The Weibull Distribution Model

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GBRA 050 303 Blatt 2

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Inhalt Blatt-Nr. Index

Titelblatt 1 Änderungsblatt 2 Inhaltsverzeichnis 3 Summary 4

1 The Weibull Distribution and the Weibull Theory 5 1.1 The Weibull Distribution 5 1.2 The Weibull Theory 5

2 Fundamental Aspects 6 2.1 Deterministic and Probabilistic Design 6 2.2 Strength Distributions 8

3 System Reliability 10 3.1 Reliability of a single component 10 3.2 Probability of Combination of Events 11 3.3 Reliability of Series Systems 11 3.4 Reliability of a Redundant System 12 3.5 Combined Systems 13 3.6 Reliability of Parallel Systems With More Than One Survivor 13 3.7 The probability of system failure is 14

4 The Significance of a Safety Factor for the Design of Ceramic Structures 16

4.1 Safety and Allowable Stress 16 4.2 Definition of a Safety Factor 17 4.3 Application to the Design of the Graphite Structures of a HTR 18

5 References 20 Appendix A: Guide to the use of the Weibull Distribution A1 - A29 1. Introduction 2. The Risk Function or Failure Rate. 3. Failure Rates for Various Functions 4. Weibull Distributions with Two and Three Parameters 5. Confidence Interval 6. Tolerance Interval




Summary Probabilistic mechanics provides the theoretical and practical tools for the design, pro-

duction, testing and maintenance of components and systems having a predictably low

probability of failure, i.e. the conservatism can be specified and the risk of failure is

predetermined.

In the other hand deterministic design can be non conservative or ultra conservative

and the risk of failure is unknown.

Especially in the design of reactor components high safety demands are required.

These demands can be met best by a probabilistic design because the probability of

failure can be determined corresponding to the effects of such a failure.

This report deals with the fundamental aspects of the deterministic design, the prob-

abilistic design, the reliability of systems, and the significance of safety factors for the

design of ceramics structures.




1 The Weibull Distribution and the Weibull Theory

In the following there are two terms that should not be mixed up, namely the „Weibull Distribution“ and the „Weibull Theory”*).

1.1 The Weibull Distribution

It is a well known procedure to interpret experimental data by relating the distribution of information to some function. Very often a histogram will be used at first. Then the in-terpreter tries to fit the data to a known statistical curve in order to facilitate predictions concerning the measured data. For this a large variety of statistical distributions, such as the Binomial, Normal, Poisson, Chi-Squared, Hypergeometric, Gamma, Beta, and others can be applied. The attempt to apply some of these distributions to the failure rate of different equipment resulted in varying success. It can be shown that the fail-ures experienced in a large number of identical equipment follows an exponential pat-tern. Among the various statistical probability density functions used in reliability stud-ies mostly the Weibull distribution that belongs to the class of exponential distribution, is used. To summing-up: The Weibull distribution, also known as extreme value distribution, is a statistical distribution especially used in reliability studies.

1.2 The Weibull Theory

In his papers „Investigation into Strength Properties of Brittle Material“ and „The Phe-nomenon of Rupture in Solids“ Weibull postulated that the random size, spacing and orientation of minute flaws in brittle materials leads to a volume dependent strength. Accordingly the probability of failure in structural components is influenced not only by the magnitude of local stresses but also by the amount of stressed volume. Let us consider a simple bar under a certain load. We divide the bar in single elements of the same volume. Each element experiences the same stress. The bar fails if the weakest element fails . Therefore the Weibull theory is also called „Weakest Link The-ory“ or „Series Chain Model“. The larger the number of elements, i.e. the bigger the bar, the smaller the stress to failure. From this it follows for a specimen which is only locally stressed: the larger the stressed region, the higher the probability of the speci-men to fail. The technique described above was used successfully to predict the failure behaviour in wide ranging applications from ceramic structures through low temperature fracture in metals to solid propellant rocket motors. *) see references /1,2,3/




