Bar  and  Wire  Drawing  

By  pulling  a  bar,  tube  or  wire  through  a  die  the  cross  sec6on  is  reduced.  The  percentage  reduc6on  of  area  (%r)  is  given  by  the  equa6on:  

r%=Ao − Af

Ao×100

Where  Ao  is  the  ini6al  area  and  Af  is  the  final  area.  It  is  usually  performed  in  round  sec6ons.  Very  small  diameters  can  be  obtained  by  successive  drawing  opera6ons  through  dies  of  progressively  small  diameters.  Annealing  before  each  set  of  reduc6ons  allows  large  reduc6on  percentages.  In  steels  the  annealing  process  is  called  paten6ng.  

When  drawing  hard  materials,  the  working  face  of  the  die  is  made  of  tungsten  carbide  (WC)  by  powder  metallurgy.  For  very  fine  diameters  a  diamond  impregnated  die  is  preferred.  In  a  mul6ple  die  configura6on,  the  pulling  force  through  the  die  is  provided  by  each  of  the  drums  (capstan  drums).  

Mechanics  of  Wire  Drawing  and  Extrusion  Direct  extrusion  with  the  diametral  area  reduced  from  Ao  to  Af.  Then  the  ideal  work  is  given  by:  

wideal = σ δε0

ε f∫ σ = Kε n ε f = εaxial = ln AoAf

"

#$$

%

&''= ln 1

1− r"

#$

%

&'= εhomogeneous

wi = Kε n δε0

ε f∫ =Kεh

n+1

n+1 if n = 0 ⇒ w =Yεh

Wi = FΔl⇒ wi =FΔlAoΔl

= PeWhere  Wi  is  the  external  work,  Pe  is  the  applied  extrusion  pressure  and  σd  is  the  applied  drawing  stress  .  In  reality:    

Pe =σ d ≥Kεh

n+1

n+1Taking  into  considera6on  fric6on  (work  due  to  inhomogeneous  deforma6on,  then  

wtotal = wideal +wfriction

Equilibrium  

σ x +δσ x( )π4D+δD( )2 −σ x

π4D2 + p πDδx

cosαsinα +µp πDδx

cosαcosα = 0

σ x +δσ x( )π4D2 + 2DδD+δD2( )−σ x

π4D2 + pπDδx tanα +µpπDδx = 0

π42σ xDδD+D

2δσ x( )+ pπDδx tanα +µpπDδx = 02σ xδD+Dδσ x + 4pδx tanα + 4µpδx = 0

tanα =δD2

δx⇒ δx = δD

2 tanα

2σ xδD+Dδσ x + 2pδD+ 2µpδDtanα

= 0

2σ xδD+Dδσ x + 2p 1+µtanα

#

$%

&

'(δD = 0

Expanding  and  elimina6ng  higher  orders  

Slab  Analysis  for  Wire  Drawing  and  Extrusion  

Using  Maximum  Shear  Stress  (Tresca  Criterium)  

σ x + p =σ flow = 2τ flow

δσ x = −δp

4τ flowδD− 2pδD+ 2p 1+µtanα

"

#$

%

&'δD = Dδp

δDD

4τ flow + p2µtanα"

#$

%

&'

(

)*

+

,-= δp

δDDDf

Do∫ =δσ x

σ x2µtanα"

#$

%

&'− 2τ flow 2+

2µtanα

"

#$

%

&'

σ xf

σ xo∫

σ xf

2τ flow

=1+ µ

tanα"

#$

%

&'

µtanα"

#$

%

&'1−

Df

Do

"

#$

%

&'

2 µtanα"

#$

%

&')

*

+++

,

-

.

.

.+σ xo

2τ flow

Df

Do

"

#$

%

&'

2 µtanα"

#$

%

&'

δDD

=δp

4τ flow + p2µtanα!

"#

$

%&

'

()

*

+,

δDD

=δσ x

σ x2µtanα!

"#

$

%&− 2τ flow 2+

2µtanα

!

"#

$

%&

σxo  is  the  back  stress  (tension)  σxf  is  the  pulling  stress  (tension)  

σ xf =σ drawing =Y 1+ tanαµ

!

"#

$

%& 1−

Af

Ao

!

"#

$

%&

µtanα!

