mechanical-properties-and-working-of-metals-and-alloys_compress

640 13 Extrusion which is the vector difference between v1 and v3 and that v3Ã2 ¼ vÃ13 ; along the plane BC is v3Ã2; which is the vector difference sin h sin w between v3 and v2: Or, vÃ32 ¼ v1Ã3 sin h From the diagram in Fig. 13.35a, we get the following: sin w AB ¼ h1 ð13:56Þ ¼ v1 sin a sin h ð13:62Þ sin sin w sinðh À aÞ ð13:63Þ 2 h In DABC; BC ¼ AB aÞ ; Again; vÃ32 ¼ v2 sinðh À aÞ sinðw þ sin a sin½180 À ðw þ aފ So, using (13.56) we can write ¼ v2 ; sinðw þ aÞ BC ¼ 2 h1 sinðh À aÞ ð13:57Þ vÃ32 sinðw þ aÞ sin h sinðw þ aÞ sin a Or, v2 ¼ Again; BC ¼ h2 ð13:58Þ ¼ v1 sin h sinðw þ aÞ 2 sin w sin w sinðh À aÞ So, equating (13.57) with (13.58) for BC; we get [from (13.62)] 2 h2 w ¼ 2 h1 sinðh À aÞ ; ) v1 ¼ sin w sinðh À aÞ ð13:64Þ sin sin h sinðw þ aÞ v2 sin h sinðw þ aÞ Or, h2 ¼ sin w sinðh À aÞ ð13:59aÞ From (13.59a) and (13.64), it is seen that h2=h1 ¼ v1=v2; h1 sin h sinðw þ aÞ which satisfies the velocity boundary conditions, because equal volumes of material must pass a given point per unit If the reduction in cross-sectional area of the work-piece time, i.e. h1v1 ¼ h2v2; for the work-piece of unit width. of unit width is given by r ¼ ðh1 À h2Þ=h1; then in terms of the reduction in area, r; (13.59a) can be written as Now from (10.61), the rate of internal energy consump- tion caused by the internal flow field is 1 À r ¼ h2 ¼ sin w sinðh À aÞ ð13:59bÞ dðW :D:Þ kÂvÃ13 vÃ32 à h1 sin h sinðw þ aÞ dt BC ¼ Á AB þ Á ð13:65Þ ) sinðh À aÞ ¼ ð1 À rÞ sinðw þ aÞ ; Again using (13.56), (13.57), (13.61) and (13.62), we can sin h sin w write (13.65) as follows: sin h cos a À cos h sin a Or, sin h ¼ ð1 À rÞ sin w cos a þ cos w sin a ; sin w Or, cos a À cot h sin a ¼ ð1 À rÞfcos a þ cot w sin ag; Or, cot h sin a ¼ cos a À ð1 À rÞ cos a À ð1 À rÞ cot w sin a ð13:66Þ ¼ r cos a À ð1 À rÞ cot w sin a; ) cot h ¼ r cot a À ð1 À rÞ cot w ð13:60Þ Again, the rate of external work done by the extrusion (ram) pressure rL; for the work-piece of unit width, is: Again from the hodograph (velocity vector diagram) in dðW :D:Þ ¼ rL h2v2 ¼ rL h1v1 ð13:67Þ Fig. 13.35b, we get the following: dt 2 2 v1Ã3 ¼ v1 aÞ ; sin a sinðh À Or, v1Ã3 ¼ v1 sin a ð13:61Þ sinðh À aÞ

13.13 Upper-Bound Solution for Plane-Strain Frictionless Extrusion 641 Equating (13.66) with (13.67) for dðW:D:Þ=dt, we obtain friction at billet–chamber and billet–die interfaces is (a) 0.05, (b) 0.15 and (c) 0.4. Neglecting the redundant deformation, determine the maximum load required at the start of the extrusion for (a), (b) and (c). Solution ð13:68Þ Given that the billet diameter before extrusion is D1 ¼ 150 mm, and that after extrusion is D2 ¼ 50 mm, semicone Example angle of die is a ¼ 60; average deformation resistance of Suppose the reduction in cross-sectional area, r; of the aluminium is r0 ¼ 60 MPa ¼ 60 N/mm2; and the length of work-piece of unit width is 1=2; i.e. r ¼ ðh1 À h2Þ=h1 ¼ the billet is L À l ¼ 380 mm (where L is the length of the 1 À ðh2=h1Þ ¼ 1=2; or ; h1=h2 ¼ 2; and semicone die angle, a ¼ 30: billet plus the length of working zone of die, measured from the exit plane of deformation zone). Hence, the length of working zone of die is: If we further assume that h ¼ 70; then from (13.60): l ¼ ½ðD1 À D2Þ=2Š cot a ¼ ½ð150 À 50Þ=2Š cot 60 ¼ 28:87 mm: cot 70 ¼ 0:5 cot 30 À ð1 À 0:5Þ cot w Or; 0:36397 ¼ 0:5  1:732 À 0:5 cot w; Therefore, L ¼ 380 þ 28:87 ¼ 408:87 mm: ) cot w ¼ 0:866 À 0:36397 ¼ 1:00411; (a) When the coefficient of friction, l ¼ 0:05; B ¼ 0:5 l cot a ¼ 0:05 cot 60 ¼ 0:0289: Now, from (13.31) Or, w ¼ 44:8825: the sticking–sliding boundary is:  D1 So; w þ a ¼ 74:8825 xS ¼ 4l and h À a ¼ 40: ln*1=p3ffiffi l\"1 þ B(D12B ) #+ À1 À 1 B D2 1 1  Now; sin h sinðh À aÞ þ sin w sinðw þ aÞ ¼ 3:12349: þ D1 À D2 cot a ¼ 150 From (13.68), we get *2  \" þ4 Â0:002:0859(1502Â0:0289 ) À #+ pffiffi 0:05 1 À1 1 rL ¼ sin 30  3:12349 0:0289 50 2k 2 ln 1= 3 ¼ 0:78087: þ 28:87 ¼ 1647:53mm: Taking different values of h; and proceeding in the way Since xS [ L; so there will be sliding friction over the similar to the above, it can be shown that the lowest value of entire billet–chamber interface. Hence, (13.22) can be taken rL=2k % 0:78; is obtained when h % 72; provided the reduction, r ¼ 50%; and the semicone die angle, a ¼ 30: to determine the maximum load, PL; required at the start of the extrusion, which is: PL ¼ pD12 *\" þ ( 2B ) À # 4 r0 1 B B D1 À1 1 D2 !) 13.14 Solved Problems exp 2B f2L tan a À ðD1 À D2Þg þ 1 D1 01::00228899À30:0578 13.14.1. A cylindrical aluminium billet of 150 mm diameter pð1502Þ ( Á ! and 380 mm length is extruded by direct extrusion process 1 1 to a final diameter of 50 mm through a conical converging ¼ 4  60  À À die with a total die angle of 120 under different frictional conditions, as mentioned below. The average deformation ! ) resistance of aluminium is 60 MPa. Assume that there is no exp 0:0578 fð2  408:87Þ tan 60 À ð150 À 50Þg þ 1 sticking friction along the die land and the coefficient of 150 ¼ 3409303:727 N ¼ 3:41 MN:

642 13 Extrusion (b) When the coefficient of friction, l ¼ 0:15; B ¼ PL ¼ r600\"Â\"1þB1Bþ0:20(3:213DD112(2BÀ155100)þ2Âp0:2ffi334ffi1DÀ11ð)L À # pD12 l cot a ¼ 0:15 cot 60 ¼ 0:0866: Now, from (13.31) ¼ lÞ 4 the sticking–sliding boundary is:  ¼ln*14=Âp15ffi30ffi0:Â150:15\"1 xS þ pffiffi 4 !   1502 380 p þ 0:0866(1502Â0:0866 ) #+   À1 1 3  150 4 À 0:0866 50 ¼ 9939331:842N ¼ 9:94MN: þ 28:87 ¼ 243:7 mm: 13.14.2. A 1-m-long cylindrical aluminium alloy billet of Since L [ xS [ l; so the mixed sliding–sticking frictional 300 mm diameter is extruded at 650 K to a rod of 75 mm condition prevails at the billet–container interface, i.e. full diameter. It is extruded at a ram speed of 25 mm/s with poor sticking friction occurs from x ¼ L (the entry to the lubrication through a square die in direct extrusion process. chamber) to x ¼ xS (sticking–sliding boundary) and the rest portion undergoes sliding friction. Hence, (13.34b) can The coefficient of friction is 0.3, and the average flow stress be taken to determine the maximum ram pressure, rL; which of the alloy at that temperature is given by r0 ¼ 200Àe_tE Á0:15 is: MPa, where e_tE ; is time-average extrusion true strain rate. Neglecting the redundant deformation and the temperature p1ffiffi þ 1 þ 4ðpL ffiÀffi xSÞ ! change of the billet, compute the following: 3l 3D1 rL ¼ r0 (a) Maximum load required at the start of the extrusion. ! (b) Work done in extruding the billet. ¼ 60  pffiffi 1 þ 1 þ pffiffi 4 ð408:87 À 243:7Þ 3  0:15 3  150 Solution ¼ 443:518 N=mm2 (a) Since extrusion occurs through a square die with poor ) Maximum load required at the start of the extrusion lubrication, so a dead zone with a semicone angle, a ¼ (13.36), PL ¼ rL  pðD1=2Þ2 45 is assumed to form in the front of billet. ¼ 443:518  p  ð75Þ2N ¼ 7837610 N ¼ 7:84 MN: Therefore, cot a ¼ 1; and B ¼ l cot a ¼ l ¼ 0:3: Given that the billet diameter before extrusion is D1 ¼ (c) When the coefficient of friction, l ¼ 0:40; B ¼ l cot a ¼ 300 mm, and that after extrusion is D2 ¼ 75 mm, the coef- 0:4 cot 60 ¼ 0:231: Now, from (13.31) the sticking– ficient of friction is l ¼ 0:3; the ram speed is vram ¼ sliding boundary is: 25 mm/s; and the length of the billet including the dead zone is L ¼ 1000 mm, in which the length of the dead zone is:  l ¼ ½ðD1 À D2Þ=2Š cot a ¼ ðD1 À D2Þ=2 150 ¼ ð300 À 75Þ=2 ¼ 112:5 mm: xS ¼ 4  0:4 * pffiffi \" þ0:203:2131(155002Â0:231À1) #+ 0:4 1 1 ln 1= 3  À Equation (13.48b) can be taken to determine the time-average extrusion true strain rate e_tE ; which is: þ 28:87 ¼ À23:5 mm: 12vramD21 tan a   D31 À D23 ln D1 e_ tE ¼ ¼ D2 Since xS\\l; so there will be full sticking friction over the  entire billet–container interface. Hence, (13.28a) can be ð12  25  3002Þ tan 45 ln 300 sÀ1 taken to determine the maximum load, PL; required at the start of the extrusion, which is: 3003 À 753 75 ¼ 1:41sÀ1

13.14 Solved Problems 643 ) Average flow stress, Given that D2 ¼ 20 mm, and the interface friction factor, r0 ¼ 200Àe_tE Á0:15¼ 200  ð1:41Þ0:15 m ¼ 0.1. ¼ 210:578 MPa or N=mm2 Reduction in apreffiffiaffiffi,ffiffiffiffirffiffi ¼ ðA1 À A2Þ=A1 ¼ 1 À ðD2=D1Þ2 Or, ðD1=D2Þ ¼ 1 1 À r: Now, from (13.42) the sticking–sliding boundary is: ) D1 ¼ D2 pffiffi1ffiffiffiffiffiffiffiffi ¼ 10  pffiffi1ffiffiffiffiffiffiffiffi mm:  ln*1=p3ffiffi l\"1 þ ( 2l ) #+ 2 2 1Àr 1Àr D1 l l D1 À1 1 xS ¼ 4l À D2 From (13.9), optimum semicone angle of die is:  þln*ðD11=pÀ2 3ffiffiDÂ2Þ0¼:3\"41Â3þ0000:30:3:3(370502Â0:3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 D1 aopt ¼ m ln ) #+ 2 D2 À1 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À ¼ 3  0:1 ln D1 þ 112:5 ¼ À106:543 mm: 2 D2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:15 ln D1 Since xS\\l; so there will be full sticking friction over the D2 entire interface from the entry plane of the dead zone to the sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi entry to the chamber. Hence, (13.41) can be taken to ¼ 0:15 ln pffiffi1ffiffiffiffiffiffiffiffi rad: determine the maximum load, PL; required at the start of the 1Àr extrusion, which is: ¼ \" þ l(D12lÀ1) If x is the distance measured from the exit plane to the r0 1 l D2 entry plane of deformation zone, then from (13.10a): PL þ pffi2ffi f2L À ðD1 À D2Þg ! x ¼ D1=2 À D2=2 tan aopt 3D1 \" þ0:30:3(370502Â0:3À1) ¼ D1=2 À 10 mm pD12 ¼ 210:578 1 tan aopt  4 ¼ qffiffiffiDffiffiffiffi1ffiffi=ffiffi2ffiffiffiÀffiÀffiffiffiffi1ffiffiffi0pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÁffiffi mm: tan 0:15 ln 1= 1 À r þ pffiffi 4 ð1000 À ! p  3002 112:5Þ 3  300 4 The values of the size of incoming stock, ÆD1=2; which ¼ 185376772:1 N ¼ 185:38 MN: are equal to the die gaps at the entry to the die, for a wide range of reductions, r; up to 50% are calculated at the cor- (b) Work done in extruding the billet is: responding values of x and presented in Table 13.3. W:D: ¼ PL  L ¼ 185376772:1  1 N-m ¼ 185376772:1 J Taking the values of x and ÆD1=2; from the last two ¼ 185:38 MJ: columns of Table 13.3, the trumpet-shaped die profile is drawn with abscissa as x along the die axis and ÆD1=2; as 13.14.3. It is desired to keep the optimum cone angle of a ordinate on two opposite sides of the die axis and shown in conical converging die for a wide range of reductions up to 50% with a product diameter of 20 mm in each case. Fig. 13.36. Considering the interface friction factor to be 0.1, draw the trumpet-shaped die profile for the above condition. To draw 13.14.4. During the direct extrusion of a 1.5-m-long billet the die profile, take coordinate points as many as possible, with diameter of 500 mm through a conical converging die, keeping the die exit at coordinate points of (0, +10) and (0, it was noted that the extrusion loads were, respectively, −10) on two opposite sides of the die axis. 100 MN and 50 MN when the ram moved one-third and two-thirds distance from the entry to the container. The Solution average deformation resistance of the billet is 200 MPa. Neglecting redundant deformation and assuming Coulomb’s Let the diameters and the cross-sectional areas are respec- law of sliding friction with a constant value of coefficient of tively D1 and A1 for incoming stock, and D2 and A2 for friction to be valid throughout the homogeneous deforma- outgoing product. tion process, calculate the value of coefficient of friction at the billet–container interface.

644 13 Extrusion Table 13.3 Values of x and ÆD1=2 at various values of r ÆD1=2 ðin mmÞ ¼ Æ10 Â ðD1=D2Þ Reduction in area, r ðD1=Dp2ffiffiÞffiffiffiffiffiffiffiffi tan aopqt ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÁffiffi x ðin mmÞ ¼ 1= 1 À r ¼ tan 0:15 ln 1= 1 À r ¼ ðD1=2 À 10Þ= tan aopt Æ10 Æ10.05 01 0 Æ10.1 1.834 Æ10.15 0.01 1.005 0.027 2.607 Æ10.21 3.208 Æ10.26 0.02 1.01 0.039 3.723 Æ10.31 4.183 Æ10.37 0.03 1.015 0.048 4.605 Æ10.43 5 Æ10.48 0.04 1.021 0.055 5.372 Æ10.54 5.728 Æ10.85 0.05 1.026 0.062 6.069 Æ11.18 7.636 Æ11.55 0.06 1.031 0.068 9.073 Æ11.95 10.46 Æ12.4 0.07 1.037 0.074 11.83 Æ12.91 13.23 Æ13.48 0.08 1.043 0.079 14.68 Æ14.14 16.21 0.09 1.048 0.084 17.85 0.1 1.054 0.089 0.15 1.085 0.111 0.2 1.118 0.13 0.25 1.155 0.148 0.3 1.195 0.165 0.35 1.24 0.182 0.4 1.291 0.198 0.45 1.348 0.215 0.5 1.414 0.232 Fig. 13.36 Trumpet-shaped die 15 profile, drawn taking the values of 10 x and ÆD1=2 from the last two columns of Table 13.3 Incoming stock radius, D1/2 (mm) 5 0 0 2 4 6 8 10 12 14 16 18 20 Axial distance from die exit, x (mm) -5 -10 -15 Solution 106 N/m2: When the ram moved one-third from the entry to the container, the length of the billet is L1 À l ¼ 1 m, and for Given that the billet diameter before extrusion is D1 ¼ 0:5 m, two-thirds distance movement of the ram from the entry to the length of the billet is L À l ¼ 1:5 m (where L is the length the container, the length of the billet will be L2 À l ¼ 0:5 m: of the billet plus the length l of the working zone of die, measured from the exit plane of deformation zone). The Given that when L1 À l ¼ 1 m, the extrusion load is PL1 ¼ average deformation resistance of the billet is r0 ¼ 200 Â 100 Â 106 N/m2; and when L2 À l ¼ 0:5 m; the extrusion

13.14 Solved Problems 645 load is PL2 ¼ 50  106 N/m2: Suppose, the initial 13.Ex.4. A round billet of 200 mm diameter undergoes an cross-ÀsectioÁnal area of the billet is A1; where indirect extrusion through a square die with poor lubrication A1 ¼ pD21 =4 ¼ fðp  0:52Þ=4g m: to produce an extrudate of 50 mm diameter. Assume that the coefficient of friction at the billet–die interface is 0.5. If the Let us consider (13.19d), which can be written in the average flow stress of the billet material is 100 MPa, cal- culate the load required for extrusion. Neglect the redundant following form: deformation.  13.Ex.5. A steel billet of 100 mm diameter and 500 mm rL1 À r0 ¼ 4l length undergoes an open-die extrusion through a lubricated ln rl À r0 D1 ðL1 À lÞ conical converging die with a die angle of 100° to produce an extrudate of 40 mm diameter. If the average flow stress of ¼ 4l  1 ¼ 8l ð13:69Þ the steel is 300 MPa and 5 min is taken for the extrusion 0:5 process, calculate the work done in extruding the billet and the power utilized in the operation, neglecting the redundant  deformation. Assume that the coefficient of friction at the rL2 À r0 ¼ 4l billet–die interface is 0.1. ln rl À r0 D1 ðL2 À lÞ 13.Ex.6. A 400-mm-long cylindrical copper billet of ¼ 4l  0:5 ¼ 4l ð13:70Þ 175 mm diameter is extruded at 850 °C by direct extrusion 0:5 process at a ram speed of 50 mm/s through a conical con- verging die with a total included angle of 90° to produce an where rL1 and rL2 are the extrusion pressures, respectively, extrudate of 75 mm diameter. The average flow stress of at a distance of L1 and L2 from the exit plane of the defor- copper at that temperature is given by r0 ¼ 240Àe_tE Á0:06 mation zone, rl is the extrusion pressure at the entry to the MPa, where e_tE ; is time-average extrusion true strain rate. die and l is the coefficient of friction at the interface, which Assume that there is no sticking friction along the die land and the coefficient of friction at billet–chamber and billet–die is to be determined. interfaces is 0.2. Determine the maximum load required for the operation, neglecting the redundant deformation and the Subtracting (13.70) from (13.69), we get temperature change of the billet.  13.Ex.7. When round billets of 100 mm diameter are rL1 À r0 extruded through a conical converging die with semicone ln rL2 À r0 ¼ 4l; angle of 60° to get extrudates of 60 mm diameter, the  chevron cracks are very frequently observed to form in A1 rL1 À A1 r0  PL1 À A1 r0  extrudates. Justify it. Determine the maximum allowable A1 rL2 À A1 r0 PL2 À A1É r0 semicone angle of the die so that the formation of these Or, ln 1 100  106 À ¼ ln 0:52Þ=4  ¼ 4l; Á! cracks may be avoided. )l ¼ ln È À 4 50  106 À 200  106 13.Ex.8. Indicate the correct or most appropriate answer(s) ðp  from the following multiple choices: fðp  0:52Þ=4g  ð200  106Þ (a) Which of the following respective combination of reduction and coefficient of friction will give the lowest ¼ 1 60:73  106 optimum semicone die angle? 4 ln 10:73  106 (A) 45% and 0.2; (B) 45% and 0.4; ¼ 1 lnð5:66Þ ¼ 0:43: (C) 30% and 0.2; (D) 30% and 0.4. 4 (b) Extrusion ratio of 200:1 corresponds to the percentage Exercise reduction in cross-sectional area of 13.Ex.1. If an alloy is extruded at 50 mm/s from 150 mm (A) 99; (B) 99.25; (C) 99.5; diameter to 50 mm diameter through a conical converging (D) 99.75; (E) none of the above. die with a semidie angle of 60°, calculate the value of time-average extrusion true strain rate. 13.Ex.2. If D1 is the diameter and v1 is the velocity of a cylindrical billet at the entry to a conical converging die having a semi-angle a; show that at any axial distance x from the entry plane of the die towards the exit plane of the deformation zone, the true strain rate e_ of extrusion through the die is e_ ¼ 4v1D12 tan a : ðD1 À 2x tan aÞ3 13.Ex.3. If the interface friction factor is 0.8 and the extru- sion stress during sound flow is the minimum for a conical converging die of angle 80°, what will be the percentage reduction in cross-sectional area during extrusion?

