The document discusses torsion and torsional loads on circular shafts. It defines torsion as twisting of a structural member when loaded by couples that produce rotation about its longitudinal axis. It describes how torsional loads produce shearing stresses on planes perpendicular to the shaft axis and axial shear stresses. Formulas are provided for shear strain, shear stress, angle of twist, and polar moment of inertia in elastic torsion applications. Examples are included to demonstrate calculating shear stresses and sizing shafts based on allowable stresses. Design considerations for transmission shafts transmitting power at specified speeds are also briefly covered.

1. AEROSTRUCTURE : TORSION
TUTORIAL BY AWWAL

2. Contents
• Introduction
• Torsional Loads on Circular Shafts
• Net Torque Due to Internal Stresses
• Axial Shear Components
• Shaft Deformations
• Shearing Strain
• Stresses in Elastic Range
• Angle of Twist
• Normal Stresses
• Torsional Failure Modes
• Sample Problem 1.1
• Statically Indeterminate Shafts
• Sample Problem 1.2
• Design of Transmission Shafts
• Introduction to mechanics of
deformable solids and concept of
strain energy
• •Thin shells subject to internal
pressure
• • Moments of area for thin- and
thick-walled sections
• • Theories of torsion for circular
cylinders (solid and hollow) and
also closed thin-walled sections of
arbitrary shapes
• • Beams subject to symmetric
bending; thick-walled and open
thin-walled sections
• • Combined loadings; two-
dimensional (2D) stress-strain
analysis and failure criteria
Torsion
Aerospace structures
Engineering

3. Introduction
Torsion : twisting of a structural
member,when it is loaded by
couples that produce rotation
about its longitudinal axis

4. Net Torque Due to Internal Stresses
 
 

 dA
dF
T 


• Resultant of internal shearing stresses is an
internal torque, equal and opposite to
applied torque,
• Although net torque due to shearing stresses is
known, the distribution of the stresses is not
• Unlike normal stress due to axial loads, the
distribution of shearing stresses due to torsional
loads cannot be assumed uniform.
• Distribution of shearing stresses is statically
indeterminate – must consider shaft
deformations

5. Axial Shear Components
• Torque applied to shaft produces
shearing stresses on planes
perpendicular to shaft axis.
• Existence of axial shear stresses is
demonstrated by considering a shaft made up
of axial slats.
Slats slide with respect to each other when
equal and opposite torques applied to shaft
ends.
• Moment equilibrium requires existence of
equal shear stresses on planes containing the
shaft axis, i.e., “axial shear stresses”.

6. • Will see that angle of twist of shaft is
proportional to applied torque and to shaft
length.
L
T




Shaft Deformations
• When subjected to torsion, every cross-
section of circular (solid or hollow) shaft
remains plane and undistorted. This is due to
axisymmetry of cross section.
• Cross-sections of noncircular ( hence non-
axisymmetric) shafts are distorted when
subjected to torsion – since no axisymm.

7. Shearing Strain
• Consider interior section of shaft. When
torsional load applied, a rectangular element
on the interior cylinder deforms into
rhombus.
so shear strain proportional to twist and
radius
max
max and 




c
L
c


L
L



 


• Thus,
• So shear strain equals angle between BA and
A
B 

8. Torsion Formulae in elastic range (shear stress, angle
of twist








J
L
GJ
dA
L
G
dA
L
G
dA
T




  

2
• Recall: sum of moments from internal
stress distribution equals internal torque at
the section,
4
2
1 c
J 

 
4
1
4
2
2
1 c
c
J 
 
GJ
TL
J
T

 

 ;
• Thus, elastic torsion formulas are
• Hooke’s law,
max
max 






c
c
G
L
G
G 



So shearing stress also varies linearly with
radial position in the section.
J= Polar moment of inertia

9. Torsion formulae in Elastic Range
• If torsional loading or shaft cross-section
changes (discretely) along length, the angle of
rotation is found as sum of segment rotations


i i
i
i
i
G
J
L
T

constant
if
constant
a
,
so
;
max
T
GJ
T
L
J
Tc
GJ
TL
J
T












10. Comparison of Axial and Torsion formulae.
AE =Axial rigidity
Axial Stiffness
Axial Flexibilty:
GJ =Torsional rigidity
Torsional
Stiffness
Torsional Flexibilty:
L
AE
kA 
1

 A
A k
f
L
GJ
kT 
1

 T
T k
f
Axial displacement


i i
i
i
i
G
J
L
T



i i
i
i
i
E
A
L
P

Torsional displacement
Axial stress
A
P


J
T
 
Torsional stress

11. (a) element in pure shear generated due to applied torque,
(b) stresses acting on inclined plane of a triangular stress
element,
(c) forces acting on the triangular stress element (FBD).
Stressed on Inclined Plane


u
Sign convention for stresses on inclined
plane (Normal stresses tensile positive,
shear stresses producing
counterclockwise rotation positive.)

