Rigid Coupling

1. 1

2. PRESENTATION TOPIC
RIGID COUPLING

3. INTRODUCTION
R I G I D C O U P L I N G :
 Rigid coupling is a unit of hardware used to joint two
shafts with a motor or mechanical system.
 A device that is used to connect two shafts together for
the purpose of power transmission.
 It may be used to connect two separate systems, such as
a motor and a generator or to repair a connection within
a single system.
 Rigid couplings are used when precise shaft alignment is
required, shaft misalignment will affect the coupling
performance as well as its life.
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4. TYPES OF RIGID COUPLING
 Sleeve or Muff Coupling
 Clamp or Compression Coupling
 Flange Coupling
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5. APPLICATIONS
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Rigid couplings are used mainly on low speed or
low torque applications where the shafts are an
identical size and are not subject to misalignment.
These are the following applications of rigid
coupling used in our daily life.
 Conveyors
 Food processing applications
 Lawn mowers
 Medical equipment's e.g. Xray machines
 Washing machines

6. PROJECT OBJECTIVE:
• Design rigid flange coupling connected between
shaft of a motor and centrifugal pump that producing
a torque.
6

7. ASSUMPTIONS
Materials:
1. Cast Iron (ASTM A-47)
2. Steel (ASTM A-36)
As for the above materials we used the following data;
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 = 𝜎𝑦 = 250 𝑀𝑃𝑎
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑓𝑜𝑟 𝑐𝑎𝑠𝑡 𝑖𝑟𝑜𝑛 = 𝜎𝑦 = 230 𝑀𝑃𝑎
𝐴𝑛𝑑, 𝑤𝑒 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 𝐹. 𝑆 = 8
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑚𝑜𝑡𝑜𝑟 𝑖𝑛 𝑟𝑒𝑣/𝑚𝑖𝑛 = 𝑁 = 2900 𝑟𝑝𝑚
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8. 8
8
Cast iron is used to design for flange and steel is used for the keys, bolts and shaft.
𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡,
𝜎𝑎𝑙𝑙,𝑠𝑡𝑒𝑒𝑙 =
𝜎𝑦
𝐹. 𝑠
=
𝜎𝑦
8
=
250
8
= 31.25 𝑀𝑃𝑎
𝜎𝑎𝑙𝑙,𝑐𝑎𝑠𝑡 𝑖𝑟𝑜𝑛 =
𝜎𝑦
8
=
230
8
= 28.75 𝑀𝑃𝑎
As from the data we collected,
𝑁 = 2900 𝑟𝑝𝑚
𝜔 =
2𝜋𝑁
60
=
2𝜋 × 2900
60
𝜔 = 303.69 𝑟𝑎𝑑/𝑠
and,
𝑃 = 59.13 𝑏ℎ𝑝
𝑃 = 59.13 × 746 ∴ 1𝑏ℎ𝑝 = 746 𝑤𝑎𝑡𝑡
𝑃 = 44110.98 𝑤𝑎𝑡𝑡
CALCULATIONS

9. 9
9
We use the service factor of safety is 1.5 so,
𝑇𝐷 = 1.5 𝑇𝑅
𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡,
𝑃 = 𝑇. 𝜔
𝑇𝑅 =
𝑃
𝜔
=
44110.98
303.69
= 145.25 𝑁𝑚
So,
𝑇𝐷 = 1.5 × 145.25 = 217.88 𝑁𝑚
CALCULATIONS

10. 10
10
SHAFT DESIGN
We use the factor of safety = F. S = 8
So, 𝜎𝑎𝑙𝑙,𝑠𝑡𝑒𝑒𝑙 = 31.25 𝑀𝑃𝑎
𝜏𝑎𝑙𝑙,𝑠𝑡𝑒𝑒𝑙 = 𝜏𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 =
𝜎𝑎𝑙𝑙
2
=
31.25
2
= 15.625 MPa
𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤,
𝜏
𝑟
=
𝑇𝐷
𝐽
𝜏 =
𝑟. 𝑇𝐷
𝜋
2 . 𝑟4
=
2𝑇𝐷
𝜋𝑟3
𝜏𝑎𝑙𝑙,𝑠 =
16𝑇𝐷
𝜋. 𝑑𝑠
3
𝑑𝑠 =
3 16 × 𝑇𝐷
𝜋 × 𝜏𝑎𝑙𝑙,𝑠
𝒅𝒔𝒉𝒂𝒇𝒕 = 𝟒. 𝟏𝟒 𝒄𝒎
In our collected data the diameter of the shaft was 4.3 𝑐𝑚
∴ 𝒅𝒔𝒉𝒂𝒇𝒕,𝒂𝒄𝒕𝒖𝒂𝒍 > 𝒅𝒔𝒉𝒂𝒇𝒕,𝒅𝒆𝒔𝒊𝒈𝒏
So, our shaft is safe.

