CH-4 Plane problems in
linear isotropic elasticity
HUMBERT Laurent
laurent.humbert@epfl.ch
laurent.humbert@ecp.fr
Thursday, march 18th 2010
Thursday, march 25th 2010
1
4.1 Introduction
Framework : linear isotropic elasticity, small strains assumptions, 2D
problems (plane strain , plane stress)
The basic equations of elasticity (appendix I) :
- Equilibrium equations (3 scalar equations) :
div σ  f  0
 ij , j  fi  0
Explicitly,
f : body forces (given)
 : Cauchy (second order) stress tensor
 xx  yx  zx


 fx  0
x
y
z
 yx  yy  yz


 fy  0
x
y
z
 zx  zy  zz


 fz  0
x
y
z
e3
R3
x, x1
e1
σT  σ
 ij   ji
z, x3
symmetric
e2
y, x2
Cartesian basis
2
- Linearized Strain-displacement relations (6 scalar equations) :
ε

1
T
u   u 
2
1   ui  u j
 ij  


2  x j xi




Equations of compatibility:
 2 11  2  22
 2 12

2
2
2
 x2
 x1
x1 x 2
   23 13 12   2 11
,




 x1   x1  x 2  x 3  x 2 x 3
 2  23
 2  22  2 33

2
2
2
 x3
 x2
x 2 x 3
,
   23 13 12   2  22




 x 2   x1  x 2  x 3  x 3 x1
,
   23 13 12   2 33




 x 3   x1  x 2  x 3  x1 x 2
 2 33  2 11
 2 12

2
2
2
 x1
 x3
x 3 x1
because the deformations are defined as partial derivative of the displacements u1 , u 2 , u 3
- Hooke’s law (6 scalar equations) :
Isotropic homogeneous stress-strain relation
 ,
σ   tr εI  2 ε
 ij    kk  ij  2  ij
Inversely,

1 
ε   tr σ I 
σ
E
E

1 
 ij    kk  ij 
 ij
E
E
15 unknowns ui ,
E
Lamé’s constants
  3  2 

with
E 0


2   
1    1 2
 ij , ij
→ well-posed problem
→ find u
4
Young’s modulus for various materials :
5
1D interpretation :
1
Before elongation
3
2
traction
 33  E  33
11   22    33
0 0 0 
σ   0 0 0 
0 0  33 
but
0
0 
   33 E
  33 E
0 
ε    0

0
 33 E 
 0
6
- Navier’s equations:
(λ+μ)u k,ki +μui,jj +fi =0
3 displacement components
ui
taken as unknowns
- Stress compatibility equations of Beltrami - Michell :
ij,kk 
1

mm,ij  fi, j  f j,i 
f n,n ij  0
1 
1 
6 stress components ij considered as unknowns
7
Boundary conditions :
Displacements imposed on Su
uu
t o
t

Surface tractions applied on St
t  σn  t
n outwards unit normal to St
f
u
  St  Su
St  Su  
→ displacement and/or traction boundary conditions to solve the
previous field equations
8
4.2 Conditions of plane strain
Assume that u 3  0
Strain components :
11 
 u1
 x1
 22 
 u2
 x2
1   u1  u 2 
12  


2   x 2  x1 
Thus,
11 12 0 
ε    21  22 0
 0 0 0 
“thick plate”
functions of x1 and x2 only
9
Associated stress components
11 
E
11 1- ν   ν 22 
1  ν 1- 2ν 
E
 22 1- ν   ν11 
22 
1  ν 1- 2ν 
12 
 11  12 0 
σ    21  22 0 
 0 0  33 
E
νE
 21 and also (!), 33 
11   22 

1 ν
1  ν 1- 2ν 
Inverse relations,
1 ν
1  ν  11  ν  22 
E
1 ν
1  ν   22  ν 11 
 22 
E
11 
12 
1 ν
12
E
rewritten as
1
11 
 11  ν *22 
E*
1
 22 
 22  ν *11 
E*
E* 
E

