1 ch-4 plane problems in linear isotropic elasticity humbert laurent thursday, march 18th 2010...
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CH-4 Plane problems in linear isotropic elasticity
HUMBERT Laurent
Thursday, march 18th 2010
Thursday, march 25th 2010
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The basic equations of elasticity (appendix I) :
div 0 σ f
, f 0ij j i
f : body forces (given)
- Equilibrium equations (3 scalar equations) :
: Cauchy (second order) stress tensor
ij ji
T σ σ
symmetric
Framework : linear isotropic elasticity, small strains assumptions, 2D problems (plane strain , plane stress)
4.1 Introduction
f 0
f 0
f 0
yxxx zxx
yx yy yzy
zyzx zzz
x y z
x y z
x y z
Explicitly, 3R
2e1e
3e
x, x1
y, x2
z, x3
Cartesian basis
- Linearized Strain-displacement relations (6 scalar equations) :
T1
2 ε u u
uu1
2ji
ijj ix x
2 2 2 223 1311 22 12 12 11
2 22 1 1 2 1 1 2 3 2 3
2 22 233 23 23 1322 12 22
2 23 2 2 3 2 1 2 3 3 1
2 2 233 2311 122 2
1 3 3 1 3
2 ,x x x x x x x x x x
2 ,x x x x x x x x x x
2 ,x x x x x x
213 3312
1 2 3 1 2x x x x
Equations of compatibility:
because the deformations are defined as partial derivative of the displacements 1 2 3u ,u ,u
44
15 unknowns , ,i ij iju
→ find u
tr 2 σ εI ε
Inversely,
1tr
E E
ε σ I σ
- Hooke’s law (6 scalar equations) :
→ well-posed problem
2ij kk ij ij
, Lamé’s constants
Isotropic homogeneous stress-strain relation
1ij kk ij ijE E
3 2E
2
0E 1 1 2 with
5
Young’s modulus for various materials :
6
1
2
11 22 33 33 33E
3
Before elongation
33
0 0 0
0 0 0
0 0
σ but 33
33
33
0 0
0 0
0 0
E
E
E
ε
1D interpretation :
traction
7
k,ki i,jj i(λ+μ)u +μu +f =0
3 displacement components taken as unknowns
ij
- Navier’s equations:
6 stress components considered as unknowns
iu
ij,kk mm,ij i, j j,i n,n ij
1f f f 0
1 1
- Stress compatibility equations of Beltrami - Michell :
8
Boundary conditions :
t
f
u
u uDisplacements imposed on Su
t σ tn
Surface tractions applied on St
S S t u
S S t u
t o
n outwards unit normal to St
→ displacement and/or traction boundary conditions to solve the previous field equations
9
4.2 Conditions of plane strain
111
1
u
x
Strain components :
222
2
u
x
1 212
2 1
u u1
2 x x
Thus, 11 12
21 22
0
0
0 0 0
ε
Assume that 3u 0
“thick plate”
functions of x1 and x2 only
10
11 11 22
E1- ν ν
1 ν 1- 2ν
Associated stress components
22 22 11
E1- ν ν
1 ν 1- 2ν
12 21
E
1 ν
33 11 22
νE
1 ν 1- 2ν
and also (!),
11 12
21 22
33
0
0
0 0
σ
Inverse relations,
11 11 22
1 ν1 ν ν
E
22 22 11
1 ν1 ν ν
E
12 12
1 ν
E
11 11 22
1ν *
E *
22 22 11
1ν *
E *
2
EE* , *
1 ν 1 ν
rewritten as
11
- 2D static equilibrium equations :
11 121
1 2
f 0x x
12 222
1 2
f 0x x
1 11 1 12 2t n n Surface forces t are also functions of x1 and x2 only
tn
: components of the unit outwards vector n
2 12 1 22 2t n n
1 2n , n
1 2x ,xf : body forces
12
- Non-zero equation of compatibility (under plane strain assumption)
2 2 211 22 122 2
2 1 1 2
2x x x x
implies for the relationship for the stresses:
2 21 2
11 222 21 2 1 2
f f1
x x 1 ν x x
2 2
11 222 21 2
0x x
That reduces to by neglecting the body forces,
Proof ?
