Askeland D.R., Fulay P.P. Essentials of Materials Science & Engineering

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There are several ways in which a crack will interact with a material’s micro-

structure. Some of these interactions can be used to enhance the fracture toughness of

materials. The intera ctions of an advancing crack include crack bridging by a ductile

phase, deﬂection of the crack in a brittle matrix by another phase, and creating a stress-

induced phase transformation at the crack tip [Figure 7-10(c)]. Other strategies include

creating crack interactions with grain boundaries and crack interactions with brittle,

second phase particles in which the second phase particles may crack or debond or

create microvoids.

EXAMPLE 7-6

Fracture in Composites

Describe the di¤erence in fracture mechanism between a boron-reinforced alu-

minum composite and a glass ﬁber-reinforced epoxy composite.

SOLUTION

In the boron-aluminum composite, the aluminum matrix is soft and ductile;

thus we expect the matrix to fail in a ductile manner. Boron ﬁbers, in contrast,

fail in a brittle manner. Both glass ﬁbers and epoxy are brittle; thus the com-

posite as a whole should display little evidence of ductile fracture.

7-5 Weibull Statistics for Failure Strength Analysis

We need a statistical approach when evaluating the strength of ceramic materials. The

strength of ceramics and glasses depends upon the size and distribution of sizes of ﬂaws.

In these materials, ﬂaws originate from the ceramic manufacturing process. The ﬂaws

also can result during machining, grinding, etc. Glasses can also develop microcracks as

a result of interaction with water vapor in air. If we test alumina or other ceramic

components of di¤erent sizes and geometry, we often ﬁnd a large scatter in the meas-

ured values—even if their nominal composition is the same. Similarly, if we are testing

the strength of glass ﬁbers of a given composition, we ﬁnd that, on average, shorter

ﬁbers are stronger than longer ﬁbers.

The strength of ceramics and glasses depends upon the probability of ﬁnding a ﬂaw

that exceeds a certain critical size. For larger components or larger ﬁbers this probabil-

ity increases. As a result, the strength of larger components or ﬁbers is likely to be lower

than that of smaller components or shorter ﬁbers. In metallic or polymeric materials,

which can exhibit relatively large plastic deformations, the e¤ect of ﬂaws and ﬂaw size

distribution is not felt to the extent it is in ceramics and glasses. In these materials,

cracks initiating from ﬂaws get blunted by plastic deformation. Thus, for ductile mate-

rials, the distribution of strength is narrow and close to a Gaussian distribution. The

strength of ceramics and glasses, however, varies considerably (i.e., if we test a large

number of identical samples of silica glass or alumina ceramic, the data will show a

wide scatter owing to changes in distribution of ﬂaw sizes). The strength of brittle ma-

terials, such as ceramics and glasses, is not Gaussian; it is given by the Weibull distri-

bution. The Weibull distribution is an indicator of the variability of strength of materials

resulting from a distribution of ﬂaw sizes. This behavior results from critical sized ﬂaws in

materials with a distribution of ﬂaw sizes (i.e., failure due to the weakest link of a chain).

C HA P T E R 7 Fracture Mechanics, Fatigue, and Creep Behavior200

The Weibull distribution shown in Figure 7-11 describes the fraction of samples

that fail at di¤erent applied stresses. At low stresses, a small fraction of samples contain

ﬂaws large enough to cause fracture; most fail at an intermediate applied stress, and a

few contain only small ﬂaws and do not fail until large stresses are applied. To provide

predictability, we prefer a very narrow distribution.

Consider a body of volume V with a distribution of ﬂaws and subjected to a stress

s. If we assumed that the volume, V, was made up of n elements with volume V

and

each element had the same ﬂaw-size distribution, it can be shown that the survival

probability, PðV

Þ, (i.e., the probability that a brittle material will not fracture under

the applied stress s) is given by:

PðV

Þ¼exp



s  s



ð7-6Þ

The probability of failure, FðV

Þ, can be written as:

FðV

Þ¼1  PðV

Þ¼1  exp



s  s



ð7-7Þ

In Equations 7-6 and 7-7, s is the applied stress, s

is cha racteristic strength (often as-

sumed to be—even though it is not—equal to the average strength), s

is the stress level

below which the probability of failure is zero (i.e., the probability of survival is 1.0). In

these equations, m is the Weibull modulus. In theory, Weibull modulus values can

range from 0 to y. The Weibull modulus is a measure of the variability of the strength

of the material.

