Lecture Notes of Mechanics of Solids, Chapter 4 1
Chapter 4 Torsion of Circular Shafts
In addition to the bars/rods under axial loads as discussed in Chapters 1 to 3, there are other
loading cases in engineering practice. In this chapter we will discuss the effects of applying a
torsional loading to a long straight circular member such as a shaft or tube, as extracted from
the machine showing in Fig. 4.1. We are going to show how to determine both the
• Shear strain and shear stress
• The angle of twist
Turbine
Machine
Generator
A
B
Shaft
Transmit
mechanical
power
Transmit
electrical
power
Wires
Driven Torque T
D
Resistant Torque T
R
F.B.D.
Fig. 4.1 Engineering example of torsional shaft
4.1 SHEAR STRESS/STRAIN RELATIONSHIP
(4
th
: 6970,106107; 5
th
:6970,106107)
Shear Stress
Let’s recall the definition of shear stress in Chapter 2. When parts of a deformable body try to
slip past another part, a shear stress is set up.
P
∆F
∆F
n
∆F
t
Cross section
∆A
Shearing in torsion
T
T
Shear force
τ
Fig. 4.2 Definition of shear stress and shearing in torsion
The shear stress equation was defined in section 2.1 as:
A
F
lim
t
A
∆
∆
=τ
→∆ 0
(4.1)
which is a shear force intensity that acts parallel to the material cross sectional plane as
shown in Fig. 4.2. It is worth pointing out that the shear stress in an element always comes
with pairs to maintain equilibrium as shown in Fig. 4.3.
Lecture Notes of Mechanics of Solids, Chapter 4 2
Shear Deformation
We call the deformation created by shear stress as Shear Strain, given the symbol γ
(gamma). It is defined as the change in angle of the element, it is a nondimensional quantity.
Unit of shear strain is radian
γ
τ
τ
τ
τ
Fig. 4.3 Element of material with applied shear stress τ and shear strain γ
Hooke’s Law for Shear
By conducting a similar material testing to normal stress (Section 2.4), there is a linear
relationship (for most engineering materials) between the shear stress and shear strain, as
shown in Fig. 4.4.
τ
γ
1
Gradient = G
Fig. 4.4 Relationship of shear stress τ – shear strain γ for linear elastic material
This relationship is called Hooke's law for Shear and is represented by equation Eq. (4.2).
γ=τ G
(4.2)
where: G = Shear Modulus of Elasticity (for short, Shear Modulus) or Modulus of Rigidity.
As the shear modulus is a material property (determined by material shear testing), it is
related to the Young's Modulus E and Poison's ratio v by the following equation (we are going
to prove this late),
( )
v
E
G
+
=
12
(4.3)
Now that we have these relationships we can examine the effect of an applied torque to the
shaft of circular cross section.
4.2 TORSION OF CIRCULAR SECTIONS (4
th
: 177188;5
th
:177188)
Assumptions
• This analysis can only be applied to solid or hollow circular sections
• The material must be homogeneous
• Torque is constant and transmitted along bar by each section trying to shear over its neighbor.
• Transverse planes remain parallel to each other.
• For small angle of rotation, the length of shaft and its radius remain unchanged.
Lecture Notes of Mechanics of Solids, Chapter 4 3
Shear Strain/Stress Distribution
Examine the deformation of a length dx between two transverse planes of a shaft with an
applied torque T.
For this differential element, assume the left end is fixed and the right end rotates by dφ due to
the applied torque T, where dφ is termed as Angle of Twist
of the element
dx
γ
dφ
Fix
End
Twisted
End
x
da
T
ρ
Fig. 4.5 Small transverse element with applied torque T rotated by an amount dφ
The surface of radius “ρ” rotates through angle γ, which is shear strain.
The arc is defined as length da, which is equal to:
dxdda γ=ϕρ=
which gives that:
dx
dϕ
ρ=γ
(4.4)
where:
dx
dϕ
= Rate of Twist (4.5)
which is constant for the crosssectional plane. Eq. (4.4) states that the magnitude of shear
strain for any of these elements varies only with its radial distance
ρ
.
By using Hooke’s law:
γ
=τ G
(4.2)
and by substituting for shear strain
γ
, Eq. (4.4), Eq. (4.2) becomes that:
dx
d
G
ϕ
ρ=τ
(4.6)
which relates the shear stress linearly to the distance ρ away from the centre of the section.
As a result, the shear stress distribution then looks as Fig. 4.6.
