduction to the slab as it sweeps out a circular path between the backing rolls and the slab. • As each pair of planetary rolls ceases to have contact with the work piece, another pair of rolls makes contact and repeat that reduction. • The overall reduction is the summation of a series of small reductions by each pair of rolls. Therefore, the planetary mill can hot reduces a slab directly to strip in one pass through the mill. • The operation requires feed rolls to introduce the slab into the mill, and a pair of planishing rolls on the exit to improve the surface finish.
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Rolling mills
Rolling mill is a machine or a factory for shaping metal by passing it through rollers A rolling mill basically consists of • rolls • bearings • a housing for containing these parts • a drive (motor) for applying power to the rolls and controlling the speed • Requires very rigid construction, large motors to supply enough power (MN). Successive stands of a large continuous mill

Modern rolling mill

• skills • engineering design • construction
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+

Huge capital investment
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Different types of rolling processes
There are different types of rolling processes as listed below;

• Continuous rolling • Transverse rolling • Shaped rolling or section rolling • Ring rolling • Powder rolling • Continuous casting and hot rolling • Thread rolling

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Conventional hot or cold-rolling
The objective is to decrease the thickness of the metal with an increase in length and with little increase in width. • The material in the centre of the sheet is constrained in the z direction (across the width of the sheet) and the constraints of undeformed shoulders of material on each side of the rolls prevent extension of the sheet in the width direction. • This condition is known as plane strain. The material therefore gets longer and not wider.

σy σx σz
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• Otherwise we would need the width of a football pitch to roll down a steel ingot to make tin plate!
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Transverse rolling

• Using circular wedge rolls. • Heated bar is cropped to length and fed in transversely between rolls. • Rolls are revolved in one direction.

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Shaped rolling or section rolling
• A special type of cold rolling in which flat slap is progressively bent into complex shapes by passing it through a series of driven rolls. • No appreciable change in the thickness of the metal during this process. • Suitable for producing moulded sections such as irregular shaped channels and trim.

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Shaped rolling or section rolling
A variety of sections can be produced by roll forming process using a series of forming rollers in a continuous method to roll the metal sheet to a specific shape Applications: - construction materials, - partition beam - ceiling panel - roofing panels. - steel pipe - automotive parts - household appliances - metal furniture, - door and window frames - other metal products. www.formtak.com A variety of rolled sections
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Ring rolling

Seamless rings

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Simulation of ring rolling
• The donut shape preform is placed between a free turning inside roll and a driven outside roll. • The ring mills make the section thinner while increasing the ring diameter.

www.qcforge.com www.rz.rwth-aachen.de

Simulation of ring rolling www.shape.co.kr Relative strain in ring rolling Suranaree University of Technology
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Seamless ring rolling

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Powder rolling
Metal powder is introduced between the rolls and compacted into a ‘green strip’, which is subsequently sintered and subjected to further hot-working and/or cold working and annealing cycles.

Advantage : - Cut down the initial hot-ingot breakdown step (reduced capital investment). - Economical - metal powder is cheaply produced during the extraction process. - Minimise contamination in hot-rolling. - Provide fine grain size with a minimum of preferred orientation.
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Continuous casting and hot rolling

• Metal is melted, cast and hot rolled continuously through a series of rolling mills within the same process. • Usually for steel sheet production.
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Thread rolling
• Dies are pressed against the surface of cylindrical blank. As the blank rolls against the in-feeding die faces, the material is displaced to form the roots of the thread, and the displaced material flows radially outward to form the thread's crest. • A blank is fed between two grooved die plates to form the threads. • The thread is formed by the axial flow of material in the work piece. The grain structure of the material is not cut, but is distorted to follow the thread form. • Rolled threads are produced in a single pass at speeds far in excess of those used to cut threads. • The resultant thread is very much stronger than a cut thread. It has a greater resistance to mechanical stress and an increase in fatigue strength. Also the surface is burnished and work hardened.
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Cut thread and rolled thread
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Hot-rolling
Plate rolling

• The first hot-working operation for most steel products is done on the primary roughing mill (blooming, slabbing or cogging mills). • These mills are normally two-high reversing mills with 0.6-1.4 m diameter rolls (designated by size).

www.msm.cam.ac.uk

• The objective is to breakdown the cast ingot into blooms or slabs for subsequent finishing into bars, plate or sheet. • In hot-rolling steel, the slabs are heated initially at 1100 -1300 oC. The temperature in the last finishing stand varies from 700 - 900 oC, but should be above the upper critical temperature to produce uniform equiaxed ferrite grains.
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Example for hot strip mill process www.nzsteel.co.nz Red hot slab 210 mm thick is ready for rolling

