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

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12.19 Solved Problems 589 For the exit zone of rolling from (12.56b): y þ ¼ R00 ¼ R 1þ Cr P0 ! ¼ w ð1h1þÀ0h:1ð22ÞÂ:16ð1Â:60150ÀÀ111Þ:Â1332754Þ 8Â941:06À3 !m 591:3 h exp ðl HÞ: 0:125 Â As found earlier, y1 ¼ 0:8165; and y2 ¼ 0:6697: With H1 ¼ 12:98; the value of Hn from (12.57a) is: ¼ 0:146 m: ¼ H1 À 1 ln y2 h1 Hn 2 2 l y1 h2 12:98 1 0:6697 Â 1:605 Â 10À3 Hence, the variation in roll radius is: 2 0:055 ln 0:8165 Â 1:1325 Â 10À3 ¼ À Â R00 À R0 Â 100 ¼ 0:146 À 0:145 Â 100 R0 0:145 2 ¼ 5:1218: ¼ 0:69 %\\1%; Hence with R0 ¼ 0:145 m and Hn ¼ 5:1218; the new that satisfies the given condition: no-slip angle from (12.57b) is: So, the cold-rolling load is: P0 ¼ 374894:6 N ¼ 374:9 kN: rﬃﬃﬃﬃ rﬃﬃﬃﬃ ! (b) Cold-rolling torque for each work roll, MT =2; in case of h2 h2 Hn elastically ﬂattened roll radius R0; can be obtained from w¼ R0 tan R0 2 (12.64): rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ! Za 1:1325 Â 10À3 1:1325 Â 10À3 5:1218 ¼ 0:145 tan MT ¼ w R R0 r00 h dh þ wR ðh1 rB À rF Þ 0:145 2 2 y 2 h2 ¼ 0:02035 rad: 0 To ﬁnd the value of Ra yh dh; let us take the data for y and 0 According to (12.8a), the new stock thickness at an angular coordinate h from Table 12.3 and make another arbitrary contact angle of h is: Table 12.4 as shown below: h ¼ h2 þ R0 h2 ¼ 1:1325 Â 10À3 þ 0:145 h2 m: Hence, by summing up the values of y h Dh from h ¼ 0 to h ¼ a ¼ 0:057084 rad from Table 12.4, we get To get arbitrarily the angular coordinates h; let us divide from the exit plane of rolling, where h ¼ 0; to the neutral Za Xh¼a plane of rolling, where h ¼ w ¼ 0:02035 rad; into two y h dh ¼ y h Dh ¼ 13:7537 Â 10À4: nearly equal angles and from the neutral plane to the entry plane of rolling, where h ¼ a ¼ 0:057084 rad; into four 0 h¼0 nearly equal angles, which are shown in the Table 12.3: Therefore, cold rolling torque for each roll in case of Hence, by summing up the values of y Dh from h ¼ 0 to elastically ﬂattened roll radius R0 is: h ¼ a ¼ 0:057084 rad from Table 12.3, we get MT Za wR 2 2 ¼ w R R0 r00 y h dh þ ðh1 rB À h2 rF Þ Za Xh¼a ¼ 0:1 Â 0 Â 0:145 Â À Â 106Á y dh ¼ y Dh ¼ 0:04744: 545 0:125 0 h¼0 Â 13:7537 Â 10À4 þ 0:1 Â 0:125 2 Therefore, new cold-rolling load of the strip is: Â Â Â 100 À 1:1325 Â 180Þ Â 103Ã N m ð1:605 Za ¼ 1087:67 N m ¼ 1:1 kN m: P0 ¼ w R0 r00 y dh (c) According to (12.11b), the length of the deformation zone ¼ Â Â 0 Â À Â 106Á Â Ã for elastically ﬂattened rolls having R0 ¼ 0:145 m is: 0:1 545 0:04744 N 0:145 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ L ¼ pﬃRﬃﬃ0ﬃﬃðﬃﬃhﬃﬃﬃ1ﬃﬃﬃÀﬃﬃﬃﬃhﬃﬃﬃ2ﬃﬃÞﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 374894:6 N: ¼ 0:145 Â ½ð1:605 À 1:1325Þ Â 10À3 m With the new load P0; the second new radius R00 of the elastically ﬂattened rolls according to (12.26b) is: ¼ 0:00828 m:

590 12 Rolling Table 12.3 Intermediate steps to calculate rolling load for elastically ﬂattened roll radius Line No. Column No. 12 3 4 5 6 7 0.0295 0.0387 0.0479 0:057084 1 Angular coordinate, h; (radian) 0 0.01017 0:02035 (N.P.) 1.2587 1.3497 1.4652 1.605 Â10À3 Â10À3 Â10À3 Â10À3 2 h ¼ 1:1325 Â 10À3 þ 0:145 h2 1.1325 1.1475 1.1925 Â10À3 1.4935 1.6717 1.8559 2.0421 (m) Â10À3 Â10À3 − − − − 0.8755 0.8387 0.8201 0.8164 3 expðl HÞ ¼ exp ½1:245 tanÀ1ð11:315 hÞ 1.0 1.1533 1.3255 0.8755 0.8387 0.8201 0.8164 4 y þ ¼ 591:3 h exp ðl HÞ 0.6696 0.7825 0.9346 0.9051 0.8571 0.8294 0.8183 0.00915 0.0092 0.0092 0.009184 5 yÀ ¼ 1038:8 h −− 0.9346 0.00828 0.00789 0.00763 0.00752 exp ðl HÞ 6y 0.6696 0.7825 0.9346 7 Consecutive means of values in line 6, y − 0.7261 0.8586 8 Increments in radians, Dh − 0.01017 0.01018 9 y Dh − 0.00738 0.00874 Table 12.4 Intermediate steps to calculate rolling torque for elastically ﬂattened roll radius Line No. Column No. 12 3 4 5 6 7 0.0387 0.0479 0:057084 1 Angular coordinate, h; (radian) 0 0.01017 0:02035 0.0295 (N.P.) 0.8387 0.8201 0.8164 0.03246 0.03928 0.0466 2y 0.6696 0.7825 0.9346 0.8755 0.02914 0.03587 0.04294 3. y h 0 0.00796 0.01902 0.02583 0.0092 0.0092 0.009184 2.681 3.3 3.944 4 Consecutive means of values in line 3, 0.00398 0.01349 0.02242 Â10À4 Â10À4 Â10À4 yh 5 Increments in radians, Dh − 0.01017 0.01018 0.00915 6 y h Dh − 4.047 1.373 2.051 Â10À5 Â10À4 Â10À4 Lever arm ratio can be obtained from (12.92) as follows: WR ¼ 4p MT N ¼ 4 p Â 1087:67 Â 150 W 2 60 60 MT 1 ¼ 34:17 Â 103 W: k¼ 2 P0 L ¼ 1087:67 ¼ 0:35: Since the bearing diameter is db ¼ 250=2 mm ¼ 0:125 m; 374894:6 Â 0:00828 and the coefﬁcient of friction at the roll neck is lb ¼ 0:01; so according to (12.96) the total power required for four bearings is: Since the lever arm or moment arm is a ¼ k L; so for WB ¼ lb P0 db 2pN L ¼ 0:00828 m; and k ¼ 0:35; the lever arm or moment arm 60 is: ¼ 0:01 Â 374894:6 Â 0:125 Â 2 p Â 150 W a ¼ k L ¼ 0:35 Â 0:00828 m ¼ 2:898 Â 10À3 m 60 ¼ 2:9 mm: ¼ 7:36 Â 103 W: (d) Given that the rotating speed of rolls is N ¼ 150 rpm: Hence, the total power requirement for two rolls with four bearings will be Hence, for elastically ﬂattened rolls, the power required to WT ¼ WR þ WB operate a pair of rolls according to (12.95) is: ¼ ð34:17 þ 7:36Þ Â 103 W ¼ 41:53 kW:

12.19 Solved Problems 591 Since the overall efﬁciency of the power unit is g ¼ 80 %; (a) From (12.11b), the length of the deformation zone is: so the overall power requirement of the cold-rolling mill is: pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ WM ¼ WT ¼ 41:53 kW ¼ 51:9 kW: L ¼ R ðh1 À h2Þ ¼ 0:5 Â ð0:025 À 0:015Þ m g 0:8 ¼ 0:0707 m: From (12.81) the no-slip angle is: 12.19.4. A 500-mm-wide and 25-mm-thick plate of alu- rﬃﬃﬃﬃ \" rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃ # minium alloy is hot reduced homogeneously by 40% in w ¼ h2 tan 1 tanÀ1 h1 À h2 À p h2 ln h1 thickness in one pass with a pair of steel rolls of diameter 1 m R2 h2 8 R h2 each. The uniaxial ﬂow stress of the alloy is expressed by Hollomon stress–strain relation with strength coefﬁcient of According to (12.8a), the plate thickness, hn; at the 140 MPa and strain-hardening coefﬁcient of 0.2. Assuming neutral plane, where the angle of contact is w; is: homogeneous deformation and using Sims’ theory, deter- mine the following considering elastic ﬂattening of rolls into hn ¼ h2 þ R w2 \" rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃ # account, till the roll radius varies by less than 1%: ¼ h2 þ h2 tan2 1 tanÀ1 h1 À h2 À p h2 ln h1 2 h2 8 R h2 (a) Rolling load. \" rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! (b) Rolling torque for the pair of work rolls. ¼ 0:015 þ 0:015 tan2 1 tanÀ1 0:025 À 0:015 (c) Lever arm ratio and the lever arm or moment arm; 2 0:015 (d) Overall power requirement of the hot-rolling mill, if the rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ # rolls rotate at 100 rpm. Assume that the power loss due Àp 0:015 0:025 to roll-neck friction is 20% and the overall efﬁciency of 8 0:5 ln 0:015 m ¼ 0:01651 m: the power unit is 80%. Again for undeformed rolls, QS from (12.87b) is: Solution rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p h2 tanÀ1 h1 À h2 QS ¼ 2 rﬃhﬃﬃ1ﬃﬃﬃÀﬃﬃﬃﬃﬃhﬃﬃﬃ2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃ h2 Given that the roll radius is R ¼ 0:5 m; and the width of the plate is w ¼ 0:5 m; which remains constant since rolling is þ rhﬃﬃ1ﬃﬃﬃÀﬃRﬃﬃﬃﬃhﬃﬃﬃ2ﬃﬃﬃlﬃnﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃhh1nh2 Àp r4ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! assumed to be plane strain deformation. The initial thickness of the plate is h1 ¼ 0:025 m; and after 40% reduction, the ¼p 0:015 tanÀ1 0:025 À 0:015 reduced thickness is h2 ¼ 0:025 Â ð1 À 0:4Þ m ¼ 0:015 m: 2 0:025 À 0:015 0:015 With the given strain-hardening coefﬁcient of n ¼ 0:2; and rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! strength coefﬁcient of K ¼ 140 MPa ¼ 140 Â 106 Pa; the 0:5 0:025 Â 0:015 Àp þ 0:025 À 0:015 ln 0:01651 4 uniaxial ﬂow stress, r0; of the aluminium alloy following HKoelnlo¼mÀo1n4s0trÂess1–0s6tÁraei0n:2rePlaa;tiownh(e1r.e90tha)e can be given by r0 ¼ ¼ 1:66: true plastic strain e varies from 0 to e1 ¼ lnðh1=h2Þ ¼ lnð0:025=0:015Þ ¼ 0:51: Hence from (12.86b), hot-rolling load is: Hence the strain average uniaxial ﬂow stress is: Âw r00 R0Re10er1 0dede R e1 P ¼ 0:5 L QÀS 106Á Ã 0R ¼ Â 117:74 1:66 r0 ¼ ¼ K ende Â Â 0:0707 Â N de e1 ¼ 6:91 Â 106 N: 0 ¼ K en1 ¼ 140 Â 106 ð0:51Þ0:2 Pa The radius of the elastically ﬂattened rolls, R0; can be ðn þ 1Þ 1 þ 0:2 obtained from Hitchcock relation given by (12.26a), which is: ¼ 101:967 Â 106 Pa: R0 R So, the strain average plane strain deformation resistance is: ¼ 1 þ w Cr P ðh1 À h2Þ pﬃﬃ r00 ¼ 2= 3 r0¼117:74 Â 106 Pa: where for steel rolls, Cr ¼ 2:16 Â 10À11PaÀ1

592 12 Rolling Hence, R00 ¼ R 1þ Cr P0 ! 10À11 À 106Á! w ðh1 À h2 Þ À 106Á! 6:91 ð2:16 Â 7:06 R0 ¼ 0:5 Â ð2:16 Â Þ Â Â 10À11 Þ Â Â 1 þ 0:5 Â ð0:025 À 0:015Þ m ¼ 0:5 Â 1 þ 0:5 Â ð0:025 À 0:015Þ m ¼ 0:5149 m: ¼ 0:5152 m: For elastically ﬂattened rolls with R0 ¼ 0:5149 m: Hence, the variation in roll radius is: R00 À R0 Â 100 ¼ Hence, the new length of the deformation zone is: R0 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 0:5152 À 0:5149 Â 100 ¼ 0:058 %\\1%; that satisﬁes the L ¼ R0 ðh1 À h2Þ ¼ 0:5149 Â ð0:025 À 0:015Þ m 0:5149 ¼ 0:0717 m: given condition. The new thickness of plate at the neutral plane is: So, the hot-rolling load is: P0 ¼ 7:06 Â 106 N ¼ 7.06 MPa. rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 tanÀ1 h1 À h2 hn ¼ h2 þ h2 tan2 2 # h2 (b) For R0 ¼ 0:5149 m; the angle of contact according to rﬃﬃﬃﬃ (12.7) is: Àp h2 ln h1 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 8 R0 h2 h1 À h2 0:025 À 0:015 a ¼ 2 sinÀ1 4 R0 ¼ 2 sinÀ1 4 Â 0:5149 rad \" rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! ¼ 0:015 þ 0:015 tan2 1 tanÀ1 0:025 À 0:015 2 0:015 ¼ 0:1395 rad: rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ # For R0 ¼ 0:5149 m; the no-slip angle from (12.81) is: Àp 0:015 0:025 8 0:5149 ln 0:015 m ¼ 0:01652 m. rﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ w¼ h2 tan 1 tanÀ1 h1 À h2 For elastically ﬂattened rolls with R0 ¼ 0:5149 m and R0 2 h2 hn ¼ 0:01652 m; the new QS is: rﬃﬃﬃﬃ # Àp h2 h1 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 8 R0 ln h2 QS ¼p rrhﬃhﬃﬃﬃﬃ1ﬃ1ﬃﬃﬃﬃﬃRÀﬃhÀﬃﬃﬃﬃ2ﬃ0ﬃﬃﬃﬃhﬃhﬃﬃﬃﬃ2ﬃ2ﬃﬃﬃlﬃtnﬃaﬃﬃnﬃﬃÀﬃpﬃﬃ1ﬃﬃﬃhﬃhﬃﬃ1nﬃﬃﬃhﬃﬃ2hﬃ 1 hÀÀ2rhp42ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ \" rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! 2 þ ¼ 0:015 Â tan 1 tanÀ1 0:025 À 0:015 0:5149 2 0:015 ¼p 0:015 0:025 À 0:015 À p rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 0:025# rad 2 0:025 À 0:015 0:015 8 0:015 ln 0:015 tanÀ1 0:5149 rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ! ¼ 0:05432 rad. þ 0:5149 0:025 Â 0:015 Àp Hence, total torque for the pair of work rolls according to 0:025 À 0:015 ln 0:01652 4 (12.89) is: ¼ 1:672: MT ¼ w R R0 r00 ða À 2wÞ \" Hence, new hot-rolling load is: 0:5 Â 0:5 Â 0:5149 Â À Â 106Á # 117:74 ¼ Nm Âf0:1395 À ð2 Â 0:05432Þg P0 ¼ Âw r00 L QÀS ¼ 0:5 Â 117:74 106Á Ã Â Â 0:0717 Â 1:672 N ¼ 467:72 kN m: ¼ 7:06 Â 106 N: With the new load P0; the second new radius R00 of the (c) For elastically ﬂattened rolls, lever arm ratio according to (12.93a) is: elastically ﬂattened rolls according to (12.26b) is:

12.19 Solved Problems 593 k ¼ w R R0 r00 a À Stands Initial thickness Thickness of stock Diameter of P0 L 2 w of stock (mm) after rolling (mm) rolls (mm) A À Á B 2 1.5 250 0:5 Â 0:5 Â 0:5149 Â 117:74 Â 106 ¼ ð7:06 Â 106Þ Â! 0:0717 1.5 1.2 200 Â 0:1395 À 0:05432 ¼ 0:46: Assume that the coefﬁcient of friction at the roll-stock 2 interface for both the stands is 0.15 and no looping of the stock occurs between the two stands. If the rolls in the stand Alternative ways to evaluate k: A rotate at 175 rpm, at what rpm would the rolls in the stand B rotate? Neglect the change in temperature and the spread Lever arm ratio can be obtained from (12.93b) as follows: of the stock, if any. Use Ekelund’s expression wherever required. ða=2Þ À w ! R QS Solution k ¼ h1 À h2 Given that for both the stands, the coefﬁcient of friction at 0:5 ð0:1395=2Þ À 0:05432 ! the roll-stock interface is l ¼ 0:15: 0:025 À 0:015 1:672 ¼ Stand A: ¼ 0:46 Lever arm ratio for elastically ﬂattened rolls, according to The roll radius is RA ¼ 0:125m; and the peripheral velocity (12.92) is: of the rolls is: k ¼ MT ¼ 467:72 Â 103 ¼ 0:46: vrA ¼ 2 p RA Â 175 ¼ 2p Â 0:125 Â 175 m sÀ1 2 P0 L ð7:06 Â 106Þ Â 0:0717 60 60 2 Â ¼ 2:29 m sÀ1: Since the lever arm or moment arm is a ¼ k L; so for Initial thickness of stock is h1A ¼ 0:002 m; and thickness L ¼ 0:0717 m; and k ¼ 0:46; the lever arm or moment arm of stock after rolling is hA2 ¼ 0:0015 m: As it is cold rolling is: having small angle of contact, the following relations may be a ¼ k L ¼ 0:46 Â 0:0717 m ¼ 0:03298 m ¼ 32:98 mm: taken. From (12.8b), the angle of contact of the stock with the roll is: (d) Given that the rotating speed of rolls is N ¼ 100 rpm: rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Hence, for elastically ﬂattened rolls, the power required h1A À h2A 0:002 À 0:0015 to operate a pair of rolls according to (12.95) is: aA ¼ RA ¼ 0:125 rad ¼ 0:0632 rad: WR ¼ 2 p MTÀðN=60Þ 103Á The no-slip angle according to (12.19b) is: ¼ 2 p Â 467:72 Â Â ð100=60Þ W ¼ aA À ðaAÞ2 0:0632 ð0:0632Þ2 2 4l 2 4 Â 0:15 ¼ 4:9 Â 106 W: wA ¼ À rad ¼ 0:0249 rad: Considering the power loss due to roll-neck friction to be Hence, the forward slip from (12.22c) is: 20%, the total power input will be: WT ¼ 1:2 Â WR ¼ 1:2 Â 4:9 Â 106 W ¼ 5:88 Â 106 W: SFA ¼ RA ÀwAÁ2¼ 0:125 Â ð0:0249Þ2¼ 0:052: h2A 0:0015 Since the overall efﬁciency of the power unit is g ¼ 80%; so the overall power requirement of the hot-rolling mill is: WM ¼ WT =g ¼ À Â 106Á=0:8 ¼ 7:35 Â 106 W If vA2 is the exit velocity of the stock from stand A, then by 5:88 ¼ 7:35 MW: deﬁnitionÀ since Á ÀÁ SAF ¼ v2A=vAr v2A=vrA À 1 [see (12.20a)], so À 1 ¼ 0:052: 12.19.5. A and B are two consecutive stands in a two-high ) vA2 ¼ ð0:052 þ 1Þ Â vAr ¼ 1:052 Â 2:29 m sÀ1 tandem continuous cold-rolling mill through which a stock is ¼ 2:41 m sÀ1: rolled. The following data are provided:

594 12 Rolling Stand B: Solution The roll radius is RB ¼ 0:1 m: Initial thickness of stock is The roll gap set prior to the entry of the strip is the passive roll gap g; and the rolled gauge of the strip is h2: h1B ¼ 0:0015 m; and thickness of stock after rolling is hB2 ¼ 0:0012 m: As it is cold rolling having small angle of contact, (a) It is given that g ¼ 3:4 mm; and h2 ¼ 3:7 mm; at a rolling load of P ¼ 337:5 kN: the following relations may be taken. From (12.8b), the From (12.101), we get the mill ‘springback’: angle of contact is: sm ¼ h2 À g ¼ ð3:7 À 3:4Þ mm ¼ 0:3 mm: rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ h1B À hB2 0:0015 À 0:0012 aB ¼ RB ¼ 0:1 rad ¼ 0:0548 rad: The no-slip angle according to (12.19b) is: From (12.103), we get the mill modulus: wB ¼ aB À ðaBÞ2 ¼ 0:0548 À ð0:0548Þ2 rad ¼ 0:0224 rad: 1 ¼ sm ¼ 0:3 mm=kN ¼ 0:889 Â 10À3 mm=kN 2 4l 2 4 Â 0:15 m P 337:5 ¼ 0:889 lm kNÀ1: Hence, the forward slip from (12.22c) is: (b) Given that rolling load P ¼ 450 kN; which produces a rolled strip of 3.5 mm thickness, i.e. h2 ¼ 3:5 mm: Since SBF ¼ RB ÀwBÁ2¼ 0:1 Â ð0:0224Þ2¼ 0:042: the rolling mill remains the same, the value of mill h2B 0:0012 modulus will remain unchanged, because it is a charac- teristic of the mill. Hence, the roll gap setting prior to the If v1B is the entry velocity of the stock into stand B and vB2 entry of the strip, i.e. the passive roll gap from (12.102) is: is the exit velocity of the stock from stand B, then from Â À 10À3Á Ã constancy of volume ﬂow rate we can write: 3:5 0:889 450 vB1 h1B ¼ vB2 hB2 ; g ¼ h2 À 1 P ¼ À Â Â mm where v1B ¼ vA2 ¼ 2:41 m sÀ1: m ¼ 3:1 mm: ) vB2 ¼ v1B hB1 ¼ 2:41 Â 0:0015 m sÀ1 ¼ 3:01 m sÀ1: hB2 0:0012 Exercise ÀÁ ÀÁ Since SBF ¼ vB2 =vBr À 1; [see (12.20a)], so vB2 =vrB À 1 12.Ex.1. In the hot rolling of a slab, if the radial force exerted by the rolls on the slab at the entry plane is 4 kN per ¼ 0:042: mm width of the slab and the tangential friction force between the rolls and the slab at the entry plane is 1.6 kN per Hence, the peripheral velocity of rolls in stand B is: mm width of the slab, calculate the angle of bite for the unaided entry of the slab into the throat of the rolls. vrB ¼ vB2 ¼ 3:01 m sÀ1 ¼ 2:89 m sÀ1 0:042 þ 1 1:042 12.Ex.2. Determine the maximum draft and reduction possi- ble per pass for a 200-mm-thick work-piece, when the coef- Now the rpm at which the rolls in the stand B would ﬁcients of friction at the interface of rolls and work-piece are: rotate is: (i) (a) 0.1 during cold rolling on a two-high mill with rolls of vrB Â 60 ¼ 2:89 Â 60 ¼ 2:89 Â 60 rpm ¼ 276 rpm: 400 mm diameter. 2 p RB 2 p RB 2 p Â 0:1 (b) 0.5 during hot rolling on the same mill. 12.17.6. The gauge of a 4.0-mm-thick metal strip after a single pass in cold rolling shows a thickness of 3.7 mm. The (ii) If the roll diameter is changed from 400 mm to 1.2 m, roll gap set prior to the entry of the strip was 3.4 mm and the what will be the effect on the maximum draft and reduction rolling load required was found to be 337.5 kN. Assume that per pass for the same work-piece maintaining the same the width of the strip remains constant. frictional condition as in (a) and (b)? (a) Calculate the mill ‘springback’ and mill modulus. 12.Ex.3. If a 30-mm-thick stock is rolled to 20 mm thick- (b) If the rolling load required is 450 kN to produce a ness with rolls of 250 mm diameter, what will be the angle between the entry plane of the stock and the centreline of the rolled strip of 3.5 mm thickness from the same metal rolls and the projected length of deformation zone? strip with the same rolling mill, what will be the roll gap setting prior to the entry of the strip?

