Fatigue Life Prediction of Composite Structures Under Constant Amplitude Loading

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Fatigue Life Prediction of CompositeStructures Under Constant

Amplitude Loading

CHRISTOS KASSAPOGLOU*

Composites consultant, 168 El. Venizelou street

Kastri 14671, Athens, Greece

ABSTRACT: An approach to construct the stress (or load) vs. cycles curve forcomposite structures under constant amplitude loading is presented. The approach isbased on the assumption that the cycles to failure are a function of the probability of failure during any given cycle. In this first investigation, this probability of failureduring any cycle is assumed to be constant and equal to the probability of failureobtained from static test results and the associated statistically quantified scatter.Expressions for the cycles to failure as a function of R ratio are derived. Theseexpressions do not require any curve fitting and do not involve any experimentallydetermined parameters. The fatigue predictions do not require any fatigue tests forcalibration. Comparisons to several test cases found in the literature show this firstsimple model to be very promising.

KEY WORDS: constant amplitude, probability of failure, fatigue, composites.

INTRODUCTION

FATIGUE OF COMPOSITES has been a much researched subject over the years. During

cyclic loading, several damage mechanisms arise such as matrix cracking, delamina-

tion, fiber damage, fiber/matrix interface failure, etc. Work in predicting fatigue life of

composite structures has attempted to model all these mechanisms and their interactions

[1–8 for example]. As the understanding of some of the micromechanisms involved

improved, the models became more elaborate and progressed to notched [9] and impact-

damaged [10,11] laminates. The out-of-plane effects with delamination growth in various

modes of loading were also examined [12–16]. All this work led to the creation of various

methods for predicting fatigue life. These are usually based on continuum damage

mechanics with internal state variables [1,17] modeling or tracking strength loss as a

function of cycles [18], the effect of the fiber-matrix inter-phase integrity [19], modeling

*E-mail: [email protected]

Journal of COMPOSITE MATERIALS, Vol. 41, No. 22/2007 2737

0021-9983/07/22 2737–18 $10.00/0 DOI: 10.1177/0021998307078735 SAGE Publications 2007

Los Angeles, London, New Delhi and Singapore

at Selcuk Universitesi on January 15, 2015 jcm.sagepub.comDownloaded from

http://jcm.sagepub.com/






stiffness loss and its relation to fatigue life [20,21], strain energy-based formulations for the

life of bolted joints [22], and using delamination propagation as a life-predictive tool [23].

Other approaches focus on residual strength and wear-out models [24,25].

From these and similar investigations, several general aspects of the fatigue life of

composites emerged: (a) The mechanisms are complex, interrelated, and dependent on

load type and ratio of minimum to maximum load in each cycle. (b) Modeling the

fatigue mechanisms requires modeling the damage creation and growth starting from

the microscopic level (fiber, matrix, voids) and progressing to the ply or laminate level.

(c) The scatter in composites fatigue data is very large and must be accounted for in

design. These aspects led to the creation of simplified, semi-empirical models

dependent on parameters that are experimentally determined. (In addition to the

previous references, see also [26–29]). The statistical treatment of fatigue data has been

covered by many researchers [30–32, for example].

The approaches mentioned above, while reasonably accurate (to very accurate in some

cases) in predicting fatigue life, require very detailed modeling of the damage processes,

and/or a number of empirically determined parameters. These require a number of staticand fatigue tests for characterization, curve-fitting, and statistical analysis. Composite

applications and certification in industry are then faced with a time-consuming and

expensive program for fatigue characterization.

An alternate approach is proposed here. It requires no fatigue testing, no empirically

determined parameters, and no detailed modeling of micro-crack formation, delamination,

damage creation, and propagation. At this, preliminary stage, the approach is shown to be

quite promising in predicting fatigue curves for a multitude of loading cases and structures

but requires further work to increase its applicability and its predictive power.

APPROACH – DETERMINATION OF FATIGUE LOAD

VERSUS CYCLES CURVES

The methodology is based on one assumption: the probability of failure p of the

structure during a single cycle is constant and independent of the current state of the

structure or the number of cycles up to that point. At face value, this assumption

appears to be quite limiting and not related to the progression of complex damage creation

and growth phenomena that take place during fatigue of composites. However, it forms

a basis for developing a ‘first order analysis model’ and can be relaxed later

(i.e., the probability p can be tied to the condition of the structure and the cycle

level through more elaborate analysis models). In addition, the agreement with test results

that will be demonstrated suggests that this assumption may not be as limiting as

it appears to be and the model can be used at least for preliminary design. This will be

discussed later.