The Weibull modulus m indicates the strength variability. For metals and alloys, the

Weibull modulus is @100. For traditional ceramics (e.g., bricks, pottery, etc.), the

Weibull modulus is less than 3. Engineered ceramics , in which the processing is better

controlled and hence the number of ﬂaws is expected to be less, have a Weibull modulus

of 5 to 10.

Note that for ceramics and other brittle solids, we can assume s

¼ 0. This is

because there is no nonzero stress level for which we can claim a brittle material will

not fail. For brittle materials, Equations 7-6 and 7-7 can be rewritten as follows:

PðV

Þ¼exp





ð7-8Þ

and

Figure 7-11 The Weibull distribution describes the fraction of the samples that fail at any

given applied stress.

7-5 Weibull Statistics for Failure Strength Analysis 201

FðV

Þ¼1  PðV

Þ¼1  exp





ð7-9Þ

From Equation 7-8, for an applied stress s of zero, the probability of surviv al is 1. As

the applied stress s increases, PðV

Þ decreases, approaching zero at very high values of

applied stresses. We can also describe another meaning of the parameter s

. In Equa-

tion 7-8, when s ¼ s

, the probability of survival becomes 1/e G 0.37. Therefore, in

brittle materials, s

is the stress level for which the survival probability is G 0.37 or

37%. We can also state that s

is the stress level for which the failure probability is

G 0.63 or 63%.

Equations 7-8 and 7-9 can be modiﬁed to account for samples with di¤erent vol-

umes. It can be shown that, for an equal probability of survival, samples with larger

volumes will have lower strengths. This is what we mentioned before (e.g., longer glass

ﬁbers will be weaker than shorter glass ﬁbers).

The following examples illustrate how the Weibull plots can be used for the ana-

lyzing of mechanical properties of materials and designing of components.

EXAMPLE 7-7 Weibull Modulus for Steel and Alumina Ceramics

Figure 7-12 shows the log-log plots of the probability of failure and strength of a

0.2% plain carbon steel and an alumina ceramic prepared using conventional

powder processing in which alumina powders are compacted in a press and sin-

tered into a dense mass at high temperature. Also included is a plot for

alumina ceramics prepared using special techniques that leads to much more

Figure 7-12 A cumulative plot (using special graph paper) of the probability that a sample will

fail at any given stress yields the Weibull modulus or slope. Alumina produced by two different

methods is compared with low carbon steel. Good reliability in design is obtained for a high

Weibull modulus. (Source: Used with permission from M.A. Meyers & K.K. Chawla, Mechanical

Behavior of Materials, 2nd ed., 2008, Cambridge University Press, UK.)

C HA P T E R 7 Fracture Mechanics, Fatigue, and Creep Behavior202

uniform and controlled particle size. This in t urn minimizes the ﬂaws. The data

for these samples are labeled as controlled particle size (CPS). Comment on the

nature of these graphs.

SOLUTION

The failure probability and strength when plotte d on a log-log scale result in

data that can be ﬁtted to a straight line. The slope of these lines provides us the

measure of variability (i.e., the Weibull modulus).

For plain carbon steel the line is almost vertical (i.e., slope or m value is

essentially approaching large values). This means that there is very little varia-

tion (5 to 10%) in the strength of di¤erent samples of the 0.2% C steel.

For alumina ceramics prepared using traditional processing, the variability

is high (i.e., m is low @4.7).

For ceramics prepared using improved and controlled processing tech-

niques the m is higher @9.7 indicating a more uniform distribution of ﬂaws.

The characteristic strength (s

) is also higher (@578 MPa) suggesting a lesser

number of ﬂaws that will lead to fracture.

EXAMPLE 7-8

Strength of Ceramics and Probability of Failure

An advanc ed engineere d ceramic has a Weibull modulus m ¼ 9. The ﬂexural

strength is 250 MPa at a probability of failure F ¼ 0:4. What is the level of

ﬂexural strength if the probability of failure has to be 0.1?

SOLUTION

We assume all samples tested had the same volume, thus the size of the sam-

ple will not be a factor in this case. We can use the symbol V for sample vol-

ume instead of V

. We are dealing with a brittle material, so we begin with

Equation 7-9.