646 13 Extrusion (c) The allowable extrusion ratio at a given working tem- 13.Ex.8. (a) (C) 30% and 0.2. (b) (C) 99.5. (c) (B) respec- perature increases if the extrusion pressure and the speed of tively increases and decreases. (d) (D) both low (e) (D) deformation in extrusion molten glass. (f) (A) Impact extrusion. (g) (B) Fir-tree cracking. (h) (B) is independent of the billet length; (C) in- (A) both increase; creases with increasing the extrusion ratio. (B) respectively increases and decreases; (C) respectively decreases and increases; References (D) both decrease. Avitzur, B.: Metal Forming: Process and Analysis, TMH edn., vol. (d) Chevron cracking during extrusion through certain high 1977, p. 165. McGraw-Hill, Inc., New York (1968) angle of conical converging die may occur most easily if the extrusion ratio and the friction at the die–job interface are: Boulger, F.W., Wilcox, B.A.: The application of high pressure in metal-deformation process. DMIC Rept., 199, 2 Mar 1964 (1964) (A) respectively high and low; (B) respectively low and high; Boulger, F.W., Gehrke, J.H.: Metal working process development. (C) both high; Third DMIC Status Report, no. 218, June 1965, pp. 41–60 (1965) (D) both low. Bryant, A.J., Dixon, W., Fielding, R.A.P., Macey, G.: Isothermal (e) Lubricant used for hot extrusion of steel in extrusion. Light Met. Age (1999) Ugine-Sejournet process is Chadwick, R.: Developments and problems in package extrusion press (A) graphite; (B) lanolin; design. Met. Mater., 162–170 (1969) (C) molybdenum sulphide; (D) molten glass. Dieter, G.E.: Mechanical Metallurgy, p. 616. McGraw-Hill Book (f) The manufacturing process for dry cell zinc cans is: Company (UK) Limited, London (1988) (A) Impact extrusion; (B) Rolling; Gurney, F.J., DePierre, V. Trans. ASME, Ser. B J. Eng. Ind. 96, 912– 916 (1974) (C) Forging; (D) Cold drawing. Harris, J.N.: Mechanical Working of Metals: Theory and Practice, (g) The type of defects observed in extrusion is: p. 160. Pergamon Press, Oxford, U.K. (1983) (A) Earing; (B) Fir-tree cracking; Hirst, S., Ursell, D.H.: Proc. Conf. Tech. Eng. Mfr, 149 (1958) Kulkarni, K.M., Schey, J.A., Wallace, P.W., DePierre, V.: J. Inst. Met. (C) Flash cracking; (D) Alligatoring. 100, 33–39 (1972) (h) The pressure required to maintain plastic flow of metal Laue, K.: Isothermal extrusion. Z. Metallkd. 51, 491 (1960). (in during indirect extrusion German) (A) increases with increasing the billet length; Laue, K., Stenger, H.: Extrusion: Processes, Machinery, Tooling. ASM, (B) is independent of the billet length; (C) increases with increasing the extrusion ratio; Metals Park, Ohio (Translated from the German) (1981) (D) is independent of the extrusion ratio. Lewandowski, J.J., Lowhaphandu, P.: Effects of hydrostatic pressure Answer to Exercise Problems on mechanical behaviour and deformation processing of materials. Int. Mater. Rev. 43(4), 145–187 (1998) 13.Ex.1. 7.9 s−1. Maier, J.: CVD coating technology for increased lifetime of aluminum 13.Ex.3. 55.6%. extrusion dies. In: Proceedings of the Eighth International Alu- 13.Ex.4. 28.27 MN. minum Extrusion Technology Seminar (ET ’04), Wauconda, IL, ET 13.Ex.5. 2.53 MJ; 8.43 kW. Foundation, 18–21 May 2004 (2004) 13.Ex.6. 44 MN. Morris, F.W.: J. Inst. Met. 90, 101–106 (1961–1962) 13.Ex.7. Since for 60° semicone angle of the die, D ¼ Pearson, C.E.: The Extrusion of Metals. Chapman & Hall, London Dm=Lc ¼ 3:46 [ 2; so the probability to form chevron (1953) cracking is high. In order to avoid the possibility of chevron Pugh, H.LI.D., Low, A.H.: The hydrostatic extrusion of difficult metals. cracking, the maximum allowable semicone angle of the die J. Inst. Met., 201–217 (1965) is 30°. Rogers, J.A., Rowe, G.W.: J. Inst. Met. 95, 257–263 (1967) Rowe, G.W.: Principles of Industrial Metal Working Processes. Edward Arnold (Publishers) Ltd., U.K., p. 84, 201 (1977) Saha, P.K.: Aluminium Extrusion Technology. ASM International, Materials Park, Ohio (2000) Sejournet, J., Delcroix, J.: Glass lubricant in the extrusion of steel. Lubr. Eng. 11, 389–396 (1955) Smith, C.S.: J. Inst. Metals 76, 429 (1949–1950) Takahashi, M., Yoneyama, T.: Isothermal extrusion of aluminum alloys. In: Proceedings of the Eighth International Aluminium Extrusion Technology Seminar (ET ’04), Wauconda, IL, ET Foundation, 18–21 May 2004 (2004) Wilcox, R.J., Whitton, P.W.: J. Inst. Met. 88, 145 (1959–1960)

Drawing: Flat Strip, Round Bar and Tube 14 Chapter Objectives • Drawing: definition, purposes, demerits and fields of application. • Drawing stress with friction for work-hardening and non-strain-hardening strip drawn through wedge-shaped dies. Frictionless ideal drawing stress. Maximum reduction of area in a single pass with and without friction. • Drawing stress with friction for strip drawn through cylindrical dies. • Treatments of work metal prior to drawing, such as heat treatment (recrystallization annealing, patenting), surface preparation (pickling, surface conditioning or coating), and pointing or chamfering. • Drawing equipments, such as draw bench, multiple-die continuous and single-die block drawing machines, stepped-cone multiple-die drawing machine. Conical converging drawing die, describing internal profile geometry and functions of dif- ferent zones of die nib. • Drawing of rod and wire: dry drawing and wet drawing. Drawing load and power with friction, and with and without back tension, with and without consideration for work hardening. Frictionless ideal drawing stress without back tension. • Maximum reduction of area in a single drawing pass: for non-strain-hardening rod or wire with and without friction and back tension, and for strain-hardening rod or wire with friction and back tension. • Redundant deformation, redundant work factor and its effect on drawing stress. • Variation of drawing stress with die-cone angle, describing optimum cone angle, dead-zone formation and shaving mode of flow. • Tube drawing using fixed plug, floating plug and mobile mandrel. Stress and maximum reduction of area in a single pass for plug drawing and mandrel drawing. Tube sinking and stress for sinking. Stability condition of a floating plug. • Application of slip-line field to frictionless plane-strain strip drawing through wedge-shaped dies. • Upper-bound solution for frictionless plane-strain strip drawing through wedge- shaped dies, with an example. • Problems and solutions. © Springer Nature Singapore Pte Ltd. 2018 647 A. Bhaduri, Mechanical Properties and Working of Metals and Alloys, Springer Series in Materials Science 264, https://doi.org/10.1007/978-981-10-7209-3_14

648 14 Drawing: Flat Strip, Round Bar and Tube 14.1 Introduction production of angles, channels and strips. However, many of the smaller-sized round bars are drawn to give larger Drawing is usually a cold working operation which involves reductions of area that may go up to 50% reduction per pass. the shape change or the reduction in the cross-sectional area Annealed wires may be drawn with 90% reduction in of a workpiece, usually a flat strip, round bar or tube, by sequential passes prior to their re-annealing. The final pulling through a die with a tensile force applied at the exit diameter of some wires is 0.025 mm or even less, when they end of the die. The reaction of material with the die under the are drawn through a large number of dies before arriving at application of tensile force develops two mutually perpen- the final size and may be re-annealed for several times dicular compressive forces that cause most of the plastic between drafts. The draft is the amount of reduction in the deformation. The drawing is somewhat similar to the process cross-sectional area of the workpiece per pass during of extrusion except that the workpiece in drawing is pulled drawing operation. The outside diameter of a hypodermic with tensile force applied at the exit end of the die, whereas needle of stainless steel produced by cold drawing is about the workpiece in extrusion is pushed with compressive force 0.25 mm. Most hot-worked (rolled or extruded) tubes are applied at the entry side of the die. Drawing is usually cold-drawn single time or several times at room temperature, performed at room temperature, but the temperature during some to wall thicknesses of only a few hundredths of a this operation will rise considerably due to heat generated millimetre. On the other hand, large hydraulic benches may from large deformations usually involved in this process. An be used to cold-draw large storage cylinders with diameter of important deviation is the warm working of tungsten to 0.5 m or more. produce incandescent lamp filament. To produce a desired shape and/or size, most often the raw stock is successively 14.2 Strip Drawing Through Wedge-Shaped reduced by drawing with several passes through a series of Dies progressively smaller dies. Intermediate heat treatment or annealing may be necessary to restore ductility so that the Strip drawing is not so common production process as tube workpiece can be further cold worked. Drawing is usually a drawing and rod or wire drawing, but it is probably the most finishing operation, and the drawn products are either often studied deformation in the field of mechanics of metal applied directly in service or subjected to further processing, working. This section will discuss about the determination of like machining or bending, to get the desired size or/and plane-strain drawing stress for wide flat strip drawn through shape. Apart from obtaining the desired shape and size of the wedge-shaped dies using slab method. At first, drawing of product, the main purposes of cold drawing are: non-strain-hardening strip with and without friction will be discussed and then drawing of strain-hardening strip with (1) To obtain closer dimensional tolerances, friction will be covered. The drawing of strip through (2) To improve surface finishes and wedge-shaped dies is shown in Fig. 14.1. (3) To improve the mechanical properties by strain hard- The exit plane of the deformation zone, i.e. the plane at ening; drawing usually increases the strength and the exit side of the die where the reduction of the stock ends hardness of the metal. (henceforth only called as the exit plane), is considered as the origin of the coordinate system, shown by x = 0 in the It is claimed (Gokyü and Ōkubo 1964) that drawing with diagram, where x is the horizontal axis that increases from superimposed vibrations reduces friction and wear of the the exit to the entry plane of the deformation zone, i.e. the dies and improves surface finish of the product. But the entry plane of the stock within the die (henceforth only demerits of cold drawing are the requirement of higher called as the entry plane). On the flat strip, the inclined dies deformation load and the limitation of ductility in metal. exert a pressure p normal to the die–strip interface, which will be called the die pressure. Outward horizontal flow of Cold drawing is applied as a finishing operation for large the material during the process of its deformation leads to quantities of rods, wires, tubes and special sections. The tangential shearing frictional stresses s at the die–strip con- starting metals for drawing are ferrous and non-ferrous tact surfaces. Hence, the direction of the friction stresses metals mostly in the form of rolled or formed sections, rolled opposing the outward horizontal flow will always be towards or extruded rods, welded pipes, and seamless rolled or the entry plane. However, the presence of frictional shear extruded pipes. Frequently, large round bars up to 150 mm stress s leads to the longitudinal stress rx, in the horizontal diameter or sometimes more are lightly cold-drawn to reduce direction, where rx increases from the plane of entry to the the diameter by 1.5 mm, with an objective to improve the plane of exit. Let us take the thicknesses of the strip at the dimensional tolerance and surface finish. Similar light cold reduction passes (sizing passes) are also given in the

14.2 Strip Drawing Through Wedge-Shaped Dies 649 Fig. 14.1 Stresses acting on an p h2 elemental slab in the deformation 1 µp α zone during plane-strain drawing σF of a wide, flat strip through dh wedge shaped dies of constant 2 (Drawing stress) angle p h1 h σx + dσx σx 2α w dx 1 dx pcos α dx dh cos α 2 µp p x x= 0 entry and exit planes to be h1 and h2, respectively. To make (ix) No back tension or back pull is applied to the strip, the analysis simpler, the following assumptions are made. i.e. at the entry plane, rx ¼ 0: (i) Material being drawn is assumed to be isotropic and (x) The strip is thin enough to assume that the die pres- homogeneous. sure, p, remains constant through the thickness. (ii) Homogeneous deformation is assumed, i.e. internal (xi) The horizontal longitudinal stress rx is constant across redundant work is neglected. the cross-section of the strip and is a principal stress since no shear stress acts on this cross-section in the (iii) Elastic strain is neglected because plastic deformation vertical direction. involved in drawing is quite large compared to insignificant elastic deformation. 14.2.1 Drawing Stress with Friction (iv) The original height of the strip, h1, is assumed to be Let us determine the pulling stress exerted by the draw very much less than its width w, resulting in a negligible bench grip for drawing a wide, flat non-hardening strip lateral spread. Then the conditions approximate closely metal through two wedge-shaped dies with a total included to plane strain deformation, i.e. the strip is reduced in angle of 2a; under sliding frictional condition. thickness and extended in length but the width of the strip remains nearly constant during deformation. Let us consider the state of stresses on a vertical slab element of thickness dx inside the strip under the die. (v) For non-strain-hardening strip, it is assumed that According to free-body equilibrium approach, the horizontal there is no strain hardening of the work-piece, i.e. the longitudinal stresses are, respectively, rx and rx þ drx at an plane strain flow stress or deformation resistance r00 of arbitrary distances x and x þ dx; from the exit plane, which is the material being drawn remains constant. taken as x ¼ 0. Figure 14.1 shows that these horizontal stresses along with the die pressure p and friction stress lp (vi) Coulomb’s law of sliding friction holds well at the are acting on this slab element of thickness dx and width w, contact surface between the dies and the flat strip, whose height varies arbitrarily from h to h + dh. Consider- wherever interface friction is considered. For cases of ing the horizontal equilibrium of forces acting on this ver- friction, s ¼ lp; where, s is the tangential frictional tical slab element, the horizontal longitudinal stress, rx, at shear stress at the contact surface, l is the coefficient the exit plane can be determined. There are three force of friction, and p is the die pressure on the flat strip components acting in horizontal direction along x-axis: normal to the contact surface. (vii) The coefficient of friction l is the same at all points on the surface of contact between the dies and the flat strip.

650 14 Drawing: Flat Strip, Round Bar and Tube 1. Due to change in horizontal longitudinal stress in the pv dx w ¼ p cos a dx w À lp sin a dx w; positive direction of x, i.e. from the exit to entry side: cos a cos a ðrx þ drxÞðh þ dhÞw À rxhw Or; pv ¼ p À lp tan a ð14:6Þ ¼ ðrx dh þ h drx þ drx dhÞw ð14:1Þ l and a are small quantities, and so l tan a ( 1; which can be explained with typical values, such as l ¼ 0:05 and 2. Due to the die pressure at two interfaces in the direction a ¼ 10; then tan a ¼ 0:176; and l tan a ¼ 0:009 ( 1: Therefore, lp tan a can be neglected and from (14.6) it can from the exit to entry side: be written as pv % p: Hence, p can be considered as the principal stress.  dx Since the stresses required for plastic flow during working 2p w cos a sin a ¼ 2pw tan a dx ð14:2Þ operations are mainly compressive in nature, the principal 3. Due to the frictional stress at two interfaces in the stresses r1; and r2; in Von Mises’ yielding criterion for plane strain condition given by (1.88b), are considered to be posi- direction from the exit to entry side: tive, if they are compressive and negative when they are  dx tensile stresses. As the die pressure p is compressive in nature, 2lp w cos a cos a ¼ 2lpw dx ð14:3Þ and the horizontal longitudinal stress, rx is tensile in nature, therefore r1 ¼ p; and r2 ¼ Àrx; and (1.88b) then becomes For steady-state drawing conditions, the horizontal equi- librium of forces acting on this vertical slab element in the r1 À r2 ¼ p þ rx ¼ r00 ð14:7Þ positive x-direction, i.e. from the exit to entry side, is obtained by equating the summation of (14.1), (14.2) and where, Àr00p= fficffiÁompressive flow stress in plane strain condi- (14.3) to zero. Eliminating the constant width w, and tion = 2 3  ÀcompprffieffiÁssive flow stress in homogeneous neglecting the term drx dh; as it is a product of two very strain condition = 2 3 r0: small quantities, this gives Now, denoting l cot a ¼ B; we get from (14.5) and rx dh þ h drx þ 2p tan a dx þ 2lp dx ¼ 0 ð14:4Þ (14.7): h d x   x   0   x 1  B dh  0 Or , h d x   x   0 1  B   x  dh    x  B x   0 1  B   x  dh  Bx d x  B  dh (14.8) h   0 1 From the geometry, we can write dx ¼ dh=2 tan a; which Equation (14.8) is the basic differential equation of wide simplifies (14.4) as follows: strip drawing. This form of equation was first given by Sachs et al. (1944) in 1944. Integrating (14.8), we get rx dh þ h drx þ p dh þ lp cot a dh ¼ 0 h drx þ ½rx þ pð1 þ l cot aފ dh ¼ 0 ð14:5Þ 1  À r00ð1 þ à ¼ ln h þ C; B ln Brx BÞ Let, the vertical pressure exerted by the wedge-shaped where C is an integration constant: dies on the flat strip is pv. The vertical pressure pv is assumed to be uniform throughout a transverse section and considered Or; Brx À r00 ð1 þ BÞ ¼ AhB ð14:9Þ as a principal stress. Equation (14.5) can be solved by integration if a relationship between rx and p would be where A ¼ eBC, is a new integration constant, which can be found. For this, let us take the equilibrium of vertical forces evaluated from the following boundary condition. This exerted by the die on the slab element, which is given by condition is: at the entry plane where h ¼ h1; rx ¼ 0; since there is no back tension (assumed). Thus, from (14.9) we get

14.2 Strip Drawing Through Wedge-Shaped Dies 651 A ¼ À r00ð1 þ BÞ : drx ¼ À dh ð14:14Þ ) \"h1B r00 h # 1  B BÞ Integrating (14.14) between the lower limit of h = h1, at B h the entry plane where rx ¼ 0; and the upper limit of h = h2, rx ¼ Àr00ð1 þ þ r00ð1 þ at the exit plane where rx ¼ ðrFÞideal; (say), we get the ideal BÞ drawing stress applied under frictionless condition for the \" h1 1  h B# reduction of height from h1 to h2, of a non-strain-hardening Or, rx ¼ 1 þB À h1 ð14:10Þ strip metal, as follows: r00 B rx is the direct pulling stress in the direction of drawing at any Zrx¼ðrF Þideal hZ¼h2 arbitrary distance x from the exit plane, considered as the dh h origin, x = 0, and rx varies with the position x. Equation drx ¼ À r00 (14.10) shows the expression for the ratio of the working rx ¼0 h¼h1 stress rx; to the plane stress yield stress r00 , of the working metal, which is very useful in plotting the results in a general r00 h1 h2 way without relating a particular metal. The deformation ) Ideal drawing stress, ðrF Þideal ¼ ln ð14:15Þ stress for any given metal can easily be determined if its yield stress is known. Further from this dimensionless ratio, one Since no additional work is required to overcome the can quickly compare with the stress required for ideal friction, so for the same reduction in thickness of the strip, deformation and get a measure of the process efficiency in the ideal drawing stress ðrFÞideal is less than the drawing stress rF in frictional condition. For frictionless ideal terms of frictional losses. Now, the die pressure, p, at any deformation, the strip drawing load, Pideal; is given by arbitrary distance x from the origin can also be evaluated from (14.10) using yielding criterion given by (14.7) as follows: Ideal drawing load, Pideal ¼ ðrFÞidealðwA2Þ p ¼ 1 À rx ¼ 1 À 1 þ B \" À  h B# ð14:11Þ ¼ r00 wA2 ln h1 ð14:16Þ r00 r00 B 1 h1 h2 Similar to rx; p also varies with the position x. Equa- 14.2.3 Maximum Reduction of Area in a Single tions (14.10) and (14.11) show that rx increases and p de- Pass With and Without Friction creases as the arbitrary height h of the strip decreases from h1 to h2. Hence, at the exit plane, h = h2, rx ¼ rFðsayÞ; If r is the reduction of area in one pass then where rF is the drawing stress applied under sliding fric- tional condition for the reduction of height from h1 to h2, of a r ¼ ðh1 À h2Þw ¼ 1 À h2 or, h2 ¼ 1 À r non-strain-hardening strip metal. h1w h1 h1 ) Drawing stress, rF ¼  \" À h2B# Or, h1 ¼ 1 : r00 1 þB h1 h2 À 1 r 1 B ð14:12Þ Equations (14.12) and (14.15) can be written in terms of reduction of area, r, for a non-strain-hardening strip metal If P is the drawing load, then with friction and without friction (ideal condition), respec- tively, as given below. Drawing load, P ¼ rFðw h2Þ \"  B# ¼ r00 w h2 1 B h2 þ À With friction: B 1 h1  B 1 ð14:13Þ rF ¼ þ 1 À ð1 À rÞBà ð14:17Þ r00 B Frictionless ideal condition:  14.2.2 Frictionless Ideal Drawing Stress ðrF Þideal ¼ ln 1 ð14:18Þ r00 1Àr Suppose the dies are perfectly lubricated so that there is no interfacial friction, i.e. the coefficient of friction, l ¼ 0, and It is possible to increase the reduction of area per pass in so, B ¼ l cot a ¼ 0. Under this condition, (14.8) reduces to strip drawing till the drawn strip does not undergo tensile fracture ahead of the die. Since we have assumed that the

652 14 Drawing: Flat Strip, Round Bar and Tube work metal does not strain harden, so the drawing stress for realistic than the flow stress considered at a mean value of true strain. But when the work metal undergoes deformation non-strain-hardening strip metal must be equal to or less than at the initial stage after annealing, its rate of strain hardening is relatively high. For such cases, or for any deformation the uniaxial flow stress of the work metal, r0; to avoid where the metal work hardens rapidly, it is preferable to tensile fracture, and not the plane-strain flow stress r00 of the incorporate the relation showing the variation of stress with material being drawn, because the strip of finite width may strain in the basic differential equation given by (14.8). Let us assume that the work-hardening strip is subjected to a freely undergo lateral contraction after drawing. Therefore, back tension, rB, applied at the entry plane of die. Back when the maximum drawing stress is equal to r0; the tension is a horizontal tensile stress applied at the back side maximum possible reduction of area, rmax, in one pass can of the deformation zone. be achieved. Since the increment of true strain, de, can be defined by If ðrFÞmax is the maximum drawing stress at the exit the reduction in height, (−dh), based on the instantaneous plane, we get the maximum possible reduction of area, rmax, height, h, i.e. de ¼ Àdh=h; so the (14.8) change to in one pass for non-strain-hardening strip metal with friction from (14.17) as follows: ðrF Þmax ¼ r0 ¼ pffiffi ¼ 0:866 ¼  þ B À ð1 À rmax ÞB à drx r00 r00 3  1 1 ; r00ð1 2 B Brx À þ BÞ ¼ Àde;  B Or; 1 À ð1 À rmaxÞB¼ 0:866 1þB Or, drx ¼ r00ð1 þ BÞ À Brx; de  !1 B B drx 1þB de ) With friction: rmax ¼ 1 À 1 À 0:866 ) þ Brx ¼ r00 ð1 þ BÞ ð14:21Þ ð14:19Þ Multiplying both sides of (14.21) by expðBeÞ; and then integrating it between the lower limit of true strain e ¼ Âà e1ðsayÞ; at the entry plane, since the work metal has not yet If ðrFÞmax ideal is the maximum ideal drawing stress at the been deformed and the upper limit of true strain e ¼ e2ðsayÞ; exit plane, the maximum possible reduction of area, ðrmaxÞideal; at the exit plane, we get in one pass for non-strain-hardening strip metal in friction- less ideal condition is obtained from (14.18) as follows: Âà pffiffi ! Ze¼e2 Ze¼e2 ðrF Þmax ideal r0 ¼ 3 ¼ 0:866 ¼ ln 1 drx ¼ r00 1 À ðrmaxÞideal ; expðBeÞ de de þ Brx expðBeÞ de r00 2 e¼e1 e¼e1 1 Or; 1 À ðrmaxÞideal ¼ expð0:866Þ Ze¼e2 ¼ r00ð1 þ BÞ expðBeÞ de ) condition:ðrmaxÞideal¼ À 1 ( e¼e1 Z drx ! & 'Z drx ! )e¼e2 Frictionless 1 expð0:866Þ de d de de de expðBeÞ de À de expðBeÞ e¼e1 1 ¼ 1 À 2:377 ¼ 0:579 ’ 58% Ze¼e2 ð14:20Þ þ Brx expðBeÞ de ) Theoretically maximum possible reduction in a single e¼e1 or, pass for non-strain-hardening strip metal is 58%. Ze¼e2 ¼ r00ð1 þ BÞ expðBeÞ de; 14.2.4 Drawing Stress for Work-Hardening Strip e¼e1 For heavily cold-worked strips, either the metal does not Ze2 Ze2 practically have the capacity to further strain harden or the ½expðBeÞrxŠee21 À ½B expðBeÞrxŠ de þ Brx expðBeÞ de rate of strain hardening is very low and steady. In such cases, the assumption of constant flow stress is practically accept- e1 e1 able. However, it would be better if strain hardening of the strip is included in the equations for drawing stress by Ze2 replacing r00 with the mean plane strain flow stress r00; cal- ¼ r00ð1 þ BÞ expðBeÞ de culated by means of (10.30), which is obviously more e1 Since at the entry plane, e ¼ e1; and rx ¼ rB (back ten- sion), and at the exit plane, e ¼ e2; and rx ¼ rF; which is the pulling or drawing stress, then

