PSI - Issue 52
Akihide Saimoto et al. / Procedia Structural Integrity 52 (2024) 323–339 A. Saimoto et al. / Structural Integrity Procedia 00 (2023) 000–000
330
8
faces of the crack can be calculated by using results by Sih et al. (1965) as, ∆ u ( x ) = 2 c 11 ℑ µ 1 µ 2 p 0 + ( µ 1 + µ 2 ) q 0 √ b 2 − x 2 √ b 2 + p 0 +
(58)
∆ v ( x ) = − 2 c 22 ℑ
1 µ 2
q 0 µ 1 µ 2
√ b 2
1 µ 1
11 ℑ µ 1 µ 2 ( µ 1 + µ 2 ) p 0 + µ 1 µ 2 q 0
− x 2 = 2 c
− x 2
(59)
where the operator ℑ stands for an imaginary part of the argument. Eq.(59) can be easily concluded using a following relation. µ 1 µ 2 µ 1 µ 2 = (60) which comes from a relationship between solutions and coe ffi cients of the characteristic equation (Lekhnitskii (1968)). c 11 µ 4 − 2 c 16 µ 3 + (2 c 12 + c 66 ) µ 2 − 2 c 26 µ + c 22 = c 11 ( µ − µ 1 )( µ − µ 2 )( µ − µ 1 )( µ − µ 2 ) = 0 (61) By multiplying the magnitude of point force doublets f ( ξ ), g ( ξ ) and h ( ξ ) with the compliance matrix C inEq.(6), the relative displacement between the upper and lower crack faces is obtained. Therefore, the density function of point force doublets f ( ξ ), g ( ξ ) and h ( ξ ) that produce the identical relative displacements in Eqs.(58) and (59) are obtained in the following manner as discussed by Aono and Noguchi (2000). 0 ∆ v ( ξ ) ∆ u ( ξ ) = 2 c 11 0 0 ℑ [ µ 1 µ 2 ( µ 1 + µ 2 )] ℑ [ µ 1 µ 2 ] ℑ [ µ 1 µ 2 ] ℑ [ µ 1 + µ 2 ] p 0 q 0 b 2 − ξ 2 = c 11 c 12 c 16 c 12 c 22 c 26 c 16 c 26 c 66 f ( ξ ) g ( ξ ) h ( ξ ) (62) By solving this relation with respect to the density of distributed point force doublets f ( ξ ), g ( ξ ) and h ( ξ ), those of corresponding to the relative displacement between the upper and lower faces of a line crack subjected to uniform internal pressure p 0 or shear stress q 0 in its plane is obtained by combining Eq.(62) with Eqs.(58) and (59) as follows. f I ( ξ ) g I ( ξ ) h I ( ξ ) = 2 c 11 c 11 c 12 c 16 c 12 c 22 c 26 c 16 c 26 c 66 − 1 0 ℑ [ µ 1 µ 2 ( µ 1 + µ 2 )] ℑ [ µ 1 µ 2 ] p 0 b 2 − ξ 2 (63) c 22 c 11
for mode I problem and f II ( ξ ) g II ( ξ ) h II ( ξ ) = 2 c 11
− 1
q 0 b
c 11 c 12 c 16 c 12 c 22 c 26 c 16 c 26 c 66
0 ℑ [ µ 1 µ 2 ] ℑ [ µ 1 + µ 2 ]
2 − ξ 2
(64)
for mode II problem. It should be noted that components of compliance matrix c i j and material parameter µ j in Eqs.(63) and (64) have to be transformed according with the crack inclination. That is, c i j and µ j have to be replaced with those transformed by accounting the k -th crack inclination β k , which will be denoted by adding the superscript k . At the same time, the internal pressure and the shear stresses acting on the crack face are replaced by not a constant but a weighting function W I ( s k ) and W II ( s k ) which may vary with a position on the crack to treat a non-uniform loading along the crack face. Therefore, they are replaced by a function of s k where s k is a local coordinate along the k -th crack line. Accordingly, the complex potential for a point force doublet considering a fundamental property of the relative displacements on the crack line becomes,
11 π i
I k ( s k ) + B 0 k − s k (cos β k + µ j sin β k ) b 2 II j ( β k ) W II k ( s k )
B I
j ( β k ) W
c k
b k
2 k ds k , ( j = 1 , 2; k = 1 , 2 , · · · , N )
Φ j ( z j ) = −
k − s
(65)
z j − ζ
− b k
ζ 0 k is a complex number expressing a central point of the k -th crack, b k is a half length of the k -th crack and N is a total number of cracks. Constants B I j ( β k ) and B II j ( β k ) are defined as follows.
( a j 1 cos β k + a j 2 sin β k )(cos β k + µ j sin β k ) ( − a j 1 sin β k + a j 2 cos β k )( − sin β k + µ j cos β k ) a j 1 ( µ j cos2 β k − sin2 β k ) + a j 2 (cos2 β k + µ j sin2 β k ) T d k 12 d k 16 d k 22 d k 26 d k 26 d k 66 ℑ [ µ k 1 µ k 2 ( µ k 1 + µ k 2 )] ℑ [ µ k 1 µ k 2 ]
j ( β k ) =
B I
(66)
Made with FlippingBook Annual report maker