12.19 Solved Problems 595 12.Ex.4. If forward slip of a cold-rolling mill is 5%, ﬁnd % the roll–strip interface and the overall efﬁciency of the power maximum reduction possible by that mill without any spread unit is 80%. using Ekelund’s expression. Given that, roll diameter ¼ 250 mm; stock thickness before the roll pass ¼ 5 mm; stock 12.Ex.11. A 1-m-wide and 500-mm-thick slab is hot rolled thickness after the roll pass ¼ 3:5 mm: in one pass to thickness of 400 mm on a two-high mill with cast iron rolls of 2 m diameter. The average plane-strain ﬂow 12.Ex.5. The thickness of a plate is to be reduced from 60 to stress of the slab is 500 MPa. Assume full sticking frictional 30 mm by multi-pass rolling. The roll radius is 339 mm and condition at the roll-slab interface. Consider elastic ﬂattening the coefﬁcient of friction is 0.15. Assuming equal draft in of rolls into account till the roll radius varies by less than each pass, what would be the minimum number of passes 1%, and determine the following using the simpliﬁed required for the above reduction? hot-rolling formula analogous to forging formula: 12.Ex.6. If rolls of 700 mm diameter are operating at (a) Rolling load. 125 rpm, calculate the mean true strain rate under sticking (b) Lever arm or moment arm and hot-rolling torque for the friction for a 200-mm-thick slab as well as for a pair of work rolls, assuming typical value of the lever arm 20-mm-thick strip, while both have undergone 25% reduc- ratio. tion in thickness. What do you conclude from the results obtained? 12.Ex.12. A 60-mm-thick slab is given the maximum pos- sible reduction by rolling with rolls of 600 mm diameter 12.Ex.7. In a planetary hot-rolling mill, each back-up roll operating at 100 rpm. If the coefﬁcient of Coulomb’s friction with a diameter of D has 20 numbers of work rolls, each at the roll-slab interface is 0.3, calculate the velocity of the with a diameter of d and the curvilinear gap between any two rolled product at the exit plane under the condition of adjacent work rolls is 9.96% of d. Find the ratio of D/d. maximum draft. Use Ekelund’s expression wherever required. 12.Ex.8. When a strip of 250 mm width is rolled using steel rolls with diameter of 250 mm and length of 500 mm, the 12.Ex.13. A 200-mm-wide strip is cold rolled by a two-high roll-separating force is found to be 750 kN. The elastic and mill in thickness from 2 to 1.5 mm, without any spread. The shear moduli of steel rolls are, respectively, 207 and 78 GPa. rolls are 300 mm in diameter and operate at 100 rpm. If the Determine the deﬂection of the rolls at their centre taking rolling load is 1.1 MN, determine the mill power required for suitable values of the constants from the text. this cold reduction, assuming typical value of lever arm ratio for cold rolling. Neglect elastic ﬂattening of the rolls for this 12.Ex.9. What is the thinnest gauge to which a fully hard reduction, the power needed for the roll-neck bearings and steel sheet with an average ﬂow stress of 800 MPa can be energy losses in the motor and transmission. rolled with steel rolls of diameter 400 mm? Assume that the coefﬁcient of friction at the job-roll interface is 0.1. 12.Ex.14. A 500-mm-wide plate is homogeneously hot rolled in thickness from 200 to 150 mm, without any spread 12.Ex.10. A 30-mm-wide 70–30 brass strip is cold reduced under full sticking frictional condition. The diameter of each homogeneously in thickness from 2.5 to 1.75 mm in one roll is 1 m. Neglecting elastic ﬂattening of the rolls and pass on a two-high mill having steel rolls. The rolls are using Sims’ theory of hot rolling determine the following: 400 mm in diameter and travels at 100 rpm. The uniaxial ﬂow stress of the alloy is expressed as true stress ðMPaÞ ¼ (a) The no-slip angle. 895 ðtrue strainÞ0:49: The strip is subjected to a back tension (b) The lever arm ratio. of 100 MPa and a front tension of 150 MPa. The coefﬁcient of friction at the roll–strip interface is 0.05. Consider elastic 12.Ex.15. A strip is cold reduced in thickness from 1.5 to ﬂattening of rolls into account till the roll radius varies by 1.2 mm in one pass on a two-high mill with steel rolls of less than 1%, and determine the following using the sim- 100 mm diameter. The resistance of the strip to plane-strain pliﬁed cold-rolling formula analogous to forging formula: deformation at the entry plane is 400 MPa, which increases linearly to 500 MPa at the exit plane. The strip is subjected (a) Rolling load. to a back tension of 50 MPa and a front tension of 100 MPa. (b) Cold-rolling torque for each work roll, assuming typical If the coefﬁcient of friction at the roll–strip interface is 0.05, value of the lever arm ratio. determine the rolling load per unit width of the strip, using (c) Power required for this cold-rolling operation, if the Ekelund’s formula, considering elastic ﬂattening of rolls into diameter of roll neck is half of that of the roll. Assume that account, till the new roll radius does not change more than the coefﬁcient of friction at the roll neck is the same as that at 1% of the previous one. Assume homogeneous deformation

596 12 Rolling and neglect elastic compression and elastic recovery of the (C) respectively, decreases and increases; strip. (D) both decrease. 12.Ex.16. Indicate the correct or most appropriate answer(s) (g) To manufacture thin sheet or foil of high-strength alloy, from the following multiple choices: select which one of the following rolling mills is usually used: (a) To obtain the thinnest possible strip by cold rolling, which of the followings should be selected? (A) Two-high tandem continuous mill; (B) Z-mill; (C) Planetary mill; (D) Three-high mill. (A) Big-diameter rolls made of steel; (B) Small-diameter rolls made of tungsten carbide; (h) To achieve a round product like rod, by rolling, the (C) Big-diameter rolls made of tungsten carbide; ‘Leader’ pass must be (D) Small-diameter rolls made of steel. (A) a box pass; (B) a square pass; (C) a diamond pass; (b) The unaided entry of the stock into the roll throat for a (D) an oval pass; (E) a round pass. given initial thickness of the stock becomes difﬁcult if the roll diameter and the reduction in cross-sectional area (i) For a given diameter of rolls, the minimum thickness up imparted to the stock to which a particular metal can be reduced by cold rolling is nearly proportional to (A) both increase; (B) respectively, increases and decreases; (A) Er=l; (B) Er l; (C) 1=ðEr lÞ; (D) l=Er; (C) both decrease; (D) respectively, decreases and increases. where Er is the elastic modulus of rolls and l is the coefﬁ- cient of friction at metal–roll interface. (c) Increase of back tension in the plane of sheet during its cold rolling operation causes the no-slip angle and the for- (j) For rolling of blooms to billets and to help in descaling of ward slip to the stock, it is usual practice to commence the reduction with a few (A) increase; (B) increase and decrease, respectively; (A) box passes; (B) square passes; (C) decrease and increase, respectively; (C) diamond passes; (D) oval passes. (D) decrease. (k) The type of defects observed in rolling is: (d) For the occurrence of ‘bite’ in rolling, the following condition should be satisﬁed: (A) Alligatoring; (B) Flash cracking; (C) Fir-tree cracking; (D) Earing. (A) The coefﬁcient of friction should exceed the tangent of contact angle; (l) During ingot breakdown by hot rolling, the roll surfaces (B) The roll-separating force should reach a maximum usually have grooves cut in them parallel to the roll axis in value; order to (C) The friction coefﬁcient should be zero; (D) The contact length should be a minimum. (A) decrease the rolling load; (B) increase the angle of bite; (C) achieve a larger reduction; (D) decrease elastic ﬂattening (e) Select, which of the following respective combinations of of rolls. roll diameter and rolling speed will produce thinner gage sheet by cold-rolling operation: Answer to Exercise Problems (A) Large and high; (B) Large and low; 12.Ex.1. 21.8°. (C) Small and low (D) Small and high. 12.Ex.2. (i) (a) 2 mm, 1%; (b) 42 mm, 21%. (ii) (a) 6 mm, 3%; (b) 126.7 mm, 63.3%. (f) The unaided entry of the stock into the roll throat for a 12.Ex.3. 16.26°, 35 mm. given roll diameter becomes easier if the initial thickness of 12.Ex.4. 76%. the stock and reduction imparted to the stock 12.Ex.5. 4. 12.Ex.6. Slab:10.15 s−1; Strip: 31.56 s−1. For equal per- (A) both increase; centage of reduction with the same diameter rolls rotating at (B), respectively, increases and decreases;

12.19 Solved Problems 597 the same speed, the mean true strain rate of a thinner Hitchcock J. H. (1935) Elastic Deformation of Rolls during Cold work-piece is much greater than that of a thicker one. Rolling. Roll Neck Bearings, ASME Res. Publ., New York, p. 61 12.Ex.7. 6. 12.Ex.8. 1.191 mm. Larke E. C. (1963). The Rolling of Strip, Sheet, and Plate, 2nd edn., 12.Ex.9. 0.65 mm. Chapman and Hall, Ltd., London, pp. 267–273, 346 12.Ex.10. (a) 137.57 kN; (b) 803.5 N m; (c) 39 kW. 12.Ex.11. (a) 194.63 MN; (b) 164.5 mm; 64 MN m. Nadai A. (1931). Plasticity, McGraw-Hill 12.Ex.12. 3.28 m s−1. Oliver, B.R., Bowers, J.E.: A Simpliﬁed Method of Deriving 12.Ex.13. 89.78 kW. 12.Ex.14. (a) 6.36°; (b) 0.49. Cold-Rolling Schedules. J. Inst. Met. 93(7), 218–222 (1965) 12.Ex.15. 1.54 kN mm−1 or MN m−1. Orowan, E.: Proc. Inst. Mech. Eng. 150, 140 (1943) 12.Ex.16. (a) (B) Small-diameter rolls made of tungsten Orowan E., and Pascoe K. J. (1946) A Simple Method of Calculating carbide. (b) (D) respectively decreases and increases. (c) (D) decrease. (d) (A) The coefﬁcient of friction should exceed Roll Pressure and Power Consumption in Hot Flat Rolling. Iron and the tangent of contact angle. (e) (D) Small and high. (f) (D) Steel Inst., Special Report No. 34, p. 124 both decrease. (g) (B) Z-mill. (h) (D) an oval pass. (i) (D) Polakowski N. H. (1949–1950). J. Inst. Met., 76: 755–757 l=Er: (j) (A) box passes. (k) (A) Alligatoring. (l) (B) Roberts, W.L.: Blast Furn. Steel Plant 56, 382–394 (1968) increase the angle of bite; (C) achieve a larger reduction. Rowe G. W. (1977). Principles of Industrial Metalworking Processes, Edward Arnold (Publishers) Ltd., U.K., Indian Reprint by CBS References Publishers & Distributors, New Delhi, India, p. 211, 228 Saxi, K.: The Pendulum Mill, Rolling Mill Design Developed Afresh. Alder, J.F., Phillips, V.A.: Inst. Metals 83, 80 (1954) Engineering 195, 494–495 (1963) Aleksandrov, P.A.: In: Samarin, A.M. (ed.) Contemporary Problems of Sendzimir, M.G.: The Sedzimir Cold Strip Mill. J. Metals 8, 1154– 1158 (1956) Metallurgy, pp. 416–422. Consultants Bureau, New York (1960) Sendzimir M. G. (1963). 16th Ann. Tech. Meeting of the Indian Institute Al-Salehi, F.A.R., Firbank, T.C., Lancaster, P.R.: Int. J. Mech. Sci. 15, of Metals, Jamshedpur, Feb., 1963 Siebel, E., Lueg, W.: Measurement of Stress in Roll Gap. Stahl Eisen 693–710 (1973) 53(14), 346–352 (1933a) Avitzur, B.: An upper bound approach to cold strip rolling. Trans. Siebel, E., Lueg, W.: Untersuchungen über die Spannungsverteilung im Walzspalt. Mitt. K. W. Inst. Eisenforschung, Düsseldorf 15, 1–14 ASME, Ser. B. 86, 31–48 (1964) (1933b) Bland, D.R., Ford, H.: The calculation of roll force and torque in cold Sims, R.B.: Calculation of Roll Force and Torque in Hot Rolling Mills. strip rolling with tensions. Proc. Inst. Mech. Eng. London 159, 144– Proc. Inst. Mech. Eng. London 168(1), 191–200 (1954) 153 (1948) Capus J. M., Cockcroft M. G.: J. Inst. Met., 90, 289–297 (1961–62) Sims, R.B., Arthur, D.F.: J. Iron Steel Inst. London 172, 285–295 Cofﬁn, L.F.: J. Met. 15, 14–22 (1967) (1952) Ekelund S. (1933) The Analysis of Factors Inﬂuencing Rolling Pressure and Power Consumption in the Hot Rolling of Steel. Steel, Vol. 93, Snee, C.E.: Iron Steel Eng. 33, 124 (1956) Nos. 8–14. (Translated from Jernkontorets Ann., Feb., 1927, By B. Sparling, L.G.M.: Calculation of Rolling Load and Torque in a Hot Blomquist) Giles, J.L., Gutteridge, C.: J. Iron Steel Inst. London 211, 9–12 (1973) Planetary Rolling Mill. J. Mech. Engg. Sci. 4(3), 257–269 (1962) Harris, J.N.: Mechanical Working of Metals, p. 123. Pergamon Press, Sturgeon, G.M., Taylor, R.L.S.: Metal Strips from Powders. Mills and Oxford (1983) Hessenberg, W.C.F., Sims, R.B.: The Effect of Tension on Torque and Boon, London (1972) Roll Force in Cold Strip Rolling. J. Iron Steel Inst. London 168, Tong, K., Sachs, G.: Roll Separating Force and Minimum Thickness of 155–164 (1951) Cold-rolled Strips. J. Mech. Phys. Solids 6, 35–46 (1957) Underwood, L.R.: The Rolling of Metals, pp. 286–296. John Wiley & Sons Inc, New York (1950) Underwood, L.R.: Roll Neck Bearings. Iron and Steel Ind. Res, Council (1943) Von Kármán, T.: Contribution to the Theory of Rolling. Zeit. Fur Angew. Math. u. Mechanik 5, 139–141 (1925) Whitton, P.W., Ford, H.: Surface Friction and Lubrication in Cold Strip Rolling. Proc. Inst. Mech. Engrs. (London) 169, 123–140 (1955)

Extrusion 13 Chapter Objectives • Advantages and drawbacks of extrusion. • Two basic methods of extrusion: direct and indirect, and their comparison. • Extrusion equipments and dies. • Metal ﬂow during extrusion. • Factors inﬂuencing extrusion process, such as type of extrusion, extrusion ratio, extrusion (ram) speed, approach angle of conical die and optimum semicone angle, friction and lubrication. • Estimation of extrusion (ram) pressure and load for: open-die, indirect and hydro- static extrusions, direct extrusions through conical converging die and square die under conditions of Coulomb’s sliding friction, full sticking friction and mixed sticking–sliding friction, prevailing, respectively, from the entry plane of conical die or from that of dead zone in square die to the entry to extrusion chamber. • Strain rate in extrusion. • Extrusion defects. • Impact extrusion: direct (Hooker process) and indirect. • Hydrostatic extrusion and its basic difference with conventional extrusion. Con- ventional hydrostatic and differential pressure hydrostatic extrusion. Advantages and disadvantages of hydrostatic extrusion. • Seamless tube production by extrusion: using an external long ﬁxed, ﬂoating or piercing type of mandrel, or without using a normal external mandrel (welding- chamber method). Extrusion of cable sheathing. • Application of slip-line ﬁeld to steady-state motion: 50% and 2=3 plane-strain frictionless extrusion. • Upper-bound solution for plane-strain frictionless extrusion with an example. • Problems and solutions. 13.1 Introduction the harder and stronger metals requires high temperatures and pressures, which must be withstood by suitable die material. Extrusion is a relatively new method for fabrication of met- The problems of getting such suitable die material could not als. Originally, it was developed in the late 1700s to produce be overcome until well into the twentieth century. Nowadays, lead pipe and later to the lead sheathing of cable. Extrusion of successful extrusion of aluminium, copper, nickel, lead and their alloys and steels have become possible. © Springer Nature Singapore Pte Ltd. 2018 599 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_13

600 13 Extrusion Extrusion is the working process by which a block of • Complex shapes, which are not possible to produce by metal enclosed in a strong container is forced to ﬂow rolling, can be made by extrusion. through a die oriﬁce under high pressure in order to reduce its cross-section. The die may be round or various other • Shapes produced by extrusion may not require machin- shapes. The initial material in extrusion is rolled or cast billet ing operation. Thus, the extrusion process helps in of usually cylindrical shape. After placement of the billet material conservation. into a strong metal container, it is compressed by means of a ram so that it is ejected through the oriﬁce of a die. The • Changes from one size or shape of product to another can ejected or extruded metal acquires the contour and dimen- be accomplished with virtually no break in production. sions of the die oriﬁce. In general, cylindrical bars or hollow tubes are produced by means of extrusion, but the most • Compared to rolling, automation of extrusion is simpler easily extrudable metals like aluminium may also be extru- as items are produced in a single passing. ded into shapes of irregular cross-section. Since a container with a constant speciﬁc volume is involved in extrusion, • Dimensional accuracy of extruded articles is normally each billet is extruded individually and hence, extrusion is superior to that of rolled ones. essentially a batch process. Few drawbacks of extrusion are: Metals may be extruded hot or cold. Since the deforma- tion resistance of metal is low at elevated temperature, most • Process waste in extrusion amounts to 10–15% against metals are extruded hot in order to reduce the large force 1–3% in rolling because extrusion waste consists of an requirement in extrusion. However, cold extrusion is possi- extruded butt (discard) and an insigniﬁcantly deformed ble for many metals like lead, tin, aluminium and copper. leading end of the extruded product. These metals are extruded cold with a high speed of defor- mation, as in impact extrusion, resulting generation of con- • The structure and properties of extruded items exhibit siderable heat, to produce short lengths of hollow shapes relatively great inhomogeneities due to differences in ﬂow such as collapsible toothpaste tubes. Extrusion of tubing can of materials at the surface and central regions of the billet. also be made by attaching a mandrel to the end of a ram, which will be discussed in Sect. 13.11. • Extrusion tooling has a shorter service life because contact stresses and slip rates are high. 13.1.1 Comparison with Rolling • Tooling costs in extrusion are relatively high since costly There are some advantages of extrusion over rolling. These alloy steels are used to manufacture extrusion tools. are: • Overall costs of extrusion process are generally higher • In extrusion, the plasticity of metals is higher and the than those of other methods of working. cracking of the materials during primary breakdown from ingot tends to reduce due to the presence of favourable • As far as productivity is concerned, extrusion is much high compressive stresses produced by the reaction of the inferior to rolling because extrusion is a batch process, extrusion billet with the container and die. This advan- whereas rolling can be carried out as a continuous tage is an important reason for the increased utilization of operation. extrusion of poorly plastic metals and alloys, like stain- less steels, nickel-based alloys and other Because of the above factors, extrusion is chieﬂy applied high-temperature materials, which are difﬁcult or in the ﬁelds of: impossible to work by rolling or other methods of working. • Working of poorly plastic and non-ferrous metals and alloys; • Reduction ratio (ratio of cross-sectional areas of unworked to worked billet) is generally less than or equal • Production of sections and pipes of complex shapes; to 2 in per pass of rolling, whereas reduction ratio is • Batch production of medium and small sizes; generally 50–60 for extrusion. • Production of components with high dimensional • It requires less time to change a press for extrusion of a accuracy. new item than a similar operation on a rolling mill. 13.2 Two Basic Methods of Extrusion There are basically two methods of extrusion process: • Direct extrusion (also called forward extrusion) and • Indirect extrusion (also called reverse, inverted or backward extrusion).

