Chapter 9

 

Modeling Material Failure

 

 

9.2 Stress and strain based fracture and fatigue criteria

 

Many 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 criterion. 

 

 Failure criteria for isotropic materials

 

The simplest brittle fracture criterion states that fracture is initiated when the greatest tensile principal stress in the solid reaches a critical magnitude,

${\sigma }_{1\max }={\sigma }_{TS}$

(The subscript TS stands for tensile strength).  To apply the criterion, you must

1.      Measure (or look up) ${\sigma }_{TS}$ for the material.  ${\sigma }_{TS}$ 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’ ${\sigma }_r$ for the material.  It is nominally equivalent to ${\sigma }_{TS}$ but in practice usually turns out to be somewhat higher.

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 ${\sigma }_{1\max }$.  If  ${\sigma }_{1\max }<{\sigma }_{TS}$ the design is safe (but be sure to use an appropriate factor of safety!).

 

 

 Failure criteria for anisotropic materials

 

More 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 ${\sigma }_{TS1},{\sigma }_{TS2},{\sigma }_{TS3}$ parallel to the three characteristic directions $\{e_1,e_2,e_3\}$ in the solid.  The tensile strength when loaded  parallel to a general direction $n=\sin \varphi \cos \theta e_1+\sin \varphi \sin \theta e_2+\cos \varphi e_3$ could be interpolated between these values as

${\sigma }_{TS}(n)=\left({\sigma }_{TS1}{\cos }^2\theta +{\sigma }_{TS2}{\sin }^2\theta \right){\sin }^2\varphi +{\sigma }_{TS3}{\cos }^2\varphi$

where $(\varphi ,\theta )$ 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.

$n_i(\theta ,\varphi ){\sigma }_{ij}{n}_j(\theta ,\varphi )={\sigma }_{TS}(\varphi ,\theta )$

where ${\sigma }_{ij}$ are the stress components in the basis $\{e_1,e_2,e_3\}$.  To use this criterion to check for failure at any point in the solid, you must

(i) Find the components of stress in the  $\{e_1,e_2,e_3\}$ basis;

(ii) Maximize the function $n_i(\theta ,\varphi ){\sigma }_{ij}{n}_j(\theta ,\varphi )/{\sigma }_{TS}(\varphi ,\theta )$ with respect to $(\varphi ,\theta )$; and

(iii) Check whether the maximum value of  $n_i(\theta ,\varphi ){\sigma }_{ij}{n}_j(\theta ,\varphi )/{\sigma }_{TS}(\varphi ,\theta )$ 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

$A_{ij}{\sigma }_{ij}+B_{ijkl}\sigma {}_{ij}{\sigma }_{kl}=1$

Here A and B are material constants: A is diagonal ( $A_{ij}=0i\ne j$ ) and $B$ has the same symmetries as the elasticity tensor, i.e. $B_{ijkl}=B_{klij}=B_{jikl}=B_{ijlk}$.  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 ${\sigma }_{11},{\sigma }_{22},{\sigma }_{12}$ denote the nonzero components of stress, with basis vectors $e_1$ and $e_2$ oriented parallel and perpendicular to the fibers in the sheet, as shown.  The Tsai-Hill failure criterion is

${\left(\frac{{\sigma }_{11}}{{\sigma }_{TS1}}\right)}^2+{\left(\frac{{\sigma }_{22}}{{\sigma }_{TS2}}\right)}^2-\frac{{\sigma }_{11}{\sigma }_{22}}{{\sigma }_{TS1}^2}+\frac{{\sigma }_{12}^2}{{\sigma }_{SS}^2}=1$

at failure, where ${\sigma }_{TS1}$, ${\sigma }_{TS2}$ and ${\sigma }_{SS}$ are material properties.  They are measured as follows:

1.      The laminate is loaded in uniaxial tension parallel to the fibers. The material fails when ${\sigma }_{11}={\sigma }_{TS1}$

2.      The laminate is loaded in uniaxial tension perpendicular to the fibers.  The material fails when  ${\sigma }_{22}={\sigma }_{TS2}$

3.      In principle, the laminate could be loaded in shear $–$ it would then fail when ${\sigma }_{12}={\sigma }_{SS}$. In practice it is preferable to pull on the laminate in uniaxial tension with stress ${\sigma }_0$ at 45 degrees to the fibers, which induces stress components ${\sigma }_{11}={\sigma }_{22}={\sigma }_{12}={\sigma }_0/2$.  A simple calculation then shows that ${\sigma }_{SS}={\sigma }_{TS2}{\sigma }_0/\sqrt{4{\sigma }_{TS2}^2-{\sigma }_0^2}$.

