Torsional Loads on Circular Shafts
• Interested in stresses and strains of
circular shafts subjected to twisting
couples or torques
• Turbine exerts torque T on the shaft
• Shaft transmits the torque to the
generator
• Generator creates
opposite torque T’
an
equal
and
1
Net Torque Due to Internal Stresses
• Net of the internal shearing stresses is an internal
torque, equal and opposite to the applied torque,
T   r dF   r  dA
• Although the net torque due to the shearing stresses
is known, the distribution of the stresses is not
• Distribution of shearing stresses is statically
indeterminate – must consider shaft deformations
• Unlike the normal stress due to axial loads, the
distribution of shearing stresses due to torsional
loads cannot be assumed uniform.
2
Shaft Deformations
• From observation, the angle of twist of the shaft is
proportional to the applied torque and to the shaft length.
 T
L
• When subjected to torsion, every cross-section of a
circular shaft remains plane and undistorted.
• Cross-sections for hollow and solid circular shafts remain
plain and undistorted because a circular shaft is
axisymmetric.
• Cross-sections of noncircular (non-axisymmetric) shafts
are distorted when subjected to torsion.
3
Shearing Strain
• Consider an interior section of the shaft. As a
torsional load is applied, an element on the interior
cylinder deforms into a rhombus.
• Since the ends of the element remain planar, the
shear strain is equal to angle of twist.
• It follows that
L  r or  
r
L
• Shear strain is proportional to angle of twist and
radius
c
r
 max 
and    max
L
c
4
Stresses in Elastic Range
• Multiplying the previous equation by the shear
modulus,
r
G  G max
c
From Hooke’s Law,   G , so
r
c
   max
J
 c4
2

The shearing stress varies linearly with the radial
position in the section.
d 4
32
• Recall that the sum of the moments from the
internal stress distribution is equal to the torque on
the shaft at the section,
T   r dA 
 max 2
 max
r
dA

J
c 
c
• The results are known as the elastic torsion
formulas,
J
 c 24  c14   d 24  d14 
2

32
 max 
Tc
Tr
and  
J
J
5
Angle of Twist in Elastic Range
• Recall that the angle of twist and maximum
shearing strain are related,
c
L
 max 
• In the elastic range, the shearing strain and shear
are related by Hooke’s Law,
 max 
 max Tc

G
JG
• Equating the expressions for shearing strain and
solving for the angle of twist,

TL
JG
• If the torsional loading or shaft cross-section
changes along the length, the angle of rotation is
found as the sum of segment rotations
 
i
Ti Li
J i Gi
6
EXAMPLE 1
Determine the maximum shearing stress caused by a torque of magnitude
T = 800 N.m.
7
EXAMPLE 1
Determine the maximum shearing stress caused by a torque of magnitude
T = 800 N.m.
8
Tr

J
 max 
J
d 4
32

 0.0364
32
 1.65  10 7 m 4
Tc 8000.018

 87.3  10 6 Pa  87.3 MPa
7
J
1.65  10
9

If the length of the rod is 600 mm and its
Modulus of Rigidity is 70 GPa, determine the
angle of twist for its free end.
10
EXAMPLE 2
Knowing that the internal diameter of the hollow shaft shown is d = 23 mm,
determine the maximum shearing stress caused by a torque of magnitude T = 1.0
kN.m.
11
EXAMPLE 2
Knowing that the internal diameter of the hollow shaft shown is d = 23 mm,
determine the maximum shearing stress caused by a torque of magnitude T = 1.0
kN.m.
12

