The document discusses the design of various types of rigid and flexible couplings. It provides steps to design a flange coupling connecting two shafts transmitting 37.5 kW power at 180 rpm. Key details include calculating torque from power, selecting shaft diameter, coupling dimensions based on standards, and checking design of key and bolts for shearing and crushing. The document also provides problems and solutions for designing flange, muff, and clamp couplings for given power and speed conditions.

1. DESIGN OF COUPLINGS UNIT II

2. DESIGN OF COUPLINGS
Types of Shafts Couplings: (A) Rigid Couplings & (B) Flexible
Couplings
Rigid Couplings are used to connect two shafts which are perfectly
aligned. These are simple and inexpensive. Rigid Couplings are of
following types.
(i) Sleeve or Muff Coupling
(ii) Clamp or Split-muff or Compression Coupling
(iii) Flang
Flexible Couplings are of following types.
(i) Bushed pin type Coupling (ii) Universal Coupling (iii) Oldham
Coupling

3. MUFF COUPLING & CLAMP
COUPLING
Outer diameter of the sleeve, D = 2d + 13
[d - the diameter of the shaft] Length of the sleeve,
L = 3.5d
(

4. FLANGE COUPLING

5. DESIGN OF KEYS
Types of Keys 1. Sunk keys 2. Saddle keys 3. Tangent keys 4.
Round keys 5. Splines

6. PROBLEM IN FLANGE COUPLING
Design a rigid type flange coupling to connect two shafts. The input
shaft transmits 37.5 kW power at 180 rpm to the output shaft
through the coupling. The service factor for the application is 1.5.
Select suitable material for various parts of coupling

7. PROBLEM IN FLANGE COUPLING
Given:
Power, P = 37.5 kW = 37500 W
Speed = 180 rpm
Service factor = 1.5 Selection:
Flange material: Cast iron
Allowable shear stress of cast iron,
[τ] = 15 N/mm2
Shaft, key & bolt material: Mild steel
Allowable shear stress of mild steel,[τ] = 40 N/mm2
Allowable normal stress of mild steel,[σ] = 80 N/mm2

8. PROBLEM IN FLANGE COUPLING
Step I: Calculate the torque from power
Calculated Torque, Mt = 1989.179 N.m = 1989179
N-mm Design Torque, Mt = Calculated Torque x
service factor = 1989179 x 1.5 = 2983768 N-mm

9. PROBLEM IN FLANGE COUPLING
Step II: Calculate the shaft diameter from torque

10. PROBLEM IN FLANGE COUPLING
Step III: Calculate the coupling dimensions as per standard proportion
Outer diameter of hub, D = 2d = 2x73 = 146 mm
Pitch Circle diameter of bolts, D1 = 3d = 3x73 = 219 mm
Outer diameter of flange, D2 = 4d = 4x73 = 292 mm
Length of hub, L = 1.5d =1.5x73 = 110 mm
Thickness of flange, tf = 0.5d =0.5x73 = 36.5 mm

11. PROBLEM IN FLANGE COUPLING
Step IV: Design of key
The size of the key is selected on the basis of shaft diameter from
data book and then it is checked for shearing and crushing to validate
the selection.
key width, b = 20 mm [PSG 5.16]
key thickness or height, h = 12 mm
key Length, L = 110 mm
{Key length is equal to hub length}

12. PROBLEM IN FLANGE COUPLING
Check for shearing of key: Mt = L x b x τ x (d/2) 2983768 = 110 x 20 x τ x
(73/2)
τ = 37.2 < 40(Allowable shear stress for key). Design is safe.
Check for crushing of key: Mt = L x (h/2) x σ x (d/2) 2983768 = 110 x
(12/2) x σ x (73/2)
σ = 123.8 > 80(Allowable normal stress for key).
Design is not safe Select a higher thickness for key to meet the crushing
requirement. 20mm thickness is selected.
For 20 mm thickness,
2983768 = 110 x (20/2) x σ x (73/2)
σ = 74.3 < 80(Allowable normal stress for key). Design is safe
Selected key - FINAL: 20 x 20 x 110

