Stress concentration factor Kt in complex structures · 2018. 11. 15. · A. Zeghloul Fracture mechanics, damage and fatigue –Stress intensity factor 20 It appears that the stress

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A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 1

Stress concentration factor Kt

in complex structures

The tubular junctions of offshore platforms are examples of complex structures

(figure below) . These platforms consist of tubes welded together, forming tubular

junctions. Complex intersections of these junctions are structural discontinuities

leading to high levels of stress in the weld seams.


Offshore wind turbines in Germany

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These tubular junctions are classified according to their form, type T, Y, K …

schematized below.

The study of these junctions requires setting show below, for a K-type junction,

which is the most general case.

The sleeve of the junction is embedded at both ends, the applied loads are also

shown below.

A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 4A. Zeghloul CFMR Concentration des contraintes près des entailles 4

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The calculation of the SCF Kt in these junctions, uses the finite elements method.

The great difficulty in modelling these junction, is the generation of meshes in the

areas of geometric discontinuities where the stress gradients are important.

An example is shown below.

Meshing of DTDK

type junction

Parametric relationships widely used in the offshore industry for

the calculation of the SCF Kt in the T, K, X, K … type junctions,

have been proposed by Efthymiou and Lloyd11.

For the Y- type junction, the relationships given the SCF Kt at

the quarter point of the sleeve, are :


Where

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Master Mécanique-Matériaux-Structures-Procédés

Chapter 4 – Stress intensity at the crack tip

Prof. Abderrahim Zeghloul Université de Lorraine

Fracture Mechanics,

Damage and Fatigue

Contents

Introduction – Stress intensity factor Concept

Cracks loading modes

Westergaard approach

Expressions of stresses and displacements fields from Westergaard stress

function

Stress intensity factor K - Expressions of stresses and displacements fields

Anti plane shear mode

Superposition principle in LEFM

Plastic zone at crack tip

Practical methods for calculation of SIF – Methods of weight functions

Toughness – Critical SIF

Griffith’s energy approach

Relationship between Griffith energy and SIF K

A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 8

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Introduction

Stress concentration in the vicinity of a notch, studied in the

previous chapter, has allowed to introduce the stress concentration

factor, called Kt . This parameter is suitable for the characterisation

of the severity of a notch.

For a plate with an elliptical hole loaded in tension, the SCF Kt is

defined by :

1 2 1 2t

a aK

b ρ= + = +

max ( ) t aA Kσ σ=

A crack can be considered as a very flattened elliptical hole, i.e.

b<<a.

Under these conditions, the Kt parameter tends to infinity and the

concept of SCF is no longer applicable.

It is therefore necessary to introduce a new parameter.

Based on Westergaard’s work14, Irwin15 proposed the stress intensity

factor (SIF). Applying the SIF concept to the description of the

stress field in the vicinity of a crack tip, is commonly called LEFM.


The use of a single parameter (the SIF K) to describe the stress

distribution near the crack tip, is justified by the similarities that can

be observed between various cracks loaded in tension, as shown in

the figure below.

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Edge crack

Central crack

Crack

near hole

The fringes of the

photoelasticity cliche are

similar for the three

different cracks.

This result suggests that the

stress distributions are the

same.

The SIF concept is presented in this chapter. To determine the stress

field and the displacement field near a crack tip, we use the

Westergaard approach based on the complex potentials introduced

previously (chapter 2).


Cracks modes loading

According to the loading direction, three modes of the crack lips

displacement can be distinguished. These mode shown below,

correspond to three kinematics of displacement of the lips of a

crack.

Opening mode

Mode IPlane shear mode

Mode II


Mode III

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Mode I or opening mode - The relative displacement of the lips

of the crack, is defined by :

Mode II or plane shear mode - The relative displacement of the

lips of the crack, is defined by :

Mode III or anti plane shear mode - The relative displacement

of the lips of the crack, is defined by :

Mode I leads to a discontinuity of displacement

according to the direction x2, while mode II

leads to a discontinuity according to direction

x1.


The cracks in service do not always appear in the configuration

shown in the previous figure.

When the cracks are sufficiently long, they generally pass through

the thin structures such plates.

In a thicker structures, cracks may be corner or surface.

The figure below shows various crack configurations, that can be

observed in the vicinity of a circular hole.

Through cracks, are usually treated as a two-dimensional problem

(planar problem), while corner cracks or surface cracks involve

three-dimensional calculations.