11. 11
11
HUB DESIGN
𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡,
𝜏ℎ𝑢𝑏 =
16 × 𝑇𝐷
𝜋 𝐷4 − 𝑑4
𝐴𝑠,
𝐻𝑢𝑏 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 𝐷 = 2𝑑𝑠 = 2 × 4.14
𝐷 = 8.28 𝑐𝑚
𝜏ℎ𝑢𝑏 =
16 × 2.17.88 × 8.28 × 10−2
𝜋 8.28 × 10−2 4 − 4.14 × 10−2 4
𝝉𝒉𝒖𝒃 = 𝟐. 𝟎𝟖 𝑴𝑷𝒂
𝐴𝑠, 𝜏ℎ𝑢𝑏,𝑎𝑙𝑙 =
𝜎𝑎𝑙𝑙
2
=
28.75
2
= 14.375 𝑀𝑃𝑎
𝝉𝒉𝒖𝒃 < 𝝉𝒉𝒖𝒃,𝒂𝒍𝒍
So, our hub is safe.

12. 12
12
KEY DESIGN
As we observed that a rectangular key was installed in practical so,
𝑑𝑠 = 4.14 𝑐𝑚 = 41.4 𝑚𝑚
As from keys and coupling table the shaft diameter "41.4 𝑚𝑚" lies between “38 𝑚𝑚”
And “44 𝑚𝑚” so we will take the dimensions of key at “44 𝑚𝑚.
𝐹𝑜𝑟 𝑘𝑒𝑦 𝑤𝑒 𝑢𝑠𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 𝐹. 𝑆 = 6
𝑠𝑜, 𝜎𝑎𝑙𝑙
=
250
6
= 41.6 𝑀𝑃𝑎 𝑎𝑛𝑑 𝜏𝑎𝑙𝑙 = 20.83 𝑀𝑃𝑎
𝐾𝑒𝑦 𝑤𝑖𝑑𝑡ℎ = 𝑊 = 14 𝑚𝑚 = 1.4 𝑐𝑚
𝐾𝑒𝑦 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 𝑡 = 9 𝑚𝑚 = 0.9 𝑐𝑚
𝜎𝑐 =
4𝑇𝐷
𝑙. 𝑡. 𝑑𝑠
𝐴𝑠 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑘𝑒𝑦 = 𝑙 = 1.5𝑑𝑠 = 1.5 × 4.14 = 6.2 𝑐𝑚
𝜎𝑐 =
4 × 217.88
6.21 × 10−2 × 0.9 × 10−2 × 4.14 × 10−2
= 37.66 𝑀𝑃𝑎 ∴ 𝜎𝑎𝑙𝑙,𝑠𝑡𝑒𝑒𝑙 = 𝜎𝑐,𝑎𝑙𝑙
so, 𝝈𝒄 < 𝝈𝒄,𝒂𝒍𝒍
It means key was safe.

13. 13
13
FLANGE DESIGN
𝐴𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤,
𝜏𝑓 =
2𝑇𝐷
𝑙. 𝐷2. 𝑡𝑓
𝑡𝑓 = 0.5𝑑𝑠 = 0.5 × 4.14
𝑡𝑓 = 2.07 𝑐𝑚
𝜏𝑓 =
2 × 217.88
𝜋 × 8.28 × 10−2 2 × 2.07 × 10−2
𝝉𝒇 = 𝟎. 𝟖𝟗 𝑴𝑷𝒂
∴ 𝝉𝒇< 𝝉𝒇,𝒂𝒍𝒍
So, our flange is safe.

