Deformation Processing - Drawingramesh/courses/ME649/drawing.pdf · in area per pass for a wire...
Transcript of 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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Overview
• Description• Characteristics• Mechanical Analysis• Thermal Analysis• Tube drawing
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Geometry
DaDb
Fa, σxaFb, σxb
α
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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)
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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)
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
+
− +
= ⋅ = + == ⇒ = +
−= = −
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Shear temperature (θs)
Ccu
ww
ss
o2.185007850
1071.4 6
=××
==ρ
θ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Q calculation
WQ
vDuQ f
143524
5.1012.0106.84
42
6
2
=
⋅⋅⋅×=
=
π
π
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Die temperature
( )DD
lkQr o
Do ln
21
⋅−
+=π
θθ
( )
( )r
r
−⋅+=
⋅⋅⋅⋅−
+=
113591201272ln
00717.042214352120
πθ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Tube drawing
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Tube Drawing
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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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
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Summary
• Description• Characteristics• Mechanical analysis• Thermal analysis• Tube drawing
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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