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A simple model for the contact problem of a finite cracked plate in bending
0013-7944/88 s3.00+.00 @ 1988 Pergamon Journals Ltd
En@eering Fnuhme Mechanics Vol. 29, No. 2, pi. 227-231,1988 Printed in Great Britain.
A SfMPLE MODEL FOR THE CONTACT PROBLEM OF A FINITE CRACKED PLATE IN BENDING C. W. WOO, Y. K. CHEUNC, Y. Z. CHEN and Y. H. WANG? University of Hong Kong, Hong Kong Abstract-In bending of a finite cracked plate, the stress intensity factor (SfF) at a cracked tip is usually negative. This means that the displacement at the vicinity of the crack tip is overlapped. However, this is not reasonable. In this paper a simple model is suggested. When the SIF is negative, some part of the crack will close. By using a coflocation method the length of the closed part of the crack, and the SIF value of the other crack tip have been calculated. The results are useful for engineering practice.
1. INTRDDUCTION To DATE, many works have done in the field of 2-D fracture analysis. Isida[l’J studied a crack problem in a finite width sheet which is subjected to a uniform moment. Later, Benthem ef al. [2] studied this problem, too. In their results, the SIF at a crack tip is negative. Clearly, the negative SIF at a crack tip means that the displacements at the vicinity of crack tip are overlapped. This is not possible in continuum mechanics, so we propose a model to deal with this problem. 2. FORMULATIONS
OF THE PROBLEM AND ANALYSIS
It is assumed that, on both ends of rectangular cracked plate a pair of moments is applied (see Fig. 1). The sheet of width 2b and height 2h contains a crack of length 2a located centrally. The SIF at the crack tips A and B are equal to each other, but with different sign, i.e. KIA > 0. To solve the problem the basic assumption in this paper is proposed as follows: If Kt value of crack tip B is negative, then some part of crack surface BBt is closed, BBt becomes an inner part of continuum body. Then the new crack tip is located at point B1. Again, if the K1 value of crack tip B1 is negative, then some part of crack will be closed and BIB2 becomes an inner part of continuum body as shown in Fig. 2. We can repeat the calculation and get a series of crack tips Bj (i = 1,2,3, . . .). Finally, when the KIB value at the right crack tip equals zero, the K tA value can be determined, which is need in practice.
Fig. I. tTo whom correspondence
should be addressed. 227
228
Fig. 2.
The proposed problem can be solved by the use of complex formulations of plane elasticity and the collocation method. In plane elasticity, the stresses, resultant forces and displacements are given by the following formulae[3]
uxx*+ uyy= 4 Re a’(z)
(1)
. =@(z)+R(F)+(z-$D’(Z) l~xxx
UYY -
(2)
-
-Y+iX=[+(z)+w(.f)+(z-Z)@(z)]-C c = [q/(z) + w(Z) + (z - z)@~]*=Z,, 2G(u+iu)=
K+(Z)-
(3) (4)
w(Z)-(z-fz)@m
where $(z), a(z) = q’(z), w(z) and a(z) = w’(z) are Muskhelishvili’s potentials. G is the shear modulus of elasticity, K = 3 - 4v (in plane strain conditions) or K = [(3 - u)/(l + v)] (in plane stress conditions), v is Poisson’s ratio. X and Y are the components of total forces per unit thickness, exerted upon the left hand-side by right-hand across an arbitrary path leading to the point 2(x, y) from some point zO(xO,yO). The traction-free condition leads to the following expansion[4] (b,(z)=
dF-2
f EKZK-‘+ f
w(z)
=m
E EKzK-’
K=l
FKZK
K=l
K=l
- f
~~~~
(5)
K=l
where the coefficients EK and FK are real constants. From eqs (5) and (3), we can get the governing equations of the undetermined coefficients EK and FK, and this linear system can be used in place of the boundary collocation technique. After the linear equation system is solved, we insert the coefficients EK obtained into equations, KIB = Lim 2-@(z) Z-Cl KIA = Lim 2ixkGfZGjQ(z) z-+-(1 then the SIF at crack tips can be calculated.
