ME8593 - DESIGN OF MACHINE

ELEMENTS

Mr.M.JAGADEESAN M.E.
ASSISTANT PROFESSOR

DEPARTMENT OF MECHANICAL ENGINEERING

VELTECH HIGHTECH DR RANGARAJAN DR SAKUNTHALA

ENGINEERING COLLEGE, AVADI

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 UNIT II DESIGN OF SHAFTS AND
COUPLINGS

Design of solid and hollow shafts based on strength, rigidity and

critical speed – Design of keys, key ways and splines - Design of

crankshafts -- Design of rigid and flexible couplings.

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 BOOKS REFERRED

1. Bhandari V.B, “Design of Machine Elements”, Tata McGraw-Hill

Book Co, 2003.

3
 INTRODUCTION
1. Transmission Shaft refers to, “A rotating machine element, circular
in cross-section, that supports transmission elements like gears,
pulleys and sprockets and transmits power”.

2. The shaft is always stepped with maximum diameter in the middle

portion and minimum diameter at the two ends, where bearings are
mounted.

4
 INTRODUCTION
3. The steps on the shaft provide shoulders for positioning transmission
elements like gears, pulleys and bearings.

4. The rounded-off portion between two cross-sections of different

diameters is called a fillet.

5. The fillet radius is provided to reduce the effect of stress

concentration due to abrupt change in cross-section.

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 CATEGORIES OF
TRANSMISSION SHAFTS
1. Axle – Shaft that supports rotating elements like wheel, fitted to the
housing by means of bearings, subjected to bending moment.

2. Spindle – Short rotating shaft, used in machine tools.

3. Countershaft – It is the secondary shaft that is driven by the main

shaft and from which the power is supplied to a machine component.
It is used in multistage gearbox.

4. Jackshaft – An auxiliary or intermediate shaft between two shafts.

5. Line Shaft – Consists of number of shafts connected in axial

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direction by coupling.
 SHAFT DESIGN – STRENGTH
BASIS
1. Transmission shafts are subjected to axial tensile force, bending
moment or torsional moment or their combinations.

2. Most of the transmission shafts are subjected to combined bending

and torsional moments.

3. When the shaft is subjected to tensile fore, the tensile stress is,

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 SHAFT DESIGN – STRENGTH
BASIS
1. When the shaft is subjected to pure bending moment, the bending
stresses are given by,
DDB – 7.1 & 7.21

2. When the shaft is subjected to pure torsional moment, the torsional

shear stress is given by,

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 SHAFT DESIGN – STRENGTH
BASIS
1. When the shaft is subjected to combination of loads, the principal
stress and principal shear stress are obtained.

2. Maximum principal stress Theory

DDB 7.2

Equivalent Bending Moment

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 SHAFT DESIGN – STRENGTH
BASIS
1. Maximum shear stress Theory
DDB 7.2

Equivalent Torsional Moment

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 SHAFT DESIGN – RIGIDITY
BASIS
1. In some applications, the shafts are designed on the basis of either
torsional rigidity or lateral rigidity.

2. A transmission shaft is said to be rigid on torsional rigidity, if it

does not twist too much under the action of external torque.

3. A transmission shaft is said to be rigid on lateral rigidity, if it does

not deflect too much under the action of external forces and bending
moment.

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SHAFT DESIGN – RIGIDITY
BASIS
DDB - 7.1

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 ASME CODE FOR SHAFT
DESIGN
1. According to this code, the permissible shear stress τmax for the shaft
without keyways is taken as 30% of yield strength in tension or 18%
of ultimate tensile strength of the material, whichever is minimum.

2. If keyways are present, the above values are be reduced by 25%.

3. According to ASME code, the bending and torsional moments are to

be multiplied by factors kb and kt respectively.
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ASME CODE FOR SHAFT
DESIGN

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PROBLEMS

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PROBLEMS

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PROBLEMS

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 KEYS

1. Key can be defined as, “A machine element which is used to connect

the transmission shaft to rotating machine elements like pulleys,
gears, sprockets or flywheels”.

2. Basic functions of key are,

1. To transmit the torque from the shaft to the hub of the mating
element and vice-versa.

2. To prevent relative rotational motion between the shaft and the

joined machine element like gear or pulley.
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KEY JOINT

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 KEYWAY

1. Keyway is defined as, “A recess or slot machined either on the shaft

or in the hub to accommodate the key”.

2. The keyway is usually cut using vertical or horizontal milling.

3. The keyway results in stress concentration.

4. During transmission of torque, the keyways should withstand shear

and compressive stresses.

