SELECTION OF FLEXIBLE FLANGE COUPLINGS
Selection procedure :
Given:
Power in kw
Speed of the shaft in RPM.
Step: 1
Calculate the power to be transmitted at 100 rpm – Refer P.7.108. Include
service factor. For service factor refer page 7.109.
Based on maximum rating at 100 rpm, select the coupling from the table on
page 7.108
Step: 2
Step: 3
Find the torque the shaft and coupling have to transmit using the power and
the speed. Include the service factor and determine the max. torque.
Step: 4
Find the diameter of the shaft required to transmit the max. torque.
Check whether the diameter of shaft is between the max. and min. values given
in the table-2. If it is, select a suitable diameter within the range.
If it is less than the min. value, raise the diameter to the minimum value.
CHECK FOR THE FLANGE AGAINST ITS SHEARING FROM THE HUB.
7.1) Determine the allowable shear stress of the flange material.
Since the flanges are usually made of cast iron, allowable shear stress [] is to
be calculated for cast iron.
Determination of [] for cast iron
Find u of the cast iron corresponding to the grade assumed.
Calculate u
u = u / 2 (according to max shear stress theory)
[] =  u / factor of safety ; Assume factor of safety 3 to 5.
7.2) Calculate allowable torque [Mt] using the equation on P.7.106 and
compare with actual torque-Mt-to be transmitted for safer design Mt[Mt]
For safer design Mt[Mt] ; If Mt > [Mt] then select next higher coupling,
Note down its dimensions and repeat the step from 7.2 until Mt  [Mt]
Step: 5
Step:6
CHECK FOR THE BEARING STRENGTH OF PIN
6.1) Calculate bearing pressure p using the equation on P.7.106. Note that
Ft = 2Mt/D
6.2) Select the allowable bearing or compressive stress [c] for steels from page
7.124.
6.3) If the bearing pressure on the pin p < [c], the pin can safely transmit the
torque. If p > [c], select the next higher coupling and repeat the step 6 till
p >[c].
Step:7
CHECK FOR THE BENDING STRENGTH OF PIN
6.1) Calculate Mb using the equation on P.7.106
6.2) Calculate the bending stress  using the following equation
32M b
 
 db 3
6.3) Calculate design stress [].
a) For the material chosen, find y from the Data book.
b) [] = y / F.O.S (Assume factor of safety 1.5 to 2)
6.4) For safer design  > []
If  > [] bending select next higher coupling. Note down its dimensions and
repeat step 6 until   [] bending.
Step:8
CHECK FOR THE CRUSHING or BEARING STRENGTH OF THE BUSH.
Assumptions to be made:
Bush material = Rubber
22.29
Allowable crushing stress for the bush =[c] = 1N/mm2
Bush Thickness = 4 to 8mm
Bush length lb = G –(2/3)F. Refer figure on page 7.106
5.1) Calculate outside diameter of bush; outside dia. of bush,Db=dia. of bolt+2
(THICKNESS OF BUSH)
5.2) Calculate the bearing pressure P using the following equation.
2Mt
P 
DDblbn
For safer design P  [c] < 1N/mm2
If P > [c], select the next higher coupling. Note down its dimensions and repeat
step No.5 until, P  [c]
Step:9
Step:10
Step: 11
SELECT KEY
Based on the diameter of the shaft, select the width and thickness of the key
from P.5.16
CHECK FOR THE SHEAR STRENGTH OF THE KEY
9.1) Select the material for the key.
9.2) Determine the allowable shear stress [] for the key
9.3) Determine the allowable torque [Mt] using the equation on P.7.106.
9.4) Compare the allowable torque [Mt] with the actual torque Mt
For safer design Mt  [Mt] ; If Mt>[Mt], then increase the width & thickness to
the next higher standard values and repeat the step from 9.3 until Mt  [Mt]
CHECK FOR THE BEARING STRENGTH OF THE KEY
Determine allowable torque [Mt] using the equation on P.7.106. Allowable
bearing stress can be taken from P.7.125. )
Compare the allowable torque [Mt] with the actual torque Mt
For safer design Mt  [Mt] ; If Mt>[Mt], then increase the width & thickness to
the next higher standard values and repeat the step from 9.3 until Mt  [Mt]
EXAMPLE PROBLEM.
Select a bush type flexible flange coupling to transmit 37.5kW at 1440 rpm from a
motor to a vacuum pump.
Select coupling No.
Power at 100 rpm (for service factor refer page 7.109)
kW. of power application  service factor  100
=
rpm of application
37.5  3  100
 7.8kW
1440
So, coupling No.7 is selected from P 7.108
=
Find the torque the shaft and coupling have to transmit:
2 NM t
 37.5
60000
60000  37.5
Mt