13.2 Two Basic Methods of Extrusion 601 Direct Extrusion: (a) to minimize friction between the ram or dummy block and the container, and The process of direct extrusion is illustrated in Fig. 13.1a. The heated billet is loaded into a chamber, usually made of a (b) also to allow scraping of the billet to form a thin skull heavy steel shell, lined with heat-resisting alloy steel. A hy- containing the oxidized surface layers to be left on the draulic ram or stem is used to apply a pushing force to the wall of the container after the billet has been extruded. billet so that the billet is upset and extruded through a die made of heat-resisting tool steel. Note that the billet slides Prior to the completion of extrusion of the billet, when a relative to the container wall (Laue and Stenger 1981). The small portion amounting to about 5–15% of the billet is left ram travels in the same direction as the extruded section, in the container and there is a very rapid rise of the load which assumes the shape of the die oriﬁce. Generally, the required for extrusion, extrusion is halted. This remaining diameter of the ram ﬁtted into the container is smaller than billet is called the discard or butt, which is left in the con- the inside bore of the container. A dummy block or a tainer as an extrusion waste and reduces the yield. This pressure plate or pad in preheated condition is placed waste is purposely made in order to avoid the development between the billet and the ram so that the back end of the hot of extrusion defects, caused by extrusion of the billet skin billet is prevented from chilling in contact with the ram and and the rear end of the billet into the extruded product, and on the other hand, the ram, which is highly loaded in to avoid the rise of extrusion load. The extruded product compression, is also protected from the hot billet. The (also called extrudate) is separated from the butt by cutting dummy block may also be afﬁxed to the ram stem and is off with a hot saw. Alternatively, complete extrusion of the called ﬁxed dummy block. Often the dummy block used is billet without a butt left in the chamber is possible to carry slightly smaller in diameter than the inside diameter of the out when a plain graphite dummy block is used, and asbestos extrusion chamber. The purpose of the clearance between pads are placed between the surfaces of the billet and the container wall and the ram or the dummy block is dummy block, and the spacing between the chamber and the twofold: billet is ﬁlled with graphite. This will eliminate the formation of rear-end piping defect and the sharp rise of pressure at the (a) Container Ram end of stroke. Finally, there must be a runout table to receive Billet the extrudate and a straightener like roller levelling or Die Dummy stretcher levelling to correct minor warpage in the extrudate. block The leading end of an extrudate is always but negligibly Extrudate deformed; when cast ingots are extruded, their cast structure Die can be retained in the central zone of the forward end of the holder extrudate even for an extrusion ratio (ratio of cross-sectional areas of unextruded to extruded billet) of 15. Therefore, it is (b) necessary to crop the leading end of the extrudate. Container The container is drawn back and separated from the die, while the butt remains still adhered to the die face and may Ram Closure be attached to the dummy block if the dummy block is not Ram plate afﬁxed to the ram stem. The butt is separated by shearing from the die face and discarded. The container, die and ram Extrudate Billet are brought back to their initial loading positions. After- wards, extruded sections are heat treated usually in electric Fig. 13.1 Extrusion of solid by a direct extrusion and b indirect furnaces, and their surface defects are eliminated by cutting extrusion out, scraping, pickling and other techniques. Then extruded items are varnished, oxide coated, anodized, coated with grease, etc., to protect them against corrosion. Indirect Extrusion: The indirect extrusion process is illustrated in Fig. 13.1b. The extrusion chamber is similar to that in the direct extrusion process, and the heated billet is loaded into the chamber as in the direct extrusion. In indirect extrusion, the die is inserted into the press at the end of an hollow ram stem; i.e., the die and the ram remain on one side (exit side) of the billet instead of on the opposite sides of the billet as in

602 13 Extrusion the direct extrusion, while a closure plate closes the chamber 13.3 Extrusion Equipments on the other side (entry side) of the billet. To form the extrudate in indirect extrusion, either the die is moved into The force required to carry out extrusion may be supplied the stationary container and pushed against the stationary via a mechanical or a hydraulic (water or oil) drive. Hence, billet by movement of the ram (Rowe 1977; Harris 1983) or extrusion presses can be classiﬁed into mechanical press and the container with the billet is caused to move jointly while hydraulic press depending upon the types of drive. The ram the hollow ram carrying the die is kept stationary (Dieter in a mechanical press may be powered by a motor. 1988). Since there is no relative motion at the container– Mechanical press having higher speed than hydraulic press billet interface (Laue and Stenger 1981), except at the die, is used in some applications, such as in the production of there is no frictional stress at the billet–container interface. small tubes, usually by impact extrusion (direct or indirect). Hence due to lower frictional forces, the extrusion load and Most extrusions are performed with hydraulic presses. The power required for indirect extrusion are less than for direct two basic types of hydraulic drives are direct and accumu- extrusion. The absence of container friction in indirect lator designs. The most widely used press in the past were extrusion also reduces the increase in temperature caused by accumulator-drive hydraulic presses, but nowadays, the friction. As in the direct extrusion process, a butt is left in direct-drive hydraulic presses are most extensively used. the chamber after the billet is extruded by indirect extrusion; the leading end of the extrudate is cropped, and the die and Higher ram speed is required for extrusion at higher the butt are separated from the extrudate. temperature in order to avoid the problem of heat transfer from the hot billet to the tools. The accumulator-drive However, there are practical limitations to indirect hydraulic press is desirable where higher ram speeds are extrusion because the requirement for using a hollow ram required for hot extrusion. This hydraulic circuit of an (which is really a hollow die holder) weakens the whole accumulator press consists of one or several air-over-water press and limits the loads which can be applied to the ram. accumulators charged by high-pressure water pumps. The This in turn limits the size or cross-section of the section accumulator vessel (series of vessels) supplies the quantity which can be produced and thus, limits the maximum of water necessary to provide the extrusion pressure reduction possible by this process. requirement during the entire extrusion cycle, with a pres- sure drop limited to roughly 10%. Apart from the pressure 13.2.1 Direct Versus Indirect Extrusion drop, the high cost of high-pressure water pumps, accumu- lators and valves as well as large ﬂoor space requirements The direct extrusion process is compared with the indirect have restricted the wide uses of the accumulator-drive one on the basis of the process, and the advantages and hydraulic press and resulted in the extensive applications of disadvantages of one over another, and presented in a tabular direct-drive hydraulic press nowadays. However, for extru- form in Table 13.1. sion of steels and refractory metals that may require a high ram speed in the range of 0.4–0.6 m s−1, the accumulator A number of advantages of the indirect extrusion over the water drives are desirable because of the advantages that direct one are given in the columns numbered from 4 to 7 in these units provide higher ram speeds and water is a Table 13.1, while the disadvantages of indirect extrusion are non-ﬂammable liquid, considered to be important in the included in the columns numbered 8 and 9. extrusion of hot billet. Table 13.1 Comparison of direct and indirect extrusion Indirect extrusion Direct extrusion 1. The ram and the die are at the exit side of the extrusion; i.e., they 1. The ram is at the entry side of the extrusion, and the die is at the are on the same side of the billet, and a closure plate is placed on the exit side of the extrusion; i.e., they are on the opposite sides of the opposite side of the billet; i.e., the entry side of the extrusion billet. There is no closure plate 2. The ram is a solid cylinder which may be attached with a preheated 2. The ram is a hollow cylinder which holds the die. There is no dummy block or pressure plate at the entry side of the extrusion dummy block or pressure plate 3. The ram and the billet move but the container is stationary 3. Either the ram carrying the die moves, while the container with the 4. Due to the movement of the billet, frictional resistance at the billet remains stationary, or the ram containing the die is stationary contact surface between the billet and the container is high. So, the and the container with the billet moves breakthrough pressure, i.e. the pressure at which the metal begins to ﬂow through the die, is higher leading to requirement of higher 4. Due to the absence of relative motion between the wall of the container and the billet, frictional resistance is low. So, the breakthrough pressure is lower leading to requirement of lower power. The maximum load is approximately 20–30% lower than that (continued)

13.3 Extrusion Equipments 603 Table 13.1 (continued) Indirect extrusion Direct extrusion in direct extrusion. During subsequent extrusion of the billet, the pressure required to deform the metal through the die remains power. As the billet extrudes through the die, the pressure required to approximately constant with ram travel relative to the container. maintain the metal ﬂow progressively decreases with ram travel Typical curve of extrusion pressure versus ram travel for indirect because the length of the billet in the container decreases causing a extrusion is shown in Fig. 13.7a decrease of the contact area and thereby of the frictional resistance between the billet and the container wall. Typical curve of extrusion 5. Extrusion pressure is not a function of billet length because of the pressure versus ram travel for direct extrusion is shown in Fig. 13.7a absence of relative motion at the container–billet interface. So, billet length is not limited by the load, but only by the length, strength, and 5. Extrusion pressure depends on the billet length because of the stability of the hollow ram needed for a container of given length presence of relative motion at the container–billet interface. Therefore, the billet length is limited by the load 6. Due to absence of frictional stress at the container–billet interface, the increase in temperature caused by the friction cannot occur at the 6. Frictional stress present at the container–billet interface produces billet surface. Therefore, the tendency to form surface defects is lower heat and consequently, surface temperature of the billet increases. and signiﬁcantly higher extrusion speeds can be applied Therefore, there is a higher tendency to form surface defects and extrusion speeds cannot be increased to a very high level, since higher 7. Due to reduction in friction and temperatures, the service life of speeds cause further increase in temperature extrusion tooling, especially the inner liner of the container, may be enhanced 7. Due to high friction and temperatures, the service life of extrusion tooling will be relatively shorter 8. The presence of a hollow ram, which is really a hollow die holder, weakens the press and during the extrusion of the billet, it is not 8. Higher load can be applied to the ram during the extrusion of the possible to apply a high load to the ram, which in turn limits the size billet, since load is applied through a solid ram. So, a section of or cross-sectional of the section that can be produced and thus, limits higher size or cross-sectional area can be produced the maximum possible reduction 9. The impurities or defects on the billet surface usually retain as a 9. Since the impurities or defects on the billet surface may not be discard or shell in the chamber. Hence, the quality of the extrudate is retained as a discard or shell in the chamber, they affect the quality of not affected and it is not needed to use scalped or machined billets the extrudate. Hence, scalped or machined billets are used in many instances Modern direct-drive hydraulic presses use oil as hydraulic 13.3.1 Extrusion Dies medium and high-pressure variable-delivery oil pumps, some of which operate at pressures over 34.5 MPa. Apart The tooling assembly for hot extrusion consists of several from the less ﬂoor space requirements and the less cost of individual components, such as containers with liners, ram these presses, their most important advantage is that there is (stem), dummy block, mandrel, if required, and die stack. no drop in the maximum available force during the entire More than one layer of wear-resistant liner are ﬁtted inside extrusion cycle. A limitation of these presses is that the ram the extrusion container to withstand the high radial stresses. speeds are slower than those in accumulator presses. In Since the liner and dummy block are subjected to many direct-drive presses, ram speeds to 50 mm/s are typical. cycles of thermal shocks, their periodic replacement will be These speeds are adequate for alloys prone to hot shortness, needed. such as copper and aluminium alloys where the ram speed must be restricted to a few mm/s in order to avoid rise in The die stack consists of die supported by a die holder, a temperature leading to hot shortness. However, ram speeds backer and a bolster, all of which are held in a die head (die to 200 mm/s can be attained by employing oil accumulators holder carrier), as shown in Fig. 13.2. The entire assembly with oil-hydraulic drives. of die stack is sealed against the container by the pressure applied by a wedge. The dies and tooling selected for hot Again, depending upon the direction of travel of ram, extrusion must be capable to withstand high stresses and there are two types of hydraulic press. These are horizontal resist oxidation and thermal shock. The dies are usually hydraulic press and vertical hydraulic press. The advantages made from high-alloyed tool steel, such as high-speed tool and disadvantages along with their applications are com- steel. Apart from improvements in steel quality and heat pared below in Table 13.2. treating practices, nitriding and hard thin-ﬁlm coatings are applied to improve die life. Nowadays, most dies are nitrided In addition to the extrusion press, (a) billet heating in which atomic nitrogen is adsorbed into the surface of steel facilities, (b) automatic transfer equipment for placing the to produce a hard and wear-resistant nitrided case. For heated billet in the container and (c) provision for heating the production of large quantities product of smaller proﬁles, extrusion container are also needed.

604 13 Extrusion Table 13.2 Comparison of horizontal and vertical hydraulic press Vertical hydraulic press 1. Hydraulic press where the ram travels vertically is known as Horizontal hydraulic press vertical hydraulic press 2. Presses are generally built with less loading capacity of 3–20 MN 1. Hydraulic press where the ram travels horizontally is known as horizontal hydraulic press 3. Presses need considerable headroom and to make extrusions of appreciable length, a ﬂoor pit is frequently necessary 2. Presses with high loading capacity of 15–50 MN are in regular 4. Alignment between the press ram and the tools is easier operation while a few presses of 160 MN loading capacity have been 5. The rate of production is higher constructed 6. For the operation of the press, less ﬂoor space is required 7. Presses will produce uniform cooling of the hot billet in the 3. For the operation of the press and to make extrusion of appreciable container and thus, symmetrically uniform deformation will result length, considerable headroom is not needed 8. Application: In commercial operations, the chief use for the press is 4. Alignment between the press ram and the tools is difﬁcult in the production of thin wall tubes where uniform wall thickness of tubes and concentricity are required. Since the loading capacity of 5. The rate of production is low press is less, so it is generally not used for the extrusion of bars and shapes 6. For the operation of the press, more ﬂoor space is required 7. The bottom surface of the hot billet which lies in contact with the container wall will cool more rapidly than the top surface, unless the extrusion container is internally heated, and therefore, the deformation will be non-uniform. As a consequence, warping of bars will result and non-uniform wall thickness will occur in tubes 8. Application: Since the loading capacity of press is high, so it is used for most commercial extrusion of bars and shapes since production of bars and shapes require higher loads than tubes. The press is generally not used for the extrusion of tubes, as non-uniform wall thickness will occur in tubes due to non-uniform cooling of the billet in the container Fig. 13.2 Schematic view of Die holder carrier Taper seal various extrusion process tooling Bolster Dummy block Extrusion Billet Extrusion stem Liner Intermediate liner Container Mantle Backer Die Die holder chemical vapour deposition (CVD) or physical vapour whose cross-sections are similar to those of desired extruded deposition (PVD) technique can be applied to coat dies with products. They are characterized by a bearing surface that is hard thin-ﬁlm multilayered coatings. Such coatings provide perpendicular to the face of the die, i.e. the semidie angle superior wear resistance, and it is claimed that a single a ¼ 90: When this die is used, the metal that enters the die CVD-coated die (Maier 2004) can be used for extrusion of forms a dead-metal zone and shears internally to form a more than 1000 aluminium billets. die-like ﬂow channel having an angle decided by the dead-metal zone. The billet ﬂows in a converging manner For solid proﬁle hot extrusions, the two most common through this die channel. There is a back taper on the exit types of dies used are ﬂat-faced die (also called square die or side of the die, which reﬁnishes the surface of the product shear die) and conical converging die (also termed tapered without increasing the diameter at the exit plane. These dies die or streamlined die or shaped die), as shown in Fig. 13.3. are easier and cheaper to design and manufacture than Flat-faced die consists of one or more apertures (openings),

13.3 Extrusion Equipments 605 Fig. 13.3 Schematic view of the (a) Back (b) Back aperture details of a a tapered die taper taper and b a ﬂat-faced die for hot Original Original extrusion bearing bearing Choke Present bearing Back taper Degree of Relief as specified choke needed Extrusion Undercut Extrusion Undercut direction direction Back taper as specified conical converging dies and are commonly used for the hot identical oriﬁces are usually cut in order to equalize metal extrusion of aluminium alloys. ﬂow. For production of hollow tubes, a bridge die (also known as a porthole, spider-mandrel or torpedo die) can be Conical converging die usually has a conical aperture at used, which will be discussed subsequently in Sect. 13.11. the entrance plane with a circular cross-section that changes progressively to the desired ﬁnal size of the extruded prod- 13.4 Metal Flow During Extrusion uct. The approach angle to the bearing surface of the die is referred to as choke. These dies are used in extrusion with Since the ﬂow pattern is axially symmetrical in extrusion, no good lubrication. As the semi-angle a of conical die shear stresses can exist on longitudinal axial planes. In other decreases, the homogeneity of deformation increases and the words, if a billet is bisected longitudinally and the two extrusion pressure is decreased, but below a critical value of halves are combined together and extruded through an axi- a; the friction in die surfaces becomes so high that the ally located die (Fig. 13.5a), the fact is that the ﬂow of the extrusion pressure begins to increase. This critical value of a cut billet will be same as that of an uncut billet, but if the die giving the minimum stress required for extrusion is called is placed asymmetrically (Fig. 13.5b), the ﬂow will the optimum semicone die angle, which varies between 45 and 60° for most extrusion operations. Optimum angle of die Fig. 13.4 Extrusion die consisting of multiple oriﬁces will be subsequently discussed in Sect. 13.5. However, it is important to note that small cone angles of die are associated with high relative speed between the billet and the die, which leads to faster wear of the die than is experienced with large cone angles. These conical dies being costly are generally used for the hot extrusion of steels, high-strength aluminium alloys, titanium alloys and other metals. It has been seen that ﬂat-faced die can be used for steel and titanium with glass lubrication, with the glass pad forming the die contour at the entrance. The conical-contoured dies are mostly applied for steel and titanium with lubrication consisting of graphite dispersed in grease, although there is evidence that conical-entry dies are also used with glass lubrication, at least in extrusion of other high-strength alloys. When the extrusion load is excessively high for extrusion of a single small area, multidie extrusion, in which a number of oriﬁces are cut as in Fig. 13.4, can be used. A number of sections of identical shapes, or of different shapes, can be extruded at the same time through these oriﬁces. Since the tendency of metal to ﬂow through a larger oriﬁce is faster than through a smaller one within the same multidie,

606 13 Extrusion (a) (b) Cut Ram Cut Extrudate Ram Extrudate Billet Billet Fig. 13.5 a Symmetrical die, in which cut billet ﬂows normally, b asymmetrical die, in which cut billet does not ﬂow normally (a) (b) (c) (d) Fig. 13.6 Four types of metal ﬂow patterns in extrusion with square dies, a ﬂow pattern S, b ﬂow pattern A, c ﬂow pattern B, d ﬂow pattern C (Lewandowski and Lowhaphandu 1998) obviously be affected by the cut. While investigating ﬂow than that of the centre, and the corresponding increase in the patterns in extrusion, advantage can be taken of the above ﬂow stress of metal at the colder surface can signiﬁcantly fact. Numerous ﬂow patterns in extrusion have been detected affect the ﬂow pattern. The different types of ﬂow patterns by many investigators, but the greatest contribution to metal observed in the extrusion of metals with square dies, where ﬂow in extrusion was made by Pearson (1953). Pearson used the semidie angle, a; is 90°, have been classiﬁed according the grid method in which a cylindrical billet was cut into to the increasing order of non-uniformity of ﬂow into four halves along the axis and a regular grid pattern was engraved types: ‘S’, ‘A’, ‘B’ and ‘C’, which are shown in Fig. 13.6 on the ﬂat interfaces and bound together with wire prior to (Lewandowski and Lowhaphandu 1998). extrusion. After the extrusion, the two parts were separated easily along the axial plane and from the extent of distortion (a) Flow Pattern S: of the grid, the ﬂow patterns were quantitatively assessed. The results have shown that the ﬂow of the billet in the Flow pattern ‘S’ is characterized by the maximum possible extrusion chamber varies in a characteristic way depending uniformity of ﬂow of a homogeneous material in the on the process used and the material being extruded. chamber due to minimal friction. Localized plastic defor- mation occurs mainly in a zone just prior to the entrance of The main cause of these variations is due to the differ- die. The major part of the non-extruded billet, pushed as a ences in the magnitude of frictional resistance between the rigid body through the die, remains undeformed in the billet and the wall of the extrusion chamber. In certain cases container, as can be visualized from the undistorted grid in hot extrusion, heat losses in the container cause the pattern (Fig. 13.6a). Hence, the front of the billet moves temperature of the billet surface to be signiﬁcantly lower

13.4 Metal Flow During Extrusion 607 evenly into the deformation zone. This very uniform ﬂow of Non-lubricated extrusion hardly shows this type of ﬂow metal can take place only when there is no friction between patterns; instead, these ﬂow patterns are observed during the container wall and the billet or at the surface of the die lubricated extrusion of soft metals and alloys, such as lead, and its holder. Frictionless extrusion is impossible in prac- tin, alpha-brasses and tin bronzes and during extrusion of tice, but ﬂow patterns of this type can be closely approxi- copper billets covered with oxide, which acts as a lubricant. mated to the conditions, where very effective lubrications are used, for example, hydrostatic extrusion in which the billet is (c) Flow Pattern B: surrounded with a pressurized ﬂuid, or indirect extrusion with a well-lubricated die, as well as direct extrusion of steel Flow pattern ‘B’ occurs in a homogeneous material if there with glass lubricant. Even in the absence of friction, the is signiﬁcant interface friction at both the container wall and strain rate and strain potential vary throughout the defor- at the surface of the die and its holder. The friction at the mation zone because of change in the direction of metal as it billet/container interface retards the ﬂow of material in the ﬂows towards the die oriﬁce. The ﬂow of metal cannot occur peripheral zones while the material at the central region of around the sharp corner between the container and the die the billet experiencing less resistance ﬂows at a rapid rate, because of material constraints, and consequently a shortest resulting in a velocity gradient along the cross-section of the path is required for metal ﬂow. Thus, the material in the billet. The velocity gradient results in the veeing of the peripheral zones shears internally and ﬂows along a funnel, horizontal lines of the grid pattern at an early stage before thereby resulting in a dead zone of stagnant metal ahead of the material reaches the die vicinity, as can be seen from the die entrance. The dead-metal zone, which undergoes Fig. 13.6c. Compared to ﬂow pattern ‘A’, the vee’s are far little or no deformation, is very small in ﬂow pattern ‘S’. more severe in this type of ﬂow and a deeper ‘rear-end hollow pipe’ is formed in the extrusion earlier in the cycle. (b) Flow Pattern A: However, the shear zone between the retarded region at Flow pattern ‘A’ occurs in a homogeneous material when the periphery and the rapidly ﬂowing material in the centre there is virtually no friction between the container wall and extends back axially into the billet to an extent that depends the billet, but signiﬁcant friction exists at the surface of the on the alloy and the extrusion parameters. This results in a die and die holder. Due to absence of friction in the con- large dead-metal zone, which starts to form on the die tainer, the metal slides up towards the die region with no shoulders. When sticking friction develops between the deformation, same as that in ﬂow pattern ‘S’. This can be billet and container wall, the dead-metal zone builds up very seen from the grid pattern in Fig. 13.6b, where the horizontal quickly and may extend from the die shoulders back to the line remains so until they come close to the die entrance. ram. At the start of extrusion, shear deformation is concen- This indicates that the grid elements at the billet centre trated in the peripheral regions, but it extends towards the undergo essentially pure elongation into the extruded prod- centre with the progress of deformation. This increases the uct. Friction at the die land retards the radial ﬂow of the danger of material ﬂowing from the billet surface with metals in peripheral zones and increases the amount of lubricant, oxides or impurities along the shear zone and shearing in this region. This results in a slightly larger dead extruding as defects into the interior of the extruded product zone of stagnant metal than that in ﬂow pattern ‘S’. How- (blistering and scale formation). Further, the dead-metal ever, deformation in the centre is still relatively uniform. It is zone can even inﬂuence the ﬂow of metals to a limited to be noted that the shear distortion requires expenditure of degree since it is not completely rigid. Flow pattern ‘B’ is energy because of additional plastic shear deformation that observed in single phase homogeneous copper alloys that do does not result in any change in the external dimension or not form a lubricating oxide ﬁlm at surface and in most shape of the product extruded from the billet, and this is aluminium alloys (Laue and Stenger 1981). termed redundant deformation. (d) Flow Pattern C: In the vicinity of the die, the outside portion of the billet is held back by the sharp shoulders of the chamber, while the Flow pattern ‘C’, illustrated in Fig. 13.6d, occurs in the hot metal in the core region ﬂows easily and rapidly into the die. extrusion of inhomogeneous materials when the interface This results in the veeing of the horizontal lines and the friction is high retarding the ﬂow of surface material, as in formation of ‘rear-end hollow pipe’, an extrusion defect. As ﬂow pattern ‘B’, and/or when the billet surface is chilled by the rear end of the billet advances towards the die, the degree the walls of a cold chamber and the ﬂow stress of the cold of veeing increases. Veeing is also produced in ﬂow pattern surface material is considerably higher than that of the hot ‘S’ but to a lesser extent. material in the core. Due to this, the stronger material in the