J
0.04
32
 max
Tc 1.0  10 3 0.02


 89.3 M Pa
7
J
2.24  10
4


 0.023 4  2.24  10 7 m 4

13
EXAMPLE 3
Under normal operating conditions, the electric motor exerts a torque of 2.4 kN.m at
A. Knowing that each shaft is solid, determine the maximum shearing stress (a) in
shaft AB, (b) in shaft BC, (c) in shaft CD.
14
EXAMPLE 3
Under normal operating conditions, the electric motor exerts a torque of 2.4 kN.m at
A. Knowing that each shaft is solid, determine the maximum shearing stress (a) in
shaft AB, (b) in shaft BC, (c) in shaft CD.
15
16
EXAMPLE 4
The allowable stress is 104 MPa in the 38 mm diameter rod AB and 55 MPa in the 46
mm diameter rod BC. Neglecting the effect of stress concentrations, determine the
largest torque that may be applied at A.
17
EXAMPLE 4
The allowable stress is 104 MPa in the 38 mm diameter rod AB and 55 MPa in the 46
mm diameter rod BC. Neglecting the effect of stress concentrations, determine the
largest torque that may be applied at A.
18
19
EXAMPLE 5
The ship at A has just started to drill for oil
on the ocean floor at a depth of 1500 m.
Knowing that the top of the 200 mm
diameter steel drill pipe (G = 77 GPa)
rotates through two complete revolutions
before the drill bit at B starts to operate,
determine the maximum shearing stress
caused in the pipe by torsion.
20
EXAMPLE 5
The ship at A has just started to drill for oil
on the ocean floor at a depth of 1500 m.
Knowing that the top of the 200 mm
diameter steel drill pipe (G = 77 GPa)
rotates through two complete revolutions
before the drill bit at B starts to operate,
determine the maximum shearing stress
caused in the pipe by torsion.
TL
GJ
T
GJ
L
Tc GJc Gc
 max 


J
JL
L
77  10 9 2  2 0.1
 max 
 64.5 M Pa
1500



21
EXAMPLE 6
The torque shown are exerted on pulleys B, C and D. Knowing that the entire
shaft is made of steel (G = 27 GPa), determine the angle of twist between (a) C
and B, (b) D and B.
22
EXAMPLE 6
The torque shown are exerted on pulleys B, C and D. Knowing that the entire
shaft is made of steel (G = 27 GPa), determine the angle of twist between (a) C
and B, (b) D and B.
23
24
25
26
27
28
29
EXAMPLE 7
The electric motor exerts a torque of 800 N.m on the steel shaft ABCD when it is
rotating at constant speed. Design specifications require that the diameter of the shaft
be uniform from A to D and that the angle of twist between A and D not to exceed
1.5°. Knowing that max= 60 MPa and G = 77 GPa, determine the minimum diameter
of shaft that may be used.
30
EXAMPLE 7
The electric motor exerts a torque of 800 N.m on the steel shaft ABCD when it is
rotating at constant speed. Design specifications require that the diameter of the shaft
be uniform from A to D and that the angle of twist between A and D not to exceed
1.5°. Knowing that max= 60 MPa and G = 77 GPa, determine the minimum diameter
of shaft that may be used.
31
32
33
Material Property Relationship
Assuming that the material is homogenious, isotropic and linearelastic:Generalized Hooke’s Law:
If a material at a point is subjected to a state of triaxial stress, sx, sy,
sz, associated normal strains ex, ey, ez are developed in the material.
Poisson’s ratio, n = -(elat./elong).
Consider normal strain in the x direction.
e’x = sx/E
due to sx
e’’x = -nsy/E due to sy
e’’’x = -nsz/E due to sz
Hence
34
ex


1
s x  s y  s z
E

ey


1
s y   s x  s z 
E
ez


1
s z  s x  s y
E

These three equations express Hooke’s Law in general form for a triaxial state
of stress. (Applicable only for linear-elastic material response and small
deformations.
If we apply a shear stress xy to the element, it will deform only due to a shear
strain gxy – i.e. xy will not cause any other strains. Likewise yz and xz will only
cause shear strains gyz and gxz respectively.
g xy
Hooke’s law for shear stress can therefore be written as
1
1
1
  xy
g yz   yz
g xz   xz
G
G
G
35
Relationship between E, n and G.
E
G
21  n 
Dilatation and Bulk Modulus
When an elastic material is subjected to
normal stress, its volume will change.
dV = (1+ex)(1+ey)(1+ez) dx dy dz – dx dy dz
Neglecting the product of strains (very small)
dV = (ex + ey + ez) dx dy dz
The change in volume is called the volumetric
strain or the dilatation e.
e
dV
dV
 ex e y ez
36
In terms of stress, the dilatation can be written as
e

