NASA Technical Memorandum 4721

Least Squares Best Fit Method for the Three

Parameter Weibull Distribution: Analysis

of Tensile and Bend Specimens With

Volume or Surface Flaw Failure

Bernard Gross

JULY 1996

National Aeronauticsand

Space Administration
NASA Technical Memorandum 4721

Least Squares Best Fit Method for the Three

Parameter Weibull Distribution: Analysis

of Tensile and Bend Specimens With

Volume or Surface Flaw Failure

Bernard Gross

Lewis Research Center

Cleveland, Ohio

National Aeronautics and

Space Administration

Office of Management

Scientific and Technical

Information Program

1996
Least Squares Best Fit Method for the Three Parameter Weibull Distribution:

Analysis of Tensile and Bend Specimens With Volume or Surface Flaw Failure

Bernard Gross

National Aeronautics and Space Administration

Lewis Research Center

Cleveland, Ohio 44135

Summary

Material characterization parameters obtained from naturally flawed specimens are necessary for reliability evaluation of nondeterministic advanced ceramic structural

components. The least squares best fit method is applied to

the three parameter uniaxial Weibull model to obtain the

material parameters from experimental tests on volume or

surface flawed specimens subjected to pure tension, pure

bending, four point or three point loading. Several illustrative

example problems are provided.

Introduction

The objective of this report is to apply the least squares

best fit (LSBF) method to evaluate the parameters used in the

uniaxial Weibull three parameter model. These parameters,

scale factor Go, Weibull modulus m, and threshold (location)

parameter Cu, are material dependent. Weibull two or three

parameter models are used to specify a probabilistic distribution for monolithic ceramic materials. The success in the use

of the two parameter model rather than the three parameter

model depends on the importance of ignoring the threshold

(location) parameter. Disregarding the threshold parameter is

conservative and simplifies matters. This simplification can

be justified only by comparing the predicted behavior of a

component with its observed performance.

Equations are developed to obtain the three material

parameters from inert volume or surface flawed data. Inert

data imply fast fracture (no subcritical crack growth). The

inert data are obtained from experimental tests on specimens

subjected to either pure tension, pure bending, and four or

three point loading (fig. 1). Ideally the data are obtained

under conditions representative of the service environment.

Several applications are presented in the section entitled

EXPERIMENTAL APPLICATIONS. Experimental data are

analyzed for volume flaw failure of silicon nitride (SNW1000) specimens tested in four point bending (ref. 1). In

addition, analysis is made of surface flaw failure data of

silicon carbide specimens, annealed in both the longitudinal

and transverse direction and tested in three point bending

(Private communication from Sung Choi and Jonathan

Salem, NASA Lewis Research Center). The four point bend

volume flaw data are also used for a four point bend surface

flaw analysis to illustrate the application of the developed

equations. It is realized that these data are not representative

of the physical problem.

Symbols

A

b

L

m

n

Pf

V

W

x,y,z

fil j,_2j,_3j

Gtj(x,y,z)

G_m_

(_ [Jmax,comp

co

Gu

GO

Subscripts

assumed

comp

tensile surface area

specimen thickness

beam length

Weibull modulus (shape parameter)

number of inert data points

probability of failure

volume in tension

specimen depth

Cartesian coordinates

lower limit of integral for jth specimen

stress distribution in the specimen j at fracture

maximum principal stress in specimen j at

fracture

computed maximum principal stress in specimen j based on P8 and assumed material

parameters

scale parameter

threshold stress (location parameter)

characteristic strength

assumed

computed
previous assumedpreviousvalueofassumed

j specimendesignation

T tensile

v volume

s surface

Analysis Based on Three Parameter

Uniaxial Weibull Model

The three parameter uniaxial Weibull model is used to

describe the material inert strength probabilistic distribution.

For both volume and surface flawed specimens, least squares

best fit (LSBF) methods are developed to obtain the three

material parameters from experimental tests on pure tension,

pure bending, and four or three point loaded specimens. The

necessary and sufficient condition for a solution is satisfied

when the three computed parameters produce the lowest

value of the sum of the residuals squared, that is, when

n 2

Z(OfJ ..... p-Oqnl_ ) =minimum, where n is the number

j=l

of specimens tested. The (Pfj, Ofjmax ) data points are obtained

from the experimental tests where P0 = (J - 0.3)/(n + 0.4).

PO andOfJma _ are, respectively, the probability of failure and

maximum principal tensile stress in the jth specimen at failure. On...... p is the computed maximum failure stress based

on the value of Pfj and the computed inert strength material

parameters.

Pure Tension (Fig. l(a)), Volume Flaws

r /mvvl , exp-[ fo <x,y,z)-oov

L .v,j _, °ov

(1)

Ofj(x,y, z) > Ouv

where VTj is the volume in tension of the j_a specimen with a

stress distribution throughout the volume denoted by

oq(x,y,z), ffuv is the threshold stress, ffov is the scale factor,

and mv is the Weibull modulus. For this case ffq(x,y,z) =

fffJma_and the tensile gage volume of specimen j is VTj =

L21_Wj. Hence

In VTj 1= m 'n(O0m -Ou )-m 'nOo (2)

The following system of n linear equations is solved in a

LSBF sense:

ln[ln( - Pfl )-1 ] _,_, j

lnI.ln( 1-Pf2)-I L I

In[ In( 1_r_n- Pfn )-1 I

 ln(Oflm --Oov)

l°(0,2o --Oov)

ln(ofn_ -Ouv)

1"

1

{mv::Oov

1

(3)

In matrix notation {Y } = [A] {X }, where the jth term in the

column vector {YI is yj =In • and vector

{ mv }• The equation that must be satisfied to

{X}= -mvlnoov

obtain the LSBF solution is

(4)

where superscript T defines the transpose.

