Jadaan_pt1_Cring_Oring_JTEVA91

Jadaan_pt1_Cring_Oring_JTEVA91 - T .._. ._,5._.._..., p l l...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: T .._. ._,5._.._..., p l l I. l 0. M. Jadaan,1 D. L. Shellemanfll. C. Conway, Jr.,3 J. J. Mecholsky, Jr.,4 and R. E. Tressler5 j Prediction of the Strength of‘Ceramic Tubular COmponents: 7 Part l—Analysis ‘ 1 Mafia 1 REFERENCE: Jadaan, O. M., Shelleman, D. L., Conway, .l. C., Jr., T Mecholsky, J. J ., Jr., and Tressler, R. E., “Prediction of the Strength 1 u of Ceramic Tubular Components: Part I—Analysis,” Journal of Testing and Evaluation, JTEVA, Vol. 19, No. 3, May 1991, pp. 181—191. ABSTRACT: The objective of this paper is to develop the analytical background for test methodologies that will enable accurate predic- tion of the strength distribution of ceramic tubular components from the strength distributions of simple specimens. Four simple specimen configurations and two tubular configurations were selected for this purpose. The simple specimen configurations were (1) four-point and (4) O—ring tested in diametral compression. In addition, a short tube tested by axially compressing rubber inside the tube and a long tube subjected to internal pressure were analyzed. These specimen configurations were for the most part selected in a tubular shape in order to simulate the shape of tubular structural components. The prediction of the strength distribution of one specimen from that of another was based on Weibull statistical theory. Effective volume and area expressions, necessary for failure prediction, were derived for these specimen configurations. KEY WORDS: strength distribution, failure probability, Weibull analysis, effective volume I l J bend, (2) C-ring tested in compression, (3) Oring tested in tension, r l l Ceramic materials are desirable in structural applications be- cause of their strength at high temperature, high thermal con— ductivity, low thermal expansion, and excellent wear resistance. However, structural ceramics are generally brittle with low frac- ture toughness which impoSes a severe restriction on the size of the critical flaw. Ceramics fail due to inherent flaws which can be introduced into the material through processing, handling, machining, or while in service. As a result, fracture strengths of identical specimens vary and must be statistically described [1]. Furthermore, as the volume and/or surface area of a stressed ceramic component increases, the probability of encountering a . severe flaw increases. This effect introduces a size dependence Paper received 3/23/90; accepted for publication 9/19/90. ‘Assistant Professor, Division of General Engineering, University of Wisconsin-Platteville, Platteville, WI 53818. 2Research Assistant, Center for Advanced Materials, The PennsyL Vania State University, University Park, PA 16802. 3Professor, Department of Engineering Science and Mechanics, The Pennsylvania State University, University Park, PA 16802. 4Professor, Department of Materials Science and Engineering, Uni- versity of Florida, Gainesville, FL 32611. sProfessor, Director of the Center for Advanced Materials, The Penn- sYlvania State University, University Park, PA 16802. )2 l, 1 © 1991 by the American Society for Testing and Materials 181 of strength [2, 3]. In other words, the strength of a brittle ceramic component depends on the stress distribution and the volume and/or surface area stressed in tension [4, 5]. Most of the tests performed to evaluate the mechanical reli- ability of ceramic materials are conducted on small laboratory specimens which in general do not simulate the size and stress field of actual structural components. This is because the testing of actual large size components subjected to complex stress fields at high temperatures is both expensive and tedious. Hence a sound test methodology is needed whereby the mechanical re- liability of standard ceramic components can be evaluated, with a high degree of confidence, from simple laboratory specimens. The objective of this paper is to develop the theoretical back- ground for the strength distributions of several test specimens and load configurations in order to determine which tests enable us to most accurately predict the strength distribution of full- scale tubular components. The selection of these specimen con- figurations was based on simple geometry, ease of machining, ease of loading, and unaltered flaw population in comparison with the tubular sections from which the specimens were .cut. The exception was the four—point bend configuration, which was selected because of its widespread use and to check for the effect of grinding on the prediction of the strength of tubular compo- nents. Six specimens and load configurations were selected for this study: (1),four-point bend bar, (2) C—ring tested in compression, (3) C-ring tested in tension, (4) O-ring tested in diametral compression, (5) a short