If the probability of failure during each cycle is p, then, the probability P that failure has

occurred between 1 and N cycles can be determined as a function of N . This is obtained on

the basis of the fact that the probability Pi that there is failure between cycle 1 and cycle i is

the product of the probability of failure p for that cycle and the probability that there has

been no failure in the remaining N 1 cycles:

Pi ¼ pð1 pÞN 1: ð1Þ

2738 C. KASSAPOGLOU








Then, the probability P of failure anywhere between 1 and N cycles is the sum of the

probabilities Pi (i.e., failure can occur during any of the N cycles):

P

¼ XN

i ¼1

Pi

¼XN

i ¼1

p

ð1

p

ÞN 1

:

ð2Þ

And since p is independent of cycles, Equation (2) simplifies to:

P ¼ Npð1 pÞN 1: ð3Þ

Equation (3) relates the probability P that the structure has failed anywhere between 1 and

N cycles to the probability of failure p during any given cycle and the number of cycles N .

It is important to stress the fact that P is not the probability of failure after N cycles but

the probability that there has been (one and only one) failure between 1 and N cycles.

The value of p will be determined later. A plot of P as a function of N for p ¼ 1 106 is

given in Figure 1.

Strictly speaking, there is an issue with P defined as the probability of failure anywhere

between 1 and N cycles. This has to do with the fact that once there is failure at some value

N c, P as a function of cycles n where n > N c is strictly meaningless because the structure

has already failed. This can be circumvented if P is defined as the probability that the

structure gets arbitrarily close to failure but never reaches failure. For example, if failure

corresponds to some critical damage level, whichever way one uses to quantify damage, P

would be the probability that the damage in the structure gets arbitrarily close to that

critical damage but not equal to it. Then, over N cycles, if N is sufficiently large, the

structure may approach that critical condition more than once (the same way over N coin

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

000.0E+0 1.0E+6 2.0E+6 3.0E+6 4.0E+6 5.0E+6 6.0E+6 7.0E+6 8.0E+6 9.0E+6

Cycles

P r o b a b i l i t y o f f a i l u r e

a f t e r N c y c l e s ,

P

Pmax

Nc

Figure 1. Probability P of failure after N cycles as a function of cycles ( p ¼ 1 106 ).

Fatigue Life Prediction of Composite Structures 2739








tosses, the outcome heads may occur more than once). This means that if P is the

probability that the critical condition is approached once and only once between 1 and N

cycles, P will have to decrease beyond a certain value of N since it will be more likely that

the critical condition is reached a second or third time.

As is seen in Figure 1, P(N ) is not a symmetric function. The number of cycles N c at

which P is maximized, can be determined by differentiating Equation (3) with respect to N ,

and setting the result equal to 0. Then,

N c ¼ 1

lnð1 pÞ : ð4Þ

For small p, ( p<0.05), N c 1/ p, which provides a very quick and accurate evaluation

of N c. Also of interest, is the value Pmax corresponding to N c. It can be shown that for

small p, ( p<0.05):

Pmax ffi 1

e , p < 0:05 ð5Þ

where e is the Euler constant (¼2.718 . . .).

It is very significant that Pmax is independent of p for small p, which, as will be shown,

covers most cases. This implies that the maximum likelihood that the structure has failed

between 1 and N cycles is, in most cases, relatively constant. Note, from Equation (3), that

the highest possible value for P is p, which occurs when N ¼ 1. This would correspond to a

static case and the highest value of p will be the probability of failure at the mean staticstrength which is 0.5 (for the case of a normally distributed static strength).

It is postulated now that the number of cycles to failure under loading with per cycle

probability of failure p, is the value of N that corresponds to Pmax the maximum likelihood

of failure, between 1 and N cycles. Other values of N can be used, such as the value

corresponding to the mean value of P, which is twice the value of N c but, at this point, N c,

given by Equation (4), is used as the prediction of cycles to failure.

Application to Fatigue of Composites – Determination of p

Depending on the value of R ( min/ max) the cyclic load can have any of the forms

shown in Figure 2.

