In case with infiltration under the dam, the angle a is given by So that

(1)

A. Zeghloul Fracture mechanics, damage and fatigue – Plane elasticity 1

In case there is no infiltration under the dam, the angle α is given by So that

γ _b =2 γ _e

From where

In case with infiltration under the dam, the angle a is given by So that

From where

A B

y

x α H

Sol O

γ

e

γ

b

The Airy stress function associated with this

loading is :

( , ) sin

A r θ = cPr θ θ

Show that A(r, θ ) is biharmonic.

Determine the components of the stress tensor.

Determine the constant c as a function of angle α ^. Calculate the stresses when α ⁼ π ^/2.

Tutorial 1 (continued) : Semi infinite plane under point load

(2)

2 2 2 2

2 2 2 2 2 2

1 1 1 1

( ) A A A

A r r r r θ r r r r θ

 ∂ ∂ ∂  ∂ ∂ ∂ 

∆ ∆ =  + +  + + 

∂ ∂ ∂ ∂ ∂ ∂

  

( , ) sin

A r θ = cPr θ θ

The stress function A(r, θ ⁾ is well biharmonic

The stresses are defined by the following expressions

( , ) sin

A r θ = cPr θ θ

The balance of forces is written to determine the constant c :

If α=π /2, the constant c is -1/π

and the stress σ r is the given by :

(3)

Tutorial 1 (continued) : Stress field in a plate loaded in tension and pierced with a tiny hole

σ

^∞

σ

^∞

x y

r θ

2 2

M

( , ) ln f

2

cos 2

A r br c r d er

θ = + + ^  + + r ^  θ

 

The Airy stress function associated with this

loading is :

Show that ( , ) is biharmonic Determine the stress , , ( is the radius of the hole)

r r

A r

a

θ θ

θ

σ σ τ

2 2

( , ) ln f

2

cos 2

A r br c r d er

θ = + + ^  + + r ^  θ

  ^σ

∞

σ

^∞

x y

r θ M

infinite plate loaded in tension and pierced by a circular hole of radius a

2 2

2 2 2

1 1

A A A

A r r r r θ

∂ ∂ ∂

∆ = + +

∂ ∂ ∂

3

2 2 2 cos 2

A c f

br er

r r r θ

∂  

= + +  − 

∂  

2

2 2 4

2 2 6 cos 2

A c f

b e

r r r θ

∂  

= − +  + 

∂  

2

4

2

cos 2

A f

d er

r θ

θ

∂  

= −  + + 

∂  

2

4 4 d cos 2

A b

r θ

∆ = −

3

8 cos 2

A d

r r θ

∂∆ = ∂

2

2 4

24 cos 2

A d

r r θ

∂ ∆ = −

∂

2

2 2

16 cos 2

A d

r θ

∂ ∆ = θ

∂

( A ) 0

∆ ∆ =

( , ) is well bi-harmonic

A r θ

(4)

, 2 , , ,

,

1 1 1

r r rr r

r

A A A A

r r

^θθ ^θ ^θ

r

^θ

σ σ τ  

= + = = −  

 

2 2

( , ) ln f

2

cos 2

A r br c r d er

θ = + + ^  + + r ^  θ

 

σ

^∞

σ

^∞

x y

r θ M

3

2 2 2 cos 2

A c f

br er

r r r θ

∂  

= + +  − 

∂  

2

2 2 4

2 2 6 cos 2

A c f

b e

r r r θ

∂  

= − +  + 

∂  

2

4

2

cos 2

A f

d er

r θ

θ

∂  

= −  + + 

∂  

2 2 4

2 2 4 6 cos 2

r

c d f

b e

r r r

σ = + − ^  + + ^  θ

 

² ⁴

2 c 2 6 f cos 2

b e

r r

σ

θ

= − + ^  + ^  θ

 

2 4

2 2 6 sin 2

r

d f

e r r

τ

θ

= ^  − − ^  θ

 

2

sin 2

A f

d er

r θ

θ

∂  

− =  + + 

∂  

Boundary conditions at ∞ , ie far from the hole

0 0 0

( , ) 0 0

0 0 0

σ x y σ

^∞

 

 

=  

 

 

( , ) r P ( , ) x y

^t

P

σ θ = ⋅ σ ⋅ ^cos _sin _cos ^sin ⁰ ₀

0 0 1

P

θ θ

 

 

= −  

 

 

2

sin (1 cos 2 )

2 cos (1 cos 2 )

2 sin cos sin 2 2

r

r θ

θ

σ σ θ σ θ

τ σ θ θ σ θ

∞ ∞

= = −

= = +

= =

at ∞ r >> a

2 2

b e

σ σ

∞

 =

  

 =



σ

^∞

σ

^∞

x y

r θ

M

(5)

