Page 187 - Analysis and Design of Machine Elements

Chain Drives
The selection of lubrication method depends on the speed of operation and the power 165
being transmitted. Commonly used methods are illustrated in Table 7.1. Where the
chain would be exposed to considerable dirt, dust or moisture, a protective casing should
be used.
References

1 Mott, R.L. (2003). Machine Elements in Mechanical Design, 4e. Prentice Hall.
2 Pu, L.G. and Ji, M.G. (2006). Mechanical Design, 8e. Beijing: Higher Education Press.
3 Juvinall, R.C. and Marshek, K.M. (2011). Fundamentals of Machine Component
Design, 5e. New York: Wiley.
4 Budynas, R.G. and Nisbett, J.K. (2011). Shigley’s Mechanical Engineering Design,9e.
New York: McGraw-Hill.
5 Hindhede, U., Zimmerman, J.R., Hopkins, R.B. et al. (1983). Machine Design Funda-
mentals: A Practical Approach.New York:Wiley.
6 ANSI/ASME B29.1-2011 (2011). Precision Power Transmission Roller Chains, Attach-
ments, and Sprocket. Three Park Avenue, New York, NY: The American Society of
Mechanical Engineers.
7 ISO 606:2015 (2015). Short-Pitch Transmission Precision Roller and Bush Chains,
Attachments and Associated Chain Sprockets. Geneva, Switzerland: International
Organization for Standards.
8 GB/T 1243-2006 (2006). Short-Pitch Transmission Precision Roller and Bush Chains,

Attachments and Associated Chain Sprockets. Beijing: Standardization Administra-
tion of the People’s Republic of China.
9 Wen, B.C. (2015). Machine Design Handbook, 5e, vol. 2. Beijing: China Machine
Press.
10 Collins, J.A. (2002). Mechanical Design of Machine Elements and Machines: A Failure
Prevention Perspective, 1e. New York: Wiley.
11 ISO 10823:2004 (2004). Guidelines for the Selection of Roller Chain Drives.Geneva,
Switzerland: International Organization for Standards.
12 American Chain Association (1982). Chains for Power Transmission and Material
Handling.New York:MarcelDekker.

Problems

Review Questions

1 List three typical failure modes of roller chains.

2 Is it possible to express the speed ratio of a chain drive as i = n 1 = z 2 = d 2 ?
n 2 z 1 d 1
Explain why.