Unit 2.3 Design of Coupling

11/5/2019 Unit II Coupling 1
COUPLING

2
Coupling
• Shafts are usually available up to 7 m length due to
inconvenience in transport. In order to have a greater length,
it becomes necessary to join two or more pieces of the shaft
by means of a coupling.
• A Coupling can be defined as a mechanical device that
rigidly joins two rotating shafts to each other.
• Uses:-
 To Provide connections of shafts
 To provide mechanical flexibility for misalignment of
shaft
 To reduce transmission of shock loads from one shaft to
another shaft.
 To introduce protection against overloads.
11/5/2019 Unit II Coupling

Unit II Coupling 3
Types of coupling
1. Rigid coupling- to connect two shafts which are
perfectly aligned.
(a) Sleeve or muff coupling
(b) Clamp or compressive coupling
(c) Flange coupling
2. Flexible coupling- to connect two shafts having both
lateral and angular misalignment.
(a) Bushed pin coupling
(b) Universal coupling
(c) Oldham coupling
11/5/2019

Unit II Coupling 5
1. Sleeve or Muff coupling
11/5/2019
Problem 1. Design a muff
coupling which is used to
connect two steel shafts
transmitting 25 kW power
at 360 rpm. The shafts and
key are made of plain
carbon steel 30C8
(Syt=Syc= 400N/mm2). The
sleeve is made of grey
cast iron FG 200 (Sut= 200
N/mm2). The factor of
safety for the shafts and
key is 4. For sleeve the
factor of safety is 6 based
on ultimate strength.
Nov-Dec 2016, April May- 2013

Unit II Coupling 6
Design of sleeve or muff coupling
1. Design for shaft:
find diameter of the shaft using equation
2. Design for sleeve:
• Find outer dia. of sleeve using relation
D = 2d +13 mm & L= 3.5 d
• Check induced shear stress in the sleeve using equation
Where Mt = Torque to be transmitted by the coupling
τ = Permissible shear stress for the material of the sleeve
which is cast iron
3
1660
2
t
t
MP
M and
n d

 

 
 44
16
dD
DMt




11/5/2019

Unit II Coupling 7
Cont…
3. Design for key:
 For Parallel (rectangular and square sunk) keys
• Refer Data book page no. 5.16 and select width(b) and height (h)
of the key for given shaft diameter (d).
• Length of the coupling key = 3.5 d
length of the key in each shaft l = L/2 = 3.5d/2
 Check for shearing and crushing stress
(considering shear of key)
(considering crushing of key)
2 tM
dbl
 
4 t
c
M
dhl
 
11/5/2019

Data Book page No. 5.16
DIMENSIONS OF PARALLEL KEYS AND KEYWAYS

Unit II Coupling 9
Problem 1. Design a muff coupling which is used to
connect two steel shafts transmitting 25 kW power at
360 rpm. The shafts and key are made of plain carbon
steel 30C8 (Syt=Syc= 400N/mm2). The sleeve is made of
grey cast iron FG 200 (Sut= 200 N/mm2). The factor of
safety for the shafts and key is 4. For sleeve the factor of
safety is 6 based on ultimate strength.
Nov-Dec 2016, April May- 2013
11/5/2019

Unit II Coupling 10
Problem 2. Design a rigid muff coupling. Use CI for
the muff. The power transmitted is 25 kW at 300
rpm. Ultimate tensile strength =200 MPa and
factor of safety =6. Use 30C8 for the shaft
considered. Yield Point stress = 330 Mpa and
factor of safety =4.
(April-May 2012, Nov- Dec 2013)
11/5/2019

Unit II Coupling 11
Problem 3. Design and make a neat dimensioned
sketch of a muff coupling, which is used to
connect two steel shafts transmitting 40 kW at
350 rpm. The material for the shaft 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 15 MPa.
(Nov-Dec 2012)
11/5/2019

Unit II Coupling 12
2. Clamp or compression coupling (split muff coupling)
11/5/2019

