Shaft coupling design Procedure/numericals

Shaft Coupling Design :Theory Questions and answers and Numerical problems

Shaft coupling design article containg design procedure and numerical problems of different types of couplings commonly used.

Design of Flanged Coupling (Unprotected and Protected type)

Unprotected type shaft coupling design

Protected type shaft coupling design

Notations used in design

d= Diameter of the shafts to be connected {Generally made up of steel}

D= Diameter Of hub {Hub and flange are one part but considered separated for failure {Generally made up of C.I}

D1= Pitch circle diameter of the Bolt circle

D2= Outer diameter Of the flange

L= Length of the flange

tf=Thickness of the flange

tp=Thickness of the protected portion.

dc=Core diameter of bolt {Threaded portion}

do= Outer diameter of bolt. {Threaded portion}

n=no of bolts connecting two flanges

t= thickness of the key

w=width of the key

l= length of the key

Design Steps

Step 1. To find torque acting on the shaft

$T = \frac{60 P}{2\times \pi \times N} \times10^6 \times OF$

Where OF is the overload or service factor, for example if the maximum torque is 20% more than mean torque then OF should be taken as 1.2 .

Using this formula the Toque acting on the shaft is calculated.

Step 2. To find diameter of shaft to transmit required torque

$T = \frac{\pi}{16} \times \tau \times d^3$

Use this equation to find the diameter of the shaft.

Step 3. Design of flanges

Emperical Relations
$D=2 \times d$
$D_1=3 \times d$
$D_2=4 \times d$
$l= 1.5 \times d$
$t_f =0.5 \times d$
$t_p =0.25 \times d$

Check shear stress in hub

$T = \frac{\pi}{16} \times \tau \times [D^3 (1-k^4)]$

Find induced stress This should be less than allowable shear stress

Check Shear stress in the Flange

Force = area resisting Stress

$\frac{T}{D/2} = {\pi} \times D \times t_f \times \tau$

Find induced stress This should be less than allowable shear stress

Step 4. Design Of Key

Empirical relation

{if crushing stress is twice the shear stress}

and {if crushing stress is not twice the shear stress}

check shear stress induced in key

Find induced stress This should be less than allowable shear stress

check crushing stress induced in key

Find induced stress This should be less than allowable shear stress

Step 5. Design Of Bolts

No of Bolts

n=4 if diameter of shaft is upto d<55 mm

n=6 if diameter is between 55 <d< 150 mm

n=8 if diameter is above d>150mm.

Shear Failure of bolts

The bolt size should be rounded to nearest even number..

Steps In short :

Part	Failure	Equation	To find
Shaft	Torque Shear	$T = \frac{60 P}{2\times \pi \times N} \times10^6 \times OF$ $T = \frac{\pi}{16} \times \tau \times d^3$	T= d=
Hub and flange	Empirical relations Shear Stress in hub Shear Stresses in Flange	$D=2 \times d$ $D_1=3 \times d$ $t_p =0.25 \times d$ $l= 1.5 \times d$ $D_2=4 \times d$ $t_f =0.5 \times d$ $T = \frac{\pi}{16} \times \tau \times [D^3 (1-k^4)]$ $\frac{T}{D/2} = {\pi} \times D \times t_f \times \tau$	Check Check
Key	Empirical relation Shear Crushing	{if crushing stress is twice the shear stress} and {if crushing stress is not twice the shear stress}	Check Check
Bolts	No of bolts	n=4 up to d= 55 mm n=6 d= 55 to d= 150 mm n=8 aboved=150mm	Find n
	Size of bolt Shear		Find dc Find do

Part

Failure

Equation

To find

Shaft

Torque

Shear

$T = \frac{60 P}{2\times \pi \times N} \times10^6 \times OF$

$T = \frac{\pi}{16} \times \tau \times d^3$

Hub and flange

Empirical relations

Shear Stress in hub

Shear Stresses in Flange

$D=2 \times d$ $D_1=3 \times d$ $t_p =0.25 \times d$

$l= 1.5 \times d$ $D_2=4 \times d$ $t_f =0.5 \times d$

$T = \frac{\pi}{16} \times \tau \times [D^3 (1-k^4)]$

$\frac{T}{D/2} = {\pi} \times D \times t_f \times \tau$

Check

Key

Empirical relation

Shear

Crushing

{if crushing stress is twice the shear stress}

and {if crushing stress is not twice the shear stress}

Check

Bolts

No of bolts

n=4 up to d= 55 mm

n=6 d= 55 to d= 150 mm

n=8 aboved=150mm

Find n

Size of bolt

Shear

Find dc

Find do

NUMERICAL PROBLEMS :

A rigid coupling is used for transmitting 20 kW power at 720 rpm. There are four bolts and the pitch circle diameter of the bolts is on 125 mm circle. The bolts are made up of plain carbon steel having ultimate tensile strength of 400 MPa . Determine the diamter of the bolts taking factor of safety as 3. Assume that there is no initial tightening load on bolts.
A rigid type flanged coupling transmits 60 kW at 350 rpm. it has out diamter of flange as 250 mm and inner diamter as 200 mm. The bolts are six in numbers and are maded up of steel having 380 MPa . Taking factor of safety as 3, determine the size of bolts.
A rigid coupling transmits 35 kW at 180 rpm . The service factor for the application is 1.5 ( take design torque as 1.5 times the mean torque ). Select the suitable material for the various parts of the coupling. Take the material for shaft as 40C8 ( syt= 380 MPa), material for bolts is 30C8 ( 400 MPa) and flanges are madeup of cast iron FG 150 ( Sut =150 MPa). Take factor of safety as 2.5 for all components. Also draw neat sketch of the coupling.
A protected type flanged coupling is required to transmit 60 kw power at 1440 rpm. Design the coupling with following materials, Material for shaft material for shaft as 40C8 ( syt= 380 MPa), material for bolts is 30C8 ( 400 MPa) and flanges are madeup of cast iron FG 150 ( Sut =150 MPa). Take factor of safety as 2.5 for all components.

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Shaft coupling design Procedure/Numericals

Shaft Coupling Design :Theory Questions and answers and Numerical problems

Design of Flanged Coupling (Unprotected and Protected type)

Notations used in design

Design Steps

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