DeformaonProcesses* Rolling*...

Deforma(on Processes Rolling Rolling –reducing the thickness or changing the cross-‐sec4on of a long workpiece by compressive forces applied through a set of rolls Accounts for 90% of all metals produced by metal working processes OAen carried out at elevated temperatures first (hot rolling).

Hot Rolling It changes the coarse-‐grained, briHle, and porous ingot structures to wrought structures with finer grain sizes and enhanced proper4es. It does not provide precise final dimensions or good surface finishes. Cold Rolling

Homologous Temperature Ranges (Kelvin)

Process T/Tmel4ng

Cold Working <0.3

Warm Working 0.3 to 0.5

Hot Working >0.6

It provides a large number of semi-‐finished products, such as sheets, strips and foils. It provides a precise control of product dimensions and good surface finishes. The material work hardens, i.e. increased strengths.

Dynamic recrystalliza4on of the deformed grains occurs during hot-‐rolling.

Terminology Cross sec4onal Area (CSA)

Semi-‐finished Products

Bloom Product of the first breakdown of ingot (CSA>230cm2)

Billet Bloom with further reduc4on by hot rolling (CSA 40x40mm2)

Slab Hot rolled ingot (CSA>100cm2 with a width >2 x thickness)

Mill Products

Plate Thickness > 6mm (examples: boiler support – 0.3m ; reactor vessels – 150mm ; baHleships and tanks – 100-‐125mm)

Sheet Thickness < 6mm and width > 600mm (examples Boeing 747 skin thickness – 1.8mm ; Aluminum beverage cans – starts as sheets – 0.28mm thick and it is reduced aAer deep drawing to 0.1mm)

Strip Thickness < 6mm and width < 600mm

Rolling Opera(ons

Deforma4on process in which work thickness is reduced by compressive forces exerted by two opposing rolls

Flat Rolling

Rota4ng rolls perform two main func4ons: §  Pull the work into the gap between them by fric4on between work-‐piece and rolls §  Compresses the work to reduce its cross sec4on

The material is constrained in the z-‐direc4on, i.e the material is not extended in the width direc4on. The material gets longer and not wider. This condi4on is known as plane strain. Fric4on plays a cri4cal role in enabling rolling µ > tan α Rolling can not occur without fric4on.

Flat Rolling Analysis

Ini4al thickness ho Final thickness hf Roll gap L Surface speed of rolls Vr Entry velocity of the strip Vo Final velocity of the strip Vf Neutral point : No slip point – point along contact length where the velocity of the strip = velocity of the roll


Rolling can not take place without fric4on, for rolling to occur µ>tan(α)

Stresses in the slab on the entry-‐zone

Stresses in the slab on the exit-‐zone

σ x +δσ x( ) ⋅ h+δh( )− 2pR ⋅δφ ⋅sinφ ± 2µpR ⋅δφ ⋅cosφ −σ xh = 0h ⋅δσ x +δσ x ⋅δh− 2pR ⋅δφ ⋅sinφ ± 2µpR ⋅δφ ⋅cosφ = 0

σ x +δσ x( ) ⋅ h+δh( )− 2pR ⋅δφ ⋅sinφ ± 2µpR ⋅δφ ⋅cosφ −σ xh = 0h ⋅δσ x +δσ x ⋅δh− 2pR ⋅δφ ⋅sinφ ± 2µpR ⋅δφ ⋅cosφ = 0δ σ xh( )δφ

= 2pR ⋅ sinφ µ cosφ( )

δ σ xh( )δφ

= 2pR ⋅ φ µ( )

The angle φ designates any point of contact between the material and the roll surface. The stresses present are the radial pressure p and the tangen4al stress generated by fric4on µp. The stress σx is assumed to be uniformly distributed in the sec4on.

Simplifying and ignoring the High Order Terms (HOT’s). For very small angles sinΦ=Φ and cosΦ=1


Simplifying and ignoring higher order terms

Restric4ng the analysis for contact angles <6degrees α<<1 , then sin φ = φ and cos φ =1

δ σ xh( )δφ

= 2pR ⋅ sinφ µ ⋅cosφ( )

δ σ xh( )δφ

= 2pR ⋅ φ µ( )For Plane Strain condi4ons and using the Von Mises Criterium

p−σ x =23

"

#$

%

&'Yflow =1.15Yflow =Yflow

'

δ p−Yflow'( )h"

#$%

δφ= 2pR ⋅ φ µ( )

Yflow' ⋅h ⋅ δ

δφp

Yflow'

"

#$$

%

&''+

pYflow' −1

"

#$$

%

&''⋅δδφ

Yflow' ⋅h( ) = 2pR ⋅ φ µ( )

δδφ

Yflow' ⋅

pYflow' −1

#

$%%

&

'((⋅h

)

*++

,

-..= 2pR ⋅ φ µ( )

δδφ

pYflow'

!

