ME16A: CHAPTER SIX
TORSION OF
CIRCULAR CROSSSECTIONS
6.1. SIMPLE TORSION THEORY
When a uniform circular shaft is subjected to
a torque, it can be shown that every section
of the shaft is subjected to a state of pure
shear (Fig. 6.1), the moment of resistance
developed by the shear stresses being
everywhere equal to the magnitude, and
opposite in sense, to the applied torque.
For the purposes of deriving a simple theory
to describe the behaviour of shafts subjected
to torque it is necessary to make the
following basic assumptions:
Shear System Set Up on an Element in
the Surface of a Shaft Subjected to
Torsion
Resisting torque, T
Applied torque T
Assumptions




(1) The material is homogeneous, i.e. of
uniform elastic properties throughout.
(2) The material is elastic, following Hooke's
law with shear stress proportional to shear
strain.
(3) The stress does not exceed the elastic
limit or limit of proportionality.
(4) Circular sections remain circular.
Assumptions Contd.



(5) Cross-sections remain plane. (This is
certainly not the case with the torsion of
non-circular sections.)
(6) Cross-sections rotate as if rigid, i.e. every
diameter rotates through the same angle.
Practical tests carried out on circular shafts
have shown that the theory developed below
on the basis of these assumptions shows
excellent correlation with experimental
results.
(a) Angle of Twist
Consider now the solid circular shaft of radius R subjected to a torque T at one end, t
other end being fixed (Fig. 6.2).
Under the action of this torque a radial line at the
free end

the shaft twists through an angle , point A moves to B, and AB subtends an angle
at'
fixed end. This is then the angle of distortion of the shaft, i.e. the shear strain.
Since:
angle in radians = arc /radius
arc AB = R = L


= R
/L
From the definition of rigidity modulus:
G
shear stress 
shear strain 
(1)
Simple Torsion Theory Contd.
 = /G
(2)
where  is the shear stress set up at radius R.
Therefore equating eqns. (1) and (2),
R


L
G

R

where

F
IJ ……………….
G
HK
G  '
L r
’ is the shear stress at any other radius r.
(3)
(b) Stresses
Let the cross-section of the shaft be considered as divided into elements
of radius r and thickness dr as shown in Fig. 6.3 each subjected to a
shear stress  ’
Fig. 6.3. Shaft cross-section.
The force set up on each element


= stress x area = ’ x 2
r dr (approximately)
This force will produce a moment about the center axis of the shaft, providing a
contribution to the torque:
Shear stress
( ’ x 2
= 2 
’ r2 dr
z
R
T
r dr) x r
2   ' r2
0

’ will vary with the radius r and will therefore be replaced in terms of
From equation (3)
G
' 
r
L
z
R
T
0
2
G 3
r dr
L
G
L
0
2 r 3 dr
z
R
The integral
z
R
0
2 r 3 dr is called the polar moment of area J, and may be evaluated as
a standard form for solid and hollow shafts as shown in the section 6.2 elow.
T
T

J
G
J
L
G
L
……………
(4)
Combining equations (3) and (4) produces the so called simple theory of torsion
equation:
T  G
 
J R
L
………………. (5)
6.2 POLAR SECOND MOMENT OF
AREA
As stated above, the polar second moment of area, J is defined as
z
R
J
=
0
2 r 3 dr
Lr O 2  R
2  MP
N4 Q 4
4
For a solid shaft
J =
R
4

 D4
0
(6)
32
For a hollow shaft of internal radius r:
z
R
J =
0
2 r 3 dr
= 2
L
r O 
 (R
M
P
N4 Q 2
4
R
4
 r4 ) 
r
Where D is the external and d is the internal diameter.

