DME-MODULE 2

DME-MODULE 2
Shafts:

A shaft is a rotating machine 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 (or twisting moment) set up within the shaft permits the power to be transferred to
various machines linked up to the shaft.

I. Commonly used rotating machine element which transmit power

II. Power is delivered to the shaft through the action of some tangential force and the
resultant torque or torsional moment set up within the shaft permits the power to be
transferred to various machines linked to the shaft

III. Various machine elements like gear, pulleys, sprockets etc. mounted on the shaft by
means of key
Specific categories of transmission shaft
Axle : used as a shaft that supports rotating elements like wheels, hoisting, drums
etc.
Spindle: a short rotating shaft, drive shaft of late, drilling machine etc.
Counter shaft : a secondary shaft which is driven by the main shaft and from
which the power is supplied to a machine component
Jack shaft: it is an auxiliary or intermediate shaft between two shaft that are used
in transmission of power
Line shaft: consist of number of shaft connected in axial direction by means of
couplings
Classification of shaft (https://youtu.be/3Hjmile-cNU )
According to their use
1). Prime mover shaft
2). Machine shaft

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3). Power transmission shaft

1).Prime mover shaft
i. connected to the source of power
ii example – engine shaft, generator shaft, motor shaft, turbine shaft
2) Machine shaft: These shafts form an integral part of the machine itself. The
crank shaft is an example of machine shaft.
3) Power transmission shaft - Used to transmit power between the source and the
machines using power

ii. Example- line shaft, counter shaft, jack shaft etc.

Desirable properties of shaft materials

i. It should have sufficient high strength

ii. It should have high machinability

iii. It should have good wear resistance

iv. It should have ability to withstand heat and case hardening treatment

v. It should have law sensitivity to stress concentration

Shaft materials

Medium carbon steel (machinery steel)

High carbon steels

Alloy steels (nickel, nickel chromium and molybdenum steel)

Low carbon steel

Standard size of transmission shaft

i.)All transmission shaft are made in standard size and length to avoid the
difficulties in manufacturing and transporting

ii). Standard length 5 m, 6 m, 7 m

iii). Standard diameter or size are

1). 25 mm to 60 mm in steps of 5 mm increment

2) 60 mm to 110 mm in steps of 10 mm increment

3) 110 mm to 140 mm in steps of 15 mm increment

4) 140 mm to 500 mm in steps of 20 mm increment

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Stress in shaft
1. Shear stress due to transmission of torque

2. Bending stress due to the force acting up on by machine elements

3. Stress due to combined torsional and bending loads

Torsion equation assumption (https://youtu.be/ehNCICDNcJk)

(i) The material of the shaft is uniform throughout

(ii) The shaft is of circular cross section remains circular after loading.ie, the
diameter of normal cross-section remains unchanged.

(iii) The plane section of a shaft normal to its axis before loading,remains plane
even after the application of torque.

(iv) The twist along the length of the shaft is uniform throughout

(v) The distance between any two normal cross-sections remains same even after
the applications of torque.

(vi) Maximum shear stress developed in the shaft under torsion does not exceed
the value of its elastic limit.

Torsion equation

T Ʈ Gθ
= =
J r l

Where

T = Torque transmitted by the shaft in N-mm

J = Polar moment of inertia of the shaft about the axis of rotation in 𝑚𝑚4

Ʈ = Torsional shear stress induced in the shaft in Mpa

r = Distance from neutral axis to the outer most fibre or radius of the shaft in mm

G = Modulus of rigidity for the shaft material in MPa.

θ = Angle of twist in radians in length l

l = Length of the shaft in mm.

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Design of Shafts
The shafts may be designed on the basis of
1. Strength, and
2. Rigidity and stiffness.

In designing shafts on the basis of strength, the following cases may be considered:
(a) Shafts subjected to twisting moment or torque only,

(b) Shafts subjected to bending moment only,

(c) Shafts subjected to combined twisting and bending moments, and

(d) Shafts subjected to axial loads in addition to combined torsional and bending loads

Shafts Subjected to Twisting Moment Only

a) Solid shaft:

T Ʈ
=
J r
Where,
T = Torque transmitted by the shaft in N-mm

J = Polar moment of inertia of the shaft about the axis of rotation in 𝑚𝑚4
Ʈ = Torsional shear stress induced in the shaft in Mpa
r = Distance from neutral axis to the outer most fibre or radius of the shaft
(a) For solid shaft
We know that for round solid shaft,
Π
Polar moment of inertia , J = 32 d4

Where d= diameter of the shaft in mm

And, radius of the shaft, r = d/2 mm
Consider the torsion equation,
T Ʈ
=
J r
Ʈ
Or , T = ×J
r

Substitute J and r in above equation

Ʈ Π Π
T = d/2 × 32 d4 = 16 Ʈd3

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Where,
T = Twisting moment or maximum torque acting upon the shaft in N-mm

