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1. D. M. Gadhave,
Lect. In Mechanical Engg.
Govt. Polytechnic, Bramhapuri
SHAFTS, KEYS,
COUPLINGS AND
GEAR

3. SHAFT
• A shaft is a rotating element which is
used to transmit power from one place
to another. The power is delivered to
the shaft by some tangential force and
the resultant torque (twisting moment)
is set up within the shaft which permits
the power to be transferred. In order to
transfer the power from one shaft to
another, the various member such as
pulleys, gears etc are mounted on it.

4. • These members along with the the
forces acting upon them causes the
shaft to bending. Hence the shaft is
used for transmission of torque and
bending moments. The various
members are mounted on the shaft by
means of keys or splines.
• The shafts are usually cylindrical but
may be square or cross shape cross-
section. They are solid in cross- section
SHAFT

5. but sometimes hollow shafts are also
used
• An axle though similar in shape to the
shaft is a stationery machine element
and is used for the transmission of
bending moment only. It simply
supports two rotating bodies such as
axle etc.
SHAFT

6. Material For shaft
• The material used for the shaft should
have the following properties.
• It should have high strength.
• It should have good machinability.
• It should have low notch sensitivity
factor.
• It should have good heat treatment
properties.
• It should have high wear resistant properties

7. • The material used for ordinary shaft is
carbon steel of grades 40C8, 45C8,
50C4 and 50C12. The mechanical
properties of steel are given in the
following table.
Material For shaft
Designation Ultimate tensile
strength MPa
Yield Strength MPa
40C8 560-670 320
45C8 610-700 350
50C4 640-760 370
50C12 700 390

8. Manufacturing of Shaft
• Shafts are generally manufactured by
hot rolling and finished size by cold
drawing or turning and grinding. The
cold rolled shafts are stronger than hot
rolled shafts but with the higher
residual stresses. The residual stresses
may cause distortion of shaft when
machined. Shafts of larger diameter are
usually forged and turned to size on
lathe.

9. Types of shaft
• The following two types of shafts are
important.
• 1. Transmission Shaft :- These shafts
transmit the power between the source and
the machine absorbing power. The counter
shaft, machine shaft and all factory shaft are
transmission shaft. Since these shafts carry
machine parts such as pulleys, gears etc,
therefore they are subjected to bending in
addition to twisting moment.

10. • 2. Machine shaft :- These shafts forms
an integral part of the machine itself.
The crank shaft is an example of the
machine shafts.
Types of shaft

11. Standard sizes of transmission shaft
• The standards sizes of transmission
shafts.
• 25 mm to 60 mm with 5 mm steps.
• 60 mm to 110 mm with 10 mm steps.
• 110 mm to 140 mm with 15 mm steps.
• 140 mm to 500 mm with 20 mm steps.
• The standard length of shaft are 5 m ,
6m and 7 m.

12. Stresses in Shafts
• The following stresses induced in the shaft.
• 1. Shear stress due to the transmission of
torque.
• 2. Bending stresses (tensile or compressive)
due to the forces acting upon the machine
element such as pulleys, gears or its own
weight.
• 3. Stresses due to combined torsional and
bending loads.

13. Maximum Permissible working stresses
for shaft
• According to ASME, the maximum
permissible working stresses in tension
or compression may be taken as
• A. 112 MPa for shaft without keyways
• B. 84 MPa for shafts with keyways.
• For shafts purchased under definite condition, the
permissible tensile stress (σt) may be taken as 60%
of the elastic limit in tension (σel) but not more than
36 % of ultimate tensile strength (σu ).
• σt = 0.6 σel= 0.36 σu (whichever is less)

14. • According to ASME, the maximum
permissible working stresses in shear
may be taken as
• A. 56 MPa for shaft without keyways
• B. 42 MPa for shafts with keyways.
• For shafts purchased under definite condition, the
permissible tensile stress (ζ) may be taken as 30% of
the elastic limit in tension (σel) but not more than 18
% of ultimate tensile strength (σu ).
• ζ= 0.6 σel= 0.36 σu (whichever is less)
Maximum Permissible working stresses
for shaft

15. Design of Shaft
• The shaft is designed on the basis of
• 1. Strength
• 2. Rigidity
• 1. Design of shaft on the basis of
strength.
• In designing the shaft on the basis of
strength, the following three cases are
considered.

