UNIVERSITY OF CANTABRIA

Transcription

1 Failue Assessment Diagam - Cack Diving Foce Diagam COMPATIBILITY UNIVERSITY OF CANTABRIA Novembe 1997 J. Ruiz Ocejo F. Gutiéez-Solana M.A. González-Posada I. Goochategui Depatamento de Ciencia e Ingenieía del Teeno y de los Mateiales E.T.S. de Ingenieos de Caminos, Canales y Puetos Univesidad de Cantabia Avda de los Castos s/n 39005, Santande (Spain) Tel , Fax

2 Failue Assessment Diagam - Cack Diving Foce Diagam COMPATIBILITY

3 Page 1 of 16 INTRODUCTION This Repot summaises the Univesity of Cantabia s poposal fo the assessment methodologies unification within the famewok of the SINTAP Pocedue. This wok has been caied out based on fome studies developed by the Univesity of Helsinki [1] and GKSS [2]. Initially, a daft epot was poduced (not ciculated), but futhe comments fom GKSS [3] have added new ideas that have been included in this vesion. All these analyses have been integated into this wok and amplified by ou own developments. We have incopoated this eseach into the initial stuctue of SINTAP (suggestion B). So, basic, assumed N and measued N levels have been defined, in which the desied FAD-CDFD compatibility has been obtained. Some comments about mateial & geomety specific level ae also pesented. Theefoe, all these thee levels could be un by the use using eithe a Failue Assessment Diagam o a Cack Diving Foce Diagam. It does not matte which gaphic epesentation is used because both methodologies have been found to be compatible. Two analysis paagaphs ae incopoated at each level studied. The fist one, descibing the Failue Assessment Line (FAL) fo a FAD; and the second one, defining the applied J-integal fo a component as a Cack Diving Foce (CDF) fo a CDFD.

4 Page 2 of 16 LEVEL 1 BASIC 1.1 OBJECTIVE Default cuves with no pevious knowledge of the mateial and component. 1.2 OPTIONS A: B: Continuously hadening mateials, without yield plateau. Mateial with yield plateau. 1.3 ASSUMPTIONS A: B: B: The mateials haden fom the stating point following a piece-wise powe law with an exponent of 0.3. The mateial s behaviou is assumed to be linea-elastic up to σy. Fom σy on, a lowe bound mateial is consideed consisting of a not-hadening o elastic-pefectly plastic mateial. 1.4 RELEVANT EQUATIONS = Eε ef L σ y + L 3 σ y 2Eε ef 1/2 [4] (1.1) ε σ=σ y ε y 0.3 (opt. A) (1.2a) σ=eε σ=σ y fo σ σy, ε εy fo ε > εy (opt. B) (1.2b)

5 Page 3 of ANALYSIS FAL Option A GKSS [3] has developed cuves (Eq. (2.3a)) which depend exclusively on the stain hadening coefficient (N). This analysis has been pefomed expeimentally fom the initial idea of R6 FAD1. We popose that the BASIC Failue Assessment Line fo continuously hadening mateials could be the cuve calculated fo N=0.3 with a little coection, i.e. change the fist facto of Equation (2.3a) by the fist section of Equation (1.3b), thus giving Equation (1.3a). = 1+ L 2 2 1/2 [ exp( 0.6L 6 )] (opt. A) (1.3a) The eason of that change can be justified because both factos ae gaphically vey simila, but the second one can be obtained fom a mechanical point of view. Figue 1 shows the compaison between FALs developed using both factos. Option B Inseting Equation (1.2b) into Equation (1.1) we get the BASIC Failue Assessment Line expession (Figue 2) fo a mateial exhibiting a yield plateau. = 1 + L 2 2 = 0 1/2 fo L 1 fo L > 1 (opt. B) (1.3b) Figue 3 pesents all options of the BASIC Failue Assessment Line: that defined by GKSS and the ones poposed by UC epesented by Equations (1.3a) and (1.3b). CDF Making use of Annex 1: Tansfomation FAL CDF. Equation (A1.6), we can obtain the CDF equivalent to the BASIC Failue Assessment Line fo both options J = K 2 I E' 1 + L exp( 0.6L [ 6 )] 2 (opt. A) (1.4a) J = K 2 I E' 1 + L 2 2 (opt. B) (1.4b)