CASE 1: R ¼ 0

A single cycle isolated from the load train in Figure 2 for R ¼ 0 is the same as a static

load cycle. Therefore, the probability of failure p during an individual cycle is the same as

the probability of failure during static loading from zero to max. The probability p during

static loading can be obtained from the statistical distribution that describes the static

failure load and the scatter around that value. For the case where the static strength is

normally distributed, the probability p is determined using the mean strength m for the

same structural configuration (lay-up, material, notched or un-notched, etc.) and thestandard deviation s (see Figure 3).

2740 C. KASSAPOGLOU








Using normal distribution statistics, the probability p that the structure fails during one

cycle from zero to max is given by:

p ¼Z max

1

1

s ffiffiffiffiffiffi

2p exp

x mð Þ2

2s2

dx: ð6Þ

Equation (6) can be evaluated using commercially available numerical methods, tables,

numerical integration, or, with accuracy 0.07%, the expression in [33] can be used:

p ¼ 1 0:5 1 ½1 þ ðA þ BZ pÞC D þ ½1 þ ðA BZ pÞC D ð7Þ

where Z p

¼ max

m/s, A

¼0.644693, B

¼0.161984, C

¼4.874, and D

¼ 6.158.

With p determined from Equations (6) or (7), the cycles to failure N c can be obtainedfrom Equation (4) for R ¼ 0.

0≤R <1

R =0

R <0

R >1

Time

s

Figure 2. Types of cyclic loading as a function of R.

Time

s m

s

Std dev s

Density distribution of staticstrength around mean X

m

s max

Figure 3. Relation of cyclic load to static load strength distribution (R ¼ 0).









CASE 2: 0 R<1 AND R>1

In this case, the load excursion does not go back to zero (see Figure 2). As

a result, compared to a cycle with the same maximum stress (in absolute value),

the structure should be credited for the fact that it has not failed in going from zero to

the starting stress and that, in subsequent cycles, there is no probability of

failure associated with that loading portion (zero to starting stress). An analogy

could be life expectancy values, which, at any time to in a person’s life, are higher

than the value when a person is born (to¼ 0) since the person has survived up to that point

(to) in time.

It is assumed here that the probability density distribution associated with loading from

zero to the maximum value can be modified by decreasing the distance between the

1 percentile value x1 and the mean strength by an amount proportional to the reduction in

the load excursion. The distance between the mean and the 99 percentile value x2 stays the

same. A new skewed distribution results which is approximated by a two-parameter

Weibull distribution with the same mode (maximum value of probability density

distribution) as the original distribution and 98% of the data contained between x2 and(the modified) x1. This is shown schematically in Figure 4.

The quantity r shown in Figure 4 is the constant of proportionality by which the

distance m x1 is reduced. For 0 R<1, r ¼ 1 R and for R > 1, r ¼ 1 1/R. The

modified probability density is a two parameter Weibull distribution:

f ¼

x1eðx=Þ

with and given by:

1 1

1=

¼ m ð8Þ

eðx1=Þ eðx2=Þ ¼ 0:98 ð9Þ

r (s m −x 1)

1st percentile(x 1)

99th percentile(x 2)

s m

‘Original’ normaldistribution for loadexcursion startingfrom 0

‘New’ 2-parameterWeibull distribution with

same mode s m and x 1 movedto x 1

*

x 1*

s m −x 1

Figure 4. Modification of probability density function for load excursions not starting at zero.

2742 C. KASSAPOGLOU








and, for the case where the ‘original’ distribution is normal,

x1 ¼ m rð2:326Þs

x2 ¼ m þ ð2:326Þs

(the value 2.326 is the one-sided tolerance limit factor for the 99th percentile value fornormal distribution from Table 9.10.1 in [33])

Equations (8) and (9) are solved iteratively for and . Then, the probability of failure p

for applied load (equal to max if 0 R<1 or | min| if R > 1), is given by standard two-

parameter Weibull statistics:

p ¼ 1 eð=Þ : ð10Þ

It is of particular interest that, for the case of a two-parameter Weibull distribution,

placing Equation (10) in Equation (4) results in the simple expression:

N c ¼

or, solving for the stress that will lead to a failure after N cycles,

¼

ðN Þ1= ð11Þ

with and the scale and shape parameters of the two-parameter Weibull distribution

describing the static strength. Equation (11) is essentially an expression for the S – N curve

of the structure for this particular loading case. It is important to note that all the

constants in Equation (11) are known as a result of the assumptions stated earlier. No

fatigue tests are needed to determine the exponent 1/ or the coefficient .