Ʈ = Maximum torsional shear stress induced in the outer most fibre of the
shaft in Mpa
d =diameter of solid shaft in mm
(b) For hollow shaft
We know that for round hollow shaft, polar moment of inertia
Π Π d 4
J = 32 (d0 4 − d1 4) = 32 d0 4(1 − (d1 ) )
0

Π
= d0 4(1 − K 4 )
32

Where,

d0 = outside Diameter of hollow shaft in mm

d1 = inside Diameter of hollow shaft in mm

K = Ratio of inside diameter and outside diameter of the hollow shaft or
d1
Diameter ratio =
d0

And, distance from the neutral axis to the outer most fibre of the shaft,

r = d0 /2
Consider the torsion equation,

T Ʈ
=
J r
Ʈ
Or, T = ×J
r

Substitute J and r in above equation

Ʈ Π Π
= × 32 d0 4(1 − K 4 ) = 16 Ʈ d0 3 (1 − K 4 )
d0 /2

Where,
T = Twisting moment or maximum torque acting upon the shaft in N-mm

Ʈ = Maximum torsional shear stress induced in the outer most fibre of the
Shaft in Mpa

d0 = outside Diameter of hollow shaft in mm

K = Ratio of inside diameter and outside diameter of the hollow shaft or

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d1
Diameter ratio =
d0

Design of shaft subjected to bending moment only (https://youtu.be/Uxrkmzmf-80)

When shaft subjected to pure bending load the principle stress induced in the shaft will be bending
stress

Using bending equation

Bending moment equation

M б E
= =
I Y R

M б
=
I Y

Where,

M = Maximum bending moment in N-mm

I = Moment of inertia of the shaft cross-sectional area about the axis of rotation in
mm4

б = Maximum bending stress induced in the outer most fibre of the shaft in Mpa

Y = distance from the neutral axis to the outer most fibre of the shaft or radius of

the shaft in mm

FOR SOLID SHAFT

Π
Moment of inertia, I = 64 d4

Where,

d =diameter of solid shaft in mm

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𝑑
And, radius of shaft, Y =
2

M б
Consider the bending equation, =
I Y
б
Or, M = ×I
Y

Substitute I and y in the above equation

б Π Π
M= 𝑑 × 64 d4 = 32
б d3
2

Where,
M = Maximum bending moment in N-mm
б = Maximum bending stress induced in the outer most fibre of the shaft in Mpa
d =diameter of solid shaft in mm

FOR HOLLOW SHAFT

Π
Moment of inertia, I = 64 (d0 4 − d1 4)

Π d 4
= d0 4 (1 − ( 1 ) )
64 d0

Π
= d0 4(1 − K 4 )
64

M б
Consider the bending equation, =
I Y
б
Or, M = ×I
Y

Substitute I and y in the above equation

б Π
M= × 64 d0 4(1 − K 4 )
d0 /2
Π
= 32 бd0 3(1 − K 4 )

Where,

d0 = outside Diameter of hollow shaft in mm

d1 = inside Diameter of hollow shaft in mm

K = Ratio of inside diameter and outside diameter of the hollow shaft or

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d1
Diameter ratio =
d0

SHAFT SUBJECTED TO BENDING MOMENT EQUATIONS

Bending equation

M б E
= =
I Y R

FOR SOLID SHAFT

Π
M= б d3
32

FOR HOLLOW SHAFT

Π
M= бd0 3(1 − K 4 )
32

DESIGN OF SHAFT SUBJECTED TO COMBINED TWISTING MOMENT AND BENDING

MOMENT

Usually based on

I. Maximum shear stress theory (Guests theory)- shaft generally made up of ductile materials

II. Maximum normal stress theory (Rankines theory)- shaft made of brittle materials

Guest’s theory

Maximum shear stress

1
Ʈmax = √б2 + 4Ʈ2
2

Where,
Ʈ = Shear stress induced due to twisting moment in Mpa

б = Bending stress induced due to bending moment in Mpa.

Ʈmax = Maximum shear stress induced due to combined twisting and bending moment in
Mpa
From strength equations

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Π
√M 2 + T 2 = 16 Ʈmax d3

√M 2 + T 2 is known as equivalent twisting moment

Π
For solid shaft , T = Ʈd3
16

Π
For hollow shaft T = Ʈ d0 3 (1 − K 4 )
16

From strength equation

32 M 16 T
б= and Ʈ =
Πd3 Πd3
Substituting this in stress equation
1 Π
(M + √M 2 + T 2 ) = 32 бmax× d3
2

1
(M + √M 2 + T 2 ) Is known as equivalent bending moment
2

For solid shaft equivalent bending moment

Π
M = 32 б × d3

Π
For hollow shaft equivalent bending moment, б × d3 (1 − K 4 )
32

Equivalent twisting moment

The twisting moments which will act alone produces the same shear stress as the actual twisting
moment