6 Page 4 of 16 LEVEL 2 ASSUMED N 2.1 OBJECTIVE Development of cuves though a simplified stess-stain cuve model. 2.2 OPTIONS A: B: Continuously hadening mateials, without yield plateau. Mateial with yield plateau. 2.3 ASSUMPTIONS A: B: B: B: The mateials haden fom the stating point following a piece-wise powe law. The mateial s behaviou is assumed to be linea-elastic up to σy. Thee could be a yield plateau. Fom σy on, the stess-stain cuve can be modelled as a piece-wise powe law. 2.4 RELEVANT EQUATIONS = Eε ef L σ y + L 3 σ y 2Eε ef 1/2 (2.1) ε σ=σ y ε y N (opt. A) (2.2a) σ=eε fo σ σy, ε εy σ=σ y ε σ=σ y ε y + ε N fo εy < ε εy+ ε fo σ > σy, ε > εy+ ε (opt. B)) (2.2b)

7 Page 5 of ANALYSIS FAL Option A Fo stesses unde σy (L 1), GKSS has developed cuves (Eq. (2.3a)) which depend exclusively on the stain hadening coefficient (N). This analysis has been pefomed expeimentally fom the initial idea of R6 FAD1. These cuves constitute the ASSUMED N Failue Assessment Line fo a continuously hadening mateial Although the Univesity of Cantabia has tied to obtain them analytically by simplifying Eq. (2.1) fo a law that satisfies Eq. (2.2a), the mathematical expessions obtained don t follow the eal mateial behaviou as they gow in the inteval fom L=0 to L=1. = 1 L exp( 2NL [ 6 )] (opt. A) (2.3a) Once again, UC poposes to change the fist facto of Eq. (2.3a) fo the fist section of Eq. (1.3b), esulting: = 1+ L 2 2 1/2 [ exp( 2NL 6 )] (opt. A) (2.4a) Fo values geate than σy (L 1), GKSS follows Annex 2: Tansfomation CDF FAL, giving: = ( L =1) L N 1 2N (opt. A) (2.5a) whee the value fo K at L=1 is obtained though Eq. (2.4a). Figue 4 shows diffeent FALs constucted following Equations (2.4a) and (2.5a) fo diffeent stain hadening coefficients. Option B Inseting the diffeent pats of Eq. (2.2b) into Eq.(2.1) we get the ASSUMED N Failue Assessment Line expession fo a mateial with yield plateau in thee diffeent sections. Fist section. Elastic behaviou up to L = 1. = 1+ L 2 2 1/ 2 fo L 1, σ σy, ε εy (opt. B) (2.3b)

8 Page 6 of 16 Second section. Yield plateau at L = 1. = λ + 1 2λ 1/2 fo L = 1, σ = σy, εy < ε εy+ ε (opt. B) (2.4b) whee: λ=1 + E σ y and λ max = 1+ E σ y ε (2.5b) being ε the extension of the yield plateau and ε. Thid section. Stain hadening behaviou fom L = 1. 1 N = λ max L 1 + L 3 1 N 2λ max 1/2 fo L > 1, σ > σy, ε > εy+ ε (2.6b) This thid section (Eq. (2.6b)) could have been developed by means of Annex 2: Tansfomation CDF FAL. In such a case, the same equation that (2.5a) would have been obtained whee K value fo L=1 is calculated though Eq. (2.4b) fo λmax, thus giving Equation (2.7b). = λ max + 1 2λ max 1/2 N 1 2N L fo L > 1, σ > σy, ε > εy+ ε (2.7b) Figue 5 shows the similaities between both options (Equations (2.6b) and (2.7b)) fo a seies of mateials. Figue 6 shows a FAL developed though this fomulation compaed to the coesponding R6 FAD2 fo a feitic steel. This mateial s stess-stain cuve is pesented in Figue 7. CDF Making use of Annex 1: Tansfomation FAL CDF. Equation (A1.6), we can obtain the CDF equivalent to both options. J = K I 2 E' f(l ) [ ] 2 (A1.6)

