Courses

# Torsion of Shafts Notes | EduRev

## : Torsion of Shafts Notes | EduRev

``` Page 1

ENGINEERING COUNCIL
CERTIFICATE LEVEL

ENGINEERING SCIENCE C103

TUTORIAL 3 - TORSION

You should judge your progress by completing the self assessment exercises.

These may be sent for marking or you may request copies of the solutions at a cost (see home
page).

On completion of this tutorial you should be able to do the following.

• Derive the torsion equation
• Derive polar second moment of area.
• Solve problems involving torque, shear stress and angle of twist.
• Derive the formula for the power transmitted by a shaft
• Relate power transmission to torsion.
• Outline the method of solution for rectangular cross sections.
• Solve problems with shafts of rectangular cross section.

It is assumed that students doing this tutorial are already familiar with the concepts of
second moments of area and shear stress.
Page 2

ENGINEERING COUNCIL
CERTIFICATE LEVEL

ENGINEERING SCIENCE C103

TUTORIAL 3 - TORSION

You should judge your progress by completing the self assessment exercises.

These may be sent for marking or you may request copies of the solutions at a cost (see home
page).

On completion of this tutorial you should be able to do the following.

• Derive the torsion equation
• Derive polar second moment of area.
• Solve problems involving torque, shear stress and angle of twist.
• Derive the formula for the power transmitted by a shaft
• Relate power transmission to torsion.
• Outline the method of solution for rectangular cross sections.
• Solve problems with shafts of rectangular cross section.

It is assumed that students doing this tutorial are already familiar with the concepts of
second moments of area and shear stress.
1. TORSION OF SHAFTS

Torsion occurs when any shaft is subjected to a torque. This is true whether the shaft is rotating (such as
drive shafts on engines, motors and turbines) or stationary (such as with a bolt or screw). The torque makes
the shaft twist and one end rotates relative to the other inducing shear stress on any cross section. Failure
might occur due to shear alone or because the shear is accompanied by stretching or bending.

1.1. TORSION EQUATION

The diagram shows a shaft fixed at one end and twisted at the other end due to the action of a torque T.

Figure 1

The radius of the shaft is R and the length  is L.

Imagine a horizontal radial line drawn on the end face. When the end is twisted, the line rotates through an
angle ?. The length of the arc produced is R ?.

Now consider a line drawn along the length of the shaft. When twisted, the line moves through an angle ?.
The length of the arc produced is L ?.

If we assume that the two arcs are the same it follows that R ? = L ? ?
Hence by equating L ? = R ? we get
L
R ?
? = ? .........................(1A)

If you refer to basic stress and strain theory, you will appreciate that ?  is the shear strain on the outer surface
of the shaft. The relationship between shear strain and shear stress is

?
t
G =  .....................(1B)
t is the shear stress and G the modulus of rigidity.

G is one of the elastic constants of a material. The equation is only true so long as the material remains
elastic.
Substituting (1A) into (1B) we get
R
t
L
G ?
=  .....................(1C)
Since the derivation could be applied to any radius, it follows that shear stress is directly proportional to
radius 'r' and is a maximum on the surface. Equation (1C) could be written as

r
t
L
G ?
=   .....................(1D)
Now let's consider how the applied torque 'T' is balanced by the internal stresses of the material.

Page 3

ENGINEERING COUNCIL
CERTIFICATE LEVEL

ENGINEERING SCIENCE C103

TUTORIAL 3 - TORSION

You should judge your progress by completing the self assessment exercises.

These may be sent for marking or you may request copies of the solutions at a cost (see home
page).

On completion of this tutorial you should be able to do the following.

• Derive the torsion equation
• Derive polar second moment of area.
• Solve problems involving torque, shear stress and angle of twist.
• Derive the formula for the power transmitted by a shaft
• Relate power transmission to torsion.
• Outline the method of solution for rectangular cross sections.
• Solve problems with shafts of rectangular cross section.

It is assumed that students doing this tutorial are already familiar with the concepts of
second moments of area and shear stress.
1. TORSION OF SHAFTS

Torsion occurs when any shaft is subjected to a torque. This is true whether the shaft is rotating (such as
drive shafts on engines, motors and turbines) or stationary (such as with a bolt or screw). The torque makes
the shaft twist and one end rotates relative to the other inducing shear stress on any cross section. Failure
might occur due to shear alone or because the shear is accompanied by stretching or bending.