Equivalent bending moment

The equivalent bending moment which when acting alone produces the same bending stress as the
actual bending moment

Shaft design on torsional rigidity basis (https://youtu.be/rl1mn-Ul1k0)

I. Shafts are mostly designed on the basis of strength

II. But in certain applications the shaft are designed on the basis stiffness or rigidity of the shaft

III. Long shafts of smaller diameter or where the vibration is not permissible the governing
consideration is stiffness or rigidity

Torsion equation
T Ʈ Gθ
= =
J r l

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Where

T = Torque transmitted by the shaft in N-mm

J = Polar moment of inertia of the shaft about the axis of rotation in 𝑚𝑚4

Ʈ = Torsional shear stress induced in the shaft in Mpa

r = Distance from neutral axis to the outer most fibre or radius of the shaft in mm

G = Modulus of rigidity for the shaft material in MPa.

θ = Angle of twist in radians in length l

l = Length of the shaft in mm

FOR SOLID SHAFT

T Gθ
=
J l
Π
Polar moment of inertia, J = d4
32

TL
Angle of twist, θ =
JG

180 TL 584TL
In degrees θ = × =
Π JG Gd4

Where,

θ = Angle of twist in degrees

l = Length of the shaft in mm

T = Twisting moment or torsional moment in N-mm
G = Modulus of rigidity for the shaft material in MPa
d =diameter of shaft in mm
For hollow shaft
Π
Polar moment of inertia = d0 4(1 − K 4 )
32

584TL
Angle of twist , θ =
Gd0 4(1−K4 )

Where,

d0 = outside Diameter of hollow shaft in mm

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K = Ratio of inside diameter and outside diameter of the hollow shaft or

d1
Diameter ratio =
d0

Torsional stiffness

Torsional stiffness of a shaft is defined as the amount of torque required to twist the shaft through one
radians

T JG
Torsional stiffness, q = =
θ L

Comparison between hollow and solid shaft

1. Comparison by strength

2. Comparison by weight

3. Comparison by stiffness

Comparison by strength (https://youtu.be/G4GGt1fogKI)

Π Π
TH = 16 Ʈ d0 3 (1 − K 4 ) and TS = 16 Ʈd3
Π
TH Ʈ d0 3 (1−K4 ) d0 3 (1−K4 )
16
= Π =
TS Ʈd3 d3
16

TH 1+K2
=
TS √1−K2

TH
›1 Or TH ›TS
TS

Comparison by weight((https://youtu.be/G4GGt1fogKI)

WH AH LH ⍴H AH
= × × = (LH = LS and ⍴H = ⍴S )
WS AS LS ⍴S AS

Π
d 2 (1−K2 )
4 0 d0 2 (1−K2 )
= Π 2 =
d d2
4

WH (1−K2 )
=
WS (1−K4 )2/3

WH
< 1 or WH < WS
WS

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Comparison by stiffness
T
From the rigidity equation, q =
θ

TH
q H θH TH
= ⁄T = ( θH = θS )
qS S TS
θS

qH 1 + K2
=
q S √1 − K 2

Power transmitted by the shaft (https://youtu.be/eMxjAN0-gSQ)

The main purpose of the shaft is to transmit power from one shaft to another

Work done on the shaft per minute = Force × Distance moved per minute.

Work done = F × 2ᴨrN = ( F×r) ×2ᴨN = 2ᴨNT ( Torque, T = F×r)

Power transmitted = Work done in N-m per second

2ᴨNT
P=
60

Where,

P = Power transmitted by the shafts in watts

N = Speed of the shaft in rpm

T = Average or mean torque in N-m

Please remember the torque T in N-m but in torque equation we use T in N-mm

T = ( T1− T2 ) R

Where,

T = Twisting moment or torque in N-mm

T1 = Tension in the tight side of the belt in N

T2 = Tension in the slack side of the belt in N

R = Radius of the pulley in mm

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2ᴨNT 2ᴨN 2ᴨN

P= = × T = Tω ( ω= )
60 60 60

Where,

P = Power transmitted by the shafts in watts

T = Mean or average torque in N-m

ω = Angular speed in rad/s

Couplings

I. A mechanical device that permanently connect the shaft of a driving machine to the shaft of
a driven machine

II. The term permanent is meant that the machine component are not disconnected or
disengaged during operation

Purpose of coupling (https://youtu.be/83222iK4QhE)

I. Used to connect one shaft to another to increase length of shaft according to requirement

II. To provide for the connection of shafts of two purchased parts or separately manufactured
units

III. To provide for misalignment of shafts or to introduce mechanical flexibility

IV. To alter the vibration and minimize shock characteristics of rotating units throughout the
length

Different type of misalignments of shaft axis

I. Aligned or collinear axis

II. Axis with lateral misalignment

III. Intersecting axis with angular misalignment

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Requirements of couplings
I. Capable of transmit torque from driving shaft to driven shaft without any loss