16. • A) Shaft subjected to a twisting
moment only.
• B) Shaft subjected to a bending
moment only
• C) Shaft subjected to combined twisting
and bending moments.
Design of Shaft

17. A) Shaft subjected to a twisting
moment only
• Let P be the power transmitted in Watt.
• T be the twisting moment acting upon
the shaft in N.mm.
• Then Power transmitted in Watt by the
shaft is given by
• P =
2Π NT
60000

18. • In case of belt drives, the twisting
moment (T) is given by
• T = (T1-T2) x R
• where T1 and T2 are Tension in tight
side and slack of belt respectively in N.
• R = Radius of pulleys in mm.
• When the shaft is subjected to twisting
moment only, the diameter of shaft is
obtained by using torsion equation.
A) Shaft subjected to a twisting
moment only

19. • T/J = ζ/r
• Where T = Twisting moment in Nmm
• J = Polar moment of inertia
• ζ = Torsional shear stress
• r = distance of neutral axis to the outermost fiber.
• For Solid Shaft
• T =
A) Shaft subjected to a twisting
moment only
Π x ζ x d3
16

20. • For Solid Shaft
• T =
• where do = Outside diameter of hollow shaft
• k = Ratio of inside diameter to the outside diameter of shaft
A) Shaft subjected to a twisting
moment only
Π x ζ x do
3 (1 – k4)
16

21. • When the shaft is subjected to twisting
moment only, the diameter of shaft is
obtained by using torsion equation.
• M / I = σb / y
• where M = Maximum Bending moment in N.mm.
• I = Moment of inertia of cross section of shaft
• σb = Bending stress
• y = Distance of the neutral axis to the
outermost fiber.
B) Shaft subjected to a bending
moment only

22. • For Solid Shaft
• M =
• For Hollow Shaft
• M =
Π x σb x d3
32
Π x σb x do
3 (1 – k4)
32
B) Shaft subjected to a bending
moment only

23. Difference between shaft and axle
SHAFT AXLE
1. Shaft is rotating member Axle is non-rotating member
2. The purpose of shaft is to
transmit the toque.
The purpose of an axle is to
support the transmission
element like wheels, pulleys etc.
3. Shaft is subjected to torque,
bending moment and or axial
force.
Axle is subjected bending
moment and or axial force.
4. It transmit the torque. It does not transmit the torque.

24. Difference between Solid Shaft and Hollow
Shaft
Solid Shaft Hollow Shaft
1. Weight of solid shaft is more
than hollow shaft for same
torque transmitted.
1. Weight of hollow shaft is 0.75
times less than the solid shaft.
2. Strength of solid shaft is
more.
2. Strength of hollow shaft is
0.9375 times less than the
hollow shaft.
3. T = 3. T =
4. M = 4. M =
Π x ζ x do
3 (1 – k4)
16
Π x ζ x d3
16
Π x ζ x d3
32
Π x ζ x do
3 (1 – k4)
32

25. • When the shaft is subjected combined
twisting and bending moment, then the
shaft is designed on the basis of two
theories.
• 1. Maximum shear stress theory
• 2. Maximum Normal Stress theory
• According to Maximum shear stress
theory,
C) Shaft subjected to combined twisting and bending moment

26. • Te =
• But Te= M2+T2
• Where Te is called as equivalent
twisting moment
• According to Maximum normal stress
theory,
• Me = ½ (M + (M2 +T2 )=
C) Shaft subjected to combined twisting and bending
moment
Π x ζ x d3
16
Π x σb x d3
32