FðVÞ¼1  PðVÞ¼1  exp





1  FðVÞ¼exp





Take the logarithm of both sides to get

ln½1  FðVÞ ¼





Take logarithms of both sides again,

lnfln½1  F ðV Þg ¼  m ðln s  ln s

Þð7-10Þ

We can eliminate the minus sign on the right-hand side of Equation 7-10 by

rewriting the equation as:

ln ln

1  F ðVÞ



¼ mðln s  ln s

Þð7-11Þ

7-5 Weibull Statistics for Failure Strength Analysis 203

For F ¼ 0:4, s ¼ 250 MPa, and m ¼ 9, so from Equation 7-11, we have

ln ln

1  0:4



¼ 9ðln 250  ln s

Þð7-12Þ

Therefore, lnfln 1 =0:6Þg ¼ lnfðln 1:66667Þg ¼ lnf0:510826g¼0 :67173 ¼

9ð5:52146  ln s

Þ.

Therefore, ln s

¼ 5:52146 þ 0:07464 ¼ 5:5961. This gives us a value of

¼ 269:4 MPa. This is the characteristic strength of the ceramic. For a stress

level of 269.4 MPa, the probability of survival is 0. 37 (or the probability of

failure is 0.63). As the required probability of failure (F ) goes down, the stress

level to which the ceramic can be subjected (s) also goes down.

Now, we want to determine the value of s for F ¼ 0:1. We know that

m ¼ 9 and s

¼ 269:4 MPa, so we need to get the value of s. We substitute

these values into Equation 7-11:

ln ln

1  0:1



¼ 9ðln s  ln 269:4Þ

ln ln

0:9



¼ 9ðln s  ln 269:4Þ

lnðln 1:11111Þ¼lnð0:105361Þ¼2:25037 ¼ 9ðln s  5:596097Þ

8 0:25004 ¼ ln s  5:596097 or

ln s ¼ 5: 346056

or s ¼ 209:8 MPa. As expected, as we lowered the probability of failure to 0.1,

we also decreased the level of stress that can be supported.

EXAMPLE 7-9

Weibull Modulus Parameter Determination

Seven silicon carbide specimens were tested and the following fracture

strengths were obtained: 23, 49, 34, 30, 55, 43, and 40 MPa. Estimate the

Weibull modulus for the data by ﬁtting the data to Equation 7-11. Discuss the

reliability of the ceramic.

SOLUTION

First, we point out that for any type of statistical analysis we need a large

number of samples. Seven samples are not enough. The purpose of this exam-

ple is to illustrate the calculation.

One simple, though not completely accurate, meth od for determinin g the

behavior of the ceramic is to assign a numerical rank (1 to 7) to the specimens,

with the specimen having the lowest fracture strength assigned the value 1. The

total number of specimens is n (in our case, 7). The probability of failure F is

then the numerical rank divided by n þ 1 (in our case, 8). We can then plot

C HA P T E R 7 Fracture Mechanics, Fatigue, and Creep Behavior204

ln½lnð1=1  F ðV

ÞÞ versus ln s. The following table and Figure 7-13 show the

results of these calculations. Note that s is plotted on a log scale.

Specimen (MPa) F (V

) ln{ln 1⁄ [1CF (V

)]}

1231=8 ¼ 0:125 2:013

2302=8 ¼ 0:250 1:246

3343=8 ¼ 0:375 0:755

4404=8 ¼ 0:500 0:367

5435=8 ¼ 0:625 0:019

6496=8 ¼ 0:750 þ0:327

7557=8 ¼ 0:875 þ0:732

The slope of the ﬁtted line, or the Weibull modulus m, is (using the two

points indicated on the curve):

m ¼

0:5 ð2:0Þ

lnð52Þlnð23:5Þ

2:5

3:951  3:157

¼ 3:15

This low Weibull modulus of 3.15 suggests that the ceramic has a highly vari-

able fracture strength, making it di‰cult to use reliably in high load-bearing

applications.

Figure 7-13

Plot of cumulative probability

of failure versus fracture stress.

Note the fracture strength is

plotted on a log scale.

