CH_6- Elastic-Plastic Fracture Mechanics

28
Chapter 6 Elastic -Plastic Fracture Mechanics The high stresses at the crack tip cannot be sustained by, practically, any material. Thus, if the material does not fracture, a plastic zone (or damage zone or process zone) is formed around the crack tip. The damage is specific to the materials but it can be said that in general terms, for a ductile material, the damage is in the form of plastic deformation and for brittle materials in the form of microcracking. Until now is has been assumed that the size of the plastic zone near the crack tip is relatively small as compared to the specimen dimensions. Thus, the effect of this zone has been neglected and the strain field surrounding the crack tip is dominated by linear elastic fracture mechanics asymptotic field derived in chapter 4. In this chapter the effects of a non-negligible plastic zone is considered in the case of Mode I only. First, in order to determine at what point it is necessary to include the influence of the plastic zone in the stress analysis, two approximations of the size of the plastic zone are presented. Afterwards, two elastic-plastic fracture criteria are discussed, the crack tip opening displacement and the J-integral. The J-integral is shown to be a generalization of the linear elastic release rate to elastic-plastic fracture. The properties of the J and the stress field near the crack tip for elastic-plastic fracture are then derived. Lastly, calculations of J for certain geometries are given as examples. 6.1 One dimensional estimation of Size of Plastic Zone In order to obtain a first estimate the size of the plastic zone, we apply the von Mises yield criterion, ( ) ( ) ( ) 2 2 2 2 Y σ ( ) ( ) ( ) 2 2 2 2 Y σ σ 1 2 2 3 3 1 2 σ −σ + σ −σ + σ −σ = σ (6.1) where (i =1,2,3) are the principle stresses. Thus, we assume that the region in which, i 1 2 2 3 3 1 2 σ −σ + σ −σ + σ −σ σ has plastically deformed. Note that the von Mises yield criterion is dependent only upon the principle stresses. This yield criterion is applicable since plastic flow is usually a shear deformation with constant volume, i.e. independent of the principle stress components. Thus, we can define a at the onset of plastic deformation and then assume that all further plastic deformation occurs at this constant . Y Y For this first approximation we consider only the stress field on the line θ = 0, as shown in Figure 6.1, and define L as the characteristic length of the plastic zone. By comparing the stress field with the von Mises yield criterion, we will determine the length L for which the material in the region x L has exceeded the yield limit. The asymptotic stress field near the crack tip are given by (4.36). On the line θ = 0 these reduce to, I rr r 0 2r θθ θ σ = σ = π K (6.2) March 2006 6-1

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Good in fracture mechanics and crack tip opening displacement and j-integral

Transcript of CH_6- Elastic-Plastic Fracture Mechanics

Page 1: CH_6- Elastic-Plastic Fracture Mechanics

Chapter 6 Elastic -Plastic Fracture Mechanics

The high stresses at the crack tip cannot be sustained by, practically, any material. Thus, if the material does not fracture, a plastic zone (or damage zone or process zone) is formed around the crack tip. The damage is specific to the materials but it can be said that in general terms, for a ductile material, the damage is in the form of plastic deformation and for brittle materials in the form of microcracking.

Until now is has been assumed that the size of the plastic zone near the crack tip is relatively small as compared to the specimen dimensions. Thus, the effect of this zone has been neglected and the strain field surrounding the crack tip is dominated by linear elastic fracture mechanics asymptotic field derived in chapter 4. In this chapter the effects of a non-negligible plastic zone is considered in the case of Mode I only. First, in order to determine at what point it is necessary to include the influence of the plastic zone in the stress analysis, two approximations of the size of the plastic zone are presented. Afterwards, two elastic-plastic fracture criteria are discussed, the crack tip opening displacement and the J-integral. The J-integral is shown to be a generalization of the linear elastic release rate to elastic-plastic fracture. The properties of the J and the stress field near the crack tip for elastic-plastic fracture are then derived. Lastly, calculations of J for certain geometries are given as examples.

6.1 One dimensional estimation of Size of Plastic Zone

In order to obtain a first estimate the size of the plastic zone, we apply the von Mises yield criterion,

( ) ( ) ( )2 2 2 2Y

σ

( ) ( ) ( )2 2 2 2Y

σ

σ

1 2 2 3 3 1 2σ −σ + σ −σ + σ −σ = σ (6.1)

where (i =1,2,3) are the principle stresses. Thus, we assume that the region in which, i

1 2 2 3 3 1 2σ −σ + σ −σ + σ −σ ≥ σ

has plastically deformed. Note that the von Mises yield criterion is dependent only upon the principle stresses. This yield criterion is applicable since plastic flow is usually a shear deformation with constant volume, i.e. independent of the principle stress components. Thus, we can define a at the onset of plastic deformation and then assume that all further plastic deformation occurs at this constant .

Y

Y

For this first approximation we consider only the stress field on the line θ = 0, as shown in Figure 6.1, and define L as the characteristic length of the plastic zone. By comparing the stress field with the von Mises yield criterion, we will determine the length L for which the material in the region x ≤ L has exceeded the yield limit. The asymptotic stress field near the crack tip are given by (4.36). On the line θ = 0 these reduce to,

Irr r 0

2 rθθ θσ = σ = σ =π

K (6.2)

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Note that and are the principle stresses θθσ σrr 1σ and 2σ for the line θ = 0. Applying the plane

strain condition (i.e., ( )3 1 2 ; ν being the material's Poisson ratio) to eq. (6.1), one has, σ = ν σ + σ

( ) ( ) ( )( ) ( ) (

( )( ) ( )