9 Page 7 of 16 LEVEL 3 MEASURED N 3.1 OBJECTIVE Development of cuves with the help of the actual tue stess-stain cuve. 3.2 RELEVANT EQUATIONS = Eε ef L σ y + L 3 σ y 2Eε ef 1/2 (3.1) 3.3 ANALYSIS FAL Equation (3.1) diectly epesents the MEASURED N Failue Assessment Line. CDF Making use of Annex 1: Tansfomation FAL CDF. Equation (A1.6), we can obtain the CDF equivalent to Eq. (3.1): J = K 2 I E' Eε ef L σ y + L 3 σ y 2Eε (3.2) ef

10 Page 8 of OBJECTIVE LEVEL 4 MATERIAL & GEOMETRY SPECIFIC Development of cuves fo a specific mateial and component. 4.2 ASSUMPTIONS The component is defined though the ETM maste-type cuve. 4.3 RELEVANT EQUATIONS J = F J Y F Y 1+N N [5] (4.1) 4.4 ANALYSIS FAL The expession of the Failue Assessment Line can be obtained following Annex 2: Tansfomation CDF FAL and is pesented in Equation (A2.8). = ( L =1) K 1+N I 2N L K Y = (L =1) L N 1 2N (A2.8) As can be seen, the Failue Assessment Line is not dependant on the geomety esulting on Level 2 once again. This is so, because the atio between KI and KY can be substituted by L as this atio is popotional to σ/σy on the Failue Assessment Line. Theefoe, the effect of the cack and the geomety dissapeas. CDF The applied J-integal fo a component comes diectly fom the assumption of this section -epesented by Equation (4.2)- giving Equation (A2.9). J = J Y L 1+ N N (A2.9)

11 Page 9 of 16 ANNEX 1 TRANSFORMATION FAL CDF In a Failue Assessment Diagam, the Failue Assessment Line should follow the equation: LINE = J e J 1/2 (A1.1) Any simplification of this equation which is consideed as the Failue Assessment Line on a FAD can be expessed as follows: LINE = f(l ) (A1.2) The bette the function f(l) fits Equation (A1.1), the moe accuate the assessment is. So, as Equation (A1.2) epesents the bounday of the safe aea, this can be taken as the mathematical expession of (A1.1). Theefoe both equations can be equalled: J e J 1/2 f(l ) (A1.3) Fom that, we obtain: J = J e [ f(l )] 2 (A1.4) whee Inseting Eq. (A1.5) into Eq. (A1.4), gives J e = K 2 I E' (A1.5) J = K 2 I E' [ f(l )] 2 (A1.6) which epesents the applied J-integal of a component as a function of the stess intensity facto and the Failue Assessment Line consideed.

12 Page 10 of 16 ANNEX 2 TRANSFORMATION CDF FAL Following the ETM-maste type cuve: J = F J Y F Y 1+N N (A2.1) we can obtain the expession of the Failue Assessment Line fo L 1 by means of an easy coodinate change: F F Y = L (A2.2) = J e = K 2 I J E'J Inseting Eq. (A2.2) into Eq. (A2.1) and then, this one into Eq. (A2.3), gives: 1/2 2 K = I E' J Y 1/2 1/2 1+ N 2N L (A2.3) (A2.4) whee JY must be calculated at L=1 fom the Failue Assessment Line expession fo L 1 in ode to give continuity to the oveall function. If we note that point by: (L =1) we can assue that we must obtain the same value if calculated though Eq. (A2.3) (A2.5) 2 K (L =1) = Y E'J Y 1/2 (A2.6) fom that, we can get the JY value J Y = K Y 2 E'((L =1) ) 2 (A2.7) Once this value is known, it can be incopoated into Eq. (A2.4) giving: = (L = 1) K I K Y 1+N N 1 2N 2N L = (L = 1)L (A2.8) which is the Failue Assessment Line compatible with a Cack Diving Foce defined by: J = J Y L 1+N N (A2.9)