 

 

 

9.2.2 Probabilistic Design Methods for Brittle Fracture  (Weibull Statistics)

 

The fracture criterion ${\sigma }_{1\max }={\sigma }_{TS}$ 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.

 

Weibull 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 $P_s$ is fit by a Weibull distribution

$P_s(V_0)=\exp \left\{-{\left(\frac{\sigma }{{\sigma }_0}\right)}^m\right\}$

where ${\sigma }_0$ 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 ${\sigma }_0$ is the stress at which the probability of survival is exp(-1), (about 37%). This critical stress ${\sigma }_0$ depends on the specimen volume $V_0$, and is smaller for larger specimens.

3.      Given m, ${\sigma }_0$ and the corresponding specimen volume $V_0$, the survival probability of a volume $V$ of material subjected to uniform uniaxial stress $\sigma$ follows as

$P_s(V)=\exp \left\{-\frac{V}{V_0}{\left(\frac{\sigma }{{\sigma }_0}\right)}^m\right\}$

To see this, note that the volume V can be thought of as containing $n=V/V_0$ specimens.  The probability that they all survive is ${\left\{P_s(V_0)\right\}}^n={\left\{P_s(V_0)\right\}}^{V/V_0}$.

4.      More generally, the survival probability of a solid subjected to an arbitrary stress distribution with principal values ${\sigma }_1,{\sigma }_2,{\sigma }_3$ can be computed as

$\log P_s=-\frac{1}{V_0{\sigma }_0^m}{\displaystyle \underset{V}{\int }\left({\langle {\sigma }_1\rangle }^m+{\langle {\sigma }_2\rangle }^m+{\langle {\sigma }_3\rangle }^m\right)}dV$

where

$\langle \sigma \rangle =\{\begin{array}{c}\sigma \sigma \ge 0\\ 0\sigma \le 0\end{array}$

 

This 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 interest.

 

 

9.2.3 Static Fatigue Criterion for Brittle Materials

 

`Static 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

${\sigma }_{TS}={\sigma }_{TS0}\left(1-\alpha {\displaystyle \int {\left(\sigma /{\sigma }_{TS0}\right)}^{m}dt}\right)$

where $\sigma (t)$ is the maximum principal stress acting on the solid, which may vary slowly with time t; ${\sigma }_{TS0}$ is the tensile strength of the solid at time t=0, and $\alpha ,m$ are two material constants.  Typically m has values between 5 and 10.  For the particular case of a constant stress, we see that

${\sigma }_{TS}={\sigma }_{TS0}\left(1-\alpha {\left(\sigma /{\sigma }_{TS0}\right)}^{m}t\right)$

Since failure occurs when $\sigma ={\sigma }_{TS}$, the time to failure follows as

$t_f=\frac{1}{\alpha }{\left(\frac{{\sigma }_{TS0}}{\sigma }\right)}^m\left(1-\frac{\sigma }{{\sigma }_{TS0}}\right)\approx \frac{1}{\alpha }{\left(\frac{{\sigma }_{TS0}}{\sigma }\right)}^m$

so that $\alpha$ 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 $\sigma$.

 

 

9.2.4 Constitutive laws for crushing failure of brittle materials

 

Brittle 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.

 

This 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.

 

The material is characterized by the following properties:

 The Young’s modulus E and Poisson ratio $\nu$

 The stress-v-plastic strain curve measured from a uniaxial compression test, which is fit by a functional relation of the form $\sigma =-Y({\overline{\epsilon }}^p)$, where ${\overline{\epsilon }}^p$ is the magnitude of the compressive strain.  Any of the functions listed in Section 3.6.5 could be used for the function Y.

 A material constant c, which controls how rapidly the strength of the material increases with hydrostatic compression.