608 13 Extrusion surface of the billet forms a relatively stiff shell (Laue and Extrusion pressure = extrusion load : Stenger 1981) and the material in the core regions of the billet ﬂows towards the die more easily than that in the cross-sectional area of billet peripheral regions. This leads to formation of a conical dead-metal zone, which is much larger than the other ﬂow Typical curves of extrusion pressure versus ram dis- patterns and extends from the front of the billet to the back. placement for direct and indirect extrusion at a relatively At the start of the extrusion, only the material inside the constant billet temperature and a constant ram speed after funnel is plastic and it is severely plastically deformed, initial acceleration are shown in Fig. 13.7a. The rapid rise in especially in the shear zone, as it ﬂows towards the die. As pressure during the initial ram travel, i.e. from O to A, is due the billet is extruded, the stiff shell and the dead-metal zone to (i) initial compression of the billet to ﬁll the extrusion are axially compressed and consequently, the displaced container, (ii) elastic deformation of the equipment and material of the outer zones follows the path of least resis- (iii) initial acceleration of the ram. The metal begins to ﬂow tance to the back of the billet, where it turns towards the through the die, at point A for indirect extrusion and at point centre and ﬂows into the funnel (Laue and Stenger 1981). B for direct extrusion, as shown in Fig. 13.7a. However, This type of ﬂow pattern causes to form an internal extrusion extrusion can commence before the pressure reaches the defect which appears as an annular ring of oxide in the maximum value where the increase in extrusion speed to the cross-section at the rear end of the extruded product due to set speed causes further increase in pressure. For direct ﬂow of the oxidized surface into the extrusion. extrusion, the rise of pressure from A to B is due to the frictional resistance at the contact surface between the billet This type of ﬂow is typically observed in the hot extru- and the container. The maximum value of pressure or peak sion of ða þ bÞ brasses. This is a result of cooling of billet pressure at point B, where metal ﬂow starts, is known as the surface, which leads to an increase in the ﬂow stress in the breakthrough pressure for direct extrusion. As the billet surface regions. The increase is because the ﬂow stress of the extrudes through the die, the pressure required to maintain a phase is much higher than that of the b phase during the ﬂow of metal progressively decreases at a constant rate hot-working. However, ﬂow pattern ‘C’ will occur when a for direct extrusion, because the length of the billet in the hard billet shell exists and at the same time, the friction at the container decreases causing a decrease of the contact area container wall is high. It can also occur without any phase that and thereby of the frictional resistance between the billet and leads to a higher ﬂow stress, if the temperature difference the container wall. As there is no relative motion between between the container and the billet is large resulting in a sig- the billet and the container wall for indirect extrusion, the niﬁcant difference in ﬂow stress between the peripheral and extrusion pressure, required to deform the metal through the central regions of the billet. This can take place in the extrusion die, remains approximately constant with increasing ram of tin and aluminium alloys (Laue and Stenger 1981). travel. When approximately 85% of the billet has been extruded, the pressure builds up rapidly from point C in 13.5 Factors Influencing Extrusion Fig. 13.7a, until it reaches the capacity of the press. So at the point C, the extrusion, i.e. the ram travel, must be halted The principal variables that inﬂuence the extrusion process causing 5–15% of the billet to be left as discard or butt in the are as follows: container, which is removed after cutting off the extrudate with a hot saw. The discard, if extruded, will produce a 1. The type of extrusion, i.e. direct versus indirect extrusion. back-end piping defect; i.e. the end of extrudate will be 2. The extrusion ratio, R; or the fractional reduction in area, hollow. Further, this discard often includes defects, which are obviously unwanted in the extrudate. r: 3. The extrusion temperature, T. The reason for the rapid rise of pressure from point C is 4. The extrusion speed, i.e. the speed of ram, vram: due to the direction of metal ﬂow becoming more and more 5. The semi-approach angle, a; of conical die. perpendicular to the line of action of the applied force with 6. The friction and lubrication at the walls of the die and the increasing ram travel beyond point C, as shown in Fig. 13.8b, whereas during the early stage of extrusion, i.e. container. from point A or B to C, the applied force must cause the metal to ﬂow to the die along a diagonal path, as shown in (1) Type of Extrusion (Direct vs. Indirect): Fig. 13.8a. By placing a load cell on the extrusion ram, it is possible to The deformation work for direct extrusion is divided into follow the change of extrusion load during the extrusion four zones—I, II, III and IV—as shown in Fig. 13.7b. Zone I cycle from which extrusion pressure is obtained as follows: involves the work of upsetting of billet, elastic deformation of the equipment and accelerating to the set extrusion speed. Zone II is the work associated with friction between the

13.5 Factors Influencing Extrusion 609 (a) (b) Direct extrusion curve II B Friction work D III Deformation work AC (ideal + redundant work) Indirect extrusion Extrusion pressure Extrusion pressure I IV O Ram displacement Ram displacement Fig. 13.7 a Typical curves of extrusion pressure versus ram displacement for direct and indirect extrusion at a relatively constant billet temperature and a constant ram speed after initial acceleration, b division of deformation work for direct extrusion into four zones—I, II, III and IV (a) Flow (b) Flow For large deformations, a relatively minor change in the Flow Flow fractional reduction in area corresponds to a large change in Force Force the extrusion ratio and hence, R is a more descriptive parameter than r. This distinction between R and r can be Fig. 13.8 Metal ﬂow during extrusion, a ﬂow along a diagonal path easily appreciated from the following example. When the during early stage of extrusion, b ﬂow normal to force direction at later change in the extrusion ratio is from R ¼ 20:1 to R ¼ 100:1; or ﬁnal stage of extrusion which is quite high, then the change in the fractional reduction in area appears to be very small, from r ¼ container and the billet. This work is dissipated as heat, 1 À ð1=20Þ ¼ 0:95 to r ¼ 1 À ð1=100Þ ¼ 0:99: Extrusion resulting in an increase in temperatures of billet, extrudate ratios may be about 40:1 for hot extrusion of steel and reach and tooling. Zone III is the work involved in carrying out the as high as 400:1 for aluminium. ideal and redundant deformation of the metal through the die. Zone IV is associated with the change in the direction of Due to the incompressibility of metal in plastic defor- metal ﬂow occurring at the end of extrusion. mation, constancy of volume ﬂow rate provides the fol- lowing relationship between the velocity of the ram and that (2) Extrusion Ratio (r) or Fractional Reduction in Area (r): of the extrudate at the exit plane of the die, vram and ve; respectively: The extrusion ratio R; and the fractional reduction in area r; are mathematically deﬁned by: vram A1 ¼ ve A2; ð13:3Þ or, ve ¼ vram Â R Equation (13.3) shows that for a constant ram speed vram; the higher the extrusion ratio R; the greater is the exit velocity ve of the extrudate. The relation between the true strain, e; in extrusion and the extrusion ratio, R; is given by Extrusion ratio, R ¼ A1 ð13:1Þ ZA2 ð13:4Þ A2 e ¼ À dA ¼ ln A1 ¼ ln R Reduction in area, r ¼ A1 À A2 ¼ 1 À 1 A A2 A1 R A1 ð13:2Þ where From (13.4), it is clear that there is a direct relation between the extrusion pressure and the natural logarithm of A1 the initial cross-sectional area of the billet prior to the extrusion ratio R: Hence, the extrusion pressure, p; and extrusion; the extrusion load, P; may be expressed in a simple form as: A2 the ﬁnal cross-sectional area of the ﬁnished item after p ¼ ke ln R ¼ ke ln A1 ð13:5Þ extrusion. A2

610 13 Extrusion P ¼ keA1 ln R ¼ keA1 ln A1 ð13:6Þ Although hot-working decreases ﬂow stress, which is an A2 advantage, it introduces following problems related to high temperature: where ke ¼ the ‘extrusion constant’, an overall factor which accounts for (a) the ﬂow stress of metal, (b) the friction at the (a) Oxidation of the billet and the extrusion tools container–billet and die–billet interfaces and (c) the inhomoge- (b) Softening of the die and tools, neous deformation, i.e. the redundant deformation. Equa- (c) Difﬁcult to provide adequate lubrication tions (13.5) and (13.6) show that if extrusion ratio R increases, (d) Chance of burning/hot shortness when deformation in pressure p and load P required to cause extrusion will increase. On the other hand, with increase of R; large deformation will extrusion is extensive, since it results in the generation cause considerable rise in the temperature of the material, of considerable internal heat which will result in the drop of ﬂow stress of the billet material for a constant ram speed. For the same frictional condition and Considering the above problems, it is advantageous to use redundant deformation, if the effect of the temperature rise to the minimum working temperature which will give suitable decrease the ﬂow stress would be more than that of large R to plasticity to the metal. Generally, the maximum temperature increase the extrusion pressure through the term lnðA1=A2Þ for hot-working is limited to 50 °C below the melting point in (13.5), then the pressure or load required to cause extrusion for pure metals, or the temperature at which hot shortness would have decreased with increase in extrusion ratio. takes place. Since the actual temperature of billet during extrusion depends on several factors as mentioned above, the (3) Extrusion Temperature (T): initial temperature of the billet must be maintained at a safe level below the melting point or hot-shortness range. As a The higher the temperature, the lower is the ﬂow stress or rough rule, the initial temperature of billet for hot extrusion deformation resistance of the material. To take advantage of is two-thirds of the melting point measured in Kelvin, for this, most metals are extruded hot. But as soon as the heated example, the initial temperature for aluminium is 600 K. billet is loaded into the preheated chamber and extrusion starts, the temperature of the billet during extrusion is If the rate of heat generated due to deformation and affected by the factors that generate heat in the billet and friction is greater than that dissipated, which often occurs, those that transfer heat from the billet. The factors that then the temperature of extrudate will increase as the billet is generate heat and tend to raise the temperature of billet are: extruded and this increase in temperature will be greater if the speed of ram is increased. The surface temperature of the • Deformation of billet from its initial size to ﬁnal size and extrudate can be considerably higher than its centre tem- shape of the extrudate. The higher the amount of defor- perature because of friction and localized shear deformation. mation, i.e. the higher extrusion ratio, the higher will be If the surface temperature of extrudate arrives at the critical the generation of heat. temperature at which hot shortness takes place, surface defects will be produced. The temperature of the extruded • Friction or shear stresses at the interface between the part as it exits from the die is the most important dependent billet and the extrusion tooling, which includes the con- process variable that affects the quality of extrudate. Thus, it tainer, die or dead-metal zone. is desirable to control the exit temperature of extrusion and maintain a constant exit extrusion temperature, which is • The increase in the extrusion speed or the speed of ram. known as isothermal extrusion. Isothermal extrusion meth- ods and press control have been extensively covered in On the other hand, the factors that transfer heat and tend several literatures (Chadwick 1969; Laue and Stenger 1981; to decrease the temperature of billet are: Bryant et al. 1999; Saha 2000; Takahashi and Yoneyama 2004). However, to achieve isothermal extrusion, the fol- • Heat transfer between the billet and the tooling which lowing methods may be used: includes container, die and ram. Higher the temperature difference between the billet and the tooling, the more will be • Taper-heated/taper-quenched billets: In this method, the drop in the temperature of the billet. Further, the heat when the extrusion commences, the initial temperature at transfer depends on the materials of the billet and the tooling. the back end of the billet is maintained at a lower level than that at the front (die) end. As the billet is extruded, Therefore, the actual extrusion temperature of billet will the rise in temperature due to generation of heat is offset be decided by the billet alloy and its initial temperature, by the decreasing temperature (or taper) in the billet and tooling material and temperature, extrusion ratio, shape of the thus, a constant exit extrusion temperature will be extruded section, frictional condition and extrusion speed. maintained. Taper-heated billets, where the temperature

13.5 Factors Influencing Extrusion 611 increases gradually from the back to the front end of the Extrusion pressure Increasing billet, can be obtained by heating different sections of the temperature billet to different temperature in an induction heater having multiple heating zones. To get taper-quenched Extrusion speed (log) billets, they are ﬁrst uniformly heated in a gas furnace and then partially water quenched to produce a thermal Fig. 13.9 Schematic illustration of the effect of extrusion speed and gradient along the length of the billet. temperature on extrusion pressure • Variable speed control: Since the exit temperature of extrudate can be decreased by decreasing the extrusion However, higher quantity of heat generated at higher ram speed, so, in this method, the extrusion ram speed is speed will be retained by the metal during extrusion, because decreased with the progress of extrusion during the cycle insufﬁcient time is available for this heat to dissipate due to in order to offset the rise in temperature due to generation higher extrusion speed. Excessive speed can cause over- of heat. For this technique, good control over the extru- heating of the billet, especially near the section surface sion ram speed is required. Laue in 1960 (Laue 1960; because of transformation of most of the frictional energy Saha 2000) presented a system for isothermal extrusion into heat. This overheating can cause temperature-related with variation in the ram speed by which extrudate surface defects or the development of circumferential surface temperature was maintained within required limits. cracks caused by hot shortness, especially in aluminium • Nitrogen cooling: To restrict the heating of the die stack alloys, and the latter phenomenon in extrusion is known as and hence, the exit temperature of extrusion, liquid speed cracking, because it occurs due to high ram speed. nitrogen has been used during extrusion to cool the entire assembly of die stack. In spite of the difﬁculties associated with the rise in the • Container cooling: There are efﬁcient cartridge heaters temperature of extruded metal, higher extrusion speeds are with container; the heaters are arranged into and con- desired for increased productivity. Hence, it is essential to trolled by different zones of container. Integral have an optimal extrusion speed for hot extrusion. For this, air-cooling passages are used to restrict the heating of the extrusion speed is maximized, but simultaneously an container and enhance the control of process temperature. isothermal extrusion, where a constant exit temperature of the extrudate is maintained, is carried out for optimal quality of (4) Extrusion Speed or Ram Speed ðvramÞ: the product. The ram speed was varied to maintain the tem- perature of extrudate within required limits in the system for The extrusion speed, i.e. the speed of ram, is an important isothermal extrusion presented by Laue in 1960. In this, the parameter in that the strain rate increases directly with time savings for high-strength alloys were claimed to be increasing ram speed in extrusion. Again, the ﬂow stress of higher than that for more easily extrudable alloys (Saha 2000). metal in any hot-working increases with the strain rate and the strain-rate sensitivity of the ﬂow stress (see Sect. 1.11.1 Obviously, the problem of overheating and development in Chap. 1). Strain-rate sensitivity is high at elevated tem- of surface defects and hot shortness can be minimized or perature (it is negligible at room temperature) and increases eliminated by reducing the extrusion speed. But if the with increasing temperature. Therefore, increasing extrusion extrusion speed is too low, there are some problems: speed at elevated temperature increases the ﬂow stress of metal and results in the requirement of a higher speciﬁc • Productivity will be reduced. pressure for extrusion. Further, higher speed will result in an • Greater cooling of the billet will occur. This will increase increase in temperature in the extruded metal. More heat will be generated due to the additional work arising from the the required extrusion pressure because of increased ﬂow increased ﬂow stress of the metal. The effects of ram speed stress as the billet cools. and temperature on extrusion pressure are illustrated • Tool life will be reduced because of prolonged contact schematically in Fig. 13.9. It has been observed that if the time between the hot billet and the tools at a slower extrusion ram speed is increased by ten times, the extrusion speed. pressure will increase by about 50%; the per cent increase in the extrusion pressure increases with increasing extrusion temperature. For example, experiment on lead extrusion shows that a tenfold increase in speed results in an increase in pressure by 38% at 17 °C, 44% at 100 °C, 50% at 166 oC and 55% at 325 °C.

612 13 Extrusion The higher the temperature of the billet, the greater is the the reduction possible in deformation with a certain set of the cooling of the billet at low extrusion speed and the greater will process variables (other than the die angle). For any com- be the rise in the ﬂow stress of the material and thereby in the bination of coefﬁcient of friction and reduction of area, there extrusion pressure. Therefore, high-strength metals/alloys like exists an optimum semicone die angle, a; designated by aopt:; steel, titanium or refractory metals, which need high extrusion which gives the minimum stress required for extrusion or temperature, are required to be extruded at high ram speed drawing operation using a conical converging die. (1.5–0.75 m/s). For example, stainless steel is hot extruded at 1040 °C at higher speed but it cannot be extruded at lower In extrusion or drawing, the total deformation load, Ptotal; speed due to the rise in extrusion pressure. On the other hand, is composed of the following three components: magnesium, aluminium and copper alloys which are prone to hot shortness have to be extruded at a low ram speed (i) Load, Pideal; for ideal frictionless homogeneous defor- (15–75 mm/s), otherwise a greater rise in temperature mation, which is: resulting from a higher speed will cause hot shortness. e2 ¼lZnðA1 =A2 Þ A1 Allowable or safe extrusion range under the combined effects A2 of billet temperature, extrusion pressure and extrusion strain rate For extrusion; Pideal ¼ A1 r0de ¼ A1r0 ln can be shown with a diagram, known as extrusion limit diagram. This diagram is schematically represented in Fig. 10.14, in e1¼0 which the ordinate may be considered as ‘extrusion ratio’ replacing ‘amount of deformation’. Hirst and Ursell (1958) ¼ A1r0 ln 1 1 r carried out their work on extrusion, and their ﬁndings, which are À applicable to all methods of working, have been presented in the form of the above diagram. The allowable or safe extrusion ð13:7Þ range is the region under the curve of constant extrusion speed and constant pressure. This diagram shows that for any given e2 ¼lZnðA1 =A2 Þ extrusion pressure, as the working temperature increases, the value of extrusion ratio that can be achieved increases due to And for drawing; Pideal ¼ A2 r0de lowering of the ﬂow stress with increase in temperature, and as the applied extrusion pressure increases, a greater extrusion ratio e1 ¼0 can be achieved at a given work-piece temperature and strain rate. However, the extrusion temperature will rise with increas- ¼ A2r0 ln A1 ¼ A2r0 ln 1 1 r ing extrusion speed and extrusion ratio. When the extrusion A2 À speed is extremely high resulting in the adiabatic condition, where none of the heat produced by deformation is dissipated, it ð13:8Þ is required to decrease the work-piece temperature for main- taining its ﬁnal temperature below the hot-shortness temperature. where, A1 and A2 are the cross-sectional areas of the As the extrusion strain rate increases and/or work-piece tem- work-piece, respectively, before and after the perature decreases, the ﬂow stress of the material will increase, deformation, r0 is the ﬂow stress of the material, which which will cause the allowable extrusion ratio to decrease under is assumed to be constant and may be replaced by the a given applied pressure. At inﬁnitely rapid strain rate leading to mean ﬂow stress r0 in case of increase in r0 during adiabatic condition, the allowable extrusion range will be the deformation due to strain hardening, e1 and e2 are the true minimum, and at any given strain rate the value of extrusion strains experienced by the material, respectively, before ratio will increase with decreasing the work-piece temperature. and after the deformation, r is the reduction of As the extrusion speed decreases there is greater heat dissipation cross-sectional area and given by r ¼ ðA1 À A2Þ=A1 ¼ and the allowable extrusion ratio increases for a given initial 1 À ðA2=A1Þ. Obviously, the ideal load, Pideal; is temperature of work-piece and extrusion pressure. independent of a; but is a function of r: (5) Semicone Die Angle, a, and Optimum Semicone Die (ii) The second component is the load required to over- Angle, aopt: come the frictional resistance at the die–work-piece interface, say Pfriction: This depends upon a combination It is of interest to operate with a die-cone angle which will of the normal pressure exerted by the die on the minimize the forces required for deformation or maximize work-piece (which increases with r), the friction coef- ﬁcient at the die–work-piece interface and the surface area of contact between work-piece and die. In extru- sion or drawing, since radialdifferecebetweentheincomingandoutgoingstock lengthof contactalongthedie-stockinterface ¼ D1 À D2 ¼ sin a; 2Lc

13.5 Factors Influencing Extrusion 613 therefore, as a increases, i.e. sin a increases, the contact and this results in the reduction of total load, Ptotal; for length Lc as well as the surface area of contact between extrusion or drawing until aopt: is reached. This is illustrated work-piece and die decreases for a given ðD1 À D2Þ; in Fig. 13.10. The curve of Ptotal versus a exhibits a mini- where D1 and D2 are the diameters of the incoming and mum at aopt:; where the decrease in Pfriction is balanced by the outgoing stock. Hence, Pfriction, is a function of a; r; increase in Predundant: The values of aopt: for most extrusion and the coefﬁcient of friction. operations lie between 45 and 60° depending on the values of coefﬁcient of friction and the amount of reduction of area. (iii) The third component is the load required to overcome As the friction or the reduction in area increases, the value of the redundant deformation, say Predundant; which is the optimum semicone die angle also increases. However, in wasted or extra work due to inhomogeneous deforma- hydrostatic extrusion, the values of aopt: drastically reduce, tion (see Sect. 10.5.2 in Chap. 10). The redundant which may be of the order of 20°, because friction is nearly deformation is greater for a die having a large included absent or drastically reduced at the die by the presence of the angle than for a long die of small angle. Because the steeper the die angle, the more the internal shear pressurized ﬂuid acting as lubricant. In conventional extru- exceeds the minimum shear required for elongation of the work-piece. Note that the internal shear occurs due sion, small cone angle of converging die can be used if the to changes in the ﬂow direction of the metal in the die billet is pushed indirectly by the ram through graphite and the redundant deformation is not only a function of a; but also of r: The higher the reduction, r, the lower is medium (Boulger and Gehrke 1965). Further, the operation the load required for redundant deformation. of conventional extrusion will be similar to that of hydro- Hence for a too small cone angle, Pfriction is signiﬁcantly static extrusion, if the spacing between the chamber and high because of high contact length Lc; whereas Predundant billet is larger and ﬁlled with graphite. becomes a predominant factor for too large a cone angle. For a given reduction and frictional condition as a is increased from a Beyond aopt:; further increase in a causes an increase in very low value, Pfriction decreases rapidly while Predundant the extrusion or drawing load, Ptotal; because the rate of increases slowly, but the component Pideal remains constant increase of Predundant is more than the rate of decrease of in the total load given by Ptotal ¼ Pideal þ Pfriction þ Predundant; Pfriction: The type of variation in the total load with a; as shown in Fig. 13.10, occurs in the region of sound ﬂow. There are two additional common modes of ﬂow: dead zone formation and shaving. These are observed at higher values of a; where the variation in total load or stress with a differs Fig. 13.10 Portions of Shear friction factor, m = constant; schematic plot of the extrusion or Reduction, r = constant. drawing load versus semicone angle of die, showing components Total deformation load of total deformation load and the in extrusion or drawing, optimum semicone die angle for Ptotal = Pideal + Pfriction + Predundant constant reduction and frictional (i.e., sum of the bottom three loads) condition Ideal load, Pideal Loads Redundant work load, Predundant Friction load, Pfriction αopt (Optimum semi cone die angle) Semi cone angle of die, α