2 Fundamental Aspects

2.1 Deterministic and Probabilistic Design

Probabilistic mechanics provides the theoretical and practical tools for the design, pro-duction, testing and maintenance of components and systems having a predictably low probability of failure, i.e. the conservatism can be specified and the risk of failure is predetermined. Deterministic design: is possibly non conservative or ultra conservative and the risk of failure is unknown. Example Let us design a simple bar, consisting of a material with a yield strength of R, having a cross section area of A, and be loaded axially by the load P. Then the applied stress S is S= P/A Usually a safety factor is used, e.g. Sf = 1.5. Now we can determine the cross section needed to limit the stress to the allowed value. But is not possible to answer the ques-tion: How safe is the bar? An other problem will arise if the considered structure consists of many single compo-nents and we only allowed a failure in one or some of these components. The structure fails if the applied stress is greater than the strength. For normal distrib-uted variables the failure probability is

fR

P S R=

−

Φ

σ

with S: applied stress R : mean value of yield strength σR standard deviation of the strength distribution Φ tabulated value of the cumulative normal distribution Very often the load and therefore the stress is not exactly known. In the following we also assume a statistical distribution for the stress. Then the failure probability corre-sponds to the common area formed by the two statistical distributions.




We define: safety range ∆ν = R – S median safety range ∆ν = −R S safety factor ν = R/S median or central safety factor νo = R / S Failure occurs if R ≤ S; i.e. R-S ≤ 0 or R/S ≤ 1 or ∆ν ≤ 0 or ν = 1. If the stress R and the strength S are both normally distributed with the mean values µR and µS and the standard deviations σS and σR respectively, i.e. R = µR and S = µS then their difference will also be normal distributed. Hence the probability of failure is given from normal tables as

Pf = P(R-S 0) = ≤ Φ −−

+

R S

R S

µ µσ σ2 2

or in terms of the central safety factor

Pf = ( )

Φ 1 0

0

2 2

−

+

νδ ν δR S

where δR and δS: are coefficients of variation:

δS = σS/µS and δR = σR/µR Example for deterministic and probabilistic design Let us assume the following values: µR = 800 N/mm2; µS = 400 N/mm2 δS = 20%, δR = 5% a) Deterministic Design: Design stress mean value plus two standard deviations, i.e. S + 2 σS = SD Design strength: mean value minus two standard deviations, i.e. R - 2 σR = RA Design criteria: SD = RA




Then

νµ σµ σ

µµ

δδ ν

δδ

=− ⋅

+ ⋅=

− ⋅

+ ⋅

=

−

+ ⋅

⋅R R

S S

R

S

R

S

R

S

2

2

1 21 2

1 2

1 20

According to the chosen design criteria νD = 1 and therefore

0

1 2

1 2140 9

156νδδ

=+

−= =S

R

..

. ⇒ Pf = 0.0005 (non conservative)

If the design criteria is 2 SD = RA, i.e. ν = 2, then

ν0 = 2.0 . 1.56 = 3.12 P⇒ f = 1.64 10-17 (ultra conservative) b) Probabilistic Design Required: Pf = 1.0 10-6 Calculated central safety factor: ν0 = 2.07 Design stress = mean strength/2.07

2.2 Strength Distributions

Usually the normal or gaussian distribution is used to describe the strength distribution of the material. Mostly this is a good approximation if the interesting probability lies between 2% of 98%. If the design requires very low failure probabilities, e.g. 10-2 %, then the gaussian distribution is not appropriate. Example: Let us consider a coarse grain graphite (standard electrode graphite, maximum grain size: 6 mm) with the following measured properties (500 specimens): mean tensile strength: 2.34 MPa standard deviation: 0.71 MPa minimum strength: 0.70 MPa maximum strength: 4.00 MPa




Then assuming a gaussian distribution of the strength distribution we obtain the follow-ing relationship between the applied stress and the probability of failure:

Applied stress < 0.0 < 0.2 < 0.3 < 0.5

failure probability 5.0 10-4 1.4 10-3 3.0 10-3 5.0 10-3

There is an certain probability to find specimens with a „negative tensile strength“. Also the value for the probability to find a specimen with a tensile strength lower than 0.5 MPa is not correct. The statistical function appropriate to describe the strength distribution should fulfil the following conditions:

a) The probability density function has only positive values or is equal to zero.

b) The failure rate increases monotonously (the higher the stress the higher the failure probability)

The normal distribution does not satisfy the condition a). The log-normal distribution does not fulfil the condition b); the risk function on the basis of the log-normal distribu-tion first increases, but eventually decreases at higher values (e.g. of stresses ). Another family of functions called extreme value distributions, also referred to as Weibull distributions, complies with both conditions:

( ) ( )L t c t me= − ( L = probability of survival)

( ) ( )λ t

L tdLdt

= −1 ⇒

( ) ( )(λ t cm ct m= −1) (risk function) where m > 1. If t = 0 then L = 1, i.e. the function has no negative values. Further detailed information about the Weibull distribution can be found in the Appen-dix A.




