(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx

Industrial Machinery Repair Part Episode 2 Part 1 pptx

Industrial Machinery Repair Part Episode 2 Part 1 pptx

... 11 57 2 8.00 6 21 6 11 13 31 2 1/ 4 11 .39 6 1/ 4 24 4 11 1 /2 1 520 2 1 /2 15 .63 6 1 /2 275 12 1 728 2 3/4 20 .80 6 3/4 308 12 1 /2 19 53 3 27 .00 7 343 13 21 97 3 1/ 4 34.33 7 1/ 4 3 81 14 27 44 3 1 /2 42. 88 7 1 /2 422 ... 11 " 2 1 /2 7/8 2 1 /2 × 13 /4 11 1/ 16 to 13 " 3 1 3 × 2 13 1/ 16 to 15 " 3 1...
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Machinery''''s Handbook 27th Episode 2 Part 1 pdf

Machinery''''s Handbook 27th Episode 2 Part 1 pdf

... 12 23 ⁄ 32 9 2 1 ⁄ 4 26 to 8 1 3 ⁄ 8 12 3 1 ⁄ 4 4 10 to 12 3 ⁄ 4 9 1 ⁄ 2 2 1 ⁄ 2 26 to 8 1 7 ⁄ 16 12 3 1 ⁄ 4 4 10 to 12 25 ⁄ 32 9 1 ⁄ 2 2 1 ⁄ 2 28 to 10 1 1 ⁄ 2 12 1 ⁄ 2 3 1 ⁄ 2 4 10 to 12 13 ⁄ 16 9 1 ⁄ 2 2 1 ⁄ 2 28 ... 22 12 15 3 ⁄ 4 1 12 4 1 ⁄ 2 2 1 ⁄ 2 24 7 3 ⁄ 4 11 4 6 1 ⁄ 2 2 1 ⁄...
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Bearing Design in Machinery Episode 2 Part 1 pps

Bearing Design in Machinery Episode 2 Part 1 pps

... 0:877 DT m ¼ 8:3P½R=Cðf Þ 10 6 Q nRCL  1 Àð0:5ÞQ s =Q ¼ 8:3 25 Â0:877 3 :29 Â0: 519 5 ¼ 10 6  C b. Maximum and Average Oil Temperatures: Maximum temperature: T max ¼ T in þ DT ¼ 20 10 6 ¼ 12 3  C Average temperature: T av ¼ ... highly dependent on the film thickness, particularly under high loads. For turbulent fluid films, the equation is T max À T 1 ¼ f p 2 N 2 D 3 2gc p 1...
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Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

... us: M F bc = − 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b 3 kN/m 4 kN 2 m 2 kN 2 m 2 m 4 m a c b 3 ... cd DF 0 0. 625 0.375 0.5 0.5 0 MEM M ab M ba M bc M cb M cd M dc EAM 30 DM +15 +15 COM +7.50 +7.50 DM −4.69 2. 81 COM 2. 35 1. 41 DM +0. 71 +0.70 COM +0.36...
Ngày tải lên : 05/08/2014, 09:20
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Fundamentals of Structural Analysis Episode 2 Part 1 pdf

Fundamentals of Structural Analysis Episode 2 Part 1 pdf

... us: M F bc = − 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b 3 kN/m 4 kN 2 m 2 kN 2 m 2 m 4 m a c b 3 ... abc Member ab bc DF 1 0.5 0.5 0 MEM M ab M ba M bc M cb EAM −4 FEM 2+ 2−4+4 DM 2 COM 1 DM +1. 5 +1. 5 COM +0.8 +0.8 DM -0.8 COM −0.4 DM +0 .2 +0 .2 COM +...
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Fundamentals of Structural Analysis Episode 2 Part 5 pptx

Fundamentals of Structural Analysis Episode 2 Part 5 pptx

... CJ. (F CJ ) max =10 [0.5 (1. 18)(0. 41) +0.5(6)(0. 41) −0.5 (1. 18)(0. 41) (1. 18/6)] = 14 .65 kN 10 kN/m 6 m 1 .25 1 .25 0. 625 0. 41 F CJ 10 kN/m 6 m 9 m 1. 18 1. 82 1 .25 1 .25 0. 625 0. 41 F CJ 6 m 9 m 1. 18 1. 82 Influence ... maximize F CJ . 1kN 1kN 2 kN 1m 2m 2 kN 10 kN/m 10 kN/m x 6 m 1 .25 1 .25 0. 625 0. 41 F CJ 2 kN 1 .25 1 .25 0. 625 0...
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Fundamentals Of Structural Analysis Episode 2 Part 1 pps

Fundamentals Of Structural Analysis Episode 2 Part 1 pps

... +7.50 DM −4.69 2. 81 COM 2. 35 1. 41 DM +0. 71 +0.70 COM +0.36 0.35 DM −0 .22 −0 .14 COM −0 .11 −0.07 DM +0.04 +0.03 COM +0. 02 +0. 02 DM −0. 01 −0. 01 COM 0.00 0.00 SUM 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 +7.87 In ... us: M F bc = − 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m...
Ngày tải lên : 05/08/2014, 11:20
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Fundamentals Of Structural Analysis Episode 2 Part 8 pptx

Fundamentals Of Structural Analysis Episode 2 Part 8 pptx

... 12 3 D’Alembert force, 29 4 Deflection, 67, 82, 12 1 , 26 0 transverse, 12 1 truss, 67 Deflection curve, 14 4 sketch of, 14 4 Deformation, 4, 6, 82, 12 1 axial, 12 1 flexural, 12 1 member, 18 shear, 12 1 Degrees ... members. 10 Influence Lines Problem 1. (1) I nflection point -59 .1 62. 5 - 31. 8 -13 .6 9 .1 -18 .2 I nflection point 2. 17 -2. 33 -1. 33 0.67 I nflecti...
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Industrial Machinery Repair Part Episode 2 Part 9 pptx

Industrial Machinery Repair Part Episode 2 Part 9 pptx

... Figure 23 .18 . 456 V-Belt Drives 5V 10 . 01 16 .0 O.D. 40 deg Over 16 .0 O.D. 42 deg Under 10 .0 O.D. 38 deg Figure 23 .20 Sheave gauge Figure 23 . 21 Final alignment 444 V-Belt Drives Figure 23 .3 Standard ... diameter × 3 .14 2 + driven diameter × 3 .14 2 + center to center × 2 Use the formula above to find the belt length. Belt length = 6" × 3 .14 2 + 12 & quot; × 3...
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