13. PROBLEM IN FLANGE COUPLING
Step V: Design of bolts
The size and number of bolts are selected from design data book on the
basis of shaft diameter and then it is checked for shearing and crushing to
validate the selection.
Selected bolt diameter, d = 18 mm [PSG 7.107]
Selected Number of bolts, n = 6
Check for shearing of bolt: Mt = (π /4) x d12 x τ x n x (D1/2)
2983768 = (π /4) x 182 x τ x 6 x (219/2)
τ = 17.8 < 40(Allowable shear stress for bolt). Design is safe.
Check for crushing of bolt: Mt = d1 x tf x σ x n x (D1/2)
2983768 = 18 x 36.5 x σ x 6 x (219/2)
σ = 6.9 < 80(Allowable normal stress for key). Design is safe

14. PROBLEM IN FLANGE COUPLING
STEP VI: Design check for Hub & flange

15. PROBLEM IN FLANGE COUPLING
Design a cast iron flange coupling for a mild steel shaft transmitting
90 kW at 250 rpm. The allowable shear stress in the shaft is 40 MPa
and the angle of twist is not to exceed 10 in a length of 20 diameters.
The allowable shear stress in the coupling bolts is 30 MPa.

16. PROBLEM IN FLANGE COUPLING
Given:
Power, P = 90 kW = 90000 W
Speed = 250 rpm
Allowable shear stress of shaft, [τ] = 40 MPa = 40 N/mm2
Allowable shear stress of bolt, [τ] = 30 MPa = 30 N/mm2
Allowable angle of twist in shaft,ϴ = 10 in a length of 20d
Selection:
Flange material: Cast iron
Allowable shear stress of cast iron,[τ] = 15 N/mm2
Allowable normal stress of shaft, key & bolt, [σ] = 80 N/mm2

17. PROBLEM IN FLANGE COUPLING
STEP I: Calculate the torque from power

18. PROBLEM IN FLANGE COUPLING
STEP II: Calculate the shaft diameter from torque

19. PROBLEM IN FLANGE COUPLING
STEP III: Calculate the coupling dimensions as per standard proportion
Outer diameter of hub,
D = 2d = 2x76 = 152 mm
Pitch Circle diameter of bolts,
D1 = 3d = 3x76 = 228 mm
Outer diameter of flange,
D2 = 4d = 4x76 = 304 mm
Length of hub, L = 1.5d =1.5x76 = 114 mm
Thickness of flange, tf = 0.5d =0.5x76 = 38 mm

20. PROBLEM IN FLANGE COUPLING
STEP IV: Design of key
The size of the key is selected on the basis of shaft diameter from data book
and then it is checked for shearing and crushing to validate the selection.
key width, b = 22 mm [PSG 5.16]
key thickness or height, h = 14 mm
key Length, L = 114 mm {Key length is equal to hub length}
Check for shearing of key:
Mt = L x b x τ x (d/2)
3437301 = 114 x 22 x τ x (76/2)
τ = 36.1 < 40(Allowable shear stress for key). Design is safe.

21. PROBLEM IN FLANGE COUPLING
Check for crushing of key: Mt = L x (h/2) x σ x (d/2)
3437301 = 114 x (14/2) x σ x (76/2)
σ = 113 > 80(Allowable normal stress for key). Design is not safe
Select a higher thickness for key to meet the crushing
requirement.20mm thickness is selected.
For 20 mm thickness,
3437301 = 114 x (20/2) x σ x (76/2)
σ = 79.3 < 80(Allowable normal stress for key).
Design is safe Selected key - FINAL: 22 x 20 x 114

22. PROBLEM IN FLANGE COUPLING
STEP V: Design of bolts
The size and number of bolts are selected from design data book on the basis
of shaft diameter and then it is checked for shearing and crushing to validate
the selection.
Selected bolt diameter, d = 18 mm [PSG 7.107]
Selected Number of bolts,
n = 6 Check for shearing of bolt:
Mt = (π /4) x d12 x τ x n x (D1/2)
3437301 = (π /4) x 182 x τ x 6 x (228/2)
τ = 19.7 < 30(Allowable shear stress for bolt). Design is safe.
Check for crushing of bolt: Mt = d1 x tf x σ x n x (D1/2)
3437301 = 18 x 38 x σ x 6 x (228/2)
σ = 7.3 < 80(Allowable normal stress for key).