Through

crack

Front

crackSurface

crack

Corner

crack

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The Westergaard approach

In plane elasticity, the stresses derive from a biharmonic stress

function, the Airy stress function A, whose expression according

to the complex potentials ϕ(z) and χ(z), is :

( )Re ( ) ( )A z z zϕ χ= +

The stress components are given by

( )4 Re '( )

2 2 ''( ) ''( )

y x

y x xy

z

i z z z

σ σ ϕσ σ σ ϕ χ

+ = − + = +

( )( )

( )

Re 2 '( ) ''( ) ''( )

Re 2 '( ) ''( ) ''( )

Im ''( ) ''( )

x

y

xy

z z z z

z z z z

z z z

σ ϕ ϕ χσ ϕ ϕ χ

σ ϕ χ

= − −

= + + = +


0 for y xy

z z

z aσ σ

== = <

The lips ( ) of the crack are free, and therefore the stress vector ( , )

is zero. The normal to the lips being , we have ( , ) 0 :

L T M L n

n y T M y yσ∈

= ± ± = ⋅ ± =

� �

��

(L)

Now consider a cracked body, with a very small crack size

compared to the dimensions of the cracked body.

The cracked body is subjected to a given loading, the crack length

is 2a (as shown below).

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( )( )

( )

Re 2 '( ) ''( ) ''( )

Re 2 '( ) ''( ) ''( )

Im ''( ) ''( )

x

y

xy

z z z z

z z z z

z z z


σ ϕ χ

= − − = + + = +

The boundary conditions on the lips of the crack, lead to :

( )( )

Re 2 '( ) ''( ) ''( ) 0 for

Im ''( ) ''( ) 0

z zz z z z

z az z z

ϕ ϕ χϕ χ

=+ + = <+ =

2 '( ) ''( ) ''( ) imaginary number on the lips

''( ) ''( ) real number

z z z zL

z z z

ϕ ϕ χϕ χ

+ + +

( )Re 2 '( ) ''( ) ''( )z z z zϕ ϕ χ = − −


( )Re 2 '( ) ''( ) ''( )z z z zϕ ϕ χ= − −

The function ϕ can be decomposed into half the sum of two

functions ϕ1 and ϕ2, as follows :

1 21

22

and its derivatives, are imaginary with on 2

and its derivatives, are real' '' ''

L

z

ϕ ϕ ϕϕϕϕ ϕ χ

+ =

= − −

2 2by integrating the relationship '' '' ' ( ') ' ' , we get :z d zχ ϕ ϕ ϕ ϕ ϕ= − − = − + −

2 2 1'( ) ' ( ) 2 ( )z z d z d zχ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ= − + − = − + − = − +

1( )z z dzχ ϕ ϕ = − + ( )Re ( ) ( )A z z zϕ χ= +

( )1ReA z z dzϕ ϕ ϕ = − +

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( )1 1 1 2Re Re Im Im

III

AA

A z z dz dz y yϕ ϕ ϕ ϕ ϕ ϕ= − + = + + ��

( )( )

( )

Re 2 ' '' ''

Re 2 ' '' ''

Im '' ''

x

y

xy

z

z

z


σ ϕ χ

= − − = + + = +

2' '' ''zϕ ϕ χ= − −

( )( )( )

1 2

1

2

Re ' 2 ' '' ''

Re ' '' ''

Im '' '' '

x

y

xy

z z

z z

z z

σ ϕ ϕ ϕ ϕσ ϕ ϕ ϕσ ϕ ϕ ϕ

= + + − = + − = − −

1 1 2 2

1 1 2

1 2 2

Re ' Im '' 2 Re ' Im ''

Re ' Im '' Im ''

Re '' Re '' Im '

x

y

xy

y y

y y

y y

σ ϕ ϕ ϕ ϕσ ϕ ϕ ϕ

σ ϕ ϕ ϕ

= − + − = + + = − − −

z x iy

z x iy

= += −


It appears that the stress field [σ] is the sum of two fields [σ1] and

[σ2] derived from the following Airy stress functions :

1 1

2

Re Im

Im

I

II

A dz y

A y

ϕ ϕ

ϕ

= +

=

Mode I

Mode II

Westergaard defines for these modes, the following

analytical functions

1

2

( ) ' ( )

( ) ' ( )

I

II

Z z z

Z z i z

ϕϕ

= =

Re Im

Im Re

I I I

II II II

A Z y Z

A y iZ y Z

= +

= − = −

Westergaard called ', '' the successive derivatives of

and , the successive primitives of

Z Z Z

Z Z Z

⋯

⋯

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Stresses and displacements expressions by the