1.1. TORSION EQUATION

The diagram shows a shaft fixed at one end and twisted at the other end due to the action of a torque T.

Figure 1

The radius of the shaft is R and the length  is L.

Imagine a horizontal radial line drawn on the end face. When the end is twisted, the line rotates through an
angle ?. The length of the arc produced is R ?.

Now consider a line drawn along the length of the shaft. When twisted, the line moves through an angle ?.
The length of the arc produced is L ?.

If we assume that the two arcs are the same it follows that R ? = L ? ?
Hence by equating L ? = R ? we get
L
R ?
? = ? .........................(1A)

If you refer to basic stress and strain theory, you will appreciate that ?  is the shear strain on the outer surface
of the shaft. The relationship between shear strain and shear stress is

?
t
G =  .....................(1B)
t is the shear stress and G the modulus of rigidity.

G is one of the elastic constants of a material. The equation is only true so long as the material remains
elastic.
Substituting (1A) into (1B) we get
R
t
L
G ?
=  .....................(1C)
Since the derivation could be applied to any radius, it follows that shear stress is directly proportional to
radius 'r' and is a maximum on the surface. Equation (1C) could be written as

r
t
L
G ?
=   .....................(1D)
Now let's consider how the applied torque 'T' is balanced by the internal stresses of the material.

Consider an elementary ring of material with a shear stress t acting on it at

The area of the ring is   dA = 2 p r dr
The shear force acting on it tangential is dF = t dA = t 2 pr dr
This force acts at radius r so the torque produced is dT = t 2 pr
2
dr
Since
L
r G ?
= t from equation (1D) then dr r 2 p
L
G ?
dT
3
=
Figure 2
The torque on the whole cross section resulting from the shear stress is
?
=
R
0
3
dr r 2 p
L
G ?
T
The expression  is called the polar second moment of area and denoted as 'J'. The Torque equation
reduces to
?
R
0
3
dr r 2 p
J
L
G ?
T = and this is usually written as
L
G ?
J
T
= ...........................(1E)
Combining (1D) and (1E) we get the torsion equation
r
t
L
G ?
J
T
= =  ................(1F)
1.2   POLAR SECOND MOMENTS OF AREA

This tutorial only covers circular sections. The formula for J is found by carrying out the integration or may
be found in standard tables.
For a shaft of diameter D the formula is
32
pD
J
4
=
This is not to be confused with the second moment of area about a diameter, used in bending of beams (I)
but it should be noted that J = 2 I.

WORKED EXAMPLE No.1

A shaft 50 mm diameter and 0.7 m long is subjected to a torque of 1200 Nm. Calculate the shear stress
and the angle of twist. Take G = 90 GPa.
SOLUTION

Important values to use are D = 0.05 m, L = 0.7 m, T = 1200 Nm, G = 90 x10
9
Pa

o
871 . 0
180
x 0.0152  ?  degrees  to Converting
0.025 x  10 x 90
0.7 x 10 x 48.89
GR
tL
? y Alternatel
10 x 613.59 x  10 x 90
0.7 x 1200
J
TL
?
MPa 48.89 or  Pa 10 x 48.89
10 x 613.59
0.025 x 1200
J
TR
t
m 10 x 613.59
32
0.05 x p
32
pD
J
9
6
9 - 9
6
9 -
max
4 9 -
4 4
= =
= = =
= = =
= = =
= = =
p

Page 4

ENGINEERING COUNCIL
CERTIFICATE LEVEL

ENGINEERING SCIENCE C103

TUTORIAL 3 - TORSION

You should judge your progress by completing the self assessment exercises.

These may be sent for marking or you may request copies of the solutions at a cost (see home
page).

On completion of this tutorial you should be able to do the following.

• Derive the torsion equation
• Derive polar second moment of area.
• Solve problems involving torque, shear stress and angle of twist.
• Derive the formula for the power transmitted by a shaft
• Relate power transmission to torsion.
• Outline the method of solution for rectangular cross sections.
• Solve problems with shafts of rectangular cross section.

It is assumed that students doing this tutorial are already familiar with the concepts of
second moments of area and shear stress.
1. TORSION OF SHAFTS

Torsion occurs when any shaft is subjected to a torque. This is true whether the shaft is rotating (such as
drive shafts on engines, motors and turbines) or stationary (such as with a bolt or screw). The torque makes
the shaft twist and one end rotates relative to the other inducing shear stress on any cross section. Failure
might occur due to shear alone or because the shear is accompanied by stretching or bending.