27. • Where Me is called as equivalent
bending moment
• Design Procedure
• A) Calculate Twisting Moment (T) in
N.mm by
• P = or T = (T1-T2) x R
• B) Determine Various forces acting on
shaft using SFD (W, T1, T2)
C) Shaft subjected to combined twisting and bending
moment
2Π NT
60000

28. • C) Calculate reaction at support of
bearing using ∑Fy = 0 and ∑M =o
• D) Calculate Bending moment at
various points and select Maximum
bending moment as M using BMD.
• E) Select maximum bending moment as
M.
• D) According to Maximum shear stress
theory,
C) Shaft subjected to combined twisting and bending
moment

29. • Calculate equivalent twisting moment
• Te = M2 +T2
• Using Te =
determine the diameter of shaft (d).
• According to Maximum normal stress
theory, calculate equivalent bending
moment Me using Me = ½ (M+ M2+T2)
C) Shaft subjected to combined twisting and bending
moment
Π x ζ x d3
16

30. • Using Me =
• Determine the diameter (d).
• Select the larger value of diameter
obtained by using both theories.
C) Shaft subjected to combined twisting and bending
moment
Π x σb x d3
32

31. Shaft subjected to fluctuating load
• In the previous article, we have
assumed that shaft is subjected to
constant torque and bending moment.
But in actual practice, the shafts are
subjected to fluctuating torque and
bending moments. In order to design
such shafts, the combined shocks and
fatigue factors may be taken into
account for calculating equivalent
twisting and bending moment.

32. • Thus for a shaft subjected fluctuating
loads, the equivalent twisting moment
is given by
• Te = (Km x M)2 +( Kt xT)2
• And the equivalent bending moment is
given by
• Me = ½ ( (KmxM)+ (KmxM)2+(Kt xT)2)
Shaft subjected to fluctuating load

33. • The following table shows the
recommended value of Km and Kt.
Shaft subjected to fluctuating load
Nature of Load Km Kt
1. Stationery Shafts
A) Gradually applied
load
1.0 1.0
B) Suddenly applied load 1.5 to 2.0 1.5 to 2.0
2. Rotating shafts
a) Steady or gradually
applied load
1.5 1.0
b) Suddenly applied
with minor shocks
1.5 to 2.0 1.5 to 2.0

34. Design of shafts on the basis of rigidity
• The torsional deflection is obtained by
using the torsion equation.
• T/J = GΘ/L
• where Θ = Torsional deflection or angle of
twist in radians.
• T = Twisting moment on the shaft
• J= Polar moment of inertia
• G= Modulus of rigidity of shaft material (84 GPa)
• L = Length of shaft

35. KEY

36. Key
• A key is a piece of mild steel insterted
between the shaft and hub or boss of
the pulley to connect these together in
order to prevent relative motion
between them. It is always inserted
parallel to the axis of shaft. Keys are
used as temporary fastening and are
subjected to crushing and shearing
stresses. A keyway is slot in the shaft
and hub of pulley to accommodate key.

37. Types of Keys
• The keys are classified as
• 1. Sunk Key
• 2. Saddle Key
• 3. Tangent Key
• 4. Round Key
• 5. Splines
• 1. Sunk Key
• The Sunk keys are provided half in the
keyways of the shaft and half in the

38. keyway of the hub or boss of the pulley.
The sunk keys are classified as
• 1. Rectangular Key
• A rectangular sunk key is shown in fig. The usual
proportion of this key are
• Width of key w = d / 4
• Thickness of key t = d/6
• Where d is the diameter of shaft
• The key has taper 1 in 100 on the
top side.
Appl. :- For preventing the rotation of gear and shaft.
Types of Keys

39. • 2. Square Key :-
• The only difference between a
rectangular key and square key is that
its width is equal to thickness.
• Thus w = t = d / 4
• 3. Parallel Sunk Key :-
• The parallel sunk keys may be rectangular or
square cross section uniform in width and
thickness throughout. It may be noted that
Types of Keys

