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DON BOSCO INSTITUTE OF TECHNOLOGY
Department Of Mechanical Engineering
DMS Course Project Report on
UNPROTECTED/PROTECTED TYPE FLANGE COUPLING
Submitted in fulfilment of the requirements Of DMS Project of Bachelor of Engineering
(Semester VIIl)
NAMES ROLL NUMBER
Aren Almeida 03
Seraj Tuscano 75
Swapnil Tuscano 76
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TABLE OF CONTENT
SR NO. TOPIC PAGE NUMBER
I INTRODUCTION 3
II ADVANTAGES & DISADVANTAGES 5
III USES 5
IV DESIGN PROCEDURE 6
V SAMPLE CALCULATION 7
VI EXCEL SHEET 11
VII ALGORITHM 14
VIII CONCLUSION 16
IX REFERENCES 15
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INTRODUCTION
Flange coupling usually applies to a coupling having two separate cast iron flanges.
Eachflange is mounted on the shaft end and keyed to it. The faces are turned up at right
angle to the axis ofthe shaft. One of the flange has a projected portion and the other flange
has a corresponding recess.
This helps to bring the shafts into lineand to maintain alignment. The twoflanges are
coupled together bymeans of bolts and nuts. The flange coupling is adopted to heavy
loadsand hence it is used on large shafting.
TYPES OF FLANGE COUPLINGS
The flange couplings are of the following three types:
1. Unprotected type flange coupling
2. Protected type flange coupling
3. Marine type flange coupling
Protected type flange coupling: In a protected type flange coupling, as shown in figure
the protruding bolts and nuts are protected by flanges on the two halves of the coupling, in
order toavoid danger to the workman.
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Unprotected type flange coupling: In an unprotected typeflange coupling, as shown
in Fig.13.12, each shaft is keyed to the bossof a flange with a counter sunk keyand the flanges
are coupled togetherby means of bolts. Generally, three,four or six bolts are used. The keysare
staggered at right angle along thecircumference of the shafts in orderto divide the weakening
effect causedby keyways.
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Advantages :
Due to its simple structure, low cost and high torque transmission, it is often used when
the speed is low, no impact, the rigidity of the shaft is large, and the centering is good. The
flange coupling is a bolt-coupled two flange (flange) disc type half coupling, and the two
coupling halves are respectively coupled with two shafts by a key to realize two shafts. Connect,
transfer torque and motion. The flange coupling has the advantages of simple structure,
convenient manufacture, low cost, reliable work, simple assembly and disassembly and
maintenance, large transmission torque, and high alignment accuracy of the two shafts, and is
generally used for stable load. Shaft drive with high speed or high transmission accuracy
requirements. The flange coupling does not have radial, axial and angular compensation
performance. If the alignment accuracy of the two shafts cannot be guaranteed during use, the
service life, transmission accuracy and transmission efficiency of the coupling will be reduced,
and vibration will be caused. And noise. According to the magnitude of the transmitted torque,
the bolts on the flange of the rigid coupling can be all bolts for reaming holes, or half of the
bolts for reaming holes, and the other half with ordinary bolts.
Disadvantages:
However, flange couplings have high requirements for the alignment of the two shafts.
When the relative displacement of the two shafts is present, additional loads are caused in the
machine parts, which deteriorates the working condition, which is its main disadvantage.
Uses:
Flange couplings are ordinarily utilized in pressurized funneling frameworks where two
pipe or tubing finishes need to meet up. It can be used in a driveshaft of a truck since it is a
mechanical component that transmits torque and rotation.
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DESIGN OF FLANGE COUPLING
Consider a flange coupling as shown,
Let
d = Diameter of shaft or inner diameter of hub
D = Outer diameter of hub
d1 = Nominal or outside diameter of bolt
D1 = Diameter of bolt circle
n = Number of bolts
Tf = Thickness of flange
τs, τb and τk = Allowable shear stress for shaft, bolt and key material respectively
τc = Allowable shear stress for the flange material i.e. cast iron,
σcb, and σck = Allowable crushing stress for bolt and key material respectively.
1. Design for hub :
The hub is designed by considering it as a hollow shaft, transmitting the same torque (T) as
that of a solid shaft.
From the above relation, the induced shearing stress in the hub may be checked.
The length of hub (L) is taken as 1.5 d.
2. Design for key:
The key is designed with usual proportions and then checked for shearing and crushing
stresses. The material of key is usually the same as that of shaft. The length of key is taken
equal to the length of the hub is under shear while transmitting the torque.
3. Design of flange
Torque transmitted,
T = Circumference of hub × Thickness of flange × Shear stress of flange × Radius of hub
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The thickness of flange is usually taken as half the diameter of shaft. Therefore from the above
relation, the induced shearing stress in the flange may be checked .