Y 1 2 2 1 1 2

2 2 2 2 21 2 1 2 1 1

2 2 2 21 2 1 2

2 1 1

4 1 2 1 2 1 1 2 1

2 1 2 2 2 1

⎡ ⎤ ⎡ ⎤σ = σ −σ + σ −ν −νσ + σ −ν −νσ⎣ ⎦ ⎣ ⎦

= ν −ν σ σ − σ σ + + ν + ν + ν + σ + + ν + ν − ν + σ

= ν −ν + σ +σ + ν − ν − σ σ

)

2 2 22

2

2

(6.3)

For the special case of 1σ = σ

( )2 2 2 2⎡ ⎤σ = ν − ν − + ν − ν + σ

(such as along the line θ = 0), eq. (6.3) is simplified to,

( )( )

Y 1

2 2 2Y 1

Y 1

2 4 4 2 2 2 2 2

2 8 8 2

1 2

⎣ ⎦

σ = ν − ν + σ

σ = − ν σ

(6.4)

Substituting the principle stress of eq. (6.2) into eq. (6.4), the yield criterion is given by,

( ) IY 1 2

2 Lσ = − ν

πK

( )

(6.5)

Thus, the length over which plastic deformation occurs is given by,

22I

Y

1 2 K2− ν ⎛ ⎞

= ⎜ ⎟π σ⎝ ⎠L (6.6)

Figure 6.1 Schematic of plastic zone ahead of crack tip.

The calculation of L, is only an approximation since the presence of damage will modify the stress field and thus the size of the zone. However, it can serve as a parameter to compare with the overall dimensions of the specimen, and determine the limits of the linear elastic solution. The four principle cases are summarized in Figure 6.2.

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Case I, elastic fracture, occurs when L is much smaller than any other of the specimen dimensions. For this case, the plastic zone is negligible and the stress field is dominated by linear elastic fracture mechanics (LEFM). This is the case studied in chapter 4 and 5.

Case II, contained yielding, the effects of the plastic zone are no longer negligible. For this case, the strain field is dominated by elastic-plastic fracture mechanics, the subject of this chapter. We will see later in this chapter that the fracture can be well described by the Jc (non-linear critical energy release rate). The final two cases will not be considered in these notes, however they are included here to show other possibilities.

Case III, full yielding, involves large deformations due to the large plastic zone. This case is less useful since typically the structure is no longer capable of supporting the applied loads.

Case IV, diffuse dissipation, is different from the other three, as there are no distinct elastic and plastic zones. Instead, the fracture is dominated by non-linear elastic, time dependent phenomena, such as creep or viscoelasticity.

Figure 6.2 Modes of fracture in specimens with different extend of plastic deformation.

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6.1.2 Two dimensional approximation (2D)

While the length L gives us an indication of the size of the plastic zone relative to the specimen dimensions, it is also useful to know the actual shape of the plastic zone around the crack tip. Using the same approach as in the previous section, it is not difficult to obtain its shape.

In cylindrical coordinates the principle stresses 1σ and 2σ for an arbitrary stress field are given by,

22rr rr

1 2 r2 2,θθ θθ

θ

σ + σ σ +σ⎛ ⎞ ⎛ ⎞σ = ± + σ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.7)

Near the crack tip, where the stress field is dominated by the terms of eq. (4.36) we obtain after some algebraic manipulation,

I1,2 cos 1 sin

2 22 rσ = ±⎜ ⎟⎜ ⎟π ⎝ ⎠⎝ ⎠

K θ θ⎛ ⎞⎛ ⎞

(

(6.8)

and )3 1 2ν σ + σσ = in the case of plane strain.

Note that for the case of θ = 0, the last equation reduces to (6.2). Substituting eq. (6.8) into (6.3) and simplifying leads to the following yield condition,

( )2K1 θ ⎡ θ ⎤⎛ ⎞ ⎛ ⎞2 2 2 2I

Y cos 4 1- -3cos2 r 2 2

σ = ν + ν⎜ ⎟ ⎜ ⎟⎢ ⎥π ⎝ ⎠ ⎝ ⎠⎣ ⎦

Thus, solving for the radius of the plastic zone at which this condition is satisfied rp(θ), gives,

( ) ( )2

2

2 ⎥2 2I

pY

K1r cos 4 1 3cos2 2

⎛ ⎞ θ ⎡ θ ⎤⎛ ⎞ ⎛ ⎞θ = −ν + ν −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢π σ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎝ ⎠ (6.9)

The shape of the plastic zone is plotted in Figure 6.3(a) for ν = 1/3 and = 1/2. Also plotted is the limiting case of plane stress (i.e.

νν = 0 ). As the state of stress changes through the thickness

of a thick specimen (i.e. plane stress → plane strain), the shape of the plastic zone also changes. Figure 6.3(b) shows a typical shape of the plastic zone across a specimen.

Note that the shapes of the plastic zone shown above do not take into account the effects of the specimen boundaries. Thus, they are derived for a crack in an infinite plate. For an actual specimen, one must often consider the finite width in order to determine the region of plastic deformation. Figure 6.4 shows examples of the shape of the plastic zone for three standard specimens.

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Figure 6.3 Plastic zone shapes; (a) as a function of ν ,. (b) through the specimen thickness.

Figure 6.4 Effect of finite width of specimen on shape of plastic zone; (a) double edge notch in tension, (b) center cracked specimen in tension, (c) edge crack in bending.

6.1.1 Irwin’s approximation of the plastic zone

It was mentioned earlier that due to the material’s yielding at the crack tip the stress distribution is not given by the asymptotic filed (4.36) and shown in Figure 6.3 by the curve (1). To account for the effects of plastic deformation on the redistribution of the stress field and obtain a simple approximation, perfect plasticity is assumed thus, the stress ahead of the crack tip equals Yσ up to a distance . After that distance, the distribution is obtained by a translation of the asymptotic field shown by the curve (2).