7-5 Weibull Statistics for Failure Strength Analysis 205

7-6 Fatigue

Fatigue is the lowering of strength or failure of a material due to repetitive stress which

may be above or below the yield strength. It is a common phenomenon in load-bearing

components in cars and airplanes, turbine blades, springs, crankshafts and other

machinery, biomedical implants, and consumer products, such as shoes or springs,

that are subjected constantly to repet itive stresses in the form of tension, compression,

bending, vibration, thermal expansion and contraction, or other stresses. These stresses

are often below the yield strength of the material! However, when the stress occurs

a su‰cient number of times, it causes failure by fatigue! Quite a large fraction of

components found in an automobile junkyard belongs to those that failed by fatigue.

The possibility of a fatigue failure is the main reason why aircraft components have a

ﬁnite life. Fatigue is an interesting phenomenon in that load-bearing components can

fail while the overall stress applied may not exceed the yield stress! Fatigue can occur

even if the components are subjected to stress above the yield strength. A component

is often subjected to the repeated application of a stress below the yield strength of the

material.

Fatigue failures typically occur in three stages. First, a tiny crack initiates or

nucleates typically at the surface, often at a time well after loading begins. Normally,

nucleation sites are at or near the surface, where the stress is at a maximum, and in-

clude surface defects such as scratches or pits, sharp corners due to poor design or

manufacture, inclusions, grain boundaries, or dislocation concentrations. Next, the

crack gradually propagates as the load continues to cycle. Finally, a sudden fracture of

the material occurs when the remaining cross-section of the material is too small to

support the applied load. Thus, components fail by fatigue because even though the

overall applied stress may remain below the yield stress, at a local length scale the stress

intensity exceeds the yield strength. For fatigue to occur, at least part of the stress in the

material has to be tensile. We are normally concerned with fatigue of metallic and pol-

ymeric materials.

In ceramics, we normally do not consider fatigue since ceramics typically fail be-

cause of their low fracture toughness. Any fatigue cracks that may form will lower the

useful life of the ceramic since it will cause the lowering of the fracture toughness. In

general, we design ceramics for static (and not cyclic) loading and we factor in the

Weibull modulus.

Polymeric materials also show fatigue failure. The mechanism of fatigue in poly-

mers is di¤erent than that in metallic materials. In polymers, as the materials are sub-

jected to repetitive stresses, considerable heating can occur near the crack tips and the

inter-relationships between fatigue and another mechanism, known as creep (discussed

in Section 7-9), a¤ect the overall behavior.

Fatigue is also important in dealing with composites. As ﬁbers or other reinforc-

ing phases begin to degrade as a result of fatigue, the overall elastic modulus of

the composite decreases and this weakening will be seen before the fracture due to

fatigue.

Fatigue failures are often easy to identify. The fracture surface—particularly near

the origin—is typically smooth. The surface becomes rougher as the original crack

increases in size and may be ﬁbrous during ﬁnal crack propagation. Microscopic and

macroscopic examinations reveal a fracture surface including a beach mark pattern and

striations (Figure 7-14). Beach or clamshell marks (Figure 7-15) are normally formed

when the load is changed during service or when the loading is intermittent, perhaps

C HA P T E R 7 Fracture Mechanics, Fatigue, and Creep Behavior206

permitting time for oxidation inside the crack. Striations, which are on a much ﬁner

scale, show the position of the crack tip after each cycle. Bea ch marks always suggest

a fatigue failure, but—unfortunately—the absence of beach marks does not rule out

fatigue failure.

Figure 7-14 Fatigue fracture surface. (a) At low magnifications, the beach mark pattern

indicates fatigue as the fracture mechanism. The arrows show the direction of growth of the

crack front, whose origin is at the bottom of the photograph. (Image (a) is from C.C. Cottell,

‘‘Fatigue Failures with Special Reference to Fracture Characteristics,’’ Failure Analysis: The

British Engine Technical Reports, American Society for Metals, 1981, p. 318.) (b) At very high

magnifications, closely spaced striations formed during fatigue are observed (1000).

Figure 7-15

Schematic representation of a

fatigue fracture surface in a steel

shaft, showing the initiation region,

the propagation of fatigue crack

(with beach markings), and

catastrophic rupture when the crack

length exceeds a critical value at the

applied stress.

EXAMPLE 7-10

Fatigue Failure Analysis of a Crankshaft

A crankshaft in a diesel engine fails. Examination of the crankshaft reveals no

plastic deformation. The fracture surface is smooth. In additi on, several other

cracks appear at other locations in the crankshaft. What type of failure mech-

anism would you expect?

