Machinery Components Maintenance And Repair Episode 2 Part 5 ppt

... [W/mK] P01–10 50 16 20 6 8 < 2 6900 16 .2 1 .2 20 P 05 25 49 16 15 8 12 < 2 7000 14 .2 1.8 20 P01– 15 48 16 20 5 11 –* 7000 15. 7 –* 20 P 05 Total carbide: 94 Total metal: 6 –* 6100 17 .2 1.8 –* P10 ... 7000 15. 7 2. 3 11 P20 Total carbide: 82 Total metal: 18 –* 7000 14 .2 2 .5 16 P01 20 Total carbide: 87 Total metal: 13 2 6600 16.7 1 .5 25 P10–30 Total ca...

Ngày tải lên: 21/07/2014, 17:20

... 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 ... s r 2 . So s r is also a stress invariant (it is known as the second stress invariant). From equation (A1 . 25 ) s r 2 = s 2 xx + s 2 yy + s 2 zz +2( s...

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 Geometrical representati...

Ngày tải lên: 21/07/2014, 17:20

... d R ) 2 – (r w – d e ) 2 = R 2 – (R – d e ) 2 } (9.9a) for a convex survace (r w + d R ) 2 – (r w + d e ) 2 = R 2 – (R – d e ) 2 Hence d e 2r w – d R for a concave surface c r = — = ———— d R 2r w – ... n 1 /n 2 ≈ 0.77 and n 1 /n 3 ≈ 0.37 for HSS tools, and exponents of the force model (equations (9.2b), (9 . 25 b)) have a relation m 1 /m 2 ≈ 0. 85 for an alloy steel,...

Ngày tải lên: 21/07/2014, 17:20

... introduced of the Cutting edge design 25 5 Fig. 8 .24 Variation of chip shape and forces at w G =2. 6 mm and h B =0.4 mm Childs Part 2 28:3 :20 00 3:18 pm Page 25 5 Vyas, A. and Shaw, M. C. (1999) Mechanics ... (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...

Ngày tải lên: 21/07/2014, 17:20

... revolution Fig 15. 3 Line elimination and resynthesis to detect small changes. Condition Monitoring 23 9 at c c o 2 * 4> 0 U 0) £ .*: (0 4) 0, n u -50 -100 - 150 -20 0 - 25 0 -300 - 350 Tooth Tooth Tooth Tooth Tooth Tooth Tooth Tooth Tooth Tooth No No No No No No No No No No 1 ... (brl*psil - brl*tan(phi2))/(pi*cos(phio)); r2 = (br2*psi2 - br2*tan(phi2))/(pi*cos(phio...

Ngày tải lên: 05/08/2014, 09:20

... appeared it was often too late and a considerable number of 23 1 22 6 Chapter 14 50 100 150 20 0 25 0 Fig 14.3 Detail of contacts for seven tooth and rack. tip base pitch roll I purfe ... (brl*psil - brl*tan(phi2))/(pi*cos(phio)); r2 = (br2*psi2 - br2*tan(phi2))/(pi*cos(phio)); conratio = rl + r2; disp( f angle rl r2 contact ratio') disp([ phi2 rl...

Ngày tải lên: 05/08/2014, 09:20

... 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 ... frequency 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 30...

Ngày tải lên: 05/08/2014, 09:20

... the rium and is not subjected to body forces; the three principal stresses. { me ¯ n ¦no ˙ rmиəl elements of the stress tensor are given by stres } ␴ 11 ϭѨ 2 ␾ 2 /Ѩx 3 2 ϩѨ 2 ␾ 3 /Ѩx 2 2 , ␴ 23 ϭϪѨ 2 ␾ 1 /Ѩx 2 Ѩx 3 , ... ␴ 11 ϭ 2 2 ␺ 1 / a blackbody in the same band at the same tem- Ѩx 2 Ѩx 3 , ␴ 23 ϭѨ 2 ␺ 2 /Ѩx 1 Ѩx 2 ϩѨ 2 ␺ 3 /Ѩx 1 Ѩx 3 , and cyclic perat...

Ngày tải lên: 21/07/2014, 15:20

... Figure 60 .2. 55 -1 55 Vulcanizate Thermoplastic Elastomers 55 .1 Introduction 55 - 1 55 .2 Properties 55 - 1 55 .3 Processing 55 - 2 55 .4 Uses of TPV 55 - 2 55 .1 Introduction ... and reasonable price. OH OH CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 H 2 C CH . CH 2 O-R OCH 2 -CH CH 2 n n n + O O Base ∆ OCH 2 CH(OH)CH 2 O-R OCH 2 -CH...

Ngày tải lên: 21/07/2014, 15:20