2r

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The two distributions, before and after yielding, should result in equal forces since equilibrium should be assured. With references to Figure 6.5, force equilibrium results in,

1r

Y 2K⎛ ⎞

r

IY

0

dx r2 x

−σ = σ⎜ ⎟π⎝ ⎠∫ (6.10)

where the distance , is the intersection of the filed (1) and the horizontal line at σ given by, 1 Y

2

IK

2r

IY 1

Y1

K 1 r 22 r

⎛ ⎞= σ ⇒ = ⎜ ⎟π σπ ⎝ ⎠

(6.11)

∞σ

Figure 6.5 A crack in an infinite plate with plastic zones ahead of the crack tip.

Integrating (6.10) and using (6.11) one obtains, r1 = . Thus, the plastic zone extends over a distance equal to given by, 1 2 1r r 2r+ =

2

0

I1

Y

K1 2r ⎛ ⎞

= ⎜ ⎟π σ⎝ ⎠ (6.12a)

The simple analysis shown above is for the case of plane stress i.e., 3σ =0

which is realistic for thin plates. In plane strain, yielding is confined due to the effects of 3σ > and a smaller plastic zone ahead of the crack tip is developed. In this case, Irwin proposed the following expression,

2

I1

Y

K1 r 6

⎛ ⎞= ⎜ ⎟π σ⎝ ⎠

(6.12b)

r1 x

yyσ (1)

y

(2)

r2

∞σ

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which is smaller than in (6.11) by a factor of 3. A simple comparison of the eqs (6.12) and (6.6) gives a plastic zone half in length of the one proposed by Irwin. This difference is due to the fact that the analysis of Irwin, being more realistic, takes into account the redistribution of stresses due to plastic deformation.

1r

( ) ( )

In addition, Irwin argued that after the plastic zone the stress redistribution due to yielding results in higher stress than that given by (6.2). This higher stress is reflected in an effective SIF which for a crack in an infinite plates is expressed as,

I 1 K ,a r a r∞ ∞σ + = σ π + 1

( )

(6.13)

6.1.2 Dudgale’s yield strip model

In thin plates of certain metallic materials with a central crack a narrow plastic zone extends from both crack tips as shown in Figure 6.6. To predict the length c, of this plastic zone, Dugdale (Hellan, 1984) assumed perfect plasticity. Thus, the plastic zones are replaced by closing tractions over the length of the zone as shown in Figure 6.6. Using the principle of superposition, the total stress intensity can be calculated as shown next.

∞σ

Figure 6.6 Dugdale’s yield strip approximation of the plastic zone.

There are two contributions to the total stress intensity factor. One comes from the remote applied load and equals to,

I ∞ ∞K ,a c (a c)σ + = σ π +

dx

(6.14)

The other comes from the closing tractions on the crack faces. Using eqs (5.3) and the principle of superposition elaborated in chapter 5, one replaces Q Y= σ as the force on the crack face, the total SIF at any of the two crack tips is given by the following integral,

∞σ

2a c c

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( )a c

x+ ⎡ ⎤σ σ+ + + −Y Y

I Ya

(a c) x (a c) xK ,a c d(a c) x (a c) x(a c) (a c)

σ + = +⎢ ⎥+ − + ++ +⎢ ⎥⎣ ⎦

∫ (6.15)

∞σ

Figure 6.7 The yield strip replaced by yield stress of the material.

Assuming perfect plasticity, the yield stress Yσ is constant. Therefore,

a c(a c) dx (a c) a++ + ⎞

⎟1

I Y Y2 2a

K 2 2 cosa c(a c) x

− ⎛= σ = σ ⎜π π +⎝ ⎠+ −∫ (6.16)

According to this model, the stress intensity due to the remote stress and the stress intensity due to the closure stresses must be zero. This is expressed as,

( ) ( )K ,a c K ,a c 0σ + − σ + =I I Y∞ (6.17)

This last equation serves as a condition to determine the length of the yield zone. Thus, introducing the expressions for the SIFs in (6.17) one easily obtains,

Y

cosa c 2

∞= ⎜+ ⎝ ⎠

a ⎛ ⎞πσ⎟σ

(6.18)

The last equation indicates that when Y∞σ → σ c→∞σ << σ

the length of the yield strip . When , eq. (6.18) can be simplified by expanding it in a Taylor’s series and keeping the first

two terms. Therefore, Y∞

22 2

⎟I2Y Y

a K ( ,a)c8 8

∞ ∞⎛ ⎞π σ σπ= = ⎜σ σ⎝ ⎠

(6.19)

∞σ

2a c

Yσ Y σ

c

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Interestingly, the predictions of the Irwin’s analysis and that of Dudgale’s are very close. Thus both models predict similar plastic zone sizes.

6.2 Crack Tip Opening Displacement as Yield Criterion

For linear elastic fracture mechanics we have developed yield criterion such as K and G. We now develop similar yield criterion for the case II, contained yielding, of Figure 6.2. An initial criterion was proposed by Wells (Anderson, 1995) who used the crack tip opening displacement (CTOD) at the crack tip as a measure of the amount of plastic deformation. His work was motivated after he attempted to measure K values for a series of steels. He found that LEFM was not sufficient because the crack tip had blunted considerably before the crack propagated. Thus, the plastic deformation was not negligible.

IC

Figure 6.8 Irwin plastic zone correction for blunted crack tip.

