Bar and Wire Drawing A − A By pulling a bar, tube or wire through a die the cross sec on r% = o f ×100 is reduced. The percentage reduc on of area (%r) is given Ao by the equa on: Where Ao is the ini al area and Af is the final area. It is usually performed in round sec ons. Very small diameters can be obtained by successive drawing opera ons through dies of progressively small diameters. Annealing before each set of reduc ons allows large reduc on percentages. In steels the annealing process is called paten ng. 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 mul ple die configura on, 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: " % ε f A " 1 % w = σ δε σ = Kε n ε = ε = ln$ o ' = ln = ε ideal ∫ 0 f axial $ ' $ ' homogeneous # Af & #1− r & n+1 ε f n Kε w = Kε δε = h if n = 0 ⇒ w = Yε i ∫ 0 n +1 h FΔl Where Wi is the external work, Pe is the applied Wi = FΔl ⇒ wi = = Pe extrusion pressure and σd is the applied drawing AoΔl stress . In reality:
Kε n+1 P = σ ≥ h e d n +1 Taking into considera on fric on (work due to inhomogeneous deforma on, then
wtotal = wideal + w friction Slab Analysis for Wire Drawing and Extrusion
Equilibrium Expanding and elimina ng higher orders
π 2 π 2 π Dδx π Dδx (σ x +δσ x ) (D +δD) −σ x D + p sinα + µp cosα = 0 4 4 cosα cosα
π 2 2 π 2 (σ x +δσ x ) D + 2DδD +δD −σ x D + pπ Dδx tanα + µpπ Dδx = 0 4 ( ) 4
π 2 2σ DδD + D δσ + pπ Dδx tanα + µpπ Dδx = 0 4 ( x x )
2σ xδD + Dδσ x + 4pδx tanα + 4µpδx = 0 D δ δD tanα = 2 ⇒ δx = δx 2tanα δD 2σ δD + Dδσ + 2pδD + 2µp = 0 x x tanα # µ & 2σ xδD + Dδσ x + 2p%1+ (δD = 0 $ tanα ' Using Maximum Shear Stress (Tresca Criterium) σ + p = σ = 2τ δD δ p x flow flow = ' ! $* δσ x = −δ p D 2µ )4τ flow + p# &, " µ % ( " tanα %+ 4τ flowδD − 2pδD + 2p$1+ 'δD = Dδ p # tanα & δD δσ = x ( + D ! 2µ $ ! 2µ $ δD " 2µ % σ − 2τ 2 + *4τ flow + p$ '- = δ p x # & flow # & D ) # tanα &, " tanα % " tanα %
Do δD σ xo δσ = x ∫ D ∫ f D σ xf " 2µ % " 2µ % σ $ '− 2τ $2 + ' x # tanα & flow # tanα &
" µ % " µ % " µ % 1+ ) 2$ ', 2$ ' $ ' " % # tanα & " % # tanα & σ xf # tanα &+ Df . σ xo Df = +1−$ ' .+ $ ' 2τ flow " µ % # Do & 2τ flow # Do & $ ' *+ -. # tanα & σxo is the back stress (tension) In wire drawing, the usual condi ons are σxf is the pulling stress (tension) σxo=0 and 2τflow=Y so the equa on becomes:
! µ $ ( # &+ " tanα % ! tanα $ ! Af $ Y 1 *1 - σ xf = σ drawing = # + &* −# & - " µ % " Ao % )* ,-
In extrusion, the usual condi ons are σxf=0 , σxo is nega ve and 2τflow=Y so the equa on becomes:
" µ % ( $ ' + " % " %# tanα & tanα * Ao - (−σ xo ) = pextrusion = Y $1+ '*$ ' −1- # µ & # Af & )* ,-
Maximum Reduc on of Area – Wire Drawing n For failure drawn stress = material flow (yield) stress. σ drawing = Kε Typical values of α=6degrees and µ=0.1 Kε n Material with a strain hardening constant K=760MPa and 2τ = σ = Y = strain hardening exponent n=0.19 flow flow n +1