14.2 Strip Drawing Through Wedge-Shaped Dies 653 rF expðBe2Þ À rB expðBe1Þ Ze2 ð14:25Þ Ze2 rF expðBe2Þ ¼ ð1 þ BÞ r00 expðBeÞ de ð14:22Þ ¼ ð1 þ BÞ r00 expðBeÞ de 0 e1 Let us assume a linear strain-hardening behaviour in the The right-hand side of (14.22) may be evaluated for uniform plastic deformation region, which is approximately different assumed stress–strain relationship. observed by many metals. Let us consider (14.25) for such (a) Non-strain-hardening Strip strain-hardening strips. Suppose the linear relation between the plane strain flow stress r00 and the plastic strain e is given by From (14.22), we can obtain the drawing stress for r00 ¼ r00 1 þ ce ð14:26Þ non-strain-hardening strip having constant plane strain flow where r00 1 ¼ the initial plane strain flow stress, when the stress r00 , as given by (14.12). Since r00 ¼ constant, so it can plastic strain e ¼ e1 ¼ 0: Now from (14.25) and (14.26), we be taken outside the integrand of (14.22), which gives get rF expðBe2Þ À rB expðBe1Þ ¼ ð1 þ BÞr00 expðBe2Þ À expðBe1Þ ð14:23Þ Ze2   B rF expðBe2Þ ¼ ð1 þ BÞ r00 1 þ ce expðBeÞ de Since there is no back pull or back tension, so the hori- Z0 Z zontal tensile stress at the entry plane is rx ¼ rB ¼ 0; and ¼ ð1 þ BÞ r001 expðBeÞ de þ ce expðBeÞ de (14.23) becomes Z &Z ' !e2 À d expðBeÞde de  de ðceÞ 1þB þ BÞ rF expðBe2Þ ¼ r00 B ½expðBe2Þ À expðBe1ފ; expðBeÞ þ ce expðBeÞ À 0Z c expðBeÞ !e2 þ BÞ B de  expðBe1 ! ¼ ð1 r001 BB 1þB expðBe2 Þ ¼ ð1 0 Or, rF ¼ r00 B 1 À Þ ; expðBeÞ!e2 expðBeÞ expðBeÞ B2 0 r001 B þ ce B À c Ze2 Zh2 À dh; h1 ! h h2 Since de ¼ or ; e2 À e1 ¼ ln ; ¼ 1þB r00 1 expðBe2Þ À r00 1 þ ce2 expðBe2 Þ À c expðBe2 Þ þ c B B B e1 h1 h   B 1 þ B r001 h1 ¼ ln h1 ) rF expðBe2Þ ¼ B þ ce2 expðBe2Þ So; Be2 À Be1 ¼ B ln h2 ; ! Or; h2 expðBe2Þ c h2B¼ À Be2Þ: Àr00 1 À c B þ B h1  expðBe1 ð14:27Þ  1þB ) rF ¼ r00 B ½1 À expðBe1 À Be2ފ Since at the exit plane, r002 ¼ r001 þ ce2, substituting this into (14.27) and dividing by expðBe2Þ; the drawing stress, ¼  þ \" À h2B# rF; in terms of true strain, e2; is obtained as: r00 1 B B h1 1 Drawing stress, rF ¼ 1 þ B h À c À  À c i B r002 B r00 1 B expðÀBe2Þ ð14:24Þ Hence, we see (14.24) is same as (14.12). ð14:28Þ (b) Linear Strain-Hardening Strip The strain-hardening behaviour for annealed metal in the uniform plastic deformation region of stress–strain curve Let us assume that there is no back tension. Since at the closely approximates an exponential relation between stress entry plane, the work metal has not yet been deformed as and strain, but the solution in such cases becomes cumber- well as there is no back pull, so we can take the true strain as some. So, it is suggested to draw an approximate chord e ¼ e1 ¼ 0; and the horizontal longitudinal stress as rB ¼ 0; through the stress–strain curve of exponential pattern so that at the entry plane. With these considerations, (14.22) can be the solution for linear strain-hardening behaviour, such as simplified as follows: given by (14.28), can be applied.

654 14 Drawing: Flat Strip, Round Bar and Tube 14.3 Drawing Stress of Strip Through To obtain the stress required to draw strip through Cylindrical Dies cylindrical dies, the horizontal equilibrium of forces acting on the vertical slab element from the exit to entry side, given The stress required for plane-strain drawing of a by (14.5), will be solved. Equation (14.5) can be rewritten as strain-hardening wide, flat strip through cylindrical dies (dies of circular profile) with Coulomb’s sliding friction will dðrxhÞ þ pð1 þ l cot aÞ dh ¼ 0 ð14:29Þ be derived in this section. The strip drawing through cylin- drical dies has been shown in Fig. 14.2. The exit plane (of To determine drawing stress, (14.29) has to be solved for the deformation zone) is considered as the origin of the coordinate system, shown by x = 0. in the diagram, where the die pressure first and for that the yielding condition given x is the horizontal axis that increases from the exit plane to the entry plane (of the deformation zone). In this, the contact by (14.7) is used to eliminate rx from (14.29) as follows: surface between the strip and die is a circular arc. In Fig. 14.2, let us consider a vertical thin slab element of metal dÀhr00 À Á þ pð1 þ l cot aÞ dh ¼ 0 ð14:30Þ of thickness dx located in the deformation zone of the strip hp (within the die gap) at an arbitrary angular position a from the exit plane. This angle a may vary from a1 at the entry The radius of the die, denoted by R, is assumed to remain plane, to a2 at the exit plane, where a2 ¼ 0: Further, the constant during deformation. Now the incremental height thickness of this infinitely thin slab element makes an angle dh of the strip in terms of R will be da at the centre of the circular die. The horizontal longitu- dinal stresses rx and rx þ drx are acting on this vertical thin dh ¼ ðR daÞ sin a; or, dh ¼ 2R sin a da slab element, respectively, at arbitrary distances x and 2 x + dx, from the exit plane. Figure 14.2 shows that these horizontal stresses along with the die pressure p and friction Substitution for dh into (14.30) gives stress lp are acting on this slab element of width w, which varies in height from h to h + dh. Suppose the heights of the dÂhÀr00 À ÁÃ ¼ À2Rp sin að1 þ l cot aÞda ð14:31Þ strip at the entry and exit planes are, respectively, h1 and h2. p All other assumptions remain the same as that made in Sect. 14.2, except that the work hardening of the strip, i.e.  ! the variation of plane–strain flow stress r00 ; has been con- d p sidered in the following analysis. Or, hr00 1 À r00 ¼ À2Rpðsin a þ l cos aÞ  da   p d Àhr00 Á d p ) 1À r00 da þ hr00 da 1 À r00 ¼ À2Rpðsin a þ l cos aÞ ð14:32Þ dα hÈÈsirÀdmp00ÀIpÂnhdrlriÈ00f001Áy1Á9À4Àtdh81ae,ÀÉÉpÈBsdoclrÀaalhu00nnrÁtdi00ÉboÁenaddniaaogdÉÃfn:oF((Tro1ehr4dhde.r3ir002n(eÂ1)da,c9Èsoo41tmhn8Àe)pbÀaetfiprehirrsismnrot 00dnÁsÈÉtuwhÀgiipdsgthaesÃrsutt00hgeaÁedgnÀedtsetthr1emtaÉdot assumption is that under most circumstances, the variation in R the die pressure p with angular position a in the die gap is R p greater than the variation in the plane strain flow stress r00. Further, the variation in the product hr00 with angular posi- μp σx h2 tion a in the deformation zone is so small that dðhr00 Þ=da dh may be considered to approach a negligible quantity, since 2 r00 increases as h decreases. Hence, this assumption is usu- h1 σx + dσx ally reasonable for strain-hardened metal. But this assump- h α tion is not valid when rate of strain hardening is relatively dh high, for example, during the first deformation pass of the 2 μp work metal after annealing, or when the work metal is p dx x sautibojnecotefdptorh00 igohvebracthketeanrscioonf because it reduces the vari- x=0 contact. For most drawing Fig. 14.2 A diagram of wide flat plane-strain drawing of strip through operations, in the second and subsequent deformation pas- cylindrical dies, showing stresses acting on a vertical thin slab element of metal of thickness dx in the deformation zone of the strip (within the ses, the accuracy of this approximation is satisfactory lead- die gap) ing to an error of only a few per cent. Thus, with the above assumption, (14.32) is modified as follows:

14.3 Drawing Stress of Strip Through Cylindrical Dies 655  It is assumed that the tensile stress applied on the d p À hr00 da r00 ¼ À2Rpðsin a þ l cos aÞ; work-piece at the exit plane is rF; and there is no back   ðsin a þ l cos aÞ tension at the entry plane. As the plane strain deformation dp p h resistance of the work-piece, r00; increases during cold Or, da r00 ¼ 2R r00 ð14:33Þ working, it is assumed that r00 ; will increase gradually from a value of r001 at the entry plane, to a higher value of r002 at the If the angle of contact, a, is small, it may be further exit plane, We further assume that the die pressure p changes assumed that sin a % a; and cos a ¼ 1 À 2 sin2 a ¼ 1 À a2 % 1: from p1 at the entry plane to p2 at th.e exit plane. As y¼ 22 pr00 r00 1 ; and at the exit ; so at the entry plane y1 ¼ p1 Again, h À h2 ¼ 2Rð1 À cos aÞ; . plane y2 ¼ p2 r002 : The integration constant, A, can be a a2 Or; h ¼ h2 þ 4R sin2 2 ¼ h2 þ 4R ¼ h2 þ Ra2: determined for the following boundary condition. At the 2 entry plane where a ¼ a1; the horizontal longitudinal stress rx1 ¼ 0; since there is no back tension (assumed). Therefore, Then, incorporation of the above approximation and from (14.7) describing the yielding criterion, we get p1 ¼ r00 1 À rx1 ¼ r001 ; With the above conditions at the substitution for h in (14.33) gives entry plane, and substitution for H = H1 and a ¼ a1 into   aþl (14.35a), we get the integration constant, A, in (14.35b) as d p p h2 þ Ra2 da r00 ¼ 2R r00 ð14:34Þ Let, y ¼ p ; then (14.34) reduces to follows: r00 y1 ¼ Ah1 expðlH1Þ dy ¼ZZ2Rddyyyyh2¼a¼þþZZRlahh22222dRþRþaaa;RdRdaaaa22þþ Z 2lR da Or, Ah1 expðlH1Þ ¼ p1 ¼ r001 ¼ 1; y 2lhR2Zþ Ra2 r001 r00 1 Or, h2 1 þ ) A ¼ expðÀlH1Þ ð14:36aÞ Or, h1 da ðRa2 =h2 Þ where ) ln y À þ Ra2Á þ 2lR rffiffiffiffi rffiffiffiffi  ¼ ln h2 rffiffiffiffi h2 R tanÀ1 R H1 ¼ 2 h2 h2 a1 ½from ð14:35aފ rffiffiffiffi h2 tanÀ1 R a þ C; R h2 ð14:36bÞ ðC is an integration constantÞ: Substitution for A from (14.36a) into (14.35b) gives Putting h2 þ Ra2 ¼ h; the above equation is written as p ¼ h expðlH À lH1Þ ð14:37Þ follows: r00 h1 y rffiffiffiffi rffiffiffiffi  In (14.37), p is the die pressure at any arbitrary distance ln 2l R R a x from the exit plane. Considering the condition of yielding ¼ tanÀ1 þ C given by (14.7), the longitudinal pulling stress rx in the h h2 rffiffiffiffi h2 rffiffiffiffi  ! direction of drawing at any arbitrary distance x from the exit Or; y ¼ h exp 2l R tanÀ1 R a þ C plane can be obtained from (14.37) as follows: h2 h2 Let us introduce a symbol H, where rx ¼ 1 À p ¼ 1 À h expðlH À lH1Þ ð14:38Þ rffiffiffiffi rffiffiffiffi  r00 r00 h1 ð14:35aÞ H ¼ 2 R tanÀ1 R a h2 h2 Again at the exit plane, where the thickness of the strip is ) y ¼ h expðlH þ CÞ ð14:35bÞ h = h2 and the angle is a ¼ a2 ¼ 0; the horizontal longitu- Or, y ¼ Ah expðlHÞ dinal stress is rx2 ¼ rF; which is the drawing stress applied for the reduction of strip height from h1 to h2. Hence, from where, a new integration constant, A = exp(C). (14.38):

656 14 Drawing: Flat Strip, Round Bar and Tube h2 ! temperature of eutectoid transformation that is 727 °C. That h1 is why this subcritical annealing treatment is also known as Drawing stress, rF ¼ r00 1 À expðlH2 À lH1Þ ‘close annealing’. Generally, the microstructure of the low- ! and medium-carbon steels prior to cold work is spheroidized h2 or largely ferritic with small amounts of pearlite, both highly ¼ r00 1 À h1 expðÀlH1Þ ð14:39Þ ductile microstructures. rffiffiffiffi rffiffiffiffi  (b) Patenting: An isothermal heat treatment operation, R tanÀ1 R called patenting, is applied to produce high-strength ropes, since H2 ¼ 2 h2 h2 a2 ¼ 0: musical strings, suspension bridge cables, etc., in rod and wire industry. In wire making, patenting is applied to 14.4 Treatments of Work Metal Prior medium-carbon steel containing above 0.45% C, or mostly to Drawing high-carbon steel before the drawing of the wire, or may be applied for several times between drafts. The process con- The starting work metal is subjected to one or more of the sists in heating to a temperature above the upper critical three following treatments prior to successful cold drawing. temperature, usually 870 to 920 °C, holding at that tem- These following preparation steps naturally depend on the perature for proper time (generally, 1 h for 1 inch diameter condition of the stock prior to drawing and on the desired of the workpiece) in order to homogenize the austenite drawing results. properly, quenching the rod / wire from that temperature into a molten lead or salt bath kept at a temperature of 450 to 1. Heat Treatment 550 °C and then holding there until the austenite is com- pletely transformed into fine pearlite with perhaps some Heat treatment usually involves annealing or softening of the upper bainite, without the formation of proeutectoid phase. deformed work metal to restore its ductility so that the annealed work metal can be further reduced by cold draw- The role of patenting in producing a high-strength state in ing. This treatment is particularly necessary for certain the metal is twofold. Firstly, it makes a wire more capable of metals that are brittle or hard in the hot-worked condition or withstanding large reductions in cold drawing without for previously cold-drawn pieces which have already been fracture which is ensured by the structure of fine-lamellar highly strain-hardened to enable further cold drawing. Initial pearlite and the absence of proeutectoid phase. Secondly, the heat treatment prior to drawing may not be required for all interplate spacing in the ferrite–cementite mixture upon cold deformed work metals, for example, hot-rolled low-carbon plastic working becomes even smaller than upon patenting. steel rods which are to be drawn into wire. Intermediate heat As a result, wire produced has both high strength and high treatment may be required repeatedly to get rid of work toughness. These combined properties are beneficial in twisting hardening that depends on the work metal and reductions and bending of wire. The strength of steels with 0.6–0.8% involved in drawing. Controlled-atmosphere furnaces are carbon upon 80 or 90% reduction is 1.8–3 GPa. Patenting of commonly used for heat treatment in drawing plants. 1% carbon steel with a heavy deformation above 95% to form thin wire can produce the highest value of real ultimate strength (a) Recrystallization Annealing: Non-ferrous wire and obtainable in industrial product: 4.8 GPa. low-carbon steel wire, which have to be repeatedly cold worked, usually undergo recrystallization annealing (also 2. Surface Preparation known as process annealing) to restore the ductility, in which a cold-worked metal is heated above its recrystal- In order to avoid surface defects of the work metal, or lization temperature in order to soften the metal for further excessive wear of the drawing die due to abrasion during working. This process of annealing has been described in cold drawing operation, and thereby to increase the die life, Sect. 10.2.1 of Chap. 10. The production of non-ferrous and the starting stock must first be cleaned of surface contami- low-carbon steel wires is made in a number of tempers nants, such as rust, glass and scale, probably developed ranging from dead soft to full hard. during previous hot working or rolling operation. This cleaning is performed by chemical, mechanical, electro- The recrystallization annealing is a subcritical heat chemical treatments or their combinations. Mechanical treatment for steel workpiece, where cold-worked steel is treatments for descaling include periodic bending of work heated above the recrystallization temperature of ferrite at 550 to 650 °C which is below but close to the critical

14.4 Treatments of Work Metal Prior to Drawing 657 metal strip between rollers, or sand or shot blasting. These (McGannon 1964). Although borax is little costlier than methods are not effective for strong scales, and the work lime, borax coating often replaces lime coating in dry metal is most often subjected to chemical treatments that drawing operation because of its advantages over liming:— involve the use of various pickling methods. (i) It dries very quickly; (ii) It does not pick up moisture; (iii) It does not flake off or form a dust in drawing room and (a) Pickling: The work piece made of steel is pickled in so, much cleaner environment in the mill is maintained sulphuric or hydrochloric acid, or a mixture of both acids; (McGannon 1964). Both the lime coating and the borax copper and brass are pickled in sulphuric acid; nickel and its coating form a mechanical bond with metal. Phosphate alloy are treated in a mixture of sulphuric acid and potassium coating (phosphating) is a quite popular kind of coating that bichromate. In electrolytic pickling, the work piece acts as is used for costlier product or variety. It forms a chemical anode, from which scale is partly removed, but the vigorous bond with metal rather than mechanical bond. The work evolution of gases forces to remove the scale mostly. metal is immersed in a zinc-phosphate, or a manganese-phosphate, or an iron-phosphate solution at In case of drawing of steel wire from a coil of conven- about 80 °C for 2–10 min. Lubricant used in drawing tionally hot-rolled steel rod, the coils of steel rod are dipped adheres well to the phosphate film, causing the coefficient of into hot dilute sulphuric acid (H2SO4) solution for 15–30 min friction to drop to 0.04–0.06. After phosphating, the coated during pickling operation. Fe3O4, the chief constituent of work piece is rinsed with hot and cold water and then the scale, is slightly soluble in H2SO4, but the acid is able to phosphate coating is neutralized with lime or some suitable penetrate to the metal beneath the scale, where it reacts with agents, followed by baking and drawing through lubricants iron (metal) forming iron (metal) sulphate, a soluble neutral (McGannon 1964). In sulling, the work metal is coated with salt and liberates a mixture of gases, mainly composed of a thin layer of iron hydroxide, Fe(OH)3, which in combi- hydrogen. This action results in loosening and detaching of nation with lime serves as filler for the lubricant. The the scale from the surface of the metal, i.e. the removal of practice of sull coating has been mostly discarded for last scale is performed mechanically rather than chemically. several years and is used now sometimes only on a few kinds of wire and for manufacture of cold-headed bolts. After pickling operation, the clean work metal is removed Coppering the work metals in a solution of vitriol reduces from pickling bath and given a thorough rinsing in a spray of the coefficient of friction to 0.08–0.12 and facilitates drawing high-pressure water. In case of continuous casting, the at large drafts and pressures. hot-rolled rod passes through a cleaning station as it leaves the hot-rolling mill and hence, a separate cleaning operation 3. Pointing or Chamfering is not required. Pointing or chamfering involves the reduction of diameter of (b) Surface Conditioning or Coating: The surface of the one end of the cylindrical feed work metal (wire, rod or tube) work metal is conditioned for receiving and retaining the to a size slightly less than the orifice of the drawing die. The drawing lubricant. In many cases, the surface of the pickled pointed end is thus ready for insertion through the die orifice and clean work metal is coated or prelubricated by liming, for gripping and pulling. The pointing operation is usually boraxing, phosphating, sulling, or coppering methods. After carried out at room temperature by means of swaging, roll- the application of coating, the work metal is dried at 230– ing or turning. However, for pointing operation, the feed 315°C in ovens or bakers. Some of the prelubricating metal can be preheated and acid etching, stretching, grinding methods allow several passes in cold drawing of the work or hammering can also be used. In some cases, in order to metal without the repeated applications of intermediate heat bypass the pointing operation, one end of the feed metal is treatment. As lime is a low cost material, so the lime coating pushed with force so that it protrudes through the die orifice. is applied to low C steel or mild steel and cheaper products. Pushing forces being much greater than pulling forces may Liming is done by dipping the clean work piece into a tank buckle the feed metals having small diameters and slender of hot milk of lime at a temperature of 87–93 °C, average sections. Although proper support can minimize this buck- being 90 °C, for long enough to bring the work piece up to ling, but pointing of feed metal with a diameter less than or the bath temperature. The purpose of lime coating is three- equal to about 9.5 mm is generally preferable. fold: (i) to protect the surface of the cleaned rods from rusting in the atmosphere, (ii) to neutralize any traces of acid left from the previous pickling operation and (iii) to serve as a carrier for the lubricant used in drawing the rod to wire