tube\filled with polyethylene rubber and then axially compressed, resulting in a uniform radial pressure being applied to the inner walls of the tube, and (6) a long tube internally pressurized with gas. 9 To use statistical techniques properly to predict the fracture behavior of one component from that of another, we must first accurately describe the stress distributions for both. The stress distributions for the four-point bend test (Fig. 1a), the C-ring tested in compression (Fig. 1b), and the internally pressurized long tube (Fig. 1f) have been discussed by previous investigators [6, 8, 9] and will not be discussed in this paper. The stress dis- tribution for the 0an specimen tested in tension (Fig. 1c) is similar to that for the C-ring specimen tested in compression with two exceptions. Firstly, a negative load should be substituted in the stress distribution expression for the C—ring specimen tested in compression; secondly, the maximum tensile stress should be evaluated at the inner surface where r = r,. The stress analyses for the O-ring specimen (Fig. 1d) and the short tube (Fig. 16) ,i i '2 ,! t . 33 t _ fit it .- - . ,‘1 i ,. t: t. .1; 3. 482 JOURNAL OF TESTING AND EVALUATION D t 0 b FIG. la—Founpoint bend specimen. are' ( read j dish V, stra . in H next effeu pre< fion I, coni tafic long P , spec ’ [1]] 1"113 the 1 ' pon . com P and FIG. 1d— O—ring irz diametral compression. 1 Stat P the flaxv cera “lat . fla“ resu , [13. of t' failt the I 1 I t 1 1 { P FIG. 1b—C-ring in Compression. \ ‘ 'whc FIG. 1e—-Short lube tested by axially compressing rubber inside the m” tube. ' cha \ cm in u . \vhi «W.MM¢_‘:£¢A&M W...” -I _. FD T_. \vhe fort P Whé FIG. 1c—C—ring in tension. ‘ FIG. 1f—-Long lube tested with internal pressure. tho A f .o‘ ’zuwsa-ud [favavJéx-JJSZY tbber inside the “if—W ——e sure. i l JADAAN ET AL. ON CERAMIC TUBULAR COMPONENTS: PART I are discussed in Appendices I and II. For the convenience of the reader, Table 1 includes the analytical expressions for the stress distributions as well as the expressions for the maximum tensile streSS for all the specimen and loading configurations considered in this paper. The stress distributions, mentioned above, will be used in the next section to derive the effective volume and area expressions for the specimen configurations considered in this paper. The effective volume and area expressions can then be utilized to predict the strength distribution of the long tube components fmrn the strength distribution of any of the simple specimen Eonfigurations used in this study. To demonstrate the compu- tational procedure for predicting the strength distribution of the long tube component from the strength distribution of a simple specimen, simulated strength data, based on actual test results [11] for the long tube and O-ring specimens, were used. Accomplishment of the objective mentioned above will enable the designer to estimate the strength distribution of tubular com- ponents by conducting a simple and inexpensive test. This ac- complishment will eliminate the need to perform such complex and expensive tests as the burst test of long tubes. Statistical Failure Analysis The strength of ceramic materials, unlike metals, is limited by the size of inherent flaws through Grifith’s relationship. These flaws can vary considerably in size. In addition, failure in a stressed ceramic component follows the weakest link theory, which states that failure is determined by the lowest stress at the most critical flaw. This phenomenon yields a size dependence of stress. As a result, strengths tend to vary and require statistical description. Weibull statistical analysis is based on the weakest link theory , [13, I4] and has been extensively used to describe these features of the strength of ceramics. In this theory, the probability of failure (F) can be related to the strength of a specimen through the two-parameter Weibull equation [14, 15]: F = 1 — (“59'3" (1) where m is the Weibull modulus (also known as the shape pa- rameter) and a measure of the scatter of the data; cr0 is the characteristic strength (also known as the scale parameter) and corresponds to the strength of a unit volume specimen stressed in uniform tension; 0 is the stress; and Fis the failure probability ; which can be estimated from n — (1/2) .F=N (2) Where N is the total number of specimens and n is the rank order Of a certain specimen when the specimens are ordered from Weakest to strongest [4]. For graphically representing the strength data the following form of Eq 1 is often used: 1 lnln<1_F> — Where KV is the effective volume under tension, which can be thought of as the volume of material being tested effectively m In <3) + in KV (3) 0'0 183 under uniform tension (K V = 1 for a unit volume under uniform tension). Equation 3 represents a straight line with slope m When ln ln (1/(1 -'F)) is plotted versus ln 0-. This parameter can then be computed