CASE 3: R<0

In this case the maximum load is positive and the minimum load is negative as shown in

Figure 5. Each load cycle is treated as consisting of two different load excursions, one from

zero to the maximum (positive) load max, and the other from zero to the minimum

(negative) load min. Each portion has its own probability p of failure.

Letting pT and pC denote the probability of failure for a single tension or compressioncycle respectively, the problem of determining P, the probability of failure between 1 and

N cycles (where a cycle consists of the positive and the negative load excursion) is reduced

to the following: determine the probability of (one and only one) occurrence P after N

trials each with probability pT and N trials each with probability pC . The values of pT and

pC are, in general, different from each other for two reasons: First, the scatter around the

tensile failure tm can be different than the scatter around the compressive failure cm.

Second, the distance of max from X tm will be different than the distance of min from X cm.

The procedure for determining P is analogous to that of Equation (3). The resulting

expression is:

P ¼ N ð pT þ pC 2 pT pC Þð1 pT ÞðN 1Þð1 pC ÞðN 1Þ: ð12Þ









To maximize P as a function of N , the right-hand side of Equation (12) is differentiated

with respect to N and the result set equal to zero. Solving for the N c, the critical value,

gives:

N c ¼ 1

lnð1 pT Þ þ lnð1 pC Þ: ð13Þ

If pC is set to zero, a tension–tension case (R ¼ 0) is recovered and Equation (13) reduces to

Equation (4). For the case where the static tension and compression strengths are

described by two-parameter Weibull distributions with respective shape and scaleparameters, T , T and C C , the equation relating stress and cycles to failure (analogous

to Equation (11) above) can be shown to simplify to:

N ¼ 1

ð max=T ÞT þ ð min=C ÞC : ð14Þ

For the case of R ¼1 ( min¼ max¼ ) Equation (14) simplifies to:

N ¼ 1

ð=T ÞT þ ð=C ÞC , R ¼ 1: ð15Þ

If T is not equal to C , which is equivalent to the tension and compression strengths

having different scatter, Equation (15) cannot be solved explicitly for and recast in the

form of Equation (11). The stress magnitude to reach a specified number of cycles N must

be determined by solving Equation (15) iteratively.

APPLICATIONS – COMPARISONS WITH EXPERIMENTAL RESULTS

The results of the previous section were applied to a number of cases found in theliterature. These cases were selected based on two criteria (a) that they cover a wide range

Time

X m t

X m c

s max

s min

Figure 5. Static failure loads and fatigue cycles for R<0.

2744 C. KASSAPOGLOU








of applications and (b) they give sufficient information on the static strength data

(in particular some way to calculate the standard deviation and/or the scatter around the

mean) so that the equations presented in the previous section can be applied. In cases

where multiple data points were present at the same load level, only the two extreme points

(lowest and highest lives) are shown here. It should be pointed out that some error in

reading off the data from the plots found in the literature is present in the figures that

follow but it does not change the conclusions.

The case of a unidirectional AS4/3501-6 laminate under repeated tension loading (R ¼ 0)

is shown in Figure 6. Here Equation (4) was applied directly. Very good agreement with

test results is observed.

The case of tension–tension fatigue for two R values (0.1 and 0.5) for a [(45/02)2]sT800/5245 bismaleimide laminate is shown in Figure 7. Here, for a given max, p was

calculated using Equations (8)–(10) and the cycles to failure were obtained from

Equation (4). Excellent agreement with the test data is observed. From the same

reference, [35], the tension–compression case (R

¼ 1) for the same laminate is shown

in Figure 8. For the predictions, Equation (15) was used. The predictions are in verygood agreement with the test data.

A compression–compression case (R ¼ 10) from the same reference [35] is shown in

Figure 9. For a given min, p was calculated using Equations (8)–(10) and the cycles to

0

0.2

0.4

0.6

0.8

1

1.2

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Cycles

N o r m a l i z e

d s t r e s s

Test [34]

__ Present

Figure 6. Unidirectional AS4/3501–6 with R ¼ 0.