4. Design for bolts:
The bolts are subjected to shear stress due to the torque transmitted. The number of bolts
(n)depends upon the diameter of shaft and the pitch circle diameter of bolts (D1) is taken as 3
d. We
know that
From this equation, the diameter of bolt (d1) may be obtained. Now the diameter of bolt may
be checked in crushing. We know that area resisting crushing of bolts
From this equation, the induced crushing stress in the bolts may be checked.
SAMPLE CALCULATION:
Design protected type of flange coupling with power =15kw and N=200rpm. Assume Mt &
suitable data
Given- P=15kw, N=200rpm
1. Selection of material:
a. The shaft is subjected to torsional shear stress on the basis of strength, plain carbon steel of
grade 40C8 (Syt=380N/mm2) is used for shaft. FOS=2.5 (PSG 1.11)
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b. The key and bolt are Subjected to shear & compressive stresses. On the basis of strength
criterion plain carbon steel of grade 30C8 (Syt=400N/mm2) is selected. FOS=2.5 (PSG 1.11)
c. Flanges have complex and hence easiest method to make flange is casting. Grey cast iron
FG200 (Syt=200N/mm2) is selected as Material for the flanges FOS=6 (PSG1.4)
2. Permissible Stresses:
A. For shaft: Syt = 380N/mm2, FOS=2.5
𝑆𝑦𝑡 380
[𝜎𝑡] = 𝐹𝑂𝑆 = = 152𝑁/𝑚𝑚2
2.5
[𝜎𝑐] = [𝜎𝑡] = 152𝑁/𝑚𝑚2
[𝜎𝑡]
[𝜏] = = 76𝑁/𝑚𝑚2
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B. For key & shaft: Syt = 400N/mm2 FOS=2.5
𝑆𝑦𝑡 400
[𝜎𝑡] = = = 160𝑁𝑚𝑚2
𝐹𝑂𝑆 2.5
[𝜎𝑐] = 𝜎𝑡 = 160𝑁/𝑚𝑚2
[𝜎𝑡]
[𝜏] = = 80𝑁/𝑚𝑚2
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C. For Flange: Syt = 200N/mm2 FOS=6
𝑆𝑦𝑡 200
[𝜎𝑡] = = = 33.33𝑁/𝑚𝑚2
𝐹𝑂𝑆 6
[𝜎𝑡] = 𝜎𝑐 = 33.33𝑁/𝑚𝑚2
STEP 3: Design Calculations:
1. Find diameter of shaft :-
2𝜋𝑁𝑇
𝑃=
60
2𝜋 ∗ 200 ∗ 𝑇
15 ∗ 103 =
60
T = 716.197 N-m
i.e T= 716.19*10^3 N-mm
𝜋
𝑇= ∗ 𝜏 ∗ 𝑑3
16
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𝜋
716.19 ∗ 103 = ∗ 766 ∗ 𝑑 3
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d = 36.34mm
(PSG 5.44) d = 39mm
2. Design of Flange:-
i. Outside diameter of hub= D= 2*d= 2*39= 78mm
ii. Length of hub= 1.5d= 1.5*39=58.5mm
iii. Pitch circle diameter of bolt= 3d=D1= 117mm
iv. Outside diameter of flange= D2= 4d= 156mm
v. Thickness of flange= tf= 0.5d= 19.5mm
vi. Thickness of Protected flange Coupling= tp= 0.25d= 9.75mm
3. No. of bolts:-
N=3 (for diameter 39mm)
4. Design of hub:-
Check for shear stress:
𝜋 𝐷4 − 𝑑4
𝑇= ∗ 𝜏𝑐 ∗ ( )
16 𝐷
3
𝜋 784 − 394
716.19 ∗ 10 = ∗ 𝜏𝑐 ∗ ( )
16 78
8.198𝑁
(𝜏) =
𝑚𝑚2
(𝜏) < [𝜏] Design is safe.
5. Design of key:
For diameter 39mm
Width= 12 mm
Thickness= 8mm ……… (PSG 5.16)
Length= 1.7d= 1.7*39= 66.3mm
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Check for shear stress,
𝑑
𝑇 = 𝑙∗𝑤∗𝜏∗
2
39
716190 = 66.3 ∗ 2 ∗ 𝜏 ∗
2
(𝜏) = 46.163𝑁/𝑚𝑚2
(𝜏) < [𝜏] Design is safe.
Check for crashing,
𝑡 𝑑
𝑇=𝑙∗ ∗ 𝜎𝑐 ∗
2 2
8 39
716190 = 66.3 ∗ ∗ 𝜎𝑐 ∗
2 2
(𝜎𝑐) = 138.49𝑁/𝑚𝑚2
(𝜏𝑐) < [𝜏𝑐] Design is safe.