3 System Reliability

3.1 Reliability of a single component

If a component is characterized by an exponential reliability function, its failure is due a chance rather than wear out.

( )TtL et = − λ ; ( ) ( )

TT tf L et

d tdt

= − = −λ λ

where λ is the failure rate (risk function) and is a constant, L is the probability of sur-vival, and fT(t) is the density function. Performing the integration, we obtain

T

te dtµ λ= −∞

∫0 or T

tt e dµ λ λλ

=−∞

∫0

T

teµλ

λ λ= =

− ∞

0

1

Hence, for an exponentially reliable component, the mean time to failure (MTTF) is 1/λ. By repeated integration it can be shown that the standard deviation is also equal to1/λ, i.e. σΤ = 1/λ. The probability that a component will reach its MTTF is

LT(µT) = exp(-λ/λ) = exp(-1) = 0.37 that means that only 37% of the components reach their mean time to failure or that the probability of exceeding the mean is .37. For the normal distribution the mean has a probability of .5. Example: If the MTTF of a component is 10,000 hrs, its reliability for a 100 hrs. opera-tion is

TL e= ≈ − =−10010000 1 100

100000 999.




3.2 Probability of Combination of Events

If A and B are two independent events with probabilities of occurrence P(A) and P(B), then the probability that both events will occur is P(AB) = P(A) • P(B) (1) If two events can occur simultaneously the probability that either A or B or both occur is (P(A∪B) = P(A) + P(B) – P(AB) (2) If the two events are mutually exclusive (P(A∪B) = P(A) + P(B) (3) IF two events are complementary and exclusive

P(A) + P(B) = 1 (4) Examples: Two dice are thrown. What is the probability of having a double six? P(A) = 1/6; P(B) = 1/6; P(6,6) = 1/36 (equ. 1) Two dice are thrown. What is the probability that at least one six appears? P(6) = 1/6 + 1/6 – 1/6 x 1/6 = 11/12 (equ. 2) A single dice is thrown. What is the probability of having a 2 or 3? P(2 or 3) = 1/6 + 1/6 = 1/3 (equ. 3) Heads and tails on a coin are complementary and exclusive events. P(H) + P(T) = 1 (equ. 4)

3.3 Reliability of Series Systems

When components are connected in such a way that the failure of a single component causes system failure, from a reliability standpoint, the components are connected in series. A chain fails if a single link breaks. For two components with reliabilities L1 and L2 in series, the reliability of the system, the two failures being independent, is LS = L1 x L2 For a system of n components the rule can be extended




( )S ii

n

L P= −=

∏ 11

If all components have exponential reliabilities with failure rates λ1, λ2, etc. then

( )St t t

ii

n

t te e e nλ λλ λ λ= • •⋅ ⋅ ⋅ ⋅ ⋅ = −

− − −

=∑1 2

1

exp .

The expression

S ii

n

λ λ=

=∑1

is called the system failure rate. The mean time to failure is µT = 1/λS and the probability that a series system will live past its MTTF is again 37%.

3.4 Reliability of a Redundant System

A system in which components are connected in such a way that each one can per-form the required function is called a redundant system. In such a system a failure of an individual component does not cause system failure. The system can be seen like a parallel arrangement of individual components. When two components with independent failure probabilities P1 and P2 are connected parallel, the system can only fail if both components fail. Using equation (1) we obtain P(system) = P1 • P2 Alternatively the system operates if either 1 or 2 or both components operate. Using equation (2) we obtain L(system) = L1 + L2 – L1 • L2 Of course the rule can be extended to several components in parallel arrangement:

( )P system and L systemii

n

ii

n

P P= == =

∏ ∏1 1

1( ) −

For identical components with exponential probabilities




( ) ( ) ( )

( )

systemt n t t

systemt

P e e e

L e

t nn n

t n

= − = − +−

+

=

− − −

−

1 11

22λ λ λ

λ

!........