658 14 Drawing: Flat Strip, Round Bar and Tube 14.5 Drawing Equipments tube is drawn through the die. Hydraulic mechanism is used to provide the pull for short-length products, while a chain The basic principles of drawing rods, wires and tubes are the drive can draw products of larger length. Drawing speeds same but the drawing equipments vary depending on the size may vary from about 0.15 to 2 m s−1. The pulling force of of the products. Drawing equipments are classified mainly draw benches may vary from about 15 kN to 1.5 MN and into two groups. One is a draw bench as schematically rods or tubes produced by draw benches may be as long as shown in Fig. 14.3, used for the production of tubes and 30 m. As soon as the drawn part exits the die, the draw head rods, which cannot be coiled. The other equipment is a block automatically releases the drawn part and ceases to move. drawing machine with a drum, for pulling and coiling the The drawn part is then removed from the draw bench, and product, as shown schematically in Fig. 14.4. For detailed the draw head is rapidly returned to the die stand for pulling description of the various types of equipment and of pro- the next rod or tube. cesses, the referred literature (Pomp 1964) is recommended. The block drawing machine is used for the cold drawing The draw bench is a chain-operated or hydraulically of rods and wires as well as tubes of small diameter, which operated, rigidly built, long, horizontal machine, which is can be uncoiled and then recoiled after drawing. Cold used for cold drawing, usually at room temperature, of drawing is usually performed at room temperature. A block cleaned, coated and pointed hot-worked straight tubes and drawing machine can generate products of long lengths rods of larger diameter. The draw bench essentially consists using a much smaller floor space, whereas a draw bench of a table of entry rollers or an elevating entry conveyor, a requires a larger floor space and thus, limits the lengths of die stand, a draw head, also known as ‘pulling dog’, and an the products. The block drawing machine mainly consists of exit rack on the upper surface of the frame or bench. Entry three parts—–a payoff stand (or payoff reels) or swift that rollers support the hot-worked feedstock and usually insert holds the coil of stock to be drawn, a die where the actual the pointed end of the stock into the drawing die hole, which reduction of feed metal takes place and a powered cylindrical is somewhat smaller than the feedstock in section. Lubricant, drawing block or capstan which delivers the load and energy if required, is provided on the entry side of the die by filling for reduction and also collects the drawn product in a coil the die box with grease or some other suitable lubricant, so form. In the drawing process, cleaned and coated coils of the that in passing through the die, the feedstock must first pass feedstock are first placed on the swift that allows free through the lubricant. The draw head has suitable jaws for unwinding of the stock. The leading end of the feedstock is gripping the pointed end of the rod or tube protruding made pointed and then inserted through the drawing die and through the die orifice and pulling it through the die. The grasped by a gripper attached to the drawing block. The draw head mounted on wheels is moved along the exit rack drawing block is rotated by means of an electric motor to either by a motor-driven chain or by a hydraulic mechanism pull and collect the drawn product in a coil form. to pull the rod or tube with tensile force till the entire rod or The block drawing may be performed on a single-die Die holder Draw head drawing machine or on a multiple-die continuous machine. A typical single-die drawing machine is called ‘Bull Block’, Die which in a single pass can produce a particular reduction corresponding to the diameter of only one existing die. Since Jaw the reduction of area per drawing pass is hardly greater than 30–35%, several passes are necessary to obtain the overall Fig. 14.3 Schematic diagram of a draw bench required reduction. Hence for multiple passes, the die of the bull block machine must be changed and replaced by a smaller diameter die after each complete pass. To eliminate Fig. 14.4 Schematic diagram of Swift or payoff stand Drawing block a single-die block drawing Rod Drawn wire machine Die Electric motor Direction Die of travel holder

14.5 Drawing Equipments 659 this difficulty, the stock can be drawn continuously on the 14.5.1 Conical Converging Die multiple-die continuous machine, in which the stock passes through a number of dies simultaneously. However, there Conical converging dies are most commonly used in rod and must be one drawing block for each die, i.e. number of dies wire drawing as well as in tube drawing and sinking. These and drawing blocks must be same in a continuous machine, types of dies can even be used in hydrostatic extrusion and rod as shown schematically in Fig. 14.5, but the diameter of the extrusion. There are two parts in the die, the casing and the die must decrease in the direction of drawing, i.e. from the nib. The main function of the casing is to protect the die nib by first to the last stand depending on the reduction of area of enclosing it with thick casing made of steel for dies of large the stock after each pass. Since the diameter of the stock is diameter and brass for dies of small diameter. Since the actual reduced after each pass, the length and velocity of the stock reduction of the workpiece occurs in the die nib, extremely will increase proportionately. For these reasons, the periph- hard material is used to make it. Drawing die nibs are man- eral speed of the drawing block must increase in the direc- ufactured using different materials. For products of larger tion of drawing, i.e. from the first to the finishing (last) draw dimensions, die nibs are made from hardened tool steels on block. This is possible to achieve in one of the following two economic grounds. The working faces of steel die nibs can be ways. In the first, each drawing block is equipped with its lined with hard and less expensive refractory material in order own electric motor with fully variable speed control which to combine the wear resistance of the refractory lining with can be regulated automatically to synchronize the peripheral the toughness of the steel support. Cemented carbides are velocity of the block with the surface velocity of the stock. If used to make die nibs for products of medium sizes. For hard these two velocities do not precisely coincide with each wires of fine diameter, die nibs are made from industrial other the stock will slide on the blocks as they rotate, diamond. The centre of the die nib has a passage, which is resulting in friction and generation of heat. The disadvan- ground and burnished and has definite profile geometry. After tages of these machines are that they are large and costly a die nib has been worn, it is ground and polished for further because of the requirement of a large number of expensive use. Figure 14.7 shows the cross section through a typical electric motors. Both the disadvantages can be overcome by conical converging drawing die. Descriptions of the internal the second type of machine, which is more compact and profile geometry and functions of different zones of the die economic. In this, only one electric motor is used to drive a nib are presented below. The die nib consists of four zones: series of stepped cones with increasing cone diameter from the first to the last or finishing cone, as shown in Fig. 14.6. 1. The entry zone, known as the bell, which is shaped in The diameters of the cones are so designed as to produce a such a way that the work-piece travelling into the die can sequence of peripheral speeds equivalent to a definite set of draw sufficient amount of lubricant with it. The reductions in size. Therefore, a specified reduction per pass bell-shaped entrance causes the hydrostatic pressure to is given by such machines. In multiple-die continuous increase and assists the flow of lubricant into the working machines, the drawing speed may attain 10 m s−1 for ferrous zone of the die. It also protects the work-piece against metal and alloys, whereas drawing speeds up to 30 m s−1 are scoring by edges of the die. The entering angle of the bell usual for non-ferrous metal and alloys. gradually tapers into the second zone. Lubrication Speed control Lubricating Finished box sheave wheels and die box product Rod Die Die Payoff reels Draw block Lubricating Draw block Die and die box Finish draw block Direction of travel Fig. 14.5 Schematic diagram of a multiple-die continuous wire drawing operation

660 14 Drawing: Flat Strip, Round Bar and Tube Fig. 14.6 Schematic diagram of a stepped-cone multiple-die Die 4 drawing machine Die 3 Fig. 14.7 Cross-section of a Die 2 conical converging drawing die Die 1 showing different zones Electric motor 4 12 3 Drawing blocks 2. The approach angle or conical drawing zone, where the on the work-piece removes the surface damage caused by work-piece deforms plastically and the actual decrease in the die wear in the conical working zone and refinishes diameter takes place. Obviously, the cone angle formed the surface of the product without changing the final size by this section of the die is smaller than that formed by of the product, which is determined by the diameter at the the bell region. One-half the included cone angle of the exit plane of the conical drawing zone. drawing zone is usually designated by a; which is an 4. The back relief or bell-shaped exit zone that allows slight important process parameter and referred to as approach expansion of the product as it exits from the die. This semi-angle or semicone angle of die. zone minimizes the possibility of damage to the die bearing zone and scoring of the finished products 3. The die bearing zone, which is a straight and short because of abrasion that takes place when the die is out cylindrical region of a few millimetres long. This cylin- of alignment or the drawing stops. drical portion of the die does not reduce the work-piece but causes additional friction loss, which is required to The individual dies are often preceded by lubrication ensure the dimensional stability of the product. The main boxes for reduction of frictional drag and prevention of wear function of this zone is that the frictional drag produced

14.5 Drawing Equipments 661 of the die nibs. The lubricant box contains dry soap, grease, internally cooled capstan. Further, external air cooling of oil or other lubricants through which the workpiece must wire and water cooling of the die holder are possible. If pass before it reaches the die. water is used to cool the wire at all, it must be withdrawn prior to the entry of the wire into the next die. In the dry 14.6 Drawing of Rod and Wire machine, the die is mounted in a holder within a box. Usual lubricant is solid dry soap powder placed in the die box and The reduction in diameter of a solid round bar by successive picked up by the surface of wire when it travels through the drawing, usually through a conical converging die, is called box. This method is applied to steel wire with diameter rod or wire drawing depending upon the diameter of the final larger than 0.5–1 mm for which the production of relatively product. There is somewhat arbitrary distinction between rod rough surface is acceptable. For serious draws, soap is often and wire. If the diameter of the final product is below 5 mm, pre-applied from a solution, if required, over a conversion it is generally called wire, which can be coiled and may be coating and must be dried prior to wire drawing. In dry rapidly drawn on multiple-die machines. Cold-drawn rods drawing, the phosphate-coated steel wire after baking is can be used as structural members, pistons, shafts and normally drawn through the lubricants consisting of calcium spindles, etc. Various applications of wire and wire products stearate or aluminium stearate and lime mixture. are high-strength ropes, electrical wirings, musical strings, springs, fencing, suspension bridge cables and welding The surface of the rod or wire of high-strength materials, electrodes, etc. such as steels, stainless steels and high temperature alloys, can have either a thin coating of softer metal or a conversion The rod drawing at room temperature starts with usually coating. Chemical deposition of copper or tin on the surface rolled or extruded straight rods of larger diameter, which of rod produces thin film of softer metal that may be used as cannot be coiled and is performed on the draw bench and a solid lubricant during drawing. Conversion coatings such subsequently, when the rod diameter decreases so that it can as oxalate or sulphate coatings may be applied to the rod. be coiled, the rod drawing is performed on the block These are further coated typically with soap to act as lubri- drawing machine. The wire drawing usually starts with a cant. In the drawing of titanium, polymers are used as solid large coil of hot-rolled rods of approximately 9 mm in lubricant. Dry wire drawing is applied to obtain product of diameter and is performed on the block drawing machine, finished size or sometimes, dry drawing passes are used for often drawn on a multiple-die continuous machine. In rod- as subsequent wet wire drawing operation. well as wire drawing, the starting stock is heat treated (if required), cleaned, coated and made pointed. If necessary, 2. Wet Drawing the cold-drawn rod or wire is intermediately heat treated between drafts in order to carry out further reduction. This technique is used to produce wire for decorative purposes Although most rod and wire drawing operations are usually or used in application where extra-clean finish is required. carried out at room temperature, but friction and large Wet drawing is more expensive than dry drawing. So, all rods deformations involved in drawing can generate huge heat are first given one pass or more passes by dry drawing to and cause temperature of the stock to rise by several hundred produce wires of process size which will be subsequently degree Kelvin. Interpass cooling technique is used for partial wet-drawn (McGannon 1964). After one or more dry drafts removal of this heat. from the rod, the process wire is then usually cleaned, rinsed and coated prior to wet drawing to finished size. Depending on the lubrication technology, there are mainly two processes for drawing wire. These are designated Prior to coating and wet drawing, cleaning of the steel as dry drawing and wet drawing. Mechanically the processes wire is performed by immersing it in a hot sulphuric acid are the same; that is, the wire is drawn through a die and (H2SO4) solution to remove all of the surface film resulting wound up in a block. The difference in the processes is in the from dry drawing so that clean metallic surface is ready for coating applied to the wire and the lubricant used plating. After acid cleaning, the wire is thoroughly rinsed (McGannon 1964). with water. For plating with copper or tin, the steel wire is immersed in dilute solution of copper sulphate or tin sul- 1. Dry Drawing phate or a mixture of both of these salts (McGannon 1964). This solution is made by mixing copper sulphate or/and tin Dry drawing does not mean the absence of lubricant but the sulphate with water and a small amount of sulphuric acid. lubricant used is only solid and not liquid. The lubricant A chemical reaction takes place which results in the depo- reduces friction and minimizes the die wear but hardly ever sition upon the steel wire of a thin metallic coating from the has a cooling function. The wire is cooled when it lies on the solution used. After plating, the wire is washed thoroughly with fresh cold water.

662 14 Drawing: Flat Strip, Round Bar and Tube After coating, the wire is usually kept under water to 14.6.1 Drawing Load and Power with Friction protect it from the influence of atmosphere until it is deliv- and Back Tension ered to the wire drawing machine. Cooling power is an important criterion for selection of a lubricant in wet draw- In tandem drawing the wire passes through several dies in ing. So, the function of a lubricant is to cool the drawing series and the drawing block remaining in between succes- tools and keep the machine clean, along with reduction of sive dies not only pulls the wire through the die, but also friction and minimization of the die wear. The wet drawing provides a small back tension on the wire entering the next lubricant is either oil-base or aqueous. For non-ferrous die. The back tension or back pull is applied to the input rod metals, the lubricant is generally an oil-in-water emulsion, or wire to keep the die contact pressure low along the which is typically a mixture of water, machine or vegetable deformation zone and the work-piece straight. This reduc- oil, oleic acid and soda ash. For wet drawing, lubricant can tion in die pressure improves the die life, which is very be applied continuously to the inlet of die, to the wire and essential in any working operation. Let us consider the also frequently to the capstan, or the entire machine con- drawing of a rod or wire subjected to back tension through a sisting of blocks and dies can be completely submerged in a conical converging die with a total included approach angle bath containing a water–lubricant solution that is constantly of 2a; as shown in Fig. 14.8. The exit plane (of the defor- recirculated by a pump from a reservoir adjacent to the mation zone) is considered as the origin of the coordinate machine. When the wire slides on the capstan, the lubricant system, shown by x = 0. in the diagram, where x is the reduces the wear of the capstan. Wet drawing is seldom horizontal axis that increases from the exit plane to the entry performed with non-water miscible oils. Oil lubricant con- plane (of the deformation zone). On the round outer surface taining an EP additive is also used in which the dies and the of the rod or wire in the deformation zone, the conical die rod are completely immersed for wet drawing. The wet exerts a radial compressive stress, rr; and a die pressure, p, drawing technique is typically used for all non-ferrous normal to the die–job interface. Outward horizontal flow of metals and for steel wires with diameter less than 0.5–1 mm. the material during the process of its deformation leads to Wet drawing usually makes it possible to give 10–14 drafts. tangential shearing frictional stress s on the circumferential interface between the die and the rod or wire. Hence, A lubrication technology intermediate between the above the direction of the friction stress opposing the outward two techniques is sometimes applied, particularly in the horizontal flow will always be towards the entry plane. low-speed drawing of round bar and tube. The die and/or However, the presence of frictional shear stress s leads to work-piece are lubricated with a high-viscosity liquid or a the horizontal axial (longitudinal) stress rx; where rx pasty or semisolid lubricant, like drawing greases and highly increases from the entry plane towards the exit plane. viscous degradable ester oil. For additional details on the To make the analysis simpler, the following assumptions are lubrication of ferrous wire, the reader is referred to literature made. (Dove 1980). Fig. 14.8 Stresses acting on an Die elemental slab in the deformation zone within a conical converging Rod μp p die during drawing of a rod or σx wire subjected to back tension dD/2 α Drawn wire σx + dσx D1 dx α D2 p σB σF (Back pull) D (Drawing stress) dD/2 dLc Die μ p x x=0

14.6 Drawing of Rod and Wire 663 (i) Material being drawn is assumed to be isotropic and D + dD. Considering the horizontal equilibrium of forces homogeneous. acting on this vertical element, the longitudinal tensile pulling stress at the exit plane can be determined. There are three (ii) Homogeneous deformation is assumed, i.e. internal force components acting in horizontal direction along x-axis: redundant work is neglected. 1. Due to change in longitudinal stress in the positive (iii) Elastic strain is neglected because plastic deformation direction of x, i.e. from the exit to entry side, the forces involved in drawing is quite large compared to over the vertical planes of the element: insignificant elastic deformation.  x  d x   D  dD2 x  D2 4 4   4  x D2   x 2 DdD   x dD2  d x D2  d x 2 DdD  d x dD2   x D2    4 2 x DdD  D2 d x (14.40) (iv) It is assumed that there is no strain hardening of the Equation (14.40) is obtained by neglecting the terms con- work-piece, i.e. the flow stress or deformation resis- taining the products of infinitesimals, such as dD2; dD drx; tance r0 of the material being drawn remains constant. and drx dD2: (v) Coulomb’s law of sliding friction holds good at the 2. Due to the die pressure on the circumference, the hori- contact surface between the dies and the rod or wire. zontal component of the die force (acting normal to the In other words, s ¼ lp; where, s is the tangential die–metal interface) in the direction from the exit to entry frictional shear stress at the contact surface, l is the side: coefficient of friction, and p is the die pressure on the rod or wire normal to the contact surface.  ¼ ppD dD dD 2 (vi) The coefficient of friction l is the same at all points on p sin aðpD dLcÞ ¼ p sin a pD 2 sin a the surface of contact between the dies and the rod or wire. ð14:41Þ (vii) The die pressure, p, is assumed to remain constant SinceðdD=2Þ=dLc ¼ sin a, ) dLc ¼ dD=2 sin a, where across the diameter of the rod or wire. dLc is the contact length between the die and the vertical element. (viii) The longitudinal stress rx acting over the cross-section of the rod or wire is assumed to be uniformly dis- 3. Due to the frictional stress at the circumference, the tributed and normal to the cross-section with no shear component. Hence, it is a principal stress. horizontal component of the friction force in the direction (ix) The radial compressive stress, rr; is also assumed to from the exit to entry side: be a principal stress.  Let us consider the state of stresses on a vertical element of dD infinitesimal thickness dx in the deformation zone of the rod lp cos aðpD dLcÞ ¼ lp cos a pD 2 sin a or wire under the die. According to free-body equilibrium approach, the longitudinal stresses are, respectively, ¼ plpD cot a dD ð14:42Þ rx and rx þ drx at an arbitrary distances x and x + dx from the 2 exit plane, which is taken as x = 0. Figure 14.8 shows that these longitudinal stresses along with the die pressure p, For steady-state drawing conditions, the horizontal equi- normal to the die-metal interface, and the friction stress lp; librium of forces acting on this vertical element in the pos- parallel to the die–metal interface, are acting on this element itive x-direction, i.e. from the exit to entry side, is obtained of thickness dx, whose diameter varies arbitrarily from D to by equating the summation of (14.40), (14.41) and (14.42) to zero. Eliminating p=4 and D, this gives

664 14 Drawing: Flat Strip, Round Bar and Tube 2rx dD þ D drx þ 2p dD þ 2pl cot a dD ¼ 0 ð14:43Þ Now, denoting l cot a ¼ B; we get from (14.43) and D drx ¼ À2½rx þ pð1 þ l cot aފ dD (14.46) Considering rr as the radial compressive stress acting on D drx ¼ À2½rx þ ðr0 À rxÞð1 þ Bފ dD ¼ 2½Brx À r0ð1 þ Bފ dD the vertical element of metal in the deformation zone, the radial equilibrium of forces exerted by the die upon the ) Brx drx þ BÞ ¼ 2 dD vertical element of the work-piece is: À r0ð1 D ð14:47Þ rr pD dx ¼ p cos a pD dLc À lp sin a pD dLc; Considering B and r0 to be constant and integrating Or, rr pD cos a dLc ¼ p cos a pD dLc (14.47), we get À lp sin a pD dLc; 1 ln½Brx À r0ð1 þ Bފ ¼ 2 ln D þ C B As pD cos a dLc ¼6 0; dividing both sides of the above ðC is an integration constantÞ: equation by pD cos a dLc , we get ) Brx À r0ð1 þ BÞ ¼ D2BeBC ¼ AD2B; rr ¼ pð1 À l tan aÞ ð14:44Þ where; A ¼ eBCis a new integration constant:   As semicone angle of die, a; usually never exceeds 15° ) rx ¼ r0 1 þ B þ A D2B ð14:48Þ and coefficient of friction, l 0:1 in drawing operation, so B B l tan a ( 1: As an example, for typical values of a = 6°, and l ¼ 0:05; l tan a ¼ 0:005: Hence, l tan a can be Let, the diameter of the rod or wire before drawing is D1 ignored in comparison with unity and (14.44) reduces to and that after drawing is D2. If the back tension applied to the rod or wire is rB; then the longitudinal stress is rx ¼ rB rr % p ð14:45Þ at the entry plane, where the rod or wire diameter is D = D1 From this entry boundary condition, the constant of inte- Hence, the die pressure p can be considered to be a gration A is found from (14.48) as: principal stress. Again, since the circumferential strain rate is  !  1þB B equal to the radial strain rate due to cylindrical symmetry, so A¼ rB À r0 B D21B ð14:49Þ the circumferential stress rh ¼ rr (radial stress). Since the stresses required for plastic flow during working operation Substituting the value of A from (14.49) into (14.48), we get are mainly compressive in nature, all the principal stresses r1; r2; r3 ; in Von Mises’ yielding criterion are considered to  þ &  þ ' D 2B be positive, if they are compressive and negative when they r0 1 B B þ rB À r0 1 B B are tensile stresses. The die pressure p, the radial stress rr; rx ¼ ; and the circumferential stress rh are compressive in nature rx ¼  þ þ D1 and equal to each other in magnitude, whereas the longitu- r0 1 B \"  2B # rBDD12B dinal stress rx is tensile in nature, therefore in the Von B Mises’ yielding criterion, ) À D ð14:50Þ 1 D1 r1 ¼ rr ¼ rh ¼ r3 ¼ p and r2 ¼ Àrx: rx is the direct pulling stress in the direction of drawing at any arbitrary distance x from the exit plane. Now, the die pressure, p, at any arbitrary distance x from the exit plane So, the Von Mises’ yielding criterion, which is can also be evaluated from (14.50) using yielding criterion qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi given by (14.46) as follows: ðr1 À r2Þ2 þ ðr2 À r3Þ2 þ ðr3 À r1Þ2 2r0; ¼ \"  (  D 2B)  D 2B# r0 1 1 B D1 D1 becomes p ¼ À þ À À rB qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B 1 r0 ðp þ rxÞ2 þ fÀðrx þ pÞg2 ¼ pffiffi ð14:51Þ 2r0; Or, p þ rx ¼ r0 Hence, if rF is the drawing stress at the exit plane applied for the reduction of rod or wire diameter from D1 to D2, then ð14:46Þ the longitudinal stress is rx ¼ rF at the exit plane, where the rod or wire diameter is D = D2. Substituting this condition where, r0 is the uniaxial compressive flow stress or defor- into (14.50), we get mation resistance of the work-piece in homogeneous strain condition.