Westergaard method

Before performing the calculations, it should be remembered that

for any analytical function g(z), we have :

It follows therefore that


Opening mode I

1 1 2 2

1 1 2

1 2 2

Re ' Im '' 2 Re ' Im ''

Re ' Im '' Im ''

Re '' Re '' Im '

x

y

xy

y y

y y

y y


σ ϕ ϕ ϕ

= − + − = + + = − − −

Stresses expressions

1as ' IZϕ =

1 1

1 1

1

Re ' Im ''

Re ' Im ''

Re ''

x

y

xy

y

y

y

σ ϕ ϕσ ϕ ϕ

σ ϕ

= − = + = −

Re Im '

Re Im '

Re '

x I I

y I I

xy I

Z y Z

Z y Z

y Z

σσ

σ

= − = + = −

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Displacements expressions

2 (1 *) *

Hooke's law 2 (1 *) *

2

x x y

y y x

xy xy

µε υ σ υ σµε υ σ υ σ

µε σ

= − − = − − =

* in plane strain

with * in plane stress

1

υ υυυ

υ

= = +

Re Im '

Re Im '

Re '

x I I

y I I

xy I

Z y Z

Z y Z

y Z

σσ

σ

= − = + = −

2 (1 2 *) Re Im ' 2 xx I I

uZ y Z

xµε υ µ ∂= − − =

∂

2 (1 2 *) Re Imx I Iu Z y Zµ υ = − −


(1 *) *2

(1 2 *) Re Im '

y x

y

I IZ y Z

υ σ υ σµε

υ− −

= − +

Re Im '

Re Im '

Re '

x I I

y I I

xy I

Z y Z

Z y Z

y Z

σσ

σ

= − = + = −

Re Im

Im ' Re

I I

I I

Z Zy

Z Zy

∂ = ∂ ∂ = − ∂

Im

Im ' Re ( Re ) Re

I

I I I I

Zy

y Z y Z y Z Zy y

∂∂

∂ ∂= − = − +∂ ∂

2 2(1 *) Im ( Re )y

I I

uZ y Z

y y yµ υ

∂ ∂ ∂= − −∂ ∂ ∂

2 2(1 *) Im Rey I I

u Z y Zµ υ = − −

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Plane shear Mode II

1 1 2 2

1 1 2

1 2 2

Re ' Im '' 2 Re ' Im ''

Re ' Im '' Im ''

Re '' Re '' Im '

x

y

xy

y y

y y

y y


σ ϕ ϕ ϕ

= − + − = + + = − − −

Stresses expressions

2as '

IIiZϕ = −

2 2

2

2 2

2 Re ' Im ''

Im ''

Re '' Im '

x

y

xy

y

y

y

σ ϕ ϕσ ϕ

σ ϕ ϕ

= − = = − −

2 Im Re '

Re '

Re Im '

x II II

y II

xy II II

Z y Z

y Z

Z y Z

σσ

σ

= + = − = −


2 (1 *) *

Hooke's Law 2 (1 *) *

2

x x y

y y x

xy xy

µε υ σ υ σµε υ σ υ σ

µε σ

= − − = − − =

2 2(1 *) Im Re ' 2 xx II II

uZ y Z

xµε υ µ ∂= − + =

∂

2 2(1 *) Im Rex II IIu Z y Zµ υ = − +

2 Im Re '

Re '

Re Im '

x II II

y II

xy II II

Z y Z

y Z

Z y Z

σσ

σ

= + = − = −

Displacements expressions

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(1 *) *2

2 *Im Re '

y x

y

II IIZ y Z

υ σ υ σµε

υ− −

= − −

2 Im Re '

Re '

Re Im '

x II II

y II

xy II II

Z y Z

y Z

Z y Z

σσ

σ

= + = − = −

Re ' Im

Im Re

II II

II II

Z Zy

Z Zy

∂ = ∂ ∂ = − ∂

Re

Re ' Im ( Im ) Im

II

II II II II

Zy

y Z y Z y Z Zy y

∂−∂

∂ ∂− = − = − +∂ ∂

2 (1 2 *) Re ( Im )y

II II

uZ y Z

y y yµ υ

∂ ∂ ∂= − − −∂ ∂ ∂

( )2 (1 2 *) Re Imy II IIu Z y Zµ υ = − − +


Stresses and displacements expressions from

the Stress Intensity Factor (SIF)

Should initially determine the analytical functions ZI and ZII

introduced by Westergaard. Consider for example the function ZI (the

reasoning is also applicable to mode II) and looking at the boundary

conditions in the vicinity of small crack length 2a, loaded in mode I.