13
Proof:
- Introduce the previous strain expressions in the compatibility equation
11 11 22
1 ν1 ν ν
E
22 22 11
1 ν1 ν ν
E
12 12
1 ν
E
- Differentiate the equilibrium equation and add
11 121
1 1 2
f 0x x x
12 222
2 1 2
f 0x x x
22 2
1211 22 22 112 2
2 1 1 2
1 ν ν 1 ν ν 2x x x x
one obtains
(1)
2 2 212 11 22 1 2
2 21 2 1 2 1 2
f f2
x x x x x x
(2)
- Introduce (2) in (1) and simplify
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4.3 Conditions of plane stress
thin plate
functions of x1 and x2 only
Condition : 33 13 23 0
11 11 22
1ν
E
22 22 11
1ν
E
From Hooke’s law,
12 12
1 ν
E
and also,
33 11 22 13 23
ν, 0
E
Similar equations obtained in plane strain : E * E *
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11 12
21 22
0
0
0 0 0
σ
Inverse relations,
11 11 222
Eν
1 ν
22 22 112
Eν
1 ν
12 21
E
1 ν
and the normal out of plane strain ,
33 11 22
ν
1 ν
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Airy’s stress function :
2
121 2x x
2
22 21x
2
11 22x
Biharmonic equation4 0
2 2
11 222 21 2
0x x
then substitute in
equations of equilibrium automatically satisfied !
4 4 4
4 2 2 41 1 2 2
2 0x x x x
leads to
introduce the function as
Same differential equation for plane stress and plane strain problems
Find Airy’s function that satisfies the boundary conditions of the elastic problem
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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
Crack when , notch otherwise
: polar coordinate system,r
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θθ rθ 0 for α, r 0 Local boundary conditions :
Find 1 2r , or x ,xσ σ
1 2r , or x ,xu u displacement field
stress field
Stress function in the form
λr,θ r θ , 0
Remote boundary conditions
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
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2 2 2
222 2 2 2
d1 1r 0
r r r r d
Biharmonic equation in cylindrical coordinates:
4 2
2 22 24 2
d d2 2 0 (1)
d d
Consider the form of solution ae , a cst
2 22 2 2λ 2 λ λ 2 01 λm m 2m awith
Solutions of the quadratic equation :
21m
222m ( )
2 i1,2a
2
3,4 2 2a i
complex conjugate roots
20
Using Euler’s formula ie cos i sin
Acos Bsin Ccos 2 Dsin 2
A, B, C and D constants
to be determined …
2 i 2 ii ie e e e1 2 3 4c c c c
Consequently, :
according to the symmetry properties of the problem !
ci (complex) constants
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Notch, crack
Modes of fracture :
Example : Compact Tension (CT) specimen :
natural crack
F
F
F
F
Mode I loading
More dangerous !
A crack may be subjected to three modes
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Mode I – loading
s Acos Ccos 2
2
r r 2 2
1 1
r r r
2
2r
Stress components in cylindrical coordinates
2
r 2
1 1
r r r
Use of boundary conditions,
r, 0
r r, 0
s 0
sd0
d
Acos C cos 2 0
A sin C 2 sin 2 0
symmetric part of
sr
with
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Non trivial solution exits for A, C if
cos cos 20
sin 2 sin 2
… that determines the unknowns
For a crack, it only remains sin 2 0
n
2 n integer
2 cos sin 2 cos 2 sin 0
2
sin 2tan 2
2 cos
or
infinite number of solutions
24n n
4 nA C
n
Relationship between A and C
For each value of n → relationship between the coefficients A and C
A C 2 1 0 1 2n n, n ..., , , , , , ...
