15/11/2018
1
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.
A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 2
Offshore wind turbines in Germany
15/11/2018
2
A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 3
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
15/11/2018
3
A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 5
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 :
A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 6
Where
15/11/2018
4
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
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
15/11/2018
5
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 9
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.
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 10
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.
15/11/2018
6
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 11
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).
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 12
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
Anti plane shear mode
Mode III
15/11/2018
7
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 13
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.
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 14
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
15/11/2018
8
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 15
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
= − −
= + + = +
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 16
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).
15/11/2018
9
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 17
( )( )
( )
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
= − − = + + = +
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 = − −
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 18
( )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 = − +
15/11/2018
10
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 19
( )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 ϕ χσ ϕ ϕ χ
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
= += −
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 20
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
⋯
⋯
15/11/2018
11
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 21
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
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 22
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 ϕσ ϕ ϕ ϕ
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
= − = + = −
15/11/2018
12
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 23
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 = − −
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 24
(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 = − −
15/11/2018
13
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 25
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 ϕσ ϕ ϕ ϕ
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
= + = − = −
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 26
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
15/11/2018
14
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 27
(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 = − − +
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 28
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
= == =
15/11/2018
15
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 29
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 :
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 30
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 :
15/11/2018
16
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 31
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 :
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 32
( )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
15/11/2018
17
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 33
( ) 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
= = =
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 34
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 :
15/11/2018
18
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 35
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
= − = − =
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 36
15/11/2018
19
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 37
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)
3x�
The displacement field is thus of the form
3 3 3 3 avec ( , )u u x u u x y= =� �
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 38
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 :
15/11/2018
20
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 39
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
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 40
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
= −
= =
15/11/2018
21
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 41
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
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 42
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
15/11/2018
22
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 43
2. Stress distribution around the crack tip
1. Calculation of ( ) for =0, 45, 90, 135 and 180°ijf θ θ
11f
22f
12f
A. Zeghloul Fracture mechanics, damage and fatigue – Stress intensity factor 44
3. Mohr construction ( angle between the directions 1 and )yα� �
1 2Using the Mohr Constrcution, the main stresses, and , are given by : σ σ