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A dugdale problem for a finite internally cracked plate
TECHNICAL NOTE A DUGDALE PROBLEM FOR A FINITE INTERNALLY CRACKED PLATE 1. INTRODUCTION ITISwellknownf I],that the Dugdale model in plastic analysis of an infinite cracked plate has been proposed early in 1960.This model not only has a theoretical significance but also an engineering interest. For example. in the plastic analysis of a thin cracked plate, using this model we can approximately estimate the plastic zone length on the crack tip and the COD value on the crack border. In an infinite plate containing a linear crack with 2a length (Fig. I), using the boundIess condition of the stress at the crack tip, one can get the following two relations 12,p. 2641 (I) ?GLVN(C. O*)= GSfN= ur( 1t ~)a cos @(In(set e))/n
(2)
0 = arc cos(c/n)
(3)
where
and Sru is the remote tension stress. UT is the cohesive or yield stress. a is the half length of a crack, r is the plastic zone length, c = a - r, Zcp~(c. 07 = SIN is the crack opening displacement. K= (3 - v)i(t + v), G is the shear modulus of elasticity, Y is the Poisson’s ratio. Nevertheless, up to now there is no solution concerning the Dugdaie problem in the case of a finite cracked plate subject to the uniaxial tension. In this paper we divide the proposed problem relating to tin&e cracked plate into three boundary value problems. Two of them belong to a problem of a finite internal,, cracked plate, which can be solved easily by the method proposed by us I3.41 (See Appendix A). The another one belongs to a problem of an infinite internally cracked plate, which has a solution in a closed form (See Appendix B). Finally, the counterpart of eqns (I) and (2), i.e. the solution corresponding to the proposed problem can be derived. In the case of a square internally cracked plate, the final numerical results can be seen from Tables 1, 2 and Fig. 3.
2. ANALYSIS AND NUMJBICAL RESULTS Suppose a square intern~iy cracked plate containing a crack with 2a length is subjected to uniaxial tension stress S (Fig. 2). Then, under the action of the cohesive stress (TTalong a part of crack border the corresponding plastic zone length is r as shown in Fig. 2. Obviously, we can separate the proposed problem into three boundary value probiems. namely I. II and III, as shown in Fig. 2. Clearly, the problem I belongs to one relating to a finite internally cracked plate (Fig. 2), which can be solved by using the eigenfunction expansion form and the variational method[3,41 (See Appendix A). We write the calculated results of stress intensity factor K, , on the crack tip and COD value CI of a point (C, 0’) on the crack border as K:J = rci.i~\/f(~ff)l
(4)
ZGo,(C,0’) = 6,.
(5)
The problem II belongs to a problem relating to an in~nite cracked plate subject to some tractions acting on the crack border (Fig. 2). which has a solution in a closed form (see Appendix B). The calculated results of SIF and COD can be expressed as follows h’l.lf = 2ma/7;)1
(6)
2Gcn(c, 0’) = &I = (I+ K)~(~sin 0 - cos 8 In (set @))/n
(2)
I sjnr 0 i
Fig. I. Dugdale’s model for an infinite cracked plate. 579
Technical Note
InfInite
-
-
plate
~
Et
II
Fig. 2. A scheme for explaining how the original problem is divided into three boundary value problems 1, II and III. where 8 = arc cos(c/a) as before. In addition, after the problem II is solved, one needs to calculate the tractions on the dash line contour in the problem If (Pig. 2). These tractions can be calculated by using the equations shown in the Appendix B. Certainly, these tractions will soon become the external force acting on the contour of cracked plate in the problem III. Problem III can be solved by using the technique as men~oned in problem I, since the external Sunday tractions has been already derived. As before, we express the obtained numerical results as K1.t = R,,11/1W)l
(8)
2G~r(c. 0’) = &ir
(9)
Suppose all the numerical results shown in eqns (4)-(9) are obtained, then we can use the boundless condition of stress at the crack tip as follows Ki = SK1.f -
UT(g1.U
-
K,.lUf
= 0.