Unit II Coupling 13
Dimensions for clamp coupling
 For sleeve halves,
Outer diameter of the sleeve D= 2.5 d,
Length of the sleeve L= 3.5 d
 For clamping bolts,
Diameter of the claming bolt may be find out either
1. By using empirical equations:
d1= 0.2 d + 10 mm (when d< 55 mm)
And d1= 0.15 d + 15 mm (when d> 55 mm)
or
2. By using following equations: (assuming that power is transmitted
by friction)
Where P1 = Tensile force on each bolt
d1 = core diameter of clamping bolt
n = total no. of bolts & f= coefficient of friction
2
1 1 1
2
4
t
t
M
P and P d
fdn

 
11/5/2019

Unit II Coupling 14
Design Procedure
Step 1. Calculate the diameter of each shaft.
Step 2. Calculate the main dimensions of the sleeve halves.
Step 3. Determine the standard cross-section of flat key from data book
page no. 5.16.
Length of the key in each shaft l = L / 2 = 3.5 d / 2
check the shear stress and compressive stress by using,
and
Step 4. calculate the diameter of clamping bolts by using
2 tM
dbl
 
4 t
c
M
dhl
 
2
1 1 1
2
4
t
t
M
P and P d
fdn

 
11/5/2019

Unit II Coupling 15
Problem 1. It is required to design a split muff coupling to
transmit 50 kW power at 120 rpm. The shafts, key and
coupling bolts are made of plain carbon steel 30C8 (Syt= 400
N/mm2). The yield strength in compression is 150% of tensile
yield strength. The factor of safety for shafts key and bolts is
5. The number of clamping bolts is 8. The coefficient of friction
between sleeve halves and the shaft is 0.3.
(i) Calculate the diameter of input and output shaft.
(ii) Specify length and outer diameter of sleeve halves.
(iii) Find out the diameter of clamping bolts assuming that the
power is transmitted by friction.
(iv) Specify bolt diameter using standard empirical relations.
(v) Specify the size of key and check the dimensions for shear
and compression criteria.
11/5/2019
April-May 2017

Unit II Coupling 16
Problem 2. design a clamp coupling to transmit 30
kW at 100 rpm. 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.
11/5/2019

Unit II Coupling 17
3. Rigid flange coupling
(A) Unprotected type flange coupling (B) Protected type flange coupling
11/5/2019

Advantages:-
1.High torque
transmitting capacity
2.Easy to assemble and
dismantle.
3.Simple construction,
easy to design and
manufacture.
Disadvantages:-
1.Does not tolerate
misalignment.
2.Unsuitable for shock
and vibration.
3.Requires more radial
spaces.
Unit II Coupling 1811/5/2019

Unit II Coupling 19
Protected type flange coupling
11/5/2019

Unit II Coupling 20
Dimensions for protected type flange coupling
• Outside diameter of hub dh= 2d
• Length of hub or effective length of key lh = 1.5 d
• Pitch circle diameter of bolts D= 3d
• Thickness of flanges t = 0.5 d
• Thickness of protective rim t1 = 0.25 d
• Diameter of spigot and recess dr = 1.5 d
• Outside diameter of flange Do= 4d+2t1
• Number of bolts = 3, for d < 40 mm
= 4, for 40 ≤ d <100 mm
= 6, for 100 ≤ d < 180 mm
11/5/2019

Design of flange coupling
Unit II Coupling 21
1. Shaft diameter
find diameter of the shaft using equations
3
1660
2
t
t
MP
M and
n d

 

 
11/5/2019

Unit II Coupling 22
2. Dimensions of flange:
dh= 2d
lh = 1.5 d
D= 3d
t = 0.5 d
 Check for torsional shear stress in the hub:-
 Check for shear stress in the flange at the junction with
the hub:-
t1 = 0.25 d
dr = 1.5 d
Do= 4d+2t1
td
M
h
t
2
2

 
 4 4
32 2
ht h
d dM r d
where J and r
J



  
11/5/2019

3. Diameter of bolts:
Number of bolts N= 3, for d < 40 mm
N= 4, for 40 ≤ d <100 mm
N= 6, for 100 ≤ d < 180 mm
diameter of bolt is determined by using equation:-
Check compressive stress in bolt:-
Unit II Coupling
23
DN
M
d t82
1 
tDNd
Mt
c
1
2

