Transcript CH_4
CH-4 Plane problems in
linear isotropic elasticity
HUMBERT Laurent
[email protected]
[email protected]
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
uu
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 ν11
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
212
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
211 222
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
ν11
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,
r0
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 d2 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 ei c2 ei c3 e
2 i
c4 e
2 i
ci (complex) constants
Using Euler’s formula ei 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 ds 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
4n
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
4n
nθ
n
n
(r,θ) A n r n 2 cos
cos 2 θ
2 n4
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 n4
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 2n 2 cos
cos 2 θ
r r
2 n4
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 n4
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 1n 2 sin sin 2 θ
r n
2 2 2
2 2
1 2
nθ n n
n n
A n r 2n 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
2r
2r
1
3
5
cos
cos
4
2 4
2
KI
KI
f
2r
2r
1
3
3
cos
cos
4
2 4
2
r
KI
KI 1
1
3
f r
sin
sin
2 4
2
2r
2r 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 )
2r
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 m3 / 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
2r
3 3
5
sin
sin
4
2 4
2
K II
2r
3 3
3
sin
sin
4
2 4
2
K II
r
2r
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
2x
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
4a
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 m3 / 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