,

*

1  ν2
1 ν
10
- 2D static equilibrium equations :
n
t
 11  12

 f1  0
 x1  x 2
 12  22

 f2  0
 x1  x 2
f  x1 ,x2  : body forces
Surface forces t are also functions of x1 and x2 only
t1  11 n1  12 n 2
t 2  12 n1  22 n 2
n1 , n 2 : components of the unit outwards vector n
11
- Non-zero equation of compatibility (under plane strain assumption)
 2 11  2  22
 2 12

2
2
2
 x2
 x1
x1 x 2
implies for the relationship for the stresses:

 2
2 
1  f1  f 2

 2  2   11  22   
1  ν  x1  x 2
  x1  x 2 

That reduces to by neglecting the body forces,
 2
2 
 2  2   11  22   0
  x1  x 2 
Proof ?
12
Proof:
- Introduce the previous strain expressions in the compatibility equation
11 
1 ν
1  ν  11  ν  22 
E
 22 
12 
1 ν
12
E
1 ν
1  ν  22  ν 11 
E 
one obtains
 212
2
2
1  ν  11  ν 22  
1  ν  22  ν 11   2
 x 22 
 x12 
 x1  x 2
(1)
- Differentiate the equilibrium equation and add

   11  12

 f1   0

x1   x1  x 2


   12   22

 f2   0

x 2   x1  x 2

  211  222
 2 12
2
 

2
 x1  x 2

x
 x 22

1
- Introduce (2) in (1) and simplify
   f1  f 2 




x
  1  x2 
(2)
13
4.3 Conditions of plane stress
Condition : 33  13  23  0
thin plate
From Hooke’s law,
1
 11  ν 22 
E
1
 22    22  ν 11 
E
11 
12 
1 ν
12
E
Similar equations obtained in plane strain : E*  E  *  
and also,
ν
33    11   22  , 13   23  0
E
functions of x1 and x2 only
14
Inverse relations,
11 
E
  ν 22 
2  11
1 ν
 22 
E
  ν11 
2  22
1 ν
E
12 
 21
1 ν
 11  12 0 
σ    21  22 0
 0 0 0 
and the normal out of plane strain ,
33  
ν
 11   22 
1 ν
15

Airy’s stress function  :
introduce the function  as
 2
11 
x 2 2
 2
 22 
x12
then substitute in
 2
12  
x1 x 2
equations of equilibrium
automatically satisfied !
 2
2 
 2  2   11  22   0
  x1  x 2 
 4
 4
 4
2 2

0
leads to
4
2
4
x1
x1 x 2
x 2
4   0
Biharmonic equation
Same differential equation for plane stress and plane strain problems
Find Airy’s function that satisfies the boundary conditions of the elastic problem
16
4.4 Local stress field in a cracked plate :
- Solution 2D derived by Williams (1957)
- Based on the Airy’s stress function
Notch / crack tip
r , : polar coordinate system
Crack when     , notch otherwise
17
Local boundary conditions : θθ  rθ  0 for    α,
r0

Remote boundary conditions 
Find
σ  r,  or σ  x1 ,x2 
stress field
u  r,  or
displacement field
u  x1 ,x2 

Concept of self-similarity of the stress field (appendix II) :
Stress field remains similar to itself when a change in the
intensity (and scale) is imposed
Stress function in the form
  r,θ   r  λ   θ  ,   0
18
Biharmonic equation in cylindrical coordinates:
  2 1  1  2   2
d 2    2 
0
 r 2  r r  r 2 2      d2  r

 


2
d 4 
d

2
2
 4     2    2  2   2    2     0 (1)

 d


d
a
Consider the form of solution      e , a  cst
1  m2    λ  2 
2
 λ2  m  λ2  λ  2  0