D
c
32
4
d4
h
(7)
6.3 Shear Stress and Shear Strain
in Shafts
The shear stresses developed in a shaft under pure torsion as shown in Fig. 6.1., their
values as given by the simple torsion theory as:
 
G
R
L
From the definition of shear or rigidity modulus, G
 G 
The two equations can be combined to relate shear stress and strain in the shaft to the
angle of twist per unit length thus:
 
G
R  G
L
(8)
or, in terms of some internal radius, r,
' 
G
r
L
 G
(9)
These equations indicate that the shear stress and shear strain vary linearly with
radius and have their maximum value at the outside radius (Fig. 6.4). The applied shear
stresses in the plane of the cross-section are accompanied by complementary stresses of
equal value on longitudinal planes as indicated in Figs. 6.1 and 6.4.
Complementary longitudinal shears
Fig. 6.4. Complementary longitudinal shear stress in a shaft subjected
to torsion
8.4 Section Modulus
It is sometimes convenient to re-write part of the torsion theory formula to obtain the
maximum shear stress in shafts as follows:
T 

J R

TR
J
With R the outside radius of the shaft the above equation yields the greatest value

possible for (Fig. 6.4),
 max 
TR
T

J
Z
(10)
where Z = J/R is termed the polar section modulus. From the preceding section:
for solid shafts:
Z
 D3
16
and for hollow shafts,
Z
 ( D4  d 4 )
16 D
(11)
6.5 Torsional Rigidity
The angle of twist per unit length of shafts is given by the torsion theory as:

L

T
GJ
The quantity G J is called the torsional rigidity of the shaft and is thus given as:
GJ
T
/L
(12)
i.e. the torsional rigidity is the torque divided by the angle of twist (in radians) per unit
length.
Power Transmitted by Shafts
If a shaft carries a torque T Newton-Metre and rotates at w rad/s, it will do work a
the rate of T w Nm/s ( or joule/s)
Now the rate at which a system works is defined as its power, the basic unit of power
being the Watt ( 1 Watt = 1 N m/s). Thus, the power transmitted by the shaft:
= T w
Watts =
2 N T
60
N = revolutions per minute
Since the Watt is a very small unit of power in engineering terms use is normally made
of S.I. multiples, i.e. kilowatts (kW) or megawatts (MW).
Combined Stress Systems-Combined
Bending and Torsion

In most practical transmission situations
shafts which carry torque are also subjected
to bending, if only by virtue of the self-weight
of the gears they carry. Many other practical
applications occur where bending and
torsion arise simultaneously so that this type
of loading represents one of the major
sources of complex stress situations.
Combined Stress Systems Contd.


In the case of shafts, bending gives rise to
tensile stress on one surface and
compressive stress on the opposite surface
whilst torsion gives rise to pure shear
throughout the shaft.
An element on the tensile surface will thus
be subjected to the stress system indicated
in Fig. 6.5 and equation or the Mohr circle
procedure derived in Chapter 4 can be
used to obtain the principal stresses present.
Combined Bending and TorsionEquivalent Bending Moment
For
shafts
subjected
to
the
simultaneous application of a bending
moment M and torque T the principal
stresses set up in the shaft can be
shown to be equal to those produced by
an equivalent bending moment,
of a certain value Me acting alone.

Combined Bending and TorsionEquivalent Bending Moment Contd.
From the simple bending theory the maximum direct stresses
set up at the outside surface of the shaft owing to the bending moment
M are given by

M ymax M D

I
2I
Similarly, from the torsion theory, the maximum shear stress in the surface of the shaft is
given by

TR TD

J
2J
Combined Bending and TorsionEquivalent Bending Moment Contd.
But for a circular shaft: J = 2 I,
 
TD
4I
The principal stress for this system can now be obtained by applying the formula
derived in the last Chapter.
1
2
 1 or  2  ( x   y ) 
and with

y
1
( x   y ) 2  4 2
2
= 0, the maximum principal stress
F
IJ
G
H K
1 MD
1
1 

2 2I
2
F
IJ M 
G
HK
1 D
= 2 2I

1
is given by:
F
FT D I
M DI
4G J
G
J
H2 I K H4 I K
2
(M 2  T2)
2
Combined Bending and Torsion-Equivalent
Bending Moment Concluded
Now, if Me is the bending moment which acting alone will produce the same maximum
stress, then:
1 
M e ymax M e D

I
2I
F
I
G
HJ
KM 
Me D 1 D

2I
2 2I
( M 2  T2)
i.e. the equivalent bending moment is given by:
Me 
1
M
2
( M 2  T2)
(13)
and it will produce the same maximum direct stress as the combined bending and torsion
effects.