The answer is obtained in the following manner: Assume a

value for Our, and solve for my and Oov. With these values,

compute the model failure stresses Ofjm_xo,,p at all of the (n)

Pfj data points, where

[ln(l- O)-I = -- _ + Ouv

Of]mu,eomp O°v L VTj

(5)

Evaluate the sum of the squares of the residuals, where

n

Sum = X(OO .... p _ Of Jmax )_ (6)

j=l

Repeat the process for another value of our. Compute the

new sum of the squares of the residuals (eq. (6)). Continue

until the parameters (mv,Oov,Ouv) produce the minimum

value of the sum of the residuals squared.
Pure Tension (Fig. l(a)), Surface Flaws

(7)

Ofj(x,y,z) > Ous

For pure tension, ot](x,y,z) = Ofjm_ , and the area in tension

is ATj = 2 L2(Wj + bj) where L2 is the gage length, W is the

width, and b is the thickness. Hence,

In = ms ln(O0m_x- Oos)-ms In Cos (8)

ATj

Equation (8) is the basis of a LSBF evaluation of the Weibull

parameters. From the inert data, a set of n linear equations is

obtained. In matrix notation {Y} = [A] { X} where the jth

term of the column vector {Y} is In [In (1 - Pfj)-I/ATj] and

f )• The matrix [A] is the same as

vector {X} is -ms In °os

that in equation (3), except that the subscript v is replaced by

s. The solution is obtained by the same method as that for the

volume flaw solution. Assume Cus, and solve for m s and _.

With these values, compute the failure stresses, (_fJmax.comp

for all n specimens. The computed failure stress for the jth

specimen is

I 1

,.(1_

Ofj ...... p = _os + ous (9)

ATj

Evaluate equation (6), the sum of the residuals squared.

Continue the process for another value of aus. Compute the

sum of the residuals squared. Continue until the parameters

(ms,aos,aus) produce the minimum value of the sum by

equation (6).

Pure Bending (Fig. l(b)), Volume Flaws

Substituting the expressions dV = L2b j dy and o 0 (x,y,z)

= 2 afjm _ y/Wj into equation (1) results in

Wj

my 1 f'--1

(1- Po) = expik .at v j 61j

:expI/'___lm 

[\Oov ) 1 + m v

20fjm_, Y

% Our)mY dy]

_l+m v

(Ot3m_ -- OUVI

Of Jmax

(10)

where

following expression is derived from equation (10):

[ln(1-PfJ)-I t =(l+mv)ln(Ofjm_ _Our)

In + In OfJr._ v. j

+In (1 + mv)O'omv_

_lj = OuvWj/(2Ofjma_) and VTj = L2bjWj/2. The

(11)

Thej_ term of the column vector {Y} is In [In (1 - Ptj)-I/

V_] + In Of Jm_x and the vector {X} is _]1@,[ +mv)Oom_(l+mv) J]-' }

The matrix [A] is the same as in equation (3). Assume auv,

and from equation (11), solve the system of n linear

equations in a LSBF sense, where j varies from 1 to n. In

matrix notation, {Y} = [A] IX}, and the solution to this set

of equations is obtained from equation (4). The final solution

is the set of parameters associated with the minimum sum of

the squares of the residuals defined by equation (6). They are

obtained by the following procedure. Assume Our, and solve

for my and Coy. With these values compute the predicted

failure stresses OfJm_ at all (n) Pfj data points. A simple

method is to assume Of Jmax,assume d, and solve for I_fJmax,comp

where

m_ (1 +mv)ln(l - -1]

Oov Pt]) l+mv

_ __ + Ouv

VTj

(12)

[

The next assumed value is ofjm_._o_, _ = 0.5 .[ofj .... _ +

ofi=_,._vio,s=_m_ ]" Repeat this process until oq ..... v is

within some specified tolerance of _t_m_,._ a. Then

compute the sum of the residuals squared by means of

equation (6). Repeat all of the previous steps until the

minimum value of the sum of the residuals squared is

obtained. The parameters (mv, ffov,_uv) associated with this

minimum are the solution•

Pure Bending (Fig. l(b)), Surface Flaws

From equation (1)

(1-Pfj) -1 :exp//--[(|L_'c°s1 )J ms _Tj(Ofj(x,y,z)-ous)m_dAj ]

(13)
where ATj is the tensile surface area of specimen j.

Therefore, considering both the side and bottom surfaces of

the specimen yields

expI/±lms(

L Oosj /2L26Uf(O0 x'Y'Z O°s) ms

where

2YOtJmax and 61j = .GusWj

6fj(x,y,z) = Wj 2Ofjm_

Thus

(14)

[ j wj j

= ms ln(afjmax - Ous)- ln[t_oms(1 + ms)] (15)

Solve in a LSBF sense the set of linear equations obtained by

means of equation (15) and denoted in matrix form by

{Y} = [A] {X}. Thej th term of the column vector {Y} is In

[In (1 -- PI_)-I/(LEWj)] - In [(! - 6m/c0,_x ) + (1 + ms)bj/_]

• The LSBF param- and the vector IXl is _ In c_ _(1 + m s

eters are obtained in the following way: Assume _us, and

keep this value fixed. To evaluate the vector {Y}, assume a

value for ms (ms,assumed) based on the two parameter

solution. Evaluate the column vector {Y I and the matrix [A].

Solve for the vector {X} by equation (4). Compare the

computed value of the Weibull modulus ms,comp with

ms, assumed. Repeat this process until both values, m s,comp and

ms,assumed, are within some specified tolerance. With these

parameters (ms,at _ aus) and Pfj, compute the n values of

crq_._o _ where

1

ms

(_fJ .... omp = aOs ( ]

[LjWjl G fj max Gus,assumed +bjLj

(16)

Assume a value for t_q._, and iterate until t_q.,_,,_o,,p is

within some specified tolerance of6q_.as_ . Obtain the

sum of the residuals squared by equation (6). Repeat the

process. The parameters that produce the minimum value of

the sum of the residuals squared are the solution.

