FMDF cours 181011 - LEM3

11/10/2018

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

γb =2 γe

From where

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

So that

From where

A B

y

x

Hα

Sol

O

γ e γ b


The Airy stress function

associated with this

loading is :

( , ) sinA r cPrθ 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

11/10/2018

2


2 2 2 2

2 2 2 2 2 2

1 1 1 1( )

A A AA

r r r r r r r rθ θ ∂ ∂ ∂ ∂ ∂ ∂∆ ∆ = + + + + ∂ ∂ ∂ ∂ ∂ ∂


The stress function A(r,θ)

is well biharmonic


The stresses are defined by the following expressions


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

If α=π/2, the constant c is -1/πand the stress σr is the given by :

11/10/2018

3


Tutorial 1 (continued) : Stress field in a plate loaded in tension

and pierced with a tiny hole

σ ∞

σ ∞

x

y

r θM

2 2

2( , ) ln cos 2

fA r br c r d er

rθ 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

θσ σ τ


2 2

2( , ) ln cos 2

fA r br c r d er

rθ θ = + + + +

σ ∞

σ ∞

x

y

r θM

infinite plate loaded in tension and pierced

by a circular hole of radius a

2 2

2 2 2

1 1A A AA

r r r r θ∂ ∂ ∂∆ = + +∂ ∂ ∂

3

22 2 cos 2

A c fbr er

r r rθ∂ = + + − ∂

2

2 2 4

62 2 cos 2

A c fb e

r r rθ∂ = − + + ∂

22

2 24 cos 2

A fd er

rθ

θ∂ = − + + ∂

2

44 cos 2

dA b

rθ∆ = −

3

8cos 2

A d

r rθ∂∆ =

∂2

2 4

24cos 2

A d

r rθ∂ ∆ = −

∂2

2 2

16cos 2

A d

rθ

θ∂ ∆ =∂

( ) 0A∆ ∆ =

( , ) is well bi-harmonicA r θ

11/10/2018

4


, , , ,2

,

1 1 1r r rr r

r

A A A Ar r r

θθ θ θ θσ σ τ = + = = −

2 2

2( , ) ln cos 2

fA r br c r d er

rθ θ = + + + +

σ ∞

σ ∞

x

y

rθ

M

3

22 2 cos 2

A c fbr er

r r rθ∂ = + + − ∂

2

2 2 4

62 2 cos 2

A c fb e

r r rθ∂ = − + + ∂

22

2 24 cos 2

A fd er

rθ

θ∂ = − + + ∂

2 2 42 2 4 6 cos 2r

c d fb e

r r rσ θ = + − + +

2 4

62 2 cos 2

c fb e

r rθσ θ = − + +

2 4

62 2 sin 2r

d fe

r rθτ θ = − −

2

22 sin 2

A fd er

rθ

θ∂ − = + + ∂


Boundary conditions at ∞ , ie far from the hole

0 0 0

( , ) 0 0

0 0 0

x yσ σ ∞

=

( , ) ( , ) tr P x y Pσ θ σ= ⋅ ⋅cos sin 0

sin cos 0

0 0 1

P

θ θθ θ

= −

2

2

sin (1 cos 2 )2

cos (1 cos 2 )2

sin cos sin 22

r

r

θ

θ

σσ σ θ θ

σσ σ θ θ

στ σ θ θ θ

∞∞

∞∞

∞∞

= = −

= = +

= =

at ∞ r >> a2

2

22

b

e

σ

σ

∞

∞

=

=

σ ∞

σ ∞

x

y

rθ

M

11/10/2018

5


Boundary conditions at r=a

( , ) ( , ) 0r a aσ θ τ θ θ= = ∀

2 2 44 6 cos 2

2 2r

c d f

a a a

σ σσ θ∞ ∞

= + − + +

2 4

62 sin 2

2r

d f

a aθ

στ θ∞

= − −

2

2 42

4

2 4

4 62 2

et2 6 3

2 62 2

d fd a

a ac a

d ff a

a a

σ σσ

σ σ

∞ ∞

∞

∞ ∞

+ = − = −

= − + = =

σ ∞

σ ∞

x

y

rθ

M


2 2 4

2 2 4

2 4

2 4

2 4

2 4

1 1 4 3 cos 22 2

1 1 3 cos 22 2

1 2 3 sin 22

r

r

a a a

r r r

a a

r r

a a

r r

θ

θ

σ σσ θ

σ σσ θ

στ θ

∞ ∞

∞ ∞

∞

= − − − +

= + + +

= + −

2 2 42 2 4 6 cos 2r

c d fb e

r r rσ θ = + − + +

2 4

62 2 cos 2

c fb e

r rθθσ θ = − + +

2 4

62 2 sin 2r

d fe

r rθτ θ = − −

22

22

b

e

σ

σ

∞

∞

=

=

( ,0) 3aθσ σ ∞=

3σ ∞

σ ∞

2

2c a

σ ∞

= −

2

4

2

36

2

d a

f a

σ

σ

∞

∞

= −

=

11/10/2018

6


Airy stress function for few loadings (1)

