### Chapter 6 - Dr. ZM Nizam

```BFC 20903 (Mechanics of Materials)
Chapter 6: Torsion
Shahrul Niza Mokhatar
[email protected]
Chapter Learning Outcome
1. Defined the relationship between stress and strain
2. Understand torsion theory and their applications
3. Analysed and calculate torsion in solid and hollow
circular bar
4. Analysed and calculate the torsion with endrestraints
5. Analysed and calculate the torsion with combined
bar
BFC 20903 (Mechanics of Materials)
Shahrul Niza Mokhatar ([email protected]
Introduction
• Torsion commonly found in mechanical engineering.
• application, for example machinery structures that has twisting member.
• In civil engineering applications;
secondary beam will
moment at connection
part of the beams.
water retaining in a
channel will produce
moment that are
distributed to the beam
as torsion
planks.
Torsion theory
• When a bar or shaft of circular section is twisted by moment, its called
pure tension & the deformed element are in a state of pure shear.
Few assumptions are taken into
account in torsion analysis,
Using Right Hand Law, torsion vector
can be determined
• The longitudinal axis of the shaft
• Planar cross-sectional parallel with
remains straight.
member axis will remains planar after
subjected to torsion.
• The shaft does not increase or
decrease in length.
• Shear strain, γ is changing linearly
along the bar.
as the cross section rotates.
• Cross sections rotate about the axis of
the member.
Shear stress due to torsion
• Shear stress in circular section, τ (tau). From Hooke’s Law,
τ = TR
J
τ : Shear stress in the shaft if
τmax=maximum shear stress occurs at the
outer surface @ tegasan ricih. (N/m2)
J : Polar moment of inertia of the cross
sectional area @ moment sifat tekun kutub
(m4)
T : Applied torque acting at the cross
section (Nm)
R : Radius of the shaft (m)
Angles of twist
Twisting angle is angle (in radian) produced when a bar is
subjected to torsion.

 = TL
JG
T = applied torque
L = length of member
G = shear modulus of material/ modulus of [email protected] ketegaran
(N/m2)
J = polar moment of inertia
Power transmission
• Circular bars or shafts are commonly used for transmission of
power.
• Power is defined as the work performed per unit of time. The work
transmitted by a rotating shaft equals to the torque applied times
the angle of rotation.
Work = Torque x Angular Displacement
Power = d/dt (Work)
• If torque is not a function of time, then the equation for power
simply becomes:
ω is the angular velocity of the shaft (rad/s) = 2π f
P=Tω
f : frequency (Hz @ hertz) (1 Hz = 1 cycle/s)
T : Applied torque acting at the cross section (Nm)
P : Power (W)(1W = 1 Nm/s)
• Use consistent units for P, T, and ω. Power is commonly specified in
horsepower, HP. Angular velocity is usually given in revolutions per
minute or RPM. It should then be converted to rad/sec. To do this
multiply the value in RPM by 2π and divide by 60.
Example 1
Example 2
• Determine the maximum torque of a hollow circular shaft with inside diameter
of 60mm and an outside diameter of 100mm without exceeding the maximum
shearing stress of 70MPa.
Solution;
• Given;
di = 60mm, do = 100mm, τmax = 70MPa
J 

32
(100
4
 60 )  8 . 55 x 10 mm
4
6
4
 max 
T max r
J
T max 
 max J
r
6

( 70 x 10 )( 8 . 55 x 10
6
)
 11 . 97 kNm
0 . 05
• Remember: τmax=maximum shear stress occurs at the outer surface/radius.
Quiz
A hollow steel shaft has an outside diameter of 150mm and an
inside diameter 100mm. The shaft is subjected to a torque of
35kNm. The modulus of rigidity for the steel is 80GPa.
Determine;
a) the shearing stress at the outside surface of the shaft.
b) the shearing stress at the inside surface of the shaft.
c) the magnitude of the angle of twist in a 2.5m length.
Quiz: Solution
Solution;
J 

(150
4
 100 )  39 . 89 x10 mm
4
6
4
32
• the shearing stress at the outside surface
of the shaft.
 out 
Tr
• the magnitude of the angle of
twist in a 2.5m length.
J

35 x 10 ( 0 . 075 )
39 . 89 x 10
6
Tr
J
3

35 x 10 ( 0 . 05 )
39 . 89 x 10
6
JG
 65 . 81 MPa
• the shearing stress at the inside surface of
the shaft
 in 
TL
 
3
 43 . 9 MPa
3
 
35 x 10 ( 2 . 5 )
9
80 x 10 ( 39 . 89 x 10
6
)
Composite Bars
• Combined bar consists of two or more materials to form a structure. An example
is shown in Figure. Superposition principle is used to solve this problem.