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 TYPES OF KEYS

1. Saddle key and Sunk key.

2. Square key and Flat key.

3. Taper key and Parallel key.

4. Key with or without Gib-head.

5. Woodruff key.

6. Kennedy key.

7. Feather key.
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 SELECTION OF TYPE OF KEY

1. Power to be transmitted.

2. Tightness of fit.

3. Stability of connection.

4. Cost.

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 SADDLE KEYS

1. A saddle key is a key which fits in the keyway of the hub only.

2. There is no keyway in the shaft.

3. A hollow saddle key has a concave surface at the bottom to match the
circular surface of the shaft.

4. A flat saddle key has a flat surface at it sits on the flat surface
machined on the shaft.

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 SADDLE KEY

1. The resistance to slip in case of flat key is slightly more than that of
hollow key with concave surface. Hence flat saddle key is superior.

2. Saddle key requires keyway only on the hub, which makes it cost
effective.

3. Advantage is, it is positive drive.

4. Main disadvantage is low power transmitting capacity.

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 SUNK KEY
1. A sunk key is a key in which the half the thickness of the key fits into
the keyway on the shaft and the remaining half in the keyway on the
hub.

2. Hence keyways are required on both the shaft and hub.

3. The sunk key may be rectangular or square cross-section.

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 SUNK KEY

1. In sunk key the power is transmitted due to shear resistance of key.

2. Sunk key is suitable for heavy duty application, since there is no

possibility of slip.

3. The thumb rule for selecting square sunk key is, “Square key with
sides equal to one-quarter of the shaft diameter and length at least 1.5
times the shaft diameter”.
b = width of key.

h = height or thickness of key

l = length of key

d = diameter of shaft 26
 SUNK KEY

1. The thumb rule for selecting flat rectangular sunk key is,

2. Sunk keys with square or rectangular cross-sections are classified

into,

1. Parallel keys

2. Taper keys.

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 SUNK KEY

1. Parallel key is uniform in width as well as height throughout the key.

2. Taper key is uniform in width but tapered in height. The standard

taper is 1 : 100.

3. The taper is provided for,

1. It becomes tight due to wedge action providing tightness of joint

in operating conditions.

2. It is easy to remove the key and dismantle the joint.

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 SUNK KEY

1. Tapered keys are often provided with Gib-head to facilitate removal,

but it is dangerous in rotating parts.

2. Taper is provided on one side of the key by machining.

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 FEATHER KEY

1. A feather key is a parallel key which is fixed either to the shaft or to

the hub and which permits relative axial movement between them.

2. It is with uniform width and height.

3. There is a clearance fit between the key and keyway, such that the
hub is free to slide over the key.

4. It transmits power and at the same time it permits some axial

movement of the hub.

5. Ex: Clutches or gear shifting devices.

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FEATHER KEY

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 WOODRUFF KEY

1. A woodruff key is a sunk key in the form of an almost semicircular

disk of uniform thickness.

2. The keyway in the shaft is in the form of a semicircular recess with

the same curvature as that of the key.

3. Used in tapered shafts in machine tools and automobiles.

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 WOODRUFF KEY

ADVANTAGES

1. Used on tapered shaft because it can align by slight rotation in the

seat.

2. The extra depth of key prevents slip over the shaft.

DISADVANTAGES

1. The extra depth of keyway increases stress concentration.

2. The key does not permit axial movement between shaft and hub.

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DESIGN OF SQUARE AND FLAT
KEYS
FORCES ACTING ON A FLAT KEY (b = h)

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 DESIGN OF SQUARE AND FLAT
KEYS
1. The transmission of torque from the shaft to the hub results in two
equal and opposite forces denoted by P.

2. The torque Mt is transmitted by means of a force P acting on the left

surface AC of the key.

3. The equal and opposite force P, acting on the right surface DB of the
key is the reaction of the hub on the key.

4. P on AC and P on DB are not in the same plane. Hence P’ (P’=P)

act as resisting couple preventing the key to roll in the keyway.
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 DESIGN OF SQUARE AND FLAT
KEYS
1. It is assumed that the force P is tangential to the shaft diameter.
Hence,
Torque = Force x Distance

2. The design of square or flat key is based on two criteria,

1. Failure due to shear stress.

2. Failure due to compressive stress.

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 DESIGN OF SQUARE AND FLAT
KEYS
1. The shear stress τ in the plane AB is given by,

For flat key

By substituting the value of P

2. Hence

3. The failure due to compressive stress will occur on surfaces AC or

DB. It is assumed that,

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 DESIGN OF SQUARE AND FLAT
KEYS
1. The compressive stress σc in the key is given by,

For flat key

By substituting the value of P

2. Hence

3. For square key, h = b.

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 DESIGN OF KENNEDY KEY
1. The Kennedy key consists of two square keys.