Boundary conditions at r=a

( , ) ( , ) 0

r a a

σ θ τ θ = = ∀ θ

2 4 2 6 4 cos 2

2 2

r

c d f

a a a

σ σ

σ = ^∞ + − ^  ^∞ + + ^  θ

 

2 4

2 6 sin 2

r 2

d f

a a

θ σ

τ = ^  ^∞ − − ^  θ

 

2

2 4

2

4

2 4

4 6

2 2

2 et 6 3

2 6

2 2

d f

d a

a a

c a

d f

f a

a a

σ σ

σ

σ σ

∞ ∞

∞

∞ ∞

 + = − = −

 = −   

 + = =



σ

^∞

σ

^∞

x y

r θ M

2 2 4

2 4

1 1 4 3 cos 2

2 2

1 1 3 cos 2

2 2

1 2 3 sin 2

2

r

a a a

r r r

a a

r r

a a

r r

θ

σ σ

σ θ

σ σ

σ θ

τ σ θ

∞ ∞

∞

   

=  −  −  − + 

   

   

=  +  +  + 

   

 

=  + − 

 

2 2 4

2 2 4 6 cos 2

r

c d f

b e

r r r

σ = + − ^  + + ^  θ

 

2 4

2 c 2 6 f cos 2

b e

r r

σ

θθ

= − + ^  + ^  θ

 

2 4

2 2 6 sin 2

r

d f

e r r

τ

θ

= ^  − − ^  θ

 

2 2

b e

σ σ

∞

=

( , 0) a 3 σ

θ

= σ

^∞

3 σ

^∞

σ

^∞

2 c _{= −} σ 2 ^∞ a

2

4

2 6 3

2 d a

f a

σ σ

∞

= −

=

(6)

Airy stress function for few loadings (1)

, , ,

x

A

yy y

A

xx xy

A

xy

σ = σ = σ = −

( , )

2

Beam in traction A x y = ay

( , ) Beam in shear A x y = axy

3

Beam subjected to ( , )

bending moment A x y ay

 

= 

 

Airy stress function for few loadings (2)

( , )

3

A x y axy bxy

 

= +  

 

2 2 3 2 3 5

( , )

A x y ax bx y cy dx y ey

 

= + + + +  

 

(7)

Airy stress function for few loadings (3)

Axisymmetric loading 1

_, _,

r

A

r

A

rr r

0 r

^θ ^θ

σ = σ = τ =

( , ) ( )

2

A r θ = A r = Cr

( , ) ( ) ln

2

A r θ = A r = a r + cr

2 2

( , ) ( ) ln ln

A r θ = A r = a r + br r + cr

Airy stress function for few loadings (4)

, 2 , , ,

,

1 1 1

r r rr r

r

A A A A

r r

^θθ ^θ ^θ

r

^θ

σ = + σ = τ = − ^ ^ 

 

( , ) sin

A r θ = cr θ θ

( , ) sin 2

A r θ = a θ + b θ

( , ) cos

A r θ = cr θ θ

(8)

Complex formulation of the Airy stress function

- Holomorphic functions (or analytical functions)

z x iy z x iy

= +

RS = −

M(x,y) T

x y

 = +

= −

R S ||

T ||

x z z y z z i 2 2 ( , ) x y ∈ Plan  → 

^g

g x y ( , ) ( , ) x y  →  ( , ) z z  → 

^g

g z z ( , )

( )

, , ,

1 The derivation rules are 2

1 2

z x y

g g ig

 = −

 

 = +



g g g

g i g g

x z z

y z z

, , ,

,

(

, ,

)

= +

= −

RS T

P P x y Q Q x y

=

RS =

T ^{( , )} ^{( , )} ^g ^{= +} ^P ^iQ

is holomorphic if g 0

g ∂ = z

∂

g z g

x i g ' ( ) = ∂ y

∂ = − ∂

∂ E

* Properties oy analytic functions

g P iQ dg

dz g

x i g

= + = ∂ y

∂ = − ∂ avec ∂

Cauchy conditions

P Q

x y

P Q P Q

i i

P Q

x x y y

y x

∂ ∂

∂ ∂ ∂ ∂

∂ ∂

∂ ∂ ∂ ∂

∂ ∂

 =

 

+ = − +  

 = −



∆ P = ∆ Q =

E

0 The real or imaginary parts of an analytic function, are harmonic

 ⇐

 

Conversely, if ( , ) and ( , )

is an analytic function verify the Cauchy conditions

P x y Q x y

g P iQ

 = +

- If g is an analytical function, its derivative and its integral are also

(9)

Examples of analytic functions

e ^inz , z ⁿ and ln z are analytic functions. Their real and imaginary parts that are harmonic, can be determined.