! µ $ ( # &+ " tanα % n σ drawing ! tanα $* ! Af $ - Kε = 1+ 1− = # &* # & - n Y " µ % " Ao % Kε )* ,- n +1 ! 0.1 $ ( # &+ " tan6 % ! tan6 $* ! Af $ - Ao − Af 0.19 +1= #1+ &*1−# & - ⇒ = RA = 0.58 " 0.1 % " Ao % Ao )* ,- The maximum reduc on of area must be solved for each µ, α and back tension Taking into considera on fric on and redundant work (work due to inhomogeneous deforma on, then wtotal = wideal + w friction + wredundant
! µ $ . ( # &+ 2 " tanα % 0! tanα $ ! Af $ 4 2 !1− r $0 Y 1 *1 - σ drawing = /# + &* −# & -+ α # &3 0" µ % " Ao % 3 3 " r %0 1 )* ,- 4 ! µ $ ! A $ Y 1 ln# o & σ d = Φ # + & # & A er simplifica ons " α % " Af % ! D $ Φ =1+ 0.12# average & L n " contact % Kε 2τ flow = σ flow = Y = Strain Hardening – Cold below Recrystalliza on temperature n +1 Take the average flow stress (Tresca) – due to shape effect m 2τ flow = σ flow = Y = Cε Strain Rate Effect – Hot above Recrystalliza on Temperature For a round part (the average strain rate ε, vo is the velocity at 6v D2 ⋅ tanα # A & o o ln% o ( the ini al area and A is the area ε = 3 3 % ( Do − Df $ Af ' 6V For poor lubrica on the expression of the true strain rate is ε = 0 ln R D0 Considering only the ideal work and the expression for Energy per Unit Volume
n+1 n ! $ Kεh Kεh Ao Pe = σ d = = εh = Yεh = Y ln# & n +1 n +1 " Af % Considering only the ideal work, then, the expression for the force required for drawing and the expression for the Fd = σ d Af power required are: Pd = Fdv f = σ d Af v f ! $ Drawing Limit (Ideal): Ao Pe = σ d = Y ln# & Max − Stress = Y " Af % For a Perfectly Plas c Material. The ideal maximum reduc on per pass is ! $ ! $ Ao Ao 63% σ d = Y = Y ln# & ⇒ ln# & =1 " Af % " Af % ! $ Ao Ao − Af # & = e ⇒ = 0.63 " Af % Ao
For a Strain Hardening Material. The ! $ n+1 ideal maximum reduc on per pass Ao Kεh σ d = Y ln# & = depends on the strain hardening " Af % n +1 coefficient. Example for n=0.19 then n+1 n n Kε the maximum reduc on per pass is Max.Stress = Kε ⇒ Kε = h 69.5% n +1 A − A ε = n +1 ⇒ o f =1− e−(n+1) Ao Example 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 fric onal and redundant work together cons tute 40% of the ideal work of deforma on. (a) Calculate the power required in this opera on, and (b) the die pressure at the exit of the die. Data: strain hardening exponent = 0.3 ; strain hardening coefficient = 1300MPa
!10 $2 ε1 = ln# & = 0.446 Calculate the true strain " 8 % Kε n Calculate the average flow stress Y = = 785 MPa n +1
!10 $2 Calculate the drawing stress σ d = Y ln# & = 785MPa × 0.446 = 350MPa " 8 %
"π 2 % "π 2 % Fd = σ d ×$ Df ' = 350MPa ×$ 0.008 ' =17598N Calculate the drawing force # 4 & # 4 &
Calculate the power P = Fd × v =17598N × 0.5m / s = 8799W
The actual power will be 40% higher, i.e. 12.319kW P 12319W Calculate the actual force F = = = 24637N d v 0.5m / s
Fd Calculate the actual stress σ d = = 490MPa Af
n 0.30 Calculate the flow stress of the material Yf = Kε1 = (1300)(0.446) =1020MPa
Calculate the die pressure p = Yf −σ d =1020 − 490 = 530MPa Example Wire of ini al diameter = 0.125 in. is drawn through two dies each providing a 0.20 area reduc on. 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 es mated coefficient of fric on 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.
