Illustrated Sourcebook of Mechanical Components Part 5 pdf

... sides of sheet. FIG. 11-For attachment of recessed, solid metallic or non-metallic parts in thick-sectioned cylindrical parts. FIG. 12-For closing the end(s) of cylin- drical parts. ... Removal of the screw permits easy removal of the clampcd part. Heavy pressure toward the side of the key out of contact with the slot may permit slight movement due to the springin...

Ngày tải lên: 11/08/2014, 08:22

... 52 .56 63 53 .6336 54 .7008 50 53 .3333 51 54 .4000 52 55 .4666 53 56 .53 33 54 57 .6000 55 58 .6666 56 59 .7333 58 61.8666 59 62.9333 57 6o.8000 55 .3333 56 .4000 57 .4666 58 .53 33 59 .6000 ... 55 Rc 55 Rc- 50 RC; 3 75- 4 25 Brinell 2 45- 280 Brinell 55 Rc- 55 Rc 50 Rc 160-200 Brinell . . . . . . . . Material Factor CRI 0. 85 1 .oo 1....

Ngày tải lên: 11/08/2014, 08:22

... angle of less than 1 15 deg ,POB = 0 .5 + 10 10 OB = 0. 25 = 15 deg The power loss at the driver is - 5. 5 in lb - (1 in. X 150 ,000 lb/in./in.) cn = 3.7 x 10 -5 where ... PSI 0.0 35 20,000 300 TO 50 0 LEATHER 0. 25 TO 0. 35 0.0 35 40,000 300 TO 400 WOVEN COTTON 0.2 TO 0.3 WOVEN HAIR 0.2 TO 0.3 0.0 35 30,000 300 TO 400 BALATA 0.3 T...

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... more projecting pins, the 3um of the radii of the pins being equal to the eccentricity of offset of the shafts. The lines of criiter between each pair of pins remain parallel as the coupling ... the event of failure of one of the load-carrying members, the ball will spin within the socket and provide radial restraint. Containment for periods in excess of 25...

Ngày tải lên: 11/08/2014, 08:22

... and Eq (2), accurate values I of torque can be easily determined. ILLUSTRATED SOURCEBOOK of MECHANICAL COMPONENTS SECTION 9 Various Methods of Locking Threaded Members Locking-Type ... centrifugal clutch shoe has a radius of 5. 25 in. and a width of lining of 2 .50 in. Lining pressurc is not to cxcccd 100 psi. Angulai- length of lining is...

Ngày tải lên: 11/08/2014, 08:22

... 0.017 0 139f0.004 0.022 0.210=1=0.0 05 0.032 0.2 75% 0.006/ 0.049 I- 0.139&0.004 0.022 I CI IC -0.0 05 I 1-0 001 -1-1- 0. 052 I 0.010 I 0. 057 0.083 0.010 0 090 0.113 1 0 ... value of 222,000 psi is acceptable, the thickness t can now be calculated from either Eq 3 or 6. From Eq 3: Ski, 5: Dctcrminc thc dish height, I?: I& = B1 = 0.3(...

Ngày tải lên: 11/08/2014, 08:22

... hole in spring end. ILLUSTRATED SOURCEBOOK of MECHANICAL COMPONENTS SECTION 15 12 Ways to put Balls to Work 15- 2 15- 4 Rubber Balls Find Many Jobs 15- 6 Multiple Use of Balls in Milk ... = 0 .57 7, from Fig 5, h = 1. 055 . Thus from Eq 5: I I 0.2 0.3 0.4 0 .50 .6 0.8. 1 ,I? 3 4 56 1 Projection of loading arc, e 4 Radial load function...

Ngày tải lên: 11/08/2014, 08:22

... deflection at F,., = 155 kg is picked off from the chart as being 6,,,1n=o.3 25 mm=0.25h. The stress at this deflection is S.,,.= 45 kg/mm2 (64,000 psi). For t= 1. 75 mm, the chart for ... spring has di- mensions of: D,= 45 mm (1.77 in.), D1=22 .5 mm, 1~1. 75 mm, h~1.3 mm. Then, hlt~0.74, and D,/D1=2. The spring is preloaded with a force F.,,,= 155 kg (343 lb.). F...

Ngày tải lên: 11/08/2014, 08:22

... COS [2~(0.3 15) ] ] 3.73 COS [Zn(0.3 15) ] V = 56 .3 in./sec 3: A= The acceleration is found from Eq 2(2r)(l - 0.134) ah 113 .5& quot; 0. 052 (1 - 0.134 COS 113 .5& apos;)' ... from Eq 5: 2 0. 052 Amax (optimum) = 5. 89 - = 4710 in./sec2 It is also of interest to notice that the maximum pressure angle of the modified cycloid is lower than t...

Ngày tải lên: 11/08/2014, 08:22

... range of 2 .5 to 250 Newtons (about 0 .50 to 55 Ib). Interferences are usually between 0. 25 to 2 .5 mm (about 0.010 to 0.100 in.). If: the coefficient of friction is 0. 15; the ... most useful if parts tend to tangle. The parts drop free of the rotating- barrel sides. By chance selection some of them fall onto the vibrating rack and ore fed out of t...

Ngày tải lên: 11/08/2014, 08:22