with the use of the least squares method. The effective volume expression is a function of the Weibull modulus and the geometric parameters of the specimen config~ uration being studied. The effective volume expression can be described analytically for any specimen configuration which pos- sesses a stress distribution that can also be described analytically. The effective volume can be derived for any configuration from the equation U m KV — [(UMX) dV (4) where (r is an appropriate expression for the stress distribution and arm, is the maximum stress. In the case where the stress distribution cannot be described analytically, but rather numer- ically, the effective volume can then be computed numerically. For example, if the finite element method is used to describe the stress distribution for a given specimen configuration, Eq 4 can be evaluated for each element and then summed over the entire body to yield the final KV value. The effective volume, KV, is used to predict the strength dis— tribution of one type of specimen from that of another. When the strengths of two different specimen configurations of the same material are to be compared, the following expression will result if equal failure probabilities are substituted in Eq 3: 1/771 2 = KZVZ ' (5) U: K1 V1 where K1 V1 is the effective volume of the first specimen config- uration and 1(sz is the effective volume of the second. Thus the information necessary for predicting the strength distribution of the first specimen includes the effective volume expressions of both specimen configurations and the Weibull modulus of the second specimen configuration. The volume—based Weibull analysis is used when the failure- initiating flaws exist in the bulk of the material as well as at the surface. However, when failure mainly occurs due to surface flaws, then the area Weibull analysis can better describe the strength distribution. In this casethe relationship which corre- sponds to Eq 5 is K A l/m fl = 2 2 0'2 (K1141) (6) where [CA is the effective area which can be derived from the eqllathIl f (I max KA = )mdA (7) The effective volume/area expressions for the fourvpoint bend specimen and the C-ring specimen tested in compression have been introduced in previous investigations [4,8]. Only the effec- tive volume/area of the C-ring tested in tension, O-ring in dia— metral compression, and tube configurations will be discussed below, since they have not been reported elsewhere. Table 1 392:1 M-"fieémmmr :~ UH rmT Izmfi GI FnTIUfiCtar Lw (IID)_.I .Ir mw C wk w: 0 .mmmemM w mmmm mv K IIIIWII II. I I ImI II II .I.III(II.I. I I I I .II; t .,, , IIII _. , . I . . I , w, I III a .II..I...I , .H x65ma< cum Acéx cam .35: Act 5.3% 6 303833 2: an .o I cm H 9 E I S . C l cg 5 IE :71 A; EV ck u C In :71 E H <on 9:8”. o% N M Q”: N x can Do H a «m T430 n 6:450 MNN .Um gm T430 Henna: Ho 2:: Macmflom n S s I s s I e a: r I E A; I E? AL N w a; u 2%: E n K”: + f n «k .mflnuoE :36}; H E dim :89? Eat wounmmuE 298 H a Jim @555: 89a 8.532: 295 H m .3335 u air—E: 55: u : £265 .850 H ex can anon €69.58 2: 8“ E; 03:8 H u .Ewcfl n 4 .523 H a .mmoaqu. H N 68“ II. R “223?. . . r g a: 0 % a? H t N + SEN TE§E§ E + $3533: % :2 E i E + a: a: o. a EU _ 5 I 2%: , + m + H 5 s N: Tu Labs“: 9 c + amszsai QET « H+ 5+ i E SE , 32 us. a as: s Mk an a $3 + «E «$3 + «as E N + s 3:85 .E no \ 2:? 523E A L 5 a NE. I cc: 1 .c v + w a. a $ N: + E: 25:5 K k I I p Biszfisa u 5?. + Q EA w E 45:53 a: <53 .5: as NI+ 2. £55 a 1_II Mg A 1_III mg 9.30 0:5 . IE I 55m IE I $5M Kw mag . N: + v w + v m E I 5: G I 3:: 3% 52:52 m w I m: 1 w I m. £5 E I 35 E I 5% N5 8:: u 2 I+H |+ k. mouIl mooII I A . Pub A? V .3: A? Q Q? . A 30 k a c I Sam @ G I 9: was 325 Bay :2; 83‘ $3 wand $55 $55 @535 £80 I 0:20P azmmuanO 9.0m .Eonisom :uEMuuam uézmpfiumg .833: 38 wzotExmeeu 5: MRS: 3: SK. ESEESS ESEEES mfifimhw N33 commasfiw EmhmIH msmxfl. JOURNAL OF TESTING AND EVALUATION 184 ‘ era-H considered in this paper. C-Ring Tested in Tension _ V~V~_ a_._wt..__..—— , T The effective volume and area expressions are brimfefr' ribfe + 2 rimfefr, ; KV= ' KA= J, Where ri is the inner radius and b is the width, and f6 and f; are defined as follows (Fig. 1c): f f6 = 2 [011/2 cosme d9 (9a) f: = m : r“"’ dr (9b) where 9 is the angle measured from the horizontal axis, and r2‘ = (ri + r0)/2 where r0 is the outer radius. 0ng Tested in Diametral Compression The exact stress distribution (see Appendix I) will not be used i in the derivation of the KV and KA expressions due to the complexity of the expression as well as the nonconvergence cri- terion occurring at the outer surface (the Fourier expression used ‘ to describe the stress field approaches infinity under the load). "' Instead, the stress distribution function obtained from strain energy— ; straight beam theory will be used. This equation is (see Appendix I and Eq 24) I ~ _ P ray _ 1 ray V 0'0 — —2 [0.637 T cos(6) <A— + I i (10) , where 1 = — _ . 