1.1. TORSION EQUATION

The diagram shows a shaft fixed at one end and twisted at the other end due to the action of a torque T.

Figure 1

The radius of the shaft is R and the length  is L.

Imagine a horizontal radial line drawn on the end face. When the end is twisted, the line rotates through an
angle ?. The length of the arc produced is R ?.

Now consider a line drawn along the length of the shaft. When twisted, the line moves through an angle ?.
The length of the arc produced is L ?.

If we assume that the two arcs are the same it follows that R ? = L ? ?
Hence by equating L ? = R ? we get
L
R ?
? = ? .........................(1A)

If you refer to basic stress and strain theory, you will appreciate that ?  is the shear strain on the outer surface
of the shaft. The relationship between shear strain and shear stress is

?
t
G =  .....................(1B)
t is the shear stress and G the modulus of rigidity.

G is one of the elastic constants of a material. The equation is only true so long as the material remains
elastic.
Substituting (1A) into (1B) we get
R
t
L
G ?
=  .....................(1C)
Since the derivation could be applied to any radius, it follows that shear stress is directly proportional to
radius 'r' and is a maximum on the surface. Equation (1C) could be written as

r
t
L
G ?
=   .....................(1D)
Now let's consider how the applied torque 'T' is balanced by the internal stresses of the material.

Consider an elementary ring of material with a shear stress t acting on it at

The area of the ring is   dA = 2 p r dr
The shear force acting on it tangential is dF = t dA = t 2 pr dr
This force acts at radius r so the torque produced is dT = t 2 pr
2
dr
Since
L
r G ?
= t from equation (1D) then dr r 2 p
L
G ?
dT
3
=
Figure 2
The torque on the whole cross section resulting from the shear stress is
?
=
R
0
3
dr r 2 p
L
G ?
T
The expression  is called the polar second moment of area and denoted as 'J'. The Torque equation
reduces to
?
R
0
3
dr r 2 p
J
L
G ?
T = and this is usually written as
L
G ?
J
T
= ...........................(1E)
Combining (1D) and (1E) we get the torsion equation
r
t
L
G ?
J
T
= =  ................(1F)
1.2   POLAR SECOND MOMENTS OF AREA

This tutorial only covers circular sections. The formula for J is found by carrying out the integration or may
be found in standard tables.
For a shaft of diameter D the formula is
32
pD
J
4
=
This is not to be confused with the second moment of area about a diameter, used in bending of beams (I)
but it should be noted that J = 2 I.

WORKED EXAMPLE No.1

A shaft 50 mm diameter and 0.7 m long is subjected to a torque of 1200 Nm. Calculate the shear stress
and the angle of twist. Take G = 90 GPa.
SOLUTION

Important values to use are D = 0.05 m, L = 0.7 m, T = 1200 Nm, G = 90 x10
9
Pa

o
871 . 0
180
x 0.0152  ?  degrees  to Converting
0.025 x  10 x 90
0.7 x 10 x 48.89
GR
tL
? y Alternatel
10 x 613.59 x  10 x 90
0.7 x 1200
J
TL
?
MPa 48.89 or  Pa 10 x 48.89
10 x 613.59
0.025 x 1200
J
TR
t
m 10 x 613.59
32
0.05 x p
32
pD
J
9
6
9 - 9
6
9 -
max
4 9 -
4 4
= =
= = =
= = =
= = =
= = =
p

1.3 HOLLOW SHAFTS

Since the shear stress is small near the middle, then if there are no other stress considerations other than
torsion, a hollow shaft may be used to reduce the weight.

The formula for the polar second moment of area is
( )
32
d D p
J
4 4
-
= .
D is the outside diameter and d the inside diameter.

WORKED EXAMPLE No.2

Repeat the previous problem but this time the shaft is hollow with an internal diameter of 30 mm.

() ( )
o
d
1
180
x 0.0152  ?  degrees  to Converting
0.025 x  10 x 90
0.7 x 10 x 56.17
GR
tL
? y Alternatel
10 x 534.07 x  10 x 90
0.7 x 1200
J
TL
?
MPa 56.17 or  Pa 10 x 56.17
10 x 534.07
0.025 x 1200
J
TR
t
m 10 x 534.07
32
03 . 0 0.05 x p
32
D p
J
9
6
9 - 9
6
9 -
max
4 9 -
4 4 4 4
= =
= = =
= = =
= = =
=
-
=
-
=
p

Note that the answers are nearly the same even though there is much less material in the shaft.