7-6 Fatigue 207

Figure 7-16 Geometry for the rotating cantilever beam specimen setup.

SOLUTION

Since the crankshaft is a rotating part, the surface experiences cyclical loading.

We should immediately suspect fatigue. The absence of plastic deformation

supports our suspicion. Furthermore, the presence of other cracks is consistent

with fatigue; the other cracks didn’t have time to grow to the size that pro-

duced catastrophic failure. Examination of the fracture surface will probably

reveal beach marks or fatigue striations.

A conventional and older method used to measure a material’s resistance to fatigue

is the rotating cantilever beam test (Figure 7-16). One end of a machined, cylindrical

specimen is mounted in a motor-driven chuck. A weight is suspended from the opposite

end. The specimen initially has a tensile force acting on the top surface, while

the bottom surface is compressed. After the specimen turns 90



, the locations that were

originally in tension and compression have no stress acting on them. After a half revo-

lution of 180



, the material that was originally in tension is now in compression. Thus,

the stress at any one point goes through a complete sinusoidal cycle from maximum

tensile stress to maximum compressive stress. The maximum stress acting on this type

of specimen is given by

Gs ¼

32 M

ð7-13aÞ

In this equation M is the bending moment at the cross-section, and d is the specimen

diameter. The bending moment M ¼ F ðL=2Þ, and therefore,

Gs ¼

16 FL

¼ 5:09

ð7-13bÞ

where L is the distance between the bending force location and the support (Figure

7-16), F is the load, and d is the diameter.

Newer machines used for fatigue testing are known as direct-load ing machines. In

these machines, a servo-hydraulic system, an actuator, and a control system, driven by

computers, applies a desired force, deﬂection, displacement or strain. In some of these

machines, temperature and atmosphere (e.g., humidity level) also can be controlled.

After a su‰cie nt number of cycles in a fatigue test, the specimen may fail. Gen-

erally, a series of specimens are tested at di¤erent applied stresses. The results are pre-

sented as an S-N curve (also known as the W—hler curve), with the stress (S) plotted

versus the number of cycles (N) to failure (Figure 7-17).

C HA P T E R 7 Fracture Mechanics, Fatigue, and Creep Behavior208

7-7 Results of the Fatigue Test

The fatigue test can tell us how long a part may survive or the maximum allowable

loads that can be applied to prevent failure. The endurance limit, which is the stress

below which there is a 50% probability that failure by fatigue will never occur, is our

preferred design criterion. To prevent a tool steel part from failing (Figure 7-17), we

must be sure that the applied stress is below 414 MPa. The assumption of existence of

an endurance limit is a relatively older concept. Recent research on man y metals has

shown that probably an endurance limit does not exist. We also need to account for the

presence of corrosion, occasional overloads, and other mechanisms that may cause the

material to fail below the endurance limit. Thus, values for an endurance limit should

be treated with caution.

Fatigue life tells us how long a component survives at a particular stress. For ex-

ample, if the tool steel (Figure 7-17) is cyclically subjected to an applied stress of

620 MPa, the fatigue life will be 100,000 cycles. Knowing the time associated with each

cycle, we can calculate a fatigue life value in years. Fatigue strength is the maximum

stress for which fatigue will not occur within a particular number of cycles, such as

500,000,000. The fatigue stren gth is necessary for designing with aluminum and poly-

mers, which have no endurance limit.

In some materials, including steels, the endurance limit is approximately half the

tensile strength. The ratio is the endurance ratio:

Endurance ratio ¼

endurance limit

tensile strength

A 0:5 ð7-14Þ

The endurance ratio allows us to estimate fatigue properties from the tensile test. The

endurance ratio values are @0.3 to 0.4 for metallic materials other than low and

medium strength steels. Again, note that research has shown that endurance limit does

not exist for many materials.

Most materials are notch sensitive, with the fatigue properties particularly sensitive

to ﬂaws at the surface. Design or manufacturing defects concentrate stresses and reduce

the enduranc e limit, fatigue strength, or fatigue life. Sometimes highly polished surfaces

Figure 7-17 The stress-number of cycles to failure (S-N) curves for a tool steel and an

aluminum alloy.

7-7 Results of the Fatigue Test 209

Askeland D.R., Fulay P.P. Essentials of Materials Science &amp; Engineering

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Askeland D.R., Fulay P.P. Essentials of Materials Science & Engineering