12. 





 cos
tan
sin
sec o
o
o A
A
A 

Equilibrium normal to plane,
Equilibrium along plane,






 sin
tan
cos
sec o
o
o A
A
A 

(1)
(2)


 2
sin



 2
cos



u
Stressed on Inclined Plane

13. Graph of  and  versus .
Maximum/minimum normal
stress occurs at =+45 or -
45o plane


 2
sin




 


 min
max ;
Maximum/Minimum shear
stress occurs at  = 0o or 90o
plane 


 


 min
max ;


 2
cos


14. Failure of Brittle material
Try on a piece of chalk!
Reason: Brittle materials are weak
in tension and maximum normal
stress (tensile) plane in this case is
45o
Remember: Ductile materials are weak in shear and
brittle materials are weak in tension. Thus, for
ductile material failure occurs on maximum shear
stress plane, and for brittle material failure occurs on
maximum normal (tensile) stress plane.

15. Shaft BC hollow, inner dia. 90 mm, outer
dia. 120mm. Shafts AB and CD solid,
dia. d. For loading shown, find (a) min.
and max. shearing stress in BC, (b)
required dia. d of AB and CD if allowable
shearing stress in them is 65 MPa.
Example 1.1

16. • Cut sections, use equilibrium to find
internal torque.
 
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
Example 1.1

17. • Apply elastic torsion formulae to
find min. and max. stress in BC
  
MPa
2
.
86
m
10
92
.
13
m
060
.
0
m
kN
20
4
6
2
2
max





 
J
c
TBC


MPa
7
.
64
mm
60
mm
45
MPa
2
.
86
min
min
2
1
max
min







c
c
MPa
7
.
64
MPa
2
.
86
min
max




• Given allowable shearing stress and
applied torque, find required dia. of
AB and CD
m
10
9
.
38
m
kN
6
65
3
3
2
4
2
max







c
c
MPa
c
Tc
J
Tc



mm
8
.
77
2 
 c
d
Example 1.1
     
 
4
6
4
4
4
1
4
2
m
10
92
.
13
045
.
0
060
.
0
2
2









c
c
J

18. Example 1.2
Two solid steel shafts
connected by gears. For each
shaft G = 77 GPa and
allowable shearing stress 55
Mpa. Find (a) largest torque T0
that can be applied to end of
AB, (b) corresponding angle
through which end A rotates.

19. • Equilibrium • Kinematic constraint of no slipping
between gears (to relate rotations)
Example 1.2
 
 
0
0
73
.
2
mm
60
0
mm
22
0
T
T
T
F
M
T
F
M
CD
CD
C
B









C
B
C
C
B
C
B
C
C
B
B
r
r
r
r







73
.
2
mm
20
mm
60





20. • Find T0 for max allowable torque
on each shaft – choose smallest
• Find the corresponding angle of twist for
each shaft and the net angular rotation of end
A
Example 1.2
3- 20
 
 
 
 
Nm
8
.
61
m
10
5
.
12
m
10
5
.
12
8
.
2
Pa
10
55
Nm
1
.
74
m
10
5
.
9
m
10
5
.
9
Pa
10
55
0
4
3
2
3
0
6
max
0
4
3
2
3
0
6
max
















T
T
J
c
T
T
T
J
c
T
CD
CD
AB
AB




Nm
8
.
61
0 
T
  
   
  
   
 
o
o
/
o
o
o
9
4
2
/
o
9
4
2
/
2.15
05
.
8
05
.
8
95
.
2
73
.
2
73
.
2
95
.
2
rad
0514
.
0
psi
10
77
m
0125
.
0
m
6
.
0
Nm
8
.
61
73
.
2
2.15
rad
0376
.
0
Pa
10
77
m
0095
.
0
m
6
.
0
Nm
8
.
61

















B
A
B
A
C
B
CD
CD
D
C
AB
AB
B
A
G
J
L
T
G
J
L
T









o
2
.
10

A


21. • Generator applies equal and
opposite torque T’ on shaft.
• Shaft transmits torque to generator
• Turbine exerts torque T on shaft
Design of Transmission Shafts

22. Design of Transmission Shafts
• Transmission shaft performance
specifications are:
- power
- speed
• Determine torque applied to shaft at
specified power and speed,
2
2
P T NT
P P
T
N
 