Four Point Bend Specimen Fig. l(c)), Volume Flaws

Substituting the inner span and outer span stress distributions cfj (x,y,z) = 2oq._y/Wj and oq_(x,y,z) =

4 cfjmax Xy/(LlW j) into equation (1) results in

G mv 2- ]+mY LlbjWj _uv

Pfj=l-exp 2(l+mv ) . _fj. "

xdy+L2( 1- t_uv /

Lit t_qm_ J

(17)

where _lj = OuvWj/(2 61]m_ )" From equation (17) we obtain

Fin(l-Po)-']_ln InL

+L2(1 - au,,

L1 _ t_t]m_

- Wj

2y auv dy

61.i Wj Gfjjm_

I+m" 1 = mv In _fj,, _ - ln[l:im" (1 + my)]

(18)

Solve the set of linear equations obtained by equation (18),

denoted in matrix form by {Y} = [A] {X}. For constant

values LI and L2, thej th term of column vector {Y} is

[in(l_ pfj) -1 _- l+m v

[ 1_ -In 1 2Y _uv dy+ L2

yj = In E __ y Wj _fJmax Ll

4
Thej th row of matrix [A] is [lnOfjm_ 1.0]and {x}

IIn Ioovmmv v (1-+my '1t. Assume a value for the threshold

stress auv and an appropriate value for the Weibull modulus

my,assumed (based on the two parameter solution). Evaluate

column vector {Y} and matrix [A]. Solve for solution vector

IX} by equation (4). With _uv fixed, solve for mv,comp.

Iterate until mv,comp is within a given tolerance of mv,assum_ •

To compute the sum of the squares of the residuals by

equation (6), we obtainO o ...... p from the following

equation:

Three Point Bend (Fig. l(d)), Volume Flaws

Substituting Ofjm,x (x,y,z) = 40fjm_x/(LlW j ) into equation

(1) results in

Wj Ll

2 2

ln(l-el_)-I :2hi ff

51j _2j

____ ¢ 4Ofjma _Llwjxy / my

Our dx dy

aov

(21)

where 51j = CuvWj/(2 ofL._ ) and 52j = L l'_auv](4 _fJmax Y)"

With V'q = L lbjWj/2, integration of equation (21) results in

O'OV

°fJmax comp =

• VTj

(1 + mv)ln(1 - Pfj) -1

/

1 2y Ouv dy + 1

• Y °fJmax,assume d

ffuv 1 l+mv

°fJmax,assumed

1

mv

(19)

Assume a value ofOq,,,x._um _, and iterate until ofj.... omp is

within some specified tolerance of oq ...... . Evaluate the

sum of the residuals squared by equation (6). Repeat the

process. The parameters that produce the minimum value of

the sum of the residuals squared are the solution.

If all failures occur within the inner span and the tensile

stress distribution outside the inner span is neglected

(L 1 = 0.0), equation (17) becomes the pure bend solution, that

is,

Po=l-exp [2(l+mv)_"

L2bjWj (_fJmax

Oov

/

J 1 --

Ouv

' +mv 1

Therefore,

,n/. L[ln(lEzbT_ - PfJ)-] ]+'nOOm = (1 + mv)ln(°l]m_ -°uv)

-'n[('+m,)_om_ ] (20)

Equation (20) is the same as equation (11).

nrn0: 0' .°uv ;my]

: mv Inofjm_ " -ln[(l + m,,)_omv _ ]

(22)

Equation (22) is the limit case of equation (18) with L 2 = 0.0.

Solve the set of linear equations denoted in matrix notation

by {Y} = [A] {X} by the LSBF method (eq. (4)). The jth

value of column vector {Y} is

yj ....In !n(l - P0)-I In Iw; l/ 2y u'---_v 0 dy

VTj /_lj y [,,Wj OtJraax L

and the vector is {X}= {-In [(l+mv)Oor_ m, v ]}.To obtain an

initial value of my, let Ouv = 0 and solve for the uniaxial

Weibull two parameter distribution satisfying equation (4).

Starting with this computed value of m v as my,assume d and a

fixed value of ouv, evaluate the integral (eq. (22)) in column

vector {Y}. The integral is evaluated numerically between
the lower limit _lj and upper limit W j/2. Obtain from solution

vector {X }, mv,comp. A solution is obtained when mv,comp is

within some specified tolerance of my,assumed- When this

does not occur, the next choice for my,assume d is 0.5 (my,corn p

+ my,previous assumed). Iterate until my,assume d is within some

specified tolerance of mv,comp. To compute the sum of the

residuals squared by equation (6), we evaluate ot'j.... p in

the following way: For the n data values (Pt),Ot]r_) where

j = 1, n, assume ot_._.._d = ofj.. The lower integration

limit is _lj = GuvW/(20fj .... ttm_ )" With this limit, solve

for 00 max,cornp•

OfJmax,comp _ OovVTj my

(l+mv)ln(1-Ps) -I

Wj

i,(.y

1

mv

(23)

Assume a new value for G 8..... _e = 0.5 ( ofj.,.,p ........

+ Ofjm.,_omp ). Repeat the process, integrating over the new

limit _lj until the previous assumed value is within some

specified tolerance of the computed new value (G 8..... p=

o8 ...... a )" Compute the sum of the residuals squared by

equation (6). Repeat the process assuming a new value for

Guy, and continue until the minimum sum of the residuals

squared by equation (6) is obtained. When this occurs, the

values of my, Guv, and Oov are the three material parameters.

