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

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

Deformation Processing -Drawing

ver. 1

Overview

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

Geometry

Fa, σxaFb, σxb

Equipment

Cold Drawing

A. Durer - Wire Drawing Mill (1489)

(copper wire)

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)

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)

Die Materials• Large diameter

– high carbon steel– high speed steel

• Moderate diameter– tungsten carbide (WC)

• Small diameter– diamond inserts

Characteristics

• Lubrication– Coatings– Oil

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

Mechanical analysis (round wire / rod)

Reduction in area (RA)

Fa, σxaFb, σxb

Slab analysis

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

σx + dσxσx

DD + dD

Equilibrium

( ) ( )

0coscos

sincos

πµαα

πσπσσ

dxDpdxDp

DdDDd xxx

Expanding

( ) ( )

0coscos

sincos

−+++

πµαα

πσπσσ

dxDpdxDp

DdDDdDDd xxx

Equilibrium

( ) ( )[ ]

0coscos

sincos4

πµαα

ππσ

σσσσσσπ

dxDpdxDpD

dDdDdDdDddDDdDD

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

Equilibrium

dD2tan =α

αtan2dDdx =

Notingdx

dD/2 α

0tan2cos

cos4tan2cos

sin42 =+++αα

αµαα

ασσ dDpdDpDddD xx

222 =+++α

µσσ dDppdDDddD xx

Equilibrium

122 =⋅

+⋅+⋅+⋅ dDpdDdD xx α

µσσ

Finally

Maximum shear stress (Tresca) criterion

flowflowx p στσ ==+ 2

B≡α

τflow

Differential form

flow 24 +=

( )BBd

142 τσσ

Integrating

( )∫∫ +−=

στσσ

Drawing stress

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

σxaσxb

+===nKY

flowflowεστ

Strain hardening (cold – below recrystallization

point)• For round parts - Tresca

average flow stress:due to shape of element

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

DDDv lntan6

2 αε&

Value for p

p = Y – σor

p = 2τflow – σ

maximum at entrance

τflow

Effect of back tension

with back tension

without back tension

drawing stress

entry exit

die pressure

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

+===nKY

flowflowεστn

xa Kεσσ ==

Maximum RA

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

−⋅

Energy / unit volume (u)

u = F V / Aa V = σxa

(with no back stress)V= volume

Rod/Wire Drawing Analysis• Ideal deformation

External work = Work of ideal plastic deformation

( ) ( )

LAuLAt

εσσ

0ntt Kεσ =

d AAYY

nK 0ln

1εεεσ

Rod/Wire Drawing Analysis• Ideal deformation

Drawing force, Fd = σdAf

Drawing power, Pd = Fd Vf

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

Drawing Limit• Ideal deformation of a perfectly

plastic material

11 0.63 63%

AY lnA

A Aln eA A

Maximum reduction per passA AA e

= ⇒ = ⇒ =

−= = − = =

Drawing Limit• Ideal deformation of a strain

hardening material

ideal+frictionideal

dσ dσ1

A KY lnA n

Maximum reduction per passA A

σ εσ σ ε

= ⋅ = + == ⇒ = +

−= = −

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

∫∫ nKdKd

tntttd

εεεεσσεε

12.1 εσ Yd = where … (3)

1 lnε

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

−=−

1 passper reduction max

2.11ln

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

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

Drawing - Ex. 1-3

+==nKY

flowετ

nKεσ =max

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

back stress) yields

−⋅

43.12min

43.143.11

119.011 aD

Drawing - Ex. 1-5

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

Drawing - Ex. 1-6

35.0013121

43.143.11

43.12=+

xaτσ

( ) MPanKY

flow 446119.0

16.07601

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

Dimensionless pressures (divided by 2τflow)

00.10.20.30.40.50.60.70.80.91

1212.212.412.612.813

diameter (mm)

Drawing - Ex. 1-8

Limits on analysis

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

DaDbα

Redundant work

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

DaDbα

L (contact length)

Redundant work factor (Backofen)(frictionless)

Temperature rise

Do/D ≈ 6 kD

(1-r)Q

kw, ρw, cwrQ

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

Specific energies

u = us + uf

u = σxa

us = 2τflow ε

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

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:

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

Shear temperature (θs)

o2.185007850

1071.4 6

Frictional heat (Q)

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

2vDuQ fπ

Q represents all heat generated by friction

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

• For the wire (moving):

lkQr o

⋅−

DlkQr w

wso 207.1 α

πθθθ

ref: Carslaw and Jaeger

Q calculation

vDuQ f

143524

5.1012.0106.84

⋅⋅⋅×=

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

Die temperature

lkQr o

⋅−

−⋅+=

⋅⋅⋅⋅−

113591201272ln

00717.042214352120

Wire temperature

DlkQr ww

so 207.1 απ

θθθ ⋅

⋅++=

⋅⋅××

⋅⋅⋅

⋅⋅++=

1812.385.12

00717.01053.1

6000717.0012.01435207.12.1820

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

Temperature in practice

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

so ⋅

⋅⋅

⋅++=

αρρθθ

Tube drawing

Tube Drawing

Plane strain / Slab analysis

βαµµ

tantan*

≡ mandreldieB

α= semicone angle of dieβ =semicone angle of plug

Tube Drawing – Special Cases

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)

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

Summary

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

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

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