Modeling Material Failure
9.2 Stress and
strain based fracture and fatigue criteria
of the most successful design procedures use simple, experimentally calibrated,
functions of stress and strain to assess the likelihood of failure in a
component. Some examples of commonly
used failure criteria are summarized in this section.
9.2.1 Stress based failure criteria for brittle
solids and composites.
Experiments show that brittle
solids (such as ceramics, glasses, and fiber-reinforced composites) tend to
fail when the stress in the solid reaches a critical magnitude. Materials such as ceramics and glasses can
be idealized using an isotropic
failure criterion. Composite materials
are stronger when loaded in some directions than others, and must be modeled
using an anisotropic failure
criteria for isotropic materials
simplest brittle fracture criterion states that fracture is initiated when the
greatest tensile principal stress in the solid reaches a critical magnitude,
(The subscript TS stands for
tensile strength). To apply the
criterion, you must
1. Measure (or look up) for the material. can be measured by conducting tensile tests on
specimens it is important to test a large number of
specimens because the failure stress is likely to show a great deal of
statistical scatter. The tensile
strength can also be measured using beam bending tests. The failure stress measured in a bending test
is referred to as the `modulus of rupture’ for the material. It is nominally equivalent to but in practice usually turns out to be
2. Calculate the anticipated stress distribution in your
component or structure (e.g. using FEM).
Finally, you plot contours of principal stress, and find the maximum
value . If the design is safe (but be sure to use an
appropriate factor of safety!).
criteria for anisotropic materials
sophisticated criteria must be used to model anisotropic materials (especially
composites). The criteria must take
account for the fact that the material is stronger in some directions than
others. For example, a fiber reinforced
composite is usually much stronger when loaded parallel to the fiber direction
than when loaded transverse to the fibers.
There are many different ways to account for this anisotropy a few approaches are summarized below.
Orientation dependent fracture strength. One approach
is to make the tensile strength of the solid orientation dependent. For example, the tensile strength of a
brittle, orthotropic solid (with three distinct, mutually perpendicular
characteristic material directions) could be characterized by its tensile
strengths parallel to the three characteristic
directions in the solid.
The tensile strength when loaded
parallel to a general direction could be interpolated between these values as
are illustrated in the figure. The material fails if the stress acting normal
to any plane in the solid exceeds the fracture stress for that plane, i.e.
are the stress components in the basis . To use this criterion to check for failure at
any point in the solid, you must
(i) Find the components of stress in the basis;
(ii) Maximize the function with respect to ;
(iii) Check whether the maximum value of exceeds 1.
If so, the material will fail; if not, it is safe.
Goldenblat-Kopnov failure criterion. A very general phenomenological failure criterion can
be constructed by simply combining the stress components in a basis oriented
with respect to material axes as polynomial function. The Goldenblat-Kopnov
criterion is one example, which states that the critical stresses required to
cause failure satisfy the equation
A and B are material constants: A
is diagonal ( ) and has the same symmetries as the elasticity
tensor, i.e. . The most general anisotropic material would
therefore be characterized by 24 independent material constants, but in
practice simplified versions have far fewer parameters. Most failure criteria for composites are in
fact special cases of the Goldenblat-Kopnov criterion, including the Tsai-Hill
criterion outlined below.
Tsai-Hill criterion: The Tsai-Hill
criterion is used to model damage in brittle laminated fiber-reinforced
composites and wood. A specimen of laminated composite subjected to in-plane
loading is sketched in the figure. The
Tsai-Hill criterion assumes that a plane stress state exists in the solid. Let denote the nonzero components of stress, with
basis vectors and oriented parallel and perpendicular to the
fibers in the sheet, as shown. The
Tsai-Hill failure criterion is
failure, where ,
and are material properties. They are measured as follows:
The laminate is
loaded in uniaxial tension parallel to the fibers. The material fails when
2. The laminate is loaded in uniaxial tension
perpendicular to the fibers. The
material fails when
3. In principle, the laminate could be loaded in shear it would then fail when .
In practice it is preferable to pull on the laminate in uniaxial tension with
stress at 45 degrees to the fibers, which induces
stress components . A simple calculation then shows that .
9.2.2 Probabilistic Design
Methods for Brittle Fracture (Weibull
fracture criterion is too crude for many applications. The tensile strength of a brittle solid
usually shows considerable statistical scatter, because the likelihood of
failure is determined by the probability of finding a large flaw in a highly
stressed region of the material. This
makes it difficult to determine an unambiguous value for tensile strength should you use the median value of your
experimental data? Pick the stress level
where 95% of specimens survive? It’s better to deal with this problem using a
more rigorous statistical approach.
statistics refers to a technique used to predict the probability of failure in
a brittle material. The following
approach is used
1. Test a large number of samples with identical size and
shape under uniform tensile stress, and determine their survival probability as
a function of stress (survival probability is approximated by the fraction of
specimens that survive a given stress level).