Four Point Bend Specimen (Fig. l(c)), Surface Flaws

Substitute into equation (7) the inner span side surface

uniaxial tensile stress distribution o 8 (x,y,z) = 20fjm_ y/W j,

the bottom surface tensile stress distribution G_(x,y,z) =

Gfjm_ and the outer span uniaxial tensile surface stress distributions Ol](x,y,z) = 4 Gfj Xy/(LIWj) and Ot](x,y,z) =

2Ofjm_xX/L 1. Normalizing the area with respect to LlW j

results in the following equation:

=l_exp I_/GfJ--/m_ LlWj Iwj I/2y

x dy + LIWj 1 °us

Gus / l+ms

(1 + ms)L2b j

+

L_Wj

/ 1 Gfjjm_GUs]l+ms]}

(24)

where 81j =Wjous/(2 Ot_max )" Thus

wj

lnI' '1,nlfl/ ou,F

L LlWj 1 LL y_Wj Ofj._

q-L2Wj+,lbj( L1Wj 1 Gus /'+m' 4 O0m_x

dy

(1 + ms)L2b j

LIWj

( °us/m' x 1 =mslnOs. _ -ln oomss(l+ms)j[ ]

(25)

Ot]m_

Solve the set of linear equations obtained from equation (25)

(denoted by {Y} = [A] {X}, in matrix form) by the LSBF

method (eq. (4)). The jth value of column vector {Y} is

wj ,nil°/' ,n

L L,Wj J IL y_Wj

L2Wj + Llbj (1 Ous

L1Wj _ OfJma_

l+m s

Ous dy

OSm_

l+m, (1 +ms)L2bj

+ LlWj

( /m'l X 1 Gus

08m.,
{[ ms )]}•Row j of matrix [A] is and vector{X} = -in ams(1 + m s

[Ct_m_ 1.0]. Assume a value for the threshold stress `'us and

an appropriate value for the Weibull modulus, ms,assume d

(based on the two parameter solution). Evaluate column

vector {Y} and matrix [A]. The vector {X} is evaluated by

equation (4). With aus fixed, solve for ms,comp. When

ms,comp is within a given tolerance of ms,assumed, a solution

results. To compute the sum of the squares of the residuals by

equation (6), the values of a n ...... p are obtained from the

following equation:

with alj = `'usWj/(2a0m _ ). _2jY= Cos I-oWj/(4ctL. _ ), and

63j = `'us Lj/2`'fjm_.Thus

Wj

-1 _-

In(1- PO),I--In[ f 2y

In LjWj J L_ j I(Wj

l+ms bj

`'us dy + --

`'_m= Wj

{ /'+ms [ ]

x 1 `'us = m s In `'fJmax -- In (1 + m s)`'o m_

`'fJ max

°fJmax,comp

ffOS

1

wj

! 2y Ous /'+ms dy +L2 +LIbJ{ L1Wj 1

151j YL Wj Of]max,assume d

l+m s

O'us +

aOmax,assumed

(1 + ms)L2b j rl oRS

L1Wj L ofj max ,assumed

1

ms

(28)

To determine `'fj ...... p, the process is the same as that

outlined for the four point bend, volume flawed specimen.

Likewise, the evaluation of the material parameters is the

same as that outlined for the four point bend, volume flawed

specimen.

Three Point Bend Specimen (Fig. l(d)), Surface Flaws

From equation (1)

Wj Lj ]

_1=( l__)ms 2 _ /4`'fJma xy ms

+2bji 83j r2,O0m= " ,,us/m'dx(27)

(26)

In matrix form {Y} = [A] {X} and thej _a term of column

vector { Y } is

I wj

1] =In -In f" l/2Y

YJ LjWj J [_j y_Wj

+m s

`'us dy

`'fjm_

and {X} = {-In [ms om'(l + ms) ]}.The solution to this set of

linear equations denoted symbolically by {Y} = [A] {X} is

solved by equation (4). Starting with `'us = 0.0, solve for m s

and the scale factor `'os. Next, assume a value for `'us. The

integrand is a function of ms. Starting with an assumed value

of m s, iterate until ms,assume d is within some specified limit of
ms,eomp. Then evaluate the scale factor. To find the value of

¢Yqm_._ompassociated with the probability of failure Pfi, and

computed material parameters (ms,Ct_t_us), satisfy the

following condition:

l+m s

(_fJ ..... P[SJ'Ij y_,Wj t_fj ..... p

1

|J t_us max,comp ll+ms ]_s-s

1

F(l+ms') m_ln(l_ )-11_-"

(29)

Assume values for t_fi_,.com p and iterate until the left side is

equal to the right side constant. Compute the sum of the

residuals squared via equation (6). Repeat the process

assuming a new value for ¢_us, and continue until the minimum value of the sum of the residuals squared is obtained.

When this occurs, the parameters (ms,t_us,aos) are the

solution.

Three Parameter Specimen Uniaxial

Weibull Model

This report deals with material properties that are

independent of the component geometry. However, a simple

model is often used to obtain the inert strength probabilistic

distribution of a given component (refs. 2 and 3). The characteristic strength parameter a0 in this model is component

dependent and is not a material property. For completeness,

this model is briefly mentioned. This model equation for

volume flaws is formulated as

' °°, )m, (30)

For surface flaws, subscript v is replaced by subscript s.

Since the characteristic strength is not a material property,

this model has its limitations. It is commonly used and is

mentioned for completeness.

Experimental Applications

The examples in this section make use of some of the

developed equations. Inert failure data are analyzed from the

following Modulus of Rupture (MOR) bar data:

(1) Four point bend room temperature failure data of

sintered silicon nitride, table I (ref. 1). All failures were due

to volume flaws and occurred within the inner span. These

data were also used for four point bend surface flaw analysis

to illustrate the application of the developed equations.