→ infinite number of coefficients that are written:
A sin C 2 sin 2 0 From,
n 2 (crack)
with
n n
n nA sin n C 2 sin n 2 0
2 2 2 2
n n
0 or 1
n nsin n A C 2 0
2 2 2
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Airy’s function for the (mode I) problem expressed by:
n 2n
n
nθ n n(r,θ) A r cos cos 2 θ
2 n 4 2
n
2n n
n
n nr A cos C cos 2
2 2
Reporting n n
nC A
4 n
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Expressions of the stress components in series form (eqs 4.32):
n
2n
n
222 n 2
n2 2n
2 n 2n
n
n nθ n nA r sin sin 2 θ
2 2 2 2
n 41 n nθ n nA r cos cos 2 θ
r 2 2 2 2 2
1 n nθ n nA r cos cos 2 θ
r r 2 2 n 4 2
2
r r 2 2
1 1
r r r
and recalling that,
n 2n
n
nθ n n(r,θ) A r cos cos 2 θ
2 n 4 2
Starting with,
2(2 n 2)
rr nn
n n nθ n n n nA r 1 cos 2 cos 2 θ
2 2 2 n 4 2 2 2
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From,2
2r
and using,2
r 2
1 1
r r r
2 n 2n2
n
21 n 2
nn
22 n 2
nn
1 n nθ n nA r sin sin 2 θ
r 2 2 2 2
n n nθ n nA r sin sin 2 θ
r 2 2 2 2 2
1 n n nθ n nA r sin sin 2 θ
r r 2 2 2 2 2
(2 n 2)θθ n
n
n n nθ n nA 1 r cos cos 2 θ
2 2 2 n 4 2
(2 n 2)r n
n
n n nθ n n nA r 1 sin 1 sin 2 θ
2 2 2 2 2 2
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The elastic energy at the crack tip has to be bounded
5... , 3, , 2 ,
2
3
2
or
Range of n for the physical problem ?
ij ijWd rdr d
(4 n)Wd r r dr d...
(2 n 2)ij
(2 n 2)ij
r ...
r ...
but,
3 n 0 is integrable if
n 3
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Singular term when n 3 or3
2
rr
3A 35cos cos
2 24 r
3A 3
3cos cos2 24 r
r
3A 3sin sin
2 24 r
I3A K 2π
Mode - I stress intensity factor (SIF) : IK
I Irr rr
K K 5 1 3
4 2 4 22 r 2 rf cos cos
I IK K 3 1 3
4 2 4 22 r 2 rf cos cos
I Ir r
K K 1 1 3
4 2 4 22 r 2 rf sin sin
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3cos 1 sin sin
2 2 22
3cos 1 sin sin
2 2 22
3cos sin cos
2 2 22
Ixx
Iyy
Ixy
K
r
K
rK
r
0
, 0zz xz yzxx yy
plane stress
plane strain
isPoisson’s ratio
In Cartesian components,
→ applicable for both plane stress and plane strain problems :
→ does not contain the elastic constants of the material
31
Asymptotic Stress field:
2
02
n( n )I
ij ij n ijn
Kf A r g
r
fij : dimensionless function of in the leading term
1
rsingularity at the crack tip + higher–order terms (depending on geometry)
An amplitude , gij dimensionless function of for the nth term
rrr
x
y
Oθ
r
, ,rr r
yy
xy
xx
x
y
Oθ
r
, ,xx yy xy
Similarly,
32
Evolution of the stress normal to the crack plane in mode I :
3cos 1 sin sin
2 2 22I
yyK
r
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)
33
Singularity dominated zone :
→ Admit the existence of a plastic zone small compared to the length of the crack
34
IK a
Units of stress length 3 2/MPa m M mo Nr
Ex : Through-thickness crack in an infinite plate loaded in mode -I:
Closed form solutions for the SIF obtained by expressing the biharmonic function in terms of analytical functions of the complex variable z=x+iy
Expressions for the SIF :
Westergaard (1939) Muskhelishvili (1953), ...