"#

$

%&(

)

***

+

,

---

In  wire  drawing,  the  usual  condi6ons  are  σxo=0  and  2τflow=Y  so  the  equa6on  becomes:  

In  extrusion,  the  usual  condi6ons  are  σxf=0  ,  σxo  is  negaCve  and  2τflow=Y  so  the  equa6on  becomes:    

−σ xo( ) = pextrusion =Y 1+ tanαµ

"

#$

%

&'

AoAf

"

#$$

%

&''

µtanα"

#$

%

&'

−1(

)

***

+

,

---

Maximum  ReducCon  of  Area  –  Wire  Drawing  

For  failure  drawn  stress  =  material  flow  (yield)  stress.  Typical  values  of  α=6degrees  and  µ=0.1    Material  with  a  strain  hardening  constant  K=760MPa  and  strain  hardening  exponent  n=0.19  

σ drawing = Kεn

2τ flow =σ flow =Y =Kε n

n+1

σ drawing

Y= 1+ tanα

µ

!

"#

$

%& 1−

Af

Ao

!

"#

$

%&

µtanα!

"#

$

%&(

)

***

+

,

---=Kε n

Kε n

n+1

0.19+1= 1+ tan60.1

!

"#

$

%& 1−

Af

Ao

!

"#

$

%&

0.1tan6!

"#

$

%&(

)

***

+

,

---⇒

Ao − Af

Ao= RA = 0.58

The  maximum  reduc6on  of  area  must  be  solved  for  each  µ,  α  and  back  tension  

Taking  into  considera6on  fric6on  and  redundant  work  (work  due  to  inhomogeneous  deforma6on,  then   wtotal = wideal +wfriction +wredundant

σ drawing =Y 1+ tanαµ

!

"#

$

%& 1−

Af

Ao

!

"#

$

%&

µtanα!

"#

$

%&(

)

***

+

,

---+43 3

α 2 1− rr

!

"#

$

%&

.

/0

10

2

30

40

σ d =ΦY 1+ µα

!

"#

$

%&ln

AoAf

!

"##

$

%&&

Φ =1+ 0.12Daverage

Lcontact

!

"#

$

%&

AXer  simplifica6ons  

Strain  Hardening  –  Cold  below  RecrystallizaCon  temperature  Take  the  average  flow  stress  (Tresca)  –  due  to  shape  effect  

2τ flow =σ flow =Y =Kε n

n+1

Strain  Rate  Effect  –  Hot  above  Recrystalliza6on  Temperature  For  a  round  part  (the  average  strain  rate  ε,  vo  is  the  velocity  at  the  ini6al  area  and  A  is  the  area  

2τ flow =σ flow =Y =C εm

ε = 6voDo2 ⋅ tanα

Do3 −Df

3 ln AoAf

#

$%%

&

'((

RDV ln6

0

0=εFor  poor  lubrica6on  the  expression  of  the  true  strain  rate  is  

Pe =σ d =Kεh

n+1

n+1=Kεh

n

n+1εh =Yεh =Y ln

AoAf

!

"##

$

%&&

Considering  only  the  ideal  work  and  the  expression  for  Energy  per  Unit  Volume  

Considering  only  the  ideal  work,  then,  the  expression  for  the  force  required  for  drawing  and  the  expression  for  the  power  required  are:  

Fd =σ dAf

Pd = Fdvf =σ dAf v f

Drawing  Limit  (Ideal):  

For  a  Perfectly  PlasCc  Material.  The  ideal  maximum  reduc6on  per  pass  is  63%  

Pe =σ d =Y ln AoAf

!

"##

$

%&& Max − Stress =Y

σ d =Y =Y ln AoAf

!

"##

$

%&&⇒ ln Ao

Af

!

"##

$

%&&=1

AoAf

!

"##

$

%&&= e⇒

Ao − Af

Ao= 0.63

For  a  Strain  Hardening  Material.  The  ideal  maximum  reduc6on  per  pass  depends  on  the  strain  hardening  coefficient.  Example  for  n=0.19  then  the  maximum  reduc6on  per  pass  is  69.5%  

σ d =Y lnAoAf

!

"##

$

%&&=

Kεhn+1

n+1

Max.Stress = Kε n ⇒ Kε n = Kεhn+1

n+1

ε = n+1⇒Ao − Af

Ao=1− e− n+1( )

A  round  rod  of  annealed  302  stainless  steel  is  being  drawn  from  a  diameter  of  10mm  to  8mm  at  a  speed  of  0.5m/s.  Assume  that  the  fric6onal  and  redundant  work  together  cons6tute  40%  of  the  ideal  work  of  deforma6on.  (a)  Calculate  the  power  required  in  this  opera6on,  and  (b)  the  die  pressure  at  the  exit  of  the  die.    Data:  strain  hardening  exponent  =  0.3  ;  strain  hardening  coefficient  =  1300MPa  

Example  

ε1 = ln108

!