40. a parallel key is taper less and is used
where the pulley , gear is required to
slide along the shaft.
• 4. Gib-Head Key
• It is a rectangular sunk key with a head
at one end known as gob head. It is
usually provided to facilitates the
removal of key.
• It is used where key is removed frequently.
Types of Keys

41. • A gib head key is shown in fig.
• The usual proportion are
• Width w = d/4 and thickness t= d/6
Types of Keys

42. • 5. Feather Key :-
• A feather key is a parallel key which is
fixed either to the shaft or to the hub
and which permits relative axial
movement. It is special type of parallel
key which transmits a turning moment
and also permits axial movement. Fig
shows the feather key which is fixed to
the shaft by means of two cap-screws.
There is clearance between the key and
Types of Keys

43. • 6. Woodruff Key :-
• A woodruff key is an easily adjustable
key. It is piece from a cylindrical disc
having segmental cross section. This
key is largely used in machine tool and
automobile construction.
Types of Keys

44. Strength of Sunk Key
• A key connecting the shaft and hub is
shown in Fig

45. • Let T= Torque transmitted by shaft
• F= Tangential force acting on the circumference of the shaft.
• d = Diameter of shaft
• l = length of key
• w = Width of key
• t = thickness of key
• ζ and σc = Shear and crushing stress of key.
• A little consideration will show that due to the
power transmitted by the shaft, the key may
due to shearing or crushing.
Strength of Sunk Key

46. COUPLING

47. Coupling
• Shafts are usually available up to 7
meters length to inconvenience in
transport. In order to have a greater
strength, it becomes necessary to join
two or more pieces of the shaft by
means of couplings.
• Shaft couplings are used for several
purposes such as
• 1. To provide for the connection of
shafts that are manufactured separately

48. such as motor and generator and to
provide for disconnections for repairs.
• 2. To provide misalignment of the
shafts or to introduce mechanical
flexibility.
• 3. To reduce transmission of shock
loads from one shaft to another.
• 4. To introduce protection against
overloads
Coupling

49. • 1. It should be easy to connect or
disconnect.
• 2. It should transmit the full power from
one shaft to another shaft without
losses.
• 3. It should hold the shaft in perfect
alignment.
• 4. It should have no projecting parts.
Requirement of good shaft
Coupling

50. • Shafts couplings are divided into two
categories.
• 1. Rigid Coupling
• 2. Flexible Coupling
• 1. Rigid Coupling
• It is used to connect two shafts which
are perfectly aligned.
• Following are types of rigid Coupling.
Types of shaft Coupling

51. • A. Sleeve or Muff coupling
• B. Split muff or compression coupling.
• C. Flange coupling.
• 2. Flexible Coupling
• It is used to connect two shafts having
both lateral and angular misalignment.
• Following are type of flexible coupling.
• A. Bushed pin type flexible coupling.
• B. Universal Coupling
Types of shaft Coupling

52. Points Rigid Coupling Flexible Coupling
1. Purpose It is used to connect two
shafts which are in perfect
alignment
It is used to connect two shafts
which are both lateral and
angular misalignment.
2. Vibration Due to absence of torsional
flexibility, they cannot
damp vibration.
Due to torsional flexibility, they
will damp out vibration.
3. Alignment Rigid coupling can not
tolerate any misalignment
between two shafts
Flexible coupling can tolerate
small amount of misalignment
between two shafts
4. Deflection Shaft deflection is less. Shaft deflection is more
5. Cost Rigid couplings are less
expensive.
Flexible couplings are more
expensive.
Difference between rigid and flexible coupling.

53. 1. Sleeve or muff coupling
• It is the simplest type of rigid coupling
made of cast iron. It consists of a
hollow cylinder (called as sleeve) whose
inner diameter is the same as that of
the shaft. It is fitted over the ends of
the two shafts by means of gib headed
key as shown in fig. The power is
transmitted from one shaft to another
by means of a key and sleeve.