13 Page 11 of 16 CONCLUSIONS In this wok, an effot has been done in ode to find the compatibility between the diffeent FAD and CDFD levels suggested within the SINTAP Pocedue initial stuctue [6]. In ou opinion, this has been done and demonstated to be possible fo all levels except those concening special cicumstances like constaint o mismatch, about which, no analysis has been studied yet. The scope of this document can be summaised in the following table: FAD CDFD BASIC LEVEL 1 Default MATERIAL LEVEL 2 Assumed N SPECIFIC LEVEL 3 Measued N MATERIAL & LEVEL 4 CDF FAL (*) GEOMETRY Special featues - - SPECIFIC (constaint, mismatch) - - (*) If a FAD is used, the same Failue Assessment Line that Level 2 is obtained, so the cack gowth effect is not consideed when evaluating a stuctue whee ductile teaing is allowed. Thus, diffeent Failue Assessment Lines fo diffeent cack sizes have to be developed if a moe exact and pecise evaluation is desied. It is ou belief that the assessment becomes too complicated [7]. Theefoe, ou ecommendation is to use a complete Cack Diving Foce analysis whee the component s applied J- integal is compaed to the full JR cuve.

14 Page 12 of 16 REFERENCES [1] Laukkanen, A. An Intoduction to ETM and Compaison to R6 Revision 3. Helsinki Univesity of Technology, Septembe [2] Zebst, U. and Kim, Y.-J. GKSS Poposal fo Possible Unified Flaw Assessment Method fo SINTAP. SINTAP/GKSS/09, Octobe [3] Kim, Y.-J. GKSS Poposal fo SINTAP mateial specific option (known YS/TS). SINTAP/GKSS/09 (evised), Novembe [4] Milne, I.; Ainswoth, R.; Dowling, A.R. and Stewat, A.T. Assessment of the Integity of Stuctues Containing Defects, CEGB, Repot R/H/R6-Rev.3, [5] Schwalbe, K.-H.; Zebst, U.; Bocks, W.; Conec, A.; Heeens, J. and Amstutz, H. The Engineeing Teatment Model fo Assessing the Significance of Cack-like Defects in Engineeing Stuctues (EFAM ETM 95), GKSS-Foschungszentum, [6] Bitish Steel, SINTAP Pocedue, Rev. 2.0, Daft fo comment, SINTAP/BS/17, Octobe [7] Ruiz Ocejo, J.; González-Posada, M.A.; Gutiéez-Solana, F. and Goochategui, I. Review of Existing Pocedues, Subtask 5.1, Repot, SINTAP/UC/04, June 1997.

15 Page 13 of 16 1 Figue 1 BASIC Continuously hadening mateial GKSS UC (A) L 1 Figue 2 BASIC Mateial with yield plateau UC (B) L

16 Page 14 of 16 1 Figue 3 BASIC All options GKSS UC (A) UC (B) L

17 Page 15 of 16 1 Figue 4 ASSUMED N Continuously hadening mateial N=0.05 N=0.1 N=0.15 N=0.2 N=0.25 N= L Figue 5 ASSUMED N Mateial with yield plateau. Compaison Eq. 2.6b (N=0.05) Eq. 2.6b (N=0.30) Eq. 2.7b (N=0.05) Eq. 2.7b (N=0.30) L

18 Page 16 of Figue 6 ASSUMED N Mateial with yield plateau. Compaison FAD2 R6 Eq. 2.6b UC L Figue 7 Stess-stain cuve Feitic steel σ linea-elastic yield plateau hadening ε