3 = 12 rl) 1 _ 3 12 bt A b(ro — ri) = bt P = load , and where I is the moment of inertia, tis the thickness, and A IS the cross-sectional area. Equation 10 represents a well-behaved function and is in good agreement with the exact solution (Eq w 20) described in Appendix I. t The effective volume expression for the compressed O-ring , (Fig. 1d) is given by i = % {1:}er I i F0, 9) ] ]’" rdrde "/2 + l 9?. r(9) L. [l F(r. e) l ]m rdrde} (11) , , a.“ . r ~ , wwv' ~ r r v u a mans“ my:«p“www‘ifimmwrwawmfivmamarwgmmwm W ’u lift‘m‘kwafixwfio flwaxfiaugmwmmwimuw 2. pr." , ".3... .. Shows the effective volume/area expressions for all configurations JADAAN ET AL. ON CERAMIC TUBULAR COMPONENTS: PART I 185 where c — 3 82 3 (12a) T ‘ bt2 r3 r3 F(r, 9) = —0.637 7r + 7r cos 6 ,2 1 . r2 _ _a _ . J 1 cose<1 +14) + 06371 (217) r§ 1 rfi cos + A) — 0.637 I r(6) = R (12c) %(cos 9 — 0.637) and where 61 is the angle below which a tensile stress exists at the outer surface region, 92 is angle above which a tensile streSs exists at the inner surface region, and r(9) is the function that describes the location of the neutral axis. The function F(r,6) represents the stress distribution in the compressed O—ring specimen. Since the effective volume/area expressions are only calculated for the regions where the stress field is tensile, then the function F(r,6) is always positive; there- fore Eq 11 will always yield real numbers. However, one must exercise caution when computer programs are used to evaluate these expressions,6 since at the neutral axis (where F(r,6) = 0) the computer will calculate a very small number, 5, instead of zero because of roundoff error. This small number, a, could be either positive or negative. In the case where s is negative and the m power is a noninteger number, the expression (F(r,(-))"') will result in a complex number. Therefore, in order to avoid problems with computer roundoff errors, an absolute value expression was utilized in Eq 11. Similarly, the effective area expression is given by 4 KA=—>< Cm \ b(ruf’[l Fore) l l”'d0 + l (fore) I We) (13) :fl +2<flfili l F(r,6) I ]"'rdrd0 +L L’(9)[ I F(r,6) f ]”'rdrd0) where F(r0,e) and F(r,,6) are obtainedby substituting r0 and ri for r in Eq 12b. ' Internally Pressurized Tube ‘The MATH CAD program was used to numerically evaluate the K V and KA expressions. This program can be operated on an IBM PC computer. ’ 186 JOURNALOFTESHNGANDEVALUAflON are much weaker when stressed in tension than when stressed in compression, only tensile stress fields are considered for the derivation of the effective volume and area expressions. The effective volume and area expressions for internally pres- surized tubes (taking into account only the hoop stress compo- nent) are given by (see Fig. 12 and 1f) m ,0 r: "I J; (1+ rdr 7? KV — 211L (n, + r?) (14) and r? m KA=2n<ri2+rg> x [Lri (1 + + 2m Lr, + 2 (1 + rdr] (15) I where L is the length of the tube. Discussion In the previous section, the effective volume (KV) and effec- tive area (KA) expressions for the various specimen configura— tions were derived. These expressions, with the exception of the four-point bend, were complex and required numerical com- putations. The O-ring effective volume expression (Eq 11) is an example. This expression contains a double-integral formulation with the integral limits (r(6)) being a function to be evaluated at each angular position. While for most specimen configurations the effective volume and area expressions can be separated into a load factor expression, K, that is independent of the specimen dimension, and a geometric expression that contains the size dimensions, the same cannot be said about the O-ring expression. The complexity of the function to be integrated, F(r,6), as well as the variable integral limit, r(8), prohibited the separation of 1E—2 EFFECTIVE VOLUME (KV) O 5 10 ‘15 20 25 afar the effective volume and area expressions into load factor and size functions. For this reason, the comparison between all the specimen configurations suggested in this paper is based on the effective volume and area expressions rather than on the load factor expressions. Figures 2 and 3 show the effective volume and effective area distributions, respectively, as a function of the Weibull modulus, m, for all the specimen configurations considered in this study‘ The effective volume and area distributions are functions of the specimen shapes and dimensions as well as the Weibull modulus Thus, in order to evaluate these expressions, the specimen di_ mensions must be provided. The dimensions used to prepare Figs. 2 and 3 were obtained from a companion paper [1]] that contained the experimental verification to the analytical expres_ sions derived in this paper. The dimensions are listed in Table 2. It is important to stress the point that Figs. 2 and 3 are plotted for certain specimen dimensions and therefore they are not uni- versal in nature. As