(2) Three point bend transverse annealed silicon carbide

data at 1300 °C, table II (Private communication from Sung

Choi and Jonathan Salem, NASA Lewis Research Center).

All failures were caused by surface flaws.

(3) Three point bend longitudinal annealed silicon carbide

data at 1300 °C, table III (Private communication from Sung

Choi and Jonathan Salem, NASA Lewis Research Center).

All failures were caused by surface flaws.

To develop confidence in the method developed in this

report, comparisons were made with the pure bend results

from reference i. The equations for the four point bend and

three point bend specimens were then developed and

programmed.

Sintered Silicon Nitride (Pure Bend Analysis, Volume

Flaws, Table I)

Monolithic silicon nitride data (SNW-1000, GTE Wesco

Division, table I) obtained from reference 1 are used to compare the results of various LSBF techniques. All of the data

in table I contain failures that occurred within the inner span.

In reference 1, pure bend loading (fig. l(b)) was therefore

assumed applicable to these data. The three material parameters were computed using Cooper's method (ref. 4), a modified LSBF approach (ref. 5), and the method developed

herein. Table IV summarizes the results of the three techniques used in the analysis of these data and the results

obtained herein of the two parameter model (t_uv = 0). Figure 2 is a plot of the data points and the cumulative Weibull

two parameter distribution curve. Figure 3 is a plot of the

data points and the cumulative three parameter Weibull

distribution curve.

Silicon Carbide (Three Point Bend Surface Flaws,

Tables II and III)

Table V summarizes the results obtained for the two and

three parameter uniaxial Weibull models. The cumulative

distribution curves for three point bend data (Private

communication from Sung Choi and Jonathan Salem, NASA

Lewis Research Center) for transverse and longitudinal

annealed silicon carbide and the data points are plotted in
figures4 to 7. Thetwoparameterdistributioncurvesare

plottedinfigures4 and6.Thethreeparameter distribution

curvesareplottedinfigures5and7.

Sintered Silicon Nitride (Four Point Bend Analysis,

Volume Flaws, Table I)

Table VI summarizes the results for the two and three

parameter uniaxial Weibull models. Figure 8 is a plot of the

two parameter cumulative distribution curve and data points.

Figure 9 is a plot of the three parameter cumulative distribution curve and data points.

Sintered Silicon Nitride (Four Point Bend Analysis,

Surface Flaws, Table I)

The four point bend volume flaw data are also used for a

four point bend surface flaw analysis to illustrate the application of the developed equations. It is realized that these data

are not representative of the physical problem. Table VII

summarizes the results for the two and three parameter

uniaxial Weibull models applied to these data. Figure 10 is a

plot of the two parameter cumulative distribution curve and

data points. Figure I1 is a plot of the three parameter

cumulative distribution curve and data points.

Discussion and Conclusions

Solutions are obtained from inert failure data based on the

minimizing of the sum of the residuals squared as a necessary

and sufficient condition. There are programs to evaluate the

three parameters fitted to the specimen uniaxial Weibull

model (eq. (30), refs. 2 and 3). The characteristic strength,

a0v, a parameter in this model, is not a material property but

component dependent. The results obtained using this model

are only applicable to that specific component made from the

same test material.

In this report, material property parameter estimation

methods are developed based on the uniaxial Weibull model

(eq. (1)). The parameters so obtained are applicable to any

component made from the same test data material. For the

sintered silicon nitride four point bend inert volume flaw

failure data (table I), Cooper's method (ref. 4), a modified

LSBF method (refs. l and 3), and the approach developed

herein were used to minimize the sum of the squares of the

residuals based on the pure bend solution (all failures

occurred within the inner span). Comparing the results

reveals that the largest variation of the sum of the squares of

the residuals (table IV) was less than 3 percent. Further

comparison of the results of the three methods indicated the

approach used herein was slightly less conservative in the

low probability of failure regions (Pf < 0.05) and slightly

more conservative in the upper region (Pf > 0.25). Figures 2

and 3 are plots of the data points and the computed

cumulative distribution curves for the two and three

parameter uniaxial Weibull models based on the pure bend

solution.

The results for the silicon carbide three point bend surface

flaw data (tables II and III) from longitudinal and transverse

annealed specimens are summarized in table V. Comparing

the two and three parameter models reveals that there are

large differences in the Weibull modulii and scale factors.

The cumulative distribution curves are plotted in figures 4

to 7. Superimposing the two and three parameter curves

reveals small but significant differences because most

designs are based on the very low probability of failure

region.

The four point bend solutions to the two and three

parameter uniaxial Weibull models applied to the data in

table I are summarized in table VI. The cumulative distribution curves are plotted in figures 8 and 9. Figures 2 and 3 are

the cumulative distribution curves for the pure bend solution.

A comparison of figure 2 with figure 8 and figure 3 with

figure 9 indicates the four point bend results for both cases

are slightly more conservative in the lower probability of

failure region and slightly less conservative in the higher

probability of failure region.

Four point bend volume flaw data in table I are used for a

four point bend, surface flaw analysis to illustrate the

application of the developed equations. It is realized that

these data are not representative of the physical problem. The

results are plotted in figures 10 and 11 and summarized in

table VII.

Justification for applying the three parameter model rather

than the two parameter model will depend on which model

better predicts the behavior of a component with its observed

performance.

Lewis Research Center

National Aeronautics and Space Administration

Cleveland, Ohio, March 1996

9
References

1. Starlinger, A.; Duffy, S. F.; Palko, J. L.: Parameter Estimation Techniques

Based On Optimizing Goodness-Of-Fit Statistics for Structural Reliability. Tenth Biennial Conference On Reliability, Stress Analysis, and

Failure Prevention. ASME, New York, DE Series, vol. 55, 1993.

Z Abernethy, R.B., et al.: Weibull Analysis Handbook. AFWAL-TR-83-

2079, 1983.