35
For more complex situations the SIF can be estimated by experiments or numerical analysis
IK Y a
Y: dimensionless function taking into account of geometry (effect of finite size) , crack shape
→ Stress intensity solutions gathered in handbooks :
P
→ Obtained usually from finite-element analysis or other numerical methods
Tada H., Paris P.C. and Irwin G.R., « The Stress Analysis of Cracks Handbook », 2nd Ed., Paris Productions, St. Louis, 1985
36
Examples for common Test Specimens
I
P aK Y
WB W
322 0 752 2 02 0 37 1
22
atan
a aW . . . sina W Wcos
W
aY
W
2
3 2
3 4
20 886 4 64 13 32
1
14 72 5 60
/
aa aW . . .
W WaW
a a. .
W W
aY
W
01 12
a W
a alim Y .
W W
B : specimen thickness
1 12I .P
K ,BW
a
37
Mode-I SIFs for elliptical / semi-elliptical cracks
Crack small compared to the plate dimension
a ≤ c
Solutions valid if
0 63IK . a
Circular:
Semi-circular:
0 72IK . a
When a = c 0,
2a
(closed-form solution)
38
Associated asymptotic mode I displacement field :
E: Young modulus : Poisson’s ratio
3plane stress
13 4 plane strain
1
2E
with shear modulus
Displacement near the crack tip varies with r
Material parameters are present in the solution
Ir
K 1 ν r θ 3θu 2κ 1 cos cos
2E 2π 2 2
Iθ
K 1 ν r θ 3θu 2κ 1 sin sin
2E 2π 2 2
Polar components :
2
2
1 22 2 2 2
1 22 2 2 2
Ix
Iy
K ru cos sin
K ru sin cos
Cartesian components :
39
Ex: Isovalues of the mode-I asymptotic displacement:
plane strain, =0.38
00.1
0.20.3
0.4
0.5
0.6
0.7
0.7
0.6
0.5
0.40.3
0.20.1
2 x Iu K
x= r cos
y=r
sin
crack
0
0.1
0.2
0.30.4
0.60.81
0.1
0.2
0.30.40.6
0.81
2 y Iu K
crack
x= r cos
y=r
sin
40
Mode II – loading
a Bsin Dsin 2
Same procedure as mode I with the antisymmetric part of
ar
Asymptotic stress field :
IIrr
K 5 3 3sin sin
4 2 4 22 r
IIK 3 3 3sin sin
4 2 4 22 r
IIr
K 1 3 3cos cos
4 2 4 22 r
41
3sin 2 cos cos
2 2 22
3sin cos cos
2 2 22
3cos 1 sin sin
2 2 22
IIxx
IIyy
IIxy
K
rK
r
K
r
Cartesian components:
2
2
1 22 2 2 2
1 22 2 2 2
IIx
IIy
K ru sin cos
K ru cos sin
Associated displacement field :
42
Mode III – loading
IIIrz
K θsin
22π r
IIIθz
K θcos
22π r
0 0
0 0
0
rz
z
rz z
σ
Stress components :
Displacement component :
IIIz
4K r θu 1 ν sin
E 2π 2
43
IIK a
Mode II-loading :
Closed form solutions for the SIF
Mode III-loading :
IIIK a
44
Principe of superposition for the SIFs:
1
ntotal (i)I I
i
K K
With n applied loads in Mode I,
Similar relations for the other modes of fracture
Principe of great importance in obtaining SIF of complicated specimen loading configuration
But SIFs of different modes cannot be added !