II. Permits easy connection and disconnection of the shafts for the purpose of repair and
alterations

III. Should keep the perfect alignment of two shafts

Should be safe from projecting parts

Types of couplings (https://youtu.be/83222iK4QhE)

Classified according to use in to two groups

1. Rigid couplings

2. Flexible couplings

1. Rigid couplings

I. Has no flexibility or resilience

II. It cannot tolerate misalignment between the axis of the shafts

III. It is suitable for accurately aligned shaft having law speeds

IV. Commonly used couplings are

1. Sleeve or muff coupling

2. Clamp or compression coupling

3. Flange coupling

4. Shrouded or protected type flange coupling

5. Solid or marine flange coupling

CLAMP OR COMPRESSION COUPLING

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FLANGE COUPLING

SHROUDED OR PROTECTED TYPE FLANGE COUPLING

Flexible couplings
I. Costlier due to additional parts compared to rigid couplings

II. Has flexibility and resilience

III. Can tolerate misalignment between the axes of the shafts

IV. Used for the shaft to provide a small amount of lateral and angular misalignment

1. Oldham flexible coupling

2 .Universal flexible coupling

3. Bushed pin type flexible coupling

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OLDHAM FLEXIBLE COUPLING

UNIVERSAL FLEXIBLE COUPLING BUSHED PIN TYPE FLEXIBLE

Sleeve or muff coupling

(box coupling)

I .Used for connecting smaller sizes of shaft

II. Consist of hollow cast iron cylinder whose inner diameter is same as that of the shaft
(sleeve or muff)

III. Muff enveloping the butting ends of shafts by means of a rectangular sunk key

IV. The torque is transmitted from input shaft to sleeve through the key and is then
transmitted from sleeve to output shaft through the key

Sleeve or muff coupling

Sleeve or muff coupling standard proportions

D = 2d+13
L= 3.5 mm
Where,
d = Diameter of the shaft in mm

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D = Outer diameter of the sleeve in mm

L = Axial length of the sleeve in mm

Design of muff coupling (https://youtu.be/HIYh7t3c1sw)

1. Design of shaft

2. Design of sleeve or muff

3. Design of key

1. Design of shaft
Calculate the diameter of each shaft by the following equations.
(i) Using power equations determine the mean or average torque
60P
Tmean =
2ᴨN

(ii) Compute the maximum torque as per the given conditions. If the condition is not
assumed mean torque is equal to maximum torque.

Tmax = (100+x) % of Tmean (if Tmax is x% greater than Tmean)

Or, Tmax = Tmean (Assume Tmax and Tmean are same)

(iii) Convert the maximum torque in N-mm

(iv) Using strength equation, determine the diameter of each shaft. i.e

3 16× Tmax
d= √
ᴨƮ

(v) Adopt standard size dimension for the transmission shaft.

2. Design of sleeve or muff
Calculate the dimensions of the sleeve by the standard empirical proportions.
(a) Outer diameter of sleeve or muff, D = 2d+13
(b) Axial length of sleeve or muff, L = 3.5 d
(c) Check torsional shear stress induced in the sleeve or muff by considering it to be as a
hollow shaft. Using strength equation for hollow shaft.
Π d
Tmax = 16 Ʈd3 (1 − K 4 ) Where k =
D
16 × Tmax
Or, Ʈ =
ᴨd3 (1−K4 )

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The calculated value of induced shear stress for muff Ʈ is less than the given permissible
or safe value of torsional shear stress for the muff material, design of sleeve is safe
otherwise it is not safe.
3. Design of key
Calculate the standard cross-section of key and length of the key in each shaft by the
following empirical equations.

(a) Compare the crushing stress бc,k and shearing stress Ʈk for the key material .If

бc,k = 2Ʈk , adopt square key otherwise adopt rectangular key.

d
(b) Width of key, W =
4

If possible adopt standard size or adopt next whole number.

(c) Thickness of the key

t =w (for square key)
d
t= (for rectangular key)
6

If possible adopt standard size or adopt next whole number.

