Deformation Processing - Drawingramesh/courses/ME649/drawing.pdf · in area per pass for a wire...

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

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Deformation Processing -Drawing

ver. 1


2

Overview

• Description• Characteristics• Mechanical Analysis• Thermal Analysis• Tube drawing


3

Geometry

DaDb

Fa, σxaFb, σxb

α


4


5

Equipment


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Cold Drawing


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A. Durer - Wire Drawing Mill (1489)

(copper wire)


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Characteristics• Product sizes:

– 0.0002” (5µm) to several inches (100-150 mm)

• Mostly cold (T < 0.4 Tmelting) – below recrystallization point

• Small diameter (wire):– uses a capstan

• Diameter > 1 inch (25 mm) (rod):– bull blocks on a draw bench– length up to 40 feet (12 m)


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Characteristics

• Fine wire done through several dies

• Speeds– large diameter: 30 feet per minute

(9 m/min)– small diameter: 300 feet per minute

(90 m/min)– fine wires: 5,000 feet per minute

(60 mph – 100 km/h)


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Die Materials• Large diameter

– high carbon steel– high speed steel

• Moderate diameter– tungsten carbide (WC)

• Small diameter– diamond inserts


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Characteristics

• Lubrication– Coatings– Oil

• Die angle (α)– typically small: 4-6o


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Mechanical analysis (round wire / rod)

Reduction in area (RA)

2

2

22

1

−=

−=

b

a

b

ab

DD

DDDRA

⋅=

−=

a

bt D

DRA

ln21

1lnε

DaDb

Fa, σxaFb, σxb

α


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Slab analysis

Assume p, σx are uniform– OK for small α, µ

σx + dσxσx

p

p

µp

µp

DD + dD

dx


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Equilibrium

( ) ( )

0coscos

sincos

4422

=⋅

+⋅

+

−++

αα

πµαα

π

πσπσσ

dxDpdxDp

DdDDd xxx

Expanding

( ) ( )

0coscos

sincos

42

4222

=⋅

+⋅

+

−+++

αα

πµαα

π

πσπσσ

dxDpdxDp

DdDDdDDd xxx


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Equilibrium

( ) ( )[ ]

0coscos

sincos4

224

2

2222

=⋅

+⋅

+−

+++++

αα

πµαα

ππσ

σσσσσσπ

dxDpdxDpD

dDdDdDdDddDDdDD

x

xxxxxx

small small small

Eliminating higher order terms, dividing by D & π, multiplying by 4 and canceling

0coscos

4sincos

42 =+++ αα

µαα

σσ dxpdxpDddD xx


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Equilibrium

dx

dD2tan =α

αtan2dDdx =

Notingdx

D

dD/2 α

0tan2cos

cos4tan2cos

sin42 =+++αα

αµαα

ασσ dDpdDpDddD xx

or

0tan

222 =+++α

µσσ dDppdDDddD xx


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Equilibrium

0tan

122 =⋅

+⋅+⋅+⋅ dDpdDdD xx α

µσσ

Finally


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Maximum shear stress (Tresca) criterion

flowflowx p στσ ==+ 2

B≡α

µtan

p σx

τflow


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Differential form

pBdp

DdD

flow 24 +=

τ

( )BBd

DdD

flowx

x

+−=

142 τσσ


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Integrating

( )∫∫ +−=

xb

xa

b

a

BBd

DdD

flowx

xD

D

σ

στσσ

142


21


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Drawing stress

• where:σxb = back stress (tension)σxa = pulling stress (tension)

B

b

a

flow

xb

B

b

a

flow

xa

DD

DD

BB

22

211

2

+

−

+=

τσ

τσ

DaDb

σxaσxb

α


23

12

+===nKY

n

flowflowεστ

Strain hardening (cold – below recrystallization

point)• For round parts - Tresca

average flow stress:due to shape of element


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Strain rate effect(hot – above recrystallization point)

• For a round part (derived for extrusion)

– average strain rate due to shape of element– vb = velocity of “b” side– A = area

mflowflow CY εστ &===2

⋅

−

⋅=

a

b

ab

bbAA

DDDv lntan6

33

2 αε&


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Value for p

p = Y – σor

p = 2τflow – σ

maximum at entrance

p σx

τflow


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Effect of back tension

with back tension

without back tension

drawing stress

entry exit

die pressure


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Maximum RA• Solve previous equations with:

α = 6o (typical value)µ = 0.1∴ B = 1σxb = 0For failure: draw stress = material flow {yield} stress

here, say K = 760 MPa, and n = 0.19

12

+===nKY

n

flowflowεστn

xa Kεσσ ==


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Maximum RA

• Yields RA = 0.6– must be solved for each µ, α, σxb

−⋅

+

=

+

B

b

an

n

DD

BB

nKK

2

11

1εε

−⋅

+

=+

2

11

111

119.0

b

aDD


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Energy / unit volume (u)

u = F V / Aa V = σxa

(with no back stress)V= volume


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Rod/Wire Drawing Analysis• Ideal deformation

External work = Work of ideal plastic deformation

for

( ) ( )

ttd

ffd

du

LAuLAt

εσσ

σε

∫==

=

0ntt Kεσ =

==

+=

fftft

nt

d AAYY

nK 0ln

1εεεσ


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Rod/Wire Drawing Analysis• Ideal deformation

Drawing force, Fd = σdAf

Drawing power, Pd = Fd Vf

Source: S. Kalpakjian & S. Schmidt, 4th ed. 2003


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Drawing Limit• Ideal deformation of a perfectly

plastic material

1

11 0.63 63%

od

f

o od

f f

o f

o

AY lnA

Y

A Aln eA A

Maximum reduction per passA AA e

ε

ε

σ

σ

σ σ

= ⋅

=

= ⇒ = ⇒ =

−= = − = =


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Drawing Limit• Ideal deformation of a strain

hardening material

ε

σ

n+1

ideal+frictionideal

εσ

dσ dσ1

( 1)

1

1

1

no

df

n

d

o f n

o

A KY lnA n

Kn

Maximum reduction per passA A

eA

ε

ε

εσ

σ εσ σ ε

+

− +

= ⋅ = + == ⇒ = +

−= = −


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Example ProblemAssuming zero redundant work and frictional work to be 20% of the ideal work, derive an expression for the maximum reduction in area per pass for a wire drawing operation for a material with a true-stress strain curve of σ=Kεn

Total work = Ideal work + frictional work + redundant workTotal work = Ideal work + 0.2 x Ideal work = 1.2 x Ideal work

Or, Total work of deformation = 1.2 [u x volume] … (1)

In drawing, external work of deformation = σd x volume … (2)Equating (1) and (2), we get

σd = 1.2u or

12.12.12.1

11

00

11

+===

+

∫∫ nKdKd

n

tntttd

εεεεσσεε

12.1 εσ Yd = where … (3)

=

fAA0

1 lnε


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Example ProblemMax reduction occurs when total drawing stress, σd = Flow stress of material at die exit, Y

+

−

+

+

−=−

=∴

=⇒+

=⇒+

=

=+

=

=

2.11

0

0

2.11

001

1

11

11

1 passper reduction max

2.11ln

2.111

2.1

2.1

nf

n

ff

nn

nd

eAAA

eAAn

AAn

KnK

KYY

ε

εε

εε

σ


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Drawing - Ex. 1-1Determine power, and plot σx and p

along die length.• Drawing steel rod from φ = 13 mm

to φ = 12 mm @ 1.5 m/s• K = 760 MPa, n = 0.19• µ = 0.1, α = 4o, σxb = 0


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Drawing - Ex. 1-2• First, we must see if we can do the

process, the limit is

• RA = 1 - (Da/Db)2 = 0.15 = 15%• εt = ln{1/(1-RA)}

= ln {1/(1-0.15)} = 0.16• B = µ/tanα = 0.1 / tan 4o = 1.43

nxa Kεσσ == max


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Drawing - Ex. 1-3

B

b

a

flow

xb

B

b

a

flow

xa

DD

DD

BB

22

211

2

+

−

+=

τσ

τσ

12

+==nKY

n

flowετ

nKεσ =max


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Drawing - Ex. 1-4• So, equating the equations (with no

back stress) yields

−⋅

+

+=

B

b

aDD

BB

n

2

111

11

−⋅

+

+=

×−

43.12min

131

43.143.11

119.011 aD


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Drawing - Ex. 1-5

• Solving gives Da-min = 8.53 mm, so we can do the process and proceed with the analysis


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Drawing - Ex. 1-6

35.0013121

43.143.11

2

43.12=+

−+

=×

flow

xaτσ

( ) MPanKY

n

flow 446119.0

16.07601

219.0

=+

⋅=

+==

ετ


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Drawing - Ex. 1-7

• σxa = 0.35 x 2τflow

= 0.35 x 446 MPa = 156 MPa• Fdraw = σxa x Area = 156 x π(12/2)2

= 17.6 kN = 3938 lbf• Power = Fdraw x speed = 17.6 kN x 1.5 m/s = 26.4 kW = 35.4 hp