WORKED EXAMPLE No.3

A shaft 40 mm diameter is made from steel and the maximum allowable shear stress for the material is
50 MPa. Calculate the maximum torque that can be safely transmitted. Take G = 90 GPa.

SOLUTION

Important values to use are:
D = 0.04 m, R = 0.02 m, t = 50 x10
6
Pa and G = 90 x10
9
Pa

r
t
L
G ?
J
T
= =

Nm 628.3
0.02
10 x 251.32 x 10 x 50
R
J t
T
term. middle  the ignore and Rearrange
R
t
L
G ?
J
T
is equation  torsion complete The
m 10 x 251.32
32
0.04 x p
32
pD
J
9 - 6
max
4 9 -
4 4
= = =
= =
= = =

Page 5

ENGINEERING COUNCIL
CERTIFICATE LEVEL

ENGINEERING SCIENCE C103

TUTORIAL 3 - TORSION

You should judge your progress by completing the self assessment exercises.

These may be sent for marking or you may request copies of the solutions at a cost (see home
page).

On completion of this tutorial you should be able to do the following.

• Derive the torsion equation
• Derive polar second moment of area.
• Solve problems involving torque, shear stress and angle of twist.
• Derive the formula for the power transmitted by a shaft
• Relate power transmission to torsion.
• Outline the method of solution for rectangular cross sections.
• Solve problems with shafts of rectangular cross section.

It is assumed that students doing this tutorial are already familiar with the concepts of
second moments of area and shear stress.
1. TORSION OF SHAFTS

Torsion occurs when any shaft is subjected to a torque. This is true whether the shaft is rotating (such as
drive shafts on engines, motors and turbines) or stationary (such as with a bolt or screw). The torque makes
the shaft twist and one end rotates relative to the other inducing shear stress on any cross section. Failure
might occur due to shear alone or because the shear is accompanied by stretching or bending.

1.1. TORSION EQUATION

The diagram shows a shaft fixed at one end and twisted at the other end due to the action of a torque T.

Figure 1

The radius of the shaft is R and the length  is L.

Imagine a horizontal radial line drawn on the end face. When the end is twisted, the line rotates through an
angle ?. The length of the arc produced is R ?.

Now consider a line drawn along the length of the shaft. When twisted, the line moves through an angle ?.
The length of the arc produced is L ?.

If we assume that the two arcs are the same it follows that R ? = L ? ?
Hence by equating L ? = R ? we get
L
R ?
? = ? .........................(1A)

If you refer to basic stress and strain theory, you will appreciate that ?  is the shear strain on the outer surface
of the shaft. The relationship between shear strain and shear stress is

?
t
G =  .....................(1B)
t is the shear stress and G the modulus of rigidity.

G is one of the elastic constants of a material. The equation is only true so long as the material remains
elastic.
Substituting (1A) into (1B) we get
R
t
L
G ?
=  .....................(1C)
Since the derivation could be applied to any radius, it follows that shear stress is directly proportional to
radius 'r' and is a maximum on the surface. Equation (1C) could be written as

r
t
L
G ?
=   .....................(1D)
Now let's consider how the applied torque 'T' is balanced by the internal stresses of the material.

Consider an elementary ring of material with a shear stress t acting on it at

The area of the ring is   dA = 2 p r dr
The shear force acting on it tangential is dF = t dA = t 2 pr dr
This force acts at radius r so the torque produced is dT = t 2 pr
2
dr
Since
L
r G ?
= t from equation (1D) then dr r 2 p
L
G ?
dT
3
=
Figure 2
The torque on the whole cross section resulting from the shear stress is
?
=
R
0
3
dr r 2 p
L
G ?
T
The expression  is called the polar second moment of area and denoted as 'J'. The Torque equation
reduces to
?
R
0
3
dr r 2 p
J
L
G ?
T = and this is usually written as
L
G ?
J
T
= ...........................(1E)
Combining (1D) and (1E) we get the torsion equation
r
t
L
G ?
J
T
= =  ................(1F)
1.2   POLAR SECOND MOMENTS OF AREA

This tutorial only covers circular sections. The formula for J is found by carrying out the integration or may
be found in standard tables.
For a shaft of diameter D the formula is
32
pD
J
4
=
This is not to be confused with the second moment of area about a diameter, used in bending of beams (I)
but it should be noted that J = 2 I.