14. 14
Add a Footer 14
BOLT DESIGN
𝜏𝑏 =
8𝑇𝐷
𝑛. 𝜋. 𝐷1𝑑1
2
𝑛 =
𝑑(𝑚𝑚)
50
+ 3 =
41.4
50
+ 3
𝑛 = 3.83 𝑜𝑟 𝑛 = 4
𝐴𝑠 𝑡ℎ𝑒 𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎 = 𝐷1 = 3𝑑𝑠 = 3 × 4.14
𝐷1 = 12.42 𝑐𝑚
𝑠𝑜, 𝜏𝑏 =
8 × 217.88
4 × 𝜋 × 12.42 × 10−2 × 𝑑1
2
𝑑1
2
=
8 × 217.88
4 × 𝜋 × 12.42 × 10−2 × 15.625 × 106
𝑑1 = 8.45 × 10−3
𝑚 = 𝟎. 𝟖𝟓 𝒄𝒎

15. 15
15
BOLT DESIGN
𝐴𝑠,
𝜎𝑐,𝑏𝑜𝑙𝑡 =
2𝑇𝐷
𝑛. 𝐷1. 𝑑1. 𝑡𝑓
𝜎𝑐,𝑏𝑜𝑙𝑡
=
2 × 217.88
4 × (12.42 × 10−2) × (0.85 × 10−2) × (2.07 × 10−2)
𝝈𝒄,𝒃𝒐𝒍𝒕 = 𝟒. 𝟗𝟖 𝑴𝑷𝒂
∴ 𝝈𝒄,𝒃𝒐𝒍𝒕,𝒂𝒍𝒍 > 𝝈𝒄,𝒃𝒐𝒍𝒕
So, our bolts are safe.

16. 16
16
STRESSES COMPARISON ON BOLTS
Theoretical Stress on Bolts Actual Stress on Bolts
𝛔𝐜,𝐛𝐨𝐥𝐭 = 𝟒. 𝟗𝟖 𝐌𝐏𝐚 𝛔𝐜,𝐛𝐨𝐥𝐭 = 𝟏. 𝟐𝟔𝟗𝟔 𝐌𝐏𝐚
𝛕𝐛 = 𝟏𝟓. 𝟔𝟐𝟓 × 𝟏𝟎𝟔 𝛕𝐛 = 𝟓. 𝟖𝟕𝟖𝟔 𝐌𝐏𝐚

17. THEORETICAL
AND
ACTUAL READING
COMPARISON
Add a Footer 17
Type Actual Readings
(cm)
Theoretical
Readings
(cm)
Percentage Errors
Diameter of Shaft 4.3 4.1 4.6
Diameter of Hub 9.6 8.28 13.7
Thickness of Hub 6 6.2 3.3
Diameter of Flange 17.8 16.4 7.8
Thickness of Flange 3 2.09 30.3
Thickness of key 0.6 0.9 50
Width of key 1.1 1.4 27.2
Core Diameter of Bolt 1.1 0.85 22.7
Number of Bolts 6 4 33

18. 18
18
DISCUSSIONS
We have collected this data from the Flange Coupling system installed at water supply
system in Wah cantt. The motor system throwing water into a big water tank present at
approximately 100 feet above from ground. As the power of motor can to throw the water up to
140 feet so if due to any reason a low power is supplied to the motor, the motor still can
perform its work. We design the rigid coupling by taking factor of safety “𝐹. 𝑆 = 8” and service
“𝐹. 𝑆 = 1.5”, results are given below;
 Shaft Dia:
The shaft we design was 4.14 𝑐𝑚 while shaft we practically measure was 4.3 𝑐𝑚
i.e., Approx. 0.16 𝑐𝑚 bigger i.e., negligible over designed.
 Hub Dia:
The design hub diameter was 8.28 𝑐𝑚 while practically measure Dia was 9.6 𝑐𝑚.
So, it is to be also negligible in our designed.

19. 19
19
DISCUSSIONS
 Key:
The factor of safety for the keys we used are two times less than shaft
i.e., for shaft we use 𝐹. 𝑆 = 8 and for keys we used 𝐹. 𝑆 = 6. So, in case of any
extraordinary load is applied on the system then key will be failed instead of
whole flange coupling system affected with the stress.
 Bolt:
We design “4” number of bolts having diameters are 0.85 cm. While in
the actual finding there are “6” number of bolts having diameters are 1.1 cm in
our design bolt.

20. 20
20
CONCLUSION
We have found an interesting result that the percentage error in the shaft design was
least as compare to the other components of the flange coupling system. It means that the
Engineer designed the flange coupling system in such a way that the motor shaft will least
be affected in case of any inconvenience scenario. Because if the shaft is failed than it
means water supply company have to change the motor which would be a costly as
compare to any other component failure.

21. CALCULATIONS
PICTORIAL VIEWS
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22. 22