The MTTF can be calculated according to the following formula

( )MTTF t dt iisystem

n

L= = =∞

∫ ∑0

1 1λ

Example: If the failure rate of the individual component is 0.02, then the failure rate of the system consisting of two such components is Psystem = 0.02 x 0.02 = 4.0 . 10-4

and the survival probability of the system is Lsystem = 1 – Psystem = 99.96 %

3.5 Combined Systems

The analysis of a system consisting of combination of series and parallel subsystems requires the breakdown of the subsystems into simpler form. For that individual com-ponents have to be combined in such a way that they form a series or a parallel sys-tem and the correspondent probabilities have to be calculated. Doing that several times one obtains finally a simple series or parallel system.

3.6 Reliability of Parallel Systems With More Than One Survivor

If it is necessary that in a parallel system, consisting of n identical components m sur-vive for safety operation, the reliability of the system can be evaluated from the bino-mial expansion of (L+Pf)n = 1.

( ) ( ) ( )L f

n

in

i

nn

fiP L P+ = =

=

−∑0

1 1




where , the binomial coefficient, is equal to n!/[(n-1)!i!]. The exponents of L indicate the number of surviving components while the exponents of P

( )inf show the number of

components failing. Example: Assuming a system consists of three components with identical reliabilities L = 0.99 and requires at least two components to work for safe operation, then

(L + Pf)3 = L3 + 3L2Pf + 3L Pf2 + Pf

3 = 1 The terms of this series are interpreted as follows: L3 is the probability that all three components operate (series arrangement) 3L2Pf is the probability that two of the three components operate and one fails.

This event can occur three ways: Component No. 1 and No. 2 operate and No. 3 does not or No. 1 and No. 3 operate and No. 2 does not or No. 2 and No. 3 operate and No. 1 does not.

3L Pf

2 is the probability that one component survives while two fail Pf

2 is the probability that all three fail (parallel arrangement) The reliability of the system, if two or more components (two or three) are required for operation is Lsystem = P[X≥2] = L3 + 3L2Pf = .993 + 3 .992 0.01 = .9997

3.7 The probability of system failure is

Pf system = P[X≤1] = 1 – Lsystem = 3L Pf2 + Pf

3 = 3.0 0.99 0.012 + 0.013 = 0.0003. 3.7 Application to Components of the High Temperature Reactor The core bottom of the THTR is supported by about 400 graphite columns. Assuming that no more than 10 of these are allowed to fail. What is the reliability of the system? Assuming that Lcolumn = 0.9, 0.95, 0.97, and 0.99 respectively, then

( ) ( )system ii

i iL l l= −=

−∑ 400

0

10400 1

The following table shows the reliabilities for different assumption.




Lcolumn Lsystem (390 out of 400

must operate) Lsystem (all must operate)

0.9 4.7 10-9 4.9 10-19 0.95 0.0094 1.2 10-9 0.97 0.344 5.1 10-6 0.99 0.99732 0.0179 0.995 0.9999924 0.135




4 The Significance of a Safety Factor for the Design of Ceramic Structures

4.1 Safety and Allowable Stress

The allowable stress σallowed for structures of materials is the limit stress up to which the structure can be loaded during operating. This value is evaluated by dividing the corresponding material property value by a safety factor greater than 1. σallowed = K/S’ The material property value K can be the yield strength, the ultimate tensile strength, the bending strength etc. The safety factor depends first of all on the safety require-ments and eventually on economic points of view. This procedure has proved to be successful for metallic structures. For ceramic struc-tures a corresponding term does not exist as a rule. But it is desirable to have such a simple procedure during the planning of a project. Let us consider the distribution of the strength of metallic materials and of materials like glass or ceramic (see sketch). For metallic materials and ductile materials respec-tively the scattering of the strength values is so that all the values are within a small region round the mean value. It is unlike to find a value outside a certain bandwidth round the mean value. Therefore if the allowed value is smaller than the mean value divided by a safety factor then the failure probability of the structure is very small. For materials like glass and ceramic the distribution of the strength differs in such a way that very small values can occur. Therefore a deterministic safety design based on a safety factor is not possible: it has to account for the strength distribution of the spe-cial material.

