Engineering Tribology Episode 2 Part 9 ppsx

... Coatings, Wear, Vol. 1 52, 19 92 , pp. 127 -150. TEAM LRN 446 ENGINEERING TRIBOLOGY 88 J.C. Angus, Diamond and Diamond-Like Films, Thin Solid Films, Vol. 21 6, 19 92 , pp. 126 -133. 89 B.C. Oberlander and ... Graphite Slider, Wear, Vol. 198 , 199 6, pp. 20 8 -21 5. 93 R. Ahmed and M. Hadfield, Rolling Contact Performance of Plasma Sprayed Coatings, Wear, Vol. 22 0, 199 8, pp. 80 -...

Ngày tải lên: 05/08/2014, 09:19

... y w 2F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x 2 = dy dx w F H x= h w 2F H a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = F H w 2h a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = tan θ max () wa 2F H = cos θ max () 2F H 4 F H 2 wa() 2 + = T max F H cos θ max () = F H 2 wa() 2 4 += wa 2 a 2 16h 2 1+= Guess ... y−()= Σ M = 0; Mwy y 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wa a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + wby− 0= My() wby 1 2 wy 2 − 1 2 wb 2 − 1 2...

Ngày tải lên: 21/07/2014, 17:20

... Handbook of Micro/Nanotribology. Ed. Bharat Bhushan Boca Raton: CRC Press LLC, 199 9 © 199 9 by CRC Press LLC © 199 9 by CRC Press LLC Preface Second Edition, 199 9 The first edition of ... (alk. paper) 1. Tribology Handbooks, manuals, etc. I. Bhushan, Bharat, 194 9- . TJ1075.H245 199 9 621 .8 ′ 9 dc21 98 -24 466 CIP This book contains information obtained from authent...

Ngày tải lên: 21/07/2014, 15:20

Ngày tải lên: 21/07/2014, 17:20

Ngày tải lên: 21/07/2014, 21:21

... -0. 690 -0.348 - - - 7 W [N] 1.7575×10 4 -0. 793 2. 033 2. 596 1.0 42 0.884 -1.510 - - 1.108 - 8 H [W] 2. 791 5× 10 3 -0.5 79 1.530 0.873 2. 500 1.6 42 -0 .22 5 - - 9 ε 1.0516× 10 2 0. 399 -1.040 1.3 72 -0.5 39 ... 2. 5 12 0.563 0. 528 -1. 090 -0.383 - - - 1.385 - 2 H [W] 3 .93 07× 10 3 -0.706 1.577 0.477 2. 240 1 .28 7 0 .24 9 -0 .20 4 - 1 . 324 - - 3 ε 1 .26 66× 10 -2...

Ngày tải lên: 05/08/2014, 09:19

... w y *sin(x*) (5 .99 ) The squeeze term ‘w*’ can be included in the parameter ‘G’ of the Vogelpohl equation, i.e.: = h* 1.5 ∂ 2 M v ∂x* 2 + ( R L ) 2 ∂ 2 M v ∂y* 2 = FM v + G FM v + ∂h* ∂x* + 2w* (5.100) Damping ... by: h* = h* static + ∆x*cos(x*) + ∆y*sin(x*) (5 .90 ) ∂ 2 h* ∂x* 2 () = ∂ 2 h* ∂x* 2 static + ∂ 2 ∂x* 2 [ ∆x*cos(x*) + ∆y*sin(x*) ] (5 .91 ) The...

Ngày tải lên: 05/08/2014, 09:19

... LRN 27 4 ENGINEERING TRIBOLOGY Equating flow through the orifice to the total lubricant flow through the bearing (6. 19) : πd 2 2 p s − p r 2 ( ( 1 /2 C d = p r h 3 η B Rearranging: πd 2 η 2h 3 B p s ... these bearings are shown in Figure 6 .2. TEAM LRN 26 8 ENGINEERING TRIBOLOGY 1 2 5 10 20  B 0 0 .2 0.4 0.6 0.8 1.0 C/B 0.4 0.5 0.6 0.7 0.8 0 .9 1...

Ngày tải lên: 05/08/2014, 09:19

... p max = 3W 2 a 2 = 3 × 5 2 (5. 799 × 10 −5 ) 2 = 7 09. 9 [MPa] p average = W πa 2 = 5 π(5. 799 × 10 −5 ) 2 = 473.3 [MPa] · Maximum Deflection δ= 1.0 397 W 2 E' 2 R' () 1/3 = 1.0 397 5 2 (2. 308 ... p max = 3W 2 a 2 = 3 × 5 2 (3.65 × 10 −5 ) 2 = 1 791 .9 [MPa] p average = W πa 2 = 5 π(3.65 × 10 −5 ) 2 = 1 194 .6 [MPa] · Maximum Deflection δ= 1...

Ngày tải lên: 05/08/2014, 09:19

... 1.03 39 R y R x () 0.03 0. 02 () 0.636 = 1.03 39 0.636 · Simplified Elliptical Integrals = 1. 39 82 = 1.0003 + 0. 596 8R x R y = 1.0003 + 0. 596 8 × 0. 02 0.03 = 1.77 19 = 1. 527 7 + 0.6 023 ln R y R x () = 1. 527 7 + 0.6 023 ln 0.03 0. 02 () · ... p max = 3W 2 ab = 3 × 50 2 (2. 32 × 10 −4 ) × (1.75 × 10 −4 ) = 588.0 [MPa] p average = W πab = 50 π (2. 32 × 10 −4 ) × (1.75...

Ngày tải lên: 05/08/2014, 09:19