COMPUTATIONAL METHODS IN ENGINEERING AND SCIENCE
EPMESC X, Aug. 21-23, 2006, Sanya, Hainan, China
©2006 Tsinghua University Press & Springer
The Solutions of Stress Field and Displacement Field of Orthogonal
Anisotropic Plate with Edge Cracks (II)
Z. R. Tian *, Z. Sun, Z. Y. Li
Department of Mechanics, Northwestern Polytechnical University, Xi’an, Shaanxi, 710072 China
Email: [email protected]
Abstract The integral equations of stress field σ and displacement field u of orthogonal anisotropic plate with edge
cracks are obtained. By the use of Weber-Schafheitlin discontinuous integral equation the Abel style integral equation
is given. It is transformed to the second kind of Fredholm integral equation. The solutions of stress field σ and
displacement field u can then be obtained by using calculation.
Key words: orthogonal anisotropic plate, edge cracks, equivalent space, Abel style integral equation, Fredholm
integral equation
THE STRESS FIELD
σ AND DISPLACEMENT FIELD u
The stress field σ and displacement field u of orthogonal anisotropic plate with edge cracks for one quarter of the
plane (x ≥ 0, y ≥ 0) under a unit uniform distributed tension at the ends of the plate or pressure along the surface of
cracks are derived based on the “equivalent space” — a creative idea proposed by the authors [1, 2]:
σ xx ( x, y ) =
+
1
π
∫
∞
0
1
π∫
1
π
∫
∞
0
ux (x, y) =
1
π∫
∞
0
y/a
− α 22
α1 α
e
α2
2
ξ y/a
⎞
⎟⎟cos(ξ x / a )dξ
⎠
1
π∫
∞
0
⎛ αξ
C (ξ )⎜⎜ e 1
⎝
y/a
−
α1 α
e
α2
2
ξ y/a
(1)
⎞
⎟⎟cos(ξ x / a )dξ
⎠
⎛ α η x / b α1 α 2 η x / b ⎞
⎟⎟cos(η y / b)dη
D (η )⎜⎜ e 1
e
−
α
2
⎠
⎝
σ xy ( x, y ) =
+
0
⎛
α ξ
C (ξ )⎜⎜ α12 e 1
⎝
⎛
α α η x/b ⎞
α η x/b
⎟⎟cos(η y / b)dη
D(η )⎜⎜α12 e 1
− α 22 1 e 2
α
2
⎝
⎠
σ yy ( x, y ) = −
−
∞
1
π∫
1
π∫
∞
0
∞
0
C (ξ )(α1e
α1 ξ y / a
+ α 2e
α2 ξ y / a
(2)
)sin(ξ x / a )dξ
⎡ ⎛ 2
⎛
⎞
α
α ξ
α1 ξ y / a
− α 22 1 C (ξ )eα2 ξ y / a ⎟ − S12 ⎜⎜ C (ξ )e 1
⎢ S11 ⎜ α1 C (ξ )e
α
⎠
2
⎣ ⎝
⎝
(3)
y/a
−
α1
α
⋅ C (ξ )e
α2
2
ξ y/a
⎞⎤
⎟⎟⎥ ξ sin(ξ x / a ) dξ
⎠⎦
⎡ ⎛ 2
⎞⎤
⎛
α
α
α1 η x / b
α η x/b ⎞
⎟⎟ − S12 ⎜⎜ D (η )eα1 η x / b − 1 ⋅ D (η )eα 2 η x / b ⎟⎟⎥ η sin(η y / b)dη
− α 22 1 D (η )e 2
⎢ S11 ⎜⎜ α 1 D (η )e
α2
α2
⎠
⎠⎦
⎝
⎣ ⎝
⎯ 395 ⎯
(4)
u y ( x, y ) =
+
1
π∫
∞
0
1
π∫
∞
11
0
{ [S α
{ [S α
3
11 1
3
1
− ( S12 + S 66 )α1 ]e
− ( S12 + S 66 )α1 ]e
α1 η x / b
α1 ξ y / a
+ [ S11α 23 − ( S12 + S 66 )α 2 ]e
− [ S11α 13 − ( S12 + S 66 )α 2 ]e
α2 η x / b
(−
α2 ξ y / a
(−
α1 ⎫
) ξ C (ξ )cos(ξ x / a ) dξ
α 2 ⎬⎭
α1 ⎫
) η D (η )cos(η y / b) dη
α 2 ⎬⎭
(5)
The key point is to solve the coefficients C (ξ ) and D(η ) in the equations of σ and u .
