CHAIN DRIVE WITH SPROCKETS
This is one of the most important parts of a SPV which gives the final push or
motion to the wheels and make the vehicle move. It primarily consists of a chain
and sprocket drive
A sprocket is a toothed wheel upon which a chain rides. Contrary to popular
opinion, a sprocket is not a gear.
Fig.
Chain Construction
Chains have a surprising number of parts. The roller turns freely on the bushing,
which is attached on each end to the inner plate. A pin passes through the bushing,
and is attached at each end to the outer plate. Bicycle chains omit the bushing,
instead using the circular ridge formed around the pin hole of the inner plate.
Fig.
Chain Dimensions
Chain types are identified by number; i.e. a number 40 chain. The rightmost digit is
0 for chain of the standard dimensions; 1 for lightweight chain; and 5 for roller less
bushing chain. The digits to the left indicate the pitch of the chain in eighths of an
inch. For example, a number 40 chain would have a pitch of four-eighths of an
inch, or 1/2", and would be of the standard dimensions in width, roller diameter,
etc.
The roller diameter is "nearest binary fraction" (32nd of an inch) to 5/8ths of the
pitch; pin diameter is half of roller diameter. The width of the chain, for "standard"
(0 series) chain, is the nearest binary fraction to 5/8ths of the pitch; for narrow
chains (1 series) width is 41% of the pitch. Sprocket thickness is approximately 85-
90% of the roller width.
Sprockets
There are four types of sprocket;
Type A: Plain Plate sprockets
Type B: Hub on one side
Type C: Hub on both sides
Type D: Detachable hub
Sprockets should be as large as possible given the application. The larger a
sprocket is, the less the working load for a given amount of transmitted power,
allowing the use of a smaller-pitch chain. However, chain speeds should be kept
less than 1200 feet per minute.
The dimensions of a sprocket can be calculated as follows, where P is the pitch of
the chain, and N is the number of teeth on the sprocket;
Pitch Diameter = P ÷ sin (180° ÷ N)
Outside Diameter = P × (0.6 + cot (180° ÷ N))
Sprocket thickness = 0.93 × Roller Width - 0.006
Application
Sprockets should be accurately aligned in a common vertical plane, with their axes
parallel. Chain should be kept clean and well lubricated with thin, light-bodied oil
that will penetrate the small clearances between pins and bushings.
5. COMPONENT DESIGN
Detailed part design
5.1 Chain and Sprocket design
(All equation nos. and table nos. in the following design correspond to Machine
Design Data Handbook-Vol. I by Prof. K. Lingaiah)
5.1 Chain and Sprocket design
Power to be transmitted, N = 0.25 HP (Losses are neglected)
= (1/4) * 746 Watts
= 0.1865 KW
Speed of driver sprocket, n1 = 150 rpm
Speed of driven sprocket, n2 = 150 rpm
Distance b/w Sprockets, C = 560 mm
For Roller Simplex Chain No. 04C (from table (21.67/Pg 355)),
Pitch, p = 6.35 mm
Measuring load, w = 0.05 KN/m
Breaking load, Fu = 3.4 KN
Transmission Ratio, i = (n1/n2) = (150/150) = 1
Taking No. of teeth on driver sprocket, Z1 = 36
No. of teeth on driven sprocket, Z2 = i*Z1
= 36
Pitch diameter of driver sprocket, d1= p ÷ sin(180/Z1) .…(21.126/pg316)
= 6.35 ÷ sin(180/36)
= 72.86 mm
Pitch diameter of driven sprocket, d2= i*d1
= 72.86 mm
Velocity of chain, V= (p*Z1*n1) ÷ (60*1000) ….(21.105/pg313)
= (6.35*36*150) ÷ (60*1000)
= .5715 m/s
From table (14.4/pg189),
load factor, Kl= 1.75
From table (21.61/pg351),
service factor, Ks= 1.2
Power, N= (Fθ*V) ÷ (1000*Kl*Ks) ….(21.115a/pg315)
=> .1865= (Fθ*.5715) ÷ (1000*1.75*1.2)
Required chain pull, Fθ= 665.72 N
Allowable pull, Fa= Fu/no ….(21.113/pg314)
Here working factor, no= 5
=>Fa= (3.4*103)/5
= 680 N
No. of strands, is= Fθ/Fa ….(21.118/pg315)
= 665.72/680
= .979 < 1
It shows that a single strand (simplex) chain is sufficient to carry the
required pull.
Length of chain in pitches,
Lp= 2*Cp*cosα + (Z1+Z2)/2 + α*(Z1-Z2)/180 ….(21.122/pg316)
Where, α= sin-1((d1-d2)/(2*C))
=0
And, Cp= C/p ….(21.119/pg315)
=> Lp= 2*560*cos0ْ /6.35 + (36+36)/2 + 0
= 212.38 pitches
= 214 pitches (nearest even no.)
Length of chain, L = p*Lp ….(21.125/pg316)
= 6.35*214
= 1358.9 mm
Substituting the value of Lp back in the above equation,
=>214= 2*C*cos0ْ /6.35 + (36+36)/2 + 0
Correct centre distance, C= 565.15 mm
Check for actual factor of safety
Chain pull, Fθ= 1000*N/V ….(21.117b/pg315)
= 1000*.1865/.5715
= 326.33 N
Centrifugal tension, Fcs= w*V2/g ….(21.117a/pg315)
= .05*1000*.57152/9.8066
= 1.665 N
From table (21.58/pg351), for horizontal drive,
Coefficient of sag, Ksg= 6
Tension due to chain sag, Fs= Ksg*w*C …..(21.117/pg315)
Fs= 6*.05*1000*560/1000
= 168 N
Actual FOS, no= Fu/(Fθ+Fcs+Fs) ….(21.117/pg315)
=>no= 3.4*1000/(326.33+1.665+168)
= 6.85 > 5
Hence, selection of chain is safe.