14.6 Drawing of Rod and Wire 665  1þ \" D22B# rBDD212B Power ¼ drawing load B B D1 Drawing stress; rF ¼ r0 À þ Â distance moved ; i:e:; drawing speed 1 time ð14:52Þ ¼PÂv ð14:55Þ If r is the fractional reduction in cross-sectional area of the work-piece, then r ¼ ðA1 À A2Þ=A1 ¼ 1 À ðA2=A1Þ ¼ 14.6.1.1 Without Back Tension 1 À ðD2=D1Þ2; where A1 and A2 are, respectively, the cross-sectional areas of the work-piece before and after If the back pull is not applied on the work-piece, i.e. rB ¼ 0; then the direct pulling stress in the direction of drawing, rx; drawing. Hence, the drawing stress in terms of the reduction and the die pressure, p, at any arbitrary distance x from the of cross-sectional area is: exit plane can be obtained, respectively, from (14.50) and  \" A2B# rBAA21B (14.51) as follows: B A1 rF ¼ r0 1þ À þ  þ \"  D 2 B# B 1 r0 1 B B D1  BÂ rx ¼ 1 À ð14:56Þ r0 1 1 ¼ þ À ð1 À rÞBÃ þ rBð1 À rÞB ð14:53Þ \"  (  D 2B)# B r0 1 1 B D1 þ Let, the drawing load is P and the load applied at the back p ¼ À B 1 À ð14:57Þ sidÀe, i.e. Áat the entry plane is PB, i.e. PB ¼ rB A1 ¼ The variations of rx and p with the position x from the rB pD12=4 : Then P in terms of diameters and reduction of entry to the exit plane of the deformation zone are shown schematically in Figs. 14.9a and b for both cases of with and cross-sectional area is expressed as: without back tension. These diagrams show that the curve for rx with back tension lies above that without back tension \" þ ( D22 ) according to (14.50) and (14.56) and the curve for p with r0 1 B B back tension lies below that without back tension according B to (14.51) and (14.57). These diagrams in accordance with 1 (14.50), (14.51), (14.56) and (14.57) further show that rx Drawing load; P ¼ rF A2 ¼ À D1 increases and p decreases from the entry to the exit plane with decreasing the arbitrary diameter D of the rod or wire þ 4PB  2B#pD22 from D1 to D2. pD12 D2 However, the rod or wire drawing stress rF in absence of D1 4 back tension is given by  1þ BÈ rÞBÉ 4PB rÞB!pD22 B 1 pD12 4 ¼ r0 À ð1 À þ ð1 À ð14:54Þ If v is the drawing speed, i.e. the exit velocity of the emerging rod or wire, then the power required for drawing operation is given by Fig. 14.9 Schematic diagrams (a) (b) showing the variations in a drawing stress and b die contact Drawing stress pressure from the entry plane to Die pressure the exit plane of the deformation zone, for both cases of with and without back tension Entry Exit Entry Exit plane plane plane plane With back tension Without back tension

666 14 Drawing: Flat Strip, Round Bar and Tube  þ \"  2B#  þ \"  B# drawing stress applied under frictionless condition for the r0 1 B B D2 r0 1 B B A2 reduction of rod or wire diameter from D1 to D2, as follows: rF ¼ À ¼ À 1 D1 1 A1 ¼  þ B À ð1 À rÞBà Zrx¼ðrF Þideal drx DZ¼D2 r0 1 B 1 r0 dD ¼ À2 D ð14:58Þ rx ¼0 D¼D1 In absence of back tension, the cÀorresponÁding drawing Or; ðrF Þideal ¼ À2 ln D2 load P can be calculated from P ¼ rF pD22=4 ; in which the r0 D1 drawing stress rF is given by (14.58). Taking the determined lnD12 value of drawing load P, the corresponding power in ) Ideal drawing stress, ðrFÞideal¼ r0 D2 absence of back tension can be found by means of (14.55). A1 1  A2 1Àr ¼ r0 ln ¼ r0 ln 14.6.1.2 Consideration for Work Hardening ð14:60Þ For heavily cold-worked rods, either the metal does not practically have the capacity to further strain harden or the where r = the reduction of cross-sectional area = (A1 − A2)/ rate of strain hardening is very low and steady. In such cases, A1 = 1 − (A2/A1) = 1 − (D2/D1)2. the assumption of constant flow stress is practically accept- able. For work-hardening rods and wires, it would be better Since no additional work is required to overcome the if strain hardening of the rod or wire is included in the equations for drawing stress. The normal procedure is to friction, so for the same reduction in diameter of the rod or assume a mean value of flow stress r0; and replace r0 with r0 in the equations for drawing stress. The mean flow stress wire, the ideal drawing stress ðrFÞideal is less than the r0 is calculated by means of (10.30). Replacement of r0 drawing stress rF in frictional condition. The idealÀ drawinÁg with mean flow stress r0 in (14.52) or (14.53) and (14.58) load, Pideal; can be obtained from Pideal ¼ ðrFÞideal pD22=4 ; gives drawing stresses at the exit plane with and without in which the ideal drawing stress ðrFÞideal is given by back tension, respectively. (14.60). Taking the calculated value of ideal drawing load Pideal; the corresponding ideal power can be found by means of (14.55). 14.6.2 Frictionless Ideal Drawing Stress 14.6.3 Maximum Reduction of Area in a Single Pass Suppose the dies are perfectly lubricated so that there is no 14.6.3.1 Non-strain-hardening Workpiece interfacial friction, i.e. the coefficient of friction, l ¼ 0; and With and Without Friction and Back so, B ¼ l cot a ¼ 0: Further assume that Tension (1) Deformation is homogeneous, i.e. internal redundant It is possible to increase the reduction of area per pass in rod work is neglected. or wire drawing till the drawn rod or wire does not undergo (2) Strain hardening of the work-piece is ignored, i.e. the flow stress r0 is constant. tensile fracture ahead of the die. Since we have assumed that (3) No back tension is applied to the work-piece, i.e. at the the work metal does not strain harden, so the drawing stress entry plane, rx ¼ rB ¼ 0: at the exit plane must be equal to or less than the uniaxial With the condition l ¼ 0; leading to B ¼ l cot a ¼ 0; flow stress of the work metal, r0; to avoid tensile fracture. (14.47) reduces to Therefore, when the maximum drawing stress is equal to r0; the maximum possible reduction of area, rmax; in one pass drx ¼ À2 dD ð14:59Þ can be achieved. r0 D If ðrFÞmax is the maximum drawing stress at the exit Integrating (14.59) between the lower limit of D = D1 at plane, considering the limiting condition of ðrFÞmax=r0 ¼ 1; the entry plane where rx ¼ 0; and the upper limit of D = D2 the maximum possible reduction of area, rmax; in one pass at the exit plane where rx ¼ ðrFÞideal (say), we get the ideal with friction and back tension is obtained from (14.53), and with friction but without any back tension from (14.58). These are respectively given by the following (14.61) and (14.62):

14.6 Drawing of Rod and Wire 667 With Friction and Back Tension:    r0S:H: 1þB 1þB rB r0 ¼ B À B ð1 À rmaxS:H: ÞB þ r0 ð1 À rmaxS:H: ÞB ! ðrF Þmax ¼ 1 ¼ 1 þB À 1 þ B ð1 À rmaxÞB þ rB ð1 À rmaxÞB rB 1þ B 1 þ B r0 B B ! r0 r0 B B ¼ ð1 À rmaxS:H: ÞB À þ ! ð1 À rmaxÞB 1 þ B À rB ¼ 1þBÀ1 ¼ 1; BrB À ð1 þ BÞr0 r0S:H: Or, B r0 BB Or, ð1 À rmaxS:H: ÞB Br0 r0 ¼ Or, ð1 À rmaxÞB¼ ð1 þ r0 À BrB À 1 þ B ¼ Br0S:H: À ð1 þ BÞr0 BÞr0 B Br0 0& BrB'!1=B ) rmax ¼ 1 À 1 ð1 þ BÞ À r0 Or, ð1 À rmaxS:H: ÞB¼ ð1 þ BÞr0 À Br0S:H: ¼ 1 À ½Br0S:H: =ð1 þ BÞr0Š ð1 þ BÞr0 À BrB 1 À ½BrB=ð1 þ BÞr0Š 1 À fBr0S:H: =ð1 þ BÞr0g!1=B ð14:61Þ ) rmaxS:H: ¼ 1 À 1 À fBrB=ð1 þ BÞr0g With Friction But Without any Back Tension: ð14:64Þ ðrF Þmax ¼ 1 ¼  þ B À ð1 À rmaxÞBÃ; Equation (14.64) shows that when ½Br0S:H: =ð1 þ BÞr0Š r0 1 B 1 approaches 1, rmaxS:H: also approaches 1 and there is no limit Or; ð1 À rmaxÞB¼ 1 À 1 B B ¼ 1 1 B to the reduction. If one would assume that ðrFÞmax¼ r0S:H: ¼ þ þ r0 (mean flow stress), then (14.64) would result in an  1 1B equation, which is same as (14.61) except that r0 in (14.61) ) rmax ¼ 1 À 1þB ð14:62Þ would have to be replaced by r0: Frictionless Condition Without Back Tension: The ideal drawing stress ðrFÞideal in a frictionless ideal 14.6.4 Redundant Deformation deformation condition without applying any back tension to a non-strain-hardened rod or wire must be equal to or lower than the uniaxial flow stress of the work-piece, r0; i.e. The redundant deformation involves inhomogeneous defor- ðrFÞideal r0: Hence, from (14.60) we can write mation without contributing to the change in the shape of the  body (see Sect. 10.5.2 in Chap. 10) and thus, increases the r0 ln 1 r0; 1 e; stress required for deformation and thereby, decreases the 1Àr or, 1 À r efficiency of a deformation process. So, it is always Or; r 1 À 1 attempted to maintain the level of redundant deformation to e a minimum in any process of deformation, if it cannot be ) Maximum ideal reduction, ðrmax Þideal ¼ 1 À 1 completely avoided. The redundant work is usually denoted e by a factor /; which is a function of the cone angle of the die ¼ 1 À 1 ¼ 0:63 ¼ 63% ð14:63Þ and the reduction of area in deformation. In practical wire 2:718 drawing, the redundant work factor is often small, because ) Theoretically maximum possible reduction in a single die angles used in industry are low and reductions given are pass for non-strain-hardening rod or wire in absence of back tension is 63%, which is greater than that for non- usually heavy, but it cannot be always neglected, particularly strain-hardening strip metal in absence of back tension, 58% [see (14.20)]. when die angles are high and reductions are low. Equation (14.58) gives the rod or wire drawing stress without back tension considering the effect of friction, in which the redundant work term can be included (Kalpakjian 14.6.3.2 Strain-Hardening Workpiece and Schimid 2011), as shown below: With Friction and Back Tension rF0 ¼ r0  þ BÈ À ð1 À rÞB É þ ! 1 1 p4 ffiffi a2 1 À r For a strain-hardened material, the maximum drawing stress B 33 r ðrFÞmax; strictly speaking, can reach the tensile flow stress of the material at the exit plane, i.e. at the end of the operation, ð14:65Þ say r0S:H: ; which is appreciably greater than the mean flow stress r0: So putting rF ¼ ðrFÞmax¼ r0S:H: ; and r0 ¼ r0; in where the drawing stress without back tension (14.53), we get the maximum possible reduction of area, r0F considering the effect of friction when cor- r ;maxS:H:: in one pass for a strain-hardened rod or wire with rected for redundant work, friction and back tension as follows:

668 14 Drawing: Flat Strip, Round Bar and Tube r0 the mean flow stress of the material, Apart from the influence of redundant deformation r the fractional reduction of area in drawing, B ¼ l cot a; in which l the coefficient of friction at the through the redundant work factor / based on the parameter die-workpiece interface and a the approach semi-angle of the conical die in D; the effect of redundant deformation can also be estimated radians. from the upper-bound analysis. An upper-bound solution for drawing stress that includes the effects of friction and back tension and also accounts for the redundant deformation, In (14.65), the redundant work component is given by the derived by Avitzur (1968) is: term p4 ffiffi a2 1 À r ; which shows that the larger the die &   a ' 33 r D1 þ p2ffiffi sin'2 !a À cot a rF0 ¼ rB þ r0 2f ðaÞ ln D2 3 angle, a; or/and the smaller the reduction r, the greater the & þ p2ffiffi mðcot aÞ ln D1 þ m 2lb redundant work. 3 D2 D2 The influence of redundant work can be taken into ð14:70Þ account by multiplying the drawing stress by the redundant work factor / (Rowe 1977). Hence, consideration of the where rF0 ¼ the drawing stress with back tension consider- ing the effect of friction when corrected for redundant work, redundant deformation modifies the equation for the draw stress rF; with friction but without any back tension, rB ¼ back pulling stress, r0 ¼ the flow stress in homoge- assuming r0 ¼ r0 (mean flow stress) in (14.58) neous strain condition, D1 ¼ the diameter of the ingoing stock, D2 ¼ the diameter of the outgoing product, a ¼ the r0F ¼ /  þ BÂ À ð1 À rÞBÃ ð14:66Þ approach semi-angle of the conical die in radians, m ¼ the r0 1 B 1 As seen from (14.66), the influence of redundant defor- interface shear friction factor, lb ¼ the length of bearing zone of the die, and mation is to raise the flow stress of the material through the redundant work factor /; which is again related to the \" rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi deformation-zone geometry parameter, D; (introduced in 1 1 À 11 sin2 a Sect. 10.6 of Chap. 10) through the following (14.68) or f ðaÞ ¼ sin2 a 1 À ðcos aÞ 12 (14.69). Thus, the parameter D provides a method to treat the redundant deformation in rod and wire drawing (Wright pffiffiffiffiffiffiffiffiffiffiffiffi 3 1976). For small semidie angle, a; the parameter, D; given by (10.77) may be modified by assuming sin a % a: þ 1 ln pffiffiffiffiffiffiffiffiffiffiffiffi 1 þ q1ffiffi1ffiffiffi=ffiffi1ffiffi2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi75 11 Á 12 11=12 cos a þ 1 À ð11=12Þ sin2 a a h pffiffiffiffiffiffiffiffiffiffii2 & In (14.70), the redundant 'deformation term is D ¼ 1þ 1Àr ð14:67Þ 2f ðaÞ ln R1 þ pÀ2a3ffiffisisnin2a2aaÁ R2 À cot a ; in which as a ! 0; r À cot a ! 0: ðaÞ ! 1; and In commercial wire drawing, often the values of a lie in f the range from 4 to 10° and drawing reductions involved are about 20%. This corresponds to D values typically ranging A relatively simple expression for drawing stress based from 1.25 to 3.1. Obviously, higher die angles and lower reductions correspond to higher values of D; while lower die on the upper-bound analysis (Hosford and Caddell 1983) angles and higher reductions correspond to lower values of D: Experimental studies show that the redundant work factor including the redundant deformation term ð2=3Þ tan a; and / is related to D by (Wistreich 1955; Caddell and Atkins 1968) the effect of friction is given by: rF0 ¼ r0  m ln D1 þ ! ð14:71Þ 1þ sin 2a D2 2 tan a 3 / ¼ C1 þ C2D ¼ 0:8 þ D ð14:68Þ where m = the interface friction factor. Equation (14.71) 4:4 predicts slightly higher drawing stresses than (14.70). Another source (Hosford and Caddell 2011) suggests that 14.6.5 Drawing Stress Versus Die-cone Angle: for various combinations of a < 30° and r < 0.5, the relation Optimum Cone Angle, Dead Zone between / and D for axisymmetric flow, such as drawing of and Shaving rod or wire or extrusion, is approximated as / % 1þ D In Chap. 13, it has been discussed that for any combination 6 ð14:69Þ of coefficient of friction and reduction of area, there exists an

14.6 Drawing of Rod and Wire 669 optimum semicone die angle, a; designated by aopt:; which drawing. For example, if aopt: ¼ 6 for wire drawing, the gives the minimum drawing stress. It has been shown in optimum die angle for strip drawing is about aopt: ¼ 15: Fig. 13.10 that the curve of extrusion or drawing load versus semicone angle of die initially decreases with increase of a Figure 14.10 shows schematically the extrusion or until aopt: is reached, where the load exhibits a minimum and drawing stress as a function of die semicone angle, a; for thereafter beyond aopt:; the load increases with further various shear friction factor, m, at constant reduction, r. In increase in a. The values of aopt: for drawing operations this figure, the upper level represents diagrammatically three usually lie between 6 and 24° depending on the values of common modes of flow: sound flow, flow through a coefficient of friction and the amounts of reduction of area. As the friction or the reduction of area decreases for a given dead-zone formation and shaving mode of flow. Fig- metal, the value of optimum semicone die angle also ure 14.10 shows that beyond aopt:; the gradual rise in the decreases. Further, it is found in practice that the higher the extrusion or drawing stress with increase of a continues until hardness of the metal to be drawn, the lower is the value of a first critical semicone die angle, acr1 ; is reached, up to aopt:; because the coefficient of friction will be lower for the which sound flow continues. At this point, normal sound harder material (see Sect. 10.3.1 in Chap. 10). For example, the value of aopt: may be 24° for aluminium, 12° for copper flow changes into flow through a dead-zone formation. With and 6° for steel. However, when all other factors are same, the change in the modes of flow from sound flow to the optimum die angle aopt: for round-bar drawing having larger redundant work factor will be lower than that for strip dead-zone formation, the slope of the characteristic curve in Fig. 14.10 describing extrusion or drawing stress undergoes a discontinuity: the stress thereafter continues unchanged with increasing semicone angle, a; of die up to a second critical semicone die angle, acr2 ; at which shaving mode of Fig. 14.10 Drawing or α extrusion stress as a function of semicone angle of die, a, for α various shear friction factor, m, at constant reduction, r α αα α α αα α α α

670 14 Drawing: Flat Strip, Round Bar and Tube flow begins and the extrusion or drawing stress starts to α drop. Note that an increase in reduction or friction increases the required stress for drawing or extrusion, except that when the cone angle is in the super critical range, further increase in friction will not affect the required stress. Dead Zone: A dead zone is a zone where the material adjoining the die αα adheres to it and is immobilized. Dead-zone formation takes place when a material, whether extruded or drawn through Fig. 14.11 Schematic plot of first critical semicone die angle, acr1 ; the dies of increasingly high cone angle, shears within itself versus reduction%, r%, for various shear friction factor, m to develop a dead zone of stagnant metal, which undergoes little deformation or does not take part in the flow. The dead Shaving Mode of Flow: zone forms a die-like flow channel through which the workpiece passes in a still-converging kind of flow. If the For the processes of wire drawing, open-die extrusion and expected dead-zone semicone angle, say a1; is small, it can hydrostatic extrusion, with increasing semicone angle, a; be approximated from the relation for optimum semicone die finally a second critical semicone die angle, acr2 ; will be angle, aopt; by considering the full sticking condition where reached at which the transition from dead-zone formation to the interface shear friction factor m = 1. Hence, substitution shaving will occur. Actually this shift is gradual: it takes of m = 1 into (13.9) gives: place over a range of die angles. When the semicone angle a increases beyond acr2 ; i.e. a [ acr2 ; the dead-zone material pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi.ffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ceases to adhere to the die and at first begins to flow back- a1 ¼ ð3=2Þ lnðD1=D2Þ ¼ ð3=2Þ ln 1 1 À r rad ward and away from the entrance at a slow rate, causing a change in the slope of the curve in Fig. 14.10. With further ð14:72Þ increase of semicone angle a; the dead zone flows backward faster and faster until it becomes chip (Avitzur 1977), as It has already been mentioned that acr1 is the limiting shown in Fig. 14.12. Simultaneously more and more of the critical semicone angle of the die, which causes the surface material goes into the dead zone and less converges dead-zone formation to start. Hence, when m ¼ 1; acr1 ¼ a1; and flows through the exit of the die. In this way when the otherwise for m\\1; acr1 [ a1: When m = 1, the optimum, converging flow terminates, the core of the workpiece moves the first critical and the dead-zone semicone angles are through the die undeformed plastically with essentially no identical, i.e. aopt ¼ acr1 ¼ a1: For a given reduction, the change in diameter. This causes the exit velocity v2 to dead-zone semicone angle, a1; is independent of frictional become the same as the entrance velocity v1, i.e. v2 ¼ condition, i.e. remains constant for all values of m. Fig- v1ðD1=D2Þ2¼ v1; since D1 (diameter of the ingoing stock)/ ure 14.11 shows schematically the variations of first critical D2 (diameter of the worked product) = 1. At this stage, all of semicone die angle, acr1 ; with percentage of reduction for the skin, i.e. all the material from the surface layer of the various shear friction factor, m. From the Fig. 14.11, several workpiece is shaved off. It is postulated that shaving will conclusions can be drawn as follows: occur for that range of variables, where less force is required for shaving. The range of dead-zone formation will decrease 1. The lowermost curve shows the variation of dead-zone with decreasing friction for a fixed amount of reduction. So, semicone angle, a1; which increases with reduction. if friction is low, with increasing the cone angle, the tran- sition from sound flow to shaving may occur without a range 2. The first critical semicone die angle, acr1; increases with where a dead zone forms. an increase of reduction and with a decrease of friction. 3. The first critical semicone die angle, acr1; is always less than 90°. If reduction approaches infinity, only in such case acr1 will approach 90°. Thus, for square dies having semicone angle a = 90°, usually applied in extrusion, for any finite reduction, a dead zone will form with an angle of a1 < 90°.

