Transcript Theory of Elasticity

Theory of Elasticity
Chapter 9
Two-Dimensional Solution
Content
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Introduction
Mathematical Preliminaries
Stress and Equilibrium
Displacements and Strains
Material Behavior- Linear Elastic Solids
Formulation and Solution Strategies
Two-Dimensional Problems
Introduction to Finite Element Method
Three-Dimensional Problems
Bending of Thin Plates
Chapter 9
Page 1
Two-Dimensional Solution in Polar Coordinate
• 9.1 Polar Coordinate Formulation(极坐标下的求解)
• 9.2 Coordinate Transformation of Stress
Components （应力分量的坐标变换）
• 9.3 Axisymmetrial stresses and corresponding
displacements（轴对称应力和位移）
• 9.4 Hollow cylinder subjected to uniform
pressures （圆环受均布压力）
• 9.5 Stress concentration of the circular hole（圆
孔的孔口应力集中）
Chapter 9
Page 2
Two-Dimensional Solution in Polar Coordinate
Some symmetry and circular structure
Aeroengine and its rotor system
Practical Engineering need solutions in polar coordinate
Chapter 9
Page 3
Two-Dimensional Solution in Polar Coordinate
Stress Concentration in Practical Engineering
First civil airline Comet: May 2nd,1953 , London-Johnston
Air disaster: Jan.10,1954, and April 8,1954
Chapter 9
Page 4
Two-Dimensional Solution in Polar Coordinate
Stress Concentration
is the main reason
that cause the air
disasters
Chapter 9
Page 5
9.1 Polar Coordinate Formulation
Polar Coordinates
y  r sin 
x  r cos 
Y
r x y
2
2
  tan 1
y
x
r

X
 r   
 sin  


 cos 

x x r x 
r
r 
 r   
 cos 


 sin 

y y r y 
r
r 
Chapter 9
Page 6
9.1 Polar Coordinate Formulation
Polar Coordinates
Y
r
X

 1  1 2 
2
2
1 2 
2
2 1 
  2 sin  cos  2

 cos  2  sin  
 2

2 
r

r


r

r


x 2
r
r


r




 1  1 2 
2
2
1 2 
2
2 1 
  2 sin  cos  2

 sin  2  cos  
 2

2 
y 2
r
r

r
r


r


r

r






 2 1  1 2 
2
2
2

 sin  cos  2 
 2

cos


sin

2 
xy

r
r

r
r





Chapter 9
Page 7
 1  1 2 
 2


r


r

r





9.1 Polar Coordinate Formulation
Basic Equations in Polar Coordinates（极坐标下的基本方程）
Stresses
1  1  2
 rr 
 2
r r r  2
 
Laplace operators
Biharmonic operators
 2
 2
r
 r  
  1  


r  r  
2
2
2 1 
1 2
  2 2  2
 2
x
y
r
r r r  2
2
 2 1 
1 2
4
   2 
 2
r r r  2
 r
2
2


1

1

 4   2 2  
 r 2  r r  r 2  2

Chapter 9
Page 8
2



2


  0

9.1 Polar Coordinate Formulation
Equilibrium Equations(平衡方程)
O

 r 1  r  r   
 kr  0


r
r r 
d
1    r 2 r


 k  0
r 
r
r
rd
y
r
 r
B
r
 r
P
dr
kr
k
C
 x
(r  dr )d
A
 r 
 r
dr
r
r 
 r
dr
r
 r
 


d
 
d r


Chapter 9
Page 9
9.1 Polar Coordinate Formulation
Geometrical Equations几何方程
u r
r 
r
ur 1 u
  
r r 
 r
Chapter 9
1 ur u u



r 
r
r
Page 10
9.1 Polar Coordinate Formulation
Physical Equations物理方程
1  2

r 
( r 
 )
E
1 
1  2

 
(  
r)
E
1 
1
2(1   )
 r   r 
 r
G
E
1
 r  ( r    )
E
1
  (    r )
E
1
2(1   )
 r   r 
 r
G
E
Plain Strain
Plain Stress
E
Chapter 9
E
1  2

Page 11

1 
9.1 Polar Coordinate Formulation
Boundary Conditions(边界条件)
   0   r q0

 r  0  0


     0
 r    0
lr
Chapter 9
Page 12
l
9.2 Coordinate Transformation of Stress
Components
Polar Coordinate  Cartesian Coordinate
x 
 r  

 r 
cos 2   r sin 2
2
2
 r    r 
y 

cos 2   r sin 2
2
2
 
 xy  r
sin 2   r cos 2
2
Cartesian Coordinate  Polar Coordinate
r 
 x  y
 x  y

cos 2   xy sin 2
2
2
 x  y  x  y
 

cos 2   xy sin 2
 2x   y 2
 r 
sin 2   xy cos 2
2
Chapter 9
Page 13
9.3 Axisymmetrial stresses and corresponding
displacements(轴对称应力和位移)
axisymmetric
field quantities are independent of the angular coordinate
√
X
Chapter 9
Page 14
9.3 Axisymmetrial stresses and corresponding
displacements

