Mechanical Engineers Handbook 2011 Part 4 ppsx

Friction and Lubrication in Mechanical Design Episode 2 Part 4 ppsx

Friction and Lubrication in Mechanical Design Episode 2 Part 4 ppsx

... alloys Invar “36” annealed H,M,80 annealed Monel C 27 0 29 6 22 4 25 2 27 0 27 0 22 4 29 6 I99 46 8 27 0 397 3 59 160 I80 3 84 27 6 24 2 26 0 22 0 21 6 746 22 0 22 6 26 2 340 27 0 29 6 ... Dry 32 1 Dry 44 0 c Dry 0. 54 0.09 0. 54 0.15 0. 54 0.1 1 0. 54 0.60 0. 54 0 .27 0. 54 0 .27 0 .20 0.70 0 .20 0 .22 0. 54 0.19 0...

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Friction and Lubrication in Mechanical Design Episode 2 Part 4 ppsx

Friction and Lubrication in Mechanical Design Episode 2 Part 4 ppsx

... alloys Invar “36” annealed H,M,80 annealed Monel C 27 0 29 6 22 4 25 2 27 0 27 0 22 4 29 6 I99 46 8 27 0 397 3 59 160 I80 3 84 27 6 24 2 26 0 22 0 21 6 746 22 0 22 6 26 2 340 27 0 29 6 ... 0. 54 0.15 0. 54 0.1 1 0. 54 0.60 0. 54 0 .27 0. 54 0 .27 0 .20 0.70 0 .20 0 .22 0. 54 0.19 0 .20 0.78 0 .20 0 .23 0. 54 0.13 0 .20 0....

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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

... z z 3  z=−ı + 2 2!  d 2 dz 2 (z 3 + z + ı) sin z z + ı  z=0 = 2 (−ı sinh(1)) + ıπ  2  3z 2 + 1 z + ı − z 3 + z + ı (z + ı) 2  cos z +  6z z + ı − 2( 3z 2 + 1) (z + ı) 2 + 2( z 3 + z + ı) (z ... formula.  C z z 2 + 1 dz =  C 1 /2 z −ı dz +  C 1 /2 z + ı dz = 1 2 2 + 1 2 2 = 2 3.  C z 2 + 1 z dz =  C  z + 1 z  dz =  C z dz +  C 1 z dz = 0 + 2 = 2...

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Hydrodynamic Lubrication 2011 Part 4 ppsx

Hydrodynamic Lubrication 2011 Part 4 ppsx

... Eq. 4. 35 [ 24] : r = 1 2à p r z(z h) + r 2 à hz 4 z 2 2 + z 3 3h z 4 12h 2 (4. 37) G c = r 2 à h 3 40 (4. 38) If G c = 0 is assumed in Eq. 4. 34, Reynolds’ equation ignoring the centrifugal force ... r sin θ (4. 41) 4. 3.2 Numerical Solution of a Sector Pad Let us solve the sector pad problem numerically using Eq. 4. 39. Namely: r h 3 p r + h 3 r p r + 1 r 2 h 3 p =...

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Materials Handbook 2011 Part 4 potx

Materials Handbook 2011 Part 4 potx

... resem- bles tantalum, is yellowish-white, has a specific gravity of 8.57, a melting point of 44 74 F ( 246 8°C), and an electrical conductivity of 13.2% relative to copper. Columbium has a body-centered-cubic ... strength of 94, 000 lb/in 2 ( 648 MPa) at 70°F and 25,000 lb/in 2 (172 MPa) at 2000°F (1093°C). After recrystal- lization at 240 0°F (1315°C), however, yield strength drops to 50,00...

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Materials Handbook 2011 Part 7 ppsx

Materials Handbook 2011 Part 7 ppsx

... of 8,200 lb/in 2 ( 57 MPa), a flex- ural strength of 11,000 lb/in 2 (76 MPa), a compressive strength of 40,000 lb/in 2 ( 276 MPa), and a modulus of 2.5 ϫ 10 6 lb/in 2 ( 172 , 375 MPa). And having ... tensile strength ranges from 200,000 to nearly 500,000 lb/in 2 (1, 379 to 3,448 MPa) and modulus of elasticity from 28ϫ10 6 to 75 ϫ10 6 lb/in 2 (193,060 to 5 17, 125 MPa). Graphite-reinforced...

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Materials Selection Deskbook 2011 Part 4 ppsx

Materials Selection Deskbook 2011 Part 4 ppsx

... 0.89 0.90 0. 94 1.03 1.06 1.08 1. 14 1.21 1.30 1.32 1.38 1 .48 1. 54 1.70 1.87 1.93 2.01 2.13 2 .47 3.07 3. 34 3 .42 3.67 4. 06 4. 40 4. 5 0 5.20 5.76 6.52 11.7 14. 2 17.5 19.3 ... 0. 047 R 0. 044 U 0. 049 R 0.100 1 0. 045 R 0.0 94 U 0.030 1 0.062 U 0.061 R 0. 048 SI 0.102 1 0.085 U 0.069 SR 0.062 1 0.05 1 1 0.106 1 0. 045 M...

