Machinery Components Maintenance And Repair Episode 2 Part 4 ppt

Metal Machining - Theory and Applications Episode 2 Part 4 pptx

Metal Machining - Theory and Applications Episode 2 Part 4 pptx

... V 0 .27 ) } (9.2b) × (VB –0.66 – 0 .23 V 0 .27 ) (VN 0.03 – 0 .23 V 0 .27 ) F c = 1862f 0. 94 d 1.11 + 26 77(VS 0 . 24 – 0.05ln V) ×(VB 0 .23 – 0.05ln V) (VN 0.16 – 0.05ln V) where F d , F f , and F c are ... conventional, (b) and (c) type I and II tools, K = 1 .2 Childs Part 2 28:3 :20 00 3:19 pm Page 26 2 where F* t , F* r and F* z are the specific cutting forces in t...

Ngày tải lên: 21/07/2014, 17:20

20 308 0
Dimensioning and Tolerancing Handbook Episode 2 Part 4 ppt

Dimensioning and Tolerancing Handbook Episode 2 Part 4 ppt

... ) ( ) 2 1 22 222 222 2 22 222 222 222 2 22 222 222 2 0300.)1(8.0060.)1(8.0070.)1(8. 0075.)1 (2. 0050.)1(8.0070.)1 (2. 0050.)1(8. 0075.)1 (2. 0030.)1(8.0 020 .)1 (2. 0155).1 (2. 0300).1 (2. 0060).1 (2. 0070).1 (2. 0075).1(8.0050).1 (2. 0070).1(8. 0050).1 (2. 0075).1(8.0030).1 (2. 0 020 ).1(8.0155).1(8.             +−+ ++++ ++++− ++−++++ +++++−= ems t t ems ... = . 04 72 1 )...

Ngày tải lên: 21/07/2014, 15:20

25 404 1
Metal Machining - Theory and Applications Episode 2 Part 11 ppt

Metal Machining - Theory and Applications Episode 2 Part 11 ppt

... [W/mK] P01–10 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 P01–15 48 16 20 5 11 –* 7000 15.7 –* 20 P05 Total carbide: 94 Total metal: 6 –* 6100 17 .2 1.8 –* P10 Total ... a 1 ) 2 a 1 + a 2 ————— a 1 ≤ x < ———— (a 2 — a 1 ) 2 2 SF(x, a 1 , a 2 ) = { (A7.4b) 2( x – a 2 ) 2 a 1 + a 2 1– ————— ———— ≤ x < a 2 (a...

Ngày tải lên: 21/07/2014, 17:20

16 475 0
Metal Machining - Theory and Applications Episode 2 Part 8 pptx

Metal Machining - Theory and Applications Episode 2 Part 8 pptx

... 3(s xx ′ 2 + s yy ′ 2 + s zz ′ 2 ) + 6(s 2 xy + s 2 yz + s 2 zy ) ≡ (s xx – s yy ) 2 + (s yy – s zz ) 2 + (s zz – s xx ) 2 + 6(s 2 xy + s 2 yz + s 2 zy ) = 6k 2 or 2Y 2 (A1 .26 b) which is the same ... the results, a 346 Appendix 1 Childs Part 3 31:3 :20 00 10: 42 am Page 346 p A + 2kf A = p B + 2kf B; p C + 2kf C = p D + 2kf D (A1.19) p A + 2kf A = p D – 2kf...

Ngày tải lên: 21/07/2014, 17:20

20 409 0
Metal Machining - Theory and Applications Episode 2 Part 7 ppt

Metal Machining - Theory and Applications Episode 2 Part 7 ppt

... 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 Geometrical representation...

Ngày tải lên: 21/07/2014, 17:20

20 341 0
Gear Noise and Vibration Episode 2 Part 4 pps

Gear Noise and Vibration Episode 2 Part 4 pps

... 5*cos(1 .44 88 -(i-l)*0.1); yl(i) - 40 .7 84 - 5*sin(1 .44 88 -(i-l)*0.1); end fori=16:N; % involute ra = (i-16)*0. 02; xl(i) =40 .7 84* (sin(raHi-16)*0. 02* cos(ra)); yl(i) =40 .7 84* (cos(ra)+(i-16)*0. 02* sin(ra)); end for ... i=(N+l) :2* N ; % Image in x=0 other flank x2(i) = - xl (2* N+l-i); y2(i) - yl (2* N+l-i); rot 1=0 .45 413; xl(i) =x2(i)*cos(rotl) +y2(i)...

Ngày tải lên: 05/08/2014, 09:20

20 256 0
Gear Noise and Vibration Episode 2 Part 4 pot

Gear Noise and Vibration Episode 2 Part 4 pot

... 5*cos(1 .44 88 -(i-l)*0.1); yl(i) - 40 .7 84 - 5*sin(1 .44 88 -(i-l)*0.1); end fori=16:N; % involute ra = (i-16)*0. 02; xl(i) =40 .7 84* (sin(raHi-16)*0. 02* cos(ra)); yl(i) =40 .7 84* (cos(ra)+(i-16)*0. 02* sin(ra)); end for ... i=(N+l) :2* N ; % Image in x=0 other flank x2(i) = - xl (2* N+l-i); y2(i) - yl (2* N+l-i); rot 1=0 .45 413; xl(i) =x2(i)*cos(rotl) +y2(i)...

Ngày tải lên: 05/08/2014, 09:20

20 335 0
Gear Noise and Vibration Episode 2 Part 6 ppt

Gear Noise and Vibration Episode 2 Part 6 ppt

... the pinion at position 2. So 8, = Fpi, + Fp 12 and 5 2 = Fp 12 + Fp 22 . As the transmission error is the sum of 81 and S 2 T.E. = F(p n +2p, 2 + p 22 ) giving F. Once F ... analysis of the input and of the output and the ratio (complete with phase) is the frequency transfer function required. 14 12 50 100 150 20 0 25 0 300 350 40 0 F...

Ngày tải lên: 05/08/2014, 09:20

20 388 0
Process Selection - From Design to Manufacture Episode 2 Part 4 pptx

Process Selection - From Design to Manufacture Episode 2 Part 4 pptx

... strategies and PRIMAs 24 3 //SYS21///INTEGRAS/B&H/PRS/FINALS_07-05-03/07506 543 76-CH0 02- 1.3D – 22 4 – [35 24 8 /21 4] 9.5 .20 03 2: 05PM . Infrared Brazing (IRB): uses quartz-iodine incandescent lamps ... jigging and fixtures. 7.13F Soldering process. 22 6 Selecting candidate processes //SYS21///INTEGRAS/B&H/PRS/FINALS_07-05-03/07506 543 76-CH0 02- 1.3D – 24 0 – [35 2...

Ngày tải lên: 21/07/2014, 16:21

20 284 0
342 TOEIC Vocabulary Tests Episode 2 part 4 pptx

342 TOEIC Vocabulary Tests Episode 2 part 4 pptx

... inattention; forgetfulness; abandonment definition (c) neglect 48 8 Questions Index PHOTOCOPIABLE © www.english-test.net Answers 1 24 TOEIC Vocabulary / Word by Meaning / Test # 1 24 (Answer Keys) A1 adj. ... portions; to cleave definition (b) slice 47 9 Questions Index PHOTOCOPIABLE © www.english-test.net Answers 122 TOEIC Vocabulary / Word by Meaning / Test # 122 (Answer Keys) A1 v. to...

Ngày tải lên: 21/07/2014, 22:20

35 286 0
w