2
with m  a 2
Solutions of the quadratic equation :
m1  2
 a1,2    2   i 
m2  (   2 )2
 a3,4       2    i    2 
complex conjugate roots
2
19
Consequently, :
     c1 ei  c2 ei  c3 e
 2 i
 c4 e
  2 i
ci (complex) constants
Using Euler’s formula ei   cos   i sin 
     A cos   Bsin   Ccos    2    Dsin    2  
A, B, C and D constants
to be determined …
according to the symmetry properties of the problem !
20
Modes of fracture :
A crack may be subjected to three modes
More dangerous !
Notch, crack
Example : Compact Tension (CT) specimen :
F
F
natural crack
Mode I loading
F
F
21
Mode I – loading
  r  s   
with s     A cos   Ccos    2  
symmetric part of    
Stress components in cylindrical coordinates
1  1  2 
r r 
 2
r r r 2
 2
  2
r
1  1  2 
r   2

r  r  r
Use of boundary conditions,
  r,  

r   r,  
0

 s     0
 0  ds     0
d

 0
 A cos   C cos    2  


A sin   C    2  sin    2    0
22
Non trivial solution exits for A, C if
cos 
 sin 

cos    2  
0
   2  sin    2  
   2
cos   sin    2     cos    2   sin    0
or
tan 2   
sin 2  
  2 cos 2   
… that determines the unknowns 
For a crack, it only remains sin 2    0

n
2
n integer
infinite number of solutions
23
Relationship between A and C
For each value of n → relationship between the coefficients A and C
→ infinite number of coefficients that are written:
An , Cn
n  ...,  2,  1, 0, 1, 2, ...
From,
 A  sin   C    2  sin    2    0
with   n 2
  
A n
n  
sin  n   Cn
2  2
(crack)
n
  


2

 sin  n  2   0
2
  2

n
  
 sin  n   A n  C n
2
 2
n


2

  0
2

0 or 1
A n  C n
4n
n
24
Airy’s function for the (mode I) problem expressed by:
  r
n

n
2

n
n
 
 A n cos 2   Cn cos  2  2   

 

Reporting Cn  A n
n
4n

nθ
n
n
 
(r,θ)   A n r  n 2 cos 
cos   2  θ 
2 n4
2
 
n

25
Expressions of the stress components in series form (eqs 4.32):
Starting with,

nθ
n
n
 
(r,θ)   A n r  n 2 cos 
cos   2  θ 
2 n4
2
 
n

1  1  2 

and recalling that, r r 
r r r 2 2
n
  n

nθ n  n
 
  A n r 2   sin  sin   2  θ 

2 2 2
 
n
 2
  n 2
1  2
nθ n  n  4 
n
 
 2 n 2 

A
r

cos

cos

2
  
n n

 θ
r 2 2
2
2
2
2
2

 
  
1 
nθ
n
 n 
n
 
  A n r  2n 2    cos 
cos   2  θ 
r r
2 n4
 2 
2
 
n
2


n
n
nθ
n
n
n




 
n
 
 (2 n 2) 
rr   A n r
    2   cos   2  θ 
   1 cos 
2 n 4 2  2
 
2
 
n

 2 2 

26
 2
From,   2
r
θθ   A n
n
n  n   (2 n 2) 
nθ
n
n
 

1
r
cos

cos

2




 θ
2 2 
2 n4
2
 

1  1  2 
and using,  r   2

r  r  r
1 
nθ n  n
 
 2 n 2   n

A
r

sin

sin

2

n

 θ
 2
r 2 
2
2
2

 
n

 2
nθ n  n
 n  n
 
  A n r 1n 2      sin  sin   2  θ 
r n
2 2 2
 2  2
 
1  2
nθ n  n
 n  n
 

  A n r  2n 2     sin  sin   2  θ 
r r n
2 2 2
 2  2
 
 n  n  nθ n  n   n
 
r   A n r  (2 n 2)    1 sin    1 sin   2  θ 
2 2 2   2
 
n
 2 2 
27
Range of n for the physical problem ?
The elastic energy at the crack tip has to be bounded
 Wd     rdr d
ij ij


but, ij
r  (2 n 2)  ...
ij
r  (2 n 2)  ...
 Wd

 (4 n)
r
  ...  r dr d


is integrable if 3  n  0
n  3
or
5
3
  ... ,  3,  ,  2 , 
2
2
28
Singular term when n  3 or   
3A 

