Extrusion

ver. 1

Instructor Ramesh Singh: Notes by Prof. Colton Georgia Tech

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2

Overview

• Equipment• Characteristics• Mechanical Analysis

– direct extrusion– indirect extrusion

• Redundant work• Defects

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Geometry (90o die)

D1

D2

p

dead zone45o angle

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Equipment

5

Extrusion

6

Equipment

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Extrusions

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Characteristics

• Similar to closed die forging• Forging

– slug (bulk) is forging– flash (extrusion) is waste

• Extrusion– extrusion (flash) is part– billet (bulk) is waste

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Types

• Direct• Indirect• Tubular• Hydrostatic• Cold Impact

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Types

1 – direct2 – indirect3 – heading (forging also)

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Flow types• “Laminar”• “Turbulent”

– redundant work– can bring outside of billet into

center– leaving the skin keeps outside

scale out of final extrusion

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Steel extrusion

• Tprocessing = 2100 to 2400oF (1150 – 1315oC) • Tmelting = 2500 - 2800oF (1370 – 1540oC)• Die » 400oF (205oC)• Obviously “Hot”

– above recyrstallization point• Lubricants

– glass (viscous lube ) 0.001” thick– MoS2

– graphite

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Mechanical Analysis

D1

D2

p

dead zone45o angle

externalfriction

internalfriction

x

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Assumptions

• Metal deforms uniformly– D1 to D2

• No redundant work• Can’t use slab analysis

– die angles too great– friction too high

• Dead zone sets up at 45 degrees

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p tfriction

Upper bound analysis• Work input by external forces

= plastic work expended

friction external

compress work toplastic

friction internalpressure

W

W

W W

!

!

!!

+

+

=

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Rate of work = Power

• Work rate = Power• Work rate = Area • stress • velocity

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Pressure work input

• Power = A • p • v– ram moves at velocity, vram

ramp vpDW ××=4

21p!

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Internal “frictional” work input• Work determined by integrating rate of

frictional work dissipation at each cross section from D2 to D1 – tfriction = tflow

– vi is in x-direction

DD + dD

dx

dL=dD/Ö245o÷

÷÷

ø

ö

ççç

è

æ×= ò

1

2

D

D

iflowf DdLvW pt!

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• Volumetric flow rate

– where D, Ai, vi are instantaneous

Internal “frictional” work input

iiram vAvAQ == 1

rami vDDv

21 ÷øö

çèæ=

21

2

121 ln

2 DDDv

W flowramf ×=

tp!

Internal “frictional” work input

ò=1

22

21D

D

flowramf D

dDDvW

tp!

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Plastic work to compress input• Power = up x Area x velocity

• hence

etees flowfp YduvolumeEnergy 2/ ==== ò

2

1ln2DD

=e

ramflowpw vD

DD

W ×÷÷ø

öççè

æ×÷÷ø

öççè

æ×=\

4ln4

21

2

1 pt!

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Total work input (without external friction)

• reducing

2

121

2

121

21

ln2

44

ln444

DDvD

DDvDvpD

flowram

flowramram

×××+

÷÷ø

öççè

æ×××=××

tp

tpp

2

1ln414.32 D

Dp

flow×=

t

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Extrusion ratio (re)

• Reduction in area (RA) is large– it is not sensitive for classification

• Use re instead

RADDre -

=÷÷ø

öççè

æ=

11

2

2

1

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Extrusion pressure (without external friction)

2

2

1

2

1 ln707.1ln414.32 ÷÷

ø

öççè

æ×=×=

DD

DDp

flowt

eflow

rp ln707.1

2×=

t

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Billet - wall friction

• Assume limiting case:friction stress = shear flow stress

tf = tflow

tflow

tflow

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Additional pressure due to billet - wall friction

xDD

p flow ×××=×

×D 1

21

4ptp

1

22 D

xp

flow=

Dt

tflow

tflow

x

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Direct extrusion pressure

1

22222 D

xpppp

flowflowflowflow

x +=D

+=tttt

1

12

1

2ln707.12

2ln414.32

Dx

rp

Dx

DDp

eflow

x

flow

x

+×=

+×=

t

t

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12

+===nKY

n

flowflowest

Strain hardening (cold – below recrystallization point)

• Not plane strain (Tresca)

average flow stress:due to shape of element

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Example – 1-1

• You are forward, cold extruding Al-1100 (K = 140 MPa, n = 0.25), 10-cm diameter billet to a diameter of 5-cm at 1 m/min. The billet is initially 25 cm long

• The ram is made of a high-strength steel with a yield stress of 1.5 GPa.

• Determine the extrusion force and power.• Determine the safety factor for indenting the

ram.

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Example – 1-2

• The equations we use are:

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1 2ln414.32 D

xDDp

flow

x +×=t

12

+==nKY

n

flowet

÷÷ø

öççè

æ=

2

1ln2DDe

1+=== ò ò nKdKdu n

peeees

because

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Example – 1-3• We need to determine the dead-zone length to

subtract from the initial billet length.

• so X = 0.25 – 0.025 = 22.5 cm45°

2.5 cm5 cm2.5 cm

2.5 cm

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Example – 1-4• Substituting values

39.1510ln2ln2

2

1 =÷øö

çèæ=÷÷

ø

öççè

æ=

DDe

( )MPa

nKY

n

flow 6.121125.039.1140

12

25.0

=+

´=

+==

et

÷÷ø

öççè

æ+×´=

12

1 2ln414.32Dx

DD

p flowx t

MPaPextrusion 83410

5.222510ln414.36.121max, =÷

øö

çèæ ´

+×´=

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Example – 1-5

• Safety factor against indenting the ram– to determine the “press-fit” failure, we would

need the dimensions of the extrusion die and its material

8.1834.05.1

max,

===GPaGPan

extrusion

y

ss

( ) MNAreaPF extrusionextrusion 6.61.04

10834 26 =´´=´=p

kWmMNspeedFPower 110sec60min/min/16.6 =´´=´=

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Redundant work

• D = dm/L• dm = (D1 + D2) / 2• p = Qr sflow

p

D1

D2L (contact length)

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Redundant work factor (Backofen)(frictionless)

Qr =

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Defects• Surface materials drawn

into center– pipe, tail pipe

• Surface materials extruded– eliminate by leaving

skin

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Chevron Cracking

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Chevron cracking defect

• Hydrostatic tension– outer layer in compression– inner layer in tension, if

entire part is not plastic• eliminate by using a fluid

– hydrostatic compression– reduces friction

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Defects

• Surface speed cracking– high friction– temperature– speed

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Summary

• Equipment• Characteristics• Mechanical Analysis

– direct extrusion• Redundant work• Defects