Annex 3 Page 217
Annex 3
Analytical calculation of [ ]
Vand [ ]
IHContent
A3.1. Basic model 218
A3.2. Analytical integration of [ ]V over a triangle 220
A3.3. Analytical integration of [ ]IH over a triangle 223 A3.4. Special case: the edge AB crosses the axis of negative X 225
Annex 3 Page 218
A3.1. Basic model
Consider a LFMVC as shown in the figure A3.1 where X and Y are the local axes associated with the crack tip.
Figure A3.1. A LFMVC divided into triangles
We divide it into several triangles. The integration of [ ]V and [ ]IH over the domain thus transforms into the sum of integrations over these triangles.
Let us consider the integration over the triangle OAB as an example.
Figure A3.2. Equation of edge AB
O
BY
X
θB θA A
rB
rA
r θ
α O
X
Y
θ r(θ)
A
B
θA
θB
O
Annex 3 Page 219
A B
A A
B A
Y Y
Y Y X X
X X
−
= −
−
−
θ θ
= r( )cos X
θ θ
= r( )sin Y
A B
A A
B
A
Y Y
Y sin ) ( r X
X
X cos ) ( r
−
− θ
= θ
−
− θ θ
Hence
θ
−
− θ
−
= −
θ (X X )sin (Y Y )cos X
Y X ) Y
( r
A B A
B
A B B A
In this formula, the term (XB−XA)sinθ−(YB −YA)cosθ can be equal to zero only if the edge AB is radial (i.e. if the line containing A and B also contains point O), which is impossible in the present context.
Let
A B B AX Y X Y
a= −
A
B X
X
s= −
A B Y Y c= − Hence
θ
−
= θ
θ ssin ccos ) a
( r
For any function F(r,θ)we have:
( ) θ θ
=
∫ ∫
∫
θθθ F(r, )rdrd dA
F B
A
r
OAB 0
Annex 3 Page 220
A3.2. Analytical integration of
[ ]
V over a triangle[ ]
[ ]
{ }
[ ]{ }
[ ]
{ }
[ ]{ }
v v dAv v dA
H D H dA
H D H
dA H D H dA
H D H V
OAB OAB
I OAB
I
OAB
I OAB
I
OAB
∫
∫
∫
∫
∫
⎥⎦
⎢ ⎤
⎣
= ⎡
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢
⎣
⎡
=
Σ Σ
Σ Σ
Σ Σ
Σ Σ
22 21
12 11 2
2 1
2
2 1
1 1
Introducing the values of
{ }{ }
HΣ1 , HΣ2 given by the equations (V.20) and (V.21) and using the expression of [ ]D appearing in (V.25), we get:[ ]
*
*
*
E r
cos ) (
cos
v π
θ ν
+ +
− θ ν
−
= 2
1 2 3
2 11
[ ]
*
*
*
E r
sin cos ) v (
v π
θ θ ν
+ +
−
= ν
= 2
1 1
21 12
*
*
*
*
E r
cos ) (
cos ) v (
π
θ ν
+ + θ
− ν + ν
= +
8
2 1
3 1
4 9
22
Consequently,
( ) ( )
⎥ θ
⎦
⎢ ⎤
⎣
= ⎡
⎥ θ
⎦
⎢ ⎤
⎣
= ⎡
⎥⎦
⎢ ⎤
⎣
⎡
∫ ∫ ∫ ∫
∫
θθθ θ
θ
θ drd
w w
w d w
dr v r
v v dA v
v v
v
v B
A B
A
r r
OAB 0 21 22
12 11
0 21 22
12 11 22
21 12 11
with:
[ ]
*
*
*
E
cos ) (
cos
w π
θ ν
+ +
− θ ν
−
= 2
1 2 3
2 11
[ ]
*
*
*
E
sin cos ) w (
w π
θ θ ν
+ +
−
= ν
= 2
1 1
21 12
*
*
*
*
E
cos ) (
cos ) w (
π
θ ν
+ + θ
− ν + ν
= +
8
2 1
3 1
4 9
22
Hence, ⎥
⎦
⎢ ⎤
⎣
⎡
22 21
12 11
w w
w
w is not a function of r.