t 10)
Substituting eqns (4), (6) and (8) into above equations we get s (21n)-_ li,.,,, -= K!.l UT
(11)
Table 1. fl-Values shown in ?qn (12) cla or ria
0.9 0.1
a/b =O.l a/b =0.2 a/b =0.3
0.98843 0.95531 0.90587 0.84668 0.78367 0.72035
a/b =0.4
a/b =0.5 a/b =0.6 a/b =0.7 a/b =0.8
a/b =0.9
0.8 0.2
0.99009 0.96160 0.918% 0.86751 0.81258 0.75753 0.65591 0.70261 0.58269 0.64305 0.483% 0.56803
0.7 0.3
0.6 0.4
0.5 0.5
0.4 0.6
0.3 0.7
0.2 0.8
0.1 0.9
0.99166 0.96760 0.93139 0.88762 0.84067 0.79371 0.74763 0.69970 0.64317
0.99315 097328 0.94330 090691 0.86776 0.82865 0.79081 0.75283 0.71076
0.99455 0.97864 0.95458 0.92528 0.89367 0.86214 0.831% 0.80257 0.77192
0.99585 0.98367 0.96520 0.94264 0.91826 0.89395 0.87088 0.84897 0.82743
0.99707 0.98834 0.97510 0.95889 0.94134 0.92386 0.90737 0.89200 0.87786
0.99818 0.99265 0.98424 0.97395 0.96278 0.95165 0.94121 0.93161 0.92357
0.99920 0.99657 0.99258 0.98770 0.982~ 0.97712 0.97218 0.%768 0.96477
581
Technical Note Comparing above equation with the eqn (l), we can rewrite eqn (11) as
where the cafculated fr values are shown in Table 1. In addition, using the eqns (5) (7), (9) and (12) the COD value on the point (c. 0’) can be expressed as 2Gv(c, O+)= S6r - (TT(8rr - &II)
(13)
ZGu(c,O+)= &jr . (29) ’ i%/n-(l + n)a(e sin e - cos BIn (set e))ln + 6rn).
(14)
or
Comparing above equation with eqn (2), we can rewrite eqn (14) as 2Gu(c, 0’) = fi(;.;)
.‘LGvr~(c,0”)
or ZGu(c,O+)= f?(;,
;)OT(I
+
K)tl
COS
tJ(tn(set e))/rr
(15)
where the calculated fi values are shown in Table 2. Thus, the Tables 1 and 2 can be considered as a tinal numerical results investigated. In all these calculations, 2M = 12 terms are truncated in the eigenfun~t~onexpansion form in Appendix A, which can be considered as a responsible value from our previous experience[4].
Table 2. fa-Values shown in eqn (15) da or r/a
0.9 0.1
a/c=O.l
l.oOoo Loo05 1.0023 1.0066 1.0144 1.0260 1.0398 1.0502 1.0418
a/b = 0.2
a/b = 0.3 a/b = 0.4 a/b = 0.5 a/b = 0.6 a/b = 0.7 a/b = 0.8 a/b= 0.9
0.8 0.2 1.0001 l.GQO8 1.0039 1.0115 1.0255 1.0471 1.0754 1.1039 1.1144
01
0.7 0.3
0.6 0.4
l.oaol 1.0010 1.0050 1.0148 1.0333 1.0628 1.1035 1.1509 1.1888
Loo01 1.0011 1.0055 1.0166 1.0378 1.0725 1.1224 1.1853 1.2492
02
03
04
0.5 0.5 1.oGil1 1.0012 1.0057 1.0170 1.0393 1.0764 1.1314 1.2043 1.2g76
05
06
0.4 0.6
0.3 0.7
l.C& 1.0011 1.0054 1.0164 1.0381 1.0748 1.1306 1.2070 1.3006
l.Wol 1.0010 1.0049 1.0148 1.0346 1.0685 1.1208 1.1941 1.2883
07
08
0.2 0.8
09
l.oool l.OOO8 1.0040 1.0124 1.0291 I .OSSO 1.1030 I.1617 1.2529
0.1 0.9 l.OMO 1.0006 1.0030 1.0092 1.0217 1.0435 1.0776 1.1266 1.1%5
I
Fig. 3. A relation between the plastic zone length ratio (r/n) and the applied stress ratio (S/ur). EFM Vol. 17. No. &F
Technical Note As an another comparison with the numerical results corresponding to the case of a infinite cracked plate. we rewrite eqn (12) as
Note that (r ii $a = 1(Fig. 2). then from above equation one can get the relative plastic zone length r/u as
follows
where fd-values are shown in Fig. 3. From this figure we see that the curve corresponding to the case a/h = 0. I is very like the curve shown in Kef.Ill. It is the aim of this paper for us to get necessary numerical results rather than a detailed explanation of these results. This is the end of our paper. ‘4~kno~~Iedgemenfs-Theauthor is greatly indebted fo the ~omputjng center of ~ortbwestern Polytecbni~al university for carmine out computations. The author also wishes to thank Mrs. Cheng Yi-shan and Lin Ju-lie for their encouragement and some help to complete this paper.