0

0.2

0.4

0.6

0.8

1

1.2

1.41.6

1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

Cycles

S t r e s s ( G P a )

R =0.1

R =0.5

Test [35]___

Present

Figure 7. Tension–tension fatigue for [( 45/02 )2 ] s T800/5245.









failure were obtained from Equation (4). Here the agreement between test and analysis is

not as good. Measured on stress, the predictions are off a best fit curve to the test data by

approx 17% at the location of greatest discrepancy. There are potential explanations for

the discrepancy including issues with the scatter calculations for the static strength data

and the validity of the approach in obtaining p via Equations (8)–(10). These will be

discussed in some detail in the next section.

A case of combined tension and torsion [36] is shown in Figure 10. The magnitude of the

tension load equals the torsion load and R is very nearly zero. The material used is woven

glass fabric. In this case, the torsion load is more critical and p was calculated for torsion

only using Equation (7). The cycles to failure were calculated from Equation (4). The

predictions have a different slope than the test data. Measured on stress, the predictions

are off a best fit curve to the data by approx. 16% at the location of greatest discrepancy.

The case of out-of-plane (pull-off) loading for skin/stiffener configurations [37] is shown

in Figure 11. The material was IM6/3501-6. The flange layup was [45/90/–45/0/90] s and the

skin layup was [90/45/02/–45/45/–45/90]s. The load shown in Figure 11 is the load causing

onset of delamination. In this case, only three data points were available for the static tests.

The standard deviation was estimated assuming that the low and high data points

contained 90% of all the test population, which was assumed to be normally distributed.

This is an approximation the validity of which is in question. For this reason, theagreement between test and predictions (maximum discrepancy on stress of 14%) should

0

0.1

0.2

0.30.4

0.5

0.6

0.7

0.8

0.9

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

Cycles

S t r e s

s ( G P a )

Test [35]

___ Present

Figure 8. Tension–compression fatigue (R ¼ 1) for [( 45/02 )2 ] s T800/5245.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.70.8

0.9

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

Cycles

S t r e s s ( G P a )

Test [35]

___ Present

Figure 9. Compresion–compression fatigue (R ¼ 10) for [( 45/02 )2 ] s T800/5245.

2746 C. KASSAPOGLOU








be viewed with caution. The data in Figure 11 correspond to a thermoset material.

Analogous results for a thermoplastic (PEEK) material are shown in Figure 12 with test

data from [38]. This shows the cycles to onset of delamination for a [352/352/02/902]slaminate using the energy release rate G as the loading parameter. It should be noted that

because the static data in [38] show that the mean is not half-way between the low and

high, a non-symmetric statistical distribution (two-parameter Weibull) was used to match

the static test data. The probability p was then calculated using Equation (10), and the

cycles to failure using Equation (4). The highest discrepancy (on stress) between test results

and predictions is 27%.

Finally, another case of delamination onset for quasi-isotropic [45/45/0/90]s glass

epoxy laminates from [39] is shown in Figure 13 for tension–tension loading (R ¼ 0.1).

Equations (8)–(10) and (4) were used. There is a difference in slope but, overall, good

agreement with test results is observed (10% maximum discrepancy of stress predictions

from test results).

The case of tension–compression (R ¼ 1) for [02/45/02/45/90]s with Celion 6000

fiber and H795E bismaleimide resin from [40] is shown in Figure 14. Equation (14) was

applied. As is seen from Figure 14, test results and analysis are in good agreement (11%maximum discrepancy on stress between predictions and test results).

0

0.05

0.1

0.150.2

0.25

0.3

0.35

0.4

0.45

0.5

1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

Cycles

M a x a p p l i e d

s t r e s s t o s t a t i c

f a i l u r e s t r e s s

Test [36]___ Present

Figure 10. Tension–torsion case (tension ¼ torsion and R ¼ 0) for woven glass fabric.

0

100

200

300

400

500

600

700

800

900

1000

1100

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

Cycles

L o a d ( N )

Test [37]

___ Present

Figure 11. Onset of delamination load for skin/stiffener configuration (R ¼ 0.1, IM6/3501-6 material).









0

50

100

150

200

250

300

350

400

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Cycles

E n e r g y r e

l e a s e r a t e G

( N - m

/ m ^ 2 )

Test [38]

___ Present

Figure 12. Onset of edge delamination for [352/352/02/902 ] s AS4/PEEK (R ¼ 0.1).

