Engineering Tribology Episode 2 Part 4 docx

... shown in Figure 7 .20 [7]. 100 20 0 500 1000 20 00 5000 10 4 2 × 10 4 5 × 10 4 10 5 2 × 10 5 5 × 10 5 10 6 2 × 10 6 5 × 10 6 10 7 20 0 500 1000 20 00 5000 10 4 2 × 10 4 5 × 10 4 10 5 2 × 10 5 5 × 10 5 10 6 100 50 20 10 Piezoviscous-elastic Piezoviscous-rigid Lubrication ... p max = 3W 2 ab = 3 × 50 2 (2. 32 × 10 4 ) × (1.75 × 10 4 ) = 588.0 [MPa] p...

Ngày tải lên: 05/08/2014, 09:19

... Steel-on-Si 3 N 4 and Systems with Vapor Phase Lubrication of Oleic Acid and TCP, Wear , Vol. 21 4, 1998, pp. 20 7 -21 1. TEAM LRN 4 02 ENGINEERING TRIBOLOGY 0 0.1 0 .2 0.3 µ 0 100 20 0 300 Temperature ... Nature, Vol. 21 7, 1968, pp. 48 1 -4 82. 51 F.G. Rounds, The Influence of Steel Composition on Additive Performance, ASLE Transactions, Vol. 15, 19 72, pp. 54- 66. TEAM...

Ngày tải lên: 05/08/2014, 09:19

... con- ditions is: ͚M ϭ 0 O 2 ϪF ϫ 14 ϩ F ϫ 12 Ϫ F ϫ 2 ϭϪF ϫ 14 ϩ 510 ϫ 12 Ϫ 1, 140 ϫ 2 ϭ 0 Bs2 B F ϭ 27 4 lb B ͚M ϭ 0 O 3 P ϫ 31. 82 Ϫ F ϫ 2. 18 ϭ P ϫ 31. 82 Ϫ 27 4 ϫ 2. 18 ϭ 0 B from which Solving for ... / 2) ␲ ϭ e ϭ e ϭ 5 .20 F 2 and F ϭ 5 .20 F 12 Solving the equations for force simultaneously, we find F ϭ 5 940 lb (26 42 1 N) F ϭ 1 140 lb (5071 N) 12 Then, ͚M ϭ...

Ngày tải lên: 05/08/2014, 09:20

... E 21 . C 22 . B 23 . D 24 . B 25 . A 26 . B 27 . C 28 . E 29 . B 30. B 31. A 32. E 33. B 34. E 35. C 36. E 37. D 38. A 39. B 40 . E 41 . C 42 . D 43 . A 44 . B 45 . C 46 . E 47 . B 48 . A 49 . D 50. E 51. B 52. B 53. ... O A O B O C O D O E 41 O A O B O C O D O E 42 O A O B O C O D O E 43 O A O B O C O D O E 44 O A O B O C O D O E 45 O A O B O C O D O E 46 O A O...

Ngày tải lên: 22/07/2014, 10:22

... coordinates. 62 2 26 2 44 ++ ,,       = () 2. (C) O is the midpoint of AB. x xx y yy + + + + 4 2 244 0 6 2 1 6 24 === === , , − A is the point (0, 4) . 3. (A) d = ()() = === 84 63 4 3 16 9 25 5 22 22 − ... < 3 2 (B) b > 3 2 (C) b <− 3 2 (D) b < 2 3 (E) b > 2 3 7. If x 2 < 4, then (A) x > 2 (B) x < 2 (C) x > 2 (D) 2 <...

Ngày tải lên: 22/07/2014, 11:20

... [°C] 135 21 0 23 0 24 0 25 0 23 0 28 0 43 0 43 0 370 370 Kinematic viscosity [cSt] at -20 °C 170 193 16 85 115 20 0 850 1000 0°C 75 75 16 38 47  100 25 0 25 00 8000 44 0 40 °C ... 44 0 40 °C 19 13 15 11 12 33 74 70 363 515 150 100°C 5.5 3.3 4. 5 4 4 11 25  6.3 13.1 35 41  20 0°C 1.1 1.3 3.8 22  1 .4 2. 1...

Ngày tải lên: 05/08/2014, 09:19

... downstream of upstream groove TEAM LRN 24 4 ENGINEERING TRIBOLOGY 1.5 2. 5 0 0.1 Groove len g th/axial bearin g len g th Dimensionless load 2 0 .2 0.3 0 .4 0.5 0.6 0.7 0.8 0.9 1 0 15 5 Dimensionless ... = 0 .2 [m] and the radial clearance of the bearing is c = 0.00 04 [m], then from equation (5.87) the value of flow ‘Q’ is: Q = 0.5 × 6.8 × 0 .2 × 10 × 0.00 04...

Ngày tải lên: 05/08/2014, 09:19

... geometries of 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 1.0 A A H B C B FIGURE ... 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...

Ngày tải lên: 05/08/2014, 09:19

... p max = 3W 2 a 2 = 3 × 5 2 (6.88 × 10 −5 ) 2 = 5 04. 4 [MPa] p average = W πa 2 = 5 π(6.88 × 10 −5 ) 2 = 336 .2 [MPa] · Maximum Deflection δ= 1.0397 W 2 E' 2 R' () 1/3 = 1.0397 5 2 (2. 308 ... p max = 3W 2 a 2 = 3 × 5 2 (5.799 × 10 −5 ) 2 = 709.9 [MPa] p average = W πa 2 = 5 π(5.799 × 10 −5 ) 2 = 47 3.3 [MPa] · Maximum Deflection δ= 1.0397...

Ngày tải lên: 05/08/2014, 09:19

... equation (7. 42 ) yields: p max = πbl W = = 1 748 .96 [MPa] π(1. 82 10 4 ) × (5 × 10 −3 ) 5 × 10 3 hence: 2. 45 × 0 .2 × (1 748 .96 × 10 6 ) 1.5 (46 .7 × 7800 × 46 0) 0.5  2 − 1  = 1 84. 91 [°C] T f maxc = 6 ... χ= ρσ K = 46 .7 7800 × 46 0 = 13. 02 × 10 −6 [m 2 /s] · Peclet Number L A = 2 U A b = 2 × (1. 82 × 10 4 ) 2 × (13. 02 × 10 −6 ) = 13.98 TEAM LRN EL...

Ngày tải lên: 05/08/2014, 09:19