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TRANSCRIPT
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.