 

The constitutive equations specify a relationship between an increment in stress $d{\sigma }_{ij}$ applied to the material and an increment in strain $d{\epsilon }_{ij}$, as follows

1.      The strain is decomposed into elastic and irreversible (damage) parts  as $d{\epsilon }_{ij}=d{\epsilon }_{ij}^e+d{\epsilon }_{ij}^p$;

2.      The elastic part of the strain is related to the stress by the linear elastic constitutive equations

$d{\epsilon }_{ij}^e=\frac{1+\nu }{E}d{\sigma }_{ij}-\frac{\nu }{E}d{\sigma }_{kk}{\delta }_{ij}$

3.      The critical stress that initiates crushing damage is given by a failure criterion (analogous to the yield criterion for a metal) of the form

$f({\sigma }_{ij})=\sqrt{\frac{3}{2}{S}_{ij}{S}_{ij}}+3cp-(1-c)Y({\overline{\epsilon }}^p)=0$

where $p={\sigma }_{kk}/3$, $S_{ij}={\sigma }_{ij}-p{\delta }_{ij}$, and ${\overline{\epsilon }}^p={\displaystyle \int \sqrt{2d{\epsilon }_{ij}^{p}d{\epsilon }_{ij}^p/3}}$ 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.

4.      The plastic strain components are determined using an associated flow rule

$d{\epsilon }_{ij}^p=\frac{d{\overline{\epsilon }}^p}{\sqrt{1+2c^2}}\frac{df}{d{\sigma }_{ij}}=\frac{d{\overline{\epsilon }}^p}{\sqrt{1+2c^2}}\left\{\frac{3}{2}\frac{S_{ij}}{\sqrt{3S_{kl}{S}_{kl}/2}}+c{\delta }_{ij}\right\}$

 

5.      The magnitude of the plastic strain increment is related to the stress increment by

$d{\overline{\epsilon }}^p=\frac{1}{h(1-c)}\langle \frac{3}{2}\frac{S_{ij}d{\sigma }_{ij}}{\sqrt{3S_{kl}{S}_{kl}/2}}+cd{\sigma }_{kk}\rangle$

where $h=dY/d{\overline{\epsilon }}^p$ is the slope of the uniaxial stress-strain curve, and $\langle x\rangle =x$ for $x>0$, while $\langle x\rangle =0$ for $x<0$.

 

HEALTH WARNING: These constitutive equations should only be used in regions where the hydrostatic stress is compressive $p<0$.  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 critical magnitude.

 

 

 

9.2.5  Ductile Fracture Criteria

 

Strain 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

${\displaystyle \underset{}{\int }\sqrt{\frac{2}{3}d{\epsilon }_{ij}^{p}d{\epsilon }_{ij}^p}}={\epsilon }_f$

at failure, where ${\epsilon }_f$ is the plastic strain to failure in a uniaxial tensile test.

 

Porous metal plasticity: Experiments 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.

 

Porous metal 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 $V_f$ 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.

 

Both 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.’  

 

 

The material is characterized by the following properties:

 The Young’s modulus E and Poisson ratio $\nu$;

 A characteristic stress Y, a characteristic strain ${\epsilon }_0$ and strain hardening exponent n, which govern the strain hardening behavior of the matrix material;

 A characteristic strain rate ${\dot{\epsilon }}_0$ and strain rate exponent m, which govern the strain rate sensitivity of the solid;

 A constant $N_v$, which controls the rate of void nucleation with plastic straining;

 The flow strength of the matrix ${\sigma }_0$, the void volume fraction, $V_f$, and the total accumulated effective plastic strain in the matrix material ${\overline{\epsilon }}_m$ which all evolve with plastic straining.