11/5/2019

THREADS FOR BOLTS AND NUTS
COARSE AND FINE SERIES

Unit II Coupling 25
4. Design for key:
 For Parallel (rectangular and square sunk) keys
• Width X height (b x h) of the key:-
Refer PSG Data book page no. 5.16
• length of the key in each shaft l = lh = 1.5 d
 Check shearing and crushing stress
2 tM
dbl
 
4 t
c
M
dhl
 
11/5/2019

Unit II Coupling 26
Problem 1. Design a cast iron protective type flange
coupling to transmit 15 kW at 900 rpm from an electric
motor to a compressor. The service factor may be
assumed as 1.35. The following permissible stresses
may be used:
Shear stress for shaft, bolt and key material = 40 MPa
Crushing stress for bolt and key = 80 MPa
Shear stress for cast iron = 8 MPa
Draw also a neat sketch of the coupling.
(April May 2015, Nov-Dec 2013)
11/5/2019

Unit II Coupling 27
Problem 2. Design and draw a protective type of
cast iron flange coupling for a steel shaft
transmitting 15 kW at 200 rpm and having an
allowable shear stress of 40 MPa. The working
stress in the bolts should not exceed 30 MPa.
Assume that the same material is used for shaft
and key and that the crushing stress is twice the
value of its shear stress. The maximum torque is
25% greater than the full load torque. The shear
stress for cast iron is 14 MPa. (2008 Supp.)
11/5/2019

Unit II Coupling 28
Problem 3. Two 35 mm shafts are connected by a flanged
coupling. The flanges are fitted with 6 bolts on 125 mm
pitch circle diameter of bolts. The shaft transmit a torque
of 800 N-m at 300 rpm. For the safe stresss mentioned
below calculate: (i) diameter of bolts (ii) thickness of
flanges (iii) Key dimensions (iv) hub length and (v) Power
transmitted
Safe shear stresses for shaft material = 63 MPa, Safe
shear stress for bolt material = 56 MPa, safe shear stress
for cast iron coupling = 10 MPa, Safe shear stress for key
material = 46 MPa
Also draw the line sketch of flange coupling.
(Nov-Dec 2011)
11/5/2019

Unit II Coupling 29
Problem 4. A rigid flange coupling is used to transmit 15
kW power at 720 rpm between two steel shaft. The shaft
key and bolt are made of plain carbon steel 30C8 (Syt=
400 N/mm2), FOS= 3. The yield strength in compression
may be taken as 150 % of tensile yield strength. The
flanges are made of grey cast iron of grade FG200 (Sut=
200 N/mm2), FOS= 6. The keys have square cross
section. Design a coupling and specify the dimensions of
its part.
11/5/2019

Unit II Coupling 30
Problem 5. A rigid flange coupling is required to transmit 50
kW at 300 rpm. There are six bolts. The outer diameter is of
flange is 200 mm and diameter of recess is 150 mm. The
coefficient of friction between the flanges is 0.15. The shaft
and bolt are made of plain carbon steel 45C8 (Syt= 380
N/mm2), fos= 3. Determine the diameter of the bolts.
Assume that the bolts are set in large clearance holes.
11/5/2019

(April-May 2016)

(April-May 2015)

(Nov-Dec 2015)

(Nov-Dec 2014)

Unit II Coupling 35
4. Bushed-pin flexible coupling
11/5/2019

36
Step 1. Shaft diameter:-
Step 2 . Dimension of flange:-
• dh= outside diameter of hub= 2d
• lh= length of hub or effective length of key = 1.5d
• D= pitch circle diameter of pins= 3d to 4d
• t= thickness of output flanges = 0.5d
• t1= thickness of protective rim= 0.25d
• The torsional shear stress in the hub is given by-
Design procedure for flexible coupling
 4 4
32 2
ht h
d dM r d
where J and r
J



  