can be seen from Figs. 2 and 3, the long tube test config- uration has the largest effective volume and area, since this spec- imen possesses the largest volume/surface area stressed in tension of the specimen and load configurations considered here. The entire tube volume and surface area are stressed in circumfer- ential tension, unlike the other Specimen configurations which are partially stressed in tension. The short tube configuration has the second greatest effective volume/area values and results in a distribution similar to that of the long tube configuration, since they have similar stress distributions. The difference in effective volume/area values between the long and short tube specimen configurations is related entirely to the size difference. The four—point bend, C-ring loaded in compression, and C—ring loaded in tension display similar effective volume/area distri- butions because they have similar stress distributions. The C—ring specimen stress distribution displays the form of half a cosine cycle along its circumference, while the stress distribution for the four-point bend specimen displays a trapezoidal shape 0 BEND . C-r1ng (comp) A C—ring (lens) A D-rtng . Cl SHORT TUBE I LONG TUBE 30 35 4O 50 45 WEIBULL MODULUS (m) FIG. Z—Effective volume versus Weibull modulus for four—point bend specimen, C-ring in compression, C-ring in tension, O—ring in diamelml compression, short tube tested by axially compressing rubber inside the lube, and long tube tested with internal pressure. WW" ‘3‘ h‘r“';..i‘“"_._."1£fm- ' JAM fl’m‘ *Mmfl’” M“ wi13.a"”j__j’:r'lfl35m’whnatal—5W warm: . 1m meme-.KE-Juhmskxe‘amfirzs' mmtnzzsafasm t‘JTLG'iWLIW43mfl21H3meQiVé? 5? ) , Fll com; ) .5 i \ "3 E 7 S . 4 J L a 1 sp 5. Eara'ximmuszzsvtaflt load factor and between all the is based on the 1am on the load .d effective area ’eibull modulus, :d in this study, functions of the ’eibull modulus he specimen d1. Ised to prepare paper [11] that ialytical expres_ : listed in Table nd 3 are plotted hey are not uni. ube test config- , since this spec- ‘essed in tension {cred here. The r :d in circumfer- gurations which e configuration lues and results a configuration, 1e difference in and short tube size difference. .ion, and C—ring ime/area distri- tributions. The form of half a :ess distribution tpezoidal shape P.) --— -,_... “11..-...— to ‘ r .___.i.‘ J“... H... 1 s...‘_ 4.... c 1,~-__’\.___.. 1 or d‘lbfl, 5 .L 'ing in diametru/ l t z I :1; . «a -. .N , . . , , _ ' Err,me :Pmm' iamhmvmmz‘a’mflw 1'::t“*‘,:."“.£1“a.£2' Ams‘eyaxsr ‘v “3252i semen :msmtaeml amwweamw- 259’ m * figs." JADAAN ET AL. ON CERAMIC TUBULAR COMPONENTS: PART 1 EFFECTlVE AREA (KA) O 5 1O 15 20 25 187 O BEND . C-rlng (comp) A C—rtng (tens) A O-rtng El SHORT TUBE I LONG TUBE 50 30 35 40 45 WEIBULL MODULUS (m) FIG. 3—Effective area versus Weibull modulus for four-point bend specimen, C-ring in compression, C-ring in tension, O-ring in diametral compression, short tube tested by axially compressing rubber inside the tube, and long tube tested with internal pressure. TABLE 2—Dimensions of specimen configurations. Dimensions (mm)" Specimen b l L ro ri Flexure Bars 4.0 3.0 50.8 . . . . . . C-n'ng in Compression 9.5 21.9 (0.4) 17.1 (0.4) C-ring in Tension 9.5 21.9 (0.4) 17.1 (0.4) O-ring 9.5 . . . 21.9 (0.4) 17.1 (0.4) Short Tube 101.6 21.9 (0.4) 17.1 (04) Long Tube 660.4 21.9 (0.4) 17.1 (0.4) "b = width,t = thickness, L = length, rL, = outer radius, ri = inner radius. 1 FRACTURE STRENGTH (MPa) 7 20 55 148 403 100.0 E] o-ring . long tube 93‘4 30.8 4.9 0.7 In In (1/(1-F)) 0.09 FAILURE PROBABILITY (F) 0.01 In S (MP3) FIG. 4—Generated strength distributions simulating those of an O—ring Specimen and long tube. along its length. Furthermore, the neutral planes are located at the geometric center along the lengths of the C-ring and four- point bend configurations. The compressed O—ring Specimen dis— plays effective volume/area distributions that are unique when compared to the other specimen configurations. The distribu- tions are steeper and do not converge to a minimal plateau as fast as the other specimen configurations do. Equations 5 and/or 6 are used to predict and distinguish the strength distribution of one specimen from that of another. An example for a volume related Weibull analysis follows. Figure 4 shows two randomly generated strength distributions simulating those of an O-ring specimen and a long tube.7 A random number generator program was used to simulate 30 strength data points for each specimen configuration. The Weibull modulus and char- acteristic strength for each specimen configuration were obtained from a companion paper [1]] to generate realistic strength dis- tributions for both specimen configurations. The dimensions for 7The program MTHSRANDOM was used to generate numbers that are uniformly distributed between 0.0 and 1.0. 