D. S. Dugdale, Yielding of steel sheets containing slit. .I Mech. Php. Solids, g(2), 100-104(1960). J. R. Rice, Mathematjcal analysis in the mechanics of fracture, in Fracture Vol. 2 (Edited by Liebowitz~ (1968). Chen Yi-zhou and Chen Yi-heng. A mixed boundary problem for a finite internally cracked plate, Enpng Fructf~re Mech. 14. 741-753 (1981). Chen Yi-zhou. An investigation of the stress intensity factor on crack tip for a finite internally cracked plate by using variational method. Engng Fracture Me&. 17.387-394 (1983). N. I. Muskhelishvili, Some Basic Prublerns itr the ~a[~~er~aficu~Theory of Elasticity. Noordhoff, Groningen, Holland (19531. (Received 18 January 1982; received for pubIicurion 20 April 1982)
APPENDIX A Sofut~u~for a finite internffItywrackedplate by means of ~uri~tiofiulmef~od Suppose a cracked plate have both geometry and loading symmetrical conditions, for example, as shown in the probiem i or ilI of Fig. 2. In this case, the stress and displacement components can be expressed as follows rr!, t gvr = 4Ra@(Z)
(All
css - & = e(Z) t Q(Z) t (2 - Z)5@
(‘A?)
2G(u + it?)= K(F(Z)- o(Z) -(Z - Z,@?j
(Ail
where a(Z) = w’(Z), Q(Z) = q’(Z), (o(Z) and w(Z) arc analytical function. K = (3- v)/(l+ v) in plane stress condition, Y is the Poisson’s ratio, u, u are the displacement in x, y direction. From the stress free condition on the crack surface and the symmetrical condition mentioned above. we can get the following solution for complex potentials[3,41
where
and XK are real. Furthermore, we define op’, u\“’ as follows err’ deriving from cp’k’(Z),o(“)(Z) by eqns (Al) and (A2) ~:k’ deriving from ~‘~‘(a, o”‘(Z) by eqn (A3). Using a variational method141we get a linear system for XK as follows
(A61
583
Technical Note AIKXK
Aix =Ak, = eoit(K)ntui(0 dS I Br =
Ic
1,2,.
I =
= BI,
,2M
(A7)
f,K=1.2 ,..., 2M
,u/” dS
1=1,2,...,2M
(48)
and Pi are the traction acting on the outer boundary, n, are the direction cosine, c is the outer boundary contour. After the linear svstem (A71is solved. we using the obtained XK the disolacement distribution on the whole cracked ulate can be easily found. In addition, the stress intensity factor on the crack tip-can be expressed as K, = 2v(?r) $
=I
X,‘o’k~‘/~/a.
(A9)
The detailed description of above equations can be found in our previous papers[3,4].
APPENDIX B Solution for problem II in Fig. 2
We take the conformal mapping function z = a(i + [_I)/2
(Bl)
which maps the region exterior to unit circle on l-plane into the region exterior to crack contour on z-plane. It is well known[S], that we can express the stress and displacement components by some complex potentials as follows oxx+ oyyI= 4 Re W3
032)
uyy- 6 + 2&,, = 2[w~S)@U/w’(D+ WY1 -2W + iv) = ~(0 - wO+d5W(0 - W,‘)
(B3) (B4)
where
W) = (4lhP”m - W”(5)(P’(5))/(O’(i))?.
IBS)
Using Muskhelishvili’s method[5], we can get the solution of the complex potentials as follows p([)=&[-yi+(Z-C)ln(s)t(Z+C)ln(s)) 4%) = P(i) - P’(i) at+ where
fl =
Mi2- 1)
036)
arc cos(c/a) and for the particular branch considered ,s~~ ln(([ - e’“)/([ - e-‘@))= 0
,$im+ ln(([ + e”)/([ t ee”)) = 0.