Slabs are organised according to rolling schedule

Oxidation scale is removed

Slab is reduced to a long strip approx 25 mm thick

Mill reverses after each pass (5 or 7) and the roll gap is reduced each time

The strip is coiled and uncoiled to make the tail end lead Leading edge and tail end are removed

The strip is progressively reduced to the required thicknesses

Plate 12 to 30 mm thick

Coiled steel 1.8 to 12 mm thk 910 mm to 1550 mm wide

Strip is coiled and up ended or passed through if heavy plate

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• Flat plate of large thickness (10-50 mm) is passed through different set of working rolls, while each set consecutively reduces thickness. www.uksteel.org.uk www.reverecopper.com

Plate rolling

• Hot strip is coiled to reduce its increasing length due to a reduction of thickness. • Reducing the complication of controlling strips of different speeds due to different thicknesses. (thinner section moves faster)
Hot rolled coil produced on strip mill
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Cold-rolling
• Cold rolling is carried out under recrystallisation temperature and introduces work hardening.

• The starting material for cold-rolled steel sheet is pickled hot-rolled breakdown coil from the continuous hot-strip mill. www.williamsonir.com • The total reduction achieved by cold-rolling generally will vary from about 50 to 90%. • The reduction in each stand should be distributed uniformly without falling much below the maximum reduction for each pass. • Generally the lowest percentage reduction is taken place in the last pass to permit better control of flatness, gage, and surface finish.
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Example for cold strip mill process.
In batches of 9 coils, cold rolled steel is annealed to reduce work hardening

www.nzsteel.co.nz

Coils are transferred to and from the annealing furnace

Cold rolling mill
20 tonne coils of cold rolled steel are dispatched to Metal Coating Line

The 6 roll configuration enables this high speed mill to produce steel of high quality with consistent shape and flatness

The combination mill has a dual function, cold rolling and single pass temper rolling Cold rolling reduces the thickness and increases the strength of hot rolled steel. The surface finish and shape improve and work hardening results Temper rolling improves the shape of the strip after its workability has been improved by annealing

Cold rolled, annealed and tempered coils are transferred to the Cold Finishing Section

Coils up to 40 tones enter on a conveyor from the pickle line

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Cold-rolling
• Cold rolling provide products with superior surface finish (due to low temperature no oxide scales) • Better dimensional tolerances compared with hot-rolled products due to less thermal expansion. • Cold-rolled nonferrous sheet may be produced from hot-rolled strip, or in the case of certain copper alloys it is cold-rolled directly from the cast state. Cold rolled metals are rated as ‘temper’
Cold rolled strips

• Skin rolled : Metal undergoes the least rolling ~ 0.5-1% harden, still more workable. • Quarter hard : Higher amount of deformation. Can be bent normal to rolling direction without fracturing • Half hard : Can be bent up to 90o. • Full hard : Metal is compressed by 50% with no cracking. Can be bent up to 45o.
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Fundamental concept of metal rolling
Assumptions
1) The arc of contact between the rolls and the metal is a part of a circle. 2) The coefficient of friction, µ, is constant in theory, but in reality µ varies along the arc of contact. o R y y

α

3) The metal is considered to deform plastically during rolling. vf hf 4) The volume of metal is constant before and after rolling. In practical the volume might decrease a little bit due to close-up of pores. 5) The velocity of the rolls is assumed to be constant. 6) The metal only extends in the rolling direction and no extension in the width of the material. 7) The cross sectional area normal to the rolling direction is not distorted.

ho

vo

x x

Lp o Suranaree University of Technology

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Forces and geometrical relationships in rolling o R

α

• A metal sheet with a thickness ho enters the rolls at the entrance plane xx with a velocity vo. vf • It passes through the roll gap and leaves the exit plane yy with a reduced thickness hf and at a velocity vf. • Given that there is no increase in width, the vertical compression of the metal is translated into an elongation in the rolling direction. • Since there is no change in metal volume at a given point per unit time throughout the process, therefore

ho

vo

x x

y y

hf

Lp o bho vo = bhv = bh f v f

…Eq.1

Where b is the width of the sheet v is the velocity at any thickness h intermediate between ho and hf.
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From Eq.1 R x ho vo x’ y’ vo < vf y vf hf Given that bo = bf

bho vo = bh f v f

h

Lo o t

=h

Lf f t

Then we have

vo ho = v f h f
When ho > hf , we then have vo < vf The velocity of the sheet must steadily increase from entrance to exit such that a vertical element in the sheet remain undistorted.