(6)
Finite cracked plate in bending
229
SIF 2ai_l 2a. . . 1
Fig. 3.
If some part of the crack is closed, then we only need to consider the other part of the crack. For example, when the BBi is closed in Fig. 2. We consider the crack length as ABi = 2a1, and instead of a use al in the above equation and move the coordinate origin to the middle of ABi. In later calculations we use this method in the same way. If we determine the SIF value KfB’ according to the crack length aim1in the i - 1 step, and K& according to crack length ai in the i step, then we can calculate the approximate crack length equal to zero. This is shown in Fig. 3; the ai+l is determined ai+i in the next step according to K fB+* by
Kfi’Ui - Kf*Ui-1 ai+l
=
Kf~l
-KfB
.
(7)
According to the crack length a,,+i, we can calculate the actual value of Kf;‘. If it is not equal to zero, we do the procedure again, until KIB differs from zero by less the given error. 3. NUMERICAL
RESULTS
Because of the symmetry of the x-axis, we only need to consider the upper part of the plate in the calculation. Several groups of specimens with different u/b and h/b ratios have been calculated. The number of boundary collocation points is taken as 31, the number of unknown coefficients (&, FK, K = 1,. . . , 10) is 20. So that 62 simultaneous’ equations are used to determine the coefficients by the Least Squares Method. First, we calculate KIB and KIA at both crack tips of Fig. 1. The SIF value can be expressed as follows:
The values of F(h/b, u/b) are listed in Table 1. From the above table we can see that KIB at crack tip B shown in Fig. 1 is negative, the other is positive. If we do not permit computer error, the absolute values at the two tips are the same. In refs [2,4] there are results of infinite crack strip in bending. Usually when h/b = 3, the specimen can be considered an infinite strip. From Table 1 we see that the results by the collocation method coincide with that of refs [2,4], so that the collocation method can be said to have enough accuracy. Then we want to determine the location of the right crack tip where KIB = 0, and calculate the KIA value for that crack length. It is enough by 10 times iteration to get the accuracy to above 10m5.At last we get the length a2 and SIF K IA. At the other point BL, KIe = 0, and the stuck crack length is BBc = 2x(u - UL). The half effective crack length uL and FiA value (KIA = FIA CT&) are listed in Table 2. From the table we can see that in the bending problem of a symmetric crack plate some part of the crack will close, so that the effective length of crack will be shorter than before, it was reduced usually by about 30%, and KIA will be larger than before, its value increased by about more than 30%. They are compared in Fig. 4.
I
3 2
hlb
0.06671 0.06672 0.~680
a_/b
0.06669 0.06672 0.06630
F;A
0.2 L
0.1338 0.1338 0.1344
aJb
S/r 0.1339 0.1338 0.1344
0.2
L
I.
0.1506
0.2015 0.2016 0.2036
atlb
0.3
0.2019 0.2020 0.2038
F‘A
Table
-0.150s -0.1503 0.1505 0.1502 -0.1508 0.1507 0.1508
R
0.3
tip, L that at the left tip.
0.1002
-0.1001 -0.1 0.1001 -0.1001 0.1001 0.1OOt
R
F at right crack
0.1
R means the value
0.0500
ref. [4]
L
-0.0500 0.0500 0.0496 -0.0500 0.0500 0.0500
R
: ref.+*]
h/b
0.1
Table
R
0.4
alb L
0.2703 0.2705 0.2749
adb
0.4 6,
values
0.2725 0.2724 0.2760
U/b
2. aL and F,,
0.2022
R
0.5 L
0.3407 0.3411 0.3488
at16
0.5 FtA
0.3468 0.3472 0.3529
0.2570
-0.2562 -0.2566 0.2569 0.2564 -0.2589 0.2599 0.2589
shown in eq. (8)
-02020 -0.2021 0.2022 -0.2031 0.2032 0.2031
F-values
0.6 L
0.4135 0.4141 0.4255
alJb
0.6 F IA
0.4300 0.4311 0.4379
0.3180
-0.3168 -0.3178 0.3178 0.3171 -0.32180.3259 0.3220
R
0.4895 0.4907 0,5052
al/b
0.7
0.5293 0.5338 0.5403
0.3902
L
~.4020 0.3958 0.4002
F;A
0.7
-0.3937 1 -0.39X -0.3992 0.4112
R
Finite cracked plate in bending
231
F
0.5
.-
F value, Table 1
I--_
FIA value, Table 2
0.4
0.3
0.2
0.1
0.0
0.1
0.2
0.3
-%
0.4
0.5
0.6
0.7
Fig. 4. Comparison of F-values.