The constitutive equations specify a relationship between the stress ${\sigma }_{ij}$ applied to the material and the resulting strain rate ${\dot{\epsilon }}_{ij}$, as follows

1.      The strain rate is decomposed into elastic and plastic parts  as ${\dot{\epsilon }}_{ij}={\dot{\epsilon }}_{ij}^e+{\dot{\epsilon }}_{ij}^p$;

2.      The elastic part of the strain rate is related to the stress rate by the linear elastic constitutive equations

${\dot{\epsilon }}_{ij}^e=\frac{1+\nu }{E}{\dot{\sigma }}_{ij}-\frac{\nu }{E}{\dot{\sigma }}_{kk}{\delta }_{ij}$

3.      The magnitude of the plastic strain rate is determined by the following plastic flow potential

${\dot{\epsilon }}_e=g({\sigma }_e,p,{\sigma }_0,V_f)={\dot{\epsilon }}_0{\left[{\left(\frac{{\sigma }_e}{{\sigma }_0}\right)}^2+2V_f\cosh \left(\frac{3p}{2{\sigma }_0}\right)-V_f^2\right]}^{m/2}$

where ${\dot{\epsilon }}_e=2{\dot{\epsilon }}_{ij}^p{\dot{\epsilon }}_{ij}^p/3$ ${\sigma }_e=\sqrt{3S_{ij}{S}_{ij}/2}$, $p={\sigma }_{kk}/3$ and $S_{ij}={\sigma }_{ij}-p{\delta }_{ij}$.  Note that for $V_f>0$ the plastic strain rate increases with hydrostatic stress p.

4.      The components of the plastic strain rate tensor are computed from an associated flow law

$\begin{array}{l}{\dot{\epsilon }}_{ij}^p=\sqrt{\frac{3}{2}}\frac{g({\sigma }_e,p,{\sigma }_0,V_f)}{{\left[\left(\partial g/\partial {\sigma }_{kl}\right)\left(\partial g/\partial {\sigma }_{kl}\right)\right]}^{1/2}}\frac{\partial g}{\partial {\sigma }_{ij}}\\ =\frac{g({\sigma }_e,p,{\sigma }_0,V_f)}{\sqrt{{\sigma }_e^2/{\sigma }_0^2+(V_f^2/2){\sinh }^2(3p/2{\sigma }_0)}}\left\{\frac{3}{2}\frac{S_{ij}}{{\sigma }_0}+\frac{V_f}{2}\sinh \left(\frac{3p}{2{\sigma }_0}\right){\delta }_{ij}\right\}\end{array}$

5.      Strain hardening in the matrix is modeled by relating its flow stress ${\sigma }_0$ to the accumulated strain in the matrix ${\overline{\epsilon }}_m$. The following power-law hardening model is often used

${\sigma }_0=Y{\left(1+{\overline{\epsilon }}_m/{\epsilon }_0\right)}^{1/n}$

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

$(1-V_f){\sigma }_0{\dot{\overline{\epsilon }}}_m={\sigma }_{ij}{\dot{\epsilon }}_{ij}^p\Rightarrow {\dot{\overline{\epsilon }}}_m=\frac{g({\sigma }_e,p,{\sigma }_0,V_f)}{(1-V_f)}\left(\frac{{\sigma }_e^2}{{\sigma }_0^2}+\frac{3p}{2{\sigma }_0}{V}_f\sinh \left(\frac{3p}{2{\sigma }_0}\right)\right)$

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

${\dot{V}}_f=(1-V_f){\dot{\epsilon }}_{kk}^p+N_v{\dot{\epsilon }}_e$

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

 

If 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.

 

Neck formation is a consequence of geometric softening.  A very simple model explains the concept of geometric softening.

1.      Consider a cylindrical specimen with initial cross sectional area $A_0$ and length $L_0$. 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 $\sigma ={\sigma }_0{\epsilon }^n$ with n<1.

3.      The true strain in the specimen is related to its length by $\epsilon =\log \frac{L}{L_0}$

4.      The force on the specimen is related to the Cauchy stress by $P=A\sigma =A{\sigma }_0{\epsilon }^n$

5.      At the point of maximum load $\frac{dP}{dL}=\frac{dA}{dL}\sigma +A\frac{d\sigma }{dL}=0$

6.      We can calculate $dA/dL$ by noting that the volume of the specimen is constant during plastic straining, which shows that

$AL=A_0{L}_0\Rightarrow \frac{dA}{dL}L+A=0\Rightarrow \frac{dA}{dL}=-\frac{A}{L}$

Notice that $dA/dL$ 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.