6. Design of flange:-
Check for Shear:-
𝐷
𝑇 = 𝜋𝐷 ∗ ∗ 𝑡𝑓 ∗ 𝜏𝑐
2
78
716190 = 𝜋 ∗ 78 ∗ ∗ 19.5 ∗ 𝜏𝑓
2
(𝜏𝑓) = 3.84𝑁/𝑚𝑚2
(𝜏𝑓) < [𝜏𝑓] Design is safe.
7. Design of bolt:
Considering Shear failure:-
𝜋 𝐷1
𝑇= ∗ (𝑑1)2 ∗ 𝜏𝑏 ∗ 𝑛 ∗
4 2
𝜋
716190 = ∗ 𝑑12 ∗ 80 ∗ 3 ∗ 117
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d1 = 8.05mm
d1 8.05
Diameter of bolt = = = 10.07mm
0.8 0.8
standard diameter = 12mm …. (PSG 5.44)
Check for crushing stress (𝜎𝑐),
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𝐷1
𝑇 = 𝑛 ∗ 𝑑1 ∗ 𝑡𝑓 ∗ 𝜎𝑐 ∗
2
𝜋
716190 = 3 ∗ 8.06 ∗ 19.5 ∗ 𝜎𝑐 ∗
2
(𝜎𝑐) = 25.96𝑁/𝑚𝑚2
(𝜎𝑐) < [𝜎𝑐] Design is safe.
Excel Sheet:
1. Put the given inputs in the table
2.For Design of safe flange
3. Make table for
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corresponding output
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4. Make the required PSG Tables
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ALGORITHM:
1.Permissible Stresses:
a. Shaft
Tensile Stress (N/mm2) = B13/B14
Crushing stress (N/mm2) = G36
Shear stress (N/mm2) = G36/2
b. Key and Bolts
Tensile stress (N/mm2) = B18/B19
Crushing stress (N/mm2) = G41
Shear stress (N/mm2) = G41/2
STEP 2
Design Of Shaft
= ((16*G53)/(3.14*G38))^(1/3)
Diameter of shaft (mm)
Shaft Standard Diameter (d)(mm) = VLOOKUP(G58,Q7:R55,2,TRUE)
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c. Flange
Tensile stress (N/mm2) = B22/B23
Crushing stress (N/mm2) = G46
Shear stress (N/mm2) = G46/2
STEP 1
Torque (N-mm) = ((60*B9*1000)/(2*3.14*B10))*1000
STEP 3 16
=2*G59
Outside diameter of hub (D) (mm)
=1.5*G59
Length of hub (mm)
=3*G59
Pitch circle diameter of bolt (D1) (mm)
Outside diameter of flange (D2) (mm) =4*G59
STEP 4
=0.5*G59
Design
Thickness of flange (Tf) (mm) Of Hub
=((G53*16*G63)/(3.14*((G63^4)-
=0.25*G59
(G59^4))))
Check shear
Thickness stress flange coupling (Tp) (mm)
protected
in Hub (N/mm2)
=IF(G77<=G48,"Safe","Unsafe")
Safe or Unsafe
STEP 5
Design Of Key
=VLOOKUP(G59,U7:W33,2,TRUE)
Key Width(mm)
=VLOOKUP(G59,U7:W33,3,TRUE)
Key Thickness
(mm)
=1.7*G59
Key Length (mm)
=(2*G53)/(G59*G83*G85)
Check for shear
stress (N/mm2)
=IF(G87<=G43,"Safe","Unsafe")
Safe or Unsafe
=(G53*4)/(G85*G84*G59)
Check for crushing
stress (N/mm2)
=IF(G90<G42,"Safe","Unsafe")
Safe or Unsafe
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STEP 7
Design Of Bolt
=((8*G53)/(3.14*G43*G72*G65))^(1/2)
Core diameter of
bolt (mm)
=G102/0.8
Diameter of Bolt
(mm)
=VLOOKUP(G104,U37:V50,2,TRUE)
Bolt Standard
Diameter (mm)
=(2*G53)/(G72*G102*G65*G67)
Check for
crushing stress
STEP 6
=VLOOKUP(G59,U54:V57,2,TRUE)
Number of Bolts
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(N/mm2)
STEP 8
=IF(G107<=G42,"Safe","Unsafe")
Safe or Unsafe Design Of Flange
=(G53*2)/(3.14*(G63^(2))*G67)
Shear stress in Flange (N/mm2)
=IF(G96<=G48,"Safe","Unsafe")
Safe and Unsafe
CONCLUSION:
The given sum solved by Algorithm successfully matches with the theoretical values.
REFERENCES:
1. A Textbook of Machine Design - R.S. Khurmi, J.K. Gupta
2. Design of Machine Elements -V.B. Bhandari