037)
Using above mentioned equations, we can get eqns (6) and (7). In addition, there is no difficulty to obtain the stress distribution along the dash line contour in the problem II as shown in Fig. 2.
(2)
0 = arc cos(c/n)
(3)
where
and Sru is the remote tension stress. UT is the cohesive or yield stress. a is the half length of a crack, r is the plastic zone length, c = a - r, Zcp~(c. 07 = SIN is the crack opening displacement. K= (3 - v)i(t + v), G is the shear modulus of elasticity, Y is the Poisson’s ratio. Nevertheless, up to now there is no solution concerning the Dugdaie problem in the case of a finite cracked plate subject to the uniaxial tension. In this paper we divide the proposed problem relating to tin&e cracked plate into three boundary value problems. Two of them belong to a problem of a finite internal,, cracked plate, which can be solved easily by the method proposed by us I3.41 (See Appendix A). The another one belongs to a problem of an infinite internally cracked plate, which has a solution in a closed form (See Appendix B). Finally, the counterpart of eqns (I) and (2), i.e. the solution corresponding to the proposed problem can be derived. In the case of a square internally cracked plate, the final numerical results can be seen from Tables 1, 2 and Fig. 3.
2. ANALYSIS AND NUMJBICAL RESULTS Suppose a square intern~iy cracked plate containing a crack with 2a length is subjected to uniaxial tension stress S (Fig. 2). Then, under the action of the cohesive stress (TTalong a part of crack border the corresponding plastic zone length is r as shown in Fig. 2. Obviously, we can separate the proposed problem into three boundary value probiems. namely I. II and III, as shown in Fig. 2. Clearly, the problem I belongs to one relating to a finite internally cracked plate (Fig. 2), which can be solved by using the eigenfunction expansion form and the variational method[3,41 (See Appendix A). We write the calculated results of stress intensity factor K, , on the crack tip and COD value CI of a point (C, 0’) on the crack border as K:J = rci.i~\/f(~ff)l
(4)
ZGo,(C,0’) = 6,.
(5)
The problem II belongs to a problem relating to an in~nite cracked plate subject to some tractions acting on the crack border (Fig. 2). which has a solution in a closed form (see Appendix B). The calculated results of SIF and COD can be expressed as follows h’l.lf = 2ma/7;)1
(6)
2Gcn(c, 0’) = &I = (I+ K)~(~sin 0 - cos 8 In (set @))/n
(2)
I sjnr 0 i
Fig. I. Dugdale’s model for an infinite cracked plate. 579
Technical Note
InfInite
-
-
plate
~
Et
II
Fig. 2. A scheme for explaining how the original problem is divided into three boundary value problems 1, II and III. where 8 = arc cos(c/a) as before. In addition, after the problem II is solved, one needs to calculate the tractions on the dash line contour in the problem If (Pig. 2). These tractions can be calculated by using the equations shown in the Appendix B. Certainly, these tractions will soon become the external force acting on the contour of cracked plate in the problem III. Problem III can be solved by using the technique as men~oned in problem I, since the external Sunday tractions has been already derived. As before, we express the obtained numerical results as K1.t = R,,11/1W)l
(8)
2G~r(c. 0’) = &ir
(9)
Suppose all the numerical results shown in eqns (4)-(9) are obtained, then we can use the boundless condition of stress at the crack tip as follows Ki = SK1.f -
UT(g1.U
-
K,.lUf
= 0.