One can estimate the size of the CTOD tδ , using the Irwin plastic zone correction. This correction is shown in Figure 6.8. In order to model a blunted crack, the length of the crack is extended from a to ry + a, as shown in Figure 6.8. The crack opening uy, at a point along the center line of the crack is given by (4.40d) for θ = 0,

(2Iy

K ru sin 1- 2cos2 2 2 2 2 2

θ=π

θ θ⎡ ⎤= κ + =⎢ ⎥µ π µ π⎣ ⎦)IK r 1κ + (6.20)

For plane stress, the CTOD is then expressed as,

( )y

yIt y r r

4K2u1 2=

δ = =r

µ ν + π (6.21)

From the Irwin plastic zone correction, the radius of the plastic zone is,

2

Iy

Y

K1r2

⎛ ⎞= ⎜ ⎟π σ⎝ ⎠

(6.22)

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where is the yield stress of the material. Substituting eq. (6.22) into eq. (6.21) we find, Yσ

2K 4

(1 )= µ + ν

It

Y2 1δ =

µπσ + ν

And finally, using E 2 ,

24K 4GIt

Y YEδ = =

πσ πσ (6.23)

The CTOD can thus be related uniquely to KI and G, allowing one to determine the critical CTOD for crack propagation for example (i.e. tδ at KIc). However, the CTOD is often difficult to measure without large errors. In the next section, a more general criterion is presented, based upon the energy released during crack growth.

6.3 The J integral as yield criterion

6.3.1 Deformation theory of plasticity

In this section we want to derive a fracture criterion for an elastic-plastic material which has a stress-strain behavior as shown in Figure 6.9. For small loads, the stress-strain curve is linear (1). Beyond a critical load, the curve is non-linear (2). As the load is released, the stress-strain curve is once again linear (3), with the same slope as during the initial loading (1). The major difficulty in modeling this material is that the behavior is no longer reversible (i.e. the stress-strain curve does not follow the same path for loading and unloading). Thus, the material behavior is now dependent upon the strain history and the stress at a given strain is no longer unique. However, this effect does not become important unless unloading occurs.

Figure 6.9 Stress-strain curves for (a) elastic-plastic and (b) non-linear elastic materials.

Thus, until unloading, the stress-strain curve is identical to that of a non-linear elastic material, also shown in Figure 6.9. Note that the non-linear elastic curve is reversible.

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Therefore, for the derivations that follow, we will replace the elastic-plastic behavior by the non-linear case, specifying that we consider only monotonic (non-cyclic) loading. This substitution is called the Deformation Theory of Plasticity. Note that this substitution is not always valid for a truly 3D case of loading, however it is valid for several cases.

6.3.2 Definition of the J-integral

In an earlier chapter, the linear elastic energy release rate G, was derived and shown to be the area under the load-displacement curve for a given body between two load states. Similarly, one can define the nonlinear elastic energy rate J, as the area under the load-displacement curve between two load states. This area is shown in Figure 6.10. The body is loaded following the nonlinear curve until the point B. At this point the load is P and the applied displacement u. Afterwards, the displacement is increased to u+δu, or the point B' in Figure 6.10. At this point the load has decreased until P-δP. The loading is then removed, following the nonlinear curve shown. The energy release rate between the points B and B' is given by the area between the two curves.

Figure 6.10 Load-displacement curve of non-linear elastic material. The shaded area gives energy, released from state B to state B', defined as J.

In the same manner as G, J is used as a fracture criterion, meaning that for a given material at a certain critical value, Jc, the crack will propagate. Note that in the case of linear fracture, Jc reduces to Gc. In practice, the nonlinear energy release rate J, is calculated using the J-integral, written as,

iiJ wdy T ds

= −⎜ ∂⎝ ⎠∫u∂⎛ ⎞

ε

(6.24)

where W is the strain energy, defined by,

ij ij0w d= σ ε∫

and the T are the boundary tractions, given by, i

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i ij jT n= σ

x yThe contour over which the integral is evaluated is shown in Figure 6.11. The - axes are defined to have their origin at the crack tip and are in the direction shown. The contour Γ is evaluated in the counter-clockwise direction, starting on one crack face and ending on the other crack face. It will be shown in the next section that the choice of Γ is arbitrary. The normal to the contour and the increment along the contour ds, are also shown in Figure 6.11. in

wdA Tu dsΠ = −

As mentioned above, the integral expression of eq. (6.24) defining J is equivalent to the area under the curve of Figure 6.10, i.e. the nonlinear elastic energy release rate. This can be derived as shown in the following.

ds in

iT

y

x

Γ

Figure 6.11 Contour J- integral to calculate nonlinear energy release rate for elastic plastic fracture.

The potential energy of a body (with or without a crack), in two dimensions, is expressed as,

i iA Γ∫ ∫

The first integral is the strain energy of deformation and the second on the potential energy of the applied loads.

The change in potential energy due to a virtual extension of the crack da is then given by,

ii

A

dA T dsa a daΓ

= −∂ ∂∫ ∫

uw ∂∂Π ∂ (6.25)

As the crack is extended by da, the coordinate axes are also moved by da. Thus,

a x= −

∂ ∂∂ ∂

and

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ii

A

dA T dsa x dxΓ

= − +∂ ∂∫ ∫

uw ∂∂Π ∂ (6.26a)

Applying the divergence theorem on the first integral one obtains,

i ii x i

A

dA T ds wn ds T dsa x dx daΓ Γ Γ

= − + = − +∂ ∂∫ ∫ ∫ ∫

u uw ∂ ∂∂Π ∂

Therefore,

i i

x i x iJ wn T ds wn T dsdx dxΓ Γ

= − = −⎜ ⎟ ⎜⎝ ⎠ ⎝∫ ∫

u u∂ ∂⎛ ⎞ ⎛ ⎞⎟⎠

iiJ wdy T

a xΓ

= − = −⎜∂ ∂⎝ ⎠∫u ds∂∂Π ⎛ ⎞

G

(6.26b)

Thus, the J-integral defined in eq. (6.15) is equivalent to the nonlinear energy release rate. When the material is linear elastic, the ERR is equal to J. This is evident by comparing (3.34) and (6.25).