Re Im '

Re Im '

Re '

x I I

y I I

xy I

Z y Z

Z y Z

y Z

σσ

σ

= − = + = −

On the crack plane (at y=0), we have :

Re at 0

0

x y I

xy

Zy

σ σσ

= == =

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The lips of the crack being unloaded, the BC are :

0

à 0 et

y xy

y z a

σ σ= = = <

From these two relationships,

we deduce that

0

at 0 and

x y xy

y z a

σ σ σ= = = = <

Re at 0

0

x y I

xy

Zy

σ σσ

= == =

2 2

( )( )

I

g zZ z

z a=

−With g(z) a real function for y=0

and finite for z=±a

Now consider the stress σy alone. From either side of each crack tip

(A or A’), σy is either zero or tends to infinity (remember that Kt

→∞). It follows that the function ZI(z) must be of the form :


The boundary conditions are then checked, since we have on the

crack plane :

2 2

1 is a pure imaginary for Re 0

Iz a Z

z a< =

−

2 2

1 is a pure real for Re I z a

z a

z a Zz a

+

−

→+

→−

> →∞−

Ends A and A’ play identical

roles. Also, we make a translation

of the Cartesian coordinate system

(x,y), so that the origin is at point

A. This translation is equivalent to

the following change of variable :

z aζ = − ire θζ =The position of point M

is represented by :

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Westergaard stress function ZI(z), is then written :

211 0 1 2

( )( ) where ( )I

gZ z g

ζ ζ α α ζ α ζζ

= = + + + ⋅⋅⋅

0( )IZαζζ

≈

0 0lim 2 ( ) ( ) lim 2 ( ) 2I z a I IK z a Z z Zζπ πζ ζ πα→ →= − = =

As we try to determine the stress field at the immediate vicinity of

the crack tip (asymptotic field), i.e. for |ζ|→0, the stress function

ZI(ζ) can be written :

The Stress Intensity Factor (SIF), called KI in opening mode I,

is defined by :


( )2

II

KZ z

πζ=

( )2

IIII

KZ z

πζ=

As the Westergaard function has the dimension of stress ( )

the SIF will have the dimension of stress length (MPa m)

I

I

Z MPa

K i

For the plane shear mode II, the same approach as before, leads to a

similar result in the immediate vicinity of the crack tip :

KII is the Stress Intensity factor for the plane shear mode II

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( ) avec 2

iII

KZ re θζ ζ

πζ= =

Re cos Im sin2 22 2

I II I

K KZ Z

r r

θ θπ π

= = −

3/ 2

1 1 3 1 3' Re ' cos Im ' sin

2 2 2 2 22 2 2

I I II I I

K K KZ Z Z

r rr r

θ θπζ π π

= − = − =

1/ 22 Re 2 cos Im 2 sin2 2 2 22

II I I I I

K r rZ Z K Z K

θ θζπ ππ

= = =


Stresses and displacements expressions

From the preceding expressions of the derivative

and the primitive of ZI, and noting that :

2 sin cos2 2

y rθ θ=

We calculate the components of the stress tensor and the

displacement, schematized on the figure below :

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Re Im '

Re Im '

Re '

x I I

y I I

xy I

Z y Z

Z y Z

y Z

σσ

σ

= − = + = −

2 (1 2 *) Re Imx I Iu Z y Zµ υ= − −

2 2(1 *) Im Rey I I

u Z y Zµ υ= − −

Re cos Im sin2 22 2

I II I

K KZ Z

r r

θ θπ π

= = −

1/ 22 Re 2 cos Im 2 sin2 2 2 22

II I I I I

K r rZ Z K Z K

θ θζπ ππ

= = =

3/ 2

1 1 3 1 3' Re ' cos Im ' sin

2 2 2 2 22 2 2

I I II I I

K K KZ Z Z

r rr r

θ θπζ π π

= − = − =

3cos 1 sin sin

2 2 22

3cos 1 sin sin

2 2 22

3cos sin cos

2 2 22

Ix

Iy

Ixy

K

r

K

r

K

r

θ θ θσπ

θ θ θσπ

θ θ θσπ

= −

= +

=

2

2

2cos 1 2 * sin

2 2 2

2sin 2(1 *) cos

2 2 2

Ix

Iy

K ru

K ru

θ θυµ π

θ θυµ π

= − +

= − −

We have in mode I, a discontinuity of the crack lips

displacement along the y axis, uy(π)=-uy(- π)