2. The hub is bored off the center and the two keys force the hub and
the shaft to concentric position.

3. They are used for heavy duty applications.

4. Failure of Kennedy keys are based on,

1. Failure due to shear stress

2. Failure due to compressive stress.

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 DESIGN OF KENNEDY KEY

Forces acting on Kennedy key

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 SPLINES

1. Splines are keys which are made integral with the shaft.

2. They are used when there is a relative axial motion between the shaft
and the hub.

3. The gear shifting mechanism in automobile gearboxes requires it.

4. Splines are cut on the shaft by milling and in hub by broaching.

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SPLINES

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 COUPLINGS

1. Coupling can be defined as a mechanical device that permanently

joints two rotating shafts to each other.

2. Ex: A coupling is used to join the output shaft of an engine to the

input shaft of a hydraulic pump.

3. Coupling is a permanent connection, while the clutch can connect or

disconnect two shafts at the will of the operator.

4. The shafts to be connected by the coupling may have collinear axes,

intersecting axes or parallel axes with a small distance in between.
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 COUPLINGS

1. Oldham coupling – Is used to connect two parallel shafts when they

are at a small distance apart.

2. Hooke’s coupling – Is used to connect two shafts having intersecting

axis.

3. Flexible or Rigid coupling - Is used when the axes are collinear.

4. Flexible coupling tolerates a small amount of misalignment between

the shafts, there is no such provision in rigid coupling.

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Sl. No Rigid Coupling Flexible Coupling

1 It cannot tolerate Due to provision of flexible elements

misalignment between the like bush or disk, it can tolerate 1.5°
axes of the shafts. Used only of angular misalignment and 0.5 mm
in precise alignment. of axial displacement between shafts.

2 It can be used where the The flexible elements absorbs shocks

motion is free from shocks and and vibrations.
vibrations since no provisions
of flexible elements.

3 Simple and inexpensive. Comparatively costlier due to

additional parts.
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 REQUIREMENT OF GOOD
COUPLING
1. The coupling should be capable of transmitting torque from the
driving shaft to the driven shaft.

2. The coupling should keep the two shafts in proper alignment.

3. The coupling should be easy to assemble and disassemble for the

purpose of repairs and alterations.

4. The failure of revolving bolt heads, nuts, key heads and other
projecting parts may cause accidents. They should be covered by
giving suitable shape to the flanges or by providing guards.
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 MUFF COUPLING

1. Muff coupling is also called as sleeve coupling or box coupling.

2. It is a type of rigid coupling, consisting of sleeve or a hollow

cylinder, which is fitted over the ends of input and output shafts by
means of a sunk key.

3. The torque is transmitted from the input shaft to the sleeve through
the key, then transmitted from the sleeve to the output shaft through
the key.

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MUFF COUPLING

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 MUFF COUPLING - Advantages

1. It is the simplest form of coupling with only two parts, sleeve and
key.

2. Simple to design and manufacture.

3. It has no projecting parts except the key-head.

4. Compact construction with small radial dimensions.

5. Cheaper than other types of coupling.

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 MUFF COUPLING - Disadvantages

1. Difficult to assemble or dismantle. The sleeve has to be either shifted

over the shaft by half of its length or the ends of the shafts have to be
drawn together or apart by half length of the sleeve.

2. It is a rigid coupling and requires accurate alignment of shafts.

3. It cannot absorb shocks and vibrations.

4. It requires more axial space compared with flange couplings.

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DESIGN PROCEDURE FOR MUFF
COUPLING
1. Calculate the diameter of each shaft by the following equations.

2 NT  16 M t
Power P  watts T  Mt  d 3
 
60 16 d3
2. Calculate the dimensions of the sleeve by the following empirical
equations.
D  (2d  13) mm and L  3.5d

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DESIGN PROCEDURE FOR MUFF
COUPLING
3. Check the torsional shear stress induced in the sleeve using the
following equation. (consider sleeve as Hollow shaft)

Mtr  (D4 - d 4 ) D
  J  r 
J 32 2

4. Determine the standard cross-section of flat sunk key using DDB –

5.16.

5. The length of the key in each shaft is one-half of the length of the
sleeve. Therefore l = L/2.
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DESIGN PROCEDURE FOR MUFF
COUPLING
6. With these dimensions of the key, check the shear and compressive
stresses in the key using,

2M t 4M t
  and c 
dbl dbl

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 RIGID FLANGE COUPLING

1. A flange coupling consists of two flanges – one keyed to the driving

shaft and the other to the driven shaft.