I

6.3.3 Properties of J-integral

The contour integral J has two important properties, each of which has a physical significance. These properties are,

1. J integrated along any closed contour is zero. This means that a discontinuity (such as a crack) which interrupts the contour is necessary for J to be non-zero. Thus J is truly a measure of the energy dissipation strictly due to the crack.

2. J is path independent. Thus, the choice of the contour to calculate J is not important. One can choose a contour which is convenient for calculation purposes.

Proof of property 1

Consider the closed contour Γ which encloses the area A, shown in Figure 6.12. Using the definition of J, we evaluate J along the contour Γ.1

i ii i

A

J wdy T ds dxdy T dsx xΓ

Γ Γ

= − = −⎜ ⎟∂ ∂ ∂⎝ ⎠∫ ∫ ∫u uw

x∂ ∂∂⎛ ⎞

jT n= σ

(6.27a)

Next use the Cauchy’s formula apply the divergence theorem to the second term, i ij

1 The notation J

Γ indicates the integral J evaluated along the contour Γ.

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i i ii ij j ij

jA

T ds n ds dxdyx x x xΓ Γ

= σ = σ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠∫ ∫ ∫u u u∂ ∂ ∂∂ ⎛ ⎞ (6.27b)

Thus,

iij

jA

uwJ dxdyx x xΓ

∂∂ ∂ ⎛ ⎞= − σ⎢ ⎥⎜ ⎟∂ ∂ ∂⎝ ⎠⎢ ⎥⎣ ⎦∫⎡ ⎤

(6.28)

The first term of eq. (6.28) is written as,

ij ijij

ijx x= ⋅ = σ

∂ ∂ε ∂ ∂W w

x∂ε ∂ε∂ ∂ (6.29)

where,

jiij

j i

u12 x x

∂ε = +⎢ ⎥

∂ ∂⎢ ⎥⎣ ⎦

u⎡ ⎤∂

Thus,

ij ji

j i

uuWx 2 x x x x

⎛ ⎞∂∂ ∂ ∂= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞σ ∂ (6.30)

Γ

A

y

x

Figure 6.12 The closed contour Γ enclosing the area A, for the evaluation of J.

Using the symmetry of the stress field, i.e. ij jiσ = σ , we can simplify (6.30) to,

iij ij

j j

u uwx x x x x

∂ ∂∂ ∂ ∂ ⎛ ⎞= σ = σ⎢ ⎥⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠⎢ ⎥⎢ ⎥⎝ ⎠ ⎣⎣ ⎦

i⎡ ⎤⎛ ⎞ ⎡ ⎤

⎦ (6.31)

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Assuming zero body forces and applying the condition of equilibrium ∂σ , one can write,

ij jx 0/ ∂ =

iji i iij ij ij

j j j j

u u ux x x x x x x x

iu⎡ ⎤ ⎡ ⎤∂σ ∂⎡ ∂ ⎤ ∂ ∂ ∂∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞σ = + σ = σ⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ (6.32)

Substituting eq. (6.32) into (6.31) we have,

iij

jx x x= σ ⎜ ⎟⎢ ⎥∂ ∂ ∂⎝ ⎠⎣

uw ⎡ ∂ ⎤∂ ∂ ⎛ ⎞

⎦ (6.33)

Replacing the first term of eq. (6.28) leads to,

i iij ij

j jA

u uJ dx x x xΓ

⎡ ∂ ⎤ ∂∂ ∂⎛ ⎞ ⎡ ⎤= σ − σ⎜ ⎟⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎣ ⎦⎣ ⎦⎝ ⎠∫ xdy⎛ ⎞

Therefore,

J 0=Γ

(6.34)

Proof of property 2

Consider the two arbitrary contours 1Γ and 2Γ , shown in Figure 6.13(a). We want to show that J, evaluated on each of the two contours, is equivalent, i.e.

1 2J JΓ Γ= . For this we define the

closed contour Γ , shown in Figure 6.13(b). The contour is composed of four individual contour segments *

4Γ = Γ ∪Γ ∪Γ1 2 3Γ ∪ . Using the property 1, one can write

Figure 6.13 (a) Independent contours Γ1 and Γ2 for the evaluation of J; (b) closed contour contaiand Γ2.

x

y

x

y 1Γ

(a) 1Γ

(b)

4Γ *2Γ

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6-15

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1 2 3 4*Γ Γ Γ Γ Γ

J J J J J= + + + (6.35)

Note that,

2 2Γ Γ*J J= − (6.36)

Γas the direction of the contour has been reversed to form the closed contour . Using the last equality and that the contour is closed one can write,

1 2 3 4Γ Γ Γ Γ ΓJ J J J J 0= − + + =

Γ

(6.37)

Considering that the crack faces are traction free, the contributions of the contours and 3 4Γ one has,

i ij jT n 0= σ = Γ on and 3 4Γ (6.38)

In addition, the contours Γ3 and Γ4 are in the dx direction,

on 3Γ and Γ (6.39) 4dy 0=

Applying eqs. (6.38) and (6.39) to eq. (6.37), we find

3 4Γ ΓJ J 0= = (6.40)

Finally, substituting eq. (6.40) into eq. (6.37) gives,

1 2Γ Γ

J J=

2u∼

(6.41)

6.3.4 J-displacement diagrams

An elastic-perfectly plastic material is a material with a load-displacement behavior shown as a solid line in Figure 6.14(a). We consider this case first because it is the easiest type of plastic behavior to model. The material behaves linear elastically until the critical displacement u0. At this point, the material continues to deform under constant load. The J-displacement curve for this material is also shown as a solid line in Figure 14(b). In the elastic region J is equivalent to G which is proportional to the displacement squared. Once the material is perfectly plastic, J is then proportional to the displacement.