Unit II Coupling 37
• The shear stress in the flange at the junction with the hub is given by:-
Step 3. Dimensions of bushes:-
• Outer diameter of rubber bush (Db) is determined from the equation.
• Where pm is pressure between bush and C.I. flange (usually 1N/mm2 )
• Effective length of the rubber bush lb= Db
Step 4. Diameter of Pins:-
• Diameter of pins , Number of pins (n) is usually 4 or 6.
• Determine the shear stress in the pins by,
11/5/2019
td
M
h
t
2
2

 

Unit II Coupling 38
Step 5. Check for bending stress in pins:
 Torque transmitted by the coupling:
From above equation force P on each rubber bush or pin is determined.
 Bending moment on the pin is given by:
 Bending stress is checked by equation:
11/5/2019

Unit II Coupling 39
Step 6. Dimensions of keys:
 Determine the standard cross-section of flat key from
data book page no. 5.16.
 Length of the key l = length of hub = lh =1.5d
 check the shear stress and compressive stress by using,
2 tM
dbl
 
4 t
c
M
dhl
 
11/5/2019

Unit II Coupling 40
Problem 1. A bushed pin type flexible coupling is used to
connect two shafts and transmit 5 kW power at 720 rpm.
Shafts, keys and pins are made of commercial steel
(Syt=Syc=240 N/mm2) and the factor of safety is 3. The flanges
are made of grey cast iron FG200 (Sut= 200 N/mm2) and the
factor of safety is 6. Assume
Ssy=0.5 Syt and Ssu= 0.5 Sut
There are 4 pins. The pitch circle diameter of the pins is four
times of shaft diameter. The permissible shear stress for pins
is 35 N/mm2. The bearing pressure for rubber bushing is 1
N/mm2. The keys have square cross section. Calculate:
(i) diameter of the shafts
(ii) dimensions of flange and hub
(iii) dimensions of the key
(iv) diameter of the pins
(v) outer diameter and effective length of the bushes.
11/5/2019

Unit II Basic Elements Design 41
Assignment-2
Que.1 Design a Knuckle joint (with fork and eye) to with stand a load of
10,000 N. The materials used for all components have the following
properties ultimate tensile strength = ultimate compressive strength
= 480 N/mm2. Shear strength = 360 N/mm2. Factor of safety =6.
After design, draw a neat proportioned sketch of joint giving all
dimensions.
Que.2 Design a cotter joint to carry a maximum load of 50,000 N. All
components are made of the same material having the following
allowable stresses. (Sleeve and cotter joint)
Tensile stress = 20 MN/m2
Compressive stress = 50 MN/m2
Shear stress = 15 MN/m2
Draw a proportioned neat sketch of the joint you have designed.
Que.3 It is required to design a square key for fixing a pulley on the
shaft which is 50 mm in diameter. A 10 kW power at 200 r.p.m. is
transmitted by the pulley to the shaft. The key is made of steel 45C8
(Syt= Syc= 380 N/mm2) and the factor of safety is 3. Determine the
dimensions of the key.
11/5/2019

42
Que. 4 Design a C.I. flange coupling to transmit 150 H.P. at 250 rpm.
The following permissible stresses may be used:
Permissible shear stress for shaft, bolt and key material = 50 N/mm2
Permissible crushing stress for bolt and key material = 150 N/mm2
Permissible shear stress for Cast iron = 8 N/mm2
(Nov-Dec 2010)
Que.5 It is required to design a bushed pin type of flexible coupling for connecting
the motor to a centrifugal pump shafts. The details of the duty required from
the pump are:
Power to be transmitted = 18.5 kW
Speed in rpm = 1000
The diameters of the motor and pump shafts are 50 mm and 45 mm
respectively. Take the bearing pressure on the rubber bush as 0.35 N/mm2
and the working shear stress in the material of the pins as 20 N/mm2.
(2009)

Unit 2.3 Design of Coupling

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Unit 2.3 Design of Coupling

Similar to Unit 2.3 Design of Coupling (20)

More from Yugal Kishor Sahu

More from Yugal Kishor Sahu (7)

Recently uploaded

Recently uploaded (20)

Unit 2.3 Design of Coupling