3 
5cos

cos
2
2 
4 r 
3A 

3 
 
3cos

cos
2
2 
4 r 
3
2
rr 
r 
3A
4 r
3A  K I
2π
3 
 
sin

sin
 2
2 
Mode - I stress intensity factor (SIF) : K I
rr 
KI
KI
f rr    
2r
2r
 1
3 
5
cos

cos
 4
2 4
2 
 
KI
KI
f    
2r
2r
 1
3 
3
cos

cos
 4
2 4
2 
r 
KI
KI  1
 1
3 
f r    
sin

sin
2 4
2 
2r
2r  4
29
In Cartesian components,
KI


3 
 xx 
cos  1  sin sin 
2
2
2 
2 r
KI


3 
 yy 
cos  1  sin sin 
2
2
2 
2 r
KI


3
 xy 
cos sin cos
2
2
2
2 r
→ does not contain the elastic constants of the material
→ applicable for both plane stress and plane strain problems :
 zz

0


  xx   yy

plane stress

plane strain
,  xz   yz  0
 is Poisson’s ratio
30
Asymptotic Stress field:
  r
y
 rr
 yy
 xy
y
 xx
r
θ
O
x
r
θ
Similarly,
 rr ,  ,  r
O
x
 xx ,  yy ,  xy
n

KI
ij 
fij      An r 2 gij( n )   
2r
n 0
1
singularity at the crack tip + higher–order terms (depending on geometry)
r
fij : dimensionless function of  in the leading term
An amplitude , gij dimensionless function of  for the nth term
31
Evolution of the stress normal to the crack plane in mode I :
 yy 
KI

cos
2
2 r

3 

1

sin
sin


2
2 


Stresses near the crack tip increase in proportion to KI
If KI is known all components of stress, strain and displacement are determined as
functions of r and  (one-parameter field)
32
Singularity dominated zone :
→ Admit the existence of a plastic zone small compared to the length of the crack
33
Expressions for the SIF :
Closed form solutions for the SIF obtained by expressing the biharmonic
function in terms of analytical functions of the complex variable z=x+iy
Westergaard (1939)
Muskhelishvili (1953), ...
Ex : Through-thickness crack in an infinite plate loaded in mode -I:
K I   a
Units of stress
length
MPa m or MN m3 / 2
34
For more complex situations the SIF can be estimated by experiments or numerical
analysis
K I  Y  a
Y: dimensionless function taking into account of geometry
(effect of finite size) , crack shape
→ Stress intensity solutions gathered in handbooks :
Tada H., Paris P.C. and Irwin G.R., « The Stress Analysis of Cracks Handbook »,
2nd Ed., Paris Productions, St. Louis, 1985
→ Obtained usually from finite-element analysis or other numerical methods
35
Examples for common Test Specimens
KI 
P
a
Y 
B W W 
B : specimen thickness
a
Y 
W 
a
lim Y 
a W 0
W
a
2W
a
cos
2W
2 tan

a
 0.752  2.02 
W

a


1
.
12

W

3
a  


  0.37  1  sin
 
2
W


 
 K I  1.12  a ,
a
a
W  0.886  4.64  a
Y 

3/ 2 

W
a


W

 
1



 W
2

P
BW

a

13
.
32

 

W 
3
4
a
a 
14.72    5.60   
W 
 W  
2
36
Mode-I SIFs for elliptical / semi-elliptical cracks
Solutions valid if
Crack small compared to the
plate dimension
a≤c
When a = c ,   0
Circular:
KI
0.63 a
2
  a (closed-form

solution)
Semi-circular:
K I  0.72 a
37
Associated asymptotic mode I displacement field :
Polar components :
ur 
uθ 
K I 1  ν 
2E
K I 1  ν 
with
2E
Cartesian components :
r 
θ
3θ 
2κ