Annex 3 Page 221 Therefore,
( ) ( ) ⎥ ( )θ
⎦
⎢ ⎤
⎣
=⎡
⎥⎦
⎢ ⎤
⎣
= ⎡
⎥⎦
⎢ ⎤
⎣
⎡
∫
∫
rθ vv vv rdr rθ ww ww dr ww ww r22 21
12 11
0 21 22
12 11
0 21 22
12 11
Finally,
[ ] ⎥ ( )θ θ
⎦
⎢ ⎤
⎣
= ⎡
⎥⎦
⎢ ⎤
⎣
=
∫
⎡∫
θθ r dw w
w dA w
v v
v
V v B
OAB A
OAB 21 22
12 11 22
21 12 11
Define the following constants and functions
2 22 2
12
2 11 2
03 0
0
2 2
3 2 3
1
1 2
3 4
s ) (
c ) (
C
s c ) (
C
s ) (
c ) (
C
E R C a
q R
s c q
*
*
*
*
*
*
ν +
−
− ν
= ν +
=
− ν +
− ν
=
= π
= +
=
[ ]
[ ]
[ ]
θ + θ
= θ
θ
− θ
= θ
θ +
θ
− ν
= θ
θ
− θ
− ν
= θ
θ
− θ
= θ
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡ + θ
= θ
sin s cos c ) ( f
cos s sin c ) ( f
) ( f c s ) (
) ( f
) ( f s c ) (
) ( f
sin s cos c ln ) ( f
R tan c s tanh arc ) ( f
*
*
6 5
2 4
2 3
2 1 0
1 1 2
2
[ ]
{ }
[ ]
{ }
[ ]
{
C f ( ) R ( ) f ( ) f ( )}
C ) ( s
) ( f ) ( f ) (
R ) ( f C C ) ( s
) ( f ) ( f ) (
R ) ( f C C
) ( s
*
*
*
θ + θ ν
+
− θ
= θ
θ + θ ν
+ +
θ
−
= θ
θ + θ ν
+ +
θ
= θ
3 5
0 1
22 0
22
4 6
0 1
12 0 12
3 5
0 1
11 0
11
1 3 2
1 2
1 2
Then
( ) ⎥
⎦
⎢ ⎤
⎣
⎡
θ θ
θ
= θ θ
⎥ θ
⎦
⎢ ⎤
⎣
∫
⎡ww ww r d ss (( )) ss (( ))22 21
12 11
22 21
12 11
Annex 3 Page 222 We finally get :
⎥⎦
⎢ ⎤
⎣
⎡
θ θ
θ
− θ
⎥⎦
⎢ ⎤
⎣
⎡
θ θ
θ
= θ
⎥⎦
⎢ ⎤
⎣
∫
⎡vv vv dA ss (( )) ss (( )) ss (( )) ss (( ))A A
A A
B B
B B
OAB 12 22
12 11
22 12
12 11
22 21
12 11
For the practical calculation, we use the following sequence to avoid the difficulties linked with the evaluation of the arctanh which is a complex number when its argument is not in the domain [−1,+1] :
[ ]
[ ]
[ ] [ ]
[ ]
[ ]
) sin (sin
s ) cos (cos
c f
) cos (cos
s ) sin (sin
c f
f c ) (
s ) (
f
f s ) (
c ) (
f
sin s cos c ln sin
s cos c ln f
c if u
ln f
c if u
ln f
sin sin
sin sin
u
) u ( ) u (
) u ( ) u u (
R tan c s u
R tan c s u
A B
A B
AB
A B
A B
AB
AB A
* B AB
AB A
* B AB
A A
B B
AB
limAB AB
AB AB
A B B
A
A B B
A limAB
B A
A AB B
B B
A A
θ
− θ +
θ
− θ
=
θ
− θ
− θ
− θ
=
+ θ
− θ
− ν
=
− θ
− θ
− ν
=
θ
− θ
− θ
− θ
=
=
=
≠
=
θ
−
− θ θ + θ
θ
− + θ
θ + θ
=
− +
−
= + + θ
= + θ
=
6 5
2 4
2 3
2 1 1
0 0
1 1 2
2 0 1 2 0 1
2 2
2 2
1 1
1 1
2 2
[ ]
{ }
[ ]
{ }
[ ]
{
* AB AB}
AB AB
AB
* AB AB
AB
AB
* AB AB
AB
f f
) (
R f
C C
s
f f
) (
R f
C C s
f f
) (
R f
C C
s
3 5
0 1
22 0
22
4 6
0 1
12 0 12
3 5
0 1
11 0
11
1 3 2
1 2
1 2
+ ν
+
−
=
+ ν
+ +
−
=
+ ν
+ +
=
[ ] ⎥
⎦
⎢ ⎤
⎣
=⎡
AB AB
AB OAB s AB s
s V s
22 12
12 11
Annex 3 Page 223
A3.3. Analytical integration of
[ ]
IH over a triangle[ ]
( )
( ) ⎪
⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
θ θ
=
⎪⎪
⎭
⎪⎪⎬
⎫
⎪⎪
⎩
⎪⎪⎨
⎧
=
∫ ∫
∫ ∫
∫
∫
θ θ
θ Σ
θ θ
θ Σ
Σ Σ
d dr r H
d dr r H dA
H dA H IH
B A B
A
r r
OAB
I OAB
I OAB
0
2 0
1
2 1
Introducing the values of
{ }{ }
HΣ1, HΣ2 given by the equations (V.20) and (V.21), we get:( )
( ) [IHr( )]
) ( IHr ) ( IHr ) ( IHr
) ( IHr ) ( IHr ) ( IHr dr
r H
dr r H
r r
θ
⎥=
⎦
⎢ ⎤
⎣
⎡
θ θ
θ
θ θ
= θ
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
∫
∫
θ Σ
θ Σ
23 22