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Dimensionless pressures (divided by 2τflow)

00.10.20.30.40.50.60.70.80.91

1212.212.412.612.813

diameter (mm)

sxp

Drawing - Ex. 1-8


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Limits on analysis

• Larger die angles– more redundant work– σ, p, u will be larger than predicted

DaDbα


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Redundant work

• ∆ = dm/L• dm = (Da + Db) / 2• p = Qr σflow

DaDbα

L (contact length)


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Redundant work factor (Backofen)(frictionless)

Qr =


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Temperature rise

D Do

Do/D ≈ 6 kD

(1-r)Q

DIE

v

kw, ρw, cwrQ

WIRE

l


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Temperatures

θ = θo + θs + θf

θo = ambient (room) temperatureθs = temperature rise in the wire

due to plastic shear energy, us

θf = interface temperature rise due to frictional energy, uf


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Specific energies

u = us + uf

u = σxa

us = 2τflow ε


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Specific energies

From the example above (steel rod):u = σxa = 156 MPaus = 2τflow ε = 446 * 0.16

= 71.4 MPa∴ uf = u - us = 156 – 71.4

= 84.6 MPa


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Shear temperature (θs)• Since the shear strain is uniform

in the wire• and all the shear energy remains

in the rod as heat• Then, we can obtain the shear

temperature in the wire:

ww

ss c

uρ

θ =


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Material properties

• For this material:– kw = 60 W/m-K– ρw = 7850 kg/m3

– cw = 500 J/kg-K– αw = 1.53 x 10-5 m2/s

• For a WC die:– kD = 42 W/m-K


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Shear temperature (θs)

Ccu

ww

ss

o2.185007850

1071.4 6

=××

==ρ

θ


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Frictional heat (Q)

• v = velocity• (1-r)Q goes into the die

4

2vDuQ fπ

=

Q represents all heat generated by friction


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Die and wire temperatures (θ)• For the die (steady):

• For the wire (moving):

( )DD

lkQr o

Do ln

21

⋅−

+=π

θθ

vl

DlkQr w

wso 207.1 α

πθθθ

++=

ref: Carslaw and Jaeger


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Q calculation

WQ

vDuQ f

143524

5.1012.0106.84

42

6

2

=

⋅⋅⋅×=

=

π

π


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Dimensions• D = 12 mm

– from Do / D ≈ 6– Do = 72 mm in this example

• l = contact length = reduction in radius / sin α= 0.5 mm / sin 4o = 7.17 mm

drlα


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Die temperature

( )DD

lkQr o

Do ln

21

⋅−

+=π

θθ

( )

( )r

r

−⋅+=

⋅⋅⋅⋅−

+=

113591201272ln

00717.042214352120

πθ


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Wire temperature

vl

DlkQr ww

so 207.1 απ

θθθ ⋅

⋅++=

r

r

⋅+=

⋅⋅××

⋅⋅⋅

⋅⋅++=

−

1812.385.12

00717.01053.1

6000717.0012.01435207.12.1820

5

πθ


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Heat flow ratio and Temperature

• Equating the previous equations yields:r = 0.99

• hence θ = 156oC = 429 K

Tmelt = 1500oC = 1723 KSo θ/Tmelt = 0.25, cold (below

recrystallization point


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Temperature in practice

• In practice, r ≈ 1– all heat goes into wire

lvD

cu

cu

www

f

ww

so ⋅

⋅⋅

⋅++=

αρρθθ

219.0


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Tube drawing


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Tube Drawing


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Plane strain / Slab analysis

−

+=

*

112 *

* B

i

f

flow

xa

tt

BB

τσ

βαµµ

tantan*

−+

≡ mandreldieB

α= semicone angle of dieβ =semicone angle of plug


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Tube Drawing – Special Cases

αµ

tan* ≡B

Fixed mandrel- same friction at both interface(plane – tubeis modeled as aflat section)

Moving mandrel – No friction at interfaceof mandrel and tube(plane and slab)

αµ

tan2* ≡B 2*

tan2 µαµ

−≡B

Moving mandrel with friction towards exit, takes into account motion between mandreland tube (B may be negative) (plane) α

µµtan

* mandreldieB −≡

Fixed mandrel(slab – circular tube)

βαµµ

tantan*

−+

≡ mandreldieB


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Summary

• Description• Characteristics• Mechanical analysis• Thermal analysis• Tube drawing


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Deformation Processing - Drawingramesh/courses/ME649/drawing.pdf · in area per pass for a wire...

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