1
cos

cos


2π 
2
2 
ux 
KI
2
r


cos    1 2 sin2 
2
2
2
r 
θ
3θ 

2κ

1
sin

sin


2π 
2
2 
uy 
KI
2
r

sin
2
2
1   

2E
shear modulus

2 


1

2
cos


2

3  
plane stress

  1  
 3  4 plane strain
E: Young modulus : Poisson’s ratio
Displacement near the crack tip varies with
r
Material parameters are present in the solution
38
Ex: Isovalues of the mode-I asymptotic displacement:
2 u x K I
0.7
0.6
plane strain, =0.38
0.2
0.3
0.4
0.1
0
0.1
crack
0.2
0.3
0.4
0.5
2 u y K I
0.6
1
0.7
0.8
0.6
0.4
0.3
x= r cos
0.2
0.1
y=r sin
y=r sin
0.5
crack
0
0.1
0.2
1
0.8
0.6
x= r cos
0.4
0.3
39
Mode II – loading
Same procedure as mode I with the antisymmetric part of    
  r  a   
a     Bsin   Dsin    2  
Asymptotic stress field :
rr 
K II
2r
 3 3 
 5

sin
 sin 
 4
2 4
2
 
K II
2r
 3 3 
 3

sin
 sin 
 4
2 4
2
K II
r 
2r
 3
3 
1
 4 cos 2  4 cos 2 
40
Cartesian components:
 K II


3 
sin  2  cos cos 
2
2
2 
2 r
K II


3

sin cos cos
2
2
2
2 r
 xx 
 yy
 xy 
K II


3 
cos  1  sin sin 
2
2
2 
2 r
Associated displacement field :
ux 
K II
2
r
 

sin    1  2 cos 2 
2
2
2
uy  
K II
2
r

cos
2
2

2 


1

2
sin


2

41
Mode III – loading
Stress components :
rz 
K III
θ
sin
2
2π r
K III
θ
θz 
cos
2
2π r
 0
σ    0
 rz
0
0
 z
 rz 
 z 

0 
Displacement component :
uz 
4K III
E
r 
θ
1

ν
sin
 
2π 
2 
42
Closed form solutions for the SIF
Mode II-loading :
K II   a
Mode III-loading :
K III   a
43
Principe of superposition for the SIFs:
n
With n applied loads in Mode I, K
total
I
  K I(i)
i 1
Similar relations for the other modes of fracture
But SIFs of different modes cannot be added !
Principe of great importance in obtaining SIF of
complicated specimen loading configuration
Example:
p
p
p
(a)
K
(a)
I
K
(b)
(b)
I
p
a
0   p
f  
Q
s
(c)
44
4.5 Relationship between KI and GI:
K I2
Mode I only : GI 
E'
E'  E
E'  E / 1  2 
Plane stress
Plane strain
When all three modes apply :
K I2 K II2 K III2
G


E' E'
2
  E / 2 1   
Self-similar crack growth
Values of G are not additive for the same mode
but can be added for the different modes
45
Proof
 yy  x  
KI a 
2x
U
a 0 a
G I  lim
in load control (ch 3)
Work done by the closing stresses :
a
U 
 dU  x 
1
with dU  x   2 u y  x   yy  x  dx
2
0
but, u y  x  
   1 KI  a  a 
2
and also dU  x  
   1
   1 K I  a  a  a  x
r

2
2
2
K I  a  K I  a  a  a  x
dx
2  2
x
slide 38, with   0
slide 32 for yy  x 
Calculating U and injecting in GI
GI
   1
 lim
a 0
K I  a  K I  a  a  a
4a
2
 GI
  1 K 2I


8
46
4.6 Mixed mode fracture
biaxial loading
e2
 2 0 
 0   in global frame ( e1 e2 )
1