T
Distribution of
shear stress
Fig. 4.6 Shear stress distribution in circular section with applied torque
T
Torque T and Rate of Twist
We now equate the applied torque
T
to the torque generated in the section by the shear stress
distribution. To do this, look at a small circumferential section
dA
, as shown in Fig. 4.7.
Lecture Notes of Mechanics of Solids, Chapter 4 4
Elemental
distribution of
shear stress
ρ
dρ
dA
Fig. 4.7 Shear stress distribution in elemental section
dρ
of cross section
The elemental torque of a thin circular strip of thickness
d
ρ is given by
( )
dAdT τ⋅ρ=
(4.7)
Integrating over the area of the circular beam gives
( )
∫∫
ρ
ρπρρτ=ρτ= ddAT
A
2
Substituting for shear stress as Eq. (4.6):
∫
ρ
ρρ
ϕ
ρπρ= d
dx
d
GT 2 (4.8)
Since the rate of twist
dxdϕ
is constant through the section, it is not a function of radius ρ. If
we assume a homogeneous material, G is also constant, so:
J
dx
d
Gd
dx
d
GT
ϕ
=ρπρ
ϕ
=
∫
ρ
3
2 (4.9a)
or
GJ
T
dx
d
=
ϕ
(4.9b)
Polar Moment of Inertia
J
We represent the integral term as the geometric rigidity of the cross section. We call this term
the
Polar (Second) Moment of Inertia
,
J
.
∫
ρ
ρπρ= dJ
3
2
(4.10)
This term indicates the cross sectional properties to withstand the applied torque.
Since this applies to circular bars, the standard terms for J are:
•
Solid Shaft of radius R, diameter D
:
322
2
44
0
3
DR
dJ
R
π
=
π
=ρπρ=
∫
(4.11)
•
Hollow Shaft with Inner Radius R
i
and Outer radius R
o
:
(
)
(
)
㌲2
2
4444
3
ioio
R
R
DDRR
dJ
o
i
−π
=
−π
=ρπρ=
∫
(4.12)
R
D
R
o
D
o
R
i
D
i
Lecture Notes of Mechanics of Solids, Chapter 4 5
•
Thin Walled Tube with t < R/10:
R
m
= mean radius of the thin walled tube
iom
RRR ≈≈ and
io
RRt
−
=
From Eq. (4.12),
( )
(
)
(
)
(
)
( )
tRtRRRRRRRRJ
mmmioioio
3222
222
22
π=
π
=−++
π
=
tRJ
m
3
2π=
(4.13)
Engineer's Theory of Torsion ( ETT )
When we equate Eq. (4.9) with the shear stress term, Eq. (4.6), gives that:
ρ
τ
=
ϕ
=
dx
d
G
J
T
(4.14)
or
ρ=τ
J
T
(4.15)
where
T = the internal torque at the analyzed crosssection;
J = the shaft’s polar moment of inertia;
G = shear modulus of elasticity for the material
ρ = radial distance from the axis (centre).
This is called
Engineer's Theory of Torsion ( ETT )
.
The Maximum Shear Stress
The maximum shear stress can be computed as
o
R
J
T
max
=τ
(4.16)
where R
o
is the radius of the outer surface of circular shaft.
Example 4.1
Compare the weight of equal lengths of hollow and solid shafts to transmit a
torque T for the same maximum shear stress. For hollow shaft, the inner and outer diameters
have relationship D
i
= 2/3 D
o
= 2/3 D
H
.
From ETT (Eq. 4.15):
D
JJ
ttancons
T 2
=
ρ
==
τ
For Solid Shaft:
32
4
S
Solid
D
J
π
=
For Hollow Shaft:
4
4
4
81
65
323
2
32
HHHHollow
DDDJ ×
π
=
−
π
=
If we then equate the RHS of the above equation (due to the same T and τ), we get:
Hollow
H
Solid
S
D
J
D
J
=
22
, i.e.
H
Hollow
S
Solid
D
J
D
J 22
=
Substituting for the J's we get:
R
m
t
Lecture Notes of Mechanics of Solids, Chapter 4 6
3
3
81
65
1616
H
S
D
D
×
π
=
π
which gives that:
0751
65
81
3
.
D
D
S
H
== or D
H
= 1.075 D
S
, an increase in size by 7.5%
Comparing the weight ratio:
6420
4
3
2
4
2
2
2
.
D
DD
A
A
V
V
S
HH
S
H
S
H
=
π
−
π
==
which is
a reduction in weight of 35.8 % if the hollowed shaft is used!