3sin 2 cos cos

2 2 22

3sin cos cos

2 2 22

3cos 1 sin sin

2 2 22

IIx

IIy

IIxy

K

r

K

r

K

r

θ θ θσπ

θ θ θσπ

θ θ θσπ

= − +

=

= −

2

2

2sin 2(1 *) cos

2 2 2

2cos 1 2 * sin

2 2 2

IIx

IIy

K ru

K ru

θ θυµ π

θ θυµ π

= − +

= − −

We have in mode II, a discontinuity of the crack

lips displacement along de x axis, ux(π)=-ux(- ̟)

Plane shear mode II

2 Im Re '

Re '

Re Im '

x II II

y II

xy II II

Z y Z

y Z

Z y Z

σσ

σ

= + = − = −

2 2(1 *) Im Rex II IIu Z y Zµ υ= − +

( )2 (1 2 *) Re Imy II IIu Z y Zµ υ= − − +

Re cos Im sin2 22 2

II IIII II

K KZ Z

r r

θ θπ π

= = −

1/ 22 Re 2 cos Im 2 sin2 2 2 22

IIII II II II II

K r rZ Z K Z K

θ θζπ ππ

= = =

3/ 2

1 1 3 1 3' Re ' cos Im ' sin

2 2 2 2 22 2 2

II II IIII II II

K K KZ Z Z

r rr r

θ θπζ π π

= − = − =


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When a cracked structure is loaded in anti plane shear mode, the

lips of the crack move, as shown on the figure below, along a

direction perpendicular to the plane (x,y)

3x�

The displacement field is thus of the form

3 3 3 3 avec ( , )u u x u u x y= =� �


In linear elasticity, the strains are given by

13 1,3 3,1 3,1

23 2,3 3,2 3,2

1 1( )

2 2

1 1( )

2 2

u u u

u u u

ε

ε

= + = = + =

13 13 3,1

23 23 3,2

2Hooke's Law

2

u

u

σ µε µσ µε µ

= = = =

13,1 23,2Equilibrium equation 0σ σ+ =

3 0u ∆ =The displacement component u3 is thus harmonic. It can be

considered as the real part or the imaginary part of an analytical

function.

Westergaard approach can be used to address the anti plane shear

stress problem. If the Westergaard function associated with this

loading is called ZIII, it is shown that the displacement u3 can be

expressed in the following form :

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3

1Im IIIu Z

µ=

As is homogeneous to a stress, is homogeneous to a stress length

so that is homogeneous to a length

III III

III

Z Z

Z µi

13 3,1

23 3,2

u

u

σ µσ µ

= =

13

23

Im

Re

III

III

Z

Z

σσ

= =

The function ZIII has the same form as ZI and ZII

2

IIIIII

KZ

πζ=

KIII is the Stress Intensity Factor in mode III


The stresses and the displacement in mode III, are then given by

13

23

Im

Re

III

III

Z

Z

σσ

= =

3

1Im

IIIu Z

µ=

2

IIIIII

KZ

πζ=

13

23

3

sin22

cos22

2sin

2

III

III

III

K

r

K

r

K ru

θσπ

θσπ

θµ π

= −

= =

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Tutorial 8 : Crack loaded in opening mode I

From the expressions, based on KI, of the stresses in the vicinity of a crack tip :

1- Calculate σx, σy and τxy for θ =0,45,90,135 and 180° at a distance

r from the end of the crack.

2- Represent the stress state around the crack tip at the current point

M for the angles considered at question 1.

3- Determine, using Mohr construction, the main stresses and the

main directions. What values take these quantities for θ =45 and 90°?

4- For which value of θ, the normal stress is maximum?

M


Crack

3cos 1 sin sin

2 2 22

3cos 1 sin sin

2 2 22

3cos sin cos

2 2 22

Ix

Iy

Ixy

K

r

K

r

K

r

θ θ θσπ

θ θ θσπ

θ θ θσπ

= −

= +

=

( )2

Iij ij

Kf

rσ θ

π=

1. Calculation of ( ) for =0, 45, 90, 135 and 180°ijf θ θ

11f

22f

12f

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2. Stress distribution around the crack tip

1. Calculation of ( ) for =0, 45, 90, 135 and 180°ijf θ θ

11f

22f

12f


3. Mohr construction ( angle between the directions 1 and )yα� �

1 2Using the Mohr Constrcution, the main stresses, and , are given by : σ σ

Documents

Stress concentration factor Kt in complex structures · 2018. 11. 15. · A. Zeghloul Fracture mechanics, damage and fatigue –Stress intensity factor 20 It appears that the stress