"##

$

%&&

pYflow'

!

"##

$

%&&

=2Rh

φ µ( )

δδφ

pYflow'

!

"##

$

%&&

pYflow'

!

"##

$

%&&

'

(

)))))

*

+

,,,,,

∫ = ln pYflow'

'

())

*

+,,=

2Rh

φ µ( )'

()*

+,φ=α

φ=0

∫

Thickness expression as a func4on of the angle

h = hf + 2 ⋅R ⋅ 1− cosφ( )

cosφ =1− φ2

2!+φ 4

4!....

h = hf + R ⋅φ2

Taylor’s series expansion

Subs4tu4ng in the above integral and integra4ng, we obtain

ln pYflow'

!

"##

$

%&&= ln h

R 2µ R

hf⋅ tan−1 φ R

hf

!

"##

$

%&&+ lnC

H = 2 Rhf⋅ tan−1 φ R

hf

!

"##

$

%&&

p =CYflow' hReµH

At entry φ=α, hence H=Hb At exit φ=0, hence H=Hf=0. Also, at entry and at exit p=Y’flow Then:

C = RhbeµHb p = h

hbYflow

' e−µ H−Hb( )

C = Rhf

p = hhfYflow

' eµH

At entry At exit

The pressure at any loca4on is a func4on of h and hence of the angular posi4on φ.

p = hhbYflow

' e−µ H−Hb( )

p = hhfYflow

' eµH

At entry At exit

H = 2 Rhf⋅ tan−1 φ R

hf

#

$%%

&

'((

Hb = 2Rhf⋅ tan−1 α R

hf

#

$%%

&

'((

If front and back tensions are applied, the expressions are modified to:

p = hhb

Yflow' −σ b( )e−µ H−Hb( )

p = hhf

Yflow' −σ f( )eµH

At entry At exit

Back tension to cause slip If the back tension is high enough the rolls will begin to slip, that is the neutral point has moved to the exit. The whole contact area becomes the entry zone. When φ=0 then H=0

p =hfhb

Yflow' −σ b( )eµ Hb( ) =Yflow

'

σ b =Yflow' 1− hb

hf

"

#$$

%

&''e

−µHb

(

)**

+

,--

At the neutral point, the neutral point must be where the two pressure equa4ons are the same , thus equa4ng both expressions

p = hhbYflow

' e−µ Hn−Hb( ) =hhfYflow

' eµHn

hbhf=e−µ Hn−Hb( )

eµHn= eµ Hb−2Hn( ) Hn =

12Hb −

1µ

ln hbhf

"

#$$

%

&''

"

#$$

%

&''

φn =hfR

tanhfR⋅Hn

2

"

#$$

%

&''

Zero slip: At entry, the material is pulled towards the neutral point. At exit, the material is pulled back towards the neutral point.

Calcula7ng Roll Forces (roll separa7ng force):

F = wpRδφ0

φn∫ + wpRδφφn

α

∫

The strip width (w), and R is the radius of the roll. If L is the length of contact and Δh is the difference between the ini4al and final thicknesses of the strip (also called the draA), then:

Another method is calcula4ng the average contact stress

F = Lwpaverage

L = RΔh haverage =hinitial + hfinal

2

h = hf + 2 ⋅R ⋅ 1− cosφ( )h = hf + R ⋅φ

2

L = R ⋅φ hb = hf + R ⋅LR#

$%

&

'(

2

Approxima4ons based on Forging (Von Mises)

paverage =1.15 ⋅Yflow 1+µL

2haverage

"

#$$

%

&''

Small rolls (R small) and small reduc4ons Δ =

haverageL

>>1

paverage

=1.15 ⋅Yflow

Fric4on is small or no significant

For large rolls or large reduc4ons Δ =haverageL

<<1

paverage

=1.15 ⋅Yflow 1+µL

2haverage

#

$%%

&

'((

Force (per unit width) Approxima4on : Low fric4on F = L ⋅w ⋅ p

average=1.15 ⋅Yflow ⋅L ⋅w

Force (per unit width) Approxima4on : High fric4on

F = L ⋅w ⋅ paverage

=1.15 ⋅Yflow 1+µL

2haverage

"

#$$

%

&''⋅L ⋅w

Roll Torque and Power

T = wµpR2 δφ −φn

a∫ wµpR2 δφ

0

φn∫The torque can also be es4mated by considering that the roll force (F – obtained with the average pressure) is perpendicular to the rolling plane and that it acts in the middle of the arc of contact.