3. Abernethy, R.B.: New Methods for Weiball and Log Normal Analysis.

ASME Paper 92-WA/DE-14, ASME, New York, 1992.

4. Cooper, N.R.: Probabilistic Failure Prediction of Rocket Motor Components. PhD Thesis, Royal Military College of Science, England,

1988.

5_ Duffy, S.F.; Powers, L.M.; and Starlinger, A.: Reliability Analysis Using

A Three Parameter Weibull Distributions. J. Eng. Gas Turbines Power,

vol. 115, 1993, pp. 109-116.

TABLE I.--FOUR POINT

BEND SILICON NITRIDE

VOLUME FLAW INERT

FAILURE DATA (fig. l(a))

[Lj = 20.8 mm, L2 = 19.6 ram,

b =4.0 mm W = 3.1 mm.]

Specimen Failure strength,

number MPa

1

2

3

4

5

6

7

8

9

10

11

12

13

14

613.9

623.4

639.3

642.1

653.8

662.4

669.5

672.8

681.3

682.0

699.0

714.5

717.4

725.5

15

16

17

18

19

20

21

22

23

24

25

26

27

741.6

744.9

751.0

761.7

763.9

774.2

791.6

795.2

829.8

838.4

856.4

868.3

882.9

TABLE II._THREE POINT BEND SILICON

CARBIDE SURFACE FLAW TRANSVERSE

ANNEALED INERT FAILURE DATA

(fig. l(c))

[Span = L 1 = 19.936 mm.]

Specimen Thickness,

number bj,

nlrn

1 2.99t

2 2.999

3 2.999

4 2.999

5 3.000

6 2.995

7 2.996

8 2.998

9 2.997

10 2.999

11 2.997

12 2.998

13 2.998

14 2.999

15 2.995

16 2.994

17 2.998

18 2.997

19 2.994

20 2.997

21 2.999

22 3.001

23 3.000

24 3.000

25 2.993

26 2.993

27 2.995

28 2.996

29 2.994

30 2.996

31 2.996

32 2.995

33 2.996

34 2.994

Depth, Failure

wj. lo_,

nun kg

1.873 14.25

1.875 15.00

1.873 16.20

1.877 14.85

1.874 15.08

1.871 13.13

1.875 14.18

1.874 15.00

1.879 14.78

1.875 11.63

1.874 12.15

1.877 14.33

1.877 14.33

1.879 13.23

1.876 12.83

1.874 15.75

1.881 16.23

1.876 12.83

1.877 12.75

1.876 13.05

1.879 16.05

1.875 14.85

1.877 16.23

1.875 13.20

1.871 17.63

1.879 12.30

1.879 16.05

1.876 15.08

1.877 13.05

1.877 10.05

1.872 12.75

1.874 17.70

1.876 15.30

1.878 11.55

TABLE Ill.--THREE POINT BEND SILICON

CARBIDE SURFACE FLAW LONGITUDINAL ANNEALED INERT

FAILURE DATA (fig. l(c))

[ Span, L t = 19.936 ram. ]

Specimen Thickness,

number bj,

1 2.999

2 2.998

3 3.001

4 2.998

5 2.998

6 3.000

7 3.000

8 3.003

9 3.001

10 3.002

11 2.991

12 2.994

13 2.992

14 2.993

15 2.995

16 2.996

17 2.996

18 2.996

19 2.997

20 2.997

21 3.002

22 3.001

23 3.000

24 3.(102

25 3.003

26 3.000

27 3.000

28 3.0(]0

29 3.002

30 3.(102

31 2.992

32 2.993

33 2.992

34 2.994

35 2.975

Depth, Failure

W j, load,

trnn kg

1.866 12.98

1.866 14.55

1.867 17.18

1.868 15.53

1.869 14.10

1.863 15.68

1.871 15.98

1.870 16.95

1.863 17.25

1.863 12.00

1.866 15.75

1.864 15.23

1.866 14.78

1.866 14.25

1.868 14.93

1.869 12.15

1.870 15.38

1.871 14.78

1.871 14.70

1.872 12.90

1.873 9.98

1.874 10.95

1.867 15.53

1.871 14.40

1.864 13.50

1.864 12.53

1.864 12.90

1.865 13.95

1.864 14.70

1.863 12.45

1.859 12.75

1.868 13.95

1.868 15.53

1.869 17.55

1.870 12.08

lO
TABLE IV.--WEIBULL PARAMETERS OBTAINED FROM SILICON NITRIDE (SNW-1000) FOUR POINT BEND VOLUME

FLAW INERT FAILURE DATA (TABLE I)

[Data in table I are analyzed as a pure bend solution over the inner span.]

The probability fo failure for a given value of OfJma x

V / \l+mv

/ i ( VTj "](°fJmax-°UV)

isdefined as PI] = 1.0- exp/----_-S-,/-- / - ___

L 6ovV _.l+mvJ afJmax

Pure bend solution

(volume flaws, fig. 1(b))

Starlinger, et al. and Cooper - LSBF (re(. 1)

Starlinger, et al. - Modified LSBF (re(. 1)

LSBF a

Two parameter LSBF methodb

Weibull

modulus,

m_

Scale factor,

(Soy,

MPa-m 3/m,

Threshold stress,

C_uv,

MPa

1.625

1.677

1.608

11.306

0.00258276

0.00370464

0.00218469

150.1733

560.84

558.08

565.195

0.0

Sum of residuals squared,

27 of-If

j=l

2684.4

2664.0

2743.7

13440

aLSBF applied to eq. (11).

bouv set equal to zero in eq. (11).