t 10)
Substituting eqns (4), (6) and (8) into above equations we get s (21n)-_ li,.,,, -= K!.l UT
(11)
Table 1. fl-Values shown in ?qn (12) cla or ria
0.9 0.1
a/b =O.l a/b =0.2 a/b =0.3
0.98843 0.95531 0.90587 0.84668 0.78367 0.72035
a/b =0.4
a/b =0.5 a/b =0.6 a/b =0.7 a/b =0.8
a/b =0.9
0.8 0.2
0.99009 0.96160 0.918% 0.86751 0.81258 0.75753 0.65591 0.70261 0.58269 0.64305 0.483% 0.56803
0.7 0.3
0.6 0.4
0.5 0.5
0.4 0.6
0.3 0.7
0.2 0.8
0.1 0.9
0.99166 0.96760 0.93139 0.88762 0.84067 0.79371 0.74763 0.69970 0.64317
0.99315 097328 0.94330 090691 0.86776 0.82865 0.79081 0.75283 0.71076
0.99455 0.97864 0.95458 0.92528 0.89367 0.86214 0.831% 0.80257 0.77192
0.99585 0.98367 0.96520 0.94264 0.91826 0.89395 0.87088 0.84897 0.82743
0.99707 0.98834 0.97510 0.95889 0.94134 0.92386 0.90737 0.89200 0.87786
0.99818 0.99265 0.98424 0.97395 0.96278 0.95165 0.94121 0.93161 0.92357
0.99920 0.99657 0.99258 0.98770 0.982~ 0.97712 0.97218 0.%768 0.96477
581
Technical Note Comparing above equation with the eqn (l), we can rewrite eqn (11) as
where the cafculated fr values are shown in Table 1. In addition, using the eqns (5) (7), (9) and (12) the COD value on the point (c. 0’) can be expressed as 2Gv(c, O+)= S6r - (TT(8rr - &II)
(13)
ZGu(c,O+)= &jr . (29) ’ i%/n-(l + n)a(e sin e - cos BIn (set e))ln + 6rn).
(14)
or
Comparing above equation with eqn (2), we can rewrite eqn (14) as 2Gu(c, 0’) = fi(;.;)
.‘LGvr~(c,0”)
or ZGu(c,O+)= f?(;,
;)OT(I
+
K)tl
COS
tJ(tn(set e))/rr
(15)
where the calculated fi values are shown in Table 2. Thus, the Tables 1 and 2 can be considered as a tinal numerical results investigated. In all these calculations, 2M = 12 terms are truncated in the eigenfun~t~onexpansion form in Appendix A, which can be considered as a responsible value from our previous experience[4].
Table 2. fa-Values shown in eqn (15) da or r/a
0.9 0.1
a/c=O.l
l.oOoo Loo05 1.0023 1.0066 1.0144 1.0260 1.0398 1.0502 1.0418
a/b = 0.2
a/b = 0.3 a/b = 0.4 a/b = 0.5 a/b = 0.6 a/b = 0.7 a/b = 0.8 a/b= 0.9
0.8 0.2 1.0001 l.GQO8 1.0039 1.0115 1.0255 1.0471 1.0754 1.1039 1.1144
01
0.7 0.3
0.6 0.4
l.oaol 1.0010 1.0050 1.0148 1.0333 1.0628 1.1035 1.1509 1.1888
Loo01 1.0011 1.0055 1.0166 1.0378 1.0725 1.1224 1.1853 1.2492
02
03
04
0.5 0.5 1.oGil1 1.0012 1.0057 1.0170 1.0393 1.0764 1.1314 1.2043 1.2g76
05
06
0.4 0.6
0.3 0.7
l.C& 1.0011 1.0054 1.0164 1.0381 1.0748 1.1306 1.2070 1.3006
l.Wol 1.0010 1.0049 1.0148 1.0346 1.0685 1.1208 1.1941 1.2883
07
08
0.2 0.8
09
l.oool l.OOO8 1.0040 1.0124 1.0291 I .OSSO 1.1030 I.1617 1.2529
0.1 0.9 l.OMO 1.0006 1.0030 1.0092 1.0217 1.0435 1.0776 1.1266 1.1%5
I
Fig. 3. A relation between the plastic zone length ratio (r/n) and the applied stress ratio (S/ur). EFM Vol. 17. No. &F
Technical Note As an another comparison with the numerical results corresponding to the case of a infinite cracked plate. we rewrite eqn (12) as
Note that (r ii $a = 1(Fig. 2). then from above equation one can get the relative plastic zone length r/u as
follows
where fd-values are shown in Fig. 3. From this figure we see that the curve corresponding to the case a/h = 0. I is very like the curve shown in Kef.Ill. It is the aim of this paper for us to get necessary numerical results rather than a detailed explanation of these results. This is the end of our paper. ‘4~kno~~Iedgemenfs-Theauthor is greatly indebted fo the ~omputjng center of ~ortbwestern Polytecbni~al university for carmine out computations. The author also wishes to thank Mrs. Cheng Yi-shan and Lin Ju-lie for their encouragement and some help to complete this paper.
