Lecture Strength of Materials I: Chapter 6 - PhD. Tran Minh Tu

STRENGTH OF MATERIALS

1/10/2013

TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam

1


CHAPTER

6
TORSION
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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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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...
• Torsional Loads on Circular Shafts: the torsional moment or couple
A

F
B

x

C

Q1
z

y

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Q2

T
1

t
1

T
2

t
2

6


6.1. Introduction
 Internal torsional moment diagram
• Using method of section
• Sign convention of Mz

- Positive: clockwise
- Negative: counterclockwise

M

z

0

Mz > 0

Mz =

y

y
z

z
x

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x

7


6.2. Torsion of Circular Shafts
6.2.1. Simplifying assumptions

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6.2. Torsion of Circular Shafts
=> In the cross-section, only shear stress exists
6.2.2. Compatibility
• 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 – the angle of twist

- Following Hooke’s law:

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   G  G 

d
dz

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6.2. Torsion of Circular Shafts
6.2.3. Equilibrium

d 2
d
M z      dA  G
 dA  G
Ip

dz A
dz
A

d M z


 – the rate of twist
dz GI p
6.2.3. Torsion formulas
– Shearing stress
Mz – internal torsional moment

Mz
 

Ip
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Ip – polar moment of inertia
 – radial position
10


6.2. Torsion of Circular Shafts
- Maximum shearing stress

 max

Mz
Mz

R
Ip
Wp

- Wp - Section modulus of torsion
– Angle of twist
c


a

b



O

A

B
L

From 6.2.3:
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A

L

M z dz
M dz
 z
GI p
GI p
B
0

 AB  

 rad 

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6.2. Torsion of Circular Shafts
Mz
 const
GI p
– Multiply torques

 AB

M zL

GI p

GIp – stiffness of torsional shaft

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.

 Mz 

  const
 GI p i

 AB
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 Mz 

 
li



i 1  GI p 
i
n

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6.2. Torsion of Circular Shafts

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6.3. Strength Condition – Stiffness condition
 Strength condition:

  

0

 max 

Mz
  

Wp

- Determine experimentally 0

n

 





- Third strength hypothesis 3

 





- Fourth strength hypothesis 4

2

3

 Stiffness condition:
 max
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 Mz 

  
 GI 
 p max

 rad / m 
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6.3. Strength Condition – Stiffness condition
 Three main problems:
For a circular shaft:

 max

Mz
Mz





  


max
3
4
0.2D

0.1GD

1. Investigating the strength condition, (stiffness condition)

 max

Mz

   ???
3
0.2D

2. Determine the diameter of circular cross-section

Mz
D 3
0.2 
3. Determine the maximum torque
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M z  0.2  D3
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6.4. Statically Indeterminate Problem

AD = 0
 AD   AB   BD

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AB
z
AB
p

BD
z
BD
p

M a M 2a


GI
GI

32
MA  M
33

MD

M
B
a

d

A

(2)

MD  M a

M D 2a
 AD 

4
4
G  0,1  2d  G  0,1 d

1
MD  M;
33

MA
2d

• Assume that the reactions at the fixed
ends MA, MD are shown in the figure.
• Equilibrium: MA + MD = M
(1)
• Compatibility condition:

D

2a
CD

M
M

BD
z
AB
z

 MD
 MD  M

Mz

MD
D

z

M/33

0

Mz

32M/33
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6.5. Strain Energy
• For a shaft subjected to a torsional

load,
2
2 2
U 

 xy

2G

dV  

T 
2GJ

2

dV

• Setting dV = dA dx,
T 2  2 
U  
dA dx  
 dA dx
2
2

2GJ
2GJ  A

0A

0
L

 xy 

T
J

T 2 2

L

L

T2

dx
2GJ
0

• In the case of a uniform shaft,
T 2L
U
2GJ
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6.6. Example Problem 1

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3M

M

B

C
2a

D

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

2D

•

D
a

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6.6. Example Problem 1

M

CD
z

B

2a

M

D
a

3M

 3M  15kNm
MzCD

Segment BC 0  z2  2a 
BC
z

C

D

1. Internal

2D

torsional moment
diagram
Segment CD 0  z1  a 

3M

M

z1
M

3M

 2M  10kNm
MzBC

z2

a
15

10

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Mz
kNm

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6.6. Example Problem 1
3M

M




max
CD

max
BC



M zBC

0, 2  2 D 

3

B

C
2a

10  102

 0,625(kN / cm2 )
3
0, 2  20
10

  max  7,5(kN / cm2 )

D

M zCD
15  102


 7,5(kN / cm2 )
3

3
0,2 D
0,2 10

2D

2. Maximum shearing stress
D
a

15
Mz
kNm

3. Angle of twist of end D

 D   BC  CD

M zCD  a M zBC  2a
D 

CD
GI p
GI pBC

15  102 102
10 102  2 102
D 

 0,02(rad )

3
4
3
4
8  10  0,110 8 10  0,1 20

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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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