TABLE V.--WEIBULL PARAMETERS OBTAINED FROM THREE POINT BEND SILICON CARBIDE

SURFACE FLAW INERT FAILURE DATA (TABLES I1 AND Ill)

The probability of failure for a given value of Ol]ma x defined as

Pfj ='.0- i-k"_os ] (,+m-"_)81j Y

OusWj

where 81j -

2°fJmax

2y °u-_s "_l+ms ( / dy + bj 1 - Ou_s

Wj Oqmax ) Wj k °ljmax

LSBF best method Weibull Scale factor,

(surface flaws, modulus,

fig. l(d)) ms MPa - m_m,

Transverse-annealed two- 9.294 114.52

parameter model

Transverse-annealed three- 4.024 11.71

parameter model

Longitudinal-annealed two- 9.161 114.14

parameter model

Longitudinal-annealed three- 5.893 39.78

parameter model

Threshold

SII_SS,

o_

MPa

0.0

190.0

0.0

120.0

Sum of residuals

squared,

P

Ofcomputed - Ofdat a

j 1

2665

1942

1417

1259

11
TABLE VI. --WEIBULL PARAMETERS OBTAINED FROM SILICON NITRIDE (SNW-1000) FOUR

POINT BEND VOLUME FLAW INERT FAILURE DATA (TABLE I)

The probability of failure for a given value of t_fjma x is defined as

o mvL,bjWj[ _!I_ 10xp 2(l+mv,L lj,lwj

OuvWj

where _lj

2afJmax

LSBF method Weibull Scale factor,

(volume flaws, modulus, oo_,

fig. l(c)) my MPa - m3,_v

Weibull two 10.841 141.713

parameter model

Weibull three 1.443 0.0006804

parameter model

l+mv : ._l+mv 1]

,_uv 1 dy+.L_Z[l__uv 1 /_

afJmax) LI_ OfJmax)JJ

Threshold stress,

(_uv,

MPa

0.0

564.0

Sum of residuals

squared,

(°0oxcom : °0oxll

11712.5

2038.5

TABLE VII.--WEIBULL PARAMETERS OBTAINED FROM SILICON NrrRIDE (SNW-1000) FOUR POINT BEND

VOLUME FLAW INERT FAILURE DATA

[For illustrative purposes the data in table I are analyzed as surface flaw inert failure data.l

The probability of failure for a given value of O'0ma_ is defined as

PfJ [ _ ¢_os J (iT_ss)I_lj;_Wjj

Cus / l+ms

Of JJmax

ausWj

where 61j

2afJma x

dy + L2Wj+Llbj L1Wj ( 1 °us / l+ms

°fJmax

+ (l+ms) L2bj (1---

- LIWj

LSBF method

(surface flaws,

fig. l(c))

Weibull two

parameter model

Weibull three

parameter model

Weibull

modulus,

ms

Scale factor,

(Yos,

MPa - rn

Threshold stress,

NIPa

0.0

575.0

Sum of residuals .squared,

27 2

_" (¢_0 ..... p - (IfJmax)j

j=l

12
(Jr

(_

M

(b)

L Iw', i i ;

i I

i I O"

i i

i-., L2 ,i ._,-bj_l i I 1 ,

X

,y

Neutral axis

.9 L2

M, I

/'

W

i

T

r

Y

A

(c)

._ Ll/2

X

Fj2

t

L2 L1/2 _!

L1 + L2

j

W

I_ L1/2 _!

V

X

Neutral axis

b

.... L1

Fj/2 F

/2

wj

i

!

Figure 1 ._Specimen loading and geometry. (a) Pure tension. (b) Pure bend silicon nitride (SNW-1000) specimen;

L2 = 19.6 mm, b = 4.0 mm, W = 3.1 mm. (c) Four point bend silicon nitride (SNW-1000) specimen; L1 = 20.8 mm,

L2 = 19.6 mm, b = 4.0 mm, W = 3.1 mm (table ]). (d) Three point bend silicon carbide specimen; L1 = 19.936 mm

(tables I] and ]]] contain Fj, bj, and Wj values).

13
1.0 p

0_0 0 0

0

0.8

_=0.8- °

e, 0.4

.1o

2

Q.

O.2

o.o o I I I I I I

600 650 700 750 800 850 900

Failure stress, ¢rf, MPa

Figure 2.reDistribution curve for two parameter Weibull model from volume flawed inert silicon

nitride (SNW-1000) data via specimens subjected to pure bend analysis over inner span (Fig. 1(b),

table I). Weibull modulus mv = 11.306; scale factor _rov = 150.173 MPa-m3/mv; threshold stress

_ruv = 0.0 MPa; sum of residuals squared = 13440.

1°0 --

O

<>©

0.8 --

aT

_0.6

•_ 0.4

2

0.2

o.o L I I

600 650 700 750 800 850 900

Failure stress, _rf, MPa

Figure 3.reDistribution curve for three parameter Weibull model from volume flawed inert silicon

nitride (SNW-1000) data via specimens subjected to pure bend analysis over inner span (Fig. 1(b),

table I). Weibull modulus mv = 1.608; scale factor _rov = 0.0021847 MPa--m3/mv; threshold stress

_ruv = 565.2 MPa; sum of residuals squared = 2744.

14
0=

°D

o

.O

co JD

2

a.

1.0

0.8

0.6

0.4

0.2

0.0

250 300 350 400 450 500

Failure stress, _f, MPa

Figure 4.--Distribution curve for two parameter Weibull model from surface flawed inert transverse

annealed silicon carbide data via specimens subjected to three point bend analysis (Fig. l(d),

table II). Weibull modulus m s = 9.294; scale factor _ros = 114.5 MPa-m2/ms; threshold stress

Grus = 0.0 MPa; sum of residuals squared = 2665.