vo h f = ho vf

…Eq.2

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• At only one point along the surface of contact between the roll and the sheet, two forces act on the metal: 1) a radial force Pr and 2) a tangential frictional force F. • If the surface velocity of the roll vr equal to the velocity of the sheet, this point is called neutral point or no-slip point. For example, point N. • Between the entrance plane (xx) and the neutral point the sheet is moving slower than the roll surface, and the tangential frictional force, F, act in the direction (see Fig) to draw the metal into the roll. • On the exit side (yy) of the neutral point, the sheet moves faster than the roll surface. The direction of the frictional fore is then reversed and oppose the delivery of the sheet from the rolls.
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N point : vroll = vsheet α θ β x

Friction acts in opposite directions

Pr x F N

y y

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Pr is the radial force, with a vertical component P (rolling load - the load with which the rolls press against the metal). The specific roll pressure, p, is the rolling load divided by the contact area.

p=

P bL p

…Eq.3

Where b is the width of the sheet. Lp is the projected length of the arc of contact.

 (ho − h f )2  L p =  R(ho − h f ) −  4     L p ≈ R∆h
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12

≈ R (ho − h f )

[

]

12

…Eq.4

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R vo p A B Friction hill in rolling N hf

• The distribution of roll pressure along the arc of contact shows that the pressure rises to a maximum at the neutral point and then falls off. • The pressure distribution does not come to a sharp peak at the neutral point, which indicates that the neutral point is not really a line on the roll surface but an area. • The area under the curve is proportional to the rolling load. • The area in shade represents the force required to overcome frictional forces between the roll and the sheet. • The area under the dashed line AB represents the force required to deform the metal in plane homogeneous compression.

ho

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Roll bite condition
The angle of bite or the angle α of contact Prsineα α Fcosα α

For the workpiece to enter the throat of the roll, the component of the friction force must be equal to or greater than the horizontal component of the normal force.

F cos α ≥ Pr sin α F sin α ≥ ≥ tan α Pr cos α
But we know

Pr

F

F = µPr

Therefore F Pr is a tangential friction force is radial force

µ = tan α

…Eq.5

• If tan α > µ, the workpiece cannot be drawn. • If µ = 0, rolling cannot occur.

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Therefore Free engagement will occur when µ > tan α

Increase the effective values of µ , for example grooving the rolls parallel to the roll axis.

+

Using big rolls to reduce tan α or if the roll diameter is fixed, reduce the ho

α1>α2 α1

α2

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The maximum reduction
The critical variables are Lp and h R B ho A α R-a C D From triangle ABC, we have

R 2 = L2p + ( R − a ) 2 L2p = R 2 − ( R 2 − 2 Ra + a 2 ) L2p = 2 Ra − a 2

a h

vo

Lp

As a is much smaller than R, we can then ignore a2.

L p ≈ 2 Ra ≈ R∆h …Eq.6
A large diameter roll will permit a thicker slab to enter the rolls than will a small-diameter roll. Where ∆h = ho – hf = 2a

R∆h µ = tan α = ≈ ≈ R − ∆h / 2 R − ∆h / 2 Lp
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∆h R
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(∆h )max = µ 2 R

…Eq.7

Jan-Mar 2007

Problem with roll flattening
When high forces generated in rolling are transmitted to the workpiece through the rolls, there are two major types of elastic distortions: 1) The rolls tends to bend along their length because the workpiece tends to separate them while they are restrained at their ends. thickness variation. 2) The rolls flatten in the region where they contact the workpiece. The radius R’. (roll flattening) of the curvature is increased R According to analysis by Hitchcock,

Rollling R R‘

 CP '  R ' = R 1 +  b(ho − h f )    

Roll flattening

Where C = 16(1-ν2)/πE = 2.16 x 10-11 Pa-1 for steel rolls. P’ = rolling load based on the deformed roll radius.
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Example: Determine the maximum possible reduction for coldrolling a 300 mm-thick slab when µ = 0.08 and the roll diameter is 600 mm. What is the maximum reduction on the same mill for hot rolling when µ = 0.5? From Eq.7, For cold-rolling

(∆h )max = µ 2 R
(∆h )max = (0.08)2 (300) = 1.92mm (∆h )max = (0.5)2 (300) = 75mm
Lp R∆h , α = tan −1 (µ ) R

For hot-rolling

Alternatively, we can use the relationship below

sin α =

R ∆h = 1.92mm

=

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Simplified analysis of rolling load
The main variables in rolling are: • The roll diameter. • The deformation resistance of the metal as influenced by metallurgy, temperature and strain rate. • The friction between the rolls and the workpiece. • The presence of the front tension and/or back tension in the plane of the sheet. We consider in three conditions:

1) No friction condition 2) Normal friction condition 3) Sticky friction condition
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1) No friction situation
In the case of no friction situation, the rolling load (P) is given by the roll pressure (p) times the area of contact between the metal and the rolls (bLp).