4. CONCLUSION (1) In the bending problem, the crack usually closes at one end, so that the usual calculation by which the SIF is negative is not suitable. Only in combining the situation of tension and bending and there is no closure, the method is correct. But for a pure bending problem in practice, and in some combined situations where the tension is not enough to overcome crack closure, we should use the method in this paper to analyse the problem. (2) In a bending problem if there is negative SIF, some part of crack will close and the SIF at other crack tip will become larger. By the model of this paper, the increase in SIF value is accurately predicted. (3) The collocation method not only can be used to solve tensile problems[5], but also used to solve bending problems. This method is rather simple but effective and has sufficient accuracy. When tensile loading and moment act on the body together, it can be analysed in the same way. Acknowledgement-The
financial support of the Croucher Foundation is gratefully acknowledged.
REFERENCES [l]M. Isida, Jqan Sot. Mech. Engrs 22, 345 (1971). [2] J. P. Benthem and W. T. Koiter, Asymptotic approximations to crack problems. In Mechanics ofFractureI (Edited by G. C. Sih). Noordhoff, Holland (1973). [3] N. I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity. Noordhoff, Holland (1953). [4] D. P. Rooke and D. J. Cartwright, Compendium of Stress Intensity Factors. HMSO, London (1973). [5] Chen Yi-zhou, An investigation of the stress intensity factor for a finite internally cracked plate by using the variational method. Int. J. Fracture Mech. 17, 387-394 (1983). (Receioed
19 November
1986)
En@eering Fnuhme Mechanics Vol. 29, No. 2, pi. 227-231,1988 Printed in Great Britain.
A SfMPLE MODEL FOR THE CONTACT PROBLEM OF A FINITE CRACKED PLATE IN BENDING C. W. WOO, Y. K. CHEUNC, Y. Z. CHEN and Y. H. WANG? University of Hong Kong, Hong Kong Abstract-In bending of a finite cracked plate, the stress intensity factor (SfF) at a cracked tip is usually negative. This means that the displacement at the vicinity of the crack tip is overlapped. However, this is not reasonable. In this paper a simple model is suggested. When the SIF is negative, some part of the crack will close. By using a coflocation method the length of the closed part of the crack, and the SIF value of the other crack tip have been calculated. The results are useful for engineering practice.
1. INTRDDUCTION To DATE, many works have done in the field of 2-D fracture analysis. Isida[l’J studied a crack problem in a finite width sheet which is subjected to a uniform moment. Later, Benthem ef al. [2] studied this problem, too. In their results, the SIF at a crack tip is negative. Clearly, the negative SIF at a crack tip means that the displacements at the vicinity of crack tip are overlapped. This is not possible in continuum mechanics, so we propose a model to deal with this problem. 2. FORMULATIONS
OF THE PROBLEM AND ANALYSIS
It is assumed that, on both ends of rectangular cracked plate a pair of moments is applied (see Fig. 1). The sheet of width 2b and height 2h contains a crack of length 2a located centrally. The SIF at the crack tips A and B are equal to each other, but with different sign, i.e. KIA > 0. To solve the problem the basic assumption in this paper is proposed as follows: If Kt value of crack tip B is negative, then some part of crack surface BBt is closed, BBt becomes an inner part of continuum body. Then the new crack tip is located at point B1. Again, if the K1 value of crack tip B1 is negative, then some part of crack will be closed and BIB2 becomes an inner part of continuum body as shown in Fig. 2. We can repeat the calculation and get a series of crack tips Bj (i = 1,2,3, . . .). Finally, when the KIB value at the right crack tip equals zero, the K tA value can be determined, which is need in practice.