"#

$

%&2

= 0.446

Y =Kε n

n+1= 785 MPa

Calculate  the  true  strain  

Calculate  the  average  flow  stress  

Calculate  the  drawing  stress   σ d =Y ln108

!

"#

$

%&2

= 785MPa×0.446 = 350MPa

Calculate  the  drawing  force   Fd =σ d ×π4Df2"

#$

%

&'= 350MPa×

π40.0082

"

#$

%

&'=17598N

Calculate  the  power   P = Fd × v =17598N ×0.5m / s = 8799W

The  actual  power  will  be  40%  higher,  i.e.  12.319kW  

Fd =Pv=12319W0.5m / s

= 24637NCalculate  the  actual  force  

Calculate  the  actual  stress   σ d =FdAf

= 490MPa

Calculate  the  flow  stress  of  the  material   Yf = Kε1n = 1300( ) 0.446( )0.30 =1020MPa

p =Yf −σ d =1020− 490 = 530MPaCalculate  the  die  pressure  

Example   Wire  of  ini6al  diameter  =  0.125  in.  is  drawn  through  two  dies  each  providing  a  0.20  area  reduc6on.  The  metal  has  a  strength  coefficient  =  40,000  lb/in.2  and  a  strain  hardening  exponent  =0.15.    The  dies  have  an  entrance  angle  of  12o,  and  the  es6mated  coefficient  of  fric6on  is  0.10.  The  motors  driving  the  capstans  can  each  deliver  1.50  HP  at  90%  efficiency.  Determine  the  maximum  possible  speed  of  the  wire  as  it  exits  the  second  die.  

r =DO2 −Df

2

DO2 = 0.2⇒ Df 1 = 0.112⇒ Df 2 = 0.1

Diameters  aXer  the  first  and  second  pass  

ε = 2 ln DO

Df

!

"##

$

%&&= 2 ln

0.1250.1118!

"#

$

%&= 2 ln

0.11180.10

!

"#

$

%&= 0.2232

Y = 11.15!

"#

$

%&Kε n =

400001.15

× 0.2232( )0.15 = 27775.8psi

Y = 400001.15

× 0.2232+ 0.2232( )0.15 = 30819psi

1+ µα

!

"#

$

%&= 1+ 0.1

12× π180

!

"

###

$

%

&&&=1.477

First  Die  

Second  Die  

Φ =1+ 0.12Daverage

Lcontact

"

#$

%

&'=1+ 0.12

0.125+ 0.11182

"

#$

%

&'

0.125− 0.11182× sin12

"

#$

%

&'

"

#

$$$$

%

&

''''

=1.33

σ d =ΦY 1+ µα

"

#$

%

&'ln

AoAf

"

#$$

%

&''=1.33×27775.8×1.477×0.2232

σ d =12182.5psi

Force =σ d × Af =12182.5×π40.1118( )2 =119.6lb

Power =1.5HP = 825 ft − lb / s

Power = Force× velocity⇒ v = 825 ft − lb− s*0.9119.6lb

= 6.2084 ft / s

First  Die  

ventry =Df

DO

!

"#

$

%&

2

vexit =0.11180.125

!

"#

$

%&2

6.2084 = 4.966 ft / s

Φ =1+ 0.12Daverage

Lcontact

"

#$

%

&'=1+ 0.12

0.1118+ 0.102

"

#$

%

&'

0.1118− 0.102× sin12

"

#$

%

&'

"

#

$$$$

%

&

''''

=1.35

σ d =ΦY 1+ µα

"

#$

%

&'ln

AoAf

"

#$$

%

&''=1.35×30819×1.477×0.2232

σ d =13720psi

Force =σ d × Af =13720×π40.10( )2 =107.75lb

Power =1.5HP = 825 ft − lb / s

Power = Force× velocity⇒ v = 825 ft − lb− s*0.9107.75lb

= 6.89 ft / s

Second  Die  

ventry =Df

DO

!

"#

$

%&

2

vexit =0.100.1118!

"#

$

%&2

6.89 = 5.51 ft / s

Then,  between  the  first  and  second  die  a  wire  storage  drum  is  necessary.  Calculate  the  power  required  for  the  second  die  if  they  exit  speed  aXer  the  first  die  matches  the  entry  speed  of  the  second  die.  

vexit =DO

Df

!

"##

$

%&&

2

ventry =0.11180.10

!