D. S. Dugdale, Yielding of steel sheets containing slit. .I Mech. Php. Solids, g(2), 100-104(1960). J. R. Rice, Mathematjcal analysis in the mechanics of fracture, in Fracture Vol. 2 (Edited by Liebowitz~ (1968). Chen Yi-zhou and Chen Yi-heng. A mixed boundary problem for a finite internally cracked plate, Enpng Fructf~re Mech. 14. 741-753 (1981). Chen Yi-zhou. An investigation of the stress intensity factor on crack tip for a finite internally cracked plate by using variational method. Engng Fracture Me&. 17.387-394 (1983). N. I. Muskhelishvili, Some Basic Prublerns itr the ~a[~~er~aficu~Theory of Elasticity. Noordhoff, Groningen, Holland (19531. (Received 18 January 1982; received for pubIicurion 20 April 1982)
APPENDIX A Sofut~u~for a finite internffItywrackedplate by means of ~uri~tiofiulmef~od Suppose a cracked plate have both geometry and loading symmetrical conditions, for example, as shown in the probiem i or ilI of Fig. 2. In this case, the stress and displacement components can be expressed as follows rr!, t gvr = 4Ra@(Z)
(All
css - & = e(Z) t Q(Z) t (2 - Z)5@
(‘A?)
2G(u + it?)= K(F(Z)- o(Z) -(Z - Z,@?j
(Ail
where a(Z) = w’(Z), Q(Z) = q’(Z), (o(Z) and w(Z) arc analytical function. K = (3- v)/(l+ v) in plane stress condition, Y is the Poisson’s ratio, u, u are the displacement in x, y direction. From the stress free condition on the crack surface and the symmetrical condition mentioned above. we can get the following solution for complex potentials[3,41
where
and XK are real. Furthermore, we define op’, u\“’ as follows err’ deriving from cp’k’(Z),o(“)(Z) by eqns (Al) and (A2) ~:k’ deriving from ~‘~‘(a, o”‘(Z) by eqn (A3). Using a variational method141we get a linear system for XK as follows
(A61
583
Technical Note AIKXK
Aix =Ak, = eoit(K)ntui(0 dS I Br =
Ic
1,2,.
I =
= BI,
,2M
(A7)
f,K=1.2 ,..., 2M
,u/” dS
1=1,2,...,2M
(48)
and Pi are the traction acting on the outer boundary, n, are the direction cosine, c is the outer boundary contour. After the linear svstem (A71is solved. we using the obtained XK the disolacement distribution on the whole cracked ulate can be easily found. In addition, the stress intensity factor on the crack tip-can be expressed as K, = 2v(?r) $
=I
X,‘o’k~‘/~/a.
(A9)
The detailed description of above equations can be found in our previous papers[3,4].
APPENDIX B Solution for problem II in Fig. 2
We take the conformal mapping function z = a(i + [_I)/2
(Bl)
which maps the region exterior to unit circle on l-plane into the region exterior to crack contour on z-plane. It is well known[S], that we can express the stress and displacement components by some complex potentials as follows oxx+ oyyI= 4 Re W3
032)
uyy- 6 + 2&,, = 2[w~S)@U/w’(D+ WY1 -2W + iv) = ~(0 - wO+d5W(0 - W,‘)
(B3) (B4)
where
W) = (4lhP”m - W”(5)(P’(5))/(O’(i))?.
IBS)
Using Muskhelishvili’s method[5], we can get the solution of the complex potentials as follows p([)=&[-yi+(Z-C)ln(s)t(Z+C)ln(s)) 4%) = P(i) - P’(i) at+ where
fl =
Mi2- 1)
036)
arc cos(c/a) and for the particular branch considered ,s~~ ln(([ - e’“)/([ - e-‘@))= 0
,$im+ ln(([ + e”)/([ t ee”)) = 0.
037)
Using above mentioned equations, we can get eqns (6) and (7). In addition, there is no difficulty to obtain the stress distribution along the dash line contour in the problem II as shown in Fig. 2.