0( a ) ( b )
I I s
aK K p f
Q
p
p
p
pExample:
(a) (b) (c)
45
Values of G are not additive for the same mode but can be added for the different modes
2 2 2
2I II IIIK K K
GE' E'
Mode I only :2I
I
KG
E'
21E' E / Plane strain
E' E Plane stress
When all three modes apply :
2 1E /
Self-similar crack growth
4.5 Relationship between KI and GI:
46
Proof
Iyy
K ax
2 x
I a 0
UG lim
a
in load control (ch 3)
Work done by the closing stresses :
y yy
1dU x 2 u x x dx
2 with
I Iy
1 K a a 1 K a ar a xu x
2 2 2 2
but,
UCalculating and injecting in GI
I I1 K a K a a a xdU x dx
2 2 x
and also
slide 38, with 0
a
0
U dU x
I II
a 0
1 K a K a a aG lim
4 a 2
2I
I
1 KG
8
slide 32 for yy x
47
biaxial loading
2
1
R
in global frame ( ) 2
1
0
0
I(0) 1K a
1e
2e
1e
2e
1 1
2
1
2
2
1 2
e e ee
ee e e
Q Q = Rotation tensor
→ expressed in local frame ( )1e
2e
1 1 2
2 1 2
e cos e sin e
e sin e cos e
cos sin
sin cos
QThus,
11 12 2
21 22 1
0
0T
Q Q
Stress tensor components :
1e
2e4.6 Mixed mode fracture
48
Mode I loading :
21
(1)I cK aos 2
I(0)K cos
22
(2)I sK ain 2
I(0)RK sin
2 2I I(0)K K cos R sin
→ Principe of superposition :
Mode II loading :
1(1)II cos siK an
II I(0)K K cosβ sinβ 1 R
2(2)II σ cosβ siK πanβ I(0)RK cosβ sinβ
I(0)K cos sin
11 12
21 22
21 1
21 1
sin sin cos
sin cos cos
22 2
22 2
cos sin cos
sin cos sin
22
11
12
1e2e
49
Crack initiation when the SIF equals to the fracture toughness
Mode I
I ICK K2IC
I IC '
KG G
E or
Mixed mode loading
Self-similar crack growth is not followed for several material
22 2 2IIII II IC
KK K K
1
Useful if the specimen is subjected to all three Modes, but 'dominated' by Mode I
I II IC IIC iK ,K ,K ,K , ,...... 0
General criteria:
explicit form obtained experimentally
Propagation criteria
50
Examples in Modes I and II
m n
I II0
IC IIC
K KC 1
K K
m , n and C0 parameters determined experimentally
Crack growth occurs on directions normal to the maximum principal stress
Erdogan / Shih criterion (1963):
I II
3 3K sin sin K cos 3cos 0
2 2 2 2
Condition to obtain the crack direction
51
Assuming a small plastic zone compared to the specimen dimensions,
a critical value of the mode-I SIF may be an appropriate fracture parameter :
plane strain KC : critical SIF, depends on thicknessplane stress
ICK
→ plane strain fracture toughness KIC
KI > KC : crack propagation
Specimen Thickness
KC
KIC : Lower limiting value of fracture toughness KC
Material constant for a specific temperature and loading speed
4.7 Fracture toughness testing
2IC
IC
KG
E Apparent fracture surface energy 2kJ / m
52 KI based test method ensures that the specimen fractures under linear elastic conditions (i.e. confined plastic zone at the crack tip)
- Many similarities to E 399, with additional specifications important for plastics.
ASTM E 399 first standardized test method for KIC :
- 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:
How to perform KIC measurements ?
→ Use of standards:
- American Society of Testing and Materials (ASTM)
CT
- International Organization of Standardization (ISO)
53
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
54
Chart of fracture toughness KIC and yield strength Y (from Ashby)
Metals are both strong and tough !
Materials towards the bottom right : high strength and low toughness
Materials towards the top left : opposite → yield before they fracture
→fracture before they yield
55
Typical KIC values:3 21 1 /MPa m MN m
56
Ex Aircraft components
Fuselage made of 2024 alloy (Al + 4% Cu + 1% Mg)
Plane stress criterion with Kc is typically used here in place of KIC
30
100 110
IC
C
K MPa m
K MPa m
AIRBUS A330
350Y MPa (elastic limit)
Thickness of the sheet ~ 3mm