Stress concentration factor Kt in complex structures

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

Stress concentration factor K

_t

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 CFMR Concentration des contraintes près des entailles 4

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The calculation of the SCF Ktin 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 SCFK_tin the T, K, X, K … type junctions, have been proposed by Efthymiou and Lloyd¹¹.

For the Y- type junction, the relationships given the SCF K_t at the quarter point of the sleeve, are :

Where

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7

Master Mécanique-Matériaux-Structures-Procédés

Chapter 4 – Stress intensity at the crack tip

Prof. Abderrahim Zeghloul Université de Lorraine

Fracture Mechanics, Damage and Fatigue

Introduction

Stress concentration in the vicinity of a notch, studied in the previous chapter, has allowed to introduce the stress concentration factor, calledK_t . 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 K_t is defined by :

1 2 1 2

t

a a

K ^{= +} b ^{= +} ρ

max

( ) A K

_t _a

σ = σ

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

b<<a.

Under these conditions, the K_tparameter tends to infinity and the concept of SCF is no longer applicable.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, Irwin¹⁵proposed 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 I

Plane shear mode Mode II

Anti plane shear mode 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 Ileads 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 crack Surface

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 potentialsϕ^{(z) and}χ^{(z), is :}

( )

Re ( ) ( )

A = z ϕ z + χ z

The stress components are given by

( )

4 Re '( )

2 2 ''( ) ''( )

y x

y x xy

z

i z z z

σ σ ϕ

σ σ σ ϕ χ

+ =



− + = +



( )

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

Im ''( ) ''( )

x y

xy

z z z z

z z z

σ ϕ ϕ χ

σ ϕ χ

 = − −

   = + +

 = +



0 for

y xy

z z z a σ = 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 σ y

∈

= ± ± = ⋅± =

(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

σ ϕ ϕ χ

σ ϕ χ

 = − −

 = + +

  = +



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

( )

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

for Im ''( ) ''( ) 0

z z

z z z z

z a

z z z

ϕ ϕ χ

ϕ χ

+ + = =

 

  

+ = <

 



2 '( ) ''( ) ''( ) imaginary number

on the lips ''( ) ''( ) real number

z z z z

z z z L

ϕ ϕ χ

ϕ χ

+ +

  

 +

( )

Re 2 '( ) ϕ z z ϕ ''( ) z χ ''( ) z

 = − −

( )

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

The function ϕ can be decomposed into half the sum of two functionsϕ₁^andϕ₂, as follows :

1 2

1 2 2

and its derivatives, are imaginary

with on

2 and its derivatives, are real

' '' ''

L z

ϕ ϕ ϕ

ϕ

ϕ ϕ χ ϕ +

 = 

 

 = − − 



2 2

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

2 2 1

'( ) z z ' d z ( ) 2 d z ( )

χ = − ϕ ϕ ϕ + − = − ϕ + ϕ ϕ − = − ϕ ϕ +

( ) z z

1

dz

χ ϕ ϕ

 = − +  _A ₌ _Re ₍ _z _ϕ _{( )} _z ₊ _χ _{( )} _z ₎

(

¹

)

Re

A z ϕ ϕ z ϕ dz

 = − + 

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(

¹

)

¹ ¹ ²

Re Re Im Im

I II

A A

A= z

ϕ ϕ

−z +

 ϕ

dz =

 ϕ

dz+y

ϕ

+ y

ϕ

( )

Re 2 ' '' '' Re 2 ' '' ''

Im '' ''

x y

xy

z z z

σ ϕ ϕ χ

σ ϕ χ

 = − −

 = + +

 = +



'

2

z '' '' ϕ = − ϕ χ −

( )

1 2

1

2

Re ' 2 ' '' ''

Re ' '' ''

Im '' '' '

x y xy

z z

σ ϕ ϕ ϕ ϕ

σ 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

σ ϕ 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

ϕ ϕ

=



 =

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

⋯

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Stresses and displacements expressions by the Westergaard method

Before performing the calculations, it should be remembered that for any analytical functiong(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

σ ϕ ϕ ϕ ϕ

σ ϕ ϕ ϕ

 = − + −

 = + +

 = − − −



Stresses expressions

as 'ϕ 1 =Z_I 

1 1

1

Re ' Im '' Re ' Im ''

Re ''

x y

xy

y y y

σ ϕ ϕ

σ ϕ

 = −

 = +

 = −



Re Im '

Re '

x I I

y I I

xy I

Z y Z

y Z

σ σ

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 σ µε υ σ υ σ

µε σ

 = − −

 = − −

 =



* in plane strain with

* in plane stress 1

υ υ υ υ

υ

=



 =

 +



Re Im '

Re '

x I I

y I I

xy I

Z y Z

y Z

σ σ

σ

 = −

 = +

 = −



2 _x (1 2 *) Re _I Im '_I 2 u^x

Z y Z

µε = − υ − = µ ^∂x

∂

2 µ u

_x

(1 2 *) Re υ Z

_I

y Im Z

_I

 = − −

(1 )