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Mechanical Engineers Handbook 2011 Part 4 ppsx

Mechanical Engineers Handbook 2011 Part 4 ppsx

... 26 9.6± 14 3.9±5.6 10 156 1.00 25 26 97 32 11.5± 14. 8 4. 6±6.0 11.5 1 74 1.05 30 31 109 40 13±16 .4 5.2±6.6 14 201 1.10 35 36.5 1 24 48.5 14. 5±17.2 5.8±6.9 16 212 1.15 40 42 .5 140 57 16±20 6 .4 7.8 18.5 ... 2 14 3 .4 RTR Dyad 215 3.5 TRT Dyad 216 4. Kinetostatics 223 4. 1 Moment of a Force about a Point 223 4. 2 Inertia Force and Inertia Moment 2 24 4.3 Free-Body Diagrams 22...

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Mechanical Engineers Handbook 2011 Part 5 potx

Mechanical Engineers Handbook 2011 Part 5 potx

... (internal gear): Pitch circle diameter d 5  mN 5  375X0mm Addendum circle diameter d a5  mN 5 À 2365X0mm Dedendum circle diameter d d5  mN 5  2X5387X5mmX Number of Planet Gears The number ... elements 0, 4, 3, 5, and 0 (clockwise path). For the velocity analysis, the following vectorial equations can be written: v 40  v 34  v 53  v 05  0Y or v 40  v 34  v 50 ...

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Mechanical Engineers Handbook 2011 Part 6 docx

Mechanical Engineers Handbook 2011 Part 6 docx

... 34 2.03 1 16. 1 164 .1 180 34 3. 96 1 16. 1 167 .1 320 95 215 47 2.54 122.9 194.1 215 47 4.75 122.4 194 .6 L21 100 160 26 2.03 1 16. 1 1 46. 8 160 26 221 100 190 36 2.03 121.9 173.5 190 36 3. 96 121.4 175.3 321 ... (mm) (mm) (mm) L12 60 95 18 1.02 66 .8 87.9 95 18 1.52 67 .1 88 .6 212 60 110 22 1.52 70 .6 99.3 110 22 2.03 69 .3 101.3 312 60 130 31 2.03 75.4 115 .6 130 31 2.54...

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Mechanical Engineers Handbook 2011 Part 10 ppt

Mechanical Engineers Handbook 2011 Part 10 ppt

... downwardX Nu  0X54Ra 1a4 L 10 4 ` Ra L ` 10 7  0X15Ra 1a4 L 10 7 ` Ra L ` 10 9 X @ Hot surface facing downwardY or cold surface facing upwardX Nu L  0X27Ra 1a4 L 10 5 ` Ra L ` 10 10 X Immersed bodies: Horizontal ... 0X6Ra y * 1a5 Nu y  0X75Ra Ã1a5 y W a Y laminarY 10 5 ` Ra y * ` 10 13 3X248a Nu y  0X568Ra Ã0X22 y Nu y  0X645Ra Ã0X22 y A turbulentY 10 13 ` Ra y * `...

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Mechanical Engineers Handbook 2011 Part 11 doc

Mechanical Engineers Handbook 2011 Part 11 doc

... entrap these particles by means of a b value. The symbol b is immediately followed by a number that denotes the diameter of the particles involved according to the relation b d  Number of particles ... principles and have similar parts, they often have design differences that make their performances better as either motors or pumps. Moreover, some motors have no pump counterparts. In this ch...

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Mechanical Engineers Handbook 2011 Part 13 pdf

Mechanical Engineers Handbook 2011 Part 13 pdf

... Appendix Differential Equations and Systems of Differential Equations HORATIU BARBULESCU Department of Mechanical Engineering, Auburn University, Auburn, Alabama 36849 Inside 1. Differential Equations ... the conditions (1.4) A solution obtained from the general solution for particular constants c 1 Y c 2 Y FFFY c n is called a particular solution.Asingular solution is a solution that canno...

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Mechanical Engineers Handbook 2011 Part 15 doc

Mechanical Engineers Handbook 2011 Part 15 doc

... constant, 150 , 287, 388 spring index, 286 stiffness, 150 , 342, 386 tension, 150 152 torsion, 150 152 , 290±293 Spur gears, 253, 425 Square threads, 247±248, 251 Stability analysis of, 414± 415 criteria ... 171 deformation and, 3, 160±163, 389 eccentric loading, 170 expression for, 157 impact analysis, 157 159 maximum values, 158 springs and, 150 151 stiffness, 149±172 strain e...

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