188 JOURNAL OF TESTING AND EVALUATION FRACTURE STRENGTH (MPa) 7 12 20 33 55 90 148 1000 El Adjusted o-rlng ‘ Q 0 long tube 93‘4 >( A E E: 30.3 d I E {\‘l <1 5 a 4.9 8 M 5 h . e E .El _ 0.7 E .D D . 009 d < C] h: 0.01 2.0 2.5 3.0 3,5 4.0 4.5 5.0 lnS(MPa) FIG. S—Adjusted and generated strength distributions corresponding to the O-ring specimen and long tube respectively. both specimens are shown in Table 2. Figure 5 shows the adjusted O-ring strength distribution which corresponds to the predicted long tube strength distribution as calculated from the O-ring strength distribution. The strength distribution of the long tube configuration is the same as that shown in Fig. 4 and is based on test data obtained from Ref 1] . The adjusted O-ring strength distribution is obtained by using Eq 5 in the following manner. The effective volume values of the O—ring and the long tube configurations were obtained from Fig. 2 and substituted into Eq 5. The value of the Weibull modulus used in Eq 5 should correspond to that of the O-ring, since it is this specimen from which the prediction is made. As can be seen from Fig. 5, the calculations for the O-ring specimen predict accurately the strength distribution of the long tube component. The only difference be- tween the generated and the predicted long tube strength dis- tributions corresponds to the dissimilar slopes of the two curves. The slope of the generated long tube data corresponds to the Weibull modulus obtained from the experimentally tested long tube specimens (m t: 4.5, see Ref 1]), while the slope of the adjusted O-ring data corresponds to the Weibull modulus ob- tained from testing O-ring specimens (m = 5.5, see Ref 11). Care must be exercised when determining whether to use the volume or area based analysis. If the sources of failure are pre- dominantly surface related, the area analysis must be utilized and Eq 6 should be used. However, if the sources of failure are a combination between volume and surface flaws, then the _vol— ume analysis (Eq 5) should be used. Conclusions Simple test specimens and load configurations were suggested for the purpose of predicting the strength distributions of ceramic tubular components from the strength distributions of simple laboratory specimens. Stress distributions as well as the effective volume and area expressions, based on Weibull statistical theory, were derived for these tests. Simulated strength data, based on actual test data obtained from a companion paper [II], for long tube and O—ring specimens were used to demonstrate the com- putational procedure of predicting the strength distribution of the long tubes from the strength distribution of a simple speci_ men. The analytical expressions derived in this paper were used in the companion paper to assess which simple test best predicted the experimentally measured strength distributions of long m- bular components. Acknowledgments This work was supported by the Gas Research Institute under Contract 5084-2384302. APPENDIX I Stress Analysis for O-Ring Specimen V The O-ring specimen loaded in compression (Fig. 1d) is a Very convenient test due to the ease of machining and loading the specimen. However, one major drawback to this configuration is the difficulty in obtaining an accurate stress solution. This difficulty is due to the lack of a convergent Fourier expression for the boundary stresses, which become infinite under the load. However, the Fourier coefficients used in the following divergent series do legitimately form a convergent series representing the stress solution at ri S r < r0. In other words, the stress solution is accurate at the interior and throughout an arbitrary cross sec tion of the ring wall, up to a small distance, 5', from the outer surface. The stress function, 4) (r, 11;), for a circular ring,in diametral compression can be written as [10] t - fillet) r0 Cu 0 5 CG A 0“ " V 300 ri/ro=0l5 200 r'i/ro=0.6 ri/ro=0.7 r1/ro=0.8 100 rt/ro=0.BS 'lOO ~200 Stress magnification factor, Q (my) 0 15 30 45 60 75 90 Angle measured from vertical axis FIG. 6—Tangentia1 stress magnification factor, Q (I',\it), for the O—ring specimen in diametral compression as a function of angular position. It! l. l . “titere ‘- l1 ‘ ’ tution tained Sions f Stress magnification factor in the 0° plane gixaeagivhasw . . , .. l f. t a Simple spec; ‘ st best predicted ions of long m ,' : )aper were used a Institute undel- ‘ ig. id) is a very nd loading the f 3 configuration solution. This rier expression _ inder the load. Ml wing divergent presenting the stress solution ‘ _ ’ :rary cross Sec- 1 'rorn the outer 4 ‘u ignn diametral cos(nrl:) (16) 75 90 1 axis for the O-ring l ir position. ‘ .V..._.._.‘ ,._._._-\v._‘ :- ‘i s an = E; n = 0,2,4,. . . (17) TWO a,, = 0, n = 1,3,5,. .. (18) Li! = 90 — 6 and D are arbitrary constants. Solving the simultaneous equations resulting from the substi- ,“ mtion of the boundary equations into the stress equations (ob- l ained from the stress function), we obtain the following expres- sions for the constants: A": iti— - 1.2514) Be ai‘z;<‘ — 1; 321”) 09> 1 + 1.