L 3.5d
(d) Length of the key in each shaft, l= =
2 2

(e) Check torsional shear stress, Ʈk and compressive or crushing stress, бc,k
For the key material by using the strength equations for key.
d
Tmax = l×w×Ʈ×
2
2 × Tmax
Or induced shear stress in the key , Ʈk =
𝑙×𝑤×𝑑
t d
And, Tmax = l× ×бc,k ×
2 2
4× Tmax
Or induced crushing stress in the key, бc,k =
𝑙×𝑡×𝑑

The calculated value of shear stress Ʈk , and crushing stress бc,k is less than the given
permissible value of shear stress and crushing stress for the key material. Design is safe,
otherwise it is not safe.
Flange coupling
i. Standard form of rigid coupling

ii. Used in heavy transmission at low speeds

iii. Consist of two cast iron flanges one keyed to driving shaft and other half to driven shaft

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iv. The flanges are connected by together by means of number of bolts and nut arranged on
pitch circle

v. The projected portion or spigot on one flange is made to fit accurately in to a

corresponding recess or socket provided in the other flange

vi. This sprocket and spigot arrangement ensures perfect alignment of axis

vii. The torque or power is transmitted from the driving shaft to the flange on driving shaft
through the key and then transmitted to the other flange on driven shaft through the bolts;
finally power or torque from this flange is transmitted to the driven shaft through the key

Types of Flange couplings

1. Un protected type or ordinary rigid flange couplings

2. Protected type flange coupling

3. Solid flange or marine type flange coupling

1. Un protected type or ordinary rigid flange couplings

The standard proportions are

Inside diameter of hub =d

Outside diameter of hub = 2d

Length of hub or effective length = L = 1.5 d

Pitch circle diameter of bolts, DP = 3d

Outside diameter of flange, D0 = DP + (DP – D)

= 2 DP − D = 6d-2d = 4d

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Protected type flange coupling

i. The bolts are not projected beyond the flanges

ii. Bolts and nut heads are covered by projecting circumferential rim

Proportions are

Thickness of flange, t f = 0.25 d

Thickness of protective flange, t p = 0.25 d

Thickness of protective rim, t r = 0.25 d

Where d is the diameter of each shaft or inner diameter of the hub in mm.

Design procedure for rigid flange coupling (https://youtu.be/iSR9wgob6pk)

Step-1
1. Design of shaft
Calculate the diameter of each shaft by the following equations.
(i) Using power equations determine the mean or average torque

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60P
Tmean =
2ᴨN

(ii) Compute the maximum torque as per the given conditions. If the condition is not
assumed mean torque is equal to maximum torque.

Tmax = (100+x) % of Tmean (if Tmax is x% greater than Tmean)

Or, Tmax = Tmean (Assume Tmax and Tmean are same)

(iii) Convert the maximum torque in N-mm

(iv) Using strength equation, determine the diameter of each shaft. i.e

3 16× Tmax
d= √
ᴨƮ

(v) Adopt standard size dimension for the transmission shaft.

STEP-2

Compute the dimensions of the hub by the standard empirical proportions.

(a) Inside diameter of hub =d
(b) Outside diameter of hub, D = 2d
(c) Axial length of hub, L = 1.5d
(d) Check the dimensions of hub is safe or not from strength point of view. To the checking

find torsional shear stress induced in the hub by considering it to be as a hollow shaft just

similar to the muff or sleeve in muff coupling. Using strength equation for hollow shaft.

Π d
Tmax = 16 Ʈd3 (1 − K 4 ) Where k =
D
16 × Tmax
Or, Ʈ =
ᴨd3 (1−K4 )

The computed value of induced shear stress in the hub Ʈ is less than the allowable or
permissible value of torsional shear stress for the hub material; design of hub is safe
otherwise it is not safe.
STEP-3

Design of flanges
Calculate the dimensions of the flanges by the following empirical equations.

(a) Overall or outside diameter of flange , D0 = 4d

(b) Thickness of flange in protective , t f = 0 .25d
Thickness of protective flange,t p = 0 .25d

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Thickness of protective rim, t r = 0 .25d

(c) Check the safety of these dimensions from strength point of view. The failure of flange
at the junction of the hub under shear will take place while transmitting torque. Using
strength equation, for torque transmitted.
Tmax =Shearing area ×Shear stress induced × Radius of hub
= Circumference of hub ×Thickness of flange × Shear stress induced×radius
of hub
= ΠDt × Ʈ × 𝐷/2

2× Tmax
Or Ʈ =
Πd2 t

The computed value of induced shear stress for flange Ʈ is less than the given allowable
or permissible torsional shear stress for the flange material; Design of flange is safe
otherwise it is not safe.
STEP-4

4. Design of key
Calculate the standard cross-section of key and length of the key in each shaft by the
following empirical equations.

(a) Compare the crushing stress бc,k and shearing stress Ʈk for the key material .If

бc,k = 2Ʈk , adopt square key otherwise adopt rectangular key.

d
(b) Width of key, W =
4

If possible adopt standard size or adopt next whole number.

(c) Thickness of the key

t =w (for square key)
d
t= (for rectangular key)
6

If possible adopt standard size or adopt next whole number.