2 2 DO − Df Diameters a er the first and second r = 2 = 0.2 ⇒ Df 1 = 0.112 ⇒ Df 2 = 0.1 pass DO ! $ DO ! 0.125 $ ! 0.1118$ ε = 2ln# & = 2ln# & = 2ln# & = 0.2232 " Df % " 0.1118% " 0.10 %
! 1 $ n 40000 0.15 First Die Y = # &Kε = ×(0.2232) = 27775.8psi "1.15% 1.15
40000 0.15 Y = ×(0.2232 + 0.2232) = 30819psi Second Die 1.15 ! $ ! µ $ # 0.1 & #1+ & = #1+ & =1.477 π " α % # 12 × & " 180 % First Die "" 0.125+ 0.1118%% $$ '' " Daverage % # 2 & Φ =1+ 0.12$ ' =1+ 0.12$ ' =1.33 # Lcontact & $" 0.125− 0.1118%' $$ '' ## 2 ×sin12 && " % " µ % Ao σ d = ΦY $1+ 'ln$ ' =1.33× 27775.8×1.477× 0.2232 # α & # Af &
σ d =12182.5psi π 2 Force = σ d × Af =12182.5× (0.1118) =119.6lb 4 Power =1.5HP = 825 ft − lb / s 825 ft − lb − s*0.9 Power = Force× velocity ⇒ v = = 6.2084 ft / s 119.6lb
2 2 ! Df $ ! 0.1118$ ventry = # & vexit = # & 6.2084 = 4.966 ft / s " DO % " 0.125 % Second Die "" 0.1118+ 0.10 %% $$ '' " Daverage % # 2 & Φ =1+ 0.12$ ' =1+ 0.12$ ' =1.35 # Lcontact & $" 0.1118− 0.10 %' $$ '' ## 2 ×sin12 && " % " µ % Ao σ d = ΦY $1+ 'ln$ ' =1.35×30819 ×1.477× 0.2232 # α & # Af &
σ d =13720psi π 2 Force = σ d × Af =13720 × (0.10) =107.75lb 4 Power =1.5HP = 825 ft − lb / s 825 ft − lb − s*0.9 Power = Force× velocity ⇒ v = = 6.89 ft / s 107.75lb
2 2 ! Df $ ! 0.10 $ ventry = # & vexit = # & 6.89 = 5.51ft / s " DO % " 0.1118% 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 a er the first die matches the entry speed of the second die. ! $2 2 DO ! 0.1118$ vexit = # & ventry = # & 6.2084 = 7.76 ft / s " Df % " 0.10 % Power =119.6lb × 7.76 ft / s = 928.1lb − ft / s
Power90% =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 direc on as the ram. Indirect: The metal is squeezed in opposite direc on to the ram. There is also hydrosta c and impact. Ram final posi on
Final shape= a hollow rod
Final shape= a solid rod
Extrusion can be carried out hot (above the recrystalliza on temperature) or cold. Hot extrusion reduces the strength of the material and increases the duc lity.
Steel Extrusion o Tprocessing=1150-1315 C Tmel ng=1370-1540oC Die~205oC above recrystalliza on Lubricants: glass. MoS2, graphite
Mechanics of Extrusion
A Extrusion ra o is defined as: R = 0 Af The shape factor is defined as: Part Perimeter Shape Factor = It describes the shape complexity of an extrusion. Part Area
Uniform deforma on. No redundant work. Fric on 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 Internal fric on work rate: It is calculated by integra on the rate of fric onal work dissipa on at each cross sec on from Do (ini al diameter) to Df (final diameter) π 2 W = D ⋅ p⋅ v p 4 o ram # Do & Wf = τ flow ⋅ π viDδL $% ∫ Df '( * - Do Constant volumetric flow Q = AOvram = Aivi ⇒ vi = , /vram + D . rate 2 2 * - τ flowπvramDo Do δD τ flowπvramDo D W = = ln, o / f ∫ D f , / 2 D 2 + Df . Plas c work rate to compress: Power= energy-per-unit-volume x area x velocity # & Energy Do = up = ∫ σ δε = 2τ flowε ⇒ ε = 2ln% ( volume $ Df ' # # && D #π 2 & W %4 ln% o (( D v ∴ p = % τ flow ⋅ % ((⋅% o (⋅ ram $ $ Df '' $ 4 ' Total work rate input (no considering the external fric on) Wp = Wf +Wp 2 " % " " %% π 2 τ flowπvramDo D D "π 2 % D p v ln$ o ' $4 ln$ o '' D v o ⋅ ⋅ ram = $ '+$ τ flow ⋅ $ ''⋅$ o '⋅ ram 4 2 # Df & # # Df && # 4 & " % p Do = 3.414ln$ ' 2τ flow # Df & p =1.707ln R 2τ flow