( )

( ) ( )

( ) (cos sin )

The associated harmonic fonctions are cos and sin

inz in x iy ny

inz in x iy in x iy

ny ny

f z e e e nx i nx

df f df f df

ine ine ne i

dz x dz y dz

e nx e nx

+ −

+ +

− −

= = = +

∂ ∂

= = = = − =

∂ ∂

i

Exchanging n by –n, it is seen that e ^ny cosnx and e ^ny sinnx are also harmonics. It follows that sinhny sinnx, coshny sinnx, sinhny cosnx and coshny cosnx, obtained by linear combination of the preceding harmonic functions, are also harmonic.

The hyperbolic sine and hyperbolic cosine functions are defined by

sinh cosh

2 2

ny ny ny ny

e e e e

ny ny

− −

− +

= =

1 1 1

( ) ( ) ( ) (cos sin )

( ) ( )

The associated harmonic fonctions are cos and sin

n n ei n n

n n n

n n

f z z x iy r r n i n

df f df f df

nz n x iy in x iy i

dz x dz y dz

r n r n

θ θ θ

θ θ

− − −

= = + = = +

∂ ∂

= = + = = + =

∂ ∂

i

( ) ln ln( ) ln( ) ln

1 1

The associated harmonic fonctions are ln and

f z z x iy re i r i

df f df f i df

dz z x x iy dz y x iy i dz

r

θ θ

θ

= = + = = +

∂ ∂

= = = = =

∂ + ∂ +

i

(10)

Expressions of the Airy stress function

∆ ∆ Α b g ⁼ ⁰ _If P = ∆ A _then ∆ = P ₀  is harmonic P

( ) is analytic with

P Q

x y

f z P iQ

P Q

y x

∂ ∂

 =

 

= + 

 = −



Calculating the harmonic function ( , ) Q x y

Q Q

dQ dx dy

x y

P P

Q dQ dx dy

y x

∂ ∂

= +

 

= =  − + 

 

 

( ) ¹ ( ) is also analytic function 4 4

4 p q

z f z dz p iq P

x y

∂ ∂

ϕ ⁼  ^{= +} ^ ⁼ ∂ ⁼ ∂

1 1 1 1

If p = Α − px − qy then ∆ = p 0  χ ( ) z = p + iq is analytic function

A px qy p

z z z

z z z z z z

= + +  = +

= + + +

R S|

1

T| 1

2 Α Α

Re ϕ χ

ϕ χ ϕ χ

b g b g b g b g b g b g

Stresses expressions

σ

x

+ i σ

xy

= Α

,yy

− i Α

,xy

= − i d Α

,x

+ i Α

,y

i

y

= − i c ² Α

,z

h

,y

= ² c Α

,zz

− Α

,zz

h

, , ,

x yy

y xx

xy xy

σ σ σ

= Α

= −Α

( ) ( ) ( ) ( )

' ' '' ''

x

i

xy

z z z z z

σ σ ϕ ϕ ϕ χ

 + = + − −

g g ig

z x y

, , ,

= −

= +

R S|

T|

1 2 1 2

d i

d i _g ^g

^x

_{i g} ^g

^z

^g _g

^z

y z z

, , ,

,

(

, ,

)

= +

= −

RS T

( ) ( )

Expression of the stress function from the potential complex ( ) and ( )

1 ( ) ( )

2 z z

z z z z z z

ϕ χ

ϕ χ ϕ χ

 

Α =  + + + 

σ

_y

σ

_xy _xx _xy _x _y

x z x zz zz

i i i

− = Α

_,

+ Α

_,

= Α

_,

+ Α

_,

= Α = Α + Α

, , , , ,

d i ² c h ² c h

( ) ( ) ( ) ( )

' ' '' ''

y

i

xy

z z z z z

σ σ ϕ ϕ ϕ χ

 − = + + +

σ

_y

+ σ

_x

= 2 e ϕ ' b g z + ϕ ' d i z j = 4 Re c ϕ ' b g z h

( ) ( )

( )

2 2 '' ''

y x

i

xy

z z z

σ σ − + σ = ϕ + χ

(11)

Displacements expressions

2 2

µ ε σ λ

λ µ σ σ λ µ

λ µ σ σ σ

µ ε σ λ

λ µ σ σ λ µ

λ µ σ σ σ

x x x y x y y

y y x y x y x

= −

+ + = +

+ + −

= −

+ + = +

+ + −

R S ||

T || b gd i b gd i

b gd i b gd i ^

= +

+ −

= +

+ −

R S ||

T |

|

2 2

2 2 2

2 µ ε λ µ

λ µ µ ε λ µ λ µ

x xx

y yy

b g

∆ Α Α

,

∆A P p

x q

= = 4 ∂ = 4 y

∂

( ) ( )