4.3 ANGLE OF TWIST (4
th
: 198212;5
th
:198212)
The maximum shear stress is one of major design constraints in relation to strength of shaft.
However, sometime the design may depend on restricting the amount of rotation or twist
when the shaft is subjected to a torque.
Angle of Twist for General Cases
In this section, we will develop a formula for determining the angle of twist φ (phi) of one end
of a shaft with respect to its other end as shown in Fig. 4.8.
From Eq. (4.4), we have
ρ
γ=ϕ
dx
d
x
y
z
J(x)
L
B
C
x
dx
T
1
T
2
T
3
φ
Fig. 4.8
Rotational shaft under general loading conditions
According to Hooke’s law,
Gτ=
γ
, and substituting Eq. (4.15) i.e.
( ) ( )
xJxT ρ
=
τ
, we have
( )
( )
dx
xGJ
xT
d =ϕ
Integrating over the entire length L of the shaft, we obtain the
angle of twist
for the whole
shaft as
(
)
( )
∫
=ϕ
L
o
dx
xGJ
xT
(4.17)
where
Lecture Notes of Mechanics of Solids, Chapter 4 7
φ = the angle of twist of one end with respect to other end, measured in radian
T(x) = the internal torque at arbitrary position x, found from the method of sections and
equation of moment equilibrium
J(x) = the shaft’s polar moment of inertia expressed as a function of position x.
G = shear modulus of elasticity for the material
Single Constant Torque and Uniform CrossSection Area
T
T
L
φ
GJ
Fig. 4.9
Uniform shaft under a constant torque T
Usually in engineering practice, the material is homogeneous and the shaft’s crosssectional
area and applied torque are constant as shown in Fig. 4.9. Eq. (4.17) becomes
GJ
TL
=ϕ
(4.18)
Multiple Torque and CrossSection Areas
If the shaft is subjected to several different torques, or consists of a number of different the
crosssectional areas or shear moduli, Eq. (4.18) can be applied to each segment of the shaft
where these quantities are all constant.
∑
=ϕ
i
ii
ii
JG
LT
(4.19)
Sign Convention of Internal Torque
+T
+T
T
T
+φ
φ
Fig. 4.10
Sign conventions for torque and angle of twist
In order to apply the above equation (Eq. (4.18)), we must develop a sign convention for
internal torque and angle of twist of one end with respect to the other end. To do this, we will
use the
righthand rule
, whereby both the torque and angle will be positive,
provided the
thumb is directed outward from the shaft when the fingers curl to give the tendency for
rotation
, as illustrated in Fig. 4.10.
Lecture Notes of Mechanics of Solids, Chapter 4 8
Example 4.2:
a) Determine the maximum shear stress and rate of twist of the given shaft if a
T = 10kNm torque is applied to it; b) if the length of the shaft is 15 m, how much would it
rotate by? c) if restrict the maximum shear stress level within τ
allow
= 60MPa, what is the
maximum torque that the shaft can transmit? Let G = 81GPa, D = 75 mm.
T=10kNm
L=15m
φ
D=75mm
a) The maximum shear stress:
( )
46
4
4
101063
32
0750
32
m.
.D
J
−
×=
π
=
π
=
( )
(
)
(
)
( )
MPa.
.
/.
J
/DT
J
TR
o
max
7120
101063
2075010102
6
3
=
×
××
===τ
−
The rate of twist:
(
)
( ) ( )
m/rad.
.
GJ
T
dx
d
039740
10106331081
1010
69
3
=
×××
×
==
ϕ
−
which equates to :
(
)
m/.
.
dx
d
o
2772
039740180
=
π
×
=
ϕ
b) If the shaft is 15
m
long, the angle of rotation at the free end is
°=×°=
ϕ
=ϕ
15534152772
..L
dx
d
or directly from Eq. (4.18)
(
)
( ) ( )
( )
π×=°==
×××
××
==ϕ
−
/..rad.
.
GJ
TL
1805960155345960
10106331081
151010
69
3
c) The maximum shear stress must be less than the allowable stress;
allow
o
max
J
TR
τ≤=τ
, i.e.
(
)
(
)
mkN.
.
.
R
J
T
o
allow
max
⋅=
×××
=
τ
≤
−
974
03750
1060101063
66
4.4 STATICALLY INDETERMINATE TORQUELOADED MEMBERS
(4
th
: 213220; 5
th
:213220)
Compound Shafts
A compound shaft is one made from more than one material property. The aim here is to
determine how much of the applied torque is carried by each material.