   (r )
axisymmetric
0

1  1  2
 rr 

r r r 2  2
axisymmetric case
 
 2
 2
r
1 d
r 
r dr

 r  
  1  


r  r  
d 2

dr 2
2
2


1

1

 4   2 2  
 r 2  r r  r 2  2

axisymmetric case
Chapter 9
2
 d
1 d 
 2 
   0
r dr 
 dr
2
Page 15
 r  0
2


  0

9.3 Axisymmetrial stresses and
corresponding displacements
Michell solution of the biharmonic equation
2
  A ln r  Br 2 ln r  Cr 2  D
 d
1 d 
 2 
   0
r dr 
 dr
2
r 
1 d
r dr
 
d 
dr 2
2
 r  0
A
 r  2  B(1  2 ln r )  2C
r A
    2  B(3  2 ln r )  2C
r
 r  r  0
Plane stress case
1
A
ur   (1   )  2(1   ) Br (ln r  1)  (1  3 ) Br
E
r
 2(1   )Cr   I cos  K sin 
4 Br 
u 
 Hr  I sin   K cos 
E
H,I,K associated with the rigid-body motion
Chapter 9
Page 16
9.3 Axisymmetrial stresses and
corresponding displacements
Function of “hole” on distribution
No hole:
A
 r  2  B(1  2 ln r )  2C
r A
    2  B(3  2 ln r )  2C
r
 r  r  0
A and B vanish.
Otherwise when r=0, stress
component become infinite
A plate without a hole with no
body forces (axissymmetrical)
 r     const.
If there is hole at the origin, we will investigate it next
Chapter 9
Page 17
9.4 Hollow cylinder subjected to uniform
pressures
plane stress conditions
Axisymmetric problem
A
 B(1  2 ln r )  2C
2
r A
    2  B(3  2 ln r )  2C
r
r 
 r  r  0
Boundary Conditions
 r
r a
0
 r r a  qa
Chapter 9
Page 18
 r
r b
0
 r r b  qb
9.4 Hollow cylinder subjected to uniform
pressures
A
 B(1  2 ln a)  2C  qa
2
a
A
 B(1  2 ln b)  2C  qb
2
b
3 unknowns, 2 Equations ?
Single or multiply connected region?
for multiply connected regions, the compatibility equations are not
sufficient to guarantee single-valued displacements.
ur 
1
A

(
1


)
 2(1   ) Br (ln r  1)  (1  3 ) Br
E 
r
 2(1   )Cr   I cos  K sin 
4 Br 
u 
 Hr  I sin   K cos 
Page 19
E9
Chapter
B=0
9.4 Hollow cylinder subjected to uniform
pressures
A
 B(1  2 ln a)  2C  qa
2
a
A
 B(1  2 ln b)  2C  qb
2
b
a 2b 2
A 2
(qb  qa ),
2
b a
2
2
b
a
1
1 2
2
r q
r   r2
qa 
b
b
a2
1
1 2
2
a
b
Chapter 9
B=0
(qa a 2  qbb 2 )
2C 
b2  a 2
b2
a2
1
1 2
2
r q
  r 2
qa 
b
2
b
a
1
1 2
2
a
b
Page 20
9.4 Hollow cylinder subjected to uniform
pressures
b2
a2
1
1 2
2
r q
r   r2
qa 
b
b
a2
1
1 2
2
a
b
b2
a2
1
1 2
2
r q
  r 2
qa 
b
b
a2
1
1 2
2
a
b
qb  0(qa  0), b  
2
a
 r   2 qa
r
a2
   2 qa
r
Demonstration of Saint-Venant’s Principle
Chapter 9
Page 21
9.5 Stress concentration of the circular hole
Review:
b2
a2
1
1 2
2
r q
r   r2
qa 
b
2
b
a
1
1 2
2
a
b
b2
a2
1
1 2
2
r q
  r 2
qa 
b
2
b
a
1
1 2
2
a
b
Chapter 9
Page 22
9.5 Stress concentration of the circular hole
What is stress concentration?
The stress concentration near a hole is a critical issue concerning
the strength of a solid structure. The stress concentration can be
measured by the stress concentration coefficients that are the
ratios between the most severe stress at the critical point (or termed
hot spot) and the remote stress.
 max
K