2. The two flanges are connected by means of four or six bolts arranged
on a circle concentric with the axes of the shafts.

3. Power is transmitted from the driving shaft to the left side flange
through the key, then to the right side flange through the bolts.

4. From right side flange to the driven shaft through the key.

5. The mating parts have spigot and recess for precise location.
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RIGID FLANGE COUPLING -
Types

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 RIGID FLANGE COUPLING -
Advantages
1. Has high torque transmitting capacity.

2. Easy to assemble and dismantle.

3. It has simple construction.

4. It is easy to design and manufacture.

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 RIGID FLANGE COUPLING -
Disadvantages
1. Since it is a rigid coupling, it cannot tolerate misalignment between
the axes of two shafts.

2. It can be used only where the motion is free from shocks and
vibrations.

3. It requires more radial space.

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RIGID FLANGE COUPLING

58
RIGID FLANGE COUPLING

59
DESIGN PROCEDURE FOR RIGID
FLANGE COUPLING
1. Calculate the shaft diameter by using the following equations.

2 NT  16 M t
Power P  watts T  Mt  d 3
 
60 16 d3
2. Calculate the dimensions of the flanges by the following empirical
equations. d h  2d t1  0.25d
lh  1.5d d r  1.5d
D  3d Do  (4d  2t1 )
t  0.5d
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DESIGN PROCEDURE FOR RIGID
FLANGE COUPLING
3. Calculate the torsional shear stress in the Hub by considering it as a
hollow shaft subjected to torsional moment Mt.
Mtr  (d h 4 - d 4 ) dh
  J  r 
J 32 2
4. The flange at the junction of the hub is under shear while
transmitting the torsional moment Mt.
1
M t   d h2t
2

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DESIGN PROCEDURE FOR RIGID
FLANGE COUPLING
5. Determine the number of bolts using the following.

6. Determine the diameter of the bolt.

7. Compressive stress in the bolt is determined by,

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DESIGN PROCEDURE FOR RIGID
FLANGE COUPLING
8. Determine the standard cross-section of the key DDB – 5.16.

9. The length of the key in each shaft is lh. Therefore

l = lh

10. Check the shear and compressive stresses in the key using,

2M t 4M t
  and c 
dbl dbl

63
 BUSHED-PIN FLEXIBLE
COUPLING
1. In practice, it is impossible to obtain perfect alignment of shafts.

2. Misalignment exists due to the following reasons,

1. Deflection of shafts due to lateral forces.

2. Error in shaft mounting due to manufacturing tolerances.

3. Use of two separately used manufactured units such as electric

motor and worm gear box.

4. Thermal expansion of parts.

64
 BUSHED-PIN FLEXIBLE
COUPLING
3. To overcome this problem, flexible couplings are used.

4. A flexible coupling employs a rubber bush between the driving and

the driven flanges.

5. This rubber bush accommodates the misalignment and absorbs

shocks and vibrations.

6. A flexible coupling can tolerate 0.5 mm of axial or lateral

misalignment and 1.5° of angular misalignment.

65
BUSHED-PIN FLEXIBLE
COUPLING

66
 BUSHED-PIN FLEXIBLE
COUPLING
7. The construction of flexible coupling is similar to that of the rigid
flange coupling except for the provisions of rubber bush and pins in
place of bolts.

8. The coupling consists of two flanges, one keyed to the input shaft and
the other to the output shaft.

9. The two flanges are connected together by means of four or six pins.

10. At one end, the pin is fixed to the output flange by means of pin.

67
 BUSHED-PIN FLEXIBLE
COUPLING
11. The diameter of the pin is enlarged in the input flange where a rubber
bush is mounted over the pin.

12. The rubber bush is provided with brass lining at the inner surface.

13. The lining reduces the wear of the inner surface of the rubber bush.

14. Power is transmitted from input shaft to input flange through the key.

15. From the input flange the power is transmitted to the pin, which
transmits it to the output flange by shear resistance. Finally the power
is transmitted from output flange to shaft through key.
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 BUSHED-PIN FLEXIBLE
COUPLING - Advantages
1. It can tolerate 0.5 mm of lateral or axial misalignment and 1.5° of
angular misalignment.

2. It prevents transmission of shock from one shaft to the other and

absorbs vibrations.

3. It can be used for transmitting high torques

69
 BUSHED-PIN FLEXIBLE
COUPLING - Disadvantages
1. The cost of the coupling is more than that of rigid coupling due to
additional parts.

2. It requires more axial space compared with other types of couplings.

70
DESIGN PROCEDURE FOR
FLEXIBLE COUPLING

71
DESIGN PROCEDURE FOR
FLEXIBLE COUPLING

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 DESIGN PROCEDURE FOR
FLEXIBLE COUPLING
1. Calculate the shaft diameter using the following formulas.

