Bar and Wire Drawing A − A By pulling a bar, tube or wire through a die the cross sec6on r% = o f ×100 is reduced. The percentage reduc6on of area (%r) is given Ao by the equaon: 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 operaons 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 configuraon, 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 consideraon fric6on (work due to inhomogeneous deformaon, then wtotal = wideal + w friction Slab Analysis for Wire Drawing and Extrusion Equilibrium Expanding and eliminang 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 condi6ons are σxf is the pulling stress (tension) σxo=0 and 2τflow=Y so the equaon becomes: ! µ $ ( # &+ " tanα % ! tanα $ ! Af $ Y 1 *1 - σ xf = σ drawing = # + &* −# & - " µ % " Ao % )* ,- In extrusion, the usual condi6ons are σxf=0 , σxo is negave and 2τflow=Y so the equaon becomes: " µ % ( $ ' + " % " %# tanα & tanα * Ao - (−σ xo ) = pextrusion = Y $1+ '*$ ' −1- # µ & # Af & )* ,- Maximum ReducCon 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 reduc6on of area must be solved for each µ, α and back tension Taking into consideraon fric6on and redundant work (work due to inhomogeneous deformaon, 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 # + & # & AXer simplificaons " α % " Af % ! D $ Φ =1+ 0.12# average & L n " contact % Kε 2τ flow = σ flow = Y = Strain Hardening – Cold below RecrystallizaCon 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 Recrystallizaon Temperature For a round part (the average strain rate ε, vo is the velocity at 6v D2 ⋅ tanα # A & o o ln% o ( the ini6al area and A is the area ε = 3 3 % ( Do − Df $ Af ' 6V For poor lubricaon 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 PlasCc Material. The ideal maximum reduc6on 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 reduc6on 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 reduc6on 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 fric6onal and redundant work together cons6tute 40% of the ideal work of deformaon. (a) Calculate the power required in this operaon, 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 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. 2 2 DO − Df Diameters aer 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 aer 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.