' P = pbL p = σ o b R∆h

…Eq.8

Where the roll pressure (p) is the yield stress in plane strain when there is no change in the width (b) of the sheet.

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2) Normal friction situation
In the normal case of friction situation in plane strain, the average pressure p can be calculated as.
_

p

σo
Where Q

_ '

1 Q = e −1 Q

(

)

…Eq.9

= µLp/h
= the mean thickness between entry and exit from the rolls.
_

h
From Eq.8,

P = p bL p
2 3
_

We have

P=

σo

1 Q  e − 1 b R∆h  Q 

(

)

…Eq.10

Roll diameter
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Rolling load
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•Therefore the rolling load P increases with the roll radius R1/2, depending on the contribution from the friction hill. • The rolling load also increases as the sheet entering the rolls becomes thinner (due to the term eQ). • At one point, no further reduction in thickness can be achieved if the deformation resistance of the sheet is greater than the roll pressure. The rolls in contact with the sheet are both severely elastically deformed. • Small-diameter rolls which are properly stiffened against deflection by backup rolls can produce a greater reduction before roll flattening become significant and no further reduction of the sheet is possible. Backup rolls Example: the rolling of aluminium cooking foil. Roll diameter < 10 mm with as many as 18 backing rolls.

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• Frictional force is needed to pull the metal into the rolls and responsible for a large portion of the rolling load.

• High friction results in high rolling load, a steep friction hill and great tendency for edge cracking. • The friction varies from point to point along the contact arc of the roll. However it is very difficult to measure this variation in µ, all theory of rolling are forced to assume a constant coefficient of friction. • For cold-rolling with lubricants, µ ~ 0.05 – 0.10. • For hot-rolling , µ ~ 0.2 up to sticky condition.

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Example: Calculate the rolling load if steel sheet is hot rolled 30% from a 40 mm-thick slab using a 900 mm-diameter roll. The slab is 760 mm wide. Assume µ = 0.30. The plane-strain flow stress is 140 MPa at entrance and 200 MPa at the exit from the roll gap due to the increasing velocity.

ho − h f ho

x100 = 30% x100 = 30

h=
Q=
' o

_

ho + h f 2 µL p
_

=

(40) + (28) = 34mm 2
_

(40) − (h f ) (40) h f = 28mm

=

µ R∆h h =

h

(0.30) 450 x12 = 0.65 (34)

∆h = ho − h f = (40) − (28) = 12mm
From Eq.10
' o

σ =

σ ' entrance + σ 'exit
2

=

140 + 200 = 170 MPa 2

1 Q  P = σ  (e − 1)b R∆h  Q   1  P = 170  e 0.65 − 1 (0.76) 0.45 x0.012  = 13.4 MN  (0.65) 

(

)

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3) Sticky friction situation
What would be the rolling load if sticky friction occurs? Continuing the analogy with compression in plane strain

 L  a  '  p p =σ  + 1 = σ 0 _ + 1    2h   4h 
_ ' o

From Eq.8, From example;
' o

P = p bL p

_

 R∆h   P =σ + 1b R∆h _    4h   0.45 x0.012   P = 170 + 1(0.76) 0.45 x0.012   4 x0.034  P = 14.6MN
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Example: The previous example neglected the influence of roll flattening under very high rolling loads. If the deformed radius R’ of a roll under load is given in Eq.11, using C = 2.16x10-11 Pa-1, P’ =13.4 MPa from previous example.

 CP ' ' R = R 1 +  b(ho − h f 

  ) 

Where C = 16(1-ν2)/πE ,P’ = Rolling load …Eq.11

based on the deformed roll radius.
We now use R’ to calculate a new value of P’ and in turn another value of R’

 2.16 x10 −11 13.4 x10 6  ' R = 0.451 +  = 0.464m 0.76 x0.012  

(

)

Q=

µ R∆h
_

The difference between h the two estimations of R’  1  P '' = 170 e 0.66 − 1 0.76 0.464 x0.012  = 13.7 MN is not large, so we stop  0.66  the calculation at this point.  2.16 x10 −11 (13.7 x10 6 )  '' R = 0.451 +  = 0.465m 0.76 x0.012  

=

0.30 464 x12 = 0.66 34

(

)

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Relationship of µ, rolling load and torque
N point : vroll = vsheet α θ β x
Friction acts in opposite directions

• We have known that the location of the neutral point N is where the direction of the friction force changes. • If back tension is applied gradually to the sheet, the neutral point N shifts toward the exit plane. • The total rolling load P and torque MT (per unit of width b) is given by

Pr x FN
Lp

y y

Where µ is obtained by measuring the torque and the rolling load at constant roll speed and reduction with the proper back tension.