“:21‘32) D [l x?! —1_x2n 2 +1_x2n+2 " 2R” )1 x 71—1 (1 _ x2n)2 _ n2 x2n—2(1 ,_ x2)2 n M O O M O O 100 I.‘ 0 “Eula-I?.a;=£a= ‘100 -200 0.9 1.0 0.7 0.8 r/ro ratio 0.5 0.6 Stress magnification factor in the 0° plane FIG. 7— Tangential stress magnification factor, O (up), for the Ofring specimen in diametral compression as a function of radial position In the plane of the load (ti! = 0"). JADAAN ET AL. ON CERAMIC TUBULAR COMPONENTS: PART I 189 where x=5 r0 The tangential stress (09) and radial stress (0-,) are given by the equations P oe= [m own (20) 0",: b1: Z(r,ilJ) (21) where Q(r7 = 2A0 _ BO<L)_ + i re 2,4,5... 2A,,(n + 1)(n + 2)<r1> + 2n(n - 1) 3.6) — +2n(n + 1)C,, x"<L> _n— 2(n — 1)(n — 2) r0 —,. r —2 ._ an" ( > r0 X n cos(ml:) (22) and Z(r,\li) = 2A0 + B°<L> + 2 i n n—~2 An(—n3 + n2 + 2n)<rL> + B,,(—n3 + n2)(rL> o o x cos(mli) (23) Q (my) and Z(r,llJ) can be thought of as the tangential and radial stress magnification factors. Equations 22 and 23 were pro- grammed into a Vax 11/780 computer to calculate the stress magnification factors. Figure 6 displays the tangential magnification factor, Q (my), as a function of angular position for different ri/ro ratios at the inner surface of the ring. The stress magnitude and distribution is very sensitive to the wall thickness of the ring. The tangential stress distribution changes from a tensile stress field to a com— pressive stress field at an angle, 11:, ranging between 35° to 38° corresponding to an inner to outer radius ratio of 0.5 and 0.85 respectively. Figure 7 shows the tangential magnification factor, Q (hilt), as a function of radial position in the plane of the load for various ri/ro ratios. As ri/ro increases (wall thickness de- creases), the stress distribution approaches that expected for straight beam theory. This observation is of great interest from 190 JOURNAL OF TESTING AND EVALUATION a fracture surface analysis point of view. For a linear stress dis- tribution (simulating that in a flexure bar), the fracture toughness and crack branching calculations can be carried out using the crack border correction factor already derived for straight beam specimens [16]. The tangential tensile stress reaches a peak value at the inner surface of the ring in the plane of the applied load. The magnitude of the tensile stress drops sharply as it: deviates from the vertical axes until it vanishes at what is known as the boundary point. Beyond the boundary point a compressive tangential stress field develops. The exact stress solution (Eqs 20 to 23) developed above for the O-ring specimen is of great importance for strength calcu— lations. However, these equations are not feasible for failure probability analysis due to the difficulty in integrating these expressions to obtain the effective volume expression necessary for the Weibull statistical analysis. Thus a simpler and more integrable form of the stress distribution is necessary for the statistical failure probability part of the analysis. This solution comes from strain energy—straight beam theory consideration. The tangential stress component is given by ray) I my — cosi3<1 (24) P 0'B = §[0.637 Z + where I = $1)“, — ri)3 = éln‘3 = r3 — r b(r(, — ri) = bt load "uihk As the wall thickness of the ring decreases (ri/r0 increases), the exact stress distribution approaches the approximate solution obtained through strain energy—straight beam theory consider- ation. For example, at ri/ro = 0.78 (same as the r,/ro ratio used in Ref 1]), the two solutions were within 9% of each other for the tangential stresses throughout the interior surface and at the surface where r/ro = 0.989. In summary, it istreasonable to use Eq 24 for the purpose of deriving the effective volume and area expressions necessary for failure prediction analysis at ri/rO ratios greater than 0.7. How— ever, for stress and strength calculations, only Eqs 20 and 21 should be used. APPENDIX II Stress Analysis for Short Tube The short tube specimen configuration was designed to be a quick, easy, and reliable procedure for testing tubular sections at room temperature. In this test, ceramic tubes of prescribed length are filled with a polyurethane polymer that can be cured to a relatively stiff, but flexible consistency. The polymer is then axially compressed with loading platens to subject the tube to internal pressure until failure commences (Fig. 1e). For a thick—walled cylinder the radial strain, 5,, at r = ri is given by [12] all, p (1 — tor? — (1 + u,)r§ '0 w ai w—e (m ’ thick wall tube (Fig. 1f), the hoop stress distribution is described ‘ K _ r. ,.x n . m: on» erm‘r:5:1$xmb“ "4W. :1.“- . . .,. . s 3.. . <..~ . 11w .. ‘\ (wag/1,, ,;7,In~‘/1H<‘!’£-\\—1 —:1Luggagrrggrggummmw .;:§;7z;gt_ {Hm 1.2m}; Meg « . r _ , v , i ,{4 5c _‘ a. $.44, tings .4. t.,.,_...z_‘.akc d where p = pressure exerted by the rubber on the tube, r EI = Young’s modulus of tube, and uI = Poisson’s ratio of tube. ' For the rubber, ‘ 1 . 