L 3.5d
(d) Length of the key in each shaft, l= =
2 2

(e) Check torsional shear stress,Ʈk and compressive or crushing stress, бc,k
For the key material by using the strength equations for key.
d
Tmax = l×w×Ʈ×
2
2 × Tmax
Or induced shear stress in the key,Ʈk =
𝑙×𝑤×𝑑

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t d
And, Tmax = l× ×бc,k ×
2 2
4× Tmax
Or induced crushing stress in the key,бc,k =
𝑙×𝑡×𝑑

The calculated value of shear stressƮk , and crushing stress бc,k is less than the given
permissible value of shear stress and crushing stress for the key material. Design is safe,
otherwise it is not safe.
STEP-5

Design of bolt
Calculate the nominal diameter of bolts and the number of bolts by using the strength
equations and empirical equations.

(i) Decide the number of bolts by using the empirical equations.

4
n= d+ 3
150

If the computed value is decimal, next integer is taken as the number of bolts.

(ii) Using the empirical relations for pitch circle diameter of bolts

DP =3d

(iii) Compute the normal diameter of bolts from the strength equation for shearing of the

Bolt. Therefore the bolts are subjected to shear stress due to transmitting torque.ie

Tmax =Force acting on the bolts× Radius of pitch circle

= Shearing area × Shear stress ×Pitch circle radius

Π Dp
= d2 × n ×Ʈ ×
4 2

Rearranging the above equation,

𝑚𝑎𝑥8𝑇
Diameter of bolt, d = √ΠƮn D P

(iv) Adopt the standard dimension of bolt.

(v) Check the safety of nominal diameter of the bolt by crushing stress in the bolt.

Tmax = Crushing area × Crushing stress × radius of pitch circle

Dp
= d1 ×t f ×n×бc,b ×
2

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2 × Tmax
Or induced crushing stress, бc,b =
d1 ×tf ×n×DP

The computed value of crushing stress in the bolt бc,b is less than the given allowable
or permissible value of crushing or compressive stress for the bolt material, Design of
bolt is safe otherwise it is not safe.

PREVIOUS QUIESTIONS

1. Find out the minimum diameter of shaft transmitting a torque of 31.5×104 N-m without

exceeding the maximum shear stress of 50 Mpa.

Solution

Given data

Torque transmitted, T = 31.5×104 N-m

= 31.5×107 N-mm

Maximum shear stress, Ʈ = 50 Mpa = 50 N/mm2

Using strength equation for solid shaft,

Π
TS = 16 Ʈd3

3 16×T 3 16×31.5×107
Or, diameter of solid shaft, d = √ ᴨ×Ʈ = √ ᴨ×50

= 317.76 Say 320 mm

Result
Standard diameter of solid shaft, d = 320 mm
2. A solid circular transmission shaft of 100 mm diameter is subjected to a torque of 10KN-

m.Calculate the maximum shear stress produced in the shaft

Solution

Given data

Diameter of shaft, d = 100 mm

Torque, T = 10 KN-m = 10 × 106 N-mm

Using strength equation for solid shaft,

Π
TS = 16 Ʈd3

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16×T 16×10 × 106

Or, maximum shear stress , Ʈ = =
ᴨd3 ᴨ×1003

= 50.93 N/mm2 = 50.93 MPa

3. A hollow shaft is used to transmit a torque of 18 KN-m. If the shear stress for the shaft

Material is no to exceed 42 MPa. Design the shaft whose internal diameter is ¾ of the

External diameter.

Solution

Given data

Torque transmitted, T = 18 KN-m = 18×106 N-mm

Safe shear stress, Ʈ = 42 MPa = 42 N/mm2

3
Inside diameter of shaft, di = d0 = 0.75 d0
4

di 0.75 d0
Diameter ratio, K = = = 0.75
d0 d0

Using strength equation,

Π 3
T = 16 Ʈd0 (1 − K 4 )

3 16× T
Or , outside diameter of hollow shaft, d0 = √ᴨƮ(1−K4 )

3 16×18×106
= √ = 147.25 say 150 mm
ᴨ×42(1−(0.75)4 )

Inside diameter of shaft, di = 0.75 d0 =0.75× 150 = 112.5 mm

4. Find the power transmitted by a shaft of 60 mm diameter at a speed of 100 rpm, if the

Permissible shear stress is 50 MPa.