2 (1 2 *) Re Im '

y x

y

I I

Z y Z

υ σ υ σ

µε υ

− −

=  

− +



Re Im '

Re '

x I I

y I I

xy I

Z y Z

y Z σ

σ σ

 = −

 = +

  = −



Re Im

Im ' Re

I I

Z Z

y

Z Z

y

∂

 =

 ∂

 

 = − ∂

 ∂



Im

Im ' Re ( Re ) Re

I

I I I I

y Z

y Z y Z y Z Z

y y

_∂

∂

∂ ∂

= − = − +

∂ ∂

2 u

^y

2(1 *) Im

_I

( Re

_I

)

Z y Z

y y y

µ ^∂ = − υ ^∂ − ^∂

∂ ∂ ∂

2 µ u

_y

2(1 υ *) Im Z

_I

y Re Z

_I

 = − −

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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

σ ϕ ϕ ϕ ϕ

σ ϕ ϕ ϕ

 = − + −

 = + +

 = − − −



Stresses expressions

as ' ϕ

2

= − iZ

_II



2 2

2

2 2

2 Re ' Im '' Im '' Re '' Im '

x y xy

y y y

σ ϕ ϕ

σ 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

σ σ σ

 = +

 = −

  = −



2 (1 *) *

Hooke's Law 2 (1 *) *

2

x x y

y y x

xy xy

µε υ σ υ σ µε υ σ υ σ

µε σ

 = − −

 = − −

 =



2

_x

2(1 *) Im

_II

Re '

_II

2 u

^x

Z y Z

µε = − υ + = µ ^∂ x

∂

2 µ u

_x

2(1 υ *) Im Z

_II

y Re Z

_II

 = − +

2 Im Re '

Re '

Re Im '

x II II

y II

xy II II

Z y Z

y Z

Z y Z

σ σ σ

 = +

 = −

  = −



Displacements expressions

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(1 )

2 2 * Im Re '

y x

y

II II

Z y Z

υ σ υ σ

µε υ

− −

=  

− −



2 Im Re '

Re '

Re Im '

x II II

y II

xy II II

Z y Z

y Z

Z y Z

σ σ σ

 = +

 = −

 = −



Re ' Im

Im Re

II II

Z Z

y

Z Z

y

∂

 =

 ∂



 = − ∂

 ∂



Re

Re ' Im ( Im ) Im

II

II II II II

y Z

y Z y Z y Z Z

y y _∂

−∂

∂ ∂

− = − = − +

∂ ∂

2 u^y (1 2 *) Re _II ( Im _II)

Z y Z

y y y

µ^∂ = − − υ ^∂ − ^∂

∂ ∂ ∂

( )

2 µ u

_y

(1 2 *) Re υ Z

_II

y Im Z

_II

 = − − +

Stresses and displacements expressions from the Stress Intensity Factor (SIF)

Should initially determine the analytical functions Z_I and Z_II introduced by Westergaard. Consider for example the functionZ_I(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 '

x I I

y I I

xy I

Z y Z

y Z

σ σ

σ

 = −

 = +

  = −



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

Re at 0 0

x y I

xy

Z y

σ σ σ

= =

 =

 =



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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

Z y

σ σ σ

= =

 =

 =



2 2

( ) ( )

I

Z z g z

z a

= −

With g(z)a real function for y=0 and finite for z=±a

Now consider the stressσ_yalone. From either side of each crack tip (A or A’), σy is either zero or tends to infinity (remember that K_t

→∞). It follows that the functionZ_I(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 z a ReZ_I 0

z a <  =

−

2 2

1 is a pure real for Re _I _z _a

z a

z a Z

z 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 a

ζ = −

The position of point M

ζ = ^re

ⁱ^θ

is represented by :

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

1 2

1 0 1 2

( ) ( ) where ( )

I

Z z g

ζ

g

ζ α α ζ α ζ

=

ζ

= + + + ⋅⋅⋅

( ) 0

ZI

ζ α

≈

ζ

0 0

lim 2 ( ) ( ) lim 2 ( ) 2

I z a I I

K =

_→

π z − a Z z =

_ζ_→

πζ Z 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

Z_I(ζ) can be written :

The Stress Intensity Factor (SIF), called K_Iin opening mode I, is defined by :

( ) 2

I I

Z z K

= πζ

( ) 2

II II

Z z K

= 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 :

K_IIis the Stress Intensity factor for the plane shear mode II

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( ) avec 2

I i I

Z

ζ

K

ζ

re^θ

=

πζ

=

Re cos Im sin

2 2

I I

K K

Z Z

r r

θ θ

π π

= = −

3 / 2

1 1 3 1 3

' Re ' cos Im ' sin

2 2 2 2 2 2 2 2

I I I

K K K

Z Z Z

r r r r

θ θ

πζ π π

= − = − =

2 1/ 2 Re 2 cos Im 2 sin

2 2 2 2

2

I

I I I I I

K r r

Z ζ Z K θ Z K θ

π π

= π = =

Stresses and displacements expressions From the preceding expressions of the derivative

and the primitive of Z_I, and noting that : 2 sin cos

2 2

y₌ r θ θ

We calculate the components of the stress tensor and the displacement, schematized on the figure below :