• The concept to solve combined bar are:
(a) Imposed external torsion is equal to total torsion formed in the bar, i.e.,
(b) Twisting angle first material is equal to twisting angle of second material at connection
part, i.e.,
(c) Total twisting angle can be calculated from formula,
Group Exercise
• The composite bars with the different material is subjected to the torque is
shown in figure. Determine the maximum shear stress and the position.
Determine the angle of twist at C.
Solution
J AB 
J BC 
d
4
2

32
d
32
 (100 )
 9 . 82 x10 mm
6
32
4
 ( 50 )
2

32
 0 . 62 x10 mm
6
4
4
• Maximum shear stress;
end of C;
 AB 
T AB rAB
 BC 
T BC rBC
6

J AB
J BC
Angle of twist at the
6 x10 ( 50 )
9 . 82 x10
6
 30 . 55 N / mm
2
 161 . 3 N / mm
2
  
6

4 x10 ( 25 )
0 . 62 x10
6
The maximum shear stress occurs in
the bar of BC.
 TL 
 TL 

 

JG
 JG  AB  JG  BC
TL
6

( 6 x 10 )( 2000 )
6
4
9 . 82 x 10 ( 3 x 10 )
(  4 x 10 )( 300 )
6

6
4
0 . 62 x 10 ( 8 x 10 )
Torsion of non-cylindrical member
• Generally, we deal with axisymmetric bodies and the shear
strain is linear through the entire body. However, non-circular
cross-sections are not axisymmetric causing complex
behaviors, which may cause bulging or warping when the
shaft is twisted.
Condition of bulging of non-circular shaft
Empirical formulas for various shapes
Example 3
The aluminum shaft shown in figure has a cross sectional area
in the shape of an equilateral triangle. Determine the largest
torque, T that can be applied to the end of the shaft if the
allowable shear stress, allow is = 56MPa and the angle of twist
at its end is restricted to  allow = 0.02 rad. Given Gal = 26GPa.
Example 3: Solution
Thin-walled Having Closed Cross
Sections
• Thin walled of noncircular shape are often used to construct
lightweight frameworks which is used in aircraft.
• Due the applied torque, T, shear stress is developed on the front
face of the element. Shear flow in a solid body is the gradient of a
shear stress through the body. Shear flow is the product of the
tube’s thickness and the average shear stress. This value is constant
at all points along the tube’s cross section. As a result, the largest
average shear stress on the cross section occurs where the tube’s
thickness is small.
q   avg t
Thin-walled Having Closed Cross
Sections
•
In non-circular thin walled shafts for closed segments. We assume that the
stress is uniformly distributed across the thickness and that we can assume an
average shear stress. The average shear stress in the body is;
 ave 
T
dF
ds
2 tA m
where,
 ave - average shear stress
t - the thickness of the shaft at the point of interest
Am - mean area enclosed within the boundary of the
centerline of the
shaft thickness.
T - the applied torque
τave
h
t
τave
T
Thin-walled Having Closed Cross
Sections
 h ( ave t ds )
 ave 
  ave t  h ds
T
2 tA m
 2 ave t  d A m
 2 ave t A m
Since, q   avg t we can determine the shear flow throughout the cross section
using the equation;
q 
T
2 Am
• Angle of twist, 
• This angle can be determined by using the energy method. The
angle given in radians, can be expressed as;
 
TL
2
4 Am G

ds
t
• Here the integration must be performed around the entire
boundary of the tube’s cross sectional area.
where,
t - thickness of the interior segment
L - length of the section
G - modulus of rigidity of the section/shear modulus
Example 4
A square aluminum tube has the dimensions as shown in figure.
a) determine the average shear stress in the tube at point A if it is
subjected to a torque of 85Nm.
b) compute the angle of twist due to this loading. Given Gal = 26GPa.
Example 4: Solution
a) The area, Am;
Am  ( 50 )( 50 )  2500 mm
2
50mm
50mm
 avg 
T
2 tA m

85 x10
3
 1 . 7 N / mm
2
2 (10 )( 2500 )
Since t is a constant because of the square tube, the average shear stress
is the same at all points on the cross section.
Example 4: Solution
b) Angle of twist;
 
TL

2
4 Am G
ds
t
  50 mm  

2
3
4
4 ( 2500 ) ( 26 x 10 )   10 mm  
3
 
( 85 x 10 )( 1500 )
 1 . 962 x 10
 3 . 92 x 10
4
3
( 20 )