T = FL2

Power = Tω

P(kW ) = πFLN60000

P(HP) = πFLN33000

Power per roll. Where: F(N or lb) ; N (RPM) and L (meters – A)

Roll Bite Condi(on The horizontal component of the fric4on force must be equal or greater than the horizontal component of the normal force for the work piece to enter the throat of the roll.

F cosα ≥ Pr sinα⇒FPr≥sinαcosα

= tanα

F = µPr ⇒ µ = tanα

Free engagement will occur when µ>tanα 1.  Increase the effec4ve value of the

coefficient of fric4on µ. For example by grooving the rolls parallel to the roll axis.

2.  Using large diameter rolls to reduce the angle α. If the roll diameter is fixed, then reduce the sheet thickness

Maximum angle of acceptance

Maximum Reduc(on

From triangle ABC, we have

R2 = Lp2 + R− a( )2

Lp2 = R2 − R2 − 2Ra+ a2( )Lp2 = 2Ra− a2 ≅ 2Ra

As a is much smaller than R, a2 can be neglected

Lp ≅ 2Ra ≈ RΔh

µ = tanα =Lp

R−Δh2≈

ΔhR

⇒Δhmax = µ2R Maximum

reduc(on in thickness

Example: Determine the maximum possible reduc4on for cold rolling a 300mm-‐thick slab when the coefficient of fric4on µ=0.08 and the roll diameter is 600mm. What is the maximum reduc4on on the same mill for hot rolling when µ=0.5?

Δhmax = µ2R = 0.08( )2 600

2"

#$

%

&'=1.92mm

Δhmax = µ2R = 0.5( )2 600

2"

#$

%

&'= 75mm

sinα =Lp

D2=

RΔhR

α = tan−1 µ( ) Δh =1.92mm

Example: Es4mate the total power required for a rolling opera4on of a 9in wide 6061-‐O aluminum strip that is rolled from a thickness of 1in to a final thickness of 0.8in. The roller radius is 12in and the roller speed is 100RPM.

Es4mate the true strain: ε = ln tot f

!

"##

$

%&&= ln

1.00.8!

"#

$

%&= 0.223

Y =k ε n dε0

ε1

∫ε1

=kε1

n

n+1

Y =30000 0.223( )0.2

1+ 0.2=18500psi

Y flow'

=1.15×18500 = 21275psi

F = LwY flow'

=1.55×9×21275= 297000lb

L = RΔh = 12× 1.0− 0.8( ) =1.55in

P(hp) = 2πFLN33000

=2×π ×297000× 1.55 12( )×100

33000= 735hp

Roll FlaMening Due to the large forces that are generated, in the contact area the rolls flaHen, and as a result, the radius of the roll is increased. To calculate the separa4ng force, the radius of the rolls are needed. To calculate the new radius due to flaHening, the separa4ng force is needed. The need radius is given by the expression by Hitchcock:

Rflatten = R 1+CF

w hb − hf( )"

#$$

%

&''

Where

C =161−ν 2( )πE

C = 2.3×10−11Pa−1 1.67×10−7 psi−1( ) for steel rolls

C = 4.57×10−11Pa−1 3.15×10−7 psi−1( ) for cast iron rolls

Example:

Suppose the average flow stress is 40 ksi and a sheet is being rolled so that the change in height is 0.25 inches. The width of the roll is 10 inches and the radius of the roller is 10 inches. What is the effec4ve radius R’ of the roller? Roller made of steel.

C =1.67×10−7 psi−1 for steel rolls paverage =1.15 ⋅Yflow =1.15 ⋅ 40000 = 46000psi

L = RΔh = 10 ⋅0.025 = 0.5F = Lwpaverage = 0.5 ⋅10 ⋅ 46000 = 230000lb

Rflatten = R 1+CF

w hb − hf( )"

#$$

%

&''=10 ⋅ 1+1.67 ⋅10

−7 ⋅23000010 ⋅ 0.025( )

"

#$$

%

&''=11.5364in

again L = RΔh = 11.5364 ⋅0.025 = 0.537F = Lwpaverage = 0.537 ⋅10 ⋅ 46000 = 247037lb

Rflatten = R 1+CF

w hb − hf( )"

#$$

%

&''=10 ⋅ 1+1.67 ⋅10

−7 ⋅24703710 ⋅ 0.025( )

"

#$$

%

&''=11.65in

Repeat…… it converges to:

Types of Rolling Process •  Flat Rolling •  Con4nuous Rolling •  Shape Rolling •  Transverse rolling •  Ring Rolling •  Powder Rolling

Residual Stresses may lead to warping aAer bar is machined

Kalpakjian

Example:

Find the separa4ng force and the rolling torque. Data: (a) the width of the rollers are 10in, (b) the ini4al and final thickness are 0.125in and 0.100in respec4vely, (c) the average flow stress is 40ksi, (d) the coefficient of fric4on is 0.1 and the roller radius is 10in.

haverage =0.125+ 0.10

2= 0.1125 Δh = 0.0250in

paverage =1.15 ⋅Yflow 1+ µL2haverage

#

$%%

&

'((=1.15 ⋅ 40000 1+

0.1⋅ 10( ) 0.0250( )2 ⋅0.1125

#

$

%%

&

'

((= 56222.1psi

F = Lwpaverage = 10( ) 0.0250( ) ⋅10 ⋅56222.2lb = 281111.1lb

Rflatten = R 1+CF

w hb − hf( )"

#$$

%

&''=10 ⋅ 1+1.67 ⋅10

−7 ⋅28111110 ⋅ 0.025( )

"

#$$

%

&''=11.878in

paverage =1.15 ⋅Yflow 1+µL

2haverage

"

#$$

%

&''=1.15 ⋅ 40000 1+

0.1⋅ 11.878( ) 0.0250( )2 ⋅0.1125

"

#

$$

%

&

''= 57140.8

F = Lwpaverage = 11.878( ) 0.0250( ) ⋅10 ⋅57140.8lb = 310183lb

Rflatten = R 1+CF

w hb − hf( )"

#$$

%

&''=10 ⋅ 1+1.67 ⋅10

−7 ⋅31018310 ⋅ 0.025( )

"

#$$

%

&''=12.072in

Hot Rolling Must consider the strain rate effect under plain strain condi4ons

AverageStrainRate = ε = εt=VRLln hb

hf

!

"##

$

%&&

2τ flow =1.15 ⋅Y flow =1.15 ⋅C ⋅ εm

Shape Rolling Work-‐piece is deformed into a contoured cross sec4on instead than a flat (rectangular sec4on). Accomplished by progressively bend the flat slab into complex shapes, by passing it through a series of driven rolls. Products include: Construc4on shapes such as I‑beams, L‑beams, and U‑channels; rails for railroad tracks; round and square bars and rods

www.cometroll.com www.safeair-dowco.com

www.mortonbuildings.com

A series of rolling stands in sequence Tandem Rolling Mill

At each stage in the mill the strip moves at a different velocity. There must be a synchroniza4on of the speed of each roll, such that the input speed of one stand must be equal to the output speed of the preceding stand. Back and front tensions are provided by the uncoiler and windup reel.

An example of a tandem/con4nuous-‐rolling opera4on.

Roll-‐Forging

Two examples of the roll-‐forging opera4on, also known as cross-‐rolling. Tapered leaf springs and knives can be made by this process. Source: AAer J. Holub.

www.sendzimir.com

www.elgiloy.com

Mul4ple backing rolls allow even smaller roll diameters Cluster Mill

Thread Rolling Mass producing process that forms threads on cylindrical parts by forcing (pressing) them between two dies. The process is cold rolled. It produces stronger and beHer fa4gue resistant threads due to cold hardening

Reciprocal method of thread forming

Cylindrical method of thread forming

Rolled treads are produced in a single pass. The thread is formed by the axial flow of the material. The root is formed by pressing the material and the displaced material flows radially to form the thread’s crest.

Ring Rolling During ring rolling a thick-‐walled ring of a small diameter is rolled into a thin-‐walled ring of larger diameter.

Applica4ons: ball and roller bearing races, steel 4res for railroad wheels, and rings for pipes, pressure vessels, and rota4ng machinery Advantages: material savings, ideal grain orienta4on, strengthening through cold working

The process can be cold-‐worked for smaller deforma4ons.

Skew Rolling

Similar to roll forging and used for making ball bearings

Another method of produc4on is to cut pieces from a round bar and upset them between two dies with hemispherical cavi4es

Rotary Tube Piercing A hot-‐working opera4on for making long, thick walled seamless tubing and piping. Also known as the Mannesmann process. The round bar is subjected to radial compressive forces, while tensile stresses develop at the center of the bar

Residual Stresses AAer the deforma4on process has concluded and all the external forces have been removed, the component is s4ll subjected to internal stresses that remain aAer the deforma4on process. The highest value of these internal stresses is the yield point of the material and therefore these internal stresses are important in high yield point materials. The internal stresses are in equilibrium in the component. These can cause unexpected deforma4on if the component is cut so that the equilibrium is lost. The residual stresses can be reduced by a thermal treatment known as stress-‐relief annealing. The treatment does not cause a change in the microstructure of the material.

DeformaonProcesses* Rolling*...

Documents

Transcript of DeformaonProcesses* Rolling*...