Chain drive, Device used for the transmission of power wherever shafts are divided at distances higher than that for which gears are practical.

1. Chain drive Selection
Problem No 1
Using PSG Design Data Book

2. Problem Statement
• A truck equipped with a 9.5 kW engine uses a roller chain as the final drive
to the rear axle. The driving sprocket runs at 900 r.p.m and the driven
sprocket at 400 r.p.m. with a centre distance of approximately 600mm. Select
the roller chain.
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Given Data:
Power = 9.5 kW
Driving speed =900 rpm
Driven Speed= 400 rpm
Approximate centre distance a = 600mm

3. Step 1:Type of Chain
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Roller Chain is selected for the application

4. Determination of Transmission Ratio
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𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑖 =
𝑍2
𝑍1
=
𝑛1
𝑛2
PSG Design Data Book P. No: 7.74
Z1 = Number of teeth on sprocket pinion
Z2 = Number of teeth on sprocket wheel
n1 = speed of rotation of pinion (rpm)
n2 = Speed of rotation of wheel (rpm)
𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑖 =
𝑍2
𝑍1
=
900
400
= 2.25
𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑖 = 𝟐. 𝟐𝟓

5. Standard Number of Teeth on Pinion Sprocket (Z1 & Z2)
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For the calculated transmission ratio (i =2.25)
from
PSG Design Data Book P. No: 7.74
recommended number of teeth on sprocket
Z1 =27
𝒁𝟐 = 𝒊 ∗ 𝒁𝟏
=2.25 * 27 =60.75 ≈ 61
Recommended Z2 = 100 to 120
PSG Data Book Pg.No 7.74
Z2 =61

6. Selection of Standard Pitch (p)
Centre distance a= (30 to 50) p
PSG Design Data Book P. No: 7.74
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑖𝑡𝑐ℎ 𝑝𝑚𝑎𝑥 =
𝑎
30
=
600
30
= 20
𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑖𝑡𝑐ℎ 𝑝𝑚𝑎𝑥 =
𝑎
50
=
600
50
= 12
Choose pitch close to maximum value is p = 15.875 mm
PSG Design Data Book P. No: 7.74
12-10-2021 Chain Drives 6

7. Selection of Chain
Chain no 10 A-2 /DR 50 Roller chain is selected
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8. Calculation of total load (Pt)
Tangential Force
𝑝𝑡 =
102 𝑁
𝑣
(𝑘𝑔𝑓)
𝑝𝑡 =
1020 𝑁
𝑣
𝑁
𝑣 =
𝑧1 ∗ 𝑛1∗ 𝑝
60 ∗ 1000
𝑣 =
27 ∗ 900 ∗ 15.857
60 ∗ 1000
𝑣 = 6.43 m/s
𝑝𝑡 =
1020 ∗ 9.5
6.43
= 1507 𝑁
Pt = 1507 N
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Centrifugal Tension (Pc)
𝑝𝑐 =
𝑤𝑣2
𝑔
𝑘𝑔𝑓
𝑝𝑐 = 𝑤𝑣2
𝑁
𝑝𝑐 = 1.78 ∗ 6.432
𝑁
𝒑𝒄 = 𝟕𝟑. 𝟓𝟗 N
Tension due to sagging (ps)
𝑝𝑠 = 𝑘 ∗ 𝑤 ∗ 𝑎 𝑘𝑔𝑓
𝑝𝑠 = 6 ∗ 1.78 ∗ 9.81 ∗ 0.6 N
𝑝𝑠 = 62.86 N
Total load (pt)
𝑝𝑡 = 𝑝𝑡 + 𝑝𝑐 + 𝑝𝑠
𝑝𝑡 = 1507 + 73.59 + 62.86
𝑝𝒕 = 𝟏𝟔𝟒𝟑. 𝟒𝟓 𝑵

9. Calculation of Design Load
𝐷𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 = 𝑝𝑡 ∗ 𝑘𝑠
=1643.45 *1.5625
𝑫𝒆𝒔𝒊𝒈𝒏 𝒍𝒐𝒂𝒅 = 𝟐𝟓𝟔𝟕. 𝟖 𝑵
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Calculation of FoS
𝐹𝑜𝑠 =
𝐵𝑟𝑒𝑎𝑘𝑖𝑛𝑔 𝐿𝑜𝑎𝑑
𝐷𝑒𝑠𝑖𝑔𝑛 𝐿𝑜𝑎𝑑
𝐵𝑟𝑒𝑎𝑘𝑖𝑛𝑔 𝐿𝑜𝑎𝑑 FoS =
44400
2567.8
𝑭𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝑺𝒂𝒇𝒆𝒕𝒚 = 𝟏𝟕. 𝟐𝟗
Referring Table Pg No 7.77 for the given speed the acceptable factor of
safety is 11.
Since the working FoS (17.29) is greater than the desirable our Design is
safe