, , ,x yy y xx xy xyA A Aσ σ σ= = = −

2( , ) Beam in tractionA x y ay=

( , ) Beam in shearA x y axy=

3Beam subjected to

( , )bending moment

A x y ay

=



3( , )A x y axy bxy

= +

2 2 3 2 3 5( , )A x y ax bx y cy dx y ey

= + + + +

11/10/2018

7



Axisymmetric loading , ,

10r r rr rA A

rθ θσ σ τ= = =

2( , ) ( )A r A r Crθ = =

2( , ) ( ) lnA r A r a r crθ = = +

2 2( , ) ( ) ln lnA r A r a r br r crθ = = + +



, , , ,2

,

1 1 1r r rr r

r

A A A Ar r r

θθ θ θ θσ σ τ = + = = −

( , ) sinA r crθ θ θ=

( , ) sin 2A r a bθ θ θ= +

( , ) cosA r crθ θ θ=

11/10/2018

8


Complex formulation of the Airy stress function

- Holomorphic functions (or analytical functions)

z x iy

z x iy

= += −RST

M(x,y)

�x

�y

= +

= −

RS||

T||

xz z

yz z

i

2

2

( , ) ( , )x y Plan g x yg∈ → ( , ) ( , ) ( , )x y z z g z z

g → →

( )

( ), , ,

, , ,

1

2The derivation rules are

1

2

z x y

z x y

g g ig

g g ig

= − = +

g g g

g i g g

x z z

y z z

, , ,

, , ,( )

= += −

RST

P P x y

Q Q x y

==RST

( , )

( , )g P iQ= +

is holomorphic if 0g

gz

∂ =∂

g zg

xi

g

y' ( ) = ∂

∂= − ∂

∂

E� ��


* Properties oy analytic functions

g P iQdg

dz

g

xi

g

y= + = ∂

∂= − ∂

∂ avec

Cauchy conditions

P Q

x yP Q P Qi i

P Qx 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 yg P iQ = +

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

11/10/2018

9


Examples of analytic functions

einz, zn 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 dfine ine ne i

dz x dz y dz

e nx e nx

+ −

+ +

− −

= = = +∂ ∂= = = = − =∂ ∂

i

Exchanging n by –n, it is seen that eny cosnx and eny 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 cosh2 2

ny ny ny nye e e eny 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 dfnz 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

if z z x iy re r i

df f df f i dfi

dz z x x iy dz y x iy dz

r

θ θ

θ

= = + = = +∂ ∂= = = = =∂ + ∂ +

i

11/10/2018

10


Expressions of the Airy stress function

∆ ∆ Αb g = 0 If then 0 is harmonicP A P P= ∆ ∆ =

( ) is analytic with

P Q

x yf z P iQ

P Q

y x

∂ ∂∂ ∂∂ ∂∂ ∂

== + = −

Calculating the harmonic function ( , )Q x y

Q QdQ dx dy

x y

P PQ dQ dx dy

y x

∂ ∂∂ ∂

∂ ∂∂ ∂

= +

= = − +

( ) ( )1 is also analytic function 4 4

4

p qz f z dz p iq P

x y

∂ ∂ϕ∂ ∂

= = + = =

1 1 1 1If then 0 ( ) is analytic functionp px qy p z p iqχ= Α − − ∆ = = +

A px qy p

z z z

z z z z z z= + +

= +

= + + +

RS|T|

1 1

2

Α

Α

Re ϕ χ

ϕ χ ϕ χ

b g b gb g b g b g b g


Stresses expressions

σ σx xy yy xy x y y z y zz zzi i i i i+ = − = − + = − = −Α Α Α Α Α Α Α, , , , , , , ,d i c h c h2 2

,

,

,

x yy

y xx

xy xy

σσσ

= Α

= Α

= −Α

( ) ( ) ( ) ( )' ' '' ''x xyi z z z z zσ σ ϕ ϕ ϕ χ + = + − −

g g ig

g g ig

z x y

z x y

, , ,

, , ,

= −

= +

RS|

T|

1

21

2

d id i

g g g

g i g g

x z z

y z z

, , ,

, , ,( )

= += −

RST( ) ( )

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 σy xy xx xy x y x z x zz zzi i i− = + = + = = +Α Α Α Α Α Α Α, , , , , , , , ,d i c h c h2 2