21
13 12
11
0
2 0
1
with:
[ ]
π
= θ
θ 3 2
2 3
1 r( )
) ( C
) ( IHr ) ( IHr
) ( IHr ) ( IHr
) cos cos
( sin ) ( C ) ( IHr
sin cos
cos ) ( C ) ( IHr
) cos (
cos ) ( C ) ( IHr
) cos cos
( cos ) ( C ) ( IHr
θ
= θ
θ
= θ
θ + θ θ +
θ
−
= θ
θ θ θ θ
= θ
θ θ −
θ
= θ
θ + θ θ −
θ
= θ
11 23
13 22
2 2 4
21
2 2 3 2 2
13
2 2 3
2 12
2 2 2
11
1 1 1 3 1
Then
[ ] ⎥
⎦
⎢ ⎤
⎣
⎡
θ θ
θ
θ θ
= θ θ
∫
IHr(θ) d pp1121(( )) pp1222(( )) pp1323(( )) withπ
= θ
θ 3 2 2
2 q
) ( r ) a
( C
Annex 3 Page 224 2
3 13 2
5
2 3 5 2
3
2 3 3 2
11
2 3 5 2
3 5 2 3 2
2 6 3
2 5 3
3 4 2
3 3 2
2 2 2
2 1 2
− θ + θ
= θ
+ θ
− θ
= θ
+ θ
− θ
= θ
+ θ
− θ
= θ
θ θ +
+
= θ
θ θ
−
−
= θ
cos q c cos ) s c c
( ) ( g
cos q c cos ) s c c ( ) ( g
cos q s cos ) s s c ( ) ( g
cos q s cos ) s s c ( ) ( g
sin ) cos q s c ( ) ( g
sin ) cos q s c ( ) ( g
[ ]
[ ]
[ ]
[ ]
) ( p ) ( p
) ( p ) ( p
) ( g s ) ( g ) ( C ) ( p
) ( g s ) ( g ) ( C ) ( p
) ( g c ) ( g ) ( C ) ( p
) ( g c ) ( g ) ( C ) ( p
θ
= θ
θ
= θ
θ
− θ θ
= θ
θ
− θ θ
= θ
θ +
θ θ
−
= θ
θ
− θ θ
= θ
11 23
13 22
2 21
2 13
2 12
2 11
2 6
2
1 5
2
1 4
2
2 3
2
Finally
[ ] ⎥
⎦
⎢ ⎤
⎣
⎡
θ θ
θ
θ θ
− θ
⎥⎦
⎢ ⎤
⎣
⎡
θ θ
θ
θ θ
= θ
) ( p ) ( p ) ( p
) ( p ) ( p ) ( p ) ( p ) ( p ) ( p
) ( p ) ( p ) ( IH p
A A
A
A A
A B
B B
B B
OAB B 21 22 23
13 12
11 23
22 21
13 12
11
Annex 3 Page 225
A3.4. Special case: the edge AB crosses the axis of negative X
In a LFMVC, the segment AB will never cross the axis of negative X but, in chapter 6, the situation of figure A3.3 is encountered.
Figure A3.3. line AB crosses the negative X axis
In this case, the calculations of the preceding sections must be corrected to take account that θ jumps from −π to +π when the line AB crosses the crack.
Consequently, for any function g(θ), we have in this case:
θ θ +
θ θ
= θ
θ
∫ ∫
∫
−θππ
θ g( )d g( )d d
) (
g B
A
B A
Let
θ θ
=
θ)
∫
g( )d( G Then
[G( ) G( )] [G( ) G( )]
) ( G ) ( G ) ( G ) ( G d ) (
g A B B A
B
A θ θ= π − θ + θ − −π = θ − θ + π − −π
∫
Hence, the previous results obtained for an edge AB not intersecting the axis of “negative X”
must be corrected.
After some calculations, the correction on [VOAB] is found to be :
[ ] ⎥
⎦
⎢ ⎤
⎣
⎡
− ν
+
= −
⎥⎦
⎢ ⎤
⎣
⎡
π
− π
−
π
− π
− −
⎥⎦
⎢ ⎤
⎣
⎡
π π
π
= π
c s
s c E q
) (
a ) ( s ) ( s
) ( s ) ( s ) ( s ) ( s
) ( s ) ( V s
on
correction *
*
OAB 1
22 12
12 11
22 12
12 11
O O B
Y
X
θA
A θB
rA
rB
r θ
α
Annex 3 Page 226 In a similar way, the correction on [IHOAB] is
[ ]
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
= π
⎥⎦
⎢ ⎤
⎣
⎡
π π
π
π π
− π
⎥⎦
⎢ ⎤
⎣
⎡
π π
π
π π
= π
2 2
3
2 3 2
3 2
2 8
23 22
21
13 12
11 23
22 21
13 12
11
s c s c s
s c c s c q
c a a
) ( p ) ( p ) ( p
) ( p ) ( p ) ( p ) ( p ) ( p ) ( p
) ( p ) ( p ) ( IH p
on
correction OAB
Note that c=YB −YA ≠0 because c=YB −YA =0 would mean that the edge is parallel to X, that is parallel to the crack.
In such a case, the edge cannot intersect the crack.