7.      We can calculate $d\sigma /dL$ from (2) and (3) as follows

$\frac{d\sigma }{dL}=\frac{d\sigma }{d\epsilon }\frac{d\epsilon }{dL}=\frac{n{\sigma }_0{\epsilon }^{n-1}}{L}$

Notice that $d\sigma /dL$ is positive $–$ strain hardening in the material tends to compensate for the effects of geometric softening.

8.      Finally, substituting the results of (6) and (7) back into (5) and recalling that  $\sigma ={\sigma }_0{\epsilon }^n$ shows that at the point of maximum load, the strain and length of the specimen are

$-\frac{A\sigma }{L}+A\frac{n{\sigma }_0{\epsilon }^{n-1}}{L}=0\Rightarrow {\epsilon }_{\max }=n\Rightarrow L_{\max }=L_0\exp (n)$

9.      Finally, note that by volume conservation the cross sectional area is $A=A_0{L}_0/L$, so the maximum load the specimen can withstand follows as

$P_{\max }=A{\sigma }_{\max }=\frac{A_0{L}_0}{L}{\sigma }_0{({\epsilon }_{\max })}^n=A_0{\sigma }_0{n}^n\exp (-n)$

It 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 critical. 

 

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.

 

Examples of geometry induced localization are

1.      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 generation

 

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.

 

Geometric 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.

 

Here we will review criteria that are used to predict fatigue life under proportional cyclic loading. A typical stress cycle is parameterized by its amplitude $({\sigma }_{\max }-{\sigma }_{\min })/2$ and the mean stress ${\sigma }_m=({\sigma }_{\max }+{\sigma }_{\min })/2$

 

For tests run in the high cycle fatigue regime with any fixed value of mean stress, the relationship between stress amplitude ${\sigma }_a$ and the number of cycles to failure N is fit well by Basquin’s Law

${\sigma }_a{N}^b=C$

where the exponent b is typically between 0.05 and 0.15.  The constant C is a function of mean stress.

 

There 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 $\sigma ={\sigma }_{UTS}$, 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:

 Goodman’s rule uses a linear interpolation, giving

${\sigma }_a{N}^b=C_0\left(1-\frac{{\sigma }_m}{{\sigma }_{UTS}}\right)$

where $C_0$ is the constant in Basquin’s law determined by testing at zero mean stress.

 Gerber’s rule uses a parabolic fit

${\sigma }_a{N}^b=C_0\left\{1-{\left(\frac{{\sigma }_m}{{\sigma }_{UTS}}\right)}^2\right\}$

In practice, experimental data seem to lie between these two limits.  Goodman’s rule gives a safe estimate.

 

These 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 ${\sigma }_a$ and ${\sigma }_m$.  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, 27 790-800 (2005).

 

 

 

9.2.8 Criteria for failure by low cycle fatigue

 

If 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

$\Delta {\epsilon }^p{N}^b=C$

gives 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

 

Fatigue 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.

 

Suppose the load history is comprised of a set of $n_1$ load cycles at a stress amplitude ${\sigma }_a^{(1)}$, followed by a set of $n_2$ cycles at load amplitude ${\sigma }_a^{(2)}$ and so on.  For the ith set of cycles at load amplitude ${\sigma }_a^{(i)}$, we could compute the number of cycles that would cause the specimen to fail using Basquin’s law

${\sigma }_a^{(i)}{N}_i^b=C$

The Miner-Palmgren failure criterion assumes a linear summation of damage due to each set of load cycles, so that at failure

${\displaystyle \sum _i^{}\frac{n_i}{N_i}=1}$

In terms of stress amplitude

${\displaystyle \sum _i^{}{n}_i{\left(\frac{{\sigma }_a^{(i)}}{C}\right)}^{1/b}=1}$

The same approach works under low cycle fatigue conditions, in which case

${\displaystyle \sum _i^{}{n}_i{\left(\frac{\Delta {\epsilon }^p{}_{}^{(i)}}{C}\right)}^{1/b}=1}$

 

The 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 $P(\sigma )$ of finding a peak at stress level $\sigma$. (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

$N{{\displaystyle \underset{0}{\overset{\infty }{\int }}P(\sigma )\left(\frac{\sigma }{C}\right)}}^{1/b}d\sigma =1$