55. • Step I:- Calculate the torque T by using
• P =
• Step II :- Design of shaft
• Considering the shaft under pure torsion
T =
From this eqn, dia. Of shaft can be obtained.
1. Sleeve or muff coupling
2Π NT
60000
Π x ζ x d3
16

56. • Step 3 :- Design of sleeve
• Let D = Outer diameter of sleeve
• d = inner diameter of sleeve
• L = Length of sleeve
• Using standard proportion
• D = 2d + 13 mm
• L = 3.5 d
• ζs = Permissible shear stress of sleeve. The
safe value of CI may be taken as 14 MPa.
1. Sleeve or muff coupling

57. • The torque transmitted by hollow
section is
• T =
• where (k = d/D)
• From this equation induced shear stress
in the sleeve can be checked.
1. Sleeve or muff coupling
Π x ζs x D3 (1 – k4)
16

58. • Step 4 :- Design of Key
• Let w = width of key
• t= thickness of key
• l = length of key
• Considering square key,
• w = d/4
• t = d/4
• l = L/2 = 3.5d / 2
1. Sleeve or muff coupling

59. • Considering the shearing failure of key,
• T = ζ x w x l x d / 2
• From this eqn, induced shear stress in
the key can be checked.
• Considering the crushing failure of key,
• T = σck x t/2 x l x d / 2
• From this eqn, induced crushing stress
in the key can be checked.
4. Design of key

60. 2. Flange Coupling
• A flange coupling usually applies to
coupling having two separate cast iron
flanges. Each flange is mounted on the
shaft and keyed to it. The faces of
couplings are turned up at right angles
to the axis of shaft. One of the flange
has a projected portion and other
flange has corresponding recess on it.
The two flanges are coupled together
by means of nuts and bolts. Consider the

62. Design of flange coupling
• Let d = Diameter of shaft or inner diameter
of hub.
• D = Outer diameter of hub.
• D1 = Pitch circle diameter of bolts.
• D2 = Outside diameter of flange
• db = Outside diameter of bolts.
• tf = Thickness of flange
• ζs, ζf, ζk, and ζb=Permissible shear stress for
shaft, flange, keys and bolts.
• σckk and σckb be the permissible crushing stress for

63. Design of flange coupling
• Step I:- Calculate the torque T by using
• P =
• Step II :- Design of shaft
• Considering the shaft under pure torsion
T =
From this eqn, dia. Of shaft can be obtained.
2Π NT
60000
Π x ζs x d3
16

64. • Step 3 :- Design of hub
• Let D = Outer diameter of hub
• d = inner diameter of hub
• L = Length of sleeve
• Using standard proportion
• D = 2d
• L = 1.5 d
• ζf = Permissible shear stress of flange.
Design of flange coupling

65. • The hub is designed by considering it as
a hollow shaft, transmitting the same
torque as that of solid shaft.
• T =
• where (k = d/D)
• From this equation induced shear stress
in the hub can be checked.
Design of flange coupling
Π x ζf x D3 (1 – k4)
16

66. • Step 4 :- Design of Key
• Let w = width of key
• t= thickness of key
• l = length of key
• Considering rectangular key,
• w = d/4
• t = d/6
• l = L (Length of hub) = 1.5d
Design of flange coupling

67. • Considering the shearing failure of key,
• T = ζk x w x l x d / 2
• From this eqn, induced shear stress in
the key can be checked.
• Considering the crushing failure of key,
• T = σckk x t/2 x l x d / 2
• From this eqn, induced crushing stress
in the key can be checked.
4. Design of key

68. 5. Design of Flange
• The thickness of flange is usually taken
as 0.5 d.
• The flange at the junction is under
shear while transmitting the torque.
• The torque transmitted
• T = x x x
• T = ∏D x tf x ζf x D/2
• From this eqn, induced shear stress in
the flange can be checked.
{Circumference}
of hub
{Thickness of}
Flange
{Shear Stress}
of flange
{Radius }
of hub

69. 6. Design of bolts
• Let D1= Pitch circle diameter of bolt
• = 3 d
• n = Number of bolts
• = 3 for d upto 40 mm
• = 4 for d upto 100 mm
• = 6 for d upto 180 mm
• The bolts are subjected to shear stress
due to torque transmitted.