614 Large reduction 13 Extrusion Shear friction factor large cone angle m = constant D2/2 α Increasing reduction, r Drawing or extrusion stress D1 2 Optimum semi-die angle, αopt Small reduction x Small cone angle 0 Semicone angle of die, α Fig. 13.12 Trumpet-shaped die Fig. 13.11 Schematic illustration of the effect of semicone die angle, deformation zone, the trumpet-shaped die proﬁle will then a; and reduction, r; on the drawing or extrusion stress and the effect of r follow the equation: on optimum semicone die angle, aopt:; for a constant shear friction factor, m from that in the sound ﬂow region. This has been discussed x ¼ D1=2 À D2=2 tan aopt in Sect. 14.6.5 of Chap. 14. ð13:10aÞ If a\\45; then the optimum semicone die angle, aopt (in ¼ pﬃﬃﬃDﬃﬃﬃ1ﬃﬃﬃÀﬃﬃﬃﬃﬃDﬃﬃﬃ2ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 tan ð3=2Þm lnðD1=D2Þ radian), can be approximated from the following relation ¼ qﬃﬃﬃﬃﬃﬃDﬃﬃﬃ1ﬃﬃﬃÀﬃﬃﬃﬃﬃDﬃﬃÀﬃﬃ2ﬃﬃﬃﬃﬃpﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃÁﬃ (Avitzur 1968): 2 tan ð3=2Þm ln 1= 1 À r pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð13:9Þ Or, x ¼ pðDﬃﬃﬃﬃ1ﬃﬃ=ﬃﬃDﬃﬃﬃﬃ2ﬃﬃÞﬃﬃÀﬃﬃﬃﬃﬃ1ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ aopt ¼ rﬃðﬃ3ﬃﬃﬃ=ﬃﬃ2ﬃﬃﬃÞﬃmﬃﬃﬃﬃﬃlﬃnﬃﬃðﬃﬃDﬃﬃﬃﬃ1ﬃﬃ=pﬃﬃDﬃﬃﬃﬃﬃ2ﬃﬃﬃﬃÞﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ D2 2 tan ð3=2Þm lnðD1=D2Þ ¼ ð3=2Þm ln 1= 1 À r rad ¼ qﬃﬃðﬃﬃDﬃﬃﬃ1ﬃﬃ=ﬃﬃDﬃﬃﬃﬃ2ﬃﬃÞﬃﬃÀﬃÀﬃﬃﬃﬃ1ﬃﬃpﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃÁﬃﬃ ð13:10bÞ 2 tan ð3=2Þm ln 1= 1 À r where D1 and D2 are the diameters of work-piece, respec- tively, before and after the deformation, and m is the interface Although the work-pieces of different diameters will be shear friction factor, and r ¼ reduction in cross-sectional subjected to different amounts of reduction, the output of this area ¼ ðA1 À A2Þ=A1 ¼ 1 À ðD2=D1Þ2: die will be an extrudate of the same diameter; i.e., D1 will vary but D2 will remain constant. Equation (13.9) shows that for a given frictional condi- tion, as D1=D2 increases, or the reduction r increases, aopt (6) Friction and Lubrication: becomes greater, which is shown schematically in Fig. 13.11 Greater the friction force, higher is the extrusion load and for a constant shear friction factor. As the reduction r power required to obtain a given extrusion ratio. Friction approaches zero, the optimum semicone die angle aopt also produces inhomogeneous deformation as well as surface tends to zero. Similarly, as m decreases, i.e. friction cracks and other defects. Further, friction increases ‘material decreases, aopt also decreases for a constant reduction [see pickup on tools’, the tool wear, and reduces the tool life. (13.9)] and as m or friction approaches zero, aopt tends to Unlubricated extrusion having high container-wall friction zero. often progresses with the formation of a dead zone of stagnant material. The dead zone usually leaves a poor If a single die is required to be used for a wide range of surface on the extruded product. Good lubrication producing reductions where a \\ 45; it may be advantageous to use a trumpet-shaped die, as shown in Fig. 13.12, where aopt is maintained for each different reductions. If x is the distance measured from the exit plane to the entry plane of

13.5 Factors Influencing Extrusion 615 low friction can entirely eliminate the formation of dead bonded with sodium silicate, or glass wool, or both is placed zone. immediately ahead of the die. The diameter of the glass pad will be the same as that of the billet. The pad contains a Most of the hot extrusions need lubrication to overcome central hole whose diameter is the same as that of the the effects of frictional force. An effective hot-extrusion extrudate. The prelubricated billet is quickly inserted into the lubricant must have low shear strength and enough stability container along with an appropriate dummy block, and then to prevent breakdown at elevated temperature. Aluminium the extrusion cycle is started. During extrusion, the face of alloys are often extruded without lubrication, mainly to hot billet contacts and steadily softens or melts the glass pad avoid transportation of lubricant and oxide to form surface placed in front of the nose of the billet. Molten glass is and subsurface defects because square dies are used. If transported with the extruding metal and forms a uniform tapered die is used, graphite or oils can be used as lubricant thin lubricant ﬁlm between the metal and the die, typically to give low extrusion force, but staining may result and about 25 lm thick. Since the thickness of lubricant ﬁlm tapered die produces large discard. Above about 600 °C, depends on its viscosity, it is natural to think that the viscosity copper forms its oxide, which is softer than the metal itself of glass is the main criterion to select lubricants, but inves- forming a good natural lubricant. Thus, copper acts as tigation (Rogers and Rowe 1967) has shown that there is no self-lubricating metal and can be satisfactorily hot extruded correlation between the glass viscosity and the coating at 750–950 °C without further lubrication. But copper thickness or extrusion pressure. This lubricant ﬁlm provides alloys, like brasses having high zinc content in their oxides, low friction and also thermally insulates the die and improves show higher friction and so require lubrication. its life appreciably. The thickness of lubricant ﬁlm depends on the rate of availability of lubricant by softening or melting. If Graphite, usually dispersed in grease, oil or resin, and the ram speed is too slow, the time available for softening or glass are widely used as lubricants for extrusion of copper melting glass pad is more and hence, a thick lubricant ﬁlm alloys, alloy steels, stainless steels, tool steels and titanium will be produced resulting in a low initial extrusion pressure, alloys. Die wear occurs signiﬁcantly in the conventional hot but this will rapidly consume the glass pad stored between the extrusion of titanium alloys and steels with graphite lubri- billet and die and limit the length of the extrudate. On the cants because graphite often provides little or no thermal other hand, at very high extrusion speed, the resulting lubri- protection to the die. The problem with grease and graphite cant ﬁlm may become dangerously thin. This technique may is to maintain sufﬁcient lubrication over the full length of the be modiﬁed to achieve hydrostatic lubrication (Kulkarni et al. extrusion. When grease-base lubricants are used, shear metal 1972) by applying thick ﬁlms of glass only to the circum- ﬂow can occur with both ﬂat-faced and conical dies. On the ference of the billet. In this case, the extrusion pressure is other hand, both ﬂat-faced and conical dies produce laminar strongly inﬂuenced by the viscosity of lubricant. or parallel metal ﬂow with glass lubrication. Further, glass lubricant tends to produce a better surface ﬁnish and die life Successfulness of lubrication depends on maintaining a than grease and graphite. Glass lubricants have performed reservoir of glass on the die face that also depends on type satisfactorily to produce long lengths by hot extrusion. Glass and design of die. Generally, ﬂat-faced die is used for most acts as a thermal insulator and minimizes cooling between glass-lubricated extrusion. Because if conical dies are used the billet, the container and the die. Its wetting characteristics with glass lubrication, glass ﬂow increases and much of the are good, and viscosity is maintained at elevated glass pad (reservoir of glass) will be exhausted early in the temperature. extrusion, although conical die lowers extrusion force required and improves the metal ﬂow. Another disadvantage The Ugine-Sejournet process (Sejournet and Delcroix of conical dies for which they are not industrially favoured is 1955) is commonly used for hot extrusion of steel including the retention of large discard in the container unless a dis- stainless steel etc., and has also been used for hot extrusion posable follower block is used. of nickel, cobalt and titanium alloys. In this method, the billet of the above alloys is heated usually at 1000–1250 °C The choice between glass and grease-based lubricants is in an inert atmosphere and coated with a layer of based chieﬂy on the extrusion temperature. The reduction of low-melting-temperature glass. Surface coating is performed friction is the main function of lubricants at low extrusion by rolling the billet over a bed of ground glass or by temperature. Since the tooling can withstand a maximum sprinkling with glass powder. The glass coating not only temperature of 500–550 °C, so at extrusion temperatures serves the purpose of lubrication, but it also acts as thermal above 1000 °C, the thermal insulation of the tooling by insulator and reduces the loss of heat from the billet. Before lubrication against overheating is equally important as the the preheated billet is inserted into the container of the press, reduction of friction, particularly with difﬁcult-to-extrude a main source of lubricant which can be a circular glass pad alloys. Further, the lubricant ﬁlm can also prevent oxidation. of 10 mm thick compacted from ﬁne glass powder and

616 13 Extrusion Lubricants can be classiﬁed into two groups, depending on 13.6.1 Open-Die, Indirect and Hydrostatic temperature: Extrusions • Below 1000 °C: Grease lubrication, which includes Let us consider a conical converging die with a semidie grease, graphite, molybdenum disulphide, talc, mica, angle of ‘a;’ through which open-die (container-less) soap, asphalt, bentonite and plastics, such as extrusion, or extrusion with frictionless container, such as high-temperature polyimides. indirect and hydrostatic extrusions of a cylindrical billet occurs. Cylindrical symmetry is assumed to prevail in this • Above 1000 °C: Glass lubrication, which includes glass analysis. A vertical slab element of metal with inﬁnitesimal and basalt. thickness, dx; and diameter, D and ðD þ dDÞ; bounded by the conical surface of the die and by two transverse surfaces Graphite is preferred to glass as a lubricant in high normal to the axis of symmetry is considered in the defor- energy-rate extrusion, because glass requires time to soften. mation zone, as shown in Fig. 13.13. The exit plane of the deformation zone, i.e. the plane at the exit side of the die Materials that have a tendency to stick to the die or the where the reduction of the stock ends (henceforth only called container need to be canned, that is, sealed in a thin-walled the exit plane), is considered as the origin of the coordinate container (known as jacket or can) of a softer metal, such as system, shown by x ¼ 0 in the diagram, where x is the copper or mild steel or other cheap container. The billet horizontal axis that increases from the exit plane to the entry enclosed in the jacket is extruded, and subsequently, the can plane of the stock within the die (henceforth called the entry is removed. The can not only provides a low-friction inter- to the die). One transverse surface of the element is at a face, but also prevents contamination between the billet distance x from the exit plane, the other at an additional material and the environment. incremental distance dx: The element is subjected to axial and radial compression, and to a sliding frictional force 13.6 Estimation of Extrusion Load acting along the die wall. The horizontal axial stress rx over the transverse surface, situated at a distance x from the exit Free-body equilibrium approach is used in the text to plane, is assumed to be uniformly distributed and normal to develop the equations for extrusion load. To simplify the the transverse surface with no shear component. Hence, it is analysis, the following assumptions are made: a principal stress. For the incremental distance, the axial stress varies by the amount drx; i.e. the stress becomes (a) Homogeneous deformation is assumed; i.e., internal rx þ drx at a distance x þ dx from the exit plane. Over the redundant deformation is neglected. surface of the slab element, which is in contact with the conical die, a die pressure p is assumed normal to interface (b) For extrusion through conical converging die, it is and a frictional drag lp parallel to the interface. The stresses assumed that at the contact surface between the conical acting on the element of metal in the deformation zone are die and the work-piece, no sticking friction occurs and shown in Fig. 13.13. only sliding friction according to Coulomb’s law is effective. In case of extrusion through square dies, where Under steady condition, horizontal equilibrium of forces metal ﬂows through dead zone formation, it is further acting on the free slab element in the axial positive direction assumed that Coulomb sliding friction is effective at the of x; i.e. from the exit to entry side, is expressed by contact surface between the conical dead zone and the work-piece. In other words, for ﬂow through conical À ðrx þ drxÞ p ðD þ dDÞ2 þ rx p D2 ð13:11Þ converging die or dead zone, s ¼ lp\\k; and where s is 4 4 the tangential frictional stress at the contact surface, k is the shearing yield stress of work-piece material, l is the þ p sin a pDdLc þ lp cos a pDdLc ¼ 0 coefﬁcient of friction, and p is the pressure on the work-piece and is normal to the contact surface. * sin a dLc ¼ dD ; 2 (c) The coefﬁcient of friction l is the same at all points on the surface of contact. ) ðrx þ drxÞ p ðD þ dDÞ2 Àrx p D2 4 4 (d) As extrusion is a hot-working operation, so it is quite logical to assume that there is no strain hardening of the À ppD dD À lppD cot a dD ¼ 0 work-piece; i.e. the ﬂow stress of the material being 22 extruded remains constant.

13.6 Estimation of Extrusion Load 617 Fig. 13.13 Open-die D1/2 dLc = μp p (container-less) extrusion or dD D2/2 extrusion with frictionless 2 sin α container (indirect and hydrostatic σl α σx + dσx σx extrusions) of a cylindrical billet α through a conical converging die with a semicone angle of ‘a;’ D/2 showing stresses acting on an dD/2 element of metal with inﬁnitesimal thickness, dx; and μp p diameter, D andðD þ dDÞ; in the deformation zone dx x x=0 x=l Or, ðrx þ drxÞ ðD þ dDÞ2Àrx D2 As pD cos adLc ¼6 0; dividing (13.12c) by pD cos adLc; À 2pDdD À 2lpD cot adD ¼ 0 we get dD2 being very small quantity, it is neglected in the fol- rr ¼ p À s tan a ¼ pð1 À l tan aÞ ð13:12dÞ lowing equation: As semicone die angle, a; is usually large for extrusion than that for drawing, so it is not generally true for extrusion ) ðrx þ drxÞÀD2 þ 2D Á unlike drawing to ignore l tan a in comparison with unity. dD For example, if a ¼ 60; tan a ¼ 1:73; and l must be less than 0.03 if the error is restricted to even 5%. However, the À rxD2 À 2pDdD À 2lpD cot adD ¼ 0 term l tan a will be neglected in comparison with unity to avoid complexity in the following analyses for extrusion Or; rxD2 þ rx2DdD þ D2drx þ 2DdDdrx ð13:12aÞ À rxD2 À 2pDdD À 2lpD cot adD ¼ 0 problems. Thus, (13.12d) is simpliﬁed as: As dD drx is a multiplication of two small quantities, so 2DdDdrx is neglected in (13.12a). As D 6¼ 0; dividing rr % p ð13:12eÞ (13.12a) by D; we get rxD þ rx2dD þ Ddrx Hence, the die pressure p can be considered to be a À rxD À 2pð1 þ l cot aÞdD ¼ 0 principal stress. Again, since the circumferential strain rate is equal to the radial strain rate due to cylindrical symmetry, so Assuming l cot a ¼ B; the circumferential stress, rh ¼ rr: Again, rr; rh; and rx; all 2rxdD þ Ddrx À 2pð1 þ BÞdD ¼ 0 are taken as compressive stresses. Hence, considering rr; rh; and rx as the principal stresses, we can write in the Von Or; Ddrx ¼ À2rxdD þ 2pð1 þ BÞdD Mises’ yielding criterion, r1 ¼ rr ¼ rh ¼ r3 ¼ p (since ¼ 2ðp À rxÞdD þ 2BpdD ð13:12bÞ reduction is occurring within die due to die pressure, so the Let rr is the compressive radial stress exerted by the die pressure p is considered to be higher than the stress rx) conical die on the vertical element, and it is considered as a and r2 ¼ rx; so the Von Mises’ yielding criterion, which is principal stress. Radial equilibrium of forces exerted by the die on the vertical element of metal in the deformation zone qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ pﬃﬃ r0; reduces to is given by ðr1 À r2Þ2 þ ðr2 À r3Þ2 þ ðr3 À r1Þ2 2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðp À rxÞ2 þ fÀðp À rxÞg2 ¼ pﬃﬃ r0; or; p À rx ¼ r0 2 ð13:13aÞ rr pDdx ¼ p cos a pDdLc À s sin a pDdLc; where r0 is the uniaxial compressive ﬂow stress or defor- mation resistance of the work-piece. Using (13.13a), Or; rrpD cos adLc ¼ ppD cos adLc À spD sin adLc (13.12b) becomes ð13:12cÞ

618 13 Extrusion Ddrx ¼ 2r0dD þ 2Bðr0 þ rxÞdD Extrusion (ram) pressure at the entry to the die, ¼ 2½r0ð1 þ BÞ þ BrxdD þ B(D12BÀ1) r0 1 B D2 Z drx Z dD rl ¼ On Integration; þ ð1 þ D Brx BÞr0 ¼ 2 ( B ) r0 1 B A1 À1 ¼ þ ð13:15Þ B A2 1 Or; B lnfBrx þ ð1 þ BÞr0g þ ( B ) r0 1 B B À1 ¼ 2 ln D þ C ðwhere C is an integration constantÞ ¼ 1 1 À r Or; lnfBrx þ ð1 þ BÞr0g ¼ 2B ln D þ BC where A1 and A2 are the cross-sectional areas of the work-piece, respectively, before and after extrusion; r ¼ the Or; Brx þ ð1 þ BÞr0 ¼ À D2B þ Á fractional reduction of the cross-sectional area of the exp ln BC work-piece ¼ ðA1 À A2Þ=A1 ¼ 1 À ðA2=A1Þ: ) Brx þ ð1 þ BÞr0 ¼ D2B eBC Or; Brx ¼ A D2B À ð1 þ BÞr0; ) Maximum load at the start of extrusion, where A ¼ eBC is a new integration constant: ( 2BÀ1)pÀD21=4Á Pl ¼ rl pÀD12=4Á ¼ þ B D1 r0 1 B D2 D2B 1þB ð13:16Þ ) rx ¼ A B À B r0 ð13:14aÞ Let the diameter of the work-piece after extrusion, i.e. at Open-die or container-less extrusion pressure is expressed by (13.15) and maximum load by (13.16). Figure 13.14 the exit plane is D2; and that before extrusion, i.e. at the shows schematically an open-die extrusion for extrusion of entry to the die is D1: Since at outer free surface of the cylindrical shapes. This process is generally performed on a extruded product, the horizontal axial stress rx is zero, i.e. at vertical press, and a conical die can be used. Since the length the exit plane where D ¼ D2; rx ¼ 0: Hence, from (13.14a) we get A ¼ ð1 þ BÞr0 ð13:14bÞ Ram ðD2Þ2B D1 2 Now, substituting A from (13.14b) into (13.14a), we get α rx ¼ ðr10þD122BBþBÞrB0D(B2B ÀDD212BþBÀB1)r0 ¼ Die ð13:14cÞ Let l ¼ the distance measured from the exit plane to the D2 entry plane of the stock within the die. Now, at the entry to 2 the die, where D ¼ D1;, we can assume that the horizontal axial stress rx ¼ rl; where, rl will be the extrusion Fig. 13.14 Open-die extrusion (ram) pressure at the entry to the die. Hence, from (13.14c) we can get

13.6 Estimation of Extrusion Load 619 of the unsupported billet is limited to prevent buckling, only of the container wall on extrusion stress is estimated. In the short length billets are usually extruded. Since the frictional following analysis, direct extrusion of a cylindrical contribution from container wall is absent in indirect extru- work-piece through a conical converging die is considered in sion as well as in hydrostatic extrusion (to be described later), which Coulomb’s law of sliding friction is assumed to hold so (13.15) and (13.16) can be used to obtain, respectively, good at the contact surface between the die or the extrusion ram pressures and ram loads for both extrusion processes. chamber and the work-piece. At the entry to the die, where rx ¼ rl; we can assume that Let L ¼ the distance measured from the exit plane to the p ¼ pl; where pl is the die pressure at the entry to the die. Now, rear plane of the billet within the extrusion chamber with the help of (13.13), it follows from (13.15) that the die (henceforth called as the entry to the chamber), i.e. L ¼ the pressure on the work-piece at entry to the die will be given by combined length of billet and working zone of die; pl ¼ r0 þ rl B(D12B )# l ¼ the distance measured from the exit plane to the entry ¼ \" À1 to the die, i.e. the length of working zone of die; r0 1 þ 1 þ L À l ¼ the length of billet only. B A transverse metal-slice of ﬁxed diameter, D1; having an D2 inﬁnitesimal thickness, dx, is considered in the deformation \" B(A1B )# zone within the extrusion chamber, i.e. within ðL À lÞ por- ¼ r0 1 þ þ À1 ð13:17Þ tion in Fig. 13.15. As mentioned in previous Sect. 13.6.1, 1 B the exit plane (of the deformation zone) is considered as the origin of the coordinate system, shown by x ¼ 0 in the \" A2 B )# diagram, where x is the horizontal axis that increases from 1 ( À1 the exit plane to the entry to the chamber. One transverse ¼ þ þ B 1 surface of the above slice located in the deformation zone r0 1 B À r within the extrusion chamber is at a distance x from the exit 1 plane, the other at an additional incremental distance dx: The slice is subjected to axial and radial compression, and to a 13.6.2 Direct Extrusion Through Conical sliding frictional force acting along the container wall. The Converging Die horizontal axial stress rx over the transverse surface, situated at a distance x from the exit plane, is assumed to be uniformly 13.6.2.1 Coulomb’s Sliding Friction in Chamber distributed and normal to the surface with no shear With the free-body equilibrium approach in the cylindrical portion of the extrusion chamber, the inﬂuence of the friction Fig. 13.15 Details of tapered die p Tapered die extrusion of a cylindrical billet, τ showing stresses acting on a transverse metal-slice of ﬁxed Ram σx diameter, D1; having an inﬁnitesimal thickness, dx, in the deformation zone within the extrusion chamber σx + dσx σ L D1/2 D2/2 α τ x dx l p x=0 L