14.6 Drawing of Rod and Wire 671 CHIP wall. For a given die geometry, the magnitude of elongation and wall thickening depend on the flow stress of the drawn WORK tube and interface friction. As there is no internal support of the tube, tube sinking results in an uneven or wavy internal TOOL surface and the inner diameter is uncontrolled, i.e. irregular. Further, the redundant deformation is higher for tube sinking Fig. 14.12 Chip formation with sharp tool during shaving process because of large internal shearing of tube at the entry and exit planes of the deformation zone and hence, the limiting Dead-zone formation and shaving may be considered as deformation for tube sinking is less than that for other pro- defects and are to be avoided when the purpose of drawing cesses of tube drawing. Tube sinking due to absence of or extrusion is to produce smaller sizes by plastic deforma- internal support involves reverse-bending at the exit that tion. But the dead-zone formation is of great advantage when may cause internal cracking in the bore of the tube and so the the quality of the original surface of the workpiece is poor. tube sinking is not widely used for metals of moderate or Because if dead zone forms the surface of the final product poor ductility. will become a fresh surface sheared at the interior over the conical surface of dead zone, although the back end is dis- There are three major tube drawing processes with carded as a scrap. Similarly, the beneficial utilization of the internal support, which are (i) fixed plug drawing, (ii) float- phenomenon of shaving can be made in process where the ing plug drawing and (iii) mobile mandrel drawing. When it purpose is to remove the undesired surfaces rapidly. is desired to obtain a tube of smooth inner surface of con- trolled diameter and greater dimensional accuracy than in the 14.7 Tube Drawing tube sinking, a plug or a mandrel is employed to draw the tube. Plug or mandrel will size the internal diameter while Tubes or hollow cylinders which are produced by hot the surrounding die shapes and sizes the outside of the tube. working process such as rolling, and extrusion or piercing Four methods of tube elongation by drawing with and often are finished by cold drawing to obtain product with without internal support are shown in Fig. 14.13. smooth surfaces, thin walls, accurate dimensions coupled with improved strength due to strain hardening. As discussed Plugs are small in size, may be cylindrical or conical in earlier, the cold drawing of heat treated (if required), shape and can be made of wear resistant materials. The plug, cleaned, coated and pointed hot-worked tubes are usually when placed in the deformation zone of the tube, is pulled performed at room temperature either on the draw bench if forward by the frictional drag produced by the sliding tubes are of large diameter and cannot be coiled, or on the movement of the tube being deformed. Therefore, the plug block drawing machine for tubes of small diameter, which must be held fixed in the proper position by means of a plug can be uncoiled and then recoiled after drawing. If neces- rod fastened to the plug. Fixed plug is widely used to sary, the cold-drawn tube is intermediately heat treated cold-draw large- to medium-diameter straight tubes. The between drafts in order to carry out further reduction. Gen- reduction in area by fixed plug drawing rarely exceeds 30% erally, lubricating oil is provided at the entry side of the tube due to the increased friction from the plug. The pickup of or in the die box and particularly in the low-speed drawing metal from the bore of the moving tube by the stationary of the tube, a high-viscosity liquid or semisolid lubricant is plug is a serious problem in fixed plug drawing. The applied to the tube or die. reduction in area by fixed plug drawing is very often limited by metal pickup than by increase in friction alone. Only When a hollow tube is cold-drawn through a die without limited length of straight tubes can be drawn with fixed plug, any plug or mandrel to support the inner side of the tube, the because the plug rod may be pulled and even broken. If a process is called tube sinking. When the inside diameter of a chain instead of a rod is used to hold the plug, curved tubes tube is supported by using a plug or mandrel as the tube is can be drawn. cold-drawn through the die, the process is called tube drawing. Tube sinking is often performed on thick-walled For drawing tube of unlimited length, a floating plug that tubes and small-diameter tubes with diameters less than positions itself in the absence of the supporting rod is 12.7 mm, where the control of mandrel or plug is difficult. In employed. A floating plug can move in axial direction tube sinking, outside diameter of the tube is reduced, between two extreme positions (foremost and rearmost accompanied usually by a slight increase in thickness of tube position), as confirmed by laboratory measurements (Pernis and Kasala 2013). The name ‘floating plug’ is given due to its movement inside the tube. The self-positioning of the floating plug is made practically possible by the proper design of the plug and the die. Floating plugs can be used to draw any length of tubing in the form of coils by using block drawing machine at speeds as high as 10 m s−1. The

672 (a) 14 Drawing: Flat Strip, Round Bar and Tube (c) Fig. 14.13 Methods of tube (b) elongation by drawing without internal support by a sinking; and (d) with internal support by b fixed plug, c floating plug, d moving mandrel reduction in area produced by properly designed floating in the tube diameter and reduces the dimensional accuracy. plug is 45% in contrast to 30% for fixed plug drawing. The process of drawing with moving mandrel is also called Further, the loads required for the same reduction in area are ironing, which is widely applied for thinning the walls of lower for floating plug drawing than for fixed plug drawing. deep-drawn shells or cups in the manufacture of artillery However, design of floating plug and lubrication can be shells or beverage cans. critical. The floating plug consists of a conical as well as a cylindrical section. The cross-sectional view of the floating In the three tube drawing processes (fixed plug, floating plug is shown in Fig. 14.14. The cone angle of the conical plug and mandrel drawing), the major part of deformation segment of the plug is slightly less than the angle of the die. involves a large reduction of tube-wall thickness, while the This causes the wall thickness of tube to reduce while the internal diameter of the tube is reduced by a small amount in tube diameter is decreased in passing through the conical most tube drawing with a plug or mandrel, since there is segment of the dies. Thus, the conical section of the plug always an initial clearance for inserting plug or mandrel into acts as working zone. The clearance between the cylindrical the undeformed tube bore. Hence in all practical drawing, segments of the plug and the die finally determines the wall there is always a small proportion of sinking before the bore thickness of the emerging tube. The floating plug is held at of the tube contacts the plug or mandrel. Large changes in its balanced position by the equilibrium of normal force and diameter involve a high proportion of sinking which frictional drag acting in the conical and cylindrical section of increases the extent of redundant deformation and decreases the plug. the efficiency of the process. To determine the drawing load, we will consider the hypothetical close-pass drawing, in Problems related to friction can be minimized if tube is drawn with a moving long mandrel. The mandrel is a hard Plug angle β rod or wire long enough to cover the entire cavity length of the tube and drawn forward through the die at the speed at Floating plug which the tube exits the die. In the mandrel drawing, the tube elongates and the tube wall thins down, while the mandrel Entry side Exit side remains undeformed. Tube drawing with a moving mandrel may produce the greatest reduction per pass, because the Conical drawing load imparted to the tube is carried partly by the section forward friction forces acting at the interface between the tube and the mandrel and partly by the pulling force applied Cylindrical at the exit section. Since the relative motion between the section tube and the mandrel is very small, the danger of metal pickup from the bore of the tube is very little in mandrel Fig. 14.14 Cross-sectional view of a floating plug drawing. On the other hand, the disadvantages of mandrel drawing are that the mandrel is expensive, difficult to handle and must be removed from the tube after drawing by rolling in a reeling mill. The reeling process causes a slight increase

14.7 Tube Drawing 673 which the plug or mandrel fits the undeformed tube bore mean diameters of the tube are, respectively, h1 and D1 at the exactly and there is no sinking at all. entry plane, and h2 and D2 at the exit plane. To make the analysis simpler, the following assumptions are made. 14.7.1 Close-Pass Plug Drawing Stress and Load (i) Material being drawn is assumed to be isotropic and homogeneous. Let us consider the close-pass drawing of a thin-walled hol- low tube with a conical plug through a conical converging (ii) Homogeneous deformation is assumed, i.e. internal die, where the total approach angle of the die at its virtual redundant work is neglected. apex is 2a, and the conical plug makes a total angle of 2b at its virtual apex. If the plug is a cylindrical one, then the (iii) Elastic strain is neglected because plastic deforma- semi-angle b = 0. Figure 14.15a shows the stresses acting on tion involved in drawing is quite large compared to an annular slab element of tube between the die and the plug insignificant elastic deformation. (Sachs et al. 1944). In the diagram, x is the horizontal axis that increases from the exit side to the entry side of the tube. (iv) The arbitrary wall thickness, h, is very small relative The annular element has a mean diameter of Dm; and a width to the mean diameter, Dm of the tube, which is of dx. Its thickness varies arbitrarily from h to h + dh. On the assumed to remain nearly constant. The tube-wall round outer surface of the tube in the deformation zone, the thickness is alone changed during drawing. The hoop conical die exerts a radial compressive stress and a pressure strain in such conditions is negligible and so the normal to the die–job interface, which is called the die deformation can be assumed to take place under pressure. Similarly, on the round inner surface of the tube in plane strain conditions. This is nearly true in many the deformation zone, the conical plug exerts a radial com- industrial passes, where the tube-wall thickness pressive stress and a pressure normal to the plug-job inter- reduces to a large extent while there is a small face, which is called the plug pressure. These stresses are reduction in the diameter. produced because the tube is drawn in the horizontal direc- tion by a tensile force Pplug. Outward horizontal flow of the (v) It is assumed that there is no strain hardening of the material during the process of its deformation leads to tan- work-piece, i.e. the plane strain flow stress or gential shearing frictional stress, s1; on the circumferential deformation resistance, r00; of the material being interface between the die and the tube and tangential shearing drawn remains constant frictional stress, s2; on the circumferential interface between the tube and the plug. Hence, the directions of the friction (vi) The die pressure on the tube is assumed to be equal stresses opposing the outward horizontal flow will always be to the plug pressure on the tube and so both will be towards the entry plane (of the deformation zone). However, designated by p. the presence of frictional shear stresses leads to the longitu- dinal stress, rx; in the horizontal direction, where rx increases (vii) The die or the plug pressure, p, is assumed to remain from the entry plane to the exit plane. Let, the thicknesses and constant in the deformation zone across the thickness of the tube, because the pressure p varies little for a thin-walled tube. (viii) The longitudinal stress rx acting over the cross-section of the tube is assumed to be uniformly distributed and normal to the cross-section with no shear component. Hence, it is a principal stress. Fig. 14.15 a Close-pass tube (a) Die p (b) drawing over a slightly tapered h1 μ1p σ x plug showing the stresses acting D1 p on an annular slab element of tube between the die and the plug. σx + dσ x μ2 p Tube σx + dσx μ1 p b The stresses acting on the p h2 x σx annular slab element of the thin-walled tube α D2 h + dh μ2 p Dm Dm β p Plug p dx μ2 p β x Tube h x=0 α μ1 p dx p dL c Die

674 14 Drawing: Flat Strip, Round Bar and Tube (ix) The radial compressive stress, designated by rr; is 3. Projection of the forces due to the plug pressure along the also assumed to be a principal stress. x-axis: (x) Coulomb’s law of sliding friction holds good at the interfaces between the conical die and the tube as  well as between the tube and the conical plug. In dx other words, s1 ¼ l1p; and s2 ¼ l2p; where, s1 and Àp sin bðpDm dLcÞ ¼ Àp sin b pDm cos b s2 are the tangential frictional shear stresses, l1 and l2 are the coefficients of friction, respectively, at the ¼ ÀppDm tan b dx ð14:75Þ die–tube interface and at the tube–plug interface, and p is the die pressure or the plug pressure on the tube Since dx ¼ cos b: The negative sign indicates that the normal to the contact surface. dLc (xi) The coefficients of friction l1 and l2 are constant at force component is acting in the direction from the entry to all points, respectively, at the die–tube interface and at the tube–plug interface. the exit side of the die. (xii) The die and plug are assumed to have straight sides 4. Projection of the forces due to the die friction along the x- and so, the semi-angles a and b are constant. axis: (xiii) It is assumed that there is no back tension, i.e. rx ¼ rx1 ¼ 0; at the plane of entry where h = h1  dx According to free-body equilibrium approach, the longi- l1p cos aðpDm dLcÞ ¼ l1p cos a pDm cos a tudinal stresses are, respectively, rx and rx þ drx; and the tube-wall thickness varies arbitrarily from h to h + dh, at an ¼ pl1pDm dx ð14:76Þ arbitrary distances x and x þ dx; along the x-axis. Figure 14.15b shows that these longitudinal stresses along 5. Projection of the forces due to the plug friction along the with the die and plug pressures p and friction stresses l1p and l2p are acting on an annular element of the tube x-axis: between the die and the plug. Considering the horizontal equilibrium of forces acting on this element, the longitudinal  stress rx; at the exit plane can be determined. There are five dx force components acting in horizontal direction along x-axis, l2p cos bðpDm dLcÞ ¼ l2p cos b pDm cos b where the direction from the exit to the entry side of the die is considered to be positive. ¼ pl2p Dm dx ð14:77Þ For steady-state drawing conditions, the horizontal equi- librium of forces acting on this annular element in the pos- itive x-direction, i.e. from the exit to entry side is obtained by equating the summation of five equations from (14.73) to (14.77) to zero. Eliminating p and Dm; this gives ðrx dh þ hdrxÞ þ pðtan a À tan bÞdx þ pðl1 þ l2Þdx ¼ 0 ð14:78Þ 1. Due to change in longitudinal forces along the x-axis: As the element moves a distance dx, the net change in the ðrx þ drxÞðh þ dhÞ pDm À rxh pDm wall thickness is given by dh ¼ ðtan a À tan bÞdx; which can % ðrx dh þ h drxÞ p Dm ð14:73Þ be substituted into (14.78) that then becomes l1 þ l2 ! tan a À tan b [Neglecting drxdh; because it is a product of two small ðrx dh þ h drxÞ þ p 1þ dh ¼ 0 ð14:79Þ quantities] Let us denote 2. Projection of the forces due to the die pressure along the l1 þ l2 b ¼ BpÃlug ð14:80Þ x-axis: tan a À tan  dx p sin a ðpDm dLcÞ ¼ p sin a pDm cos a ¼ ppDm tan a dx ð14:74Þ Considering the parameter given by (14.80), (14.79) may be written as: dx ¼ cos a; dx h i dLc cos a h drx þ rx þ p 1 þ BÃplug dh ¼ 0 Since ) dLc ¼ : ð14:81Þ

14.7 Tube Drawing 675 As discussed in Sect. 14.6.1, consideration of radial \" 1 þ BÃplug !(  h BpÃlug )# p ¼ r00 1 À BÃplug h1 equilibrium shows that the frictional contribution to the die 1 À ð14:87Þ pressure is negligible, i.e. rr % p; [see (14.45)]. Hence, p is a principal stress. Similar to rx; p also varies with the position x. Equa- tions (14.86) and (14.87) show that rx increases and p de- As the die or plug pressure p, and the radial stress, rr; are creases as the arbitrary wall thickness h of the tube decreases compressive in nature and equal to each other in magnitude, whereas the longitudinal stress, rx is tensile in nature, from h1 to h2. Hence at the exit plane, where the wall therefore r1 ¼ rr ¼ p; and r2 ¼ Àrx; and the Von Mises’ yielding criterion for plane strain given by (1.88b) then thickness is h = h2, let the longitudinal stress is rx ¼ rF; becomes where rF is the plug drawing stress applied for the reduction of tube wall thickness from h1 to h2. r1 À r2 ¼ p þ rx ¼ r00 ð14:82Þ 1 þ BÃplug !\"  BÃplug # BÃplug h2 ) Plug-drawing stress; rF ¼ r00 1 À where r00 = compressive flow stress in plane strain h1 conÀditiponffiffi Á ¼ 2 3 ÂÀcomppffirffieÁssive flow stress in homogeneous strain ð14:88aÞ condition ¼ 2 3 r0: Assuming the mean diameter Dm to remain constant, the Substituting for p from (14.82) into (14.81) gives cross-sectional area of the tube at the entry plane is A1 ¼ pDmh1; and that at the exit plane is A2 ¼ pDmh2; so (14.88a) drÁx dh rx 1 h can also be expressed as Àr00  ¼ À þ BpÃlug !\"  BpÃlug # BpÃlug BÃplug A2 rx þ À þ rF ¼ r00 1 1 À ð14:88bÞ drx  ¼ dh ð14:83Þ A1 BpÃlugrx À r00 1 þ BpÃlug h In most cases, the mean diameter of the tube does not Equation (14.83) is valid for any value of r00 and BÃplug: remain completely constant at the value of Dm; rather the With the assumptions of r00 ; l1; l2;a and b being constants, mean diameter of the drawn tube changes from that of the (14.83) can be integrated that gives undrawn one. The change in the outer diameter of tube may be appreciable. Whenever the outer or mean diameter chan- 1 h  i ges, it is more appropriate to use (14.88b) rather than BÃplug ln BÃplugrx r00 1 BÃplug À þ ¼ ln h þ C; (14.88a). For close-pass drawing where there is no change in the inner diameter of the tube, i.e. the inner diameter remains where C is an integration constant: ð14:84Þ constant at a value of Di (say), suppose outer and mean  diameters of the tube are, respectively, Do1 and Dm1 at the entry plane, and Do2 and Dm2 at the exit plane of the defor- Or; BÃplugrx À r00 1 þ BÃplug ¼ AhBÃplug mation zone. If the outer diameter of tube changes during where A ¼ eBÃplugC is a new integration constant. This inte- plug or mandrel drawing, the cross-sectional areas of the tube gration constant A can be found as follows from the entry at entry and exit planes are considered respectively as: boundary condition, which is at the entry plane where h ¼ h1; rx ¼ 0; as there is no back tension. A1 ¼ p ÀDo21 À D2i Á ¼ p h À ðDo1 À i 4 4 D2o1 2h1Þ2  ¼ pðDo1 À h1Þh1 ¼ pDm1h1 ð14:89aÞ r00 1 þ BÃplug A¼À ð14:85Þ And hBpÃlug A2 ¼ p ÀD2o2 À Di2Á ¼ p h À ðDo2 À i 1 4 4 Do22 2h2Þ2 Substituting the value of A, from (14.85) into (14.84), we ¼ pðDo2 À h2Þh2 ¼ pDm2h2 ð14:89bÞ get 1 þ BpÃlug !\"  h BÃplug # The load required for plug drawing is the product of plug BpÃlug h1 drawing stress rF and the cross-sectional area, A2; of the rx ¼ r00 1 À ð14:86Þ tube at the exit plane. Hence, Now, the die or plug pressure, p; can also be evaluated Plug drawing load is: from (14.86) using yielding criterion given by (14.82) as follows: P ¼ rF A2 ð14:90Þ

676 14 Drawing: Flat Strip, Round Bar and Tube 14.7.2 Close-Pass Mandrel Drawing Stress Sect. 14.7.1 cannot be used. For Bmà andrel ¼ 0; the differential (14.81) simply reduces to In fixed plug drawing, the frictional forces at the tube–die hdrx þ ðrx þ pÞdh ¼ 0 ð14:93Þ interface as well as at the tube–plug interface act in the backward direction of drawing. In mandrel drawing, Substituting the condition of yielding from (14.82) into the velocity of the mandrel is equal to the exit velocity of the tube and higher than the velocity of the tube enclosed in (14.93), and then integrating with the boundary condition the die passage. Due to this, the mandrel will exert a forward frictional drag on the inner contact surface of the tube. This between the lower limit of h = h1, at the entry plane where tends to nullify the backward frictional drag exerted by the rx ¼ 0; and the upper limit of h = h2, at the exit plane where stationary die on the outer contact surface of the tube. rx ¼ rF; we get the stress for tube drawing with a mobile Hence, the direction of the friction force at the mandrel–tube cylindrical mandrel for the reduction of wall thickness from interface is opposite to that at the tube–die interface, as shown in Fig. 14.16. The stress equation and its solution are h1 to h2, as follows: exactly same as those derived for plug drawing except that the parameter BpÃlug is changed to rZx ¼rF drx hZ¼h2 r00 dh ¼ À h rx ¼0 h¼h1 l1 À l2 l1 À l2 ) Mandrel-drawing stress, tan a À tan tan a Bmà andrel ¼ b ¼ ð14:91Þ rF ¼ Àr00 ln h2 ¼ r00 ln h1 ¼ r00 ln 1 ð14:94Þ h1 h2 À 1 r Since for a cylindrical-shaped mandrel, b ¼ tan b ¼ 0: where r ¼ reduction of cross-sectional area ¼ pDmðh1 À h2Þ Hence as long as l1 [ l2; the stress, rF; (say), for tube pDmh1 drawing with a mobile cylindrical mandrel for the reduction ¼ 1 À h2 or, h2 ¼ 1 À r: of wall thickness from h1 to h2, and that for the reduction of h1 h1 cross-sectional area from A1 to A2 will be given respectively by:  þBÃmBaÃmndarnedl rel\"1  Bmà andrel # Equation (14.94) is a familiar expression for frictionless 1 h2 Mandrel-drawing stress; rF ¼ r00 À ideal homogeneous deformation in plane strain and is also h1 applicable for plug drawing stress under frictionless ð14:92aÞ condition.  þBmÃBamÃndarnedl rel\"1  BÃmandrel # However, it is possible that the friction coefficient at the 1 A2 Mandrel-drawing stress; rF ¼ r00 À tube–mandrel interface is greater than the friction coefficient A1 at the die–tube interface, i.e. l2 [ l1; so that Bmà andrel\\0; i.e. BÃmandrel is a negative quantity. For this, the analysis will not ð14:92bÞ be otherwise changed and (14.92) will still remain valid but If l1 ! l2; or; l1 ¼ l2; which is often true, then for the same reduction of area, r, the mandrel drawing stress BÃmandrel ! 0; or; Bmà andrel ¼ 0: Under these conditions, the solution of the stress equation obtained by integration in can be lower than the frictionless ideal drawing stress. Fig. 14.16 a Close-pass tube (a) Die μ1 p p (b) drawing with a moving mandrel. σx b The stresses acting on the Tube μ1p p annular slab element of the thin-walled tube drawn between σx + dσx μ2p σx the die and the mandrel σx + dσx μ2p p μ2p Mandrel p xp dx h + dh h α x Tube α μ1 p dx x =0 dL c p Die

14.7 Tube Drawing 677 To consider the strain hardening of the tube, the mean rF ¼ ðrFÞmax¼ r0: But strictly speaking, the value of the value of plane strain flow stress r00 ; is calculated by means of maximum drawing stress ðrFÞmax can reach the value of (10.30) and substituted for r00 into the drawing stress equa- the tensile flow stress of the material at the exit plane, i.e. at tions for drawing with plug or moving mandrel. the end of the operation, which is appreciably greater than 14.7.3 Maximum Reduction of Area in a Single Pass the mean flow stress r0: In such case, the maximum possible reduction per pass will increase and approach 1. 14.7.3.2 Mandrel Drawing Since r = the fractional reduction of area in one pass In mandrel drawing, the mandrel is moving with the tube A1 À A2 pDmðh1 À h2Þ h2 and hence restricts the circumferential contraction of the tube A1 pDmh1 h1 ¼ ¼ ¼ 1 À ; both during deformation through the die and after the completion of deformation when the tube leaves the exit ) A2 ¼ 1 À r; or, h2 ¼ 1 À r: plane of the deformation zone. Therefore, the plane strain A1 h1 conditions exist throughout, i.e. in the actual deformation 14.7.3.1 Plug Drawing zone as well as after the completion of deformation. Hence, when the maximum drawing stress, ðrFÞmax; is equal to r00; Equations (14.88) can be written in terms of reduction of the maximum possible reduction of area, rmax; in one pass area, r, as can be achieved without causing the tensile fracture of the rF ¼ 1 þBÃpBluÃpglug!h1 À ð1 À rÞBÃplug i ð14:95Þ tube. r00 Now, by substituting rF ¼ ðrFÞmax¼ r00; into (14.92), for the friction coefficient at the die–tube interface l1 [ l2 (that at the tube–mandrel interface), we get the maximum It is possible to increase the reduction of area per pass in possible reduction of area, rmax; per pass in mandrel drawing as: plug drawing of tube till the plug-drawn tube does not undergo tensile fracture ahead of the die. If the tube does not ðrF Þmax r00  þ BÃmandrel h i r00 r00 1 BÃmandrel 1 r Þ ;max Bmà andrel strain harden appreciably, the drawing stress will be equal to ¼ ¼ 1 ¼ À ð1 À or less than the uniaxial flow stress of the work metal, r0; to Or; ð1 À r Þ ¼max Bmà andrel 1À1=B1mà aþBndrmÃeBlaÃmndarnedl rel ¼ 1 avoid tensile fracture and not the plane strain flow stress r00 of  1 þ BÃmandrel the material, though the plane strain conditions exist in the 1 actual deformation zone. Because there is no plug to support ) rmax ¼ 1 À 1 þ BÃmandrel the inner side of the tube after the tube leaves the exit plane of the deformation zone and so, the circumference of the tube ð14:97Þ can contract freely under the action of axially applied tensile If the friction coefficient at the die–tube interface is the force. Therefore, when the maximum drawing stress, ðrFÞmax; same as that at the tube–mandrel interface, i.e. l1 ¼ l2; then is equal to r0; the maximum possible reduction of area, rmax; the parameter BÃmandrel ¼ 0; and the mandrel drawing stress is given by (14.94), which is the same as the frictionless strip in one pass can be achieved. drawing stress given by (14.18). Now, by substituting rF ¼ À NpoffiwffiÁ, by substituting rF ¼ ðrFÞmax¼ r0; and r00 ¼ ðrFÞmax¼ r00; into (14.94) for l1 ¼ l2; we get the maximum 2= 3 r0; into (14.95), we get the maximum possible possible reduction of area, rmax; per pass in mandrel drawing reduction of area, rmax; per pass in plug drawing as: as: ðrF Þmax r0 pffiffi 1 þBpÃBluÃpglu!g!h1 i r00 r00 3 ¼ 0:866 ¼ ; ¼ ¼ 2 À ð1 À r Þmax BÃplug BÃplug ðrF Þmax ¼ r00 ¼ 1 ¼ ln 1 1 ; or, 1 À rmax ¼ 1 ; 1 þ BpÃlug r00 r00 À rmax e Or; ð1 À rmaxÞBpÃlug ¼ 1 À 0:866 ; 1 1 \" BÃplug !#1=BpÃlug rmax ¼ 1 À e ¼ 1 À 2:718 ¼ 0:63 ¼ 63% ) rmax ¼ 1 À 1 À 0:866 1 þ BpÃlug ð14:98Þ ð14:96Þ Equation (14.98) shows that the limiting reduction per pass in mandrel drawing is 63%, which is greater than 58%, For a strain-hardened metal, rmax may be obtained by the maximum possible reduction in a single pass in fric- taking rF ¼ ðrFÞmax¼ r0 (mean flow stress), instead of tionless strip drawing [see (14.20)], but it is of course the