Solution

Given data

Diameter of shaft, d = 60 mm

Speed of shaft, N = 100 RPM

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Permissible shear stress, Ʈ = 50 MPa = 50 N/mm2

Using strength equation, Maximum torque

Π Π
TS = 16 Ʈd3 = 16 50 × 603

= 2120575.04 N-mm = 2120575.04×10−3 N-m

Using the power equation, power transmitted by the shaft,

2ᴨNT 2ᴨ×100×2120575.04×10−3
P= == = 22206. 6 W =22.2 KW
60 60

5. A solid steel shaft will be required to transmit 60 KW at 70 rpm. If the maximum torque is

30% greater than the mean torque and the limit of torsional stress is to be 55 MPa. What is

the suitable diameter of the shaft

Solution

Given data

Power transmitted, P = 60 KW = 60×103

Speed of the shaft, N = 70 rpm

Maximum torque, Tmax = 30 % greater than the mean torque

= 1.3 Tmean

Limiting stress, Ʈ = 55 MPa = 55 N/mm2

2ᴨNTmean
Using the power equation, P =
60

60P 60×60×103
Or, Mean torque Tmean = = = 8185. 11 N-m
2ᴨN 2ᴨ×70

= 8185. 11× 103 N-mm

Maximum torque, Tmax = 1.3 Tmean = 1.3× 8185. 11× 103

= 10640643 N-mm
Π
Using strength equation, TS = 16 Ʈd3

3 16×Tmax 3 16×10640643
Or, diameter of shaft d = √ =√ = 99.51 say 100mm
ᴨ×Ʈ ᴨ×55

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6. A shaft has to transmit 105 KW at 160 rpm. If the stress is not to exceed 65 MPa and the

twist in a length of 3.5 m must not exceed 10 .Find suitable diameter. Take G = 80 GPa

Solution

Given data

Power transmitted, P = 105 KW = 105× 103 W

Speed of shaft, N = 160 rpm

Allowable shear stress, Ʈ = 65 Mpa

Length of shaft, l = 3.5 m = 3500 mm

Angle of twist, θ = 10

Modulus of rigidity, G = 80 GPa = 80× 103 MPa

= 8×104 N/mm2

Using power equation,

60P 60×105× 103

Torque, T = = =6266.73 N-m =6266.73×103 N-mm
2ᴨN 2ᴨ×160

Case (1) considering strength

Π
Using strength equation, T = 16 Ʈd3

3 16×T 3 16×6266.73×103
Or, diameter of shaft d = √ =√
ᴨ×Ʈ ᴨ×65

= 78.89 mm

Case (2) considering torsional rigidity

584TL
Using the rigidity equation, θ =
Gd4

4 584TL
Or, diameter of shaft d = √
Gθ

4 584×6266.73×103 ×3500
=√ = 112.49 mm
8×104 ×1

Take the larger of the two values, we have

Diameter of shaft, d = 112.49 say 125 mm

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7. A line shaft is to transmit 40 KW at 150 rpm. The shaft pieces are connected by muff

Coupling. Design the coupling assuming an overload of 20 percent. The following

Permissible stresses may be used.

Shear stress for shaft and key material, = 40 MPa

Crushing stress for key material, = 90 MPa

Shear stress for muff material, = 15 MPa

Solution

Given data

Power transmitted, P = 40 KW = 40×103 W

Speed of shaft, N = 150 rpm

Maximum torque, Tmax = 20% more than Tmean

= 1.2 Tmean
Permissible shear stress for shaft, Ʈs = 40 MPa= 40 N/mm2

Permissible shear stress for key, Ʈk = 40 MPa= 40 N/mm2

Permissible crushing stress for key, бc,k = 90 MPa = 90 N/mm2

Permissible crushing stress for muff, Ʈm = 15 MPa = 15 N/mm2

Step-1

Design of shaft

60P 60×40×103
Mean torque transmitted, Tmean = = = 2546.48 N-m
2ᴨN 2ᴨ×150

Maximum torque transmitted, Tmax = 1.2 Tmean

= 1.2 × 2546.48 = 3055. 776×103 N-mm

3 16×Tmax 3
16×3055.776×10 3
Diameter of shaft d = √ =√
ᴨ×Ʈs ᴨ×40

= 73 mm

Adopt the standard dimension for shaft

The diameter of transmission shaft,d = 80 mm

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Step-2

Design of sleeve

(a) Inside diameter of sleeve,d = 80 mm

(b) Outside diameter of sleeve, 2d+13 = 2×80+13 = 173 mm (D)
(c) Length of sleeve, L = 3.5d = 3.5× 80 = 280 mm
Checking the shear stress induced in the sleeve by considering it as a hollow shaft
Π
Tmax = 16 Ʈm D3 (1 − K 4 )

Or, induced shear stress in the sleeve,

16×Tmax 16×3055.776×103
Ʈm =
ᴨD3 (1−K4 )
= 80 = 3.15 N/mm2 =3.15 MPa
ᴨ×1733 (1−( )4 )
73

The induced shear stress in the sleeve (3.15) is less than permissible shear stress of 15
MPa. Hence the above designed dimensions of sleeve is safe.
Step-3

Design of key

There is no relationship between the crushing stress and shearing stress for the material

therefore adopt rectangular sunk key.

d 80
Width of key , W= = = 20 mm
4 4

d 80
Thickness of key, t = = = 13.3 say 14 mm
6 6
L 280
Length of the key in each shaft,l = = =140 mm
2 2

Checking the shearing and crushing stress induced in the key by using the corresponding

strength equations

2Tmax 2×3055.776×103
Induced shear stress in the key, Ʈk = = = 27.28 N/mm2
lwd 140×20×80

= 27.28 MPa

4Tmax 4×3055.776×103
Induced crushing stress in the key, бc,k = = = 77.95 N/mm2
ltd 140×14×80

= 77.95 MPa

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The induced shear stress ( 27.28 MPa) and crushing stress ( 77.95 MPa) are less than safe

shear stress (40 MPa) and safe crushing stress ( 90 MPa).Hence the designed dimensions

of key is safe.