2. Fit the survival probability of these specimens is fit by a Weibull distribution
where and m are material constants. The index m is typically of the order
5-10 for ceramics, and is independent of specimen volume. The parameter is the stress at which the probability of
survival is exp(-1), (about 37%). This critical stress depends on the specimen volume ,
and is smaller for larger specimens.
3. Given m, and the corresponding specimen volume ,
the survival probability of a volume of material subjected to uniform uniaxial
stress follows as
To see this, note that the volume V can be
thought of as containing specimens.
The probability that they all survive is .
4. More generally, the survival probability of a solid
subjected to an arbitrary stress distribution with principal values can be computed as
approach is quite successful in some applications: for example, it explains why
brittle materials appear to be stronger in bending than in uniaxial
tension. Like many statistical
approaches it has some limitations as a design tool. The method can predict accurately the stress
that gives 30% probability of failure.
But who wants to buy a product that has a 30% probability of
failure? For design applications we need
to predict the probability of 1 failure in a million or so. It is very difficult to measure the tail
of a statistical distribution accurately, and a distribution that was fit to
predict 63% failure probability may be wildly inaccurate in the region of
9.2.3 Static Fatigue Criterion for Brittle
fatigue’ refers to the progressive reduction in tensile strength of a stressed
brittle material with time. The simplest
way to model static fatigue is to make the tensile strength of the material a
function of time and applied stress. The
usual approach is to set
is the maximum principal stress acting on the
solid, which may vary slowly with time t;
is the tensile strength of the solid at time t=0, and are two material constants. Typically m
has values between 5 and 10. For the
particular case of a constant stress,
we see that
Since failure occurs when ,
the time to failure follows as
that and m
can easily be determined by measuring the time to failure in uniaxial tension
as a function of applied stress.
Under multi-axial loading, the maximum principal tensile stress should be
used for .
9.2.4 Constitutive laws for crushing failure of
materials are generally used in applications where they are subjected primarily
to compressive stress. Brittle materials are very strong in compression, but
they will fail if subjected to combined hydrostatic compression and shear (e.g.
by loading in uniaxial compression).
Failure in compression is a consequence of distributed microcracking
in the solid large numbers of small cracks nucleate,
propagate for a short while and then arrest.
Failure occurs as a result of coalescence of these cracks. A typical stress-strain curve during
compression of a brittle material, such as concrete, is illustrated in the
figure. Failure in compression is less
catastrophic than tension, and in some respects qualitatively resembles metal
plasticity. For plastically deforming
metals, however, the stress-strain curve is independent of hydrostatic
stress. In contrast, the crushing
resistance of a brittle material increases with hydrostatic compression.
type of crushing is often modeled using constitutive equations based on
small-strain metal plasticity. The
governing equations for a simple, small-strain, constitutive model of this form
will be summarized briefly here. A more detailed discussion of plasticity
theory is given in Section 3.6.
material is characterized by the following properties:
Young’s modulus E and Poisson ratio
stress-v-plastic strain curve measured from a uniaxial compression test, which
is fit by a functional relation of the form ,
where is the magnitude of the compressive
strain. Any of the functions listed in
Section 3.6.5 could be used for the function Y.
material constant c, which controls
how rapidly the strength of the material increases with hydrostatic
constitutive equations specify a relationship between an increment in stress applied to the material and an increment in
The strain is decomposed
into elastic and irreversible (damage) parts as ;
The elastic part
of the strain is related to the stress by the linear elastic constitutive
3. The critical stress that initiates crushing damage is
given by a failure criterion (analogous to the yield criterion for a metal) of
and is the accumulated irreversible strain. Notice
that the failure criterion depends on the hydrostatic part of the stress:
unlike yield in metals, the material becomes more resistant to fracture if p<0.
strain components are determined using an associated flow rule
The magnitude of
the plastic strain increment is related to the stress increment by
is the slope of the uniaxial stress-strain
curve, and for ,
while for .
HEALTH WARNING: These constitutive equations should only be used in
regions where the hydrostatic stress is compressive . In regions of hydrostatic tension, a tensile
brittle fracture criterion should be used for example, the material could be assumed to
lose all load bearing capacity if the principal tensile stress exceeds a
Ductile Fracture Criteria
to failure approach: Ductile
fracture in tension occurs by the nucleation, growth and coalescence of voids
in the material. A crude criterion for
ductile failure could be based on the accumulated plastic strain, for example
at failure, where is the plastic strain to failure in a uniaxial
show that the strain to cause ductile failure in a material depends on the
hydrostatic component of tensile stress acting on the specimen, as shown in the
figure. For example, the strain to
failure under torsional loading (which subjects the material to shear with no
hydrostatic stress) is much greater than under uniaxial tension. The critical
strain is influenced by hydrostatic stress because ductile failure occurs as a
result of the nucleation and growth of cavities in the solid. A hydrostatic stress greatly increases the
rate of growth of the cavities. The simple strain-to-failure approach cannot
account for this behavior.
plasticity was developed to address this issue.