Fig. I. tTo whom correspondence
should be addressed. 227
228
Fig. 2.
The proposed problem can be solved by the use of complex formulations of plane elasticity and the collocation method. In plane elasticity, the stresses, resultant forces and displacements are given by the following formulae[3]
uxx*+ uyy= 4 Re a’(z)
(1)
. =@(z)+R(F)+(z-$D’(Z) l~xxx
UYY -
(2)
-
-Y+iX=[+(z)+w(.f)+(z-Z)@(z)]-C c = [q/(z) + w(Z) + (z - z)@~]*=Z,, 2G(u+iu)=
K+(Z)-
(3) (4)
w(Z)-(z-fz)@m
where $(z), a(z) = q’(z), w(z) and a(z) = w’(z) are Muskhelishvili’s potentials. G is the shear modulus of elasticity, K = 3 - 4v (in plane strain conditions) or K = [(3 - u)/(l + v)] (in plane stress conditions), v is Poisson’s ratio. X and Y are the components of total forces per unit thickness, exerted upon the left hand-side by right-hand across an arbitrary path leading to the point 2(x, y) from some point zO(xO,yO). The traction-free condition leads to the following expansion[4] (b,(z)=
dF-2
f EKZK-‘+ f
w(z)
=m
E EKzK-’
K=l
FKZK
K=l
K=l
- f
~~~~
(5)
K=l
where the coefficients EK and FK are real constants. From eqs (5) and (3), we can get the governing equations of the undetermined coefficients EK and FK, and this linear system can be used in place of the boundary collocation technique. After the linear equation system is solved, we insert the coefficients EK obtained into equations, KIB = Lim 2-@(z) Z-Cl KIA = Lim 2ixkGfZGjQ(z) z-+-(1 then the SIF at crack tips can be calculated.
(6)
Finite cracked plate in bending
229
SIF 2ai_l 2a. . . 1
Fig. 3.
If some part of the crack is closed, then we only need to consider the other part of the crack. For example, when the BBi is closed in Fig. 2. We consider the crack length as ABi = 2a1, and instead of a use al in the above equation and move the coordinate origin to the middle of ABi. In later calculations we use this method in the same way. If we determine the SIF value KfB’ according to the crack length aim1in the i - 1 step, and K& according to crack length ai in the i step, then we can calculate the approximate crack length equal to zero. This is shown in Fig. 3; the ai+l is determined ai+i in the next step according to K fB+* by
Kfi’Ui - Kf*Ui-1 ai+l
=
Kf~l
-KfB
.
(7)
According to the crack length a,,+i, we can calculate the actual value of Kf;‘. If it is not equal to zero, we do the procedure again, until KIB differs from zero by less the given error. 3. NUMERICAL
RESULTS
Because of the symmetry of the x-axis, we only need to consider the upper part of the plate in the calculation. Several groups of specimens with different u/b and h/b ratios have been calculated. The number of boundary collocation points is taken as 31, the number of unknown coefficients (&, FK, K = 1,. . . , 10) is 20. So that 62 simultaneous’ equations are used to determine the coefficients by the Least Squares Method. First, we calculate KIB and KIA at both crack tips of Fig. 1. The SIF value can be expressed as follows:
The values of F(h/b, u/b) are listed in Table 1. From the above table we can see that KIB at crack tip B shown in Fig. 1 is negative, the other is positive. If we do not permit computer error, the absolute values at the two tips are the same. In refs [2,4] there are results of infinite crack strip in bending. Usually when h/b = 3, the specimen can be considered an infinite strip. From Table 1 we see that the results by the collocation method coincide with that of refs [2,4], so that the collocation method can be said to have enough accuracy. Then we want to determine the location of the right crack tip where KIB = 0, and calculate the KIA value for that crack length. It is enough by 10 times iteration to get the accuracy to above 10m5.At last we get the length a2 and SIF K IA. At the other point BL, KIe = 0, and the stuck crack length is BBc = 2x(u - UL). The half effective crack length uL and FiA value (KIA = FIA CT&) are listed in Table 2. From the table we can see that in the bending problem of a symmetric crack plate some part of the crack will close, so that the effective length of crack will be shorter than before, it was reduced usually by about 30%, and KIA will be larger than before, its value increased by about more than 30%. They are compared in Fig. 4.