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Re Im '

Re '

x I I

y I I

xy I

Z y Z

y Z σ

σ σ

 = −

 = +

 = −



2µu_x = −(1 2 *) Reυ Z_I −yImZ_I 2µu_y =2(1−υ*) ImZ_I −yReZ_I

Re cos Im sin

2 2

I I

K K

Z Z

r r

θ θ

π π

= = −

2 2 2 2

2

I

I I I I I

K r r

Z ζ Z K θ Z K θ

π π

= π = =

3 / 2

1 1 3 1 3

' Re ' cos Im ' sin

2 2 2 2 2 2 2 2

I I I

K K K

Z Z Z

r r r r

θ θ

πζ π π

= − = − =

cos 1 sin sin3

2 2 2

2

cos 1 sin sin3

2 2 2

2

cos sin cos3

2 2 2

2

I x

I y

I xy

K r K

r K

r

θ θ θ

σ π

θ θ θ

σ π

θ θ θ

σ π

  

=  − 

  

  

= +

  

 



 =



2

2 cos 1 2 * sin

2 2 2

2 sin 2(1 *) cos

2 2 2

I x

I y

K r

u

K r

u

θ υ θ

µ π

θ υ θ

µ π

 =  − + 

  

 



 

 =  − − 

  



We have in mode I, a discontinuity of the crack lips displacement along the yaxis, u_y(π)=-u_y(-π)

sin 2 cos cos3

2 2 2

2

sin cos cos3

2 2 2

2

cos 1 sin sin3

2 2 2

2

II x

II y

II xy

K r K r K

r

θ θ θ

σ π

θ θ θ

σ π

θ θ θ

σ π

  

= −  + 

  



=



  

= −

  

 



2

2 sin 2(1 *) cos

2 2 2

2 cos 1 2 * sin

2 2 2

II x

II y

K r

u

K r

u

θ υ θ

µ π

θ υ θ

µ π

 =  − + 

  

 



 

 =  − − 

  



We have in mode II, a discontinuity of the crack lips displacement along de xaxis, u_x(π)=-u_x(-̟)

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

σ σ σ

 = +

 = −

 = −



2µu_x =2(1−υ*) ImZ_II +yReZ_II

( )

2µu_y = − (1 2 *) Re− υ Z_II +yImZ_II

Re cos Im sin

2 2

II II

K K

Z Z

r r

θ θ

π π

= = −

2 2 2 2

2

II

II II II II II

K r r

Z ⁼ πζ Z ⁼ K π θ Z ⁼ K π θ

3 / 2

1 1 3 1 3

' Re ' cos Im ' sin

2 2 2 2 2 2 2 2

II II II

K K K

Z Z Z

r r r r

θ θ

πζ π π

= − = − =

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Anti plane shear mode

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)

x3

The displacement field is thus of the form

3 3

avec

3 3

( , ) 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

ε ε

 = + =



 = + =



13 13 3,1

23 23 3,2

Hooke's Law 2

2

u u

σ µε µ

= =



= =



13,1 23,2

Equilibrium equation σ +σ =0

3

0  ∆ = u

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 Z_III, it is shown that the displacement u₃ can be expressed in the following form :

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3

1 Im

_III

u 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 µ

i

13 3,1

23 3,2

u u

σ µ

=



 =

13 23

Im Re

III III

Z Z σ

σ

=

  

 =

The function Z_IIIhas the same form as Z_I and Z_II

2

III III

Z K

= πζ

K_IIIis 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 σ

σ

=

 

 =

3

1 Im

_III

u Z

= µ

2

III III

Z K

= πζ

13

23

3

sin 2 2

cos 2 2

2 sin 2

III

K r K

r

K r

u σ θ

π σ θ

π

θ

µ π

 = −

 

 =

 

  =



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Tutorial 8 : Crack loaded in opening mode I

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

1- Calculate σx, σy and τxyfor θ =0,45,90,135 and 180°at a distance rfrom 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 forθ^{=45 and 90}°? 4- For which value ofθ, the normal stress is maximum?

M

Crack

cos 1 sin sin3

2 2 2

2

cos 1 sin sin3

2 2 2

2

cos sin cos3

2 2 2

2

I x

I y

I xy

K r K

r K

r

θ θ θ

σ π

θ θ θ

σ π

θ θ θ

σ π

  

=  − 

  



  

= +

  

 



=



( ) 2

I

ij ij

K f

σ r θ

= π

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

f11

f22

f12

(22)

2. Stress distribution around the crack tip 1. Calculation of ( ) for =0, 45, 90, 135 and 180°f_ij θ θ

f11

f22

f12

3. Mohr construction ( angle between the directions 1 and )α y

1 2

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

Stress concentration factor Kt in complex structures