0

50

100

150

200

250

300

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Cycles

M a x s t r e s s ( M P a )

Test [39]

___ Present

Figure 13. Onset of delamination for quasi-isotropic glass/epoxy (R ¼ 0.1).

0

100

200

300

400

500

600

700

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

Cycles

M a x c o m p r s t r

( M P a )

Figure 14. Tension–compression fatigue (R ¼ 1) of [02/ 45/02/ 45/90] s BMI laminate.

2748 C. KASSAPOGLOU








The case of a more complex structure, [41], is shown in Figure 15. This is a bolted joint

with T300/914 material and layup: [02/45/02/45/90]s base plate with [02/45/9/45/90]sdoublers. Loading was tension–compression with R ¼ 1.66. Equation (14) was used. It is

interesting that the predictions track the test data well up to almost 100,000 cycles but,

after that, the predictions depart significantly from the test results. This suggests that for

relatively low applied loads, after a large number of cycles, the joints develop failure modes

(delaminations, hole elongation, etc.) that are not present in the static test or interact

differently than during the static test. As a result, using a constant value of pT and pC obtained from static tests does not adequately represent the damage growth, which takes

place at low applied loads. On the other hand, at higher applied loads, there is little

damage growth and the static values for pT and pC are adequate to quantify the fatigue

behavior.

DISCUSSION

The approach presented is very promising in that it gives reasonable to excellent

predictions of fatigue life. It is simple, does not require any fatigue testing for calibration,

and it does not depend on complex models for damage creation and evolution. This makes

it useful for preliminary design where fatigue lives can be estimated and alternate designs

compared to each other on the basis of their expected fatigue performance. The use

of the mean and standard deviation of static tests for the same loading and failure mode as

the loading and failure mode expected in the fatigue environment are sufficient for the

proposed model to work well. Despite its simplicity, the model was shown to work for a

large variety of materials (thermosets, thermoplastics, bismaleimides, glass, and graphite

fibers), in-plane and out-of-plane loads, combined loading, and a range of lay-ups from

laminates to bolted joints. However, it is important to underline the fact that the model, at

this point, is considered preliminary and more work is necessary to improve it and evaluate

its range of applicability. Several characteristics and implications of the model as well assome of its limitations and need for further work are briefly discussed in this section.

0

50

100

150

200

250

300

350

400

450

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Cycles

M a x c o m p r s t r e s s

( N / m m s q

)

Test [40]

___ Present

Figure 15. Tension–compression (R ¼ 1.66) fatigue of T300/914 bolted joints.









The curve shape implied by the present approach is a monotonically decreasing convex

shape. This means that, in principle, there is no endurance limit, i.e., load level below

which the fatigue life is infinite. However, in practice the model does reproduce an

endurance limit. The reason is that the probability p of failure during any given cycle

becomes negligibly small below some applied load, which then becomes the endurance

limit. The user can either select a value (e.g., p ¼ 1012) that represents zero probability of

failure, or let the inherent numerical limits of the computation equipment used define that

value. Another aspect of the convex shape of the resulting S – N curve in the present model

is that there is no plateau at very low cycles where, according to test data, the S – N curve in

some cases starts nearly horizontal.

Another characteristic of the model is its pronounced sensitivity to the magnitude of the

scatter around the static failure load. For example, for R ¼ 0, a 10% increase to the

standard deviation s can decrease the life by a factor of 5–10 (depending on the case). This

makes it necessary to have reliable static test data that define the distribution around the

mean as accurately as possible. It also suggests that, if the static strength distribution can

be described more accurately by a two-parameter Weibull or other non-normaldistribution, the statistics associated with that other distribution should be preferred.

In what was presented here, every attempt was made to accurately quantify the static test

results on the basis of the published data. However, it is recognized that, in some cases,

more data would be necessary for more reliable results. In particular, the approach

presented for determining p for 0<R<1 and for R > 1 should be revisited with a more

rigorous approach.

In contrast to the sensitivity to the magnitude of the scatter (value of standard deviation

s for example), the model predictions have very low sensitivity to the value of p. Reducing

p by a factor of 2 doubles the life prediction, which is a small effect compared to the typical

scatter on life for composites which spans several orders of magnitude.Besides obtaining a prediction for the fatigue life, the approach implicitly permits the

calculation of statistically meaningful (e.g., A- or B-basis) fatigue lives. If it is assumed that

the fatigue test data were to follow the likelihood of failure as expressed by the probability

P of failure, then, the cycles corresponding to the various percentile values of P (1% for

A-basis and 10% for B-basis) can be used to define corresponding fatigue lives. Letting p

range from 1E-7 to 0.4, the corresponding graph of P vs. N (as in Figure 1) can be created.