8. Design a muff coupling to connect two shafts transmitting 100 KW at 200 rpm.The

Permissible shearing and crushing stresses for shaft and key material are 50 MPa and 100

MPa respectively. The material of muff is cast iron with permissible shear stress of 15

MPa .Assume that the maximum torque transmitted is equal to the mean torque.

Solution

Given data

Power transmitted, P = 1000 KW = 1×105 W

Speed of shaft, N = 200 rpm

Maximum torque, Tmax = Tmean

Permissible shear stress for shaft, Ʈs = 50 MPa= 50 N/mm2

Permissible shear stress for key, Ʈk = 50 MPa= 50 N/mm2

Permissible crushing stress for key, бc,k = 100 MPa = 100 N/mm2

Permissible shear stress for muff, Ʈm = 15 MPa = 15 N/mm2

Step-1

Design of shaft

60P 60×1×105
Mean torque transmitted, Tmean = = = 4774.65 N-m
2ᴨN 2ᴨ×200

Maximum torque transmitted, Tmax = Tmean = 4774.65 N-m

= 4774.65×103 N-mm

3 16×Tmax 3
16× 4774.65×10 3
Diameter of shaft d = √ =√ = 78.64 say 80 mm
ᴨ×Ʈs ᴨ×50

Step-2

Design of muff

(a) Inside diameter of muff, d = 80 mm

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(b) Outside diameter of muff , 2d+13 = 2×80+13 = 173 mm (D)

(c) Length of muff, L = 3.5d = 3.5× 80 = 280 mm
Check for safety
Checking the shear stress induced in the sleeve by considering it as a hollow shaft
Π
Tmax = 16 Ʈm D3 (1 − K 4 )

Or, induced shear stress in the sleeve,

16×Tmax 16×4774.65×103
Ʈm =
ᴨD3 (1−K4 )
= 80 = 4.92 N/mm2 = 4.92 MPa
ᴨ×1733 (1−(73)4 )

Hence the induced stress 4.92 MPa is less than permissible shear stress of 15
MPa.Therefore design of muff is safe
Step-3

Design of key

Since crushing stress of key material 100 MPa is two times of shearing stress of key

Material 50 MPa. бc,k = 2 × Ʈk , therefore adopt square sunk key

d 80
Width of key, W= = = 20 mm
4 4

Thickness of key, t = W = 20 mm

L 280
Length of the key in each shaft,l = = = 140 mm
2 2

Check for safety

2Tmax 2×4774.65×103
Induced shear stress in the key, Ʈk = = = 42.63 N/mm2
lwd 140×20×80

= 42.63 MPa

4Tmax 4×4774.65×103
Induced crushing stress in the key, бc,k = = =85.26 N/mm2
ltd 140×20×80

= 85.26 MPa

Induced stresses are less than the permissible stresses for the key material. Hence the

Design of key is safe.

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Result

Diameter of each shaft = 80 mm

Inside diameter of muff = 80 mm

Outside diameter of muff = 173 mm

Length of muff = 280 mm

Dimensions of square key = 20 mm×20 mm×140 mm

9. A flange coupling has 8 equally spaced bolts on a pitch circle diameter of 120 mm. The

Maximum torque to be transmitted is 2500 N-m. If the ultimate shear stress of the bolt

Material is 350 MPa. Estimate the minimum diameter of bolts required. Take factor of

Safety as 5

Solution

Given data

Number of bolts, n =8

Pitch circle diameter of bolts, DP = 120 mm

Maximum torque transmitted, Tmax = 2500 N − m = 25 × 105 N − mm

Ultimate shear stress for the bolt material,Ʈu,b = 350 MPa =350 N/mm2

Factor of safety, F.S = 5

Convert the ultimate stress into permissible or safe stress

Ultimate stress 350
Permissible shear stress , Ʈ = = = 70 N/mm2 = 70MPa
Factor of safety 5

Π Dp
Consider shearing failure of bolts, Tmax = d2 × n ×Ʈ ×
4 2

8 Tmax 8 ×25×105
Or , nominal diameter of bolts, d =√ =√ = 9.73 say
ΠƮnDP Π×70×8×120

= 10mm

Result

Minimum diameter of bolt required 10mm.Use M10 bolts

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