→ expressed in local frame ( e1 e2 )
e1  cos  e1  sin  e2
e2   sin  e1  cos  e2
e2
e1
e1
 e1  e1 e1  e2 
Q   

 e2  e1 e2  e2 
Thus,
Q = Rotation tensor
 cos   sin  
Q

  

 sin  cos  
Stress tensor components :
2
1
K I(0)  1 a
R
0
  11  12 
T  2
      Q   0   Q 
22 
1
 21

47
 2 cos 2 
2 sin  cos  
1 sin  cos  
  11  12   1 sin2 

    sin  cos   sin2 
      
2
2
 2

22 
 21
 1 sin  cos  1 cos  
Mode I loading :

 22
2
K (1)


c
os
 a
I
1
 K I(0) cos 2 
K (2)
  2 sin 2  a
I
 RK I(0) sin 2 
 12
e2
e1
 11
→ Principe of superposition :
K I  K I(0)  cos 2   R sin 2  
Mode II loading :
 K I(0) cos sin
K (1)
II  1cos  sin a
 RK I(0) cosβ sinβ
K (2)
II   σ 2 cosβ sinβ πa
K II  K I(0) cosβ sinβ 1  R 
48
Propagation criteria
Mode I
Crack initiation when the SIF equals to the fracture toughness
K 2IC
K I  K IC or G I  G IC  '
E
Mixed mode loading
Self-similar crack growth is not followed for
several material
2
K
K 2I  K 2II  III  K 2IC
1 
Useful if the specimen is subjected to all
three Modes, but 'dominated' by Mode I
General criteria:
  K I , K II , K IC , K IIC , i ,......  0
explicit form obtained experimentally
49
Examples in Modes I and II
m
n
 KI 
 K II 

  C0 
 1
 K IC 
 K IIC 
m , n and C0 parameters determined experimentally
Erdogan / Shih criterion (1963):
Crack growth occurs on directions normal to the maximum
principal stress


3 
 3

K I sin  sin   K II cos  3cos   0
2
2
2
2


Condition to obtain the crack direction
50
4.7 Fracture toughness testing
Assuming a small plastic zone compared to the specimen dimensions,
a critical value of the mode-I SIF may be an appropriate fracture parameter :
KC
→ plane strain fracture toughness KIC
plane stress
plane strain
KC : critical SIF, depends on thickness
KI > KC : crack propagation
K IC
Specimen Thickness
KIC : Lower limiting value of fracture toughness KC
Material constant for a specific temperature and loading speed
2
K IC
GIC 
E
Apparent fracture surface energy kJ / m 2
51
How to perform KIC measurements ?
→ Use of standards:
- American Society of Testing and Materials (ASTM)
- International Organization of Standardization (ISO)
 ASTM E 399 first standardized test method for KIC :
CT
- was originally published in 1970
- is intended for metallic materials
- has undergone a number of revisions over the years
- gives specimen size requirements to ensure measurements in the plateau region
 ASTM D 5045 -99 is used for plastic materials:
- Many similarities to E 399, with additional specifications important for plastics.
 KI based test method ensures that the specimen fractures under linear elastic
conditions (i.e. confined plastic zone at the crack tip)
52
Chart of fracture toughness KIC and modulus E (from Ashby)
Large range of KIC 0.01->100 MPa.m1/2
At lower end, brittle materials that remain elastic until they fracture
53
Chart of fracture toughness KIC and yield strength Y (from Ashby)
Materials towards the bottom right : high strength and low toughness →fracture before they yield
Materials towards the top left : opposite → yield before they fracture
Metals are both strong and tough !
54
Typical KIC values:
1 MPa m  1 MN m3 / 2
55
Ex Aircraft components
Fuselage made of 2024 alloy (Al + 4% Cu + 1% Mg)
Thickness of the sheet ~ 3mm
Y
K IC
KC
350 MPa (elastic limit)
30 MPa m
AIRBUS A330
100  110 MPa m
Plane stress criterion with Kc is typically used here in place of KIC
56