I
3 2
hlb
0.06671 0.06672 0.~680
a_/b
0.06669 0.06672 0.06630
F;A
0.2 L
0.1338 0.1338 0.1344
aJb
S/r 0.1339 0.1338 0.1344
0.2
L
I.
0.1506
0.2015 0.2016 0.2036
atlb
0.3
0.2019 0.2020 0.2038
F‘A
Table
-0.150s -0.1503 0.1505 0.1502 -0.1508 0.1507 0.1508
R
0.3
tip, L that at the left tip.
0.1002
-0.1001 -0.1 0.1001 -0.1001 0.1001 0.1OOt
R
F at right crack
0.1
R means the value
0.0500
ref. [4]
L
-0.0500 0.0500 0.0496 -0.0500 0.0500 0.0500
R
: ref.+*]
h/b
0.1
Table
R
0.4
alb L
0.2703 0.2705 0.2749
adb
0.4 6,
values
0.2725 0.2724 0.2760
U/b
2. aL and F,,
0.2022
R
0.5 L
0.3407 0.3411 0.3488
at16
0.5 FtA
0.3468 0.3472 0.3529
0.2570
-0.2562 -0.2566 0.2569 0.2564 -0.2589 0.2599 0.2589
shown in eq. (8)
-02020 -0.2021 0.2022 -0.2031 0.2032 0.2031
F-values
0.6 L
0.4135 0.4141 0.4255
alJb
0.6 F IA
0.4300 0.4311 0.4379
0.3180
-0.3168 -0.3178 0.3178 0.3171 -0.32180.3259 0.3220
R
0.4895 0.4907 0,5052
al/b
0.7
0.5293 0.5338 0.5403
0.3902
L
~.4020 0.3958 0.4002
F;A
0.7
-0.3937 1 -0.39X -0.3992 0.4112
R
Finite cracked plate in bending
231
F
0.5
.-
F value, Table 1
I--_
FIA value, Table 2
0.4
0.3
0.2
0.1
0.0
0.1
0.2
0.3
-%
0.4
0.5
0.6
0.7
Fig. 4. Comparison of F-values.
4. CONCLUSION (1) In the bending problem, the crack usually closes at one end, so that the usual calculation by which the SIF is negative is not suitable. Only in combining the situation of tension and bending and there is no closure, the method is correct. But for a pure bending problem in practice, and in some combined situations where the tension is not enough to overcome crack closure, we should use the method in this paper to analyse the problem. (2) In a bending problem if there is negative SIF, some part of crack will close and the SIF at other crack tip will become larger. By the model of this paper, the increase in SIF value is accurately predicted. (3) The collocation method not only can be used to solve tensile problems[5], but also used to solve bending problems. This method is rather simple but effective and has sufficient accuracy. When tensile loading and moment act on the body together, it can be analysed in the same way. Acknowledgement-The
financial support of the Croucher Foundation is gratefully acknowledged.
REFERENCES [l]M. Isida, Jqan Sot. Mech. Engrs 22, 345 (1971). [2] J. P. Benthem and W. T. Koiter, Asymptotic approximations to crack problems. In Mechanics ofFractureI (Edited by G. C. Sih). Noordhoff, Holland (1973). [3] N. I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity. Noordhoff, Holland (1953). [4] D. P. Rooke and D. J. Cartwright, Compendium of Stress Intensity Factors. HMSO, London (1973). [5] Chen Yi-zhou, An investigation of the stress intensity factor for a finite internally cracked plate by using the variational method. Int. J. Fracture Mech. 17, 387-394 (1983). (Receioed
19 November
1986)