1  2n
s x s y sz
E

A volume of material subjected to the
uniform pressure p of a liquid experience the
same pressure all round and is always
normal to the surface on which it acts. This
state of hydrostatic loading requires the
normal stress to be equal in all directions.
sx = sy = sz = -p
Substituting and rearranging terms yields
p
E

e
31  2n 
37
p
E

e
31  2n 
The term on the right consists only of the material’s properties E
and n and is equal to the ratio of uniform normal stress to the
dilatation or volumetric strain. It is called the volume modulus of
elasticity or bulk modulus, k
E
k
31  2n 
Note: For most metals
so
1
n 
3
kE
38
Non Uniform Torsion
39
40
41
POWER TRANSMISSION
• Power can be expressed as,
P  T
where P
=
power (watt)
T =
torque (newton-meters)
 =
angular velocity (radians per second)
• Power can also be expressed as,
P  2fT
where P
=
power (watt)
f
=
frequency (revolutions per second or
hertz)
T =
torque (newton-meters)
42
EXAMPLE 10
A solid steel shaft AB as shown in the figure is to be used to transmit 3750 W from
the motor M to which it is attached. If the shaft rotates at  = 175 rpm and the steel
has an allowable shear stress of allow = 100 MPa, determine the required diameter of
the shaft to the nearest mm.
43
EXAMPLE 10
A solid steel shaft AB as shown in the figure is to be used to transmit 3750 W from
the motor M to which it is attached. If the shaft rotates at  = 175 rpm and the steel
has an allowable shear stress of allow = 100 MPa, determine the required diameter of
the shaft to the nearest mm.
44
EXAMPLE 8
Two solid steel shafts are fitted with flanges which are then connected by fitted bolts
so that there is no relative rotation between the flanges. Knowing that G = 77 GPa,
determine the maximum shearing stress in each shaft when a 500 N.m torque is
applied to flange B.
45
EXAMPLE 8
Two solid steel shafts are fitted with flanges which are then connected by fitted bolts
so that there is no relative rotation between the flanges. Knowing that G = 77 GPa,
determine the maximum shearing stress in each shaft when a 500 N.m torque is
applied to flange B.
46
47
9.8  10 5 T A   7.09  10 5 T A  500  0
 AB   CD  0
 T A  210 N  m
 TL 
 TL 

 
 0
 GJ  AB  GJ  CD
T AB 0.6 
77  10 9 
 0.03
32
4

2100.015
 Tr 
 39.6 MPa
 
4
 0.03
 J  AB
32
 max AB  
TCD 0.9 
77  10 9 
 0.036
4
0
2900.018 
 Tr 
 31.7 MPa
 
4
 J  CD  0.036
32
 max CD  
32
9.8  10  5 T AB  7.09  10  5 TCD  0
(1)
T AB  T A
( 2)
TCD  T A  500
( 3)
48
EXAMPLE 9
A 4 kN.m torque T is applied at end A of the composite shaft shown. Knowing that the
modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminium, determine (a)
the maximum shearing stress in the steel core, (b) the maximum shearing stress in the
aluminium jacket, (c) the angle of twist at A.
49
EXAMPLE 9
A 4 kN.m torque T is applied at end A of the composite shaft shown. Knowing that the
modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminium, determine (a)
the maximum shearing stress in the steel core, (b) the maximum shearing stress in the
aluminium jacket, (c) the angle of twist at A.
50
51
T alTst  4  10 3
J st 
J al
 0.054
( 2)
Tal  1.32Tal  4  10 3
4
 8.3  10  7 m 4
32
 0.072 4  0.054 4

 1.8  10  6 m 4
32


Tal  1724 N  m
Tst  2276 N  m
(a)
 st   al
(b)
 TL 
 TL 

 

 GJ  st  GJ  al
(c)
Tr 2276  0.027

 74 MPa
J
8.3  10  7
Tr 1724  0.036
 max al 

 34.5 MPa
J
1.8  10  6
TL
2276  2.5
A 

 8.9  10  2 rad  5.1
GJ 77  10 9  8.3  10  7
 max st 
Tst
Tal

77  10 9  8.3  10  7 27  10 9  1.8  10  6
Tst  1.32Tal
(1)
52
53
54
THIS IS WRONG
55