70. • Shear load on each bolt,
• F =
• Total load on all the bolts,
• Ftotal =
• Torque transmitted
• T =
6. Design of bolts
Π x db
2x ζb x n
4
Π x db
2x ζb
4
Π x db
2x ζb x n x D1/2
4

71. 7. Other parameter
• The outside diameter of flange D2= 4d
• The thickness of protective flange ,
tp = 0.25 d

72. GEAR

73. Gear
• In case of belt or rope drives, slipping
of belt is common phenomenon which
reduces the velocity ratio of system. In
precision machine, in which definite
velocity ratio is required, gears are
used. Gears are toothed wheel which
are used for transmitting power from
one shaft to another.

74. Design consideration in Gear Drive
• In the design of gear drive, the
following data is usually given.
• 1. The power to be transmitted
• 2. The speed of driving gear
• 3. The speed of driven gear or the
velocity ratio.
• 4. The centre distance
• The following requirement must be met
in the design of gear drive

75. • 1. The gear tooth have sufficient strength so that
they will not fail under static loading or dynamic
loading during normal running condition.
• 2. The gear teeth should have wear characteristics so
that their life is satisfactory.
• 3. The use of space and material should be
economical.
• 4. The alignment of the gears and deflections of the
shafts must be considered because they effect on the
performance of the gears.
• 5. The lubrications of the gears must be satisfactory.
Design consideration in Gear Drive

76. Beam Strength of Gear Teeth : ( Lewis Equation)
• Lewis derived an equation for
determining the approximate stress in a
gear tooth by considering each tooth as
a cantilever beam of uniform strength.
• Consider each tooth as cantilever beam
loaded by normal load (WN) as shown in
fig. It is resolved in two mutually
perpendicular components i.e tangential
component (WT) and radial component
(WR) acting perpendicular and parallel

77. Beam Strength of Gear Teeth : ( Lewis Equation)

78. to the centre line of tooth respectively.
The tangential component (WT) induces
a bending stress which tends to break
the tooth. The radial component (WR)
induces a compressive stress of
relatively small magnitude, therefore
the effect on the tooth may be
neglected. Hence the bending stress is
used as the basis for design calculation.
Beam Strength of Gear Teeth : ( Lewis Equation)

79. • The critical section or section of
maximum bending stress may be
calculated by drawing a parabola
through A and tangential to the tooth
curve B and C. But the tooth is larger
than parabola at every cross section
except BC. Therefore we can say that
section BC is critical section or section
of maximum stress. The maximum
value of bending stress at BC is given
Beam Strength of Gear Teeth : ( Lewis Equation)

80. • σb /y = M / I ------ (eqn (1) )
• Where M = Maximum bending moment at BC
• = WT x h
• WT = Tangential load acting on tooth
• h = Length of the tooth
• y = Half of thickness of tooth at section BC
• = t / 2
• I = Moment of inertia about centre line of tooth
• = b t3 / 12
Beam Strength of Gear Teeth : ( Lewis Equation)

81. • b = Width of gear face
• Substituting the value of M, y, I in eqn (1)
• σb = =
• WT = σb x b x t2 / 6h
• In this expression t and h are variable
depending upon the circular pitch and its
profile.
• Let t = x x Pc and h = k x Pc
Beam Strength of Gear Teeth : ( Lewis Equation)
(WT x h ) x t/2
b t3 /12
(WT x h ) x 6
b t2

82. • where x and k constants
• Where Pc = ∏ x m
• This is the Lewis equation. The
quantity y is called as Lewis form factor
an
Beam Strength of Gear Teeth : ( Lewis Equation)
WT = σb x b x Pc x y = σb x b x ∏ x m x y