°_.Q

¢0 .O

2

n

1°0 --

0.8 m

0.6 --

0.4 --

0.2 --

0.0

250 300 350 400 450 500

Failure stress, _rf, MPa

Figure 5.--Distribution curve for three parameter Weibull model from surface flawed inert transverse

annealed silicon carbide data via specimens subjected to three point bend analysis (Fig. l(d),

table I]). Weibull modulus m s = 4.024; scale factor _ros = 11.708 MPa-m2/ms; threshold stress

_rus = 190.0 MPa; sum of residuals squared = 1942.

15
1.0

0.8 --

if

g

N 0.6 --

•_ 0.4 --

0.2 --

250 300 350 400 450 500

Failure stress, ¢rf, MPa

Figure 6.--Distribution curve for two parameter Weibull model from surface flawed inert longitudinal annealed silicon carbide data via specimens subjected to three point bend analysis

(Fig. l(d), table IT0. Weibull modulus ms = 9.161; scale factor Gros= 114.14 MPa--m2/ms;

threshold stress _rus = 0.0 MPa; sum of residuals squared = 1417.

1.0

0.8

2 0.6

.o 0.4

.0

2

ft.

0.2 --

0.0

25O 300 350 400 450

Failure stress, _rf, MPa

I

500

Figure 7.--Distribution curve for three parameter Weibull model from surface flawed inert longitudinal annealed silicon carbide data via specimens subjected to three point bend analysis

(Fig. l(d), table III). Weibull modulus ms = 5.893; scale factor _os = 39.78 MPa-m2/ms;

threshold stress ¢rus= 120.0 MPa; sum of residuals squared = 1259.

16
_=

O

0_

p

0.

1.0

0.8

0.6

0.4

0.2

0.0

<)

600 650 700 750 800 850 900

Failure stress, (rf, MPa

Figure 8.--Distribution curve for two parameter Weibull model from volume flawed inert silicon

nitride (SNW-1000) data via specimens subjected to four point bend analysis (Fig. 1 (c), table I).

Weibull modulus mv = 10.84; scale factor _ruv = 141.7 MPa-m3/mv; threshold stress

_ruv = 0.0 MPa; sum of residuals squared = 11713.

"6

D

.O

.Q

o

n

1.0

0.8

0.6

0.4

0.2

0.0

600 650 700 750 800 850 900

Failure stress, _rf, MPa

Figure 9.reDistribution curve for three parameter Weibull model from volume flawed inert silicon

nitride (SNW-1000) data via specimens subjected to four point bend analysis (Fig. 1(c), table I).

Weibull modulus mv = 1.443; scale factor Crov = 0.000680 MPa-m3/mv; threshold stress

_ruv = 564.0 MPa; sum of residuals squared = 2039.

17
1.0 ©

° ° ©

0.8 --

0.6

_ 0.4 -

.1o

2

0.2

o I I I I I I 0.0

600 650 700 750 800 850 900

Failure stress, ¢rf, MPa

Figure 10.--Distribution curve for two parameter Weibull model from volume flawed inert silicon nitride (SNW1000) data via specimens subjected to four point bend analysis. As an illustrative example these data were

analyzed as if they came from surface flawed specimens (Fig. 1(c), table I). Weibull modulus ms = 10.84;

scale factor eros= 325.2 MPa-m2/ms; threshold stress ¢rus= 0.0 MPa; sum of residuals squared = 12067.

1.0

0.8

_= 0.6

.--

o

°--

•_ 0.4

.12

2

0.2

0.0

©

©

I I I I I

600 650 700 750 800 850 900

Failure stress, _rf, MPa

Figure 11 .--Distribution curve for three parameter Weibull model from volume flawed inert silicon nitride (SNW1000) data via specimens subjected to four point bend analysis. As an illustrative example these data were

analyzed as if they came from surface flawed specimens (Fig. 1(c), table I). Weibull modulus ms = 1.997;

scale factor _ros = 1.6895 MPa-m2/ms; threshold stress ¢rus= 575.0 MPa; sum of residuals squared = 1928.

]8
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1. AGENCY USE ONLY (Leave b/ank) 2. P,_.PORT DATE 3. Ha=PORT TYPE AND DATES COVi;Pi,_u

July 1996 Technical Memorandum

4. TITLE AND SUBTITLE 5. FUNDING NUIMB_J_S

Least Squares Best Fit Method for the Three Parameter Weibull Distribution:

Analysis of Tensile and Bend Specimens With Volume or Surface Flaw Failure

6. AUTHOR(S)

Bernard Gross

7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES)

National Aeronautics and Space Administration

Lewis Research Center

Cleveland, Ohio 44135-3191

9. SPONSORING/MONITORINGAGENCYNAME(S)ANDADDRESS{ES)

National Aeronautics and Space Administration

Washington, D.C. 20546-0001

WU-505--63-5B

8. P_FORMING ORGANIZATION

REPORT NUMBER

E-9980

10. SPONSORING/I_IONITORING

AGENCY REPORT NUMBER

NASA TM-4721

11. SUPPLEMENTARY NOTES

Bernard Gross, Distinguished Research Associate. Responsible person, John P. Gyekenyesi, organization code 5250,

(216) 433-3210.

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Unclassified - Unlimited

Subject Category 39

This publication is available from the NASA Center for Aerospace Information, (301) 621--0390.

12b. Dib'_PIIBUTION CODE

13. ABSTRACT (Maximum 200 words)

Material characterization parameters obtained from naturally flawed specimens are necessary for reliability evaluation

of nondeterministic advanced ceramic structural components. The least squares best fit method is applied to the three

paramenter uniaxial Weibull model to obtain the material parameters from experimental tests on volume or surface

flawed specimens subjected to pure tension, pure bending, four point or three point loading. Several illustrative

example problems are provided.

14. SUBJECT TERMS

Weibull distribution; Three parameter weibull distribution; Ceramic surface flaws;

Ceramic volume flaws

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