The basic idea is simple: the solid is idealized as a plastic matrix
which contains a volume fraction of cavities.
To model the solid, the plastic stress-strain laws outlined in Sections
3.6 and 3.7 are extended to calculate the volume fraction of voids in the
material as part of the solution, and also to account for the weakening effect
of the voids. Failure is modeled by constructing
the plastic stress-strain law so that the material loses all its strength at a
critical void volume fraction.
rate independent and viscoplastic versions of porous plasticity exist. The viscoplastic models have some advantages
for finite element computations, because the rate dependence can stabilize the
effects of strain softening. A simple
small-strain viscoplastic constitutive law with power-law hardening and
power-law rate dependence will be outlined here to illustrate the main features
of these models. The constitutive law is
known as the `Gurson model.’
material is characterized by the following properties:
Young’s modulus E and Poisson ratio ;
characteristic stress Y, a
characteristic strain and
strain hardening exponent n, which
govern the strain hardening behavior of the matrix material;
characteristic strain rate and strain rate exponent m, which govern the strain rate sensitivity of the solid;
which controls the rate of void nucleation with plastic straining;
strength of the matrix ,
the void volume fraction, ,
and the total accumulated effective plastic strain in the matrix material which all evolve with plastic straining.
constitutive equations specify a relationship between the stress applied to the material and the resulting
strain rate ,
The strain rate
is decomposed into elastic and plastic parts as ;
The elastic part
of the strain rate is related to the stress rate by the linear elastic constitutive
3. The magnitude of the plastic strain rate is determined
by the following plastic flow potential
and . Note that for the plastic strain rate increases with
hydrostatic stress p.
The components of the
plastic strain rate tensor are computed from an associated flow law
5. Strain hardening in the matrix is modeled by relating
its flow stress to the accumulated strain in the matrix .
The following power-law hardening model is often used
6. The effective plastic strain in the matrix is
calculated from the condition that the plastic dissipation in the matrix must
equal the rate of work done by stresses, which requires that
7. Finally, the model is completed by specifying the void
volume fraction as a function of strain.
The void volume fraction can increase due to growth of existing voids,
or nucleation of new ones. To account for both effects, one can set
where the first term accounts for void growth, and the
second accounts for strain controlled void nucleation.
9.2.6 Ductile failure by strain localization
you test a cylindrical specimen of a very ductile material in uniaxial tension,
it will initially deform uniformly, and remain cylindrical. At a critical load
(or strain) the specimen will start to neck, as shown in the picture. Necking, once it starts, is usually unstable there is a concentration in stress near the
necked region, increasing the rate of plastic flow near the neck compared with
the rest of the specimen, and so increasing the rate of neck formation. The strains in the necked region rapidly
become very large, and quickly lead to failure.
formation is a consequence of geometric softening. A very simple model explains the concept of
1. Consider a cylindrical specimen with initial cross
sectional area and
The specimen is subjected to a load P, which deforms the material
plastically. After straining, the length of the specimen increases to L,
and its cross-sectional area decreases to A.
2. Assume that the material is perfectly plastic and has
a true stress-strain curve (Cauchy stress v- logarithmic
strain) that can be approximated by a power-law with n<1.
3. The true strain in the specimen is related to its
4. The force on the specimen is related to the Cauchy
5. At the point of maximum load
6. We can calculate by noting that the volume of the specimen is
constant during plastic straining, which shows that
that is negative this means that the specimen tends to soften
as a result of the change in its cross sectional area. This is what is meant by geometric softening.
We can calculate from (2) and (3) as follows
that is positive strain hardening in the material tends to
compensate for the effects of geometric softening.
substituting the results of (6) and (7) back into (5) and recalling that shows that at the point of maximum load, the
strain and length of the specimen are
that by volume conservation the cross sectional area is ,
so the maximum load the specimen can withstand follows as
turns out that the point of maximum load coincides with the condition for
unstable neck formation in the bar. This
is plausible a falling load displacement curve is always a
sign that there might be a possibility of non-unique solutions but a rather sophisticated calculation is
required to show this rigorously.
There are two important
points to take away from this discussion.