WORKED EXAMPLE No.1

A shaft 50 mm diameter and 0.7 m long is subjected to a torque of 1200 Nm. Calculate the shear stress
and the angle of twist. Take G = 90 GPa.
SOLUTION

Important values to use are D = 0.05 m, L = 0.7 m, T = 1200 Nm, G = 90 x10
9
Pa

o
871 . 0
180
x 0.0152  ?  degrees  to Converting
0.025 x  10 x 90
0.7 x 10 x 48.89
GR
tL
? y Alternatel
10 x 613.59 x  10 x 90
0.7 x 1200
J
TL
?
MPa 48.89 or  Pa 10 x 48.89
10 x 613.59
0.025 x 1200
J
TR
t
m 10 x 613.59
32
0.05 x p
32
pD
J
9
6
9 - 9
6
9 -
max
4 9 -
4 4
= =
= = =
= = =
= = =
= = =
p

1.3 HOLLOW SHAFTS

Since the shear stress is small near the middle, then if there are no other stress considerations other than
torsion, a hollow shaft may be used to reduce the weight.

The formula for the polar second moment of area is
( )
32
d D p
J
4 4
-
= .
D is the outside diameter and d the inside diameter.

WORKED EXAMPLE No.2

Repeat the previous problem but this time the shaft is hollow with an internal diameter of 30 mm.

() ( )
o
d
1
180
x 0.0152  ?  degrees  to Converting
0.025 x  10 x 90
0.7 x 10 x 56.17
GR
tL
? y Alternatel
10 x 534.07 x  10 x 90
0.7 x 1200
J
TL
?
MPa 56.17 or  Pa 10 x 56.17
10 x 534.07
0.025 x 1200
J
TR
t
m 10 x 534.07
32
03 . 0 0.05 x p
32
D p
J
9
6
9 - 9
6
9 -
max
4 9 -
4 4 4 4
= =
= = =
= = =
= = =
=
-
=
-
=
p

Note that the answers are nearly the same even though there is much less material in the shaft.

WORKED EXAMPLE No.3

A shaft 40 mm diameter is made from steel and the maximum allowable shear stress for the material is
50 MPa. Calculate the maximum torque that can be safely transmitted. Take G = 90 GPa.

SOLUTION

Important values to use are:
D = 0.04 m, R = 0.02 m, t = 50 x10
6
Pa and G = 90 x10
9
Pa

r
t
L
G ?
J
T
= =

Nm 628.3
0.02
10 x 251.32 x 10 x 50
R
J t
T
term. middle  the ignore and Rearrange
R
t
L
G ?
J
T
is equation  torsion complete The
m 10 x 251.32
32
0.04 x p
32
pD
J
9 - 6
max
4 9 -
4 4
= = =
= =
= = =

SELF ASSESSMENT EXERCISE No.1

1.  A shaft is made from tube 25 mm outer diameter and 20 mm inner diameter. The shear stress must not
exceed 150 MPa. Calculate the maximum torque that should be placed on it.
(Ans. 271.69 Nm).

2.  A shaft is made of solid round bar 30 mm diameter and 0.5 m long. The shear stress must not exceed 200
MPa. Calculate the following.

i. The maximum torque that should be transmitted.
ii.  The angle of twist which will occur.

Take G = 90 GPa.
(Ans. 1060 Nm and 4.2
o
)

1.4  MECHANICAL POWER TRANSMISSION BY A SHAFT

In this section you will derive the formula for the power transmitted by a shaft and combine it with torsion
theory.

Mechanical power is defined as work done per second. Work done is defined as force times distance moved.
Hence
P = Fx/t  where   P is the Power
F is the force
x is distance moved.
t is the time taken.

Since distance moved/time taken is the velocity of the force we may write

P = F v ........(2A) where v is the velocity.

When a force rotates at radius R it travels one circumference in the time of one revolution. Hence the
distance moved in one revolution is  x = 2 pR

If the speed is N rev/second then the time of one revolution is 1/N seconds. The mechanical power is hence
P = F 2 pR/(1/N) = 2 pNFR

Since  FR is the torque produced by the force this reduces to
P = 2 pNT ....................(2B)

Since 2 pN is the angular velocity ? radians/s it further reduces to
P = ?T.................(2C)

Note that equations (2C) is the angular equivalent of equation (2A) and all three equations should be
remembered.
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!