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Unformatted text preview: T .._. ._,5._.._..., p l
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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) fourpoint 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) Cring 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
WisconsinPlatteville, 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),fourpoint bend bar, (2) C—ring tested in compression,
(3) Cring tested in tension, (4) Oring 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 fourpoint bend test (Fig. 1a), the Cring
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 Cring 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 Oring specimen (Fig. 1d) and the short tube (Fig. 16) ,i
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t 1 1 { P FIG. 1b—Cring in Compression. \ ‘ 'whc FIG. 1e—Short lube tested by axially compressing rubber inside the m”
tube. ' cha
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Whé FIG. 1c—C—ring in tension. ‘ FIG. 1f—Long lube tested with internal pressure. tho A f .o‘ ’zuwsaud [favavJéxJJSZY 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 Oring 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 twoparameter 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 Cring specimen tested in compression have
been introduced in previous investigations [4,8]. Only the effec
tive volume/area of the Cring tested in tension, Oring in dia—
metral compression, and tube configurations will be discussed
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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 ‘ eraH considered in this paper. CRing 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 crosssectional area. Equation 10 represents a wellbehaved 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 Oring
, (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
fourpoint bend, were complex and required numerical com
putations. The Oring effective volume expression (Eq 11) is an
example. This expression contains a doubleintegral 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 Oring 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, Cring 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 fourpoint bend specimen displays a trapezoidal shape 0 BEND
. Cr1ng (comp)
A C—ring (lens)
A Drtng . 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, Cring in compression, Cring 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.KEJuhmskxe‘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 . Crlng (comp)
A C—rtng (tens)
A Ortng El SHORT TUBE I LONG TUBE 50 30 35 40 45 WEIBULL MODULUS (m) FIG. 3—Effective area versus Weibull modulus for fourpoint bend specimen, Cring in compression, Cring in tension, Oring 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 . . . . . .
Cn'ng in Compression 9.5 21.9 (0.4) 17.1 (0.4)
Cring in Tension 9.5 21.9 (0.4) 17.1 (0.4)
Oring 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] oring . long tube 93‘4 30.8 4.9 0.7 In In (1/(1F)) 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 Cring 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 Oring 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 orlng ‘ 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 Oring specimen and long tube respectively. both specimens are shown in Table 2. Figure 5 shows the adjusted
Oring strength distribution which corresponds to the predicted
long tube strength distribution as calculated from the Oring
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 Oring 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 Oring, since it is this specimen from
which the prediction is made. As can be seen from Fig. 5, the
calculations for the Oring 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 Oring data corresponds to the Weibull modulus ob
tained from testing Oring 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 50842384302. APPENDIX I
Stress Analysis for ORing Specimen V The Oring 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 Oring 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 “EulaI?.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 Oring 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. 5780. [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 Cring Geometry,” Journal of Materials Science, Vol. 8,
1986, pp. 2628—2632. [9] Landini, D. J. , “Development of Slit Ring Test Method for Ceramic
Tubes," EGGMS7531, 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 ‘ ...
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 Fall '08
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