External fric on work rate: Assume is due to only wall fric on
Friction − Stress = Shear − Flow − Stress ⇒ τ friction = τ flow 2 π Do Δp 2x Δp⋅ = τ flow ⋅π Do ⋅ x ⇒ = 4 2τ flow Do p p Δp p 2x x = + = + 2τ flow 2τ flow 2τ flow 2τ flow Do % ( px Do 2x px 2x = 3.414ln' *+ =1.707ln R + 2τ flow & Df ) Do 2τ flow Do 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 interac on, but there is a wall-dead zone interac on. Fric onal-stress = Shear Flow Stress Wp = Wf +Wp 2 " % " " %% π 2 τ flowπvramDo D D "π 2 % D p v ln$ o ' $4 ln$ o '' D v o ⋅ ⋅ ram = $ '+$ τ flow ⋅ $ ''⋅$ o '⋅ ram 4 2 # Df & # # Df && # 4 & Dead zone p " D % p Addi onal Pressure due to billet = 3.414ln$ o ' =1.707ln R 2τ $ D ' 2τ Dead Zone Interac on: flow # f & flow
π 2 $ Do − Df ' Δp Df Δp Do = τ flow ⋅π ⋅ Do ⋅& ) ⇒ =1− 4 % 2 ( 2τ flow Do $ ' $ ' $ ' p Do Df p Df = 3.414ln& )+&1− ) =1.707ln R +&1− ) 2τ flow % Df ( % Do ( 2τ flow % Do ( Cold deforma on – Strain Hot deforma on – Strain rate effect Above recrystalliza on Hardening Average flow stress temperature m 2τ flow = σ flow = Y = Cε Kε n 2 Y 2 2 τ flow = σ flow = = 6v D tanα 12v D tanα " D % n +1 ram o ln R ram o ln$ o ' ε = 3 3 = 3 3 $ ' Do − Df Do − Df # Df & 6v 12v " D % ε = ram ln R = ram ln$ o ' Approxima on for a large dead $ ' zone and an angle of 45degrees Do Do # Df & 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 lubrica on, determine the force required for the opera on
!150 $2 Total work rate input (no considering R = # & = 9 the external fric on) " 50 % p σ 250MPa =1.707ln R ⇒ τ flow = = =144MPa 6vln R 6(50mm / s)ln9 −1 2τ 3 3 ε = = = 4.39s flow Do 150mm p = 2 ×144 ×1.707ln9 =1080.2MPa σ = Cεm = 200 × 4.390.15 = 250MPa
Total work rate input (considering the p 2x 2x 2 ×380 x =1.707ln R + ⇒ = = 5.067 external fric on, for x=380mm 2τ D D 150 (maximum pressure due to container flow o o wall fric on). px =1080.2 +1459.2 = 2539.4MPa
The compressive force: !π $ 2 !π $ 2 Force = px # &D0 = 2539.4 ×# &(0.150) = 44875kN " 4 % " 4 % Example 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 lubrica on, es mate the force required in this opera on if the extruded diameter is 5.08 cm. Data:C=131MPa and m=0.06
12.72 Calculate the extrusion ra o R = = 6.25 5.082
6v0 6(25.4) -1 Calculate the average true strain rate ε = ln R = ln(6.25) = 22s D0 (12.7) Calculate the stress required σ = Cεm = (131)(22)0.06 =157.69 MPa
Assuming that: Y = σ
! 2L$ ! 2(25.4)$ p = Y #1.7ln R + & = (157.69)#1.7ln6.25+ & =1121.3335 MPa " D0 % "# (12.7) %&
2 π (12.7) F = (p)(A0 ) = (1121.3335) =14204.7kN 4 Defects Chevron Cracking: 1- Develops at the center of the extruded piece. 2 – It can occur at low extrusion ra os due to low fric onal condi ons on the zone of deforma on.
Surface Cracking: Due to longitudinal tensile stresses generated as the extrusion passes through the die. It ranges from a badly roughened surface to repe ve transverse cracking. Inhomogeneous Deforma on: A er 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, resul ng 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 mel ng on different points of the extruded pieces due to high temperature.