Metal Machining - Theory and Applications Episode 2 Part 1 potx

... cutting process (Part 1) . Trans. ASME J. Eng. Ind. 10 0, 22 2 22 8. Usui, E. and Hirota, A. (19 78) Analytical prediction of three dimensional cutting process (Part 2) . Trans. ASME J. Eng. Ind. 10 0, 22 9 23 5. Usui, ... developments 21 1 Fig. 7 .11 Three-dimensional non-steady chip formation by rigid plastic finite element method (Ueda et al ., 19 96): (a) initial model and...

Ngày tải lên: 21/07/2014, 17:20

... [W/mK] P 01 10 50 16 20 6 8 < 2 6900 16 .2 1 .2 20 P05 25 49 16 15 8 12 < 2 7000 14 .2 1. 8 20 P 01 15 48 16 20 5 11 –* 7000 15 .7 –* 20 P05 Total carbide: 94 Total metal: 6 –* 610 0 17 .2 1. 8 –* P10 ... [10 -6 K -1 ] K 01 10 √* √* √* 325 0 94.5 1 0.9 – – – K20 96 2 2 316 0 15 .7 0.9 – – – K05 20 √* √* 325 0 93 .2 1 0.85 – – – – √* √* √* 330...

Ngày tải lên: 21/07/2014, 17:20

... 46 Superalloys** 11 – 12 . 5 11 14 13 16 16 20 20 24 Titanium pure titanium* 22 21 21 21 – α , α – β , β alloys 5.5–8 8– 12 10 17 12 . 5 21 15 25 Ti-6Al-4V 6.6–6.8 8.5–9 .1 10.5– 12 . 5 13 16 16 19 *: high and commercial ... copper* 11 5– 12 0 10 0 11 0 90 10 0 – – 60/70Cu–40/30Zn 25 –35 28 –35 27 –33 25 –30 – 90/95Cu 10 /5Sn 15 25 20 25 23 –30 – – 60/90Cu–40 /10...

Ngày tải lên: 21/07/2014, 17:20

... u˘ z d l hD ikj 21 10 + —— [ 12 1 0 ] 12 1 12 0 0000 (A2 .26 ) and 11 1 q*V e 1 qD ikj 1 hT 0 D ikj 1 {F} e = ——— {} – ——— {} – ——— {} (A2 .27 ) 4 1 3 1 3 1 10 0 Global assembly of equations (A2 .20 a), with ... [H] e {T} = {F e } (A2.30) ∂t with rCV e 21 11 [C] e = ——— | 12 1 1 | 20 1 12 1 11 12 ([C] is given here for a four-node tetrahedron). Nume...

Ngày tải lên: 21/07/2014, 17:20

... (A1.48a) where, from equation (A1. 42) 1 – nn n ——— ——— ——— 0 0 0 1 – 2n 1 – 2n 1 – 2n n 1 – nn ——— ——— ——— 0 0 0 E 1 – 2n 1 – 2n 1 – 2n [D e ] = ——— [ 00 0 ] 1 + nn n1 – n ——— ——— ——— 0 0 0 1 ... 339 Fig. A1.8 Metal machining slip-line fields (left) and their velocity diagrams (right), due to (1) Lee and Shaffer (19 51) , (2 4) Kudo (19 65) and (5) Dewhurst (19...

Ngày tải lên: 21/07/2014, 17:20

... addition s r ′ 2 = s r 2 –3s m 2 =(s 1 2 + s 2 2 + s 3 2 ) – 3s m 2 (A1.5) After substituting for s m from equation (A1 .1) , 3s r ′ 2 =(s 1 – s 2 ) 2 +(s 2 – s 3 ) 2 +(s 3 – s 1 ) 2 (A1.6) The yield ... stresses: 1 s – 2 =— [ (s 1 – s 2 ) 2 +(s 2 – s 3 ) 2 +(s 3 – s 1 ) 2 ] = Y 2 or 3k 2 (A1.7) 2 330 Appendix 1 Fig. A1 .2 Geometrica...

Ngày tải lên: 21/07/2014, 17:20

... VBMM160404 53 P10 49 0 .10 0.33 5 12 17 35 0.4 MVVNN2 020 M16 VNMG160404 53 P10 45 0 .20 0.48 4 11 17 35 0.4 MVVNN2 020 M16 VNMG160408 53 P10 38 0.40 0.70 4 11 17 35 0.8 Childs Part 3 31: 3 :20 00 10 :39 ... drill 19 .5) 5: (SHAPE through-hole 11 .7 ∇) 2: (TOOL reamer 20 .0) 4: (TOOL drill 19 .5) 6: (TOOL drill 11 .7) 7: (SHAPE centre hole 2. 0) 2: (TOOL reamer 20 .0) 4: (TO...

Ngày tải lên: 21/07/2014, 17:20

... 3 31: 3 :20 00 10 :38 am Page 28 2 f mach (1 – n 1 ) n 1 1 C c t ct + C t n 1 C opt = C c t load + pDLd a ( —————— )( ————— ) f opt (n 1 –n 2 )/n 2 d opt (n 1 –n 3 )/n 3 C c C ′n 1 (9 .28 d) By replacing ... of f in equation (9 .29 a), which is negative, may satisfy the relation n 3 (n 1 – n 2 )1 – n 1 /n 2 m 1 | —–——— | = ———— < —— < 1 (9 .29 b) n 2...

Ngày tải lên: 21/07/2014, 17:20

... 0.007ln V) F f = 629 f 0.30 d 0. 720 + 11 99(VS 3.58 – 0. 023 V 0 .27 ) } (9.2b) × (VB –0.66 – 0 .23 V 0 .27 ) (VN 0.03 – 0 .23 V 0 .27 ) F c = 18 62f 0.94 d 1. 11 + 26 77(VS 0 .24 – 0.05ln V) ×(VB 0 .23 – 0.05ln ... mm Childs Part 2 28:3 :20 00 3 :18 pm Page 25 5 Vyas, A. and Shaw, M. C. (19 99) Mechanics of saw-tooth chip formation in metal cutting. Trans ASME J. Manuf....

Ngày tải lên: 21/07/2014, 17:20

... 0. 028 5 + 0.000 014 T N = 0 .18 e –0.0005T + 0.1e –0.000 015 (T–430) 2 a = 0.00 020 , m = 0.00 52 Steel Y A = 920 e –0.0 011 T + 12 0 e –0.00003(T 16 0) 2 + 11 0e –0.00004(T–340) 2 + 12 0 e –0.00 01( T–650) 2 M ... 0.0 315 + 0.000 016 T N = 0 .19 e –0.0005T + 0.085e –0.000 015 (T–430) 2 a = 0.000 32, m = 0.0 018 Steel L A = 910 e –0.0 011 T + 19 0e –0.00 011 (T 13 5)...

Ngày tải lên: 21/07/2014, 17:20