P = b

Lp

∫ pdx
0 Lp

MT = b thus

(µpdx )R = µR ∫ pdx = µR P ∫ o o

Lp

b

µ=
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MT PR
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Back and front tensions in sheet
• The presence of back and front tensions in the plane of the sheet reduces the rolling load. • Back tension may be produced by controlling the speed of the uncoiler relative to the roll speed. • Front tension may be created by controlling the coiler. • Back tension is ~ twice as effective in reducing the rolling load P as front tension. • The effect of sheet tension on reducing rolling pressure p can be shown simply by
Back tension, σb Uncoiler Front tension, σf Coiler

p

σ’o σ’o- σb

σ ’o

σ’o- σf

• If a high enough back tension is applied, the neutral point moves toward the roll exit –> rolls are moving faster than the metal. • If the front tension is used, the neutral point will move toward the roll entrance.

p = σ −σ h =
' o

2 3

σ o − σ h …Eq.11

_

Where σh = horizontal sheet tension.
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Problems and defects in rolled products
Defects from cast ingot before rolling
Defects other than cracks can result from defects introduced during the ingot stage of production. • Porosity, cavity, blow hole occurred in the cast ingot will be closed up during the rolling process. • Longitudinal stringers of non-metallic inclusions or pearlite banding are related to melting and solidification practices. In severe cases, these defects can lead to laminations which drastically reduce the strength in the thickness direction.

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Defects during rolling
There are two aspects to the problem of the shape of a sheet. 1) Uniform thickness over the width and thickness – can be precisely controlled with modern gage control system.

h

h

2) Flatness – difficult to measure accurately.

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Jan-Mar 2007

Uniform thickness
• Under high rolling forces, the rolls flatten and bend, and the entire mill is elastically distorted. • Mill spring causes the thickness of the sheet exiting from the rolling mill to be greater than the roll gap set under no-load conditions. • Precise thickness rolling requires the elastic constant of the mill. Calibration curves are needed, see Fig. (1–3 GNm-1 for screw-loaded rolling mills, 4 GNm-1 for hydraulically loaded mills).

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

• Roll flattening increases the roll pressure and eventually causes the rolls to deform more easily than the metal. •The limiting thickness is nearly proportional to µ, R, σ’o but inversely proportional to E. For example in steel rolls the limiting thickness is given by

hmin =

µR σ o
12.8

_ '

…Eq.12

In general, problems with limiting gauge can be expected when the sheet thickness is below 1/400 to 1/600 of the roll diameter.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Flatness
• The roll gap must be perfectly parallel to produce sheets/plates with equal thickness at both ends. • The rolling speed is very sensitive to flatness. A difference in elongation of one part in 10,000 between different locations in the sheet can cause waviness. g n lin tio ol R irec d

Perfectly flat

More elongated along edges

More elongated in the centre

Suranaree University of Technology

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Jan-Mar 2007

Solutions to flatness problems
• Camber and crown can be used to correct the roll deflection (at only one value of the roll force). Or use rolling mill equipped with hydraulic jacks to permit the elastic distortion of the rolls to correct deflection.

(a)

(b)

(a) The use of cambered rolls to compensate for roll bending. (b) Uncambered rolls give variation of thickness.

Suranaree University of Technology

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Jan-Mar 2007

•The roll cross angle of rolls incorporated in a stand of each rolling mill is set at a predetermined value beforehand. • If there is a roll cross angle that will enable a target sheet crown to be applied to each sheet and the roll bender load of each stand is adjusted on-line, thereby effecting sheet crown control.

• Hot mill can be provided with facilities for crown control to improve the control of the profile of hot strip mill. • For example work roll bending with continuous variable crown and pair cross mills.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Possible effects when rolling with insufficient camber

(a) (c) (d) (e)

(b)

• Thicker centre means the edges would be plastically elongated more than the centre, normally called long edges. • This induces the residual stress pattern of compression at the edges and tension along the centreline. • This can cause centreline cracking (c), warping (d) or edge wrinkling or crepe-paper effect or wavy edge (e).

