GMAT exam success Episode 1 Part 7 ppt

... -1 27 Electroless Plating 27 .1 Introduction 27- 1 27. 2 Plating Systems 27- 2 27. 3 Electroless Plating Solutions 27- 3 27. 4 Practical Applications 27- 4 27. 5 27. 6 ... of Plating Solutions 27- 7 27. 7 Electroless Plating 27- 7 27. 8 Properties of Chemically Deposited Metal Coatings 27- 10 References 27- 11 27 .1 Introduction In electroless ... Page 10 Monday, April 25, 2005 12 :18 PM© 2006 by Taylor & Francis Group, LLC 25 -1 25 Ink-Jet Printing 25 .1 Introduction 25- 1 25.2 Continuous Jet Printing 25- 1 25.3...

... process.Sheet-metal forming 11 7 //SYS 21/ //INTEGRAS/B&H/PRS/FINALS_ 07- 05-03/ 075 0654 376 -CH002 -1. 3D – 11 6 – [35–248/ 214 ] 9.5.2003 2:05PM3.8CC Sheet-metal shearing process capability chart. 11 6 Selecting candidate ... charts. 10 4 Selecting candidate processes//SYS 21/ //INTEGRAS/B&H/PRS/FINALS_ 07- 05-03/ 075 0654 376 -CH002 -1. 3D – 11 1 – [35–248/ 214 ] 9.5.2003 2:05PM3.6CC Swaging process capability chart.Swaging 11 1 ... tolerances are provided (see 3.9CC).Sheet-metal forming 11 9//SYS 21/ //INTEGRAS/B&H/PRS/FINALS_ 07- 05-03/ 075 0654 376 -CH002 -1. 3D – 11 3 – [35–248/ 214 ] 9.5.2003 2:05PM.Lead times moderate, typically...

... years, by Dawihl (19 41) ,Trent (19 52), Trigger and Chao (19 56), Takeyama and Murata (19 63), Gregory (19 65),Cook (19 73 ), Uehara (19 76 ), Narutaki and Yamane (19 76 ), Usui et al. (19 78 ) and others.The ... classification 12 9800900 10 00 11 00 12 00 13 00 14 0000.5 1. 0 1. 5Temperature at DOC (°C)Notch wear at DOC (mm)Fig. 4 .11 Al2O3/TiC DOC notch wear variation with temperature, turning Inconel 71 8 for ... speed 12 6 Tool damage0 5 10 15 0 5 10 15 00 .1 0.20.30.4020406080Real cutting time (min) Real cutting time (min) 15 0 m min -1 150 m min -1 236 m min -1 TurningMilling236 m min -1 Crater...

... couple moment.Given:M 1 40 lb ft⋅=M230 lb ft⋅=θ 1 20 deg=θ2 15 deg=θ330 deg=Solution:M 1 M 1 cosθ 1 ()sinθ2()M 1 cosθ 1 ()cosθ2()M 1 − sinθ 1 ()⎛⎜⎜⎜⎝⎞⎟⎟⎟⎠= ... couplemoment.Given:F 1 35 N=θ60 deg=F225 N=r 1 175 mm=r2 17 5 mm=Solution:MF 1 − 2 r 1 00⎛⎜⎜⎜⎝⎞⎟⎟⎟⎠F2− 2r2cosθ()F2− 2 r2sinθ()0⎛⎜⎜⎜⎝⎞⎟⎟⎟⎠+= M 16 .63− 7. 58−0⎛⎜⎜⎝⎞⎟⎟⎠Nm⋅= ... k00 1 ⎛⎜⎜⎝⎞⎟⎟⎠=M1zr 1 F 1 ×()k⋅= M1z38.2 lb in⋅=M2zr2F2×()k⋅= M2z 14 .1 lb in⋅=Problem 4-69Determine the magnitude and sense of the couple moment.Units Used:kN 10 3N=2 61 ©...