8r = E 0'1' — 1,140.6 + 0’2) for the hydrostatic case of ,, m=%=-p (m o, = -q (28) ‘ where Er = Young’s modulus of rubber, vr = Poisson’s ratio of rubber, and , q = P/A = applied pressure. Substituting Eqs 27 and 28 into Eq 26, [and then equating the radial strain for the rubber and tube at the interface (r r), we obtain the following expression for the pressure exerted by 5- the rubber on the tube: pl qu Er (1 — var? — (1 + 1).) r3, {Qii (e—e i+“.“% Since ceramic materials are very rigid compared to rubber (E,/E, = 0), then Eq 29 will simplify to the relationship ,_ 1M1 _ v,P 1—u, ‘ «Tran —v,) p (30) Equations 29 and 30 were derived for the case of a thick wall tube. From elasticity theory and for an internally pressurized by [12] ‘ (31) where p’ is the internal pressure given by Eqs 29 and 30. The hoop stress reaches its maximum value at the inner surface. Then by substituting r = r,-, the maximum hoop stress is given by the expression 7 fl r;p’ Equations 31 and 32 also apply in the case of the long tube configuration with one exception, namely, that the pressnre term is simply the pressure level of the gas inside the tube. References [I] Batdorf, S. B., “Fundamentals ofthe Statistical Theory ofFailure," in Fracture Mechanics of Ceramics, Vol. 3, R. C. Bradt, D. P. H- g l m We? as.“ ., , .EI-‘fim‘b :23,” vi “my; {in uflfifiéfittammxzemse .l tube, 4, 3n equating the I erface (r = ri), J’ sure exerted by ,~., l _ ,3} red to rubber :ionship (29) 37 (30) i of a thick wall I lly pressurized “ )n is described . 9 and 30. The surface. Then a given by the (32) u be long tube aressure term I the. ry ofFailure,” radt, D. P. H. JADAAN ET AL. ON CERAMIC TUBULAR COMPONENTS: PART I Hasselman, A. G. Evans, and F. F. Lange, Eds, Plenum Press, New York, 1978, pp. 1—29. [2] shelty, D. K, Rosenfield, A. R., and Duckworth, W. H., “Statis- tical Analysrs of.Size_and Stress State Effects on the Strength of an Alumina Ceramic,” in Methods for Assessing the Structural Relia- bility of Brittle Materials, AS TM STP 844, American Society for Testing and Materials, Philadelphia, 1984, pp. 57-80. [3] Bansal, GI. K. and Duckworth, W. H., “Effect of Specimen Size on Ceramic Strength,” in Fracture Mechanics of Ceramics, Vol. 3, R, C. Bradt, D. P. H. Hasselman, A. G. Evans, and F. F. Lange. Eds, Plenum Press, New York, 1978, pp. 189—204. [4] Deslavo, G. J_., “Theory and Structural Design Applications of Weiémll Statistics," WANL—TME—2688, Westinghouse Electric Corp., 197 . [5] Weaver, G., “Engineering with Ceramics: Part I—The Weibull Model,” Journal of Materials Education, Vol. 5, No. 5, 1983. [6] Timoshenko, S. P. and Goodier, J. N., Theory of Elasticity, McGraw— Hill, New York, 3rd ed., 1970, pp. 81—85. [7] Qassem, W. and Raftopoulus, D. D., “Stresses and Stress Factors in Curved Beams Having Initial Curvature,” presented to 22nd Annual Technical Meeting Society of Engineering Science, The Pennsylvania State University, University Park, Pa., 1985. [8] Ferber, M. K.,Tennery, V. J., Waters, S. and Ogle, J. C., “Fracture Strength Characterization of Tubular Ceramic Materials Using a 191 Simple C-ring Geometry,” Journal of Materials Science, Vol. 8, 1986, pp. 2628—2632. [9] Landini, D. J. , “Development of Slit Ring Test Method for Ceramic Tubes," EGG-MS-7531, Idaho National Engineering Laboratory, EG&G Idaho, Idaho Falls, Id., 1987. [10] Ripperger, E. A. and Davids, N., “Critical Stresses in a Circular Ring,” Transactions ofASCE, Vol. 112, 1947, pp. 619—628. [11] Shelleman, D. L., Jadaan, O.M., Conway, J. C., Jr., and Mech— olsky, J. 1., Jr., “Prediction of the Strength of Ceramic Tubular Components: II—Experimental Verification,” Journal of Testing and Evaluation, Vol. 19, No. 3, May 1991, pp. 192—200. [12] Timoshenko, 5., Strength of Materials: Part II, Robert E. Krieger Publishing Company, Malabar, Fla., pp. 205—210. [13] Weibull, W., “A Statistical Theory of the Strength of Materials," Proceedings of the Royal Swedish Institute of Engineering Research, No. 151, 1939, pp. 1—45. [14] Weibull, W., “A Statistical Distribution Function of Wide Appli- cability,” Journal of Applied Mechanics, Vol. 18, 1951, pp. 293—297. [15] Shih, T. T. , “An Evaluation of the Probabilistic Approach to Brittle Design,” Engineering Fracture Mechanics, Vol. 13, 1980, pp. 257—271. [16] Newman, J. 0, Jr., and Raju, I. 5., “Analyses of Surface Cracks in Finite Plates under Tension or Bending Load,” NASA Technical Paper 1578, 1979. f?” #55:: 3’ kafifimmfiifia m “same :‘iégr; . .- ' 3' uwmzmmmmemmsmwamnmm .r ‘ ...
View Full Document

This note was uploaded on 06/10/2011 for the course EMA 6715 taught by Professor Mecholsky during the Fall '08 term at University of Florida.

Page1 / 11

Jadaan_pt1_Cring_Oring_JTEVA91 - T .._. ._,5._.._..., p l l...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online