"#

$

%&2

6.2084 = 7.76 ft / s

Power =119.6lb× 7.76 ft / s = 928.1lb− ft / sPower90% =1.875HP

Residual  Stresses  in  Drawing  

Extrusion  The  metal  is  forced  (squeeze)  to  flow  through  a  die  opening  by  compression  forces  to  produce  the  desired  shape.  Two  basic  types:  Direct  and  Indirect.  Direct:  The  metal  is  squeezed  in  the  same  direc6on  as  the  ram.  Indirect:  The  metal  is  squeezed  in  opposite  direc6on  to  the  ram.  There  is  also  hydrosta6c  and  impact.   Ram  final  posi6on  

Final  shape=  a  solid  rod  

Final  shape=  a  hollow  rod  

Extrusion  can  be  carried  out  hot  (above  the  recrystalliza6on  temperature)  or  cold.  Hot  extrusion  reduces  the  strength  of  the  material  and  increases  the  duc6lity.  

Steel  Extrusion  Tprocessing=1150-­‐1315oC  Tmel6ng=1370-­‐1540oC  Die~205oC      above  recrystalliza6on  Lubricants:  glass.  MoS2,  graphite    

Extrusion  ra6o  is  defined  as:  fAAR 0=

The  shape  factor  is  defined  as:  It  describes  the  shape  complexity  of  an  extrusion.   Shape Factor = Part Perimeter

Part Area

Mechanics  of  Extrusion  

Uniform  deforma6on.  No  redundant  work.  Fric6on  is  high  and  dies  angle  are  high  –  No  slab  analysis.  Dead  zone  is  assume  to  set  up  at  45degrees  

Wpressure = Winternal− friction + Wplastic−work−compression + Wexternal− friction

Rate  of  work  =  Power  =  Area  x  stress  x  ram-­‐velocity  

Wp =π4Do2 ⋅ p ⋅ vram

Wf = τ flow ⋅ π viDδLDf

Do∫#$%

&'(

Q = AOvram = Aivi ⇒ vi =Do

D*

+,

-

./vram

Wf =τ flowπvramDo

2

2δDD

=Df

Do∫τ flowπvramDo

2

2ln Do

Df

*

+,,

-

.//

Internal  fricCon  work  rate:  It  is  calculated  by  integra6on  the  rate  of  fric6onal  work  dissipa6on  at  each  cross  sec6on  from  Do  (ini6al  diameter)  to  Df  (final  diameter)  

Constant  volumetric  flow  rate  

PlasCc  work  rate  to  compress:  Power=  energy-­‐per-­‐unit-­‐volume  x  area  x  velocity  

Energyvolume

= up = σ δε∫ = 2τ flowε⇒ ε = 2 ln Do

Df

#

$%%

&

'((

∴ Wp = 4τ flow ⋅ lnDo

Df

#

$%%

&

'((

#

$%%

&

'((⋅

π4Do2#

$%

&

'(⋅ vram

Total  work  rate  input  (no  considering  the  external  fric6on)  Wp = Wf + Wp

π4Do2 ⋅ p ⋅ vram =

τ flowπvramDo2

2ln Do

Df

"

#$$

%

&''+ 4τ flow ⋅ ln

Do

Df

"

#$$

%

&''

"

#$$

%

&''⋅

π4Do2"

#$

%

&'⋅ vram

p2τ flow

= 3.414 ln Do

Df

"

#$$

%

&''

p2τ flow

=1.707lnR

External  fricCon  work  rate:  Assume  is  due  to  only  wall  fric6on  

Friction− Stress = Shear −Flow− Stress⇒ τ friction = τ flow

Δp ⋅ πDo2

4= τ flow ⋅πDo ⋅ x⇒

Δp2τ flow

=2xDo

px2τ flow

=p

2τ flow

+Δp

2τ flow

=p

2τ flow

+2xDo

px2τ flow

= 3.414 ln Do

Df

%

&''

(

)**+

2xDo

px2τ flow

=1.707lnR+ 2xDo

Indirect  Extrusion   Same  result  as  direct  extrusion  if  external  work  rate  is  not  considered.  For  external  work  rate  consider  that  there  is  no  wall-­‐billet  interac6on,  but  there  is  a  wall-­‐dead  zone  interac6on.  Fric6onal-­‐stress  =  Shear  Flow  Stress    

Wp = Wf + Wp

π4Do

2 ⋅ p ⋅ vram =τ flowπvramDo

2

2ln Do

Df

"

#$$

%

&''+ 4τ flow ⋅ ln

Do

Df

"

#$$

%

&''

"

#$$

%

&''⋅

π4Do

2"

#$

%

&'⋅ vram

p2τ flow

= 3.414 ln Do

Df

"

#$$

%

&'' p

2τ flow

=1.707lnRAddiConal  Pressure  due  to  billet  Dead  Zone  InteracCon:  