620 13 Extrusion component. Hence, it is a principal stress. For the incre- Z drx Z 4l dx; mental distance, the axial stress varies by the amount drx; rx À r0 D1 i.e. the stress becomes rx þ drx at a distance x þ dx from the ¼ exit plane. Let rr is the compressive radial stress exerted by the wall of the extrusion chamber on the cylindrical Or, lnðrx À r0Þ ¼ 4l x þ C1 ð13:19bÞ work-piece, and it is considered as a principal stress. From D1 radial equilibrium of forces, we get rr ¼ p; where p is the pressure exerted by extrusion chamber on the work-piece. where C1 is integration constant, when x ! l: Hence, p will also be treated as a principal stress. Over the Since at the entry to the die, i.e. at x ¼ l; we have already surface of the slice element, which is in contact with the assumed in the previous Sect. 13.6.1 that the horizontal axial container wall, a frictional drag s is acting parallel to the stress rx ¼ rl (the ram pressure at the entry to the die), then interface where s ¼ lp: The coefﬁcient of friction l at the from (13.19b) we get billet–container interface is assumed to be the same as that at the billet–die interface. The stresses acting on the slice element of C1 ¼ lnðrl À r0Þ À 4ll ð13:19cÞ metal in the deformation zone within the extrusion chamber are D1 shown in Fig. 13.15. Another thin element of metal within the Inserting the value of C1 from (13.19c) back into (13.19b) tapered die region with diameter varying from D to D þ dD and rearranging, one can write as follows: with inﬁnitesimal thickness, dx, as discussed in previous Sect. 13.6.1, is also shown in Fig. 13.15. ln ðrx À r0Þ ¼ 4l ðx À lÞ ðrl À r0Þ D1 Under steady condition, horizontal equilibrium of forces ! 4l ðx À lÞ acting on the free metal-slice in the axial positive direction rx ¼ ðrl À r0Þ exp D1 þ r0 ð13:19dÞ of x; i.e. from the exit to entry side, is expressed by Now, substituting the value of rl from (13.15) into (13.19d), we get Àðrx þ drxÞ pD21 þ rx p D12 þ s pD1dx ¼ 0 ð13:18Þ *\" þ B(D12B ) # 4 4 r0 1 B À1 1 rx ¼ À D2 As D1 6¼ 0; dividing (13.18) by ðpD1Þ=4 and substituting & ') ð13:20aÞ s ¼ lp; we get exp 4l ðx À lÞ þ 1 D1 D1rx þ D1drx ¼ D1rx þ 4lpdx ð13:19aÞ As ðD1=2Þ À ðD2=2Þ ¼ tan a; ) D1drx ¼ 4lpdx l Cylindrical symmetry is assumed in this analysis, as in the ) l ¼ D1 À D2 cot a; 2 previous Sect. 13.6.1, where the compressive circumferential stress rh ¼ rr: rr; rh; and rx are considered as the principal Now, putting the value of l in (13.20a): stresses. Since all stresses are compressive, considering all of them to be positive we can write in the Von Mises’ yielding rx ¼ *\" þ ( 2B ) À # criterion, r1 ¼ rx; and r2 ¼ rr ¼ rh ¼ r3 ¼ p. Since there r0 1 B B D1 À1 1 is no reduction of the billet within chamber and the billet is D2 !) forced into the die by the movement of ram, so the ram exp 2l f2x À ðD1 À D2Þ cot ag þ 1 pressure ðrxÞ on the billet must be higher than the pressure *\"D1 ðpÞ exerted by the container on the billet. Hence, the Von 1þ ( 2B ) # Mises’ yielding criterion, which is ¼ r0 B B D1 À1 À 1 D2 !) qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ pﬃﬃ r0; reduces to exp 2l cot a f2x tan a À ðD1 À D2Þg þ1 ðr1 À r2Þ2 þ ðr2 À r3Þ2 þ ðr3 À r1Þ2 2 *\"D1 ( 2B ) # qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðrx À pÞ2 þ fÀðrx À pÞg2 ¼ pﬃﬃ r0; or, rx À p ¼ r0 ) rx ¼ r0 1þB D1 À1 À1 2 B D2 ð13:13bÞ !) where r0 is the compressive ﬂow stress. Substituting p ¼ exp 2B f2x tan a À ðD1 À D2 Þg þ1 ð13:20bÞ rx À r0 from (13.13b) into (13.19a), and on integrating, we D1 obtain the following: where B ¼ l cot a:

13.6 Estimation of Extrusion Load 621 Let us assume that rL ¼ the maximum extrusion pressure l at the billet–container interface than at the bpillﬃeﬃ t–die interface. However, when lp ! k; where k ¼ r0= 3; full or ram pressure at the entry to the extrusion chamber, i.e. at sticking friction develops at the contact surface between the extrusion chamber and the work-piece; and the frictional the start of the extrusion. Hence, the horizontal axial stress is strespsﬃﬃalong the chamber wall, s; is given by s ¼ k ¼ rx ¼ rL; at x ¼ L: Therefore, (13.20b) leads to r0= 3; where k is the yield stress in shear and r0 is the uniaxial compressive ﬂow stress of the work-piece. rL ¼ *\" þ B(D12B ) À # r0 1 B À1 1 With the free-body equilibrium approach in the cylin- D2 drical portion of the extrusion chamber, the inﬂuence of full !) sticking friction along the container wall on the extrusion 2B stress is estimated. Let us consider the same transverse slice exp D1 f2L tan a À ðD1 À D2Þg þ1 ð13:21Þ of ﬁxed diameter, D1; with inﬁnitesimal thickness, dx, within ðL À lÞ portion in Fig. 13.15. Under steady condition, hori- ) Maximum ram load at the start of extrusion; zontal equilibrium of forces acting on the free metal slice in prD4L 12pÀrD021*=\"4Á1 the axial positive direction of x; i.e. from the exit to entry PL ¼ B(D12B ) # side, is given by (13.18). Under sticking friction, since s ¼ ¼ À1 1 k; so lp in (13.19a) will be replaced by k: as follows: þ D2 À B 2B !) D1 exp f2L tan a À ðD1 À D2Þg þ1 ð13:22Þ Equations (13.21) and (13.22) are applicable when Cou- D1drx ¼ 4kdx ð13:24Þ lomb’s law holds well. From yielding criterion given by Integrating (13.24) ZZ (13.13b), one can get the pressure exerted on work-piece by D1drx ¼ 4kdx wall of the extrusion chamber at its entry, i.e. at the start of the extrusion as pL ¼ rL À r0; since rx ¼ rL: Hence, the Or; rx ¼ 4kx þ C2 ð13:25aÞ container pressure pL can be expressed from (13.21) as D1 pL ¼ \" þ ( 2B ) À # r0 1 B B D1 À1 1 D2 ! where C2 is an integration constant, when x ! l: Since at the entry to the die, i.e. at x ¼ l; the horizontal axial stress rx ¼ exp 2B f2L tan a À ðD1 À D2Þg ð13:23Þ rl (the ram pressure at the entry to the die), then from D1 (13.25a) we get 13.6.2.2 Full Sticking Friction in Chamber C2 ¼ rl À 4kl ð13:25bÞ High coefﬁcient of friction, l; between the work-piece D1 material and the container wall arising due to hot-working and high compressive pressure, p; exerted by the wall of the Considering (13.15) for rl and substituting C2 from extrusion chamber on the work-piece sometimes result in (13.25b) into (13.25a), we get, full sticking friction at the contact surface between the extrusion chamber and the work-piece throughout the rx ¼ rl þ 4k ðx À lÞ extrusion process. Sticking friction may be developed when ¼ D1 there is no lubrication or breakdown of lubrication at the ( 2B ) billet–container interface. In the following analysis, direct 1þ B D1 À1 4k extrusion of a cylindrical work-piece through a conical r0 B þ D1 ðx À lÞ ð13:25cÞ converging die is considered, in which Coulomb’s law of D2 sliding friction is assumed along the die wall but full sticking friction is assumed to develop at the contact surface between Hence, at the entry to the chamber, where x ¼ L; sub- the extrusion chamber and the work-piece. It is further stitution of horizontal axial stress, rx ¼ rL (maximum assumed that coefﬁcient of friction, l; along the die wall is extrusion pressure or ram pressure), into (13.25c) gives the same as that at the billet–container interface, and that full sticking friction is developed due to high chamber pressure, Ram pressure, p; although sticking friction may also develop due to higher rL ¼ þ ( 2B ) þ 4k ðL À lÞ ð13:26Þ r0 1 B B D1 À1 D1 D2 pﬃﬃ Since k ¼ r0= 3; and l ¼ ½ðD1 À D2Þ=2 cot a; so we can substitute k and l into (13.26) and get the maximum

622 13 Extrusion extrusion pressure or ram pressure, rL; at the entry to the As before, assume that the exit plane is at x ¼ 0; and x chamber, i.e. at the start of the extrusion as follows: increases from the exit plane to the entry plane to the rL ¼ \" þ B(D12B ) chamber. Suppose xS is the distance measured from the exit r0 1 B À1 plane towards the entry to the chamber, where the transition D2 ! ð13:27Þ from sliding to sticking occurs. To get the mixed mode of þ pﬃ2ﬃ f2L À ðD1 À D2Þ cot ag friction along the chamber wall, the distance xS must lie 3D1 between the entry to the die and that to the chamber, i.e. L [ xS [ l: Thus, frictional condition within the extrusion ) Maximum ram load at the start of extrusion, chamber is such that sliding friction prevails from x ¼ rL\"pÀD21=4Á( l to x ¼ xS; where s ¼ lp; and full sticking friction occurs PL ¼ 2B ) # from x ¼ xS to x ¼ L; where s ¼ k: ¼ À1 lÞ pD12 r0 1þB D1 þ pﬃ4ﬃ ðL À So; at the point of x ¼ xS; B D2 3D1 4 s ¼ lp ¼ k ¼ pr0ﬃﬃ ; ð13:28aÞ 3 Or, p ¼ prﬃﬃ0 \" þ ( 2B ) ð13:29aÞ 1 B B D1 À1 3l Or; PL ¼ r0 D2 Let the horizontal axial stress rx ¼ rS at x ¼ xS; so, from þ pﬃ2ﬃ f2L À ðD1 À D2Þ cot ! pD21 (13.19d), it follows that 3D1 ag 4 &' 4l ð13:28bÞ rS À r0 ¼ ðrl À r0Þ exp D1 ðxS À lÞ ð13:29bÞ &' 13.6.2.3 Mixed Sticking–Sliding Friction 4l in Chamber Or; rS ¼ ðrl À r0Þ exp D1 ðxS À lÞ þ r0 ð13:29cÞ The extrusion or ram pressure, rL; and correspondingly, the From Von Mises’ yielding criterion within the chamber pressure exerted by the wall of the extrusion chamber on the work-piece, p; gradually decreases as the ram travels from [see (13.13b)], p ¼ rS À r0: Hence, from (13.29b) we get the entry to the chamber towards the entry to the die. Fre- quently, the frictional conditions along the wall of the &' extrusion chamber are intermediate between full sticking and 4l sliding so that there may be sliding friction at and near the p ¼ ðrl À r0Þ exp D1 ðxS À lÞ ð13:29dÞ entry to the die where the pressure, p; is lower, but at some distance away from the entry to the die towards the entry to Now, equating (13.29a) and (13.29d) for p gives a solu- the chamber the pressure increases tpo ﬃﬃa point where tan- gential frictional stress, s ¼ k ¼ r0= 3: In the following tion for the location of the boundary between slipping and analysis, direct extrusion of a cylindrical work-piece through a conical converging die is considered, in which Coulomb’s sticking friction, which is derived below: law of sliding friction is assumed to act along the die wall, but full sticking friction is assumed to develop starting from &' the entry to the chamber up to a certain distance within the 4l ¼ prﬃ0ﬃ chamber in the forward direction of extrusion creating a ðrl À r0Þ exp D1 ðxS À lÞ 3l ; mixed sticking–slipping frictional condition at the contact & ' surface between the extrusion chamber and the work-piece. )Orx, Se¼xp4Dl1D4l1lnðx1S =Àpl3ﬃÞﬃ l¼rr0lpÀﬃ3ﬃ 1lðrr!l0þÀ r0Þ 1=p3ﬃﬃ lrr0l It is further assumed that coefﬁcient of friction, l; along the l ¼ À 1 die wall is the same as that at the billet–chamber interface, and that the mixed sticking–slipping frictional condition is ð13:30Þ developed due to high container pressure, p; although such frictional condition may also develop due to higher l at the Substituting rl from (13.15) and l ¼ ½ðD1 À D2Þ=2 cot a; billet–chamber interface than at the billet–die interface. into (13.30), we get xS ¼ ln*1=p3ﬃﬃ l\"1 þ B(D12B ) À #+ D1 À1 1 4l B D2 þ D1 À D2 cot a 2 ð13:31Þ

13.6 Estimation of Extrusion Load 623 Again, substituting xS from (13.30) into (13.29c), we get Now, substituting xS from (13.31) into (13.34b), we get rr00ÞÞ&ex1p=pD43ﬃlﬃ1lD4lr1r0llnÀ&11=p'ﬃ3ﬃþlrr0r0l '! ( p1ﬃﬃ þ 1 þ p4ﬃﬃL À1 3l 3D1 rS ¼ ðrl À þ r0 rL ¼ r0 ¼ ðrl À ! p1ﬃﬃ n1 o À 3l ln pﬃﬃ h BÞ=Bg ðD1=D2Þ2BÀ1 i 3l fð1 1 þ À ) rS ¼ prﬃ0ﬃ ½ðrl=r0Þ À 1 þ r0 ¼ r0 p1ﬃﬃ þ 1 À 2ðD1 pÀﬃﬃD2Þ cot a ) 3l½ðrl=r0Þ À 1 3l 3D1 ð13:32Þ ð13:35Þ With the free-body equilibrium approach in the cylin- ) Maximum ram load at the start of the extrusion, drical portion of the extrusion chamber, the inﬂuence of pÀD21 Á partly sliding and partly sticking friction of the chamber on PL ¼ rL =4 extrusion stress is estimated. As before, let us consider the transverse slice of ﬁxed diameter, D1; with an inﬁnitesimal ð13:36Þ thickness, dx, within the extrusion chamber, i.e. within ðL À lÞ portion in Fig. 13.15. The slice is subjected to axial 13.6.3 Direct Extrusion Through Square Die and radial compression, and to either sliding frictional force, according to Coulomb’s law, or sticking frictional force With ﬂat-faced or square die, where semidie angle is very acting along the chamber wall. Now for x ! xS; where full sticking friction prevails, horizontal equilibrium of forces on large, and high friction between the material and the chamber the free metal slice in the axial direction x; is expressed by (13.24), from which the following (13.33a), similar to wall, a dead zone as shown in Fig. 13.16a, develops where no (13.25a), can be obtained as shown below: ﬂow of material takes place. Assumption in this analysis is 4kx that the dead zone can be approximated by a semicone angle D1 of 45°, i.e. a ¼ 45: Further, it is assumed that l tan a is rx ¼ þ C3 ð13:33aÞ neglected as in the previous Sect. 13.6.1, and the Von Mises’ yielding criterion will be given by (13.13). This assumption where C3 is an integration constant, for x ! xS: introduces some inaccuracy in the following analysis, but the Since at the plane of transition from sliding to sticking, solution becomes unwieldy without the assumption. i.e. at x ¼ xS; rx ¼ rS; then from (13.33a) we get The material which is being extruded can be divided into two deformation zones as shown in Fig. 13.16a: (i) section C3 ¼ rS À 4kxS ð13:33bÞ AA to BB, where the ﬂow of material is considered as a rigid D1 body motion, and (ii) section BB to CC, where the ﬂow is After substituting C3 from (13.33b) into (13.33a), we get analogous to that in extrusion through conical converging die, as discussed in Sect. 13.6.1. The entry edge ‘CC’ of the rx ¼ rS þ 4k ðx À xSÞ square die is considered as the origin of the coordinate D1 system, shown by x ¼ 0 in the diagram shown in p4ﬃrﬃ 0 ð13:33cÞ Fig. 13.16a, where x is the horizontal axis that increases 3D1 ¼ rS þ ðx À xSÞ from the entry edge of the die, i.e. section CC, to the entry to the chamber, i.e. section AA. Figure 13.16b shows a trans- At the entry to the chamber where, x ¼ L; since the verse element in the region BB–CC (in l portion), along with horizontal axial stress, rx ¼ rL (the maximum ram pressure at the start of extrusion), so from (13.33c) it follows: the stresses acting on it. The dead zone formed in extrusion through ﬂat-faced or rL ¼ rS þ p4ﬃrﬃ 0 ðL À xSÞ ð13:34aÞ square die makes a die-like channel with a semicone angle 3D1 which is approximated to be 45°, i.e. a % 45: As a ¼ 45; so B ¼ l cot a ¼ l cot 45 ¼ l: Hence, from (13.15), the Substituting rS from (13.32) into (13.34a), we get the ram pressure rl at the entry plane of the dead zone, i.e. at the section BB in Fig. 13.16a, can be written as equation for rL as: Ram pressure at entry plane of the dead zone, ! p1ﬃﬃ þ 1 þ 4ðpL ﬃÀﬃ xSÞ ( 2l ) rL ¼ r0 3l 3D1 ð13:34bÞ þ l D1 À1 r0 1 l rl ¼ D2 ð13:37Þ

624 13 Extrusion p (a) Extrusion chamber (b) (c) Ram μp τ 45º σL A B D + dD p D1 dx 45º σx + dσx σx D1 D2 σx D σx + dσx A C Dead zone μp p dx C B 45º l τ L p x=0 Fig. 13.16 a Details of square die extrusion of a cylindrical billet, BB to CC [see (a)], c stresses on a transverse metal-slice of ﬁxed b stresses on a transverse trapezoidal shaped metal-slice with diameter, D1; having an inﬁnitesimal thickness, dx, in section AA to BB inﬁnitesimal thickness, dx; and diameter, D and ðD þ dDÞ; in section [see (a)] Note that (13.37) will also give the ram or extrusion ) Maximum ram load at the start of extrusion, prD4L 12pÀrD012*=\"4Á1 pressure for indirect extrusion through square dies, where a PL ¼ l(D12l ) # dead zone with a semi-cone angle of 45 is assumed to form ¼ À1 1 þ D2 À ð13:39Þ on the die shoulders. l The stresses acting on a transverse slice of ﬁxed diameter, !) 2l D1; with inﬁnitesimal thickness, dx, in the region between exp D1 f2L À ðD1 À D2Þg þ1 the section BB and AA, i.e. within ðL À lÞ portion are shown in Fig. 13.16c. As before there may be three frictional 13.6.3.2 Full Sticking Frictional Condition conditions at the billet–container interface prior to the entry Considering full sticking friction at the billet–chamber plane (‘BB’) of dead zone, i.e. within ðL À lÞ portion. In the interface prior to the entry plane (‘BB’) of dead zone, i.e. following analyses, note that within ðL À lÞ portion, the maximum extrusion pressure or ram pressure, rL; at the entry to the chamber, i.e. at the start L ¼ the distance measured from the entry edge of the of the extrusion can be obtained by substituting rl from square die ‘CC’ to the entry to the chamber ‘AA’, i.e. the (13.37), and l ¼ ðD1 À D2Þ=2; and rx ¼ rL at x ¼ L; into length of the billet, and (13.25c): l ¼ the distance measured from the entry edge of the square die ‘CC’ to the entry plane of the dead zone ‘BB’, i.e. the length of the dead zone. 13.6.3.1 Coulomb’s Sliding Frictional Condition rL ¼ rl þ 4k ðL À lÞ ¼ D1 ( Considering Coulomb’s sliding friction at the billet–chamber l D1 2l ) À1 L D2 interface prior to the entry plane (‘BB’) of dead zone, i.e. r0 1þ D2 þ 4k À D1 À l D1 2 within ðL À lÞ portion, the maximum extrusion pressure or ram pressure, rL; at the entry to the chamber, i.e. at the start ð13:40aÞ of the extrusion can be obtained by substituting rl from pﬃﬃ Since k ¼ r0= 3 where r0 is the uniaxial ﬂow stress in (13.37), and l ¼ ½ðD1 À D2Þ=2 cot a ¼ ½ðD1 À D2Þ=2 compression, (13.40a) becomes cot 45 ¼ ðD1 À D2Þ=2; and rx ¼ rL at x ¼ L; into (13.19d): *\" 1þ ( 2l ) # \" l(D12l ) # l l D1 À1 1 r0 1 À1 D2Þg rL ¼ r0 À þ pﬃ2ﬃ D2 rL ¼ l þ 3D1 f2L À ðD1 À D2 !) 2l exp D1 f2L À ðD1 À D2Þg þ1 ð13:38Þ ð13:40bÞ

13.6 Estimation of Extrusion Load 625 ) Maximum ram load at the start of extrusion, estimated value of xS; there may be three different conditions rr0L\"pÀ1Dþ21l=l4Á (DD212lÀ1) of friction at the billet–container interface, and knowing the PL ¼ correct frictional condition at interface, selection of proper ¼ equations for ram load and/or stress is made, as explained ð13:41Þ below: þ pﬃ2ﬃ f2L À ðD1 À ! pD12 (a) If xS ! L; for extrusion through conical converging die, 3D1 D2Þg 4 the sliding friction exists over the entire billet–chamber interface, i.e. from x ¼ l to x ¼ L; in which case maxi- 13.6.3.3 Mixed Sticking–Sliding Frictional mum ram load and pressure are to be calculated, Condition respectively, from (13.22) and (13.21). For extrusion through square die, if xS ! L; the sliding friction exists Suppose xS is the distance measured from the entry edge over the entire interface from the entry plane of the dead ‘CC’ (Fig. 13.16a) of the square die towards the entry to the zone to the entry to the chamber, i.e. from x ¼ l to x ¼ chamber, where the transition from sliding to sticking L; in which case maximum ram load and pressure are to occurs. To get the mixed mode of friction at the billet– chamber interface within the region between the section BB be calculated, respectively, from (13.39) and (13.38). and AA, i.e. within ðL À lÞ portion (Fig. 13.16a), the distance (b) If xS l; for extrusion through conical converging die, xS must lie between the entry plane of the dead zone and the the full sticking friction exists over the entire billet– entry to the chamber, i.e. L [ xS [ l: Substituting rl from container interface, i.e. from x ¼ l to x ¼ L; in which (13.37), and l ¼ ðD1 À D2Þ=2; into (13.30), we get the case maximum ram load and pressure are to be calcu- sliding–sticking transition boundary as lated, respectively, from (13.28) and (13.27). For xS ¼ ln*1=p3ﬃﬃ l\"1 þ ( 2l ) À #+ þ ðD1 À D2Þ extrusion through square die, if xS l; the full sticking D1 l l D1 À1 1 2 friction exists over the entire interface from the entry 4l plane of the dead zone to the entry to the chamber, i.e. D2 ð13:42Þ from x ¼ l to x ¼ L; in which case maximum ram load and pressure are to be calculated, respectively, from Substituting xS from (13.42) into (13.34b), we get the (13.41) and (13.40). equation for the maximum extrusion pressure or ram pres- (c) If L [ xS [ l; for extrusion through conical converging sure, rL; at the entry to the chamber, i.e. at the start of the die, the mixed sliding–sticking frictional condition pre- extrusion as: vails at the billet–container interface, i.e. full sticking ( p1ﬃﬃ þ 1 þ p4ﬃﬃL friction occurs from x ¼ L (the entry to the chamber) to 3l 3D1 x ¼ xS (sticking–sliding boundary) and the rest portion rL ¼ r0 is subjected to sliding friction, in which case maximum ram load and pressure are to be calculated, respectively, + from (13.36) and (13.35). For extrusion through square À p1ﬃﬃ ln pﬃﬃ h n1 o i À 2ðDp1ﬃﬃÀ D2Þ 3l 3l fð1 lÞ=lg ðD1=D2 Þ2lÀ1 1 3D1 þ À ð13:43Þ die, if L [ xS [ l; sliding friction occurs from x ¼ l (the entry plane of the dead zone) to x ¼ xS (sliding–sticking ) Maximum ram load at the start of the extrusion, boundary) and full sticking friction occurs from x ¼ xS pÀD21 Á to x ¼ L (the entry to the chamber), thus creating a PL ¼ rL =4 mixed sliding–sticking frictional condition along the ð13:44Þ container wall from the entry plane of the dead zone to the entry to the chamber. In this case, maximum ram load and pressure are to be calculated, respectively, from 13.6.4 Selection of Proper Equation for Ram (13.44) and (13.43). Load and Stress To select proper equations for ram load and/or pressure, the 13.7 Strain Rate in Extrusion value of xS has to be ﬁrst evaluated using either (13.30) for extrusion through conical converging as well as square die, The ﬂow stress of metal in hot extrusion depends upon the or (13.31) for extrusion through conical converging die and strain rate, and the strain rate increases with increasing ram (13.42) for extrusion through square die from the given speed in extrusion. It must be noted that the strain-rate value of the coefﬁcient of friction l at the interface. sensitivity of the ﬂow stress (see Sect. 1.11.1 in Chap. 1) is Depending on the given values of l; and L; and the