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Possible effects when rolls are over-cambered.

(a) (c) (d) (e)

(b)

• Thicker edges than the centre means the centre would be plastically elongated more than the edges, resulting in lateral spread. •The residual stress pattern is now under compression in the centreline and tension at the edges (b). • This may cause edge cracking (c), centre splitting (d), centreline wrinkling (e).

Suranaree University of Technology

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Jan-Mar 2007

• Shape problems are greatest when rolling in thin strip (<0.01 in) because fractional errors in the roll gap profile increase with decrease in thickness, producing larger internal stress. • Thin sheet is also less resistant to buckling. • Mild shape problems may be corrected by stretch levelling the sheet in tension or by bend flexing the sheet in a roller-leveller, see Fig. two high Material flow

adjustable

work rolls

Roller-leveller

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Jan-Mar 2007

• Edging can also be caused by inhomogeneous deformation in the thickness direction. • If only the surface of the workpiece is deformed (as in a light reduction on a thick slab), the edges are concaved (a). The overhanging material is not compressed in the subsequent step of rolling, causing this area under tensile stress and leading to edge cracking. This has been observed in initial breakdown of hot-rolling when h/Lp > 2 • With heavy reduction, the centre tends to expand more laterally than the surface to produced barrelled edges (b). This causes secondary tensile stresses by barrelling, which are susceptible to edge cracking. • Alligatoring (c) will occur when lateral spread is greater in the centre than the surface (surface in tension, centre in compression) and with the presence of metallurgical weakness along the centreline.
Suranaree University of Technology
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Jan-Mar 2007

• Surface defects are more easily in rolling due to high surface to volume ratio. Grinding , chipping or descaling of defects on the surface of cast ingots or billets are recommended before being rolled. • Laps due to misplace of rolls can cause undesired shapes. Roll misplacement

• Flakes or cooling cracks along edges result in decreased ductility in hot rolling such as blooming of extra coarse grained ingot. • Scratches due to tooling and handling. • Variation in thickness due to deflection of rolls or rolling speed.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Rolling mill control
• Modern continuous hot-strip and cold rolling mills operated under automatic control provide high throughput and production rate. • Of all the metal working processes, rolling is the best suited for the adoption of automatic control because it is an essentially steady-state process in which the tooling geometry (roll gap) may be changed readily during the process. • Automatic control in rolling such as the development of online sensors to continuously measure sheet thickness. The most widely used instruments are 1) flying micrometer 2) x-ray or isotope, gauges which measure thickness by monitoring the amount of radiation transmitted through the sheet. • More recently control procedures have been aimed at controlling strip shape as well as thickness.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Problem of gauge control
Normal situation Plastic curve Elastic curve P • For a given set of rolling conditions, the rolling load varies with the final sheet thickness, according to the plastic curve, which can be obtained in Eq.10.

Rolling load

1 Q  P= σ o  e − 1 b R∆h  3 Q  2
_

(

)

Roll gap

hf

ho

Sheet thickness

Characteristic elastic and plastic curves for a rolling mill

• The elastic curve for mill spring indicates that a sheet of initial thickness ho will have a final thickness hf and the load on the mill would be P in normal situation.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Situation where µ or flow stress increase Increase in µ or σ
P3 P2 P1

µ σ

due to lubricant breakdown due to Temp

Rolling load

Elastic curve The plastic curve will be raised. Therefore rolling load P1 final thickness hf1

P2 hf2

New roll gap

hf1hf2 ho
Sheet thickness

Use of characteristic curves to show changes in rolling condition.

In order to maintain a constant thickness hf1, under these new condition, the roll gap would have to be decreased. This moves the elastic curve to the left and further increases the rolling load to P3.

Example: • If the sheet thickness increases, the plastic curve will move to the right relative to the elastic curve. • If the there is an increase in strip tension, the plastic curve will move to the left.
Suranaree University of Technology
Tapany Udomphol

Jan-Mar 2007

Thickness measurement in continuous hot mill

• In a continuous hot mill, the strip thickness is measured indirectly by measuring the rolling load and using the characteristic curve of the mill to establish the thickness. • The error signal is fedback to the rolling mill screws to reposition them so as to minimise the error. • An x-ray gauge is used after the last stand to provide an absolute measurement of sheet gauge.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Thickness measurement in continuous cold strip mills
• Thickness is measured by x-ray gauges while the error in the thickness following the first stand is usually fedback to adjust the gap sitting on the first stand. • Gauge control in subsequent stands usually is achieved by controlling the strip tension through controlling the relative roll speed in successive stands or the coiler speed. • Gauge control through control of strip tension has faster response time than control through change in roll setting.