Δp π4Do

2 = τ flow ⋅π ⋅Do ⋅Do −Df

2$

%&

'

()⇒

Δp2τ flow

=1−Df

Do

p2τ flow

= 3.414 ln Do

Df

$

%&&

'

())+ 1−

Df

Do

$

%&

'

() p

2τ flow

=1.707lnR+ 1−Df

Do

$

%&

'

()

Dead  zone  

Cold  deformaCon  –  Strain  Hardening    Average  flow  stress  

2τ flow =σ flow =Y =Kε n

n+1

Hot  deformaCon  –  Strain  rate  effect  Above  recrystalliza6on  temperature  

2τ flow =σ flow =Y =C εm

ε = 6vramDo2 tanα

Do3 −Df

3 lnR = 12vramDo2 tanα

Do3 −Df

3 ln Do

Df

"

#$$

%

&''

ε = 6vramDo

lnR = 12vramDo

ln Do

Df

"

#$$

%

&''

ApproximaCon  for  a  large  dead  zone  and  an  angle  of  45degrees  

Extrusion  pressure  vs  ram  travel  for  direct  and  indirect  extrusion  

Example:   An  aluminum  alloy  is  hot  extruded  from  a  150mm  diameter  to  50mm  diameter  at  400oC  at  50mm/s.  The  flow  stress  at  the  working  temperature  is  given  by  σ=200Ė0.15  MPa.  If  the  billet  is  380mm  long  and  the  extrusion  is  done  without  lubrica6on,  determine  the  force  required  for  the  opera6on  

R = 15050

!

"#

$

%&2

= 9

ε = 6v lnRDo

=6 50mm / s( ) ln9

150mm= 4.39s−1

σ =C εm = 200× 4.390.15 = 250MPa

Total  work  rate  input  (no  considering  the  external  fric6on)  

p2τ flow

=1.707lnR⇒ τ flow =σ3=250MPa

3=144MPa

p = 2×144×1.707ln9 =1080.2MPa

Total  work  rate  input  (considering  the  external  fric6on,  for  x=380mm  (maximum  pressure  due  to  container  wall  fric6on).  

px2τ flow

=1.707lnR+ 2xDo

⇒2xDo

=2×380150

= 5.067

px =1080.2+1459.2 = 2539.4MPa

The  compressive  force:    Force = px

π4

!

"#

$

%&D0

2 = 2539.4× π4

!

"#

$

%& 0.150( )2 = 44875kN

A  copper  billet  12.7cm  in  diameter  and  25.4  cm  long  is  extruded  at  1088.7  K  at  a  speed  of  25.4  cm/s.  Using  square  dies  and  assuming  poor  lubrica6on,  es6mate  the  force  required  in  this  opera6on  if  the  extruded  diameter  is  5.08  cm.    Data:C=131MPa    and    m=0.06  

Example  

25.608.57.122

2

==R

ε =6v0D0

lnR =6 25.4( )12.7( )

ln 6.25( ) = 22s-1

( )( ) MPa 69.15722131 06.0 === mCεσ

Calculate  the  extrusion  ra6o  

Calculate  the  average  true  strain  rate  

Calculate  the  stress  required  

p =Y 1.7lnR+ 2LD0

!

"#

$

%&= 157.69( ) 1.7ln6.25+

2 25.4( )12.7( )

!

"##

$

%&&=1121.3335 MPa

F = p( ) A0( ) = 1121.3335( )π 12.7( )2

4=14204.7kN

σ=YAssuming  that:  

Defects  Chevron  Cracking:  1-­‐  Develops  at  the  center  of  the  extruded  piece.  2  –  It  can  occur  at  low  extrusion  ra6os  due  to  low  fric6onal  condi6ons  on  the  zone  of  deforma6on.  

Surface  Cracking:  Due  to  longitudinal  tensile  stresses  generated  as  the  extrusion  passes  through  the  die.  It  ranges  from  a  badly  roughened  surface  to  repe66ve  transverse  cracking.  Inhomogeneous  DeformaCon:  AXer  2/3  of  the  billet  is  extruded,  the  outer  surface  of  the  billet  (usually  an  oxidised  skin)  moves  towards  the  center  and  extrudes  to  the  through  die,  resul6ng  in  internal  oxide  stringers.    Some  lubricant  film  can  also  be  carried  into  the  interior  of  the  extrusion  along  the  shear  bands.  This  will  show  as  longitudinal  inclusions  in  the  extruded  part.  

Hot  Shortness:  In  aluminum  extrusion:  Incipient  mel6ng  on  different  points  of  the  extruded  pieces  due  to  high  temperature.