 248.68N  m  248680N  mm
2  1440
With a service factor of 3, the torque will also be service factor times the normal torque.
Therefore Mt = 3 x 248.68 N-m =3 x 248680 N-mm = 746040 N-mm.
Find the diameter of the shaft required to transmit the torque.
Selecting material C45 for the shaft
y = 36 kgf/mm2 = 360 N/mm2
 y 360
According to the maximum shear stress theory,  max =

 180 N/mm2
2
2
22.30
Factor of safety (F.O.S) is taken as 2.
[] = 180/F.O.S = 180/2 = 90 N/mm 2
16M t 16  746040

 42217.23mm 3
d3 =
  [ ]
  90
d = 34.82 mm
Rounded off to standard value = 35.5 mm
For the coupling 7, shaft dia. range = 75 mm to 85 mm. 80 mm dia. is selected. This is
greater than the dia. 72.43 mm, selected based on strength. Therefore the diameter
80mm is on safe side.
Check for the flange against its shearing from hub.
Material selected for flange coupling = GCI20
u = 200 N/mm2 (refer P. 1.4)
200
u 
 100N /mm 2 (according to the maximum shear stress theory)
2
100
 20N /mm 2
Design stress =   
5
  C 2 G  H  
Mt  
  
2


  C 2 G  H  
  140   23
  1402  (45  22)
 20 
 20

   
2
2
2


=14162299.58 N-mm > Mt (=746040 N-mm)
So the flange can withstand the shear stress near the hub and transmit the torque.
CHECKING OF PIN
2
Check for bearing strength
Material selected for pin = C60
p = Bearing pressure on the pin
b   2100 to 2400 kgf /cm2 Design bearing stress for steels  (P 7.124)
b   210 to 240 N /mm2
p
Ft
2
F )n
3
2M t
D

2
db (G  F )n
3
2M t
Ft 
 7853N
D
7853
p
2
40  (45  16)  6
3
7853

 0.953N /mm 2  [ b ]
8240
So the pin can withstand the bearing pressure.
db (G 
Check for bending strength
y for C60 = 42 kgf/mm2 = 420 N/mm2
420
420
[ ] 

 210N /mm 2
F .O .S
2
22.31
Mb 

Ft t  1/2 G  (2/3)F  
n

2M t t  1/2 G  (2/3)F  
D
n

 2  746040  5  1/2  45  (2/3)  16  



190
6

 

1492080  22.17
 29017N  mm
190  6
M y 32M b 32  29017
  b 

3
I
 db 3
  40 

32  29017 928544

 4.62N /mm 2   
  64000 201062
So the pin has sufficient bending strength.

Check for the bearing strength of rubber bush.
2
2
Length of the bush = lb  G  F  45   16  34.33mm
3
3
Outer diameter of the bush = Db = db + 2 x thickness of the bush
= 40 + 2 x 4 = 48 mm (4 mm thickness is assumed)
2Mt
2  746040
1492080
P


 0.79N /mm 2  1N /mm 2
DDblbn 190  48  34.33  6 1878537.60
So the bush can withstand the bearing pressure.
CHECKING OF KEY
For the shaft size 80 mm, key size is selected from P 5.16.
W = 22 mm, T = 14 mm
Assume
L = Length of the key = length of the flange coupling = E
= 63 mm
Material selected for key = C40
y = 33 kgf/mm2 = 330 N/mm2
330
 165N /mm 2 (according to the maximum shear stress theory)
y =
2
165
165

 82.5N /mm 2
Design stress = [] =
F .O .S
2
Check for shear strength
WLA
Mt

 s 
2
WLA
22  63  80
[ ] 
 82.5  4573800N  mm > Mt (= 746040 N-mm)
2
2
So the pin can withstand the shear stress and transmit the torque.
Check for bearing strength
b   2100 to 2400 kgf /cm2 Design bearing stress for steels  (P 7.124)
b   210 to 240 N /mm2
b   210 N /mm2 is selected.
TLA
 b 
2
TLA
14  63  80

b  
 210  7408800N  mm > Mt(=746040 N-mm)

2
2
So the pin has sufficient bearing strength and transmit the torque.
The flange coupling satisfies all the strength requirements. Therefore it is selected for
the application.
Mt

22.32