Then, the values of N corresponding to the mean and the 10 percentile (implying 90% of

the values of N are higher than the 10 percentile) are calculated and the ratio of the mean

to the 10 percentile (or B-basis) value is calculated. A plot of this ratio as a function of p is

shown in Figure 16.

As is shown in Figure 16, over a large range of values of p the ratio of mean to B-basis

life is constant and equal to 17.86. This compares favorably with the value of 13.6

calculated for the same ratio in [42] on the basis of a large number of fatigue tests.

By contrast, the corresponding number for metal structures is around 4. The current

method therefore, provides an analytical framework for calculating a statistically

significant number of fatigue lives for certification or qualification testing.

The model, as proposed, implicitly assumes that the static and fatigue failure modes are

the same. As has been pointed out by several investigators (e.g., [31]), this is not always the

case as, for example, onset of delamination and final failure after delamination growth are

two substantially different mechanisms and are expected to have different probability of

failure values p. In fact, for the case of delamination growth, p is expected to be a functionof the state of the structure at any given time, i.e., a function of N and therefore, the

2750 C. KASSAPOGLOU








assumption of constant p made here is not expected to hold. This is the reason why, results

in Figures 11–13 dealt exclusively with cycles to onset of delamination and not cycles tofinal failure. For better representation of the fatigue process, the model should be modified

so that the probability of failure at any given cycle is calculated as a function of the

damage present (matrix cracks, voids, delaminations), the current laminate properties

(reduced stiffness and strength), etc. An indication of this effect may be found in Figure 15

where, for large values of N the predictions depart significantly from the test data

suggesting the presence of damage mechanisms not captured by a single constant value of

p. On the other hand, a constant value of p can be associated with constant residual

strength. In many cases, (e.g. [3,11]) residual strength varies little over a large range of N .

Cases where residual strength is relatively constant are expected to work well with a

constant value of p at least for a range of N values. In addition, this ‘first order model’ withconstant p may be as accurate as it appears partly because the fatigue curves for

composites are relatively flat. As a result, it would take gross errors in the calculation of p

to show up as significant changes in the resulting fatigue curve.

Even for some of the cases with what was termed good agreement of analysis with tests,

(Figures 10 and 13), the slope of the predicted curve does not track the slope of the test

data closely. Besides the modeling issue just mentioned on the determination of p, other

factors may be important in these cases such as the accurate definition of static scatter, and

whether the mean or median value of the P vs. N curve should be used instead of the

modal value that was used here.

SUMMARY AND CONCLUSIONS

An approach to analytically determine the fatigue curve of composite structures

under constant amplitude loading was presented. It was applied to a wide variety of

materials, loading conditions, lay-ups, and structural configurations and was found to be

a good tool for preliminary design predictions. The approach is based on

accurate definition of the static failure strength and the scatter around it and on a

subsequent calculation of the probability of failure p during a single cycle. This probability

of failure, corrected for R ratio, is used to determine the probability P that there has

been failure between 1 and N cycles. The value of N that maximizes P is used as theprediction of cycles to failure.

5.0

7.0

9.0

11.0

13.015.0

17.0

19.0

21.0

23.0

25.0

1.E−07 1.E−06 1.E−05 1.E−04 1.E−03 1.E−02 1.E−01 1.E+00

Probability of static failure p

R a t i o o f m e a

n t o B - B a s i s l i f e

P

Mean=2Nc

Nc

10

Percentile

17.86

Figure 16. Ratio of mean to B-Basis life.









The approach allows analytical determination of the ratio of mean to B- or A-basis life

which can be used in designing certification or qualification programs.

The approach as was presented here is not expected to work well with situations where

the creation and progression of damage essentially change the structural response and

failure mode from their static counterparts. Correct modeling of such situations

(e.g., delamination growth) would require a model where the static probability of failure

p is not constant as was assumed here but depends on the current state of the structure.

The approach can be extended to such cases as well as to spectrum loading situations.

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