1 ch-4 plane problems in linear isotropic elasticity humbert laurent thursday, march 18th 2010...

1

CH-4 Plane problems in linear isotropic elasticity

HUMBERT Laurent

Thursday, march 18th 2010

laurent.humbert@epfl.ch

laurent.humbert@ecp.fr

Thursday, march 25th 2010

2

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

14

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 *

15

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

16

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

17

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

18

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

19

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

21

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

22

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

23

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

25

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

26

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

27

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

28

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

29

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

30

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