2 NT  16 M t
Power P  watts T  Mt  d 3
 
60 16 d3
2. Dimensions of flange: Calculate the dimensions of the flange by the
following empirical relationships.
d h  2d t  0.5d
lh  1.5d t1  0.25d
D  3d to 4d

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 DESIGN PROCEDURE FOR
FLEXIBLE COUPLING
3. The torsional shear stress in the hub can be calculated by considering
it as a hollow shaft subjected to torsional moment Mt.

Mtr  (d h 4 - d 4 ) dh
  J  r 
J 32 2

4. The shear stress in the flange at the junction with the hub is
calculated by, 1
M t   d h2t
2
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 DESIGN PROCEDURE FOR
FLEXIBLE COUPLING
5. Diameter of Pins: The number of pins is usually 4 or 6. The diameter
of the pins is calculated by the following empirical formula.
0.5d
d1 
N
6. Shear stress in the pins is determined by,
8M t

 DNd12
7. The shear stresses calculated by the above equation should be less
than 35 N/mm2.

75
 DESIGN PROCEDURE FOR
FLEXIBLE COUPLING
8. Dimensions of Bushes: Calculate the outer diameter of rubber bush
1 2
by using, M t  Db DN
2
9. Calculate the effective length of the rubber bush by,
lb  Db
10. Dimensions of Key: Determine the standard dimensions of the flat
from DDB – 5.16.

11. The length of the key in each shaft is lh. Therefore

l = lh 76
 DESIGN PROCEDURE FOR
FLEXIBLE COUPLING
12. Check the shear and compressive stresses in the key using,

2M t 4M t
  and c 
dbl dbl

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 DESIGN FOR LATERAL
RIGIDITY
1. A component is considered as rigid when it does not deflect or bend
too much due to external forces or moments.

2. Permissible deflection is the criteria for these cases.

3. When a transmission shaft supporting a gear is deflected, the

meshing of the gear teeth is not proper, resulting in uneven
distribution of load over the face width of the gear.

4. It also results in a misalignment between the axes of shaft and

bearing causing uneven wear of bearing surfaces.
78
 DESIGN FOR LATERAL
RIGIDITY
1. The maximum permissible deflection for the transmission shaft is
generally taken as,   (0.001) L to (0.003) L
L = span length between two bearings
2. The maximum permissible deflection at gear is taken as,   (0.001)m
m = module of the gear teeth
3. The lateral rigidity of the structure is improved by,

1. Reduce the span length.

2. Increase the number of supports.

3. Reduce the number of joints and lubricate the contact surfaces.

4. Assemble the components with pre-load. 79

 DDB – 6.12
CRITICAL SPEED OF SHAFTS
1. The critical speed of the shaft is, “The speed at which the rotating
shaft becomes dynamically unstable and starts to vibrate violently in
a transverse manner”.

2. It is a very dangerous condition because the amplitude of vibration is

so high that the shaft may break into pieces.

3. The critical speed of the shaft is also called the ‘whirling’ speed or
‘natural frequency of vibrations’.

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 CRITICAL SPEED OF SHAFTS

1. The center of mass of a transmission shaft carrying gears, pulleys or

sprockets never coincides with the center of rotation. This is due to,
1. It is practically impossible to machine a shaft and assemble its attachments
in such a way that the mass is uniformly distributed about its geometrical
axis.

2. The shaft is subjected to forces such as gear forces, belt tension, which
causes lateral deflection. This moves the center of mass away from true
axis of rotation.

2. The above factors cause the mass to rotate with small eccentricity.
81
 CRITICAL SPEED OF SHAFTS

3. As the speed of shaft increases, a centrifugal force acts on the mass

due to eccentricity, increasing the deflection of shaft.

4. When the speed is further increased, a stage is reached when the

elastic forces within the shaft no longer balance the centrifugal force.

5. This starts violent vibrations of the shaft and the speed at which this
phenomenon occurs is called ‘critical speed’.

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CRITICAL SPEED OF SHAFTS

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 CRITICAL SPEED OF SHAFTS

1. A transmission shaft of negligible mass and supported on two end

bearings is as shown. It is carrying several concentrated masses.
Rayleigh-Ritz Equation
2. The first or lowest critical speed is given by,
g (W11  W2 2  ...)
ωn = first or lower critical speed (rad/s)
n 
g = gravitational force (W112  W2 22  ...)
W1 = m1g and W2 = m2g

m1, m2 = rotating masses

δ1, δ2 = static deflections at resp. masses

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THANK YOU

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