Elastic- Plastic Fracture in “Method of Finite Elements II” · •J-integral •Definition of...

Elastic-Plastic Fracture in “Method of Finite Elements II”

Elyas Ghafoori

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2Autumn Semester 2019/20 Method of Finite Elements II

Dr. Elyas GhafooriLeader of “Sustainable Metallic Structures (SUMS)” Group Empa, Swiss Federal Laboratories for Materials Science and TechnologyÜberlandstrasse 129, 8600 Dübendorf, SwitzerlandE-Mail: [email protected] Homepage

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• Review of linear elastic fracture mechanics (LEFM)

• Elastic-plastic fracture mechanics• 1-D approximation• 2-D approximation• Irwin’s approximation of the plastic zone

• J-integral• Definition of the J-integral• Properties of the J-integral• Example 1: the J-integral along a specific contour• Example 2: double cantilever beam under constant displacement• Example 3: double cantilever beam under end load

• Summary and recommendations

Contents


Review of linear elastic fracture mechanics (LEFM)


Review of LEFM


Review of LEFM

Pictures courtesy of Dr. K. Agathos, ETH Zurich

3D and contour plot of σyy stress component


• There are three fracture modes :

Stress field in a plate with a crack


Elastic -Plastic Fracture Mechanics


Elastic-plastic fracture mechanics


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.


Estimation of plastic zone size: One dimensional approximation

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

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

has plastically deformed.



• For this first approximation, we consider only the stress field on the line θ=0, as shown in the Figure, and define L as the characteristic length of the plastic zone. By introducing the stresses in the von Mises yield criterion, we can 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,

• Note that σθθ and σrr

are the principle stresses σ1 and σ2 for the line θ=0.



• Applying the plane strain condition (i.e., σ3 = ν(σ1+σ2); ν being the material's Poisson ratio) to eq. (6.1), one has,

• For the special case of σ1=σ2 (such as along the line θ=0), eq. (6.3) is simplified to,



• Substituting the principle stress ofeq. (6.2) into eq. (6.4), the yieldcriterion is given by,

• Thus, the length L, over which plasticdeformation occurs is given by,

• Note that the value L (referred alsoas ry or r1) is often called the plasticzone radius.

Figure 6.1 Schematic of plastic zone ahead of crack tip.



• The calculation of L is only an approximation since the presence ofdamage will modify the stress field and thus the size of the zone.

• However, it can serve as an approximate parameter to comparewith the overall dimensions of the specimen, and determine thelimits of the linear elastic solution.


Estimation of plastic zone size: Two dimensional approximation



• 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 the plastic zone shape.

• In cylindrical coordinates the principle stresses σ1 and σ2 for an arbitrary stress field are given by,

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

and σ3 = ν(σ1+σ2) 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 the von Mises yield criterion (6.3) and simplifying leads to the following yield condition,

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



(a)

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

(b)



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.

• For an actual specimen, one mustoften 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.


Estimation of plastic zone size: 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 field (4.36) and shown in Figure 6.5 by the curve (1).

• To obtain a simple approximation of the effects of plastic deformation on the redistribution of the asymptotic stress field (shown below), perfect plasticity is assumed and consequently the stress ahead of the crack tip equals σY up to a distance r2 to be determined.

• After that distance, the distribution is obtained by a translation of the asymptotic field shown by the curve (2).

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

the crack tip.



• The two distributions, before andafter yielding, should result inequal forces since equilibriumshould be assured. With referencesto Figure 6.5, force equilibriumresults in,

where the distance r1, is theintersection of the field (1) and thehorizontal line at σY given by, 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=r2. Thus, the plastic zone extends over a distance equal to r1 + r2 = 2 r1 given by,

• The simple analysis shown above is for the case of plane stress i.e., σ3 = 0, which is realistic for thin plates.

• In plane strain, yielding is confined due to the effects of σ3 > 0 and a smaller plastic zone ahead of the crack tip is developed. In this case, Irwin proposed the following expression,

which is smaller than r1 in (6.11) by a factor of 3.• The value r1 (know also as ry or L) is often called the plastic zone radius.Broek, D., Elementary Engineering Fracture Mechanics, Martinus Nijhoff Publishers, 4th Edition, 1986.


• 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, Keff , which for a crack in an infinite plate is expressed as,


New stress filed

Old stress field

Effective crack

Real crack


• For a general case with finite geometry, the effective stress intensity is obtained by inserting aeff into the K expression for the geometry of interest:

• Since the effective crack size is taken into account in the geometry correction factor F( Τa eff b), an iterative solution is usually required to solve for Keff. That is, K is first determined in the absence of a plasticity correction; a first-order estimate of aeff is then obtained from Equation (6.