One can then consider the non-linear elastic load-displacement curve, shown as a dashed line in Figure 6.14(a), as a deviation from the elastic-perfectly plastic curve. When integrated, the corresponding J-displacement curve is shown as a dashed line in Figure 6.14(b). Thus, in the limits of small displacements J , and large displacements J ~ u. However, in between these limits the curve depends on the non-linearity.

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Figure 6.14 (a) Load-displacement and (b) J-displacement curves for an elastic-perfectly plastic material. Dashed lines show equivalent load-displacement and J-displacement curves for a non-linear elastic

material.

6.4 Stress Field Near Crack Tip

6.4.1 HRR Theory

As for the elastic case, it is useful to derive the form of the stress field near the crack tip for elastic-plastic fracture. This section presents the most commonly used description of the stress field. Hutchinson, Rice and Rosengren (HRR) (Anderson, 1995) assumed a power law relationship between the plastic strain and stress,

n

0 0 0

⎛ ⎞ε σ σ= +α⎜ ⎟ε σ σ⎝ ⎠

(6.42)

where 0

σ is the yield strength, a is a dimensionless constant, 0 0 Eε = σ , and n is the strain hardening exponent. Thus α and n are material properties. For a linear elastic material, n = 1 . For a strain hardening material n > 1. The material following (6.42) is called a Ramberg-Osgood material. Now, for J to remain path independent, it is necessary that the stress field vary as l/r near the crack tip. Thus,

( ) ( )1 n 1 n n 1Jij 1 ij 2

JA Ar r

; + +

⎛ ⎞ ⎛ ⎞σ = ε =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.43)

where A1 and A2 are constants. The singular stress-strain field of eq. (6.43) is called the HRR singularity. Note that for a strain hardening exponent n = 1, we recover the linear elastic case.

1 2r, ~ −σ ε

1 (n 1)/− +

ij ij (6.44)

Thus a cracked body can have two singularity dominated zones. In the plastic zone, closest to the crack tip, the stress varies as r (the HRR singularity). In the elastic zone, surrounding the plastic zone, the stress varies as 1 r .

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As seen in eq. (6.43), J defines the amplitude of the HRR singularity, just as K does for the linear elastic case. Thus, J is an effective ‘stress intensity factor’ for elastic-plastic fracture. Figure 6.15 shows a graph of a typical stress field due to elastic-plastic fracture, as compared to the HRR theory and linear elastic fracture mechanics.

Also shown in Figure 6.15 is a small region near the crack tip where crack blunting occurs (large strain region). As before in the linear elastic case, we ignore this region in assuming that it does not affect the global behavior of the regions surrounding the crack tip. The HRR theory is based upon small displacements which is not applicable in this region. To study this region, one would need to apply a large deformation analysis, for instance using finite elements.

Figure 6.15 (a) Singularity dominated regions near crack tip in elastic-plastic fracture; (b) stress field near crack tip, dashed lines show stress calculated from LEFM and HRR theories.

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6.5 Applications of the J-integral

6.5.1 The J-integral along a specific contour

To illustrate the use of the J-integral, consider first a cracked specimen loaded in Mode I and crack free surfaces. Without the need to know the form of the loading, we can proceed to determine the explicit forms of the J-integral along a symmetric contour ABCDC´B´A´ shown in Figure 6.16. The contour shown can be either along the border of the specimen or the contour around the crack in a larger specimen or component. The crack faces A´O and AO are traction free. It is also understood that the chosen contour can be chosen along the boundaries of the specimen or in it interior.

B´ C´ iT

Figure 6.16 A symmetric contour for the calculation of J.

For 1-dimensional crack growth, along x-axis,

iij jJ wdy-σ n ds i,j = x,y (for a plane problem)

xΓ= ⎜ ⎟∂⎝ ⎠∫

u∂⎛ ⎞ (6.45)

Here W is the strain energy density, ijσ is the stress tensor, is the outward normal to the element of ds of the contour and is the displacement. The stress vector and the outward normal are related by the following relation,

inu

T = σ n

T

i

i ij j (6.46)

where is the stress vector. We further assume symmetry of the specimen and loads with respect to the x-axis. For a plane stress, linear elastic problem, the strain energy density is given by,

i

(ij ij ij ij xx xx yy yy xy xyw= σ dε = σ ε = σ ε +σ ε +2σ ε2 2∫ )1 1 (6.47)

A´ A

B

D

C

in

xn

yn

ds

x

y

O

y

x

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Using the stress-strain relations for plane stress (eqs 2.15), W is expressed as,

( )2 2 2xx yy xx yy xyw= σ +σ -2νσ σ + σ

2E E1 1+ν

in cos(x, n ), n cos(y, n )

(6.48)

The components of the outward normal defined as, x i y= = are determined when dx and dy are positive, i.e., on contour BC (dx>0) and CD (dy>0). Thus,

y xn 1, nds ds

= = − = = dx dy 1

Next, one can calculate each term of the integral (6.45) along the chosen contour shown in Figure 6.16. In this example, W is considered known and we deduce the second expression of (6.45),

yi x xij j xx x xy y yx x yy yσ n ds=σ n ds+σ n ds+σ n ds+σ n

x x x xyu uu u u ds

x∂ ∂∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ (6.49a)

or,

yi x xij j xx xy yx yyσ n ds=σ dy-σ dx+σ dy-σ dx

x x x x xyu uu u u ∂ ∂∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ (6.49b)

Therefore, for each part of the contour the result is,

Along AB or A´ B´ dx = 0 and dy ≠ 0

yi xij j xx yxσ n ds=σ dy+σ dy

x x x∂ ∂ ∂

uu u ∂∂ ∂

Along BC or B´C´ dx ≠ 0 and dy = 0

yi xij j xy yyσ n ds=-σ dx-σ dx

x x∂ ∂ ∂

uu ux

∂∂ ∂

Along CD or C´D´ dx = 0 and dy ≠ 0

yi xij j xx yxσ n ds=σ dy+σ dx

x x x∂ ∂ ∂

uu u ∂∂ ∂

Finally, along AO or A´O the value of J is zero since dy = 0 and Ti = 0.