( ) ( ) ( ) ( )' ' '' ''y xyi z z z z zσ σ ϕ ϕ ϕ χ − = + + +

σ σ ϕ ϕ ϕy x z z z+ = + =2 4' ' Re 'b g d ie j b gc h

( ) ( )( )2 2 '' ''y x xyi z z zσ σ σ ϕ χ− + = +

11/10/2018

11


Displacements expressions

22

2

2

22

2

2

µ ε σ λλ µ

σ σ λ µλ µ

σ σ σ

µ ε σ λλ µ

σ σ λ µλ µ

σ σ σ

x x x y x y y

y y x y x y x

= −+

+ =++

+ −

= −+

+ =++

+ −

RS||

T||

b g d i b g d i

b g d i b g d i

= ++

−

= ++

−

RS||

T||

22

2

22

2

µ ε λ µλ µ

µ ε λ µλ µ

x xx

y yy

b g

b g

∆ Α Α

∆ Α Α

,

,

∆A Pp

x

q

y= = =4 4

∂∂

∂∂

( ) ( )

( ) ( )

,

,

22 2

22 2

x x

y y

u p y

u q x

λ µµ α

λ µλ µ

µ βλ µ

+= − Α + +

+ = − Α + +

avec αβ

y cy d

x cx d

b gb g

= += − +

RST1

2

( ) ( ) ( ), ,

22 2x y x yu iu p iq i

λ µµλ µ++ = + − Α + Α+

g g ig

g g ig

z x y

z x y

, , ,

, , ,

= −

= +

RS|

T|

1

21

2

d id i

+ = ∂∂

Α Α, ,x yiA

zd i 2

( ) ( ) ( ) ( ) ( ) ( )2 22 2 2 2 ' 'x yu iu z z z z z z

z

λ µ ∂ λ µµ ϕ ϕ ϕ ϕ χλ µ λ µ∂+ Α ++ = − = − − −+ +

( ) ( ) ( ) ( )2 ' 'x y

U iU z z z zµ κ ϕ ϕ χ+ = − −

33 4 for plane strain

with 3

for stress plane1

vλ µκλ µ

νκν

+= = −+

−=+


�eθ

�x

�er

θ

θ

�y

� � �

� � �e x y

e x y

r = += − +

RSTcos . sin .

sin . cos .

θ θθ θθ

�

�

�

�e

e

x

y

r

θ

θ θθ θ

FHGIKJ =FHGIKJ −

FHG

IKJP avec P :

cos sin

sin cos

u

u

u

u

u u

u u

r x

y

x y

x yθ

θ θθ θ

FHGIKJ =FHGIKJ =

+− +FHG

IKJP

cos sin

sin cos

u iu e u iur

i

x y+ = +−θ

θ ( ) ( ) ( ) ( ) ( )( )2 ' 'i

ru iu e z z z zθθµ κ ϕ ϕ χ−+ = − −

σ σθh h� � �

e e x y

t

r

P P, ,

= ( )22 2

r x y

i

r r y x xyi e i

θθ

θ θ

σ σ σ σσ σ σ σ σ σ

+ = + − + = − +

( ) ( )( )22 2 '' ''i

r ri e z z zθθ θσ 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.

11/10/2018

12


3 3Let ( ; , , ) denote the Cartesian coordinates system and ( ; , , )

a coordinate system associated with curvilinear coordinates , .

O x y x M xα βα β

��

The complex number is associated with the , coordinates and

the complex is associated with the curvilinear coordinates , .

z x iy x y

iζ α β α β= +

= +

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

( ) and '( )

x x y y

dzz f f

d

α β α β

ζ ζζ

= =

= =


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 , .

if f e

x y

θζ ζθ

α β

=

� arg '( ) arg arg arg , ,dz

f dz d u x ud

ζ ζ α θζ

= = − = − =��

So we have '( ) | '( ) | and '( ) | '( ) | so thati if f e f f eθ θζ ζ ζ ζ −= =

2'( )=

'( )

ife

f

θζζ

( ) and '( )dz

z f fd

ζ ζζ

= =

11/10/2018

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 bi-

harmonic, that is to say ∆(∆A)=0.

The expression of this stress function, from the complex potentials ϕand 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 :

( ) ( ) ( ) ( ) ( ) ( )1Re

2z z z z z z z z zϕ χ ϕ χ ϕ χ Α = + = + + +


In a Cartesian coordinates system (x,y)

In a curvilinear coordinates system associated to varaibles (α,β)

3 4 for plane strain

with 3 for stress plane

1

vκνκν

= −−=+

FMDF cours 181011 - LEM3

Documents

Transcript of FMDF cours 181011 - LEM3