Sketch 1: Typical Strength Distributions for Metallic and Ceramic Materials




4.2 Definition of a Safety Factor

Point of departure for the following consideration is the mean value of the tensile, compressive or bending strength, because this value is mostly given in the material data sheet. Also we assume a Weibull distribution of the strength data. The general form is

F dm

V

= − −

∫1

0

exp σ

σV

where F = failure probability

V = volume σ0, m = Weibull parameters

σ0 is related to the magnitude of the strength and m is related to the scattering of the strength. Small values of m ( m<< 10) means large scattering and large values ( m >> 10) means small scattering. In the following we only consider a uniaxial stress state. Therefore we can write

F Vm

= − −

10

exp σ

σ (5)

and obtain the following relationship between the mean value of the strength σ and the Weibull parameter σ0 :

(σ )σ= +01 1 1/ /mV

mΓ (6)

where Γ(m) is the Gamma function. We solve equation (6) with respect to V and insert V in equation (5). We obtain

( )F mm

= − − +

1 1exp σ

σΓ

In analogy to the safety factor for metals S‘ we define now a safety factor S for ceramic materials according to the following equation




allowed Sσ σ= (S > 1) (7)

Inserting S in equation (5) and solving this equation, after looking up the logarithm, with respect to S results in

( )Sm

F

m=+

−

Γ 1 1

11

1

/ )

ln/ (8)

The following table presents the failure probability for different Weibull parameters and safety factors. S for

m F = 10-2 F = 10-3 F = 10-4 F = 10-5 F = 10-7

2 8.8 28.0 88.6 280.3 2802.5 5 2.3 3.7 4.6 9.2 23.1

10 1.5 1.9 2.4 3.0 4.8 15 1.3 1.5 1.8 2.1 2.8 20 1.2 1.4 1.5 1.7 2.2 30 1.1 1.2 1.3 1.4 1.7

Safety Factor S for Some Failure Probabilities F

According to equation (8) we establish: • S does not depend on the volume • S comes up to 1 with increasing Weibull parameter m for all failure probabilities • S comes up to 1 with increasing failure probability for all values of the Weibull pa-

rameter m • Even for m= 10, the safety factor can reach a value of 3.0 or more.

4.3 Application to the Design of the Graphite Structures of a HTR

Typical reactor graphite material has a Weibull parameter m of about 10. The allowed failure probability of the side reflector blocks at the end of the reactor life is equal to 1%: Therefore the safety factor needed is 1.5. This means that the allowable stress is the mean strength divided by 1.5. Therefore in the case of a membrane tensile or compressive stress situation the allowed stress is the tensile and the compressive strength respectively divided by 1.5. In the case of local tensile stresses and in the case of bending stresses the corresponding strength may be the bending strength. In first approximation the bending strength is higher than the tensile strength by a factor




of 1.5. Therefore the tensile strength may be used as allowed stress in the design to evaluated local tensile stresses or bending stresses. The allowed failure probability for graphite blocks whose failure cause high damage or represents a risk for the further operation of the reactor is 10-4. In this case the allow-able stress is the corresponding strength divided by a factor of about 2.5.




5 References

/1/ J. Hartung, B. Elpelt, K.-H. Klösener Statistik- Lehr- und Handbuch der angewandten Statistik R. Oldenbourg Verlag München – Wien (1986) /2/ K.C. Kapur; L.R. Lamberson

Reliability in Engineering Design John Wiley & Sons NexYork

/3/ S.M. Selby Standard Mathematical Tables The Chemical Rubber co., Cleveland, Ohio, 1972



Appendix A GBRA 050 303 Blatt A1

Appendix A

Guide to the use of the Weibull Distribution

by

Robert A. Heller

Professor of Engineering Science and Mechannics Virginia Polytechnic Institute and State University

HRB-Report BF 0684 – 20.10.81, revised by A. Schmidt






















































































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