23. PROBLEM IN FLANGE COUPLING

24. PROBLEM IN BUSHED-TYPE
FLEXIBLE COUPLINGDesign a bushed-type flexible coupling to connect a pump shaft to a
motor shaft transmitting 32 kW at 960 rpm. The overall torque is 20
percent more than mean torque. The allowable shear and crushing
stress for shaft and key material is 40 MPa and 80 MPa respectively.
Allowable shear for cast iron is 15 MPa. The allowable bearing
pressure for rubber bush is 0.8 N/mm2. Material of the pin is same
as that of shaft and key

25. PROBLEM IN BUSHED-TYPE
FLEXIBLE COUPLING
Given:
Power, P = 32 kW = 32000 W
Speed = 960 rpm
Allowable shear stress of shaft, key & pin,
[τ] = 40 MPa = 40 N/mm2
Allowable crushing (normal) stress of shaft, key & pin,
[σ] = 80 N/mm2
Allowable shear stress of cast iron,[τ] = 15 N/mm2
Allowable bearing pressure in rubber bush = 0.8 N/mm2

26. PROBLEM IN BUSHED-TYPE FLEXIBLE
COUPLING

27. PROBLEM IN BUSHED-TYPE
FLEXIBLE COUPLING

28. PROBLEM IN BUSHED-TYPE
FLEXIBLE COUPLINGSTEP IV: DESIGN CHECKS FOR SELECTED COUPLING

29. PROBLEM IN BUSHED-TYPE
FLEXIBLE COUPLING

30. PROBLEM IN BUSHED-TYPE
FLEXIBLE COUPLING

31. PROBLEM IN BUSHED-TYPE FLEXIBLE
COUPLING

32. PROBLEM IN MUFF COUPLING
Design and make a neat dimensioned sketch of a muff coupling
which is used to connect two steel shafts transmitting 40 kW at 350
r.p.m. The material for the shafts and key is plain carbon steel for
which allowable shear and crushing stresses may be taken as 40 MPa
and 80 MPa respectively. The material for the muff is cast iron for
which the allowable shear stress may be assumed as 15 MPa.
Solution. Given : P = 40 kW = 40 × 103 W; N = 350 r.p.m.; τs = 40
MPa = 40 N/mm2; σcs = 80 MPa = 80 N/mm2; τc = 15 MPa = 15
N/mm2

33. PROBLEM IN MUFF COUPLING

34. PROBLEM IN MUFF COUPLING
Design for key From Table 13.1, we find that for a shaft of 55 mm
diameter,
Width of key, w = 18 mm Ans. Since the crushing stress for the key
material is twice the shearing stress, therefore a square key may be
used.
∴ Thickness of key, t = w = 18 mm Ans.
We know that length of key in each shaft,
l = L / 2 = 195 / 2 = 97.5 mm Ans. Let us now check the induced
shear and crushing stresses in the key.
First of all, let us consider shearing of the key.
We know that torque transmitted (T),

35. PROBLEM IN MUFF COUPLING

36. PROBLEM IN CLAMP COUPLING
Design a clamp coupling to transmit 30 kW at 100 r.p.m. The
allowable shear stress for the shaft and key is 40 MPa and the
number of bolts connecting the two halves are six. The permissible
tensile stress for the bolts is 70 MPa. The coefficient of friction
between the muff and the shaft surface may be taken as 0.3.
Solution. Given : P = 30 kW = 30 × 103 W; N = 100 r.p.m. ; τ = 40 MPa
= 40 N/mm2 ; n = 6 ; σt = 70 MPa = 70 N/mm2; μ = 0.3 ff and the
shaft surface may be taken as 0.3.

37. PROBLEM IN CLAMP COUPLING

38. PROBLEM IN CLAMP COUPLING