THE SOLUTIONS OF THE COEFFICIENTS C (ξ ) AND D(η ) IN
σ
AND
u
In order to obtain the coefficients C (ξ ) and D(η ) , it needs to solve Weber-Schafheislin discontinuous integral
equation [3]. By the use of variable substitution the Abel style integral equations which contain a(t) and b(t) can then
be obtained. Here a(t) and b(t) are relevant coefficients of C (ξ ) and D(η ) . In order to obtain a(t) and b(t) it is
necessary to transfer the Abel style integral into the second kind Fredholm integral equation [4]:
a (t ) +
=
1
4α 2
αa(α )dα
∫
π (α 2 − α1 ) 0
2
∫
∞
0
⎤
⎤ ⎡
⎡
α1
1
1
1
du
⋅
− 2
−
⎢ 2 2
2 2
2
2 2 ⎥ ⎢ 2
2 2
2 2⎥
⎢⎣α 2 (u + α 2 t ) α1 (u + α1 t ) ⎥⎦ ⎢⎣α + α1 u α + α1 u ⎥⎦
t f (u )
2a 2 t
du
∫
π (α 2 − α 1 ) 0 t 2 − u 2
(6)
It is seen that the equation (6) is the following style of the second kind of Fredholm integral equation:
1
a (t ) + ∫ L(t , a)a (α )d (α ) = g (t )
(7)
0
The integral nuclear function L(t, a) is given by
⎛ 1 α1 ⎞
⎛ 1 α2 ⎞
2α1α
2α 2α
⎜
⎜
− ⎟⎟ −
− ⎟
2
2 2 ⎜
2
2 2 2 ⎜
πα 2 (α 2 − α1 )(α − α 2 t ) ⎝ α 2t α ⎠ πα1 (α 2 − α1 )(α − α1 α 2 t ) ⎝ α1t α ⎟⎠
L(t , a) =
−
⎛ 1 α1 ⎞
⎛ 1 α1 ⎞
2α1α
2α 2α
⎜
⎜
− ⎟⎟ +
− ⎟
2
2 2 2 ⎜
2
2 2 2 ⎜
πα 2 (α 2 − α1 )(α − α1 α 2 t ) ⎝ α 2t α ⎠ πα1 (α 2 − α1 )(α − α1 α 1 t ) ⎝ α1t α ⎟⎠
1
a (t ) + ∫ L(t ,α )a (α ) dα = g (t ) ，
0
g (t ) =
0＜ t ≤1
(8)
(9)
t f (u )du
2α 2 t
∫
π (α 2 − α1 ) 0 t 2 − u 2
(10)
THE SOLUTION a(t) OF THE SECOND KIND OF FREDHOLM EQUATION
By using B spline function the best approximation a (t ) of the solution a(t) of Fredholm integral equation is
a (t ) =
2
∑b B
j = −3
j
j ,4
(t )
(11)
where B j ,4 (t ) is the third order B spline function.
The coefficient b(t) can also be obtained similarly. By the use of Abel integral equation we can get b(t):
b(t ) =
2 d t ydy
π dt ∫ 0 t 2 − y 2
⎧⎪ 1
⎡
1
1
1
⎢
a
u
⋅
−
(
)
⎨
∫
0
⎢⎣ u 2 + α 12 y 2
u 2 + α 22 y 2
⎪⎩α 2 − α 1
⎯ 396 ⎯
⎤ ⎫⎪
⎥ du ⎬
⎥⎦ ⎪⎭
(12)
Substituting a(t) and b(t) into the expressions of C (ξ ) and D(η )
1
C (ξ ) = π p0 ∫ ξ a(t ) J 0 (ξ t )dt ⎫
0
⎪
⎬
∞
D(η ) = π p0 ∫ η b(t ) J 0 (η t )dt ⎪
0
⎭
(13)
and then substituting the results of C (ξ ) and D(η ) into equations (1) to (5), the stress field σ and displacement field
u of the orthogonal anisotropic plate with edge cracks are derived.
REFERENCES
1.
Tian ZR. Mathematic mechanical methods of fibre composite. National Defence Press, 2004.
2.
Tian ZR. The boundary element method of composite material. Xi’an, Northwestern Polytechnical University
Press, 1992.
3.
Tian ZR. Mathematische Methoden zur Bruchmechanik von Verbundenwerkstoffen. Berlin, Germany, 1995 (in
German).
4.
Erde’lyi A. Higher transcendental functions. McGraw-Hill, New York, USA, 1959.
⎯ 397 ⎯