Recalling the properties of the line integrals and the symmetry of the contour one can write,

B C Dyu u uu u u∂ ∂ ∂⎡ ⎤ ⎡ ⎤ ⎡ ⎤∂ ∂ ∂⎥

y yx x xxx xy xy yy xx xy

A B C

J=2 w-σ -σ dy+2 σ +σ dx+2 w-σ -σ dyx x x x x x⎢ ⎥ ⎢ ⎥ ⎢∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∫ ∫ ∫ (6.50)

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It is important to notice that the stresses in (6.50) are the tractions along the contour and related to eq. (6.46).

6.5.2 Example

The double cantilever beam of unit thickness with a>>h is loaded under constant displacement , along the two horizontal borders as shown in Figure 6.17. Determine the J-integral along the

contour OABCDEO. Note that the crack tip is not included in the contour. yu

u / h

T 0

To simplify the calculations, it is assumed that the material above and below the crack is stress free and the rest of the specimen is subjected to a constant stress corresponding to a vertical displacement strain ε = . y

yu

Figure 6.17 Double cantilever beam subjected to remote vertical displacement with the contour of integration.

Solution

Contour OA

It is free of traction i = and dy = 0 . Therefore OAJ 0= .

Contour OF

For the same reasons as before . ODJ 0=

0=J 0

Contours AB

It is traction free T . It is further assumed that due to the presence of a long crack these parts of the specimen (above and below the crack) are unloaded and thus w = 0. Therefore

i

AB = .

Contour EF

For the same reasons as before . EFJ 0=

A

B C

D E

F O

y

x

yu

a

2h

B

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Contour BC

With these values, (6.49b) results in,

y yi x xij j xy y yy y xy yyσ n ds=σ n ds+σ n ds=σ dx+σ dx

x x x xu uu u u

x∂ ∂∂ ∂ ∂

0

∂ ∂ ∂ ∂ ∂

Since there is no shear stress on this contour σxy = . Also yu / x=0∂ ∂

uJ 0=

J 0=

0=

D +h

T 0=

because the vertical displacement there is constant. In addition. dy=0 along the contour and wdy=0. Therefore,

. y

BC

Contour DE

For the same reasons as before, . DE

Contour CD

It is free of traction T . It is assumed that the state of stress far ahead of the crack tip is uniform and the plate is thin with a constant strain energy w. Thus,

i

CDC -h

J=J = wdy w dy 2hw= =∫ ∫ (6.51a)

The exact form of the w depend on the state of stress (plane strain or plane stress).

- Note that the stress-stain relations could be linear or non-linear elastic.

- The contours OA and OF could be eliminated from the start since and dy = 0. i

Assuming linear elastic behavior, plane strain and the ligament size very long, one can write,

2y

yy yy

u1 1 (1-ν)EW= σ ε2 2 (1+ν)(1-2ν) h

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Here the stress component given by (4.3b) has been recalled and the Poisson effect has been neglected. Therefore,

2(1-ν)EuyJ=2hw=(1+ν)(1-2ν)h

(6.51b)

6.5.3 Example

Consider the double cantilevered beam specimen shown in Figure 6.18. The beam is loaded by dipole forces P, applied on the end faces. With the simple beam theory and assuming that a >> h, evaluate the energy release rate as a function of crack length a using,

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(a) the compliance method

(b) the J-integral method using a contour along the specimen boundary as shown by the arrows in Figure 6.18.

To simplify the calculations, it is assumed that the material above and below the crack is subjected to bending due to the force P applied at the center of gravity of the section hB with the corresponding displacement . The other part of the specimen is assumed stress free with w=0. yu

D

Figure 6.18 Double cantilevered beam subjected to end loads with the contour of integration.

Solution

(a) Compliance method

The energy release rate G, is given by,

2 2G P P2 A 2B

= =∂

1 C 1 Ca

∂ ∂∂

(6.52)

From beam theory (Del Pedro, Gmür & Botsis, 2001),

3 2 3P x x Bh⎛ ⎞y

3 3

y 3x a

u a IEI 6 2 12

Pa 4au P3EI EBh=

= − + =⎜ ⎟⎝ ⎠

= =

(6.53)

The total compliance (due to both beams),

32u 8ay3C

P EBh= = (6.54)

Substituting (6.54) into (6.52) the G is,

A

B C

E

D

y

x a

P, yu

yuP,

2h

B

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2 2 2

3

1 24a 12P a23 2G P

2B EBh EB h= = (6.55)

(b) J-integral method

Since the material is linear elastic, J = G. Thus, we can use the J-integral to calculate G. Divide the contour shown into 5 segments, as numbered in Figure 6.18. We want to evaluate,

iij jJ wdy n ds

= −σ⎜ ∂⎝ ⎠∫u∂⎛ ⎞

T 0= J J 0= =

T 0=J 0

(6.56)

Where Γ is the contour ABCDEF.