,

2 2 2

x x

y y

u p y

u q x

λ µ

µ α

λ µ λ µ

µ β

λ µ

 = + − Α +

 +

  

 = + − Α +

 +



avec α β

y cy d

x cx d

b g b g ^{= − +} ⁼ ⁺

RS T

¹²

( ) ² ⁽ ⁾ (

^, ^,

)

2 µ u

_x

iu

_y

2 λ µ p iq

_x

i

_y

λ µ

+ = + + − Α + Α

+

g g ig

z x y

, , ,

= −

= +

R S|

T|

1 2 1 2

d i

d i ^ ^Α

^,^x

⁺ ⁱ ^Α

^,^y

^{= ∂} ^∂

A

d i ² z

( ) ² ^{( )} ² ^{( ) ( )} ( ) ( )

2 u

x

iu

y

2 z 2 2 z z z ' z ' z

z

λ µ ∂ λ µ

µ ϕ ϕ ϕ ϕ χ

λ µ ∂ λ µ

+ Α +

+ = − = − − −

+ +

( ) ^{( )} ( ) ( )

2 µ U

_x

+ iU

_y

= κ ϕ z − z ϕ ' z − χ ' z

3 3 4 for plane strain with

3 for stress plane 1

λ µ v κ λ µ κ ν

ν

= + = − +

= − +

e_θ

x

e_r

θ θ

y

e x y

r

= +

= − +

RS T

^θ

^{cos .} ^{sin .} ^θ ^θ ^{sin .} ^{cos .} ^θ ^θ ^e ^e

θ^r

^x ^y

θ θ

F HG I KJ ⁼ F

HG I

KJ −

F HG I

KJ

P avec P : cos sin sin cos u

u u u

u u

r x

y

x y

θ

θ θ

F HG I KJ ⁼ F

HG I

KJ ⁼ F ⁻ ⁺ ⁺

HG I

KJ

P cos sin

sin cos

u

_r

+ iu

_θ

= e

⁻ⁱ^θ

( u

_x

+ iu

_y

) ² ^µ ( ^u

^r

⁺ ^iu

^θ

) ⁼ ^e

⁻ⁱ^θ

( ^{κ ϕ} ( ) ^z ⁻ ^z ^ϕ ^' ( ) ( ) ^z ⁻ ^χ ^' ^z )

σ σ

h

_{e e}θ

h

_{x y}^t

r

P P

,

=

,

^{ } ^  ^{σ σ}

^θ

⁻

^r

⁺ ² ^{σ σ} ⁱ ^σ

^r^r

⁺

^θ

⁼

^θ

^e ⁼

²ⁱ^θ

^{σ σ} (

^x

^{σ σ} ⁺

^y

⁻

^y ^x

⁺ ² ⁱ ^σ

^xy

)

( ) ( )

( )

2 2

2ⁱ

'' ''

r

i

r

e

^θ

z z z

θ θ

σ σ − + σ = ϕ + χ

* System coordinates change

Because we will use of polar coordinates in the solution of many

problems in elasticity, the previous governing equations will now be

developed in this curvilinear system.

(12)

3 3

Let ( ; , , ) denote the Cartesian coordinates system and ( ; , , ) a coordinate system associated with curvilinear coordinates , .

O x y x M α β x

α β

The complex number is associated with the , coordinates and the complex is associated with the curvilinear coordinates , .

z x iy x y

ζ α i β α β

= +

As ( , ) and ( , ), then we have : ( ) and '( )

x x y y

z f dz f

d

α β α β

ζ ζ

ζ

= =

We easily show that '( ) | '( ) | , in other words that the argument of the complex number is equal to , the angle between the two coordinate

systems respectively associated with , and , .

f f e

i

x y ζ ζ

θ

α β

=

arg '( ) arg dz arg arg , ,

f dz d u x u

ζ d ζ α θ

= ζ = − = − =

So we have '( ) | f ζ = f '( ) | ζ e

ⁱ^θ

and '( ) | f ζ = f '( ) | ζ e

⁻ⁱ^θ

so that '( )

2

'( ) = f

i

f e ζ θ

ζ

( ) and dz '( )

z f f

ζ d ζ

= ζ =

(13)

Summary of key findings

The resolution of a plane elasticity problem comes down to the search for a stress function, called the Airy function A, which is biharmonic, that is to say ∆(∆A)=0.

The expression of this stress function, from the complex potentials ϕ and χ which are analytical functions of the complex variable z, is given by :

The search for the Airy stress function is therefore to find these complex potentials. The components of the stress tensor and the displacement vector are then determined by the following relationships :

In case with infiltration under the dam, the angle a is given by So that

In case there is no infiltration under the dam, the angle α is given by So that

γ b =2 γ e

From where