Equilibrium:
00
21
=−−==
∑
TTTM
x
(4.19)
Compatibility:
21
ϕ
=
ϕ
=
ϕ
(4.20)
+
+
Lecture Notes of Mechanics of Solids, Chapter 4 9
J
1
J
2
T
1
T
2
T
Distribution of
Shear Stress (G1<G2)
Distribution of
Shear Strain
G
1
G
2
Fully
bonded
y
z
Substituting Eq. (4.18) into Eq. (4.20) and equating with Eq. (4.19), we can find T
1
and T
2
,
hence the rate of twist and shear stresses carried by each material.
22
2
11
1
JG
LT
JG
LT
==ϕ
, i.e
2
22
11
1
T
JG
JG
T =
T
JGJG
JG
T
2211
22
2
+
=
;
T
JGJG
JG
T
2211
11
1
+
=
(4.21)
( )
T
JGJG
G
2211
2
2
+
ρ
=τ
;
( )
T
JGJG
G
2211
1
1
+
ρ
=τ
(4.22)
Indeterminate Shafts
L
L
AC
A
B
C
T
L
BC
T
T
B
T
A
R
o
A
C
B
Global Equilibrium (for ground reactions):
00 =−−==
∑
BAx
TTTM
(4.23)
Since only one equilibrium equation is relevant and there are two unknowns, this problem is
statically indeterminate. However, the angle of twist of one end of the shaft with respect to
other end is zero. We can give compatibility condition as
Compatibility:
0==ϕ
∑
i
ii
ii
BA
JG
LT
and note that the internal torque in segment CB is negative by using the righthand rule.
( )
0
=
−
+
JG
LT
JG
LT
BCBACA
(4.24)
From Eqs. (4.22) and Eq. (4.23), we have
T
L
L
T
BC
A
=;
T
L
L
T
AC
B
= (4.25)
Therefore,
T
LJ
RL
BC
max
A
o
=τ
;
T
LJ
RL
AC
max
B
o
=τ
(4.26)
+
+
Lecture Notes of Mechanics of Solids, Chapter 4 10
COMPARISON OF TORSIONAL SHAFTS WITH AXIALLY LOADED BARS/RODS
Axial Loaded Bar/Rod
(Chapters 1, 2&3)
Torsional Shaft
(Chapters 4)
Load Type
Force F (N) Torque T (Nm)
Sign Convention
Tension or Compression
+

+F
+F
F
F
T
e
n
s
i
o
n
C
o
m
p
r
e
s
s
i
o
n
+T
+T
T
T
+φ
φ
+

RightHand Rule
Geometric Property
A – Area (m
2
) J – Polar Moment of Inertia(m
4
)
Material Property
E – Young’s Modulus (Pa) G – Shear Modulus (Pa)
Stress
Normal stress:
A
F
avg
=σ
(Pa)
Uniform
Distribution
Shear stress:
ρ=τ
J
T
(Pa)
T
Distribution of
shear stress
Strain
Normal strain:
L
L
∆
=ε (ms)
Shear strain
γ
 radian
Hooke’s Law
ε=
σ
E
γ
=
τ
G
Deformation
Deflection (m)
(Elongation/Contraction)
General:
(
)
( ) ( )
dx
xAxE
xF
L
∫
=δ
0
Single uniform:
EA
FL
=δ
Multisegments:
∑
=δ
i
ii
ii
AE
LF
Angle of Twist (Radian)
General:
( )
( ) ( )
dx
xJxG
xT
L
∫
=ϕ
0
Single uniform:
GJ
TL
=ϕ
Multisegments:
∑
=ϕ
i
ii
ii
JG
LT
Work
Strain Energy
p
PW
∆
2
1
=
∑
=
i
ii
ii
AE
LF
U
2
2
(for a Truss)
T
TW ϕ=
2
1
∑
=
i
ii
ii
JG
LT
U
2
2
(Multisegments)
WorkStrain
Energy Method
P
U
P
2
=∆
T
U
T
2
=
ϕ
Castigilinao’s
Method
( )
∂
∂
=∆
∑
ii
i
i
i
iP
AE
L
P
F
F
By introducing a virtual force Q:
( )
0=
∂
∂
=
∑
Q
ii
i
i
i
iQ
AE
L
Q
F
F∆
( )
∂
∂
=
∑
ii
i
i
i
iT
JG
L
T
T
T
ϕ
By introducing a virtual torque S:
( )
0=
∂
∂
=
∑
S
ii
i
i
i
iS
AE
L
S
T
Tϕ
Lecture Notes of Mechanics of Solids, Chapter 4
11
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