Plastic localization, as opposed to material
failure, may limit load bearing capacity;
If you measure the strain to failure of a
material in uniaxial tension, it is possible that you have not measured the
inherent strength of the material your specimen may have failed due to a
geometric effect. Material behavior does
influence the strain to failure, of course: the simple analysis of geometric
softening shows that the strain hardening behavior of the material is
Plastic localization can occur for many reasons. There are two general classes of localization
it may occur as a consequence of changes in
specimen geometry (i.e. geometric softening); or it may occur due to a natural
tendency of the material itself to soften at large strains.
of geometry induced localization are
Neck formation in
a bar under uniaxial tension;
2. Shear band formation in torsional or shear loading at
high strain rate due to thermal softening as a result of plastic heat
Examples of material
induced localization are
1. Localization in a Gurson solid due to the softening
effect of voids at large strains;
2. Localization in a single crystal due to the softening
effect of lattice rotations;
3. Localization in a brittle microcracking material due
to the increase in elastic compliance caused by the cracks.
localization can be modeled quite easily, because it does not rely on any
empirical failure criteria. A
straightforward FEM computation, with an appropriate constitutive law and
proper consideration of finite strains, will predict localization if it is going
to occur the only thing you need to worry about is to
be sure you understand what triggered the localization. Localization can start at a geometric
imperfection in the model, in which case your prediction is meaningful (but may
be sensitive to the nature of the imperfection). It may also be triggered by numerical errors,
in which case the predicted failure load is meaningless. It is usually exceedingly difficult to
compute what happens after localization.
Fortunately it’s rather rare to need to do this for design purposes.
9.2.7 Criteria for
failure by high cycle fatigue under constant amplitude cyclic loading
Empirical stress or strain based life
prediction methods are extensively used in design applications. The approach is straightforward subject a sample of the material to a cycle of
stress (or strain) that resembles service loading, in an environment
representative of service conditions, and measure its life as a function of
stress (or strain) amplitude, then fit the data with a curve.
we will review criteria that are used to predict fatigue life under
proportional cyclic loading. A typical stress cycle is parameterized by its
amplitude and the mean stress
tests run in the high cycle fatigue regime with any fixed value of mean stress,
the relationship between stress amplitude and the number of cycles to failure N
is fit well by Basquin’s Law
where the exponent b
is typically between 0.05 and 0.15. The
constant C is a function of mean stress.
are two ways to account for the effects of mean stress. Both are based on the same idea: we know that
if the mean stress is equal to the tensile strength of the material ,
it will fail in 0 cycles of loading. We
also know that for zero mean stress, the fatigue life obeys Basquin’s law. We can interpolate between these two points. There are two ways to do this:
uses a linear interpolation, giving
is the constant in Basquin’s law determined by
testing at zero mean stress.
uses a parabolic fit
practice, experimental data seem to lie between these two limits. Goodman’s rule gives a safe estimate.
criteria are intended to be used for components that are subjected to uniaxial
tensile stress. The criteria can still
be used if the loading is proportional
(i.e. with fixed directions of principal
stress). In this case, the maximum
principal stress should be used to calculate and . They do not work under non-proportional
loading. A very large number of fatigue
models have been developed for more general loading conditions a review can be found in Liu
and Mahadevan, Int J. Fatigue, 27
9.2.8 Criteria for
failure by low cycle fatigue
a fatigue test is run with a high stress level (sufficient to cause plastic
flow in a large section of the solid) the specimen fails very quickly (less
than 10 000 cycles). This regime of
behavior is known as `low cycle fatigue’.
The fatigue life correlates best with the plastic strain amplitude
rather than stress amplitude, and it is found that the Coffin Manson Law
a good fit to empirical data (the constants C and b do not have
the same values as for Basquin’s law, of course)
9.2.9 Criteria for
failure under variable amplitude cyclic loading
tests are usually done at constant stress (or strain) amplitude. Service loading usually involves cycles with
variable (and often random) amplitude.
Fortunately, there’s a remarkably simple way to estimate fatigue life
under variable loading using constant stress data.
the load history is comprised of a set of load cycles at a stress amplitude ,
followed by a set of cycles at load amplitude and so on.
For the ith set of cycles at load amplitude ,
we could compute the number of cycles that would cause the specimen to fail
using Basquin’s law
The Miner-Palmgren failure criterion assumes a linear summation of
damage due to each set of load cycles, so that at failure
In terms of stress
The same approach works
under low cycle fatigue conditions, in which case
criterion is often used under random loading. To do so, we need to find a way
to estimate the number of cycles of load at a given stress level. There are various ways to do this one approach is to count the peaks in the load
history, and compute the probability of finding a peak at stress level .
(Of course, this only works if the signal has well defined peaks - this is not
the case for white noise, for example).
Miner’s rule then predicts that the number
of cycles to failure satisfies