Contours BC or DE

i and dy=0. Thus, BC DE

Contour CD

i and W=0 due to the assumption that the part of the beam far ahead of the crack tip does not deform. Thus, CD =

Contour AB or EF

Consider first the term with the W. To evaluate the integral along these contours we proceed as follows

xx yy xyP0

hBσ = σ = σ ≈, , yx

xyu1

2 y xε = +⎜ ⎟∂ ∂⎝ ⎠

u∂⎛ ⎞∂

Thus ij ij ij ij xy xyw= σ dε = σ ε =σ ε2∫1

To determine the shear strain ε , one can use the displacement field from the theory of elasticity of a cantilever beam (Del Pedro, Gmür & Botsis, 2001) to obtain

xy

yxxy

x a y h 2

u1 02 y x

= =

∂ε = + =⎜ ⎟∂ ∂⎝ ⎠ , /

u∂⎛ ⎞

w=0

Therefore, . Now, consider the term, on AB or EF,

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y yi x xij j xx x xy y yx x yy y

y yx xxx x yx x xx yx

σ n ds=σ n ds+σ n ds+σ n ds+σ n dx x x x

u uu u =σ n ds+σ n ds=σ dy+σ dyx x x x

u uu u u sx

∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂∂ ∂ ∂ ∂

The contribution of each arm of the specimen is calculated as follows. From beam theory one obtains the vertical displacement of the gravity center of the section hB (Del Pedro, Gmür & Botsis, 2001),

For the upper arm For the lower arm

xx xyP0

HBσ = σ ≈; , j y

ij jx a

u uPnx HB x

=

∂ ∂σ =

∂ ∂ xx xy

P0HB

σ = σ ≈ −; , j yij j

x a

u uPnx HB x

=

∂ ∂σ = −

∂ ∂

3 2

yP x axuEI 6 2

⎛ ⎞−= +⎜ ⎟

⎝ ⎠

3 2

yP x axuEI 6 2

⎛ ⎞−= +⎜ ⎟

⎝ ⎠

2 2y y

x a

u uP x Paaxx EI 2 x 2E

=

∂ ∂⎛ ⎞−= + =⎜ ⎟∂ ∂⎝ ⎠

; I

2 2

y y

x a

u uP x Paaxx EI 2 x 2E

=

∂ ∂⎛ ⎞−= + =⎜ ⎟∂ ∂⎝ ⎠

; I

2 2j

ij j

u P anx 2EIHB

∂σ =

2 2j

ij j

u P anx 2EIHB

∂σ = −

Finally,

0 h2 2 2 2 2 2uIB

−∂ ⎛ ⎞jij j

h 0

P a P a P an ds dy dyx 2IEhB 2IEhB EΓ

− σ = − − − =⎜ ⎟∂ ⎝ ⎠∫ ∫ ∫

or

2u 12P a∂ 2j

ij j 2 3n dsx EB hΓ

− σ =∂∫ and

212P a2

2 3JEB h

= (6.57)

Comparing the result with the expression (6.55) it is clear that the two methods above are equivalent means to calculate the energy release rate.

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6.4.4 Relationship between J and CTOD

A simple yet important relationship can be derived between the CTOD and the J-integral for the yield strip model. The bases of the calculations are shown in Figure 6.19.

∞σ

Figure 6.19 The J-integral contour for the crack and plastic zone of the yield strip model.

The contour is chosen around the crack tip as shown in Figure 6.19. Note that the singularity is cancelled due to the presence of the plastic zone. Therefore, the only stresses applied are the remote stress and the yield stress at the crack faces over the distance c.

To calculate the J-integral along the contour Γ it is noticed that when , tc >> δ dy 0≈ . Accordingly, ( 6.45) reduces to,

i iij j ij jJ n ds n

x xΓ Γ

= − σ = − σ∂ ∂∫ ∫u u ds∂ ∂

With the only non-zero component of stress being yy Yσ = σ , the remaining of the integrand becomes,

y yiij j yy y Yn ds n ds dx

x xσ = σ = −σ

∂ ∂

u uux

∂ ∂∂∂

and

t t2

Y t

u /δ δ∂

J = σ δ

t

yY Y y Y y

2 0

J dx du dux /Γ −δ

= σ = σ = σ = σ δ∂∫ ∫ ∫ ,

Y t (6.58)

∞σ

c

a

tδ y

Γ

x

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In comparing (6.58) and (6.23) one sees that the latter result is higher by about 25%.

For a linear elastic materials, or for small scale yielding (case 1 in Figure 2), the J-integral, ERR, SIF and the CTOD are equivalent parameters.

2

Y tK

J n= σ δ

IIJ G

E'= = = σ δ (6.58)

If the material cannon be approximated by a perfect plastic material, a similar relation can be obtained between J and δ by using the Ramberg-Osgood material (6.42) (Anderson 1995), t

Y t

where σ is the yield stress and n is the strain hardening exponent in (6.42). Y

6.4.5 The J-integral along a rectilinear interface

Consider two homogeneous bodies 1Π and 2Π which are perfectly bonded along the axis x as shown in Figure 6.20. a biomaterial interface and a crack along it (Figure 6.20). The integral along the contour expressed by (6.45) is, 1 2Γ = Γ ∪Γ

iij jJ= wdy-σ n ds i,j = x,y (for a plane problem)

xΓ⎜ ⎟∂⎝ ⎠∫

u∂⎛ ⎞ (6.45)

y

x 1Π

Figure 6.20 A crack along a bi-material interface and the contours for the J-integral.

It is also path independent provided that the interface line is straight. Thus, the standard J-integral is extended to a biomaterial interface when the latter is straight. Its interpretation is also

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as the energy release rate during crack advance. For a curvilinear crack the J-integral is not path independent because the conservation law given by (6.34) is not valid2.

2 R. E. Smelser and M. E. Curtin, On the J-integral for bi-material bodies, International Journal of Fracture, Vol. 13, 1977, pp. 382-384.

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