 
 
 
• Find cross-section so that max
allowable shearing stress not
exceeded,
 
   
shafts
hollow
2
shafts
solid
2
max
4
1
4
2
2
2
max
3
max





T
c
c
c
c
J
T
c
c
J
J
Tc






• Designer must select shaft
material and cross-section to
meet performance specs.
without exceeding allowable
shearing stress.
P= Power (Watt)
T= torque (N-m)
= angular speed
(rad/s)
N= revolution per
sec

23. • Given applied torque, find torque reactions at A
and B.
Statically Indeterminate Shafts
• Equilibrium,
So problem is SID.
N.m
120

 B
A T
T
m
N
120
1
1
2
2
1











J
L
J
L
TA
• Solve equil. and compat,
0
2
2
1
1
2
1 




G
J
L
T
G
J
L
T B
A



• Compatibility,

24. Thin-Walled Hollow Shafts
• Since wall is thin, assume shear stress
constant thru wall thickness. For AB,
summing forces in x(shaft-axis)-direction,
So shear flow constant and shear stress at
section varies inversely with thickness
   
flow
shear
0










q
t
t
t
x
t
x
t
F
B
B
A
A
B
B
A
A
x





   
A
T
q
tA
T
qA
dA
q
dM
T
dA
q
q
ds
r
q
pds
q
ds
t
p
dF
p
dM
2
2
2
2
2
)
(
)
sin
(
0
0


















ds
r
• Compute shaft torque from integral of
the moments due to shear stress


t
ds
G
A
TL
2
4

• Angle of twist (from Chapt 11)
A
T
q
q
t
2
constant; 




25. Example 1.3
Extruded aluminum tubing with
rectangular cross-section has torque
loading 2.7 kNm. Find shearing
stress in each of four walls
considering (a) uniform wall
thickness of 4 mm and (b) wall
thicknesses of 3 mm on AB and AC
and 5 mm on CD and BD.

26. 3 -
26
Example 1.3
Find corresponding shearing stress
for each wall thickness.
With uniform wall thickness,
m
004
.
0
m
N
10
12
.
251 3



t
q

MPa
8
.
62


With variable wall thickness
m
003
.
0
m
N
10
12
.
251 3


 AC
AB 

m
005
.
0
m
N
10
12
.
251 3


 CD
BD 

MPa
7
.
83

 BC
AB 

MPa
2
.
50

 CD
BC 

Find shear flow q.
  
  m
N
10
12
.
251
mm
10
5376
2
Nm
2700
2
mm
5376
mm
56
mm
96
3
2
6
2








A
T
q
A

27. Stress Concentrations
• The derivation of the torsion formula,
assumed a circular shaft with uniform
cross-section loaded through rigid end
plates.
J
Tc

max

J
Tc
K

max

• Experimental or numerically
determined concentration factors are
applied as
• The use of flange couplings, gears and
pulleys attached to shafts by keys in
keyways, and cross-section discontinuities
can cause stress concentrations

28. Torsion of Noncircular Members
• At large values of a/b, the maximum
shear stress and angle of twist for other
open sections are the same as a
rectangular bar.
G
ab
c
TL
ab
c
T
3
2
2
1
max 
 

• For uniform rectangular cross-sections,
• Previous torsion formulas are valid for
axisymmetric or circular shafts
• Planar cross-sections of noncircular
shafts do not remain planar and stress
and strain distribution do not vary
linearly

43. Example 1.4
3 - 43

44. Example 1.4
3 - 44

45. Example 1.5
3 - 45

46. Example 1.5
3 - 46

47. Example 1.6
3 - 47

48. Example 1.6
3 - 48
27 0.53

49. Example 1.6
3 - 49

50. Example 1.7
3 - 50

51. Example 1.7
3 - 51
22.09
(0.5)

52. Example 1.8
3 - 52

53. Example 1.8
3 - 53

54. Example 1.8
3 - 54

55. Example 1.9
3 - 55

56. Example 1.9
3 - 56

57. Example 1.9
3 - 57
rB rC

58. Example 1.9
3 - 58
rB rC
1829.39 868.25
1829.39
868.25
43.13 MPa
48.53

59. Example 1.9
3 - 59
45
45
-894.62
CD
dCD
AB
1884.96 N.m
4121.50
<
CD

60. Example 1.10
3 - 60

61. Example 1.10
3 - 61

62. Example 1.11
3 - 62

63. Example 1.11
3 - 63

64. Example 1.12
3 - 64

65. Example 1.12
3 - 65

66. Example 1.12
3 - 66