10. Bearing Stress
𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑟𝑜𝑙𝑙𝑒𝑟 =
𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
𝜎𝑟𝑜𝑙𝑙𝑒𝑟 =
𝐷𝑒𝑠𝑖𝑔𝑛 𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
=
2567.8
1.4 ∗ 100
= 16.8 𝑁/𝑚𝑚2
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Referring Table Pg No 7.77 for
the given speed the allowable
bearing stress is 22.4
Since the working bearing
stress is FoS (16.8) is lesser
than the desirable our
Design is safe

11. Length of the Chain (L)
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑙𝑖𝑛𝑘𝑠
𝒍𝒑 = 𝟐𝒂𝒑 +
𝒛𝟏 + 𝒛𝟐
𝟐
+
𝒛𝟐 − 𝒛𝟏
𝟐𝝅
𝟐
𝒂𝒑
𝑎𝑝 =
𝑎0
𝑝
=
𝐶𝑒𝑛𝑡𝑟𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑃𝑖𝑡𝑐ℎ
=
600
15.875
= 37.795
𝑙𝑝 = (2 ∗ 37.795) +
27 + 61
2
+
61 − 27
2𝜋
2
37.795
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𝑙𝑝 = 120.36 ≈ 122 𝐿𝑖𝑛𝑘𝑠
𝐴𝑐𝑡𝑢𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝐿 = 𝑙𝑝 ∗ 𝑝
=122*15.875
=1936.75mm

12. Exact Centre Distance (a)
𝒂 =
𝒆+ 𝒆𝟐−𝟖𝑴
𝟒
∗ 𝒑
( PSG Data Book Pg No 7.75)
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𝑒 = 𝑙𝑝 −
𝑧2 + 𝑧1
2
𝑒 = 122 −
27+61
2
=78
𝑀 =
𝑧2−𝑧1
2𝜋
2
=
61−27
2𝜋
2
= 29.28
𝑎 =
78 + 782 − (8 ∗ 29.28)
4
∗ 15.875
a=613.11mm

13. Sprocket Diameters
PCD of Smaller Sprocket
𝑑1 =
𝑝
𝑠𝑖𝑛
180
𝑧1
=
15.875
sin
180
27
= 136.74
𝑑𝑜1 = 𝑑1 + 0.8 𝑑𝑟
𝑑𝑟 = 10.16 𝑓𝑟𝑜𝑚 𝑃𝑆𝐺 𝐷𝐵
𝑑𝑜1 = 136.74 + (0.8 ∗ 10.16) = 144.868
𝒅𝒐𝟏 = 𝟏𝟒𝟒. 𝟖𝟔𝟖𝒎𝒎
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PCD of Larger Sprocket
𝑑1 =
𝑝
𝑠𝑖𝑛
180
𝑧2
=
15.875
sin
180
61
= 308.38
𝑑𝑜2 = 𝑑2 + 0.8 𝑑𝑟
𝑑𝑟 = 10.16 𝑓𝑟𝑜𝑚 𝑃𝑆𝐺 𝐷𝐵
𝑑𝑜1 = 308.38 + (0.8 ∗ 10.16) = 316.51
𝒅𝒐𝟏 = 𝟑𝟏𝟔. 𝟓𝟏𝒎𝒎

14. Problems for Practice
1. Design a chain drive to actuate a compressor from 15 kW electric motor running at 1000
r.p.m., the compressor speed being 350 r.p.m. The minimum centre distance is 500 mm.
The compressor operates 16 hours per day. The chain tension may be adjusted by shifting
the motor on slides.
2. Design a roller chain to transmit power from a 20 kW motor to a reciprocating pump. The
pump is to operate continuously 24 hours per day. The speed of the motor is 600 r.p.m.
and that of the pump is 200 r.p.m. Find: 1. number of teeth on each sprocket; 2. pitch and
width of the chain.
3. Design a chain drive to run a blower at 600 r.p.m. The power to the blower is available
from a 8 kW motor at 1500 r.p.m. The centre distance is to be kept at 800 mm.
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15. Reference
• Machine Design by RS Khurmi and GK Gupta
12-10-2021 15

16. B.Varun
9787608011
bvarun.me@gmail.com
varun.me@srit.org
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