626 13 Extrusion high at elevated temperature (it is negligible at room tem- Since the extrusion ratio R ¼ A1=A2 ¼ ðD1=D2Þ2; then (13.48a) changes to perature) and increases with increasing temperature. There- fore, the ﬂow stress of metal increases more with increasing ln D1 strain rate at higher temperature than at lower temperature, e_ tE ¼ 12vramD21 tan a ð13:48bÞ D31 À D32 D2 which results in the requirement of a higher speciﬁc pressure for hot extrusion. Hence, in calculating extrusion load, one When extrusion ratio R is very high, i.e. D1 ) D2; then we can neglect D2 in (13.48a). For a square die with poor must obtain the ﬂow stress of the material corresponding to lubrication, where a dead zone develops, the semicone angle a % 45: Thus, for high extrusion ratios and square dies with the strain rate prevailing at the time of hot extrusion. Before poor lubrications, (13.48a) reduces to give a simpliﬁed form the ﬂow stress is used to calculate the hot-extrusion load, the of mean true strain rate for extrusion: strain rate in the actual extrusion process must be calculated. In extrusion, local true strain rate varies. Extrusion true strain rates are found to have local maxima near the exit from the die on the surface, and along the centre-line of the e_ tE 6vram D1 extrusion. Therefore, it is necessary to calculate an effective ¼ ln R ð13:49Þ or average true strain rate for extrusion. In case of a con- verging die, it is more usual to evaluate the average true strain rate for extrusion in terms of time, tE; required for 13.8 Extrusion Defects material to travel through a truncated conical volume of Extrusion defects are the result of non-uniform ﬂow of deformation zone. The deformation zone in extrusion at the metals in the container. It is always desired to have parallel metal ﬂow, in which the surface skin of the billet becomes entry plane of the stock within die is bounded by a circle of the surface skin of the extrudate. Another type of metal ﬂow is shear metal ﬂow, in which the surface skin of the billet diameter equal to the original diameter, D1; of the cylindrical penetrates into the mass of the billet and a stagnant stock before extrusion and that at the exit plane of die by a dead-metal zone is formed at the die shoulder. This dead-metal zone is held back in the chamber as discard. smaller circle of diameter, D2; of the product after extrusion. Shear metal ﬂow is not desirable, because it hinders effective If semicone angle of the die is a; the volume of the truncated lubrication of the die and can cause surface defects and interior laminations in the extrudate. Susceptibility to defects conical deformation zone in extrusion with virtual semi-cone increases with the degree of shear metal ﬂow, i.e. with the heterogeneous nature of the ﬂow patterns—that is in the apex angle, a; will be order of ‘S’, ‘A’, ‘B’ and ‘C’ types of ﬂow patterns (see Sect. 13.4). Various defects in extrusion are described p \"D12 D1=2 2 # below: 3 2 tan a D2 D2=2 VE ¼ À (1) Rear-end Hollow Pipe: 2 tan a ð13:45Þ During extrusion, when the length of the billet left in the p ÀD31 D32Á container is about one-fourth of its diameter, a hollow pipe ¼ 24 tan a À or channel is created axially in the rear end of the extruded product. When extrusion is carried out to the point men- If the ram velocity is vram; the volume of material tioned above, the back-end axial hole or funnel is created extruded per unit time is due to the rapid radial ﬂow of metal into the die. The back-end hollow pipe almost always occurs in the extruded Vt ¼ vram pD21 ð13:46Þ metals. This pipe may extend some distance inside the 4 extrusion from its back end. So, the pipe must be discarded from the extrusion and hence the productivity falls. Friction Now, the time required to ﬁll the volume VE of the between billet and container causes a deeper pipe to form in deformation zone is obtained from (13.45) and (13.46) as the extrusion earlier in the cycle. The length of this piping follows: defect can be minimized by inclining the face of the ram at an angle to the ram axis. tE ¼ VE ¼ D13 À D23 ð13:47Þ Vt 6 tan avramD21 If A1 is the initial cross-sectional area of the billet before extrusion and A2 is the ﬁnal cross-sectional area of the billet after extrusion, then the true strain in extrusion is given by (13.4). Hence, the time-average extrusion true strain rate, e_tE; is obtained from (13.47) and (13.4) as follows: e_ tE ¼ e ¼ 6vramD12 tan a ln R ð13:48aÞ tE D13 À D32

13.8 Extrusion Defects 627 (2) Back-end Internal Oxidized Pipe: Friction between the billet and container wall in the direct Fig. 13.17 Fir-tree cracking (repetitive transverse surface cracking) extrusion of a billet resists the ﬂow of metal at the outer from heavy die friction in extrusion surface of the billet, which causes the centre of the billet to move faster than its periphery. After the extrusion of about Fig. 13.18 Development of peripheral large grains two-thirds of the billet is over, the metal at the peripheral region ﬂows towards the central region of the billet under the product useless. In hot extrusion, this type of crack is usually action of ram and is extruded through the die oriﬁce near the intercrystalline and is associated with hot shortness, the most axis of the extruded product. Since the outer surface metals common cause of which is the use of higher ram speeds for of the billet are often oxidized, this type of ﬂow causes the that extrusion temperature. If the ram speed, i.e. the strain oxide layers to enter into the extrusion and produces internal rate, is too high, it can produce adiabatic deformation con- oxide stringers at the back end, which can be regarded as a ditions. This, in turn, results in the rise of temperature that back-end internal oxidized pipe. On viewing the transverse can exceed the solidus of the metal and produce incipient section of the product, this defect will appear as an annular melting. The ﬁr-tree cracking is more likely to occur in those ring of oxide which renders the product unacceptable. The alloys where the difference between the lower limit of greater the friction between the billet and container wall, the hot-working temperature and the solidus is not large. If the more is the tendency towards the formation of this defect. If ram speeds are reduced in order to avoid incipient melting, a hot billet is placed in a cold chamber for extrusion, the there may be problem associated with formation of very outer surfaces will be chilled and the deformation resistance coarse grains at the surface, as shown in Fig. 13.18. This of this section will rise. Therefore, the central region of the may occur if the surface of the billet is caused to cool just billet will have a greater tendency to extrude before the below the recrystallization temperature by contact with the surface layers, which will increase the tendency to form this colder wall of the chamber. Heat produced by extrusion extrusion defect. ﬂows from the central region of the extruded product to the surface layers and raises the surface temperature from just One way to eliminate this extrusion defect is to halt the below to just above the recrystallization temperature. The extrusion operation at that point where the oxide layers of extent of cold work imparted to the surface layers is such surface start to enter into the die and then discard the rest of critical that it results in the formation of extremely coarse the billet remaining in the container. But this procedure is not suitable on economic grounds because the amount of the remainder billet to be discarded in order to avoid this extrusion defect may be as much as 30%. An alternative method, which is frequently used in order to avoid the for- mation of this extrusion defect, is to use a pressure pad (a follower block) between the ram and the billet which is slightly smaller in diameter than the inside diameter of the extrusion chamber. As the ram pushes the pressure pad forward, it scalps the billet and a thin skull of the billet, which contains the oxidized surface layers, is left on the wall of the container. (3) Fir-tree Surface Cracking: In extrusion, when interface friction is high due to break- down or discontinuity of the lubricant ﬁlm or any other reasons, the ﬂow of the periphery of the billet is restrained and the centre of the billet moving relatively fast through the die stretches the periphery in the forward direction. This results in the development of longitudinal tensile stresses at the surface which causes surface cracking. The severity of surface cracking ranges from a badly roughened surface like snake skin or ﬁsh skin to repetitive transverse cracking called ﬁr-tree cracking (Fig. 13.17), which makes the

628 13 Extrusion Lc is the contact length between the die and the work-piece in the deformation zone, as shown in Fig. 13.20a, b. This can be explained by (10.77), which is: D ¼ ÀDL1mcþ¼pDD1ﬃﬃﬃ11ﬃÀﬃﬃÀþﬃﬃﬃrﬃDDÁ222 sin a ¼ sin a r Fig. 13.19 Central bursting or chevron cracking where the reduction of area, r ¼ 1 À ð1=RÞ: This shows that as a increases, sin a increases and so, Dm=Lc increases. grains at the surface on recrystallization. Very low ductility Again, as R decreases r decreases, which causes Dm=Lc to of this coarse-grained metal again makes it unacceptable. increase. For a large Dm=Lc, the deformation is localized This problem in the extrusion of aluminium alloys was near the surface region and the relatively undeformed central investigated by C.S. Smith (Smith 1949–1950) who sug- part is stretched in the forward direction by the surface gested for maintaining a very close control jointly over the layers. This causes to develop the secondary tensile stresses ram speed and the extrusion temperature to produce an at the centre which results in the development of centre-burst acceptable material. At lower temperatures where hot defect. The danger of chevron cracking is less when shortness cannot take place, ﬁr-tree cracking is believed D ¼ Dm=Lc 2: (Wilcox and Whitton 1959–1960) to occur due to momen- tary sticking in the die land and the sudden development of It has been further shown (Gurney and DePierre 1974) that pressure, and then breakaway. if the friction at the tool–billet interface is low chevron cracking occurs, while high interface friction will result in the (4) Centre Burst or Chevron Cracks: production of a sound product. Since high friction restrains the movement of relatively more deformed surface layers and Centre burst or chevron cracking is an internal crack which reduces the velocity gradient along the cross-section of a can occur when too large a deformation gradient along the billet, the formation of secondary tensile stresses at the centre cross-section of a billet develops ‘secondary tensile stresses’ responsible for the centre-burst defect is avoided. in the centre of the billet. Chevron cracking or central bursting is shown in Fig. 13.19, and how centre-burst (5) Longitudinal Laminations: defects can develop during extrusion as well as during severe cold drawing operations and form a series of Discontinuity in the lubrication ﬁlm will assist to initiate cone-shaped internal ruptures is shown in Fig. 13.20a, b. shear zones during extrusion. Lubricant ﬁlm (and also oxide This defect can typically occur where the semicone die layer or impurities) may ﬂow from the billet surface along angle, a; is large and the extrusion ratio, R; or the reduction this shear zone and be migrated into the interior of the ratio in drawing, R; (the ratio of the initial cross-sectional extrusion to appear as longitudinal laminations in the area of the work-piece to the ﬁnal cross-sectional area after product. drawing) is low, i.e. Dm=Lc is large, in which Dm is the mean diameter of the work-piece (arithmetic average of initial (6) Impurity Streaks: diameter D1 of billet and ﬁnal diameter D2 of extrudate) and This defect consists of impurities aligned along the axis of the ﬁnal product. This impurity streaks arise due to the surface impurities in the original billet. (a) (b) Lc Dm α D2 α D1 Lc Dm Fig. 13.20 Development of chevron crack, which is related to semicone die angle, a; and the ratio Dm=Lc. a In drawing, b in extrusion

13.9 Impact Extrusion 629 13.9 Impact Extrusion die and extrudes the metal as a thin ﬁlm in the opposite direction, thus forming thin tube round the punch as the Impact extrusion is a cold extrusion process. In this process, metal has no other method of escape. This process is anal- compressive force is used to form very thin-walled hollow ogous to inverted extrusion. tube or pipe of short length by striking a metal blank placed on a die, with a punch at a high velocity, e.g. collapsible Direct impact extrusion, known as the ‘Hooker process’, tooth paste tubes from soft metals and aluminium cans with as shown in Fig. 13.21b, is a forward extrusion technique very thin walls. basically similar to the tube extrusion with a mandrel attached to the pressure pad, but it operates at a very high speed. The There are two basic methods of impact extrusion. In metal blank is usually desired to be of cup shaped. The punch indirect impact extrusion, as shown in Fig. 13.21a, the die with a mandrel ﬁts the inside diameter of the cup. As the has the correct outside diameter of a tube, which is to be punch strikes the cup at a high velocity, pressure is exerted on produced, and the punch has the correct inside diameter of the cup by the shoulder of the punch and the cup is extruded that tube. A metal blank, usually made of aluminium or other into a tube. To produce thick-walled tubing, this process softer metal, coated with a lubricant is placed in a shallow requires less pressure than the inverted extrusion. This process die and a cylindrical punch strikes it at a high velocity. The is used for copper and brass tubes and can produce tubes with punch descends at a high speed, compresses the metal in the closed ends and tapering wall thickness. Fig. 13.21 a Indirect impact (a) extrusion; b Hooker process (direct impact extrusion) Punch Blank Die Prior to extrusion After extrusion (b) Punch Blank Die Prior to extrusion After extrusion

630 13 Extrusion Impact extrusion (direct or indirect) is performed typi- Maximum extrusion pressure Direct extrusion cally on high-speed mechanical vertical presses at produc- with square die tion rates as high as two parts per second. Though impact extrusion is generally performed cold, considerable heating Hydrostatic extrusion results from the high speed of deformation. Impact extrusion process is typically used to produce very thin-walled tubular Decreasing semi-cone angle of die shapes, in which tube-wall thicknesses may be as thin as for hydrostatic extrusion 0.005 times of their diameters. To achieve a uniform tube-wall thickness, the concentricity of the punch and the Extrusion ratio (R) metal blank must be maintained. The mechanical properties Fig. 13.22 Schematic plot of maximum extrusion pressure versus of impact extruded products are good. One important extrusion ratio for direct and hydrostatic extrusion and the effect of advantage of impact extrusion is that complex ﬁnished parts decreasing semicone die angle, a; on hydrostatic extrusion pressure can be produced in a single operation, without the need of expensive machining operation. Although this process is restricted to a variety of softer non-ferrous metals such as Pb, Sn, Al and Cu, but low-carbon steel, and even some alloy steels, can be cold-impact-extruded in useful sizes having good mechanical properties. Compressed air or even actual explosives have also been introduced to enhance the speed in impact extrusion, which appears to be suitable for cold-impact-extrusion even of very brittle alloys. 13.10 Hydrostatic Extrusion deformation is carried out under high hydrostatic pressure. On the other hand, hydrostatic pressure in ﬂuid extrusion ‘Hydrostatic’ extrusion, also known as ‘ramless’ or ‘ﬂuid’ or process cannot be raised to a very high level, because ﬂuids ‘hydraulic’ extrusion, was developed in the early 1950s. In usually oils, which act as pressure media in hydrostatic the process, the billet is forced through the die by means of a extrusion, start to solidify at very high pressures. Moreover, liquid acting as a pressure medium instead of the direct the container must be strong enough to withstand high application of the load with a ram. Friction between the billet pressures. Due to the above reasons, the maximum hydraulic and the wall of container is totally absent, because there is no pressure is currently around 1.7 GPa. Hydrostatic extrusion contact between them. The pressurized ﬂuid is transmitted to is usually carried out at room temperature, though the pos- the die–billet interface and often creates hydrodynamic sibility of warm and hot hydrostatic extrusion has been cited lubrication, which signiﬁcantly reduces friction. Due to in literature (Rowe 1977). The typical ﬂuid media used to signiﬁcant reduction in friction at the die–billet interface, transfer the pressure at room temperature are vegetables oils, conical converging die with a very low semicone angle, a % particularly castor oil because its viscosity is not inﬂuenced 20; is generally used in hydrostatic extrusion. Low semi- signiﬁcantly by the applied pressure and it is a good lubri- cone angle of die reduces the load required for internal cant. If hydrostatic extrusion is carried out at elevated tem- redundant deformation (see Fig. 13.10). Lower friction and perature, the ﬂuid media used to transfer the pressure are reduction in redundant deformation both cause to reduce the molten glass, polymers and waxes; these lubricants also act extrusion pressure. The reduction of maximum extrusion as thermal insulators and assist to maintain the temperature pressure in hydrostatic extrusion with conical die compared of the billet during extrusion. Hydrostatic extrusion has been to direct extrusion process with square die is shown in successfully applied to a variety of metals and polymers, Fig. 13.22. This ﬁgure further shows that the hydrostatic solid shapes, tubes and other hollow shapes, and honeycomb extrusion pressure decreases with decreasing semicone die proﬁles. angle, a; this is due to the decrease in stress required for the internal redundant deformation. Typical pressure in this There are two ways to provide the pressure using liquid process is normally on the order of 1.4 GPa. When a high lubricant: hydrostatic pressure is present the tensile stresses are reduced below the critical value for cracking, but at the same (1) Ram-type hydraulic extrusion: The pressure is provided time the ﬂow stress of material is not affected. Further, less with the ram of a press forced into the extrusion damage is created to the material during deformation when chamber, as shown in Fig. 13.23. The ram type is a constant rate extrusion due to the constant speed of the ram.

13.10 Hydrostatic Extrusion 631 Top cover Ram O ring Air-release Oil Mitre ring Pressure from pump O ring chamber screw O ring Chamber Die Bottom Billet cover Table Die Oil at high pressure, Fig. 13.24 Pump-type or ramless hydraulic extrusion chamber, acting as pressure- showing conventional hydrostatic extrusion process transmiting fluid (b) In conventional extrusion, the radial pressure on the Oil at Pressure- container, depending on the alloy and the extrusion low relief valve conditions, is 20–80% lower than the axial pressure pressure applied by the ram. In hydrostatic extrusion, the axial and radial pressures are equal. So, in hydrostatic extru- Fig. 13.23 Ram-type hydraulic extrusion chamber, showing differen- sion, the tooling has to withstand a signiﬁcantly higher tial pressure hydrostatic extrusion process pressure and more stringent requirements have to be placed on the design of tooling and the material used for tooling. (2) Pump-type or ramless hydraulic extrusion: The pressure 13.10.2 Conventional Hydrostatic Extrusion is provided through a pump, with or without a pressure intensiﬁer, by supplying oil from pump through an inlet It is the simplest method of hydrostatic extrusion, also in the extrusion container surrounding the billet and the known as ‘hydrostatic extrusion without back pressure’, die as shown in Fig. 13.24. The pump type is a shown in Fig. 13.24, in which the metals extruded through constant-pressure extrusion. the die into the atmosphere, as performed in conventional extrusion. Extrusion begins as soon as the hydrostatic 13.10.1 Basic Difference Between Hydrostatic pressure has reached a high value depending on the ﬂow and Conventional Extrusion stress of the billet material and the extrusion ratio. A conical die is used to allow a ﬁlm of lubricant to develop. The (a) In conventional extrusion, the billet is in contact with container with the pressure-transmitting ﬂuid is sealed with three components—die, chamber, ram or dummy block, high-pressure seals at the ram and the die. on the other hand, in hydrostatic extrusion, the billet is surrounded by the liquid and is in contact with the die This process is suitable for applications to (a) ductile only. When hydrodynamic lubrication prevails, even materials like aluminium and copper and (b) moderately contact with the die is avoided. brittle materials. This process is not suitable for brittle materials, because the following factors may cause cracking

632 13 Extrusion of the brittle material when it is extruded by conventional pressure instead of being extruded into the atmosphere, as hydrostatic extrusion process: performed in conventional hydrostatic extrusion. Due to the presence of the high ﬂuid pressure in the product receiving (i) Hydrostatic extrusions are most frequently performed chamber, the ductility of a brittle material is improved. Thus, at room temperature, although preheated billets are also a crack-free extrusion can be produced from a brittle material used. by this process. The ﬂuid pressure in the receiving chamber must be lower than that in the extrusion chamber, so that a (ii) Fall in the hydrostatic pressure towards zero at the die resultant force is exerted on the material, forcing it to ﬂow exit. out through the die. The back pressure in the receiving chamber causes to increase the ﬂuid pressure in the extrusion 13.10.3 Differential Pressure Hydrostatic chamber needed to carry out the extrusion of billet. Hence Extrusion for the same extrusion ratio, die angle, pressure-transmitting ﬂuid and frictional condition, the same billet material would When tensile tests are conducted on a number of specimens require higher ﬂuid pressure in the extrusion chamber for the of a material in a chamber under varying environmental high differential pressure hydrostatic extrusion than for the con- pressure and reduction in area at fracture is measured for ventional hydrostatic extrusion. those specimens as a function of mean environmental pres- sure, the measured reduction in area at fracture for that 13.10.4 Advantages of Hydrostatic Extrusion material can be plotted against mean pressure. Such kinds of plot for different material are shown in Fig. 13.25 (Boulger • Since there is no container–billet friction, so the absence and Wilcox 1964). The pressure at which the drastic change of friction reduces the extrusion pressure in hydrostatic in ductility occurs and improved ductility is obtained is extrusion, where after attaining the breakthrough pres- different for different materials (Fig. 13.25). The pressure, sure, the curve of the extrusion pressure versus ram travel not the temperature, is used here to cause the brittle-to- is nearly ﬂat like that for indirect extrusion. ductile transition. Figure 13.25 shows that the ductility of material A > that of material B > that of material C. Hence, • Since the billet is subjected to uniform ﬂuid pressure, it we can conclude that the effect of hydrostatic pressure is to does not upset to ﬁll the bore of the container as it would improve the ductility of the material being deformed. in conventional extrusion. This means that billets of a large length-to-diameter ratio may be extruded by ‘Hydrostatic extrusion with back pressure’ or ‘differential hydrostatic extrusion in contrast to conventional pressure hydrostatic extrusion’ (also known as ‘ﬂuid-to-ﬂuid extrusion. extrusion’) process is an extension of hydrostatic extrusion process which was developed to extrude brittle material or • Since friction is nearly absent or drastically reduced at less ductile material. By this process, brittle materials have the die–billet interface by the presence of the pressurized been successfully extruded (Pugh and Low 1965). In this ﬂuid acting as lubricant, it is possible to use die with a process, shown in Fig. 13.23, metal is extruded into a very low semicone angle, a % 20; which greatly reduces hydraulic pressure chamber with a sufﬁciently high ﬂuid the internal redundant deformation, and thus, the extru- sion pressure is reduced. Reduction in area at fracture Material A 100 • There is no deformation of the billet like billet buckling or barrelling in hydrostatic extrusion due to uniform Material B hydrostatic pressure in the extrusion chamber. For this reason, billets of irregular cross-section may be used; Material C even coils of wire can be extruded by hydrostatic extrusion. 0 Mean pressure • As the hydrostatic extrusion is most frequently performed at room temperature, the extruded product has a good Fig. 13.25 Schematic illustration of the effect of mean pressure on surface ﬁnish and dimensional accuracy. ductility • Very effective lubrication by the presence of the pres- surized ﬂuid causes very less wear of the tools. • In differential pressure hydrostatic extrusion process, a crack-free extrusion can be achieved from a low ductility or brittle billet due to improvement in the ductility by the presence of the high ﬂuid pressure in the receiving chamber.