678 14 Drawing: Flat Strip, Round Bar and Tube same as that for drawing of rod or wire under frictionless 14.7.5 Equilibrium Condition of Forces Acting condition [see (14.63)]. If strain hardening of the tube is on a Floating Plug considered, the maximum possible reduction per pass in mandrel drawing will be more than that given by (14.98). The frictional forces exerted by the moving tube on the floating plug acts towards the exit, which must be balanced However, still greater reduction in mandrel drawing is by the components of normal forces acting towards the actually possible to achieve, if the friction coefficient at the entry. For horizontal equilibrium, the summation of the tube–mandrel interface is increased, i.e. l2 [ l1; since the components of these normal and frictional forces operating mandrel will carry some of the load. on the conical and cylindrical section of floating plug in the axial direction of drawing will be zero, as shown by the 14.7.4 Tube Sinking following expression according to Fig. 14.17, in which only upper half of the floating plug is shown. In deriving load for tube sinking operation through conical l2Pk cos b þ l2Pv À Pk sin b ¼ 0 ð14:101Þ converging die, any thickening or thinning of the tube wall that may occur is ignored. It is assumed that the outer where l2 is the coefficient of friction at the plug–tube diameter of the tube will be reduced by the sinking operation interface, Pk and Pv are compressive loads exerted by maintaining the tube-wall-thickness constant. For this oper- the tube and acting, respectively, normal to the conical ation, the differential equation derived by Sachs and Baldwin and the cylindrical sections of the plug surface in contact (1946) is similar to that for rod and wire drawing operation with the inside surface of the tube and b is the semicone except that a modified yielding criterion is considered due to angle of the conical working section of plug. the more complex stress system in tube sinking. This yielding criterion is: If the plug temporarily moves towards the entry, the pressure on the conical surface of the plug drops and the r1 À r2 ¼ m r0 ð14:99aÞ friction at the cylindrical segment drags the plug back to its balanced position. If the plug is momentarily pulled forward, where r0 is the uniaxial compressive flow stress or defor- the pressure on the conical surface of the plug rises so that mation resistance of the material. The best value for the the plug is brought back to its balanced position. Thus, the constant multiplier m is 1.10, so that (14.99a) is given by: plug maintains its position and floats in the die throat. r1 À r2 ¼ 1:10 r0 ð14:99bÞ If the length of the cylindrical zone of plug is zero, then the frictional force l2Pv ¼ 0; and (14.101) can be written as If the outer diameter of the tube before sinking is D1 and follows: that after sinking is D2, the equation for the drawing stress in sinking, assuming no wall-thickness change, is given by Pkðl2 cos b À sin bÞ ¼ 0 ð14:102aÞ 1þ \" D2B# Or; tan b ¼ l2 ¼ tan f2 or; b ¼ f2 ¼ tanÀ 1 l2 B B D1 ð14:102bÞ Stress for sinking; rF ¼ 1:10 r0 À 1 ð14:100Þ where f2 is the angle of friction at the plug–tube interface. In the absence of the cylindrical zone of plug, the condition of where all the terms are the same as that defined in stability of the floating plug can also be considered with Sect. 14.6.1, in connection with rod and wire drawing. In (14.100) the ratio D2/D1 is raised to the power B, not 2B as Plug angle PK in rod and wire drawing, since the cross-sectional area of Pv tube in sinking is approximated by pDh; and not by ðp=4ÞD2 as in rod and wire drawing. Equation (14.100) shows that for μ 2 PK μ 2 Pv given values of reduction of area, die angle and coefficient of β friction, the drawing stress for tube sinking is 10% greater due to higher yield stress than that for rod or wire drawing, Conical Cylindrical and this has been experimentally verified by Sachs and section section Baldwin (1946). In most practical sinking operations, the thickness of tube wall slightly increases, but this does not Entry side Top half of floating plug Exit side seriously affect the stress calculated from (14.100), which is based on no change in wall thickness. Fig. 14.17 Normal and frictional forces acting on a floating plug

14.7 Tube Drawing 679 respect to Fig. 14.15b. This stability condition is obtained ZDk dD=2 p D2!Dk ¼ p D2k À D2v when the horizontal component of the forces due to the plug sin b sin b 2 Dv 4 sin b pressure on the tube is equal to the horizontal component of Ak ¼ pD ¼ 2 the forces due to the friction exerted by the plug on the tube, i.e. when the summation of (14.75) and (14.77) will be equal Dv to zero, i.e. ð14:107Þ ÀppDm tan b dx þ pl2pDm dx ¼ 0 ð14:103aÞ Substitution of contact surfaces Av and Ak from (14.106) and (14.107) into (14.105) gives l2 ¼ tan b ¼ tan f2 ð14:103bÞ pOkrð,l2pkcos1bÀÀtaslni2nbbÞÀDp4k2DÀk2siDÀn2vbDÁ 2v¼4þl2lp2vpDvðvpLvDv LvÞ ¼ 0 In the absence of cylindrical zone of plug, both (14.102a) Or, D2k ÀsDffiffiv2ffiffiffi¼ffiffiffiffiffippffiffivkffiffiffiDffiffiffiffivffiLffiffiffivffiffiffitffia4ffiffinffilffiffi2bffiffiffitffiÀaffiffinffiffiffilffibffiffi2ffiffiffi¼ffiffiffiffiffippffiffivkffiffiffiDffiffi vLv 1 À 4l2 b l2 cot and (14.103a) show the same result. ) Dk ¼ Dv Dv þ pv Lv 1 4l2 When the length of the cylindrical zone of plug is greater pk À l2 cot b than zero, the frictional force l2Pv must be compensated by the horizontal component of normal force Pk acting in the ð14:108Þ conical zone of the plug. That means the semicone angle b where Dv is the diameter of cylindrical plug land. Again, the of the conical working section must be greater than the angle straight length (not the slant length) of conical section, say of friction f2 at the plug–tube interface. Again for the Lk; of the plug [see Fig. 14.18] in contact with the inner side reduction of the wall thickness of the tube, the semicone of the tube can be determined from Dk as follows: angle, b; of the conical segment must be slightly smaller than the approach semicone angle, a; of the die, i.e. b\\a: Then the relationship between angles is as follows: a [ b [ f2 ð14:104Þ It is clear from (14.104) that for maintaining the equi- Lk ¼ ðDk À DvÞ=2 librium of forces acting on the floating plug during tube \"sffitffiaffiffinffiffiffibffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi drawing, the die angle a and the plug angle b cannot be # changed arbitrarily. Further, it can be shown from the fol- lowing analysis that the lengths of both conical and cylin- ¼ Dv Dv þ pv Lv 1 4l2 À Dv =2 tan b drical zones of the plug are dependent on each other. For pk À l2 cot b this, let us express (14.101) in terms of stress. ð14:109Þ l2pk cos b Ak þ l2pv Av À pk sin b Ak ¼ 0 ð14:105Þ Die angle α Lk Tube wall where pk and pv are pressures (compressive stresses) exerted pk Lv by the tube and acting, respectively, normal to the conical β μ pk pv and the cylindrical sections of the plug surface in contact with the inside surface of the tube; Ak and Av are, respec- μ pv tively, surface areas of the conical and the cylindrical sec- tions of the plug in contact with the inside surface of the Entry side D dD Exit side tube. Dk D + In the cylindrical section of the plug, if Dv is the inner diameter of the drawn tube land and Lv is the length in Floating Dv (Inner diameter contact with the inner side of the tube, then plug of drawn tube) μ Pk μ Pv Av ¼ p Dv Lv ð14:106Þ Pk Pv Let, Dk is the largest diameter of the conical working Tube wall section of the plug which is in contact with the inner surface of the tube, as shown by point X in Fig. 14.18. In the conical X deformation zone of plug, considering an infinitesimal thin slab whose diameter varies from D to D + dD, we get the Fig. 14.18 Tube drawing over a floating plug showing respective relation for Ak as follows: lengths Lk and Lv of conical and cylindrical sections of the plug in contact with inside diameter of tube and normal and frictional stresses acting on the plug

680 14 Drawing: Flat Strip, Round Bar and Tube Equation (14.109) shows that the length of conical zone, figure. The shown one and its mirror image lying below the Lk; of the plug which is in contact with the inner surface of horizontal centre line will constitute the complete slip-line the tube is dependent on the length of cylindrical zone, Lv; of field. The following solution is valid for the conditions stated the plug, although it is necessary to know the value of ratio below: pv=pk for actual calculation. The ratio of pv=pk for Cu–38% Zn brass tubes reaches value of 2/3 (Pernis and Kasala (1) The semi-angle of the die, a ¼ constant. (2) The reduction of area of the strip of unit width, 2013). r ¼ 2 sin a=ð1 þ 2 sin aÞ: 14.8 Application of Slip-Line Field to Strip The slip-line net shown in Fig. 14.19a consists of an Drawing isosceles triangle ABC, where AC ¼ BC and a circular sector BCD. Since the die surface is assumed to be smooth, so slip Figure 14.19a shows the slip-line field for frictionless lines CA and CB must meet the die surface AB at 45°. Again, plane-strain strip drawing through wedge-shaped dies of slip line BD must meet the horizontal centre line at 45° for included angle 2a: Only half of the field is shown in the the reasons of symmetry. Fig. 14.19 a Slip-line field and b the corresponding hodograph for plane-strain frictionless strip drawing through wedge-shaped dies α s n π s s sp α s β s ss α ps n n pa p p a O a p

14.8 Application of Slip-Line Field to Strip Drawing 681 It may be recalled that within the slip field, the shear yield /H ¼ Àðp=4Þ À a; and /G ¼ Àp=4: stress k is constant everywhere and the hydrostatic pressure rm varies from point to point with change in direction of the where / is the angle of inclination of the tangent to a-lines slip lines. measured in the anticlockwise direction from the x-axis: From Fig. 14.19a, we find Since ðr1ÞG¼ rF; hence at point G, from (10.40a) we get AC ¼ BC ¼ BD ¼ ph2ffiffi ; ) AB ¼ pffiffi ¼ ph2ffiffi pffiffi ¼ h2 ðr1ÞG ¼ ðrmÞG þ k ¼ 0; 2 AC 2 2 2 ) ðrmÞG ¼ ðr1ÞGÀk ¼ rF À k: ð14:110Þ Again along the a-line GH, using (10.47a) one can write Again AB sin a ¼ ðh1 À h2Þ=2; ðrmÞH À /H ¼ ðrmÞG À /G; 2k 2k ) h1 À h2 ¼ 2 h2 sin a; ½From ð14:110ފ; Or, h1 ¼ h2ð1 þ 2 sin aÞ; or, h1 ¼ 1 þ 2 sin a ð14:111Þ Or, ðrmÞH þ p þ a ¼ rF À k þ p ; h2 2k 4 2k 4 Since for a unit width of the strip, the reduction of area is ) ðrmÞH ¼ rF À 1 À a; r ¼ ðh1 À h2Þ=h1 ¼ 1 À ðh2=h1Þ; 2k 2k 2 Or, ðrmÞH¼ rF À k À 2ka ¼ rF À kð1 þ 2aÞ: ) From ð14:111Þ; r ¼ 1 À 1 þ 1 a ¼ 1 2 sin a a At point H, from (10.40b): 2 sin þ 2 sin ð14:112Þ ðr2ÞH¼ ðrmÞHÀk ¼ rF À 2kð1 þ aÞ: Since p ¼ Àðr2ÞH; ) p ¼ 2kð1 þ aÞ À rF ð14:114Þ From the hodograph shown in Fig. 14.19(b), we find Ox ¼ xc ; or, pffiffi ¼ xc ; since Ox ¼ v1; Again equilibrium of forces for a unit width of the strip sinðp=4Þ sin a 2v1 sin a will give: pffiffi h2 h1 À h2 ) xc q¼ ffiffiffiffiffi2ffiffiffivffiffi1ffiffiffisffiiffiffinffiffiffiaffiffiffiffi¼ffiffiffi xd ¼ qd; and, 2 2 rF ¼ p ¼ p AB sin a; xq ¼ ðxdÞ2 þ ðqdÞ2 ¼ 2v1 sin a; Or, rF h2 ¼ p h2 sin a; ½From ð14:110ފ : 2 ) v2 ¼ Oq ¼ Ox þ xq ¼ v1 þ 2v1 sin a ) rF ¼ 2 p sin a : ¼ v1ð1 þ 2 sin aÞ; which gives v2 ¼ 1 þ 2 sin a Hence, from ð14:114Þ : v1 rF ¼ 2f2kð1 þ aÞ À rFg sin a; ð14:113Þ Or; rFð1 þ 2 sin aÞ ¼ 4kð1 þ aÞ sin a; From (14.111) and (14.113), it is seen that h1=h2 ¼ ) rF ¼ 2ð1 þ aÞ sin a ð14:115Þ v2=v1; which satisfies the velocity boundary conditions, 2k ð1 þ 2 sin aÞ ð14:116Þ because equal volumes of material must pass a given point per unit time, i.e. h1v1 ¼ h2v2: From (14.112) and (14.115), we get ) rF ¼ rð1 þ aÞ Since here the slip lines are at 45° to the horizontal, the 2k principal stresses r1 and r2 are horizontal and vertical. Let us consider a point G on the boundary slip line BD. At point 14.9 Upper-Bound Solution for Strip Drawing G, the algebraically largest principal stress acting horizon- tally is ðr1ÞG¼ rF; where rF is the tensile drawing stress. Figure 14.20 shows velocity field and hodograph for fric- tionless plane-strain strip drawing through wedge-shaped The boundary slip line BD is a b-line because it lies at dies of included angle 2a: Only half of the field is shown in the figure. Let, the initial entry velocity of the material into 45° to the left from the direction of the algebraically largest the die is v1; and the exit velocity of the material from the die principal stress r1; whereas the slip line GH is an a-line, is v2; where v2 [ v1; because the exit velocity is increased since it departs to the right from the direction of r1 by the due to reduction in the initial cross-sectional area of the same angle. One can easily note that r1; passes through the first and third quadrants formed by a right-handed aÀb coordinate system. The values of angle / at points H and G in the slip-line field diagram are respectively

682 14 Drawing: Flat Strip, Round Bar and Tube Fig. 14.20 a Velocity field and (a) b the corresponding hodograph for an upper-bound solution of Die frictionless plane-strain strip drawing through wedge-shaped dies A α θ –α C θ–α Ψ +α h1 ν1 Ψ +α E Drawn metal 2 Ψ ν2 h2 2 Workpiece θΨ θ B D Pulling stress σ Velocity field F (b) ν2 e ν1 a c Ψ o Ψ +α α θΨ θ θ –α b Ψ +α d θ –α Hodograph workpiece. In the construction of the velocity field, the In a similar way, number of triangular elements (or the number of divisions of the platen surface) n and the angle of inclination of the first DE aÞ ¼ CD aÞ ; velocity discontinuity h are the independent variables. These sinðh À sinðw þ variables have been selected so that the average drawing ) DE ¼ CD sinðh À aÞ stress, say rF; is minimized. From the diagram in Fig. 14.20, sinðw þ aÞ we get the following: h1 sin w sin2ðh À aÞ AB ¼ h1=2 sin h ð14:117Þ ¼ 2 sin2 h sin2ðw þ aÞ ð14:120Þ Again, since Finally we find, BC aÞ ¼ AB aÞ ; h2 ¼ DE sin w; sinðh À sinðw þ 2 ) BC ¼ h1 sinðh À aÞ ð14:118Þ ) h2 ¼ h1 sin2 w sin2ðh À aÞ ; or; sin h sinðw þ aÞ 2 2 sin2 h sin2ðw þ aÞ 2 Since h2 ¼ sin2 w sin2ðh À aÞ ð14:121aÞ h1 sin2 h sin2ðw þ aÞ CD ¼ BC ; If the reduction of area of the strip of unit width is given sin w sin h by r ¼ ðh1 À h2Þ=h1; then ) CD ¼ BC sin w sin h h2 sin2 w sin2ðh À aÞ ¼ h1 sin w sinðh À aÞ ð14:119Þ 1 À r ¼ h1 ¼ sin2 h sin2ðw þ aÞ ð14:121bÞ 2 sin2 h sinðw þ aÞ

14.9 Upper-Bound Solution for Strip Drawing 683 For n divisions of platen surface, cd ¼ bc ; sinðw þ sinðh À h2 sinn w sinnðh À aÞ aÞ aÞ h1 sinn h sinnðw þ aÞ 1 À r ¼ ¼ ð14:122Þ Or; cd ¼ bc sinðw þ aÞ sinðh À aÞ ) ð1 À r Þ1 ¼ sin w sinðh À aÞ ; ¼ v1 sin a sin h sinðw þ aÞ ð14:126Þ n sin w sin2ðh À aÞ sin h sinðw þ aÞ Or, sinðh À aÞ ¼ 1 sinðw þ aÞ ; de cd cd sin h sin h sin w sin h sin w sin w ð1 À rÞn ¼ ; or, de ¼ sin h cos a À cos h sin a Or, sinh v1 sin a sin2 h sinðw þ aÞ  ¼ sin2 w sin2ðh À aÞ ð14:127Þ ¼ ð1 À rÞn1 sin w cos a þ cos w sin a ; sin w de oe 1 sin a ¼ sinð180 À w À aÞ Or, cos a À cot h sin a ¼ ð1 À rÞnðcos a þ cot w sin aÞ; Or, cot h sin a ¼ cos a ¼ oe ; sinðw þ À ð1 À rÞn1cos a À ð1 À rÞn1cot w sin a aÞ no ¼ 1 À ð1 À rÞ1n cos a À ð1 À rÞ1ncot w sin a; Or, oe ¼ de sinðw þ aÞ sin a no ) cot h ¼ 1 À ð1 À rÞn1 cot a À ð1 À rÞn1cot w ¼ v1 sin2 h sin2ðw þ aÞ ð14:128Þ sin2 w sin2ðh À aÞ ð14:123Þ Again from the hodograph (velocity vector diagram) in ) v2 ¼ sin2 h sin2ðw þ aÞ ð14:129Þ Fig. 14.20, we get the followings: v1 sin2 w sin2ðh À aÞ ab ¼ oa aÞ ; or, ab ¼ v1 sin a ð14:124Þ If there are n number of triangular elements, i.e. for n sin a sinðh À sinðh À aÞ divisions of the platen surface, bc ¼ ab ; or; v1 ¼ sinn w sinnðh À aÞ ð14:130Þ sin h sin w v2 sinn h sinnðw þ aÞ bc ¼ ab sin h ¼ v1 sin a sin h ð14:125Þ From (14.122) and (14.130), it is seen that h1=h2 ¼ sin w sin w sinðh À aÞ v2=v1; which satisfies the velocity boundary conditions, because equal volumes of material must pass a given point per unit time, i.e. h1v1 ¼ h2v2; for strip of unit width. Now, from (10.61), the rate of internal energy con- sumption caused by the internal flow field is: d (W .D.) = k [ AB ⋅ ab + BC ⋅bc + CD ⋅ cd + DE ⋅de ] dt ⎡ h1 ⋅ v1 sinα ) + 2 h1 sin (θ −α) ) ⋅ v1 sinα sinθ ) ⎤ ⎢ 2 sin θ (ψ + α ⎥ ⎢ sin (θ − α sinθ sin sinψ sin (θ − α ⎥ ⎢⎥ ⎢ ⎥ = k ⎢ + h1 sinψ sin (θ − α ) ⋅ v1 sinα sinθ sin (ψ + α ) ⎥ ⎢ 2sin 2 θ sin (ψ + α ) sinψ sin 2 (θ − α ) ⎥ ⎢⎥ ⎢ ⎥ ⎢ + h1 sinψ sin2 (θ − α ) ⋅ v1 sinα sin2 θ sin (ψ + α ) ⎥ ⎣⎢ 2 sin2 θ sin 2 (ψ + α ) sin 2 ψ sin2 (θ − α ) ⎦⎥ = k ⎡ h1 v1 sinα + h1 v1 sin α + h1 v1 sin α + h1 v1 sin α ⎤ ⎢ sin θ 2 sinψ 2 sinψ ⎥ ⎣ 2 sin (θ − α ) sin (ψ + α ) 2sinθ sin (θ − α ) sin (ψ + α ) ⎦ ∴ d (W.D.) = k h1 v1 sin α ⎡ 1 + 1 ⎤ ⎢⎣ sin θ ⎥ dt sin (θ −α) sinψ sin (ψ + α ) ⎦ (14.131)

