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1. Mechanics of Materials
TORSION
Chapter III

2. Contents
 Introduction
 Torsion of Linearly Elastic Circular Bars
 Stress Distribution in Circular Torsion Bars
 Torsional Deformation of Circular Bars
2

3. Compression Tension (stretched) Shearing Torsion (twisted)
Torsional Loads
3

4. Shaft AB is subjected to equal and opposite torques of magnitude T that twist
one end relative to the other, and, as shown in Fig. d, the torque T acting on a
cross section between A and B is the resultant of distributed shear stresses.
Several names are applied to torsion members, depending on the application:
shaft, torque tube, torsion rod, torsion bar, or simply torsion member.
Introduction
4

5. Introduction
5
𝛿

6. • Net of the internal shearing stresses is an internal
torque, equal and opposite to the applied torque:
• Although the net torque due to the shearing
stresses is known, the distribution of the stresses
is not.
• Unlike the normal stress due to axial loads, the
distribution of shearing stresses due to torsional
loads cannot be assumed uniform.
Net Torque Due to Internal Stresses
6

7. • From observation, the angle of twist of the
shaft is proportional to the applied torque
and to the shaft length.
L
T




• 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.
Shaft Deformations
7
Comparison of deformations in circular
(a) and square (b) shafts.

8. • Consider an interior section of the shaft. As a
torsional load is applied, an element on the
interior cylinder deforms into a rhombus.
• Shear strain is proportional to twist and radius
max
max and 




c
L
c


L
L



 
 or
• It follows that
• Since the ends of the element remain planar,
the shear strain is equal to angle of twist.
Shearing Strain
8

9. Sign convention for resisting torque
• A positive torque, T(x), is a
moment that acts on the cross
section at x in a right-hand-rule
sense about the outer normal to
the cross section. On a cross-
sectional cut at x there will be
equal and opposite torques T(x),
as indicated in Fig. b.
• A positive angle of rotation,
φ(x), is a rotation of the cross
section at x in a right-hand-rule
sense about the x axis, as
illustrated in Fig. c.
9

10. J
c
dA
c
dA
T max
2
max 


 
 


• Recall that the sum of the moments of the elementary
forces exerted on any cross section of the shaft must
be equal to the magnitude T of the torque:
and
max
J
T
J
Tc 

 

• The results are known as the elastic torsion formulas
• Multiplying the previous equation by the shear
modulus,
max


 G
c
G 
max



c

From Hooke’s Law, 
 G

The shearing stress varies linearly with the distance r
from the axis of the shaft.
J : Polar Moment of Inertia
Stresses in Elastic Range
10

11. • Recall that the angle of twist and maximum shearing
strain are related,
L
c
 
max
• In the elastic range, the shearing strain and stress are
related by Hooke’s Law,
JG
Tc
G

 max
max


• Equating the expressions for shearing strain and solving
for the angle of twist,
• 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 i
i
i
i
G
J
L
T

Angle of Twist in Elastic Range
11
JG
TL



12. 12
Example

13. 13
Example

14. Shaft BC is hollow with inner and outer
diameters of 90 mm and 120 mm,
respectively. Shafts AB and CD are solid
and of diameter d. For the loading
shown, determine (a) the minimum and
maximum shearing stress in shaft BC, (b)
the required diameter d of shafts AB and
CD if the allowable shearing stress in
these shafts is 65 MPa.
SOLUTION:
• Cut sections through shafts AB
and BC and perform static
equilibrium analyses to find
torque loadings.
• Given allowable shearing stress
and applied torque, invert the
elastic torsion formula to find the
required diameter.
• Apply elastic torsion formulas to
find minimum and maximum
stress on shaft BC.
Sample Problem 3.1
14

15. SOLUTION:
• Cut sections through shafts AB and BC
and perform static equilibrium
analysis to find torque loadings.
 
CD
AB
AB
x
T
T
T
M








m
kN
6
m
kN
6
0    
m
kN
20
m
kN
14
m
kN
6
0









BC
BC
x
T
T
M
Free-body diagram for section between A and B. Free-body diagram for section between B and C.
Sample Problem 3.1
15

16. • Apply elastic torsion formulas to
find minimum and maximum
stress on shaft BC.
     
 
4
6
4
4
4
1
4
2
m
10
92
.
13
045
.
0
060
.
0
2
2









c
c
IP
  
MPa
2
.
86
m
10
92
.
13
m
060
.
0
m
kN
20
max
4
6
2
2
max





 



P
BC
I
c
T
MPa
7
.
64
mm
60
mm
45
MPa
2
.
86
min
min
2
1
max
min







c
c
• Given allowable shearing stress and applied
torque, invert the elastic torsion formula to
find the required diameter.
m
10
9
.
38
m
kN
6
65
3
3
2
4
2
max







c
c
MPa
c
Tc
I
Tc
P



mm
8
.
77
2 
 c
d
Shearing stress distribution on cross section. Free-body diagram of shaft portion AB.
Sample Problem 3.1
16

17. Shaft AD is attached to a fixed wall at D
and is subjected to the torques shown. A
44-mm hole is drilled in the CD portion of
the shaft. Given that the shaft is made of
steel for which G = 77 GPa, find the angle
of twist at end A.
Sample Problem 3.3
17

18. Sample Problem 3.3
18

19. Two solid steel shafts are connected as shown. Knowing that
G=77 GPa and the allowable shearing stress is 55 MPa,
determine:
a) the largest torque that can be applied to the shaft at A.
b) the corresponding angle of twist at point A.
Sample Problem 3.4
19

20. Sample Problem 3.4
20

21. Sample Problem 3.4
21

22. 22

23. 23