13.10 Hydrostatic Extrusion 633 13.10.5 Disadvantages of Hydrostatic Extrusion • The differential pressure hydrostatic extrusion process is expensive and not suitable for industrial application. • Compared to the conventional extrusion, the tooling has to withstand a signiﬁcantly higher radial pressure in the hydrostatic extrusion and more stringent requirements have to be placed on the design of tooling and the 13.11 Seamless Tube Production material used for tooling. by Extrusion • Extrusion pressures up to 3.1 GPa and up have been utilized for conventional extrusion. Oils, acting as pres- For producing seamless tubes, extrusion is an ideal process, sure media in hydrostatic extrusion, start to solidify at because it may produce tubes with dimensional tolerances as such high pressure of 3.1 GPa. So, the maximum ﬂuid close as those obtained by cold drawing (Morris 1961– pressure of around 1.7 GPa currently exists because of 1962). There are two ways to produce tubes by extrusion: the requirement (a) of the strong container that can sus- tain high pressures and (b) that the ﬂuid does not solidify (1) By means of an external long mandrel; or at high pressures. (2) Without using a normal external mandrel. • The pressurized oil limits the hot-working with the hydrostatic extrusion process. When a mandrel is used, there are three possible kinds of • Inherent limitation to hot-working and high ﬂuid pressure mandrel arrangement, any one of which can be used in this in the hydrostatic extrusion limit the obtainable extrusion method: ratio, for example, less than 20:1 for steel and about 200:1 for soft metals like aluminium, whereas extrusion • Fixed. ratios reach about 40:1 for steel and may be as high as • Floating. 400:1 for aluminium in case of conventional extrusion. • Piercing. • Every billet has to be tapered to match the die angle for sealing at the start of the hydrostatic extrusion. So, Fixed and ﬂoating types of mandrel are attached to the laborious preparation or shaping of each billet is end of the extrusion ram, as shown in Fig. 13.26a, b. When required. these types of mandrel are used, a hollow billet of round • Length of extruded product is limited by the length of cross-section is used as a starting material and placed in the back pressure chamber in the differential pressure extrusion chamber. Prior to extrusion, an axial hole in the hydrostatic extrusion process. billet may be created by casting, by drilling or by hot Fig. 13.26 Mandrel attached to (a) the end of the extrusion ram, a ﬁxed mandrel, b Floating Fixed mandrel mandrel Dummy bloack Extrusion ram (b) Floating mandrel

634 13 Extrusion 21 3 4 56 Piercing mandrel Dummy block Extrusion ram Fig. 13.28 Piercing mandrel Fig. 13.27 Direct extrusion of a hollow round billet for formation of is that its internal wall will be oxidized during heating, seamless tube; 1 wall of extrusion chamber; 2 ram; 3 hollow round which may produce a tube having an oxidized inner surface. billet; 4 die; 5 extruded tube; 6 front end of ﬁxed or ﬂoating mandrel A more satisfactory method to produce extruded tube is to piercing using piercing mandrel in a separate press. The use a solid ingot or billet with a corrugated or round deformation resistance of the billet should be the same cross-section, which is pierced and extruded in one step everywhere over its cross-section and the axial hole must be using a piercing mandrel in a double-action extrusion press. concentric. The front end of the ﬁxed or ﬂoating mandrel The piercing mandrel is shown in Fig. 13.28. This technique passes freely through the whole hollow billet and die is widely used for non-ferrous metals and has been applied opening and projects a deﬁnite distance past die. This forms to metals exhibiting particularly poor plasticity. Most fre- an annular clearance between the mandrel and the wall of die quently, the piercing mandrel is driven by a separate oriﬁce. Upon travel of the ram together with the dummy hydraulic cylinder from the one which operates the ram. The block in the direction towards the die, the billet is forced piercing mandrel moves independently but coaxially with through the annular clearance and forms a tube, as shown in the ram. After the hot ingot or billet is placed into the Fig. 13.27. This annular clearance determines the wall extrusion chamber, the ﬁrst step in the operation is to upset it thickness of the extruded tube. The outside diameter of the with the ram while the piercing mandrel is pulled back into extruded tube equals the diameter of the die hole, and the the ram. Next, the pointed front end of the mandrel is forced inside diameter equals the diameter of the mandrel. Care into and through the billet to pierce it, as shown in must be taken to move the mandrel and ram in axial align- Fig. 13.29a. Subsequent advancement of the mandrel into ment with the extrusion chamber and the die so that con- the die ejects a metal plug through the die. Then the ram centric tubes are produced. Fixed mandrel tends to produce together with the dummy block travels in the direction eccentric tubes unless proper care is taken for accurate towards the die and extrudes the pierced billet over the alignment; on the other hand, ﬂoating mandrel centres itself, mandrel to form a tube, as shown in Fig. 13.29b. The main which leads to production of tubes with concentricity within advantages of this method are: 1%. Hence, the ﬂoating mandrel is preferred to the ﬁxed one in modern applications. The drawback of using hollow billet • Speedy operation. • Non-requirement of hollow billets, which obviously will cost more than the solid ones. • Elimination of separate piercing operation, which makes the process cheaper. • Minimization of the oxidation problem. (a) Piercing (b) mandrel Billet Fig. 13.29 Extrusion of a solid round billet to produce a seamless tube using a piercing mandrel, a piercing by forcing the pointed front end of the mandrel, b advancement of the ram for extrusion of the pierced billet over the mandrel to form a tube

13.11 Seamless Tube Production by Extrusion 635 Mandrel seamless tube under the high pressures that are present in the die-exit zone. The mandrel forms the inner surface of the Spider arms tube, while the die opening forms the outer surface. Since the separate metal segments are pressure welded within the Fig. 13.30 A three-aperture bridge die consisting of three thin spider die, where they are not exposed to air and remain clean, a arms holding a short mandrel in position in the die oriﬁce perfectly sound weld will be obtained. This solid-state weld, if properly made, is undetectable in either appearance or The disadvantages are that the piercing presses are very performance. It has been seen that seamless tubes produced much larger in size and much more expensive than the by porthole die extrusion are at least equivalent if not non-piercing type presses. Further, severe piercing operation superior to those extruded by other normal methods. sometimes develops cracks and tears in the internal wall of Unfortunately, lubricants cannot be applied in these the billet, leading to defects in the extruded tube. hot-working operations, because they will contaminate the surfaces to be welded and prevent to obtain sound welding The other method of extruding tube without using an in the die-exit zone. The welding-chamber method is external long mandrel is called welding-chamber method. In therefore restricted to materials, such as aluminium and this method, a solid billet and a bridge die (also known as a magnesium alloys that can be extruded without lubrication porthole, spider-mandrel, or torpedo die) with a standard and that can readily be pressure welded. In addition to the ram are used. Figure 13.30 shows a three-aperture bridge die formation of seamless tube, bridge-die extrusion is used for consisting of three thin spider arms, which holds a short production of hollow unsymmetrical shapes in aluminium mandrel in position in the die oriﬁce. The mandrel is sup- alloys. ported by the bridge. The normal external long mandrel used in the above-mentioned two methods has been actually 13.11.1 Extrusion of Cable Sheathing replaced by this short built-in mandrel. The four-aperture bridge die is also available, as shown in Fig. 13.31. To enter In extrusion of cable sheathing, the cable replaces the the die, the hot metal is forced to ﬂow around the central external long mandrel used for the production of normally bridge and sliced into separate segments while ﬂowing extruded seamless tubes. Hence, the principle of cable around the arms of the spider. The separate segments of sheathing is that the cable extends through the centre of a metal ﬂowing through the ports of die are brought together circular die oriﬁce and forms an annular space through in a welding chamber surrounding the mandrel, where these which the metal being extruded ﬂows to form the sheathing. segments are immediately compressed by the tapered bear- ing of the die and welded in the solid state to form a Let us consider lead sheathing of cable by direct extrusion process with a vertical press, as shown in Fig. 13.32. Liquid lead metal is poured into the extrusion chamber, and steam is passed through the square cut holes to cool and solidify the metal. The ram is brought downwards to contact the top surface of the solidiﬁed lead. Then the ram is further moved to apply pressure on the metal but its downward vertical movement must be synchronized with forward horizontal Fig. 13.31 A four-aperture Welding A bridge die chamber Mandrel Bridge ++ Exit face Cross section A-A A Entrance face

636 13 Extrusion Square cut holes Piston Steam the billet through square-faced dies in plane-strain condition for passing steam Lead in which frictionless condition along the container wall is assumed. The following analyses will be the same for indi- Steam rect and direct extrusion processes as long as there is no friction at the billet–container interface. Deformation is considered under plane strain condition, because it is the assumption of slip-line ﬁeld theory. In the following analy- ses, the plane-strain extrusion pressure in frictionless con- dition is estimated using slip-line ﬁeld theory and compared with that required for an ideal homogeneous deformation, so that the effect of redundant deformation becomes apparent. Cable 13.12.1 50% Plane-Strain Frictionless Extrusion Fig. 13.32 Lead sheathing of cable by a direct vertical extrusion press Figure 13.33a shows direct extrusion of the billet through square-faced dies with the formation of dead zone just prior to the entry plane of the die, in which frictionless condition along the container wall is assumed. The reduction in cross-sectional area of the work-piece, r; is assumed to be 50%; i.e. r ¼ 0:5: Let w is the width of the work-piece, which remains constant in plane strain, i.e. w ¼ constant. If movement of the cable through the die oriﬁce, from left to (a) α - line β - line right in Fig. 13.32. Hence, lead is extruded over the cable as a hollow tube through the annular clearance between the φ= 0 AH Dead metal zone cable and the wall of circular die oriﬁce. When lead is still left in the chamber, extrusion is halted and liquid lead is MO further poured into the chamber. This addition will cause partial melting of the remaining solid lead, which allows h1 BG σ1 = 0 h2 surface oxides to ﬂoat to the top. Subsequently, the metal is resolidiﬁed to produce a continuous billet, which can readily σL v1 = 1 45º v2 = 2v1 be extruded to produce the next element of tubular sheath for cable. In this manner, an endless continuously long sheath- O' ing without any joint can be produced. Rigid 45º 13.12 Application of Slip-Line Field A' 45º to Steady-State Motion Dead metal zone The signiﬁcance of velocity in slip-line ﬁeld evaluation of (b) d metal working processes has already been told in Sect. 10.5.3 in Chap. 10. In the following text, we shall consider v2 the application of slip-line ﬁeld theory to plane strain fric- v1 a tionless extrusion through square die for two different con- O ditions of reductions. In the following analyses, it is assumed that there is no friction at the interface between the billet and bc the container wall. Such frictionless condition along the container wall may easily be achieved by inverted extrusion m process using square-faced dies and no lubrication, because there is no relative movement between the billet and the Fig. 13.33 a Slip-line ﬁeld and b the corresponding hodograph for container. However, we have considered direct extrusion of 50% reduction in cross-sectional area of the work-piece in plane–strain frictionless extrusion through square dies

13.12 Application of Slip-Line Field to Steady-State Motion 637 h1 is the thickness of billet before extrusion and h2 is the thae!mve¼locitay!bdi:scTohnetirneusiutyltainstgviveleoncibtyy the vector a!m ; where thickness of the product after extrusion, then of this particle is O!m: r ¼ A1 À A2 ¼ wðh1 À h2Þ ¼ 1 À h2 ¼ 0:5: When the particle crosses the line AB at B, the velocity A1 w h1 h1 discontinuity is given by the vector !ac; since the tangent at ð13:50Þ ) h1 ¼ 2 B has rotated through 90°. The resultant velocity of this h2 particle is O!c : However, as soon as this particle crosses the boundary AB; it encounters shearing along BO and then (a) Construction of the Slip-Line Field: again moves in the horizontal direction. In crossing the slip The slip lines must meet the frictionless container wall at line BO, the particle undergoes another velocity disconti- 45: They must meet the centre-line of the billet also at 45°. nuity, which is parallel to BO, and given by the vector c!d : Thus, it is seen that two radial fans centred on O and O0 The ﬁnal velocity of the particle is in the direction of Oa would complete the ﬁeld, as shown in Fig. 13.33a. eOx!dte:ndOedbvaionudslgyi,venO!bdy=thOe!avec¼to2r :O!Sdin; cseinctheeO!vcelþocict!yd ¼ of (b) Velocity Vector Diagram or Hodograph: billet at the entry plane is v1 and that at the exit plane is v2; so O!a ¼ v1 and O!d ¼ v2; and we get The velocity at each point in a plastically deforming region can be graphically represented usually by drawing a hodo- v2=v1 ¼ 2 ð13:51Þ graph. A hodograph is a diagram giving velocity vectors appropriate to any point in the slip-line ﬁeld. Since the The lower half of the extrusion will also contribute to the velocity of the tool movement is known, i.e. the boundary ﬁnal velocity in extrusion, which must be v2=v1 ¼ 2. From velocity, a vector diagram giving velocities at all points (13.50) and (13.51), it is seen that h1=h2 ¼ v2=v1; from inside the deforming metal can be drawn. From the hodo- which we conclude that the chosen slip-line ﬁeld is com- graph, the compatibility of the chosen slip-line ﬁeld with patible with the velocity boundary conditions, because equal velocity boundary conditions can be seen directly. It may be noted that the slip line, which is boundary between rigid and volumes of material must pass a given point per unit time, plastic regions, is a line of velocity discontinuity and that the i.e. wh1v1 ¼ wh2v2: magnitude of velocity discontinuity is constant along this slip line. (c) Stress Determination from the Slip-Line Field: With reference to the upper half of the deformation pro- It may be recalled that within the slip ﬁeld, the shear yield cess (Fig. 13.33a), metal to the left of AB moving along the billet–container interface towards A, with velocity v1 ¼ 1; stress k is constant everywhere and the hydrostatic pressure ! rm varies from point to point with change in direction of the i.e. with unit velocity, is represented by the vector Oa on the slip lines. Since here the slip lines are at 45 to the hori- hodograph, as shown in Fig. 13.33b. As a metal particle crosses the velocity discontinuity line AB, it undergoes a zontal, the principal stresses r1 and r2 are, respectively, sudden shearing in a direction tangential to the AB line at the horizontal and vertical. Let us consider a point G on the point of crossing. It no more moves in the horizontal direction, but slides along the line AO; because it is con- boundary slip line OB; and at point G the algebraically strained by the dead metal. The change in velocity at any largest principal stress acting horizontally is ðr1ÞG¼ 0; point is represented by a vector that is tangential to the because the point G is on the exit plane (although r1 ¼ 0; velocity discontinuity line at that point. The velocity dis- but the other principal stress r2 is negative). The boundary continuity at point A is represented by the vector a!b ; since slip line OB is a b-line because it lies at 45 to the left from this vector is tangential to AB at A. Now, O!a þ a!b ¼ O!b ; the direction of the algebraically largest principal stress r1; which is the resultant velocity of the particle moving along whereas the curved slip line GH is an a-line, since it departs the billet–container interface and must be parallel to the to the right from the direction of r1 by the same angle. This shear plane AO: The magnitude of the velocity discontinuity means that in the centred fan AOB; the a-lines are circum- is constant at all points on the line AB and equal to ab. ferential, and the b-lines radial. One can easily note that r1; Hence, a particle crossing the line AB at any other point will passes through the ﬁrst and third quadrants formed by a undergo a velocity change equal in magnitude to ab with a right-handed aÀb coordinate system. direction tangential to the point of crossing. For example, for a particle moving horizontally and crossing the line AB at M, At point G, from (10.40a) we get ðr1ÞG¼ ðrmÞG þ k ¼ 0; ) ðrmÞG¼ Àk: /; which is the angle of inclination of the tangent to a-lines measured in the anticlockwise direction from the x-axis; is chosen to be zero in the direction OB: Since OB is

638 13 Extrusion straight, / does not vary between O and B: Hence, the value The inﬂuence of the ﬂow–constraint or the effect of redundant deformation is apparent from (13.53). Since plane of / at point G lying on OB is /G ¼ 0: strain extrusion through frictionless container wall was Since the tangent to the a-line rotates in a clockwise assumed in deriving (13.52), therefore the increase in actual extrusion pressure (13.52) compared to the ideal extrusion direction through an angle of p=2 radians between OB and pressure given by (13.53) is solely due to the effect of ﬂow– constraint. The actual extrusion pressure (13.52) is more OA; so at any point H on OA the value of / is /H ¼ À p : than twice that which would be expected if the inﬂuence of 2 the constraint was not taken into account (13.53). Now, along the a-line GH, from (10.47a) one can write 13.12.2 2/3 Plane-Strain Frictionless Extrusion ðrmÞH À /H ¼ ðrmÞG À /G; Here, the reduction in cross-sectional area of the work-piece, 2k 2k r; is assumed to be 2=3; i.e. r ¼ ðh1 À h2Þ=h1 ¼ 2=3; since for plane-strain condition, the stock width w ¼ constant. Or, ðrmÞH þ p ¼ À 1 À 0; Figure 13.34a, b show respectively the slip-line ﬁeld and Â 2k 2 Ã 2 hodograph for plane strain extrusion through frictionless * ðrmÞG¼ Àk : container wall. À1Àp ! Let us consider a point G on the boundary slip line, as 22 seen in Fig. 13.34a. At point G, the algebraically largest ) ðrmÞH¼ 2k ¼ Àk À pk: principal stress is ðr1ÞG¼ 0; because the point G is on the exit plane. At point G, from (10.40a) we get At point H, from (10.40b): ðr1ÞG ¼ ðrmÞG þ k ¼ 0; ðr2ÞH ¼ ðrm ÞH Àk ¼ À2k À pk ) ðrmÞG ¼ Àk: ¼ À2k 1 þ p 2 ) The pressure on the boundary of dead-metal zone (DMZ) is: pDMZ ¼ Àðr2 ÞH ¼ p: 2k 1 þ 2 Let the extrusion (ram) pressure be rL: Hence, equilib- rium of forces will give (since the width, w ¼ constant): rL h1 ¼ pDMZ þðh1pÀ h2Þ (a) ¼ 2k 1 2 ðh1 À h2Þ: σL 45º Ι h1 p (b) * From (13.50) h2 ¼ 2; ) rL 2h2 ¼ 2k 1 þ 2 h2; β- line Or, rL ¼ þ p; α- line σ =0 h2 k1 2 h1 G1 ð13:52Þ v1 45º v2 ) rL ¼ 1 þ p % 1:29 y 2k 2 4 45º x The extrusion (ram) pressure being rL derived by (13.52) v1 a can be compared to the pressure required to deform a metal in an ideal homogeneous deformation without considering O d the contribution due to friction and redundant deformation, 45º for which the work formula is given by (10.35b). Since the reduction in cross-sectional area is 50%, so Ideal ram pressure, ðrLÞideal ¼ r0 ln 1 r ¼ r0 ln 2 bc 1 À v2 2pk ﬃﬃ * k ¼ pr0ﬃﬃ : ¼ 0:693r0 ¼ 0:693 Â 2= 3 ; 3 Fig. 13.34 a Slip-line ﬁeld and b the corresponding hodograph for two-third reduction in cross-sectional area of the work-piece in plane– ) ðrLÞideal ¼ 0:693 ¼ 0:6 strain frictionless extrusion through square dies 2k 1:155 ð13:53Þ

13.12 Application of Slip-Line Field to Steady-State Motion 639 The values of angle / at points I and G in the slip-line for larger reduction in cross-sectional area, the ratio of ﬁeld diagram are respectively rL=ðrLÞideal decreases. It implies that higher the reduction r; the lower is the load required or energy consumed for /I ¼ p; and /G ¼ 3p : redundant deformation and higher will be the efﬁciency of 4 4 the working process. where / is the angle of inclination of the tangent to a-lines 13.13 Upper-Bound Solution measured in the anticlockwise direction from the x-axis: for Plane-Strain Frictionless Extrusion Here, IG is an a-line, since it departs to the right from the Let us obtain an upper-bound solution for extrusion under direction of the algebraically largest principal stress r1: plane-strain frictionless condition. Figure 13.35a, b shows Hence, along the a-line IG, from (10.47a) one can write respectively an upper-bound ﬁeld and the corresponding hodograph for plane strain frictionless extrusion through ðrmÞI À /I ¼ ðrmÞG À /G; conical converging die of included angle 2a; which is 2k 2k assumed to remain constant. Only half of the upper-bound ﬁeld is shown in Fig. 13.35a, which shows two planes of Or; ðrmÞI Àp ¼ À 1 À 3p ; Â ðrmÞG¼ Ã velocity discontinuity, AB and BC. Let, the initial entry 2k 4 24 * Àk: : velocity of the metal into the die is v1; and the exit velocity of the metal from the die is v2; where v2 [ v1; because the À 1 À 3p þ p ! exit velocity is increased due to reduction in cross-sectional 2 4! 4 area of the initial billet. Let, the velocity in triangle ABC is ) ðrmÞI ¼ 2k represented by v3: The hodograph is constructed by drawing horizontal vectors that represent v2 and v1; and a vector ¼ 2k À 1 À p ¼ Àk À pk: parallel to AC representing v3; where all of them start at the 22 origin. The velocity discontinuity along the plane AB is vÃ13; At point I, þfropm (10.40b): ðr2ÞI¼ ðrmÞIÀk ¼ À2k À pk ¼ À2k 1 2 If the extrusion (ram) pressure is rL; the equilibrium of forces will give rL ÈÉ Or; h1 ¼ 22kkÀð11rþþ2ÞIpp2ðhð1hh1À1ÀÀh2hhÞ22Þ; rL ¼ (a) ¼ α ¼ 2 h1 ! 2k 1 p 2: þ Â 2 ; *r¼ h1 À h2 ¼ A θ–α 23 h1 3 p ) rL ¼1 þ Â 2 ¼ 1:71 σL h1 v1 α +ψ v2 2 C h2 2 2k 2 3 ð13:54Þ θψ Again according to (10.35b), the extrusion pressure in an B ideal homogeneous deformation without considering the contribution due to friction and redundant deformation, (b) when the reduction in cross-sectional area is r ¼ 2=3; is α v1 v2 180º – θ θ Ideal ram pressure, ψ v* 32 ðrLÞideal¼ r0 ln 1 À 1 ¼ r0 ln 3 ¼ 1:0986 r0 ð2=3Þ ¼ 1:0986 Â 2pk ﬃﬃ θ –α 2= 3 v1*3 ) ðrLÞideal ¼ 1:0986 ¼ 0:95 ð13:55Þ v3 2k 1:155 Fig. 13.35 a An upper-bound ﬁeld for plane–strain frictionless The actual extrusion pressure (13.54) is 1.8 times that extrusion through conical converging die of included angle 2a; and which would be expected if the inﬂuence of the constraint b the corresponding hodograph was not taken into account (13.55). Further, we notice that

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