Bài giảng Strength of materials - Chapter 6: Torsion - Trần Minh Tú

STRENGTH OF MATERIALS
TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
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6
CHAPTER
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TORSION
Contents
6.1. Introduction
6.2. Torsional Loads on Circular Shafts
6.3. Strength Condition and stiffness condition
6.4. Statically Indeterminate Problem
6.5. Strain Energy
6.6. Examples
Home’s works
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6.1. Introduction
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6.1. Introduction
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• Torsional Loads on Circular Shafts: the torsional moment or couple
6.1. Introduction
Torsion members – the slender members subjected to torsional
loading, that is loaded by couple that produce twisting of the member
about its axis
Examples – A torsional moment (torque) is applied to the lug-wrench
shaft, the shaft transmits the torque to the generator, the drive shaft of
an automobile...
y
x
z
F
A
B C Q1
Q2
1
T
1
t 2
T
t
2
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6.1. Introduction
 Internal torsional moment diagram
• Using method of section
zM
> 0
y
z
x
• Sign convention of Mz
- Positive: clockwise
- Negative: counterclockwise
0zM  Mz =
z
y
x
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6.2.1. Simplifying assumptions
6.2. Torsion of Circular Shafts
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6.2. Torsion of Circular Shafts
• Consider the portion of the shaft
shown in the figure
• CD – before deformation
• CD’ – after deformation
- From the geometry 'DD d dz  
=> The Shear strain:
d
dz
 
d
G G
dz
  
- Following Hooke’s law:
6.2.2. Compatibility
- d – the angle of twist
=> In the cross-section, only shear stress exists
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2
z p
A A
d d
M dA G dA G I
dz dz
 
z
p
Md
dz GI
 
z
p
M
I
  
 – radial position
Ip – polar moment of inertia
Mz – internal torsional moment
6.2. Torsion of Circular Shafts
6.2.3. Equilibrium
6.2.3. Torsion formulas
– Shearing stress
 – the rate of twist
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6.2. Torsion of Circular Shafts
max
z z
p p
M M
R
I W
 
- Maximum shearing stress
– Angle of twist
From 6.2.3:
 
0
A L
z z
AB
p pB
M dz M dz
rad
GI GI
L

A B
O
a b
c
- Wp - Section modulus of torsion
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z
AB
p
M L
GI
 z
p
M
const
GI
6.2. Torsion of Circular Shafts
If the shaft is subjected to several different torques or cross-sectional
area, or shear modulus changes abruptly from one region of the shaft to
the next.
– Multiply torques
constz
p i
M
GI
1
n
z
AB i
i p i
M
l
GI

GIp – stiffness of torsional shaft
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6.2. Torsion of Circular Shafts
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6.3. Strength Condition – Stiffness condition
 ax
pW
z
m
M
  
  0
n

 - Determine experimentally 0
 
 
2

 - Third strength hypothesis 3
 
 
3

 - Fourth strength hypothesis 4
 Strength condition:
 Stiffness condition:
  ax
ax
 /zm
p m
M
rad m
GI
 
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6.3. Strength Condition – Stiffness condition
 Three main problems:
 max 30.2D
zM  For a circular shaft:  ax 4 0.1
z
m
M
GD
  
1. Investigating the strength condition, (stiffness condition) 
 max 3 ???0.2D
zM  
2. Determine the diameter of circular cross-section
 
3D
0.2
zM

3. Determine the maximum torque   3M 0.2z D 
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• Assume that the reactions at the fixed
ends MA, MD are shown in the figure.
• Equilibrium: MA + MD = M (1)
• Compatibility condition:
 AD = 0 (2)
d
a 2a
D
M MA D
A
M
B2
d
D
M DM
z
z
CD
2AB BDz z
AD AB BD AB BD
p p
M a M a
GI GI
BD
z DM M 
AB
z DM M M 
4 4
2
0
0,10,1 2
D D
AD
M M a M a
G dG d
1 32
; 
33 33
D AM M M M 
Mz
M/33
32M/33
6.4. Statically Indeterminate Problem
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J
T
xy
 
 dV
GJ
T
dV
G
U
xy
2
222
22
 
• For a shaft subjected to a torsional
load,
• Setting dV = dA dx,
L
L
A
L
A
dx
GJ
T
dxdA
GJ
T
dxdA
GJ
T
U
0
2
0
2
2
2
0
2
22
2
22
• In the case of a uniform shaft,
GJ
LT
U
2
2
6.5. Strain Energy
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6.6. Example Problem 1
• A Circular shaft made from two
segments, each having diameter
of D and 2D. The Shaft is
subjected to the torques shown
in the figure.
1. Draw the internal torsional
moment diagram
2. Determine the maximum
shearing stress
3. Determine the angle of twist of
the end D with respect to B
With M=5kNm; a=1m;
D=10cm; G=8.103 kN/cm2
2a
B
a
C D
D
M 3M
2
D
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1. Internal torsional moment 
diagram
Segment CD 
Segment BC
2a
B
a
C D
D
M 3M
2
D
Mz
kNm
15
10
 10 z a
3 15CDzM M kNm 
2 10BCzM M kNm 
 20 2 z a
Mz
CD
3M
z1
3MM
z2 a
Mz
BC
6.6. Example Problem 1
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6.6. Example Problem 1
2. Maximum shearing stress
3. Angle of twist of end D
2a
B
a
C D
D
M 3M
2
D
Mz
kNm
15
10
2
max 2
3 3
15 10
7,5( / )
0,2 0,2 10CD
CD
zM kN cm
D

2
max 2
3 3
10 10
0,625( / )
0,2 200,2 2
BC
BC
zM kN cm
D

D BC CD 
2CD BCz z
D CD BC
p p
M a M a
GI GI
2 2 2 2
3 4 3 4
15 10 10 10 10 2 10
0,02( )
8 10 0,1 10 8 10 0,1 20
D rad
2
max 7,5( / )kN cm 
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6.6. Example Problem 2
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6.6. Example Problem 2
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6.6. Example Problem 2
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Homework
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Homework
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Homeworks
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THANK YOU FOR 
ATTENTION !
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