sensors Thickness gauging is achieved by using two opposing sensors with laser spots aimed at opposite sides of a target. The sensor readings are subtracted from the sensor separation distance to yield a real-time thickness measurement.
Jan-Mar 2007

Suranaree University of Technology

Tapany Udomphol

Theory of cold rolling
A theory of rolling is aimed at expressing the external forces, such as the rolling load and the rolling torque, in terms of the geometry of the deformation and the strength properties of the material being rolled. Assumptions 1) The arc of the contact is circular – no elastic deformation of the roll. 2) The coefficient of friction is constant at all points on the arc of contact. 3) There is no lateral spread, so that rolling can be considered a problem in plain strain. 4) Plane vertical section remain plane: i.e., the deformation is homogeneous. 5) The peripheral velocity of the rolls is constant 6) The elastic deformation of the sheet is negligible in comparison with the plastic deformation. 7) The distortion-energy criterion of yielding, for plane strain, holds.

σ1 − σ 3 =
Suranaree University of Technology

2 3

' σo =σo

Yield stress in plane strain condition
Jan-Mar 2007

Tapany Udomphol

The stresses acting on an element of strip in the roll gap o µPr
Prcosθ

ho

h+dh h

hf

µPrcosθ
Pr

θ

Prsinθ

B

R dθ

• At any point of contact between the strip and the roll surface, designated by the angle θ, the stresses are the radial pressure pr and the tangential shearing stress τ = µpr. These stresses are resolved into their horizontal and vertical components (b). • The stress σx is assumed to be uniformly distributed over the vertical faces of the element.
Suranaree University of Technology Jan-Mar 2007

Tapany Udomphol

µPrsinθ



θ

σx+dσx

σx

Rd

θ
Rd

Prcosθ

Prsinθ

θ

h+dh σx+dσx

µPrsinθ

µPr
Prcosθ

θ Pr

µPrcosθ

µPrcosθRdθ

Prsinθ
R dθ

Taking summation of the horizontal forces on the element results in

(σ x + dσ x )(h + dh ) + 2µp r cos θRdθ = σ x h + 2 p r sin θRdθ
Which simplifies to

d (σ x h ) = 2 p r R(sin θ ± µ cos θ ) dθ
Tapany Udomphol

U ni tl en gt h

Suranaree University of Technology

θ θ θ θ

Pr

µPrcosθ

Pr

µ µ θ θ θ µPrsinθ µ

µPrcosθRdθ
(σx+dσx)(h+dh)

PrsinθRdθ θ θ σxh

σx

h

PrsinθRdθ θ θ

…Eq.14
Jan-Mar 2007

The forces acting in the vertical direction are balanced by the specific roll pressure p. Taking the equilibrium of forces in the vertical direction results in a relationship between the normal pressure and the radial pressure.

p = p r (1 m µ tan θ )

…Eq.14

The relationship between the normal pressure and the horizontal compressive stress σx is given by the distortion energy criterion of yielding for plane strain.

σ1 − σ 3 =

2 3

' σo =σo

' p −σ x = σo

…Eq.15

Where p is the greater of the two compressive principal stresses.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

The solution of problems in cold rolling are complicated. Some simplification to this problem has been provided by Bland and Ford. By restricting the analysis to cold rolling under conditions of low friction and for angles of contact < 6o, then we can put sin θ ~ θ and cos θ ~ 1. Thus Eq.14 can be written

d (σ d h )h) = 2 p R (θ '± µ ) x (σ dθ = 2 p r R (θ ± µ ) dθ x ' r

…Eq.16

It is also assumed that pr ~ p , so that Eq.15 can be written σx = pr-σ’o. By substituting into Eq.16 and integrating, relatively simple equations of the radial pressure result. Roll entrance to neutral point: Neutral point to roll exit:

σ  1 − xb e µ ( H − H ) …Eq.17 pr = ' ho  σ 01   
1

' σ oh 

σ xf  µH 1 − '  e pr = h f  σ 02   
' σ oh 

…Eq.18

Where

 R' H = 2 h  f

   

1

2

 ' −1  R tan  h f 

1  2   θ    

and σxb = back tension σxf = front tension
Tapany Udomphol

Suranaree University of Technology

Jan-Mar 2007

The rolling load or total force P is the integral of the specific roll pressure over the arc of contact. θ =α

P = R 'b

∫ pdθ o …Eq.19

Where

b = width of sheet

α = contact angle

The solution is replaced by the modern digital computer.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

Theory of hot-rolling
In hot working processes, the flow stress for hot-rolling is a function of both temperature and strain rate (speed of rolls)

Calculation of rolling load by Sims

P = σ b R (ho − h f
' o

[

)]

1

2

Qp

…Eq.20

Where Qp is a complex function of the reduction in thickness and the ratio R/hf. Values of Qp may be obtained from

Qp =

2 ho  hn R ∆h −1 − ln π tan 4∆h  hf hf ho h f 

 π −  4 

…Eq.21
Suranaree University of Technology
Tapany Udomphol

Jan-Mar 2007

Torque and power
Torque is the measure of the force applied to a member to produce rotational motion. Power is applied to a rolling mill by applying a torque to the rolls and by means of strip tension.

The power is spent principally in four ways 1) 2) 3) 4) The energy needed to deform the metal. The energy needed to overcome the frictional force. The power lost in the pinions and power-transmission system. Electrical losses in the various motors and generators.

Remarks: Losses in the windup reel and uncoiler must also be considered.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

The total rolling load is distributed over the arc of contact in the typical friction-hill pressure distribution. However the total rolling load can be assumed to be concentrated at a point along the act of contact at a distance a from the line of centres of the rolls. The ratio of the moment arm a to the projected length of the act of contact Lp can be given as

a R P ho P a R

MT = Pa

hf

λ=

a = Lp

a R∆h

…Eq.22

Schematic diagram illustrating roll torque

Where λ is 0.5 for hot-rolling and 0.45 for cold-rolling.
Suranaree University of Technology
Tapany Udomphol

Jan-Mar 2007

The torque MT is equal to the total rolling load P multiplied by the effective moment arm a. Since there are two work rolls, the torque is given by

MT = Pa a R P ho P a R hf

M T = 2 Pa

…Eq.23

During one revolution of the top roll the resultant rolling load P moves along the circumference of a circle equal to 2πa. Since there are two work rolls, the work done W is equal to

Work = 2(2πa) P

…Eq.24

Since power is defined as the rate of doing work, i.e., 1 W = 1 J s-1, the power (in watts) needed to operated a pair of rolls revolving at N Hz (s-1) in deforming metal as it flows through the roll gap is given by

W = 4πaPN
Where P is in newtons and a is in metres.
Suranaree University of Technology
Tapany Udomphol

…Eq.25

Jan-Mar 2007

Example: A 300 mm-wide aluminium alloy strip is hot-rolled in thickness from
20 to 15 mm. The rolls are 1 m in diameter and operate at 100 rpm. The uniaxial flow stress for aluminium alloy can be expressed as σ = 140ε0.2 (MPa). Determine the rolling load and the power required for this hot reduction. From Eq.20
' P = σ o b R (ho − h f ε1 [

)] ε1 1

2

Qp

b = 0.3 m, R = 0.5 m, ho = 0.02 m and hf = 0.015 m, we need to know σ’o and Qp. ε1 o

 20  ε 1 = ln  = 0.288  15  20 − 15 r= = 0.25 20 R 500 = = 33.3 hf 15

' σo =

k ∫ ε n dε o kε n +1 = ε 1 (n + 1)
0.2

kε 1n = n +1

Qp can be found from graph(~1.5) when reduction r and R/hf are known.

140(0.288) σ = 1.2
' o

= 91MPa
1

P=

2 3

(91)(0.3)[0.5(0.020 − 0.015)]

2

(1.5) = 2.36MN

a = 0.5 R∆h = 0.5 0.5 x0.005 = 0.025m N = 100 / 60 = 1.67 s −1 W = 4π (0.025) 2.36 x10 −6 (1.67 )Js −1 = 1.24MW
Suranaree University of Technology
Tapany Udomphol

(

)

Jan-Mar 2007

References
• Dieter, G.E., Mechanical metallurgy, 1988, SI metric edition, McGraw-Hill, ISBN 0-07-100406-8. • Edwards, L. and Endean, M., Manufacturing with materials, 1990, Butterworth Heinemann, ISBN 0-7506-2754-9. • Beddoes, J. and Bibbly M.J., Principles of metal manufacturing process, 1999, Arnold, ISBN 0-470-35241-8. • Lecture note, 2003. • Firth Rixson leavelets. • Metal forming processes, Prof Manas. • Metal forming lecture, Ass Prof. P. Srichareonchai.

Suranaree University of Technology

Tapany Udomphol

Jan-Mar 2007

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