Chapter 9
Page 23

9.5 Stress concentration of the circular hole
Examples:
Chapter 9
Page 24
9.5 Stress concentration of the circular hole
Solution:
Selection of coordinate
To analyse stress concentration
near the hole, it is convenient to
use polar coordinate.
Problem in polar coordinate
Boundary conditions in polar coordinate:
A
 x q r 

A
 r  r
Chapter 9
 x  y
2
 r  

 x  y
 x  y
2
2
cos 2   xy sin 2 
q q
 cos 2
2 2
q
sin 2   xy cos 2   sin 2
2
Page 25
9.5 Stress concentration of the circular hole
Problem in polar coordinate

b
 r r a  0
 r
 r
r
0
q q
 r r b   cos 2
2 2
q
 r r b   sin 2
2
Chapter 9
Page 26
r a
9.5 Stress concentration of the circular hole
Problem 1
Problem in polar coordinate
a

b

b
 r r a
 r

 0 r
r
＝
Problem 2
a
0
q q
 r r b   cos 2
2 2
q
 r r b   sin 2
2
r a
Chapter 9
＋
q
r 
2

b
Page 27
q
 r   sin 2
2
q
 r  cos 2
2
9.5 Stress concentration of the circular hole
Solution of Problem 1
B.C.
q
2
0
 r r a  0
 r r b 
 r
 r
r a
0
a2
1 2
r q
r 
a2 2
1 2
b
r b
a2
1 2
r q
 
a2 2
1 2
b
a

b
 r  0
q
r 
2
when b>>a
 a
 r  1  2
 r
2
Chapter 9
q

2
 a2  q
   1  2 
 r 2
Page 28
 r  0
9.5 Stress concentration of the circular hole
Problem 2
Solution of Problem 2
由边界条件可假设：
某一函数乘以cos2θ
σr为 r
；τr
θ
的
a
为r

的某一函数乘sin2θ。
1  1  
 r
r 
 2
2
r r r 
2
Assume：
  1  
 

r  r  
b
  f (r ) cos 2
 
1 
1 



2
2
 r 2
r

r
r



2
Chapter 9
2
Page 29
2


  0

q
 r   sin 2
2
q
 r  cos 2
2
9.5 Stress concentration of the circular hole
 
1 
1 



2
2
 r 2
r

r
r



2
2
2


  0

 d 4 f (r ) 2 d 3 f (r ) 9 d 2 f (r ) 9 df (r ) 

 cos 2  0

 2
 3
4
3
2
r dr
r dr
r dr 
 dr
d 4 f (r ) 2 d 3 f (r ) 9 d 2 f (r ) 9 df (r )

 2
 3
0
4
3
2
dr
r dr
r dr
r dr
1
  f (r ) cos 2
f (r )  Ar  Br  C  D 2
r
1
 4
2
   Ar  Br  C  D 2  cos 2
r 

Page
4
Chapter 9
2
30
9.5 Stress concentration of the circular hole


   Ar 4  Br 2  C  D
1
cos 2
2 
r 
1  1  2
4C 6 D
r 
 2 2  (2 B  2  4 ) cos 2
r r r 
r
r
2

6D
   2  (12 Ar 2  2 B  4 ) cos 2
r
r
  1  
2C 6 D
 r   
  (6 Ar 2  2 B  2  4 ) sin 2
r  r  
r
r
q
a2
3a 2
 r  (1  2 )(1  2 ) cos 2
2
r
r
4
q
a 
    1  3 4  cos 2
2
r 
q
a2
3a 2
 r   r   (1  2 )(1  2 ) sin 2
2
r
r
Chapter 9
Page 31
a

b
B.C.
q
 r   sin 2
2
q
 r  cos 2
2
q
B


,
A  0,
4 4
qa
C  qa 2 , D  
4
9.5 Stress concentration of the circular hole
Superposition of Solution 1and 2
q
a2 q
a2
3a 2
 r  (1  2 )  (1  2 )(1  2 ) cos 2
2
r
2
r
r
q  a2  q 
a4 
   1  2   1  3 4  cos 2
2 r  2
r 
q
a2
3a 2
 r   r   (1  2 )(1  2 ) sin 2
2
r
r
Chapter 9
Page 32
G. Kirsch Solution
9.5 Stress concentration of the circular hole
q1
q2
q1

q1

q2
x
y
x
q2

y
q2
Chapter 9
q1
Page 33
y
x
9.5 Stress concentration of the circular hole
Stress concentration of ellipse hole
q
q
2a
x
2b
y
(  ) max
Chapter 9
2a
 q(1  )
b
Page 34
Homework
•
•
•
•
4-8
4-13
4-15
4-16
Chapter 9
Page 35
期中考试
• 2005年11月23日，下午6：00－8：00
• 地点：（三）218 （一班、二班、三班前15号）
•
（三）202 (三班16号以后，四班，七班)
Chapter
Page