684 14 Drawing: Flat Strip, Round Bar and Tube For n divisions of the platen surface, From the last column of Table 14.1, considering the lowest value of ½f1= sin h sinðh À aÞg þ f1= sin w sinðw þ aÞgŠ; dðW:D:Þ ¼ nkh1v1 sin a 1 1 ! which is 4.6946, we get from (14.134) dt 2 h sinðh sinðw þ aÞ sin À aÞ þ sin w ð14:132Þ rF ¼ sinð10Þ Â 4:6946 ¼ 0:4076 ð14:136Þ 2k 2 Again, the rate of external work done with the tensile drawing stress of rF is dðW :D:Þ ¼ rF h2v2 ¼ rF h1v1 ð14:133Þ Case II when the number of divisions of the platen surface dt 2 2 n ¼ 2; then, from (14.123), we get Equating (14.132) with (14.133) for dðW:D:Þ=dt; we obtain σF h1 v1 = nk h1 v1 sin α ⎡ 1 + 1 ⎤ 2 2 ⎣⎢ sin θ sin (θ −α) sinψ sin (ψ + α ) ⎥ ⎦ ∴ σF = n sinα ⎡ 1 + 1 ⎤ 2k 2 ⎢ ⎥ ⎣ sin θ sin (θ −α) sinψ sin (ψ + α ) ⎦ (14.134) Example no cot h ¼ 1 À ð1 À rÞ1n cot a À ð1 À rÞn1 cot w Suppose, the reduction r ¼ 0:3303; and semi-angle of the die a = 10°. no ¼ 1 À ð0:6697Þ21 cotð10Þ À ð0:6697Þ12 cot w ¼ 1:0299 À 0:8184 cot w Case I when the number of divisions of the platen surface ð14:137Þ n ¼ 1; then, from (14.123) we get cot h ¼ r cot a À ð1 À rÞ cot w ð14:135Þ Again for different values of w; the values of h using ¼ 0:3303 cotð10Þ À ð1 À 0:3303Þ cot w (14.137) and also the values of the bracketed part of (14.134) are calculated and given in Table 14.2. ¼ 1:8732 À 0:6697 cot w From the last column of Table 14.2, considering the lowest For different values of w; let us calculate the values of h value of ½f1= sin h sinðh À aÞg þ f1= sin w sinðw þ aÞgŠ; using (14.135) and also the values of the bracketed part of which is 2.7046, we get from (14.134) (14.134), all of which are produced below in the tabular form in Table 14.1. rF ¼ 2 sinð10Þ Â 2:7046 ¼ 0:4696 ð14:138Þ 2k 2 Table 14.1 Intermediate steps w h wþa hÀa sin w sin h sinðw þ aÞ sinðh À aÞ to compute tensile drawing stress, (Degree) (Degree) (Degree) (Degree) when the number of divisions of the platen surface is one 45 39.7235 55 29.7235 0.7071 0.6391 0.8192 0.4958 40 42.9278 50 32.9278 0.6428 0.6811 0.7660 0.5436 35 47.4864 45 37.4864 0.5736 0.7371 0.7071 0.6086 30 54.5018 40 44.5018 0.5000 0.8141 0.6428 0.7009 sin h sinðh À aÞ sin w sinðw þ aÞ 1 þ 1 sinðh sinðw sin h À aÞ sin w þ aÞ 0.3169 0.5793 4.8818 0.3702 0.4924 0.4486 0.4056 4.7321 0.5706 0.3214 4.6946 4.8639

14.9 Upper-Bound Solution for Strip Drawing 685 Table 14.2 Intermediate steps to compute tensile drawing stress, when the number of divisions of the platen surface is two w h wþa hÀa sin h 1 À aÞ þ sin w 1 þ aÞ (Degree) (Degree) (Degree) (Degree) sinðh sinðw 50 71.0588 60 61.0588 2.7145 60 60.8649 70 50.8649 2.7046 70 53.7949 80 43.7949 2.8711 Table 14.3 Summary of computation for tensile drawing stress when after the drawing operation, the true strain up to which the the numbers of divisions of the platen surface are one and two rod has been deformed is: r a n rF=2k e1 ¼ lnðA1=A2Þ ¼ lnÂðp=4Þ112=ðp=4Þ92à ¼ 0:4: 0.3303 10 1 0:4076 It is given that at R ¼ 1:65; i.e. at eR ¼ 0:5; the defor- 2 0:4696 mation resistance of the metal is r0R ¼ 400 N=mm2; and at e ¼ 0; the initial deformation resistance is r00 ¼ Summary of the above computation is shown in 200 N=mm2: Since the flow stress varies linearly, so at e1 ¼ 0:4; the flow stress will be: Table 14.3. Hence, the answer will be the lowest value of rF ; i.e. r01 ¼ 200 þ ½ð400 À 200Þ=0:5Š  0:4 ¼ 360 N=mm2: 2k rF ¼ 0:4076: Hence, average flow stress of the metal is: 2k 14.10 Solved Problems r0 ¼ ðr00 þ r01 Þ=2 ¼ ð200 þ 360Þ=2 ¼ 280 N=mm2: 14.10.1. A round steel rod of 11 mm diameter is drawn As the semicone angle of die is a = 5°, and the coefficient homogeneously at a speed of 2 m=s through a conical converging die with a total included die angle of 10° to of friction at the rod–die interface is l ¼ 0:07; so, produce rod of 9 mm diameter. The coefficient of friction at B ¼ l cot a ¼ 0:07 cot 5 ¼ 0:8: the rod–die interface is 0.07. A tensile test on the same steel shows that its flow stress varies linearly from an initial value (a) As there is no back tension, the drawing stress rF of 200 to 400 MPa at a reduction ratio of 1.65. Assume that according to (14.58) is: there is no back tension and determine the following:  þ \"  2B# (a) Drawing load. r0 1 B B D2 (b) Power required for the above drawing operation, rF ¼ À assuming 100% efficiency of the electric motor supplying 1 D1 the power. (c) If all the work due to plastic deformation and friction is ÀÁ converted into heat, what would be the temperature rise of Hence, the drawing load P is given by P ¼ rF  pD22=4 the wire? The properties of the steel are as follows: ) P ¼ \" þ ( À  2B)#pD22 Density ¼ 7850 kg mÀ3; and specific heat ¼ 486 J kgÀ1KÀ1: r0 1 B B D2 Solution 1 D1 4 The true strain corresponding to the given reduction ratio ¼ \"     ( À  9 2Â0:8)#p  92 R ¼ 1:65 is: 280 1 þ 0:8 1 11 4 N 0:8 eR ¼ ln R ¼ ln 1:65 ¼ 0:5: ¼ 11006:85 N ¼ 11 kN: Given that the diameter of the rod before drawing is D1 ¼ 11 mm; and that after drawing is D2 ¼ 9 mm: If A1 and A2 (b) As the drawing speed is v ¼ 2 m=s; so the drawing are cross-sectional areas of the rod, respectively, before and power assuming 100% efficiency will be [see (14.55)]: P  v ¼ ð11006:85  2Þ J sÀ1 or W ¼ 22013:7 W ¼ 22 kW: (c) If we assume the length of the drawn wire is L m, then work done by the load in drawing a length of L m of wire is: P  L ¼ ð11006:85  LÞ N m:

686 14 Drawing: Flat Strip, Round Bar and Tube Given that the density of steel is q ¼ 7850 kg mÀ3; and its a mean flow stress r0 between initial strain of 0 and the specific heat is C ¼ 486 J KgÀ1KÀ1: Hence, the mass of the maximum theoretical true strain ðemaxÞideal¼ 1: Hence, according to Hollomon true stress–strain relation, the mean wire is: flow stress is: m ¼ volume of the drawn wire  its density p  ! ¼  0:0092 L  7850 kg ¼ ð0:5 LÞkg: Rðemax Þideal R1 Ke0:5 de r0 de 4 r0 ¼ 0 ¼ 0 R1 e de If DT is the temperature rise of the wire, then Rðemax Þideal e de 00 P  L ¼ m  C  DT; ¼ K ð1Þ0:5 þ 1 ¼ K : 0:5 þ 1 1:5 Or; 11006:85  L ¼ ð0:5 LÞ Â 486  DT 1 ) DT ¼ 11006:85 ÂL K ¼ 45:3C: Âà ð0:5 LÞ Â 486 Substituting ðrFÞideal¼ ðrFÞmax ideal¼ K; and r0 ¼ r0 ¼ K=1:5; into (14.60), we get the maximum reduction, rmax; as follows: 14.10.2. Assume that a wire strain-hardens according to K  Hollomon true stress–strain relation with strain-hardening 1:5 1 1 exponent of 0.5 and there is no back tension. If the hardened K ¼ ln 1 À rmax ; or; ln 1 À rmax ¼ 1:5; wire can sustain a maximum tensile stress corresponding to the maximum ideal reduction per pass for a non-hardened ) rmax ¼ 1 À 1 ¼ 0:777 ¼ 77:7%: frictionless wire drawing without back tension, what will be expð1:5Þ the maximum possible reduction per pass for the above strain-hardened wire, during its frictionless homogeneous Since n ¼ strain-hardening exponent ¼ 0:5; so the drawing operation? Show an expression for the maximum expression for the maximum possible reduction per pass in possible reduction per pass in terms of strain-hardening terms of strain-hardening exponent is: exponent. rmax ¼ 1 À 1 1Þ : Solution expðn þ The Hollomon true stress–strain relation given by (1.90a) is: 14.10.3. The cross-sectional area of a metal wire with r ¼ K en; where K ¼ strength coefficient, and n ¼ diameter of 4 mm, subjected to a back tensile force of 1 kN, strain-hardening exponent ¼ 0:5 (given), i.e. r ¼ K e0:5: is reduced by 50% by drawing homogeneously through a conical converging die with a total included die angle of 12°. The maximum ideal reduction of a non-hardened fric- The flow curve of the metal is described by Hollomon parabolic stress–strain relation with strength coefficient of tionless wire drawing according to (14.63) is: ðrmaxÞideal¼ 1200 MPa and strain-hardening exponent of 0.4. If the 1 À ð1=eÞ: Hence, the maximum ideal true strain in fric- coefficient of friction at the wire–die interface is 0.053, tionless condition corresponding to ðrmaxÞideal is: determine the maximum possible reduction for the hardened wire. ðemaxÞideal ¼ ln 1 À 1 ¼ ln 1 À ½1 1 ð1=eފ ðrmaxÞideal À Solution ¼ ln e ¼ 1: Given that the strain-hardened wire can sustain a maxi- Since the diameter of the wire before drawing is D1 ¼ 4 mm; mum tensile stress corresponding to ðemaxÞideal: Therefore, so the back tension is: the maximum drawing stress for the strain-hardened wire during a frictionless homogeneous drawing operation is: rB ¼ backÀtensiÁle force ¼ 1000  4 N=mm2 pD21 =4 p  42 Âà ðrFÞmax ideal ¼ the value of r atðemaxÞideal ¼ 79:6 N=mm2: ¼ K ðemaxÞi0d:e5al¼ K ð1Þ0:5 ¼ K: The true strain corresponding to the given fractional reduction in area r ¼ 0:5 is: The maximum possible reduction during a frictionless homogeneous drawing operation for the above er ¼ ln 1 1 r ¼ ln 1 1 ¼ 0:69: strain-hardened wire can be obtained by replacing the uni- À À 0:5 form flow stress r0 in (14.60) for a non-hardened wire, with

14.10 Solved Problems 687 Hence, the mean flow stress between initial strain of 0 14.10.4. A copper tube with outside diameter of 30 mm and and strain after deformation er ¼ 0:69 is: thickness of 2.5 mm is close-pass drawn homogeneously using a conical fixed plug with a semi-angle of 3° through a Rer 0R:69 1200e0:4 de conical die with a semi-angle of 15° to produce outside r0 de diameter of 27 mm and thickness of 2 mm. The plane strain flow stress of copper is 300 MPa. Assuming plane-strain r0 ¼ 0 ¼0 0R:69 MPa condition and the coefficient of friction at the tube–die e de interface to be 0.06 and that at the tube–plug interface to be Rer e de 0.09, determine the drawing load and power required at a drawing speed of 1.5 m=s; assuming 90% efficiency of the 00 power unit. 1200 ð0:69Þ0:4 þ 1 Nmm2 738:9 Nmm2: 0:4 þ 1 0:69 ¼ Á ¼ As the semicone angle of die is a = 6°, and the coefficient of friction at the wire–die interface is l ¼ 0:053; so, Solution B ¼ l cot a ¼ 0:053 cot 6 ¼ 0:5: As a first approximation if we assume that the maximum drawing stress ðrFÞmax¼ r0S:H: ¼ r0 ðmean flow stressÞ; then Gr00iv¼en30th0aNt thmemp2l:aTnehestcroaeinfficflioewnt stress of copper tube is (14.64) reduces to of friction at the tube–die 1 À fBr0=ð1 þ BÞr0g!1=B interface is l1 ¼ 0:06; and that at the tube–plug interface is rmaxS:H: ¼ 1 À 1 À fBrB=ð1 þ BÞr0 g!1=B l2 ¼ 0:09: The semicone angle of the die is a = 15° and that ¼1À 1=ð1 þ BÞ of the plug is b = 3°. Hence, from (14.80), BÃplug ¼ 1 À fBrB=ð1 þ BÞr0g l1 þ l2 b ¼ 0:06 þ 0:09 ¼ 0:7: tan a À tan tan 15 À tan 3 Hence, the above equation can be applied to get an The initial outside diameter of the tube is Do1 ¼ 30 mm; and the thickness is h1 ¼ 2:5 mm: After drawing, the outside approximate maximum reduction of area, ðrmaxS:H: Þapprox:; as diameter is Do2 ¼ 27 mm; and the thickness is h2 ¼ 2 mm: follows: & BrB'!1=B Hence from (14.89), the cross-sectional areas of the tube ðrmaxS:H: Þapprox: ¼ 1 À 1= ð1 r0 þ BÞ À before and after drawing are respectively: ¼1À & þ 0:5Þ À 0:5 Â 79:6'!1=0:5¼ 0:522: A1 ¼ pðDo1 À h1Þh1 ¼ pð27:5 Â 2:5Þmm2 ¼ 216 mm2; 1= ð1 and 738:9 A2 ¼ pðDo2 À h2Þh2 ¼ pð25 Â 2Þmm2 ¼ 157:08 mm2: The true strain corresponding to the approximate maxi- Using (14.88b), we get the plug drawing stress: mum reduction in area, ðrmaxÞapprox:; is: emax ¼ ln 1 ¼ ln 1 ¼ 0:74: 1 þ BÃplug !\"  BÃplug # 1 ðrmax Þapprox: 0:522 A2 À 1 À rF ¼ r00 BÃplug 1 À ¼ 300  \" À The value of the tensile flow stress of the strain-hardened A1 # metal at the exit plane, where the maximum true strain is N=mm2 emax ¼ 0:74; is: 1 þ 0:7 157:080:7 0:7 1 r0S:H: ¼ 1200e0m:a4x ¼ 1200 ð0:74Þ0:4N=mm2 216 ¼ 1063:8 N=mm2: ¼ 145:61 N/mm2: Hence from (14.90) the required plug drawing load is: For a strain-hardened metal, the value of the maximum P ¼ rF A2 ¼ 145:61 Â 157:08 N ¼ 22872:42 N ¼ 22:87 kN: drawing stress, ðrFÞmax; can reach the value of r0S:H: ; i.e. ðrFÞmax¼ r0S:H: : Hence, the maximum possible reduction of As the drawing speed is v ¼ 1:5 m=s; so the drawing area according to (14.64) is: power assuming 100% efficiency will be: rmaxS:H: ¼ 1 À 1 À fBr0S:H: =ð1 þ BÞr0g!1=B P Â v ¼ ð22872:42 Â 1:5Þ J sÀ1 1 À fBrB=ð1 þ BÞr0g Or, the minimum drawing power ¼ 34308:63 W: ¼ 1 À 1 À f0:5 Â 1063:8=ð1 þ 0:5Þ Â 738:9g!1=0:5 1 À f0:5 Â 79:6=ð1 þ 0:5Þ Â 738:9g ¼ 0:709 ¼ 70:9%:

688 14 Drawing: Flat Strip, Round Bar and Tube Since the efficiency of the power unit is g ¼ 90%; so the Drawing stress; ! actual drawing power required is: rF ¼ r00 h2 P  v 34308:63 1 À h1 expðlH2 À lH1Þ g 0:9 W ! ¼ ¼ 38120:7 W ¼ 38:12 kW: h2 h1 ¼ r00 1 À expðÀlH1Þ ð14:139Þ [Note that if we would use (14.88a) to get the plug drawing stress, then Using (14.39), we get the drawing stress for cylindrical 1 þ BÃplug !\"  BÃplug # dies: h2 rF ¼ r00 BpÃlug 1 À h2 ! ¼  \" À h1 h1 0:7# ðrF Þcylindrical ¼ r00 1 À expðÀlH1Þ N/mm2 1 þ 0:7  2 ! ¼ 500 1 À 2:5 expðÀ0:09  3:1615Þ N=mm2 300 1 0:7 2:5 5 ¼ 105:36 N=mm2:à ¼ 311:91N=mm2: The cross-sectional area after reduction is: 14.10.5. Calculate the drawing load required for 50% A2 ¼ h2  width of the strip ¼ 2:5  30 mm2 ¼ 75 mm2: reduction in area of a 30 mm wide by 5-mm-thick steel strip Hence, the drawing load required for cylindrical dies is: using cylindrical dies of 10 mm radius under plane strain condition without any back tension. Compare this load with Pcylindrical ¼ ðrFÞcylindricalÂA2 ¼¼ ð311:91  75ÞN the loads required when wedge-shaped dies with a ¼ 23393:25N ¼ 23:4 kN: semi-angle (a) equal to the entry angle and (b) equal to the mean angle of cylindrical dies are used for the same strip (a) When the entry angle to the wedge-shaped die is a ¼ drawing under plane strain condition without any back ten- a1 ¼ 0:505 rad; then sion. Assume that the average plane strain flow stress of steel is 500 MPa and the coefficient of friction at the job–die B ¼ l cot a ¼ 0:09 cotð0:505Þ ¼ 0:163: interface is 0.09. Neglect the redundant deformation, if any. Solution Using (14.12), we get the drawing stress for Given that the average plane strain flow stress of steel is wedge-shaped dies: r00 ¼ 500 N=mm2; and the coefficient of friction at the job– die interface is l ¼ 0:09: The cylindrical die radius is R ¼  þ \"  B# r00 1 B B h2 10 mm; the thickness of the strip before drawing is h1 ¼ ðrF Þwedge ¼ À 5 mm; and that after fractional reduction of 0.5 is h2 ¼ 1 h1 ð1 À 0:5Þ Â 5 mm ¼ 2:5 mm: Hence, according to (12.7) the ¼  þ 0:163\" À 2:50:163#N=mm2 1 0:163 1 5 500 initial angle of contact of the strip at the entry to the cylin- ¼ 381:13 N=mm2: drical die is: rffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffi Hence, the drawing load required for wedge-shaped dies is: a1 ¼ 2 sinÀ1 h1 À h2 ¼ 2 sinÀ1 5 À 2:5 Pwedge ¼ ðrFÞwedgeÂA2 ¼ ð381:13  75ÞN 4R 4  10 rad ¼ 28584:75N ¼ 28:58 kN: ¼ 0:505 rad: According to (14.36b): * Pwedge À Pcylindrical  100 Pcylindrical rffiffiffiffi rffiffiffiffi  2 R tanÀ1 \"rhRffiffi2ffiffiffiaffi 1 H1 ¼ rhffiffiffi2ffiffiffi ¼ 28584:75 À 23393:25  100 ¼ 22:2%; 23393:25 # ¼2 10 tanÀ1 10 ð0:505Þ So, the load required for wedge-shaped dies is 22% 2:5 2:5 higher than that required for cylindrical dies when the entry angle is the same for both dies. ¼ 3:1615:

14.10 Solved Problems 689 (b) When the entry angle to the wedge-shaped die is a ¼ (a) Since the coefficient of friction at the tube–plug interface ða1 þ 0Þ=2 ¼ ð0:505 þ 0Þ=2 rad ¼ 0:2525 rad; then B ¼ l is l2 ¼ 0:05; so according to either (14.102b) or (14.103b) cot a ¼ 0:09 cotð0:2525Þ ¼ 0:349: the semicone angle of the floating plug is: Using (14.12), we get the drawing stress for b ¼ tanÀ1 l2 ¼ tanÀ1ð0:05Þ ¼ 2:86 ¼ 25103600: wedge-shaped dies: (b) Since tan b ¼ l2 ¼ 0:05; so from (14.80): ðrF Þwedge ¼  þ \" À h2B# BpÃlug ¼ l1 þ l2 b ¼ 0:05 þ 0:05 ¼ 0:502: r00 1 B B h1 tan a À tan tan 14 À 0:05 1 ¼  \" À 2:50:349 # Using (14.88a), we get the plug drawing stress: 1 þ 0:349 N=mm2 0:349 1 5 !\" # 500 r00 1 þ BpÃlug  BÃplug 280 h2 ¼ 415:27 N=mm2: rF ¼ BpÃlug 1À ¼  \" h1 Hence, the drawing load required for wedge-shaped dies 1 þ 0:502 À 2:40:502#N=mm2 is: 1 ¼ 88:78 N=mm2: 0:502 3 Pwedge ¼ ðrFÞwedgeÂA2 ¼ ð415:27 Â 75ÞN Since the outside diameter of the drawn tube is Do2 ¼ ¼ 31145:25N ¼ 31:15kN: 30 mm; therefore the cross-sectional area of the tube after Pwedge À Pcylindrical 31145:25 À 23393:25 drawing (14.89b) is: Pcylindrical 23393:25 * Â 100 ¼ Â 100 ¼ p ÀDo22 À Di2Á ¼ pðDo2 À h2Þh2 ¼ pð27:6 Â 2:4Þmm2 4 A2 ¼ 33:14%; ¼ 208:1 mm2: So, the load required for wedge-shaped dies is 33% Hence, from (14.90) the plug drawing load is: higher than that required for cylindrical dies when the entry P ¼ rF A2 ¼ ð88:78 Â 208:1ÞN ¼ 18475:12 N: angle of the former is equal to the mean angle of the latter. 14.10.6. A copper tube having an average plane strain flow As the drawing speed is v ¼ 1:3 m=s; so the drawing power stress of 280 MPa is close-pass drawn homogeneously assuming 100% efficiency will be: P Â v ¼ 18475:12Â through a conical die with an included total angle of 28° 1:3 J sÀ1 i:e:; the minimum drawing power ¼ 24017:656 W: using a suitable conical floating plug (without having any cylindrical zone) at a speed of 1:3 m=s under plane-strain Since the efficiency of the motor supplying power is g ¼ condition. The thickness of the tube is reduced from 3 to 88 %; so the actual drawing power required is: 2.4 mm while the outside diameter of the tube remains unaltered at a value of 30 mm. If the coefficient of friction is P Â v ¼ 24017:656 W ¼ 27292:8 W ¼ 27:3 kW: 0.05 at the tube–die as well as the tube–plug interfaces, g 0:88 (a) What will be the semicone angle of the floating plug? Exercise (b) What is the power required for the above drawing operation assuming 88% efficiency of the motor supplying 14.Ex.1. For the solved problem in Sect. 14.10.4, calculate power? the die pressure at the entry as well as at the exit plane of the deformation zone. Solution 14.Ex.2. A steel tube of outer diameter of 110 mm and wall thickness of 10 mm is homogeneously reduced to outer Given that the average plane strain flow stress of copper is diameter of 98 mm without any change in the wall thickness r00 ¼ 280 N=mm2; the semicone angle of the die is a = 14°, by sinking operation, using a conical die of semi-angle 12°, the coefficient of friction at the tube–die interface is l1 ¼ and a lubricant providing the coefficient of friction = 0.1. If 0:05: The thickness of the tube before drawing is h1 ¼ the mean uniaxial flow stress of the steel is 400 MPa, cal- 3 mm; and that after drawing is h2 ¼ 2:4 mm: culate the load required for sinking.