Rolling - me.iitb.ac.inramesh/courses/ME206/Rolling.pdf · reversing rolling mill (shown below) with 750 mm diameter rolls made of tool steel is available for this task. The rolling

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Rolling

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Outline

• Rolling Introduction• Rolling Analysis• Rolling Defects

ME 206: Manufacturing Processes IInstructor: Ramesh Singh; Notes by: Prof. S.N. Melkote / Dr. Colton

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Rolling Introduction

• Process typically used to produce basic shapes e.g. flat plates, sheets, railway tracks, I-sections, etc.

Foil: < 0.2 mmSheet: 0.2 ~ 6 mm

Plate: > 6 mm

Source: http://www/world-aluminum.org/


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• More complex geometries also possible e.g. rings, threads, etc.

Ring Rolling


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• Types of rolling mills2-High Non-Reversing Mill

2-High Reversing Mill

3-High Mill

4-High Mill Cluster Mill


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• Roller materials used– Cast iron– Forged steel

• Important properties of roller materials– Strength– Wear resistance– Modulus of elasticity


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Rolling Analysis

• Objectives

– Find distribution of roll pressure

– Calculate roll separation force (“rolling force”) and torque

– Calculate rolling power


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Flat Rolling Analysis

• Consider rolling of a flat plate in a 2-roll mill

Width of plate w is large à plane strain

h0 hf

R

V0 Vf (> V0)

q


Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton 9

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).

Dh/2

FnFf

Ra

aa

aa sincos nf FF ³

µa =tan

µ = friction coefficient

nf FF ×= µ

Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton

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Processing limits

Rh

R

hR

212cos

D-=

D-

=a

Rh <<D aa 2cos1sin -=

Also

and

2

2211sin ÷

øö

çèæ D-

D+-=

Rh

Rha

0

RhD

»asin

Rh

hRh

Rh

RhRh

D»

D-D

@

÷øö

çèæ D+

D-

D

= 2

21

tana

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So, approximately( )

RhD

== 22tan µa

Hence, maximum draft

Rh 2max µ=D

Maximum angle of acceptance

µaf 1max tan-==

Maximum Draft

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• Friction plays a critical role in enabling rolling à cannot roll without friction; for rolling to occur

• Reversal of frictional forces at neutral plane (NN)

h0 hf

V0 Vf (> V0)a

L

Entry Zone Exit Zone

N

N

( )0 fL R h h R h- = D!

Projected contact length, L

tanµ a³


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h0 hf

fN

Ndx

p

p

µp

µp

sxsx + dsx

Stresses on Slab in Entry Zone

Stresses on Slab in Exit Zone

p

pµp

µp

sxsx + dsx


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p

p

µp

µp

sxsx + dsx

Stresses on Slab in Entry Zone Stresses on Slab in Exit Zone

p

pµp

µp

sxsx + dsx

Using slab analysis we can derive roll pressure distributions for the entry and exit zones as:

12 tanf f

R RHh h

f-æ ö

= ç ÷ç ÷è ø

( )0

0

23

H Hfhp Y eh

µ -=23

Hf

f

hp Y eh

µ=

Entry Zone Exit Zoneaf == @0 HH


ME 4210: Manufacturing Processes & EngineeringInstructor: Ramesh Singh Notes by: Prof. S.N. Melkote

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Effect of Coeff. of Friction Effect of % Reduction

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• Rolling force (also called roll-separating force), F = (area under the pressure vs. contact length curve) x (width of sheet)

• Mathematically,

Where fN is the location of the neutral plane

0 0

N

N

exit entryF pdA wp Rd wp Rdfa a

f

f f= = +ò ò ò


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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

materialpull-in

vovf

vrvrvr

materialpull-back


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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

n

n

f

HHexpHexp

)HH(exphh 20

00 -=-

= µµ

µ


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Neutral Point

÷÷ø

öççè

æ-=÷

÷ø

öççè

æ=

fN

fNfN h

hHHRhH

Rh 0

0 ln121 ,

2tan

µf


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Flat Rolling Analysis• Average rolling force,

Where w is the width of sheet and havg is the average thickness of sheet = 0.5(h0 + hf)

úúû

ù

êêë

é÷÷ø

öççè

æ+==

avgfavgavg h

LYLwLwpF2

132 µ

average flow stress:due to shape of element

1

10

1

+=

+== ò

+

nKY

nKdKY

n

f

nn

f

e

eeeee


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Torque

R

Froller

L/2

paveA

hRL D»

Dh = hb - hf

ApF

F

averoller

y

=\

=å 0

22/ LF

FL

FrrollerTorque rollerrollerroller =×=×=

fµfµfa

f

dpRwdpRwTn

n

òò -=0

22

entry exit


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Torque

• Rolling torque per roll,T = Fa

where a is the moment arm for the roll forcea » L/2 for hot rolling, a » 0.4L for cold rolling


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• Rolling power in KW per roll,

Where N is the roll speed in rpm, F is in N and L, a are in m.

( )( )60000600002

600002/LFN/aFNFa/NTP

pppw

====


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Example Problem

A 75 mm thick by 250 mm wide slab of AISI 4135 steel is being cold-rolled to a thickness of 60 mm in a single pass. A two-high non-reversing rolling mill (shown below) with 750 mm diameter rolls made of tool steel is available for this task. The rolling mill has a power capacity of 5 MW per roll. The rolls rotate at a constant angular speed of 100 rev/min. The steel work material has the following flow curve at the rolling temperature: MPa. Assume the coefficient of friction µ = 0.2. Is the available rolling mill adequate for the desired operation?

w = 250 mm

hi=75 mm

hf = 60 mm

14.01100 tt es =


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Example Problem

The rolling mill is adequate if the required power for the operation is less than or equal to the available power.

Power required per roll, 60000

NLFProll

¢p= kW

Roll separating force, ( ) ÷÷ø

öççè

æ+÷÷

ø

öççè

æ=

avgfavg h

LYLwF

21

32 µ

( )( ) 075.006.0075.0375.0 =-=D= hRL m

0675.02

06.0075.0=

+=avgh m


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Example Problem

96.77814.1

6075ln1100

1

14.0

0

=÷øö

çèæ

=ese

= òe t

ttt

f dY MPa

( )( ) ( )( )( ) 74.18

0675.02075.02.0196.778

3225.0075.0 =÷÷

ø

öççè

æ+÷

ø

öçè

æ=avgF MN

Therefore, ( )( )( ) 36.7

60000100075.01074.18 6

=´p

=rollP MW

Since 5>rollP MW (the available power), the rolling mill is not adequate for the operation.


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Rolling Forces• Ways to reduce rolling forces

– Lowering friction via use of lubricants

– Using smaller diameter rolls

– Taking smaller reductions

– Increasing workpiece temperature

– Use of front and/or back tension


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Front & Back Tension

Lowers apparent yield strength of material à lowers roll force


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Rolling Defects• Roll bending• Remedy: roll camber

Cambered Rolls Un-Cambered Rolls

Source: Metal Forming, W.F. Hosford & R.M. Caddell, 1993 ME 206: Manufacturing Processes I

Instructor: Ramesh Singh; Notes by: Prof. S.N. Melkote / Dr. Colton

ME 4210: Manufacturing Processes & EngineeringInstructor: Ramesh Singh Notes by: Prof. S.N. Melkote

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Rolling Defects

• Un-cambered or under-cambered rolls

Residual Stresses

Centerline Cracking

Warping

Edge Wrinkling

Un-cambered/Under-cambered Rolls

Source: Metal Forming, W.F. Hosford & R.M. Caddell, 1993

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Rolling Defects

• Over-cambered rolls

Un-cambered/Under-cambered Rolls

Source: Metal Forming, W.F. Hosford & R.M. Caddell, 1993

Residual Stresses

Edge Cracking

Splitting

Centerline Wrinkling


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Rolling Defects

• Roll flattening: similar to tyre flattening

– Due to elasticity of rolls

– Increases roll radius and hence rolling force

• Remedy– Choose rolls with high elastic modulus

– Reduce rolling force


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Summary

• Rolling Introduction– Rolling mill types

• Rolling analysis– Roll pressure, rolling force, torque

• Rolling defects– Roll bending, flattening, etc.


Rolling - me.iitb.ac.inramesh/courses/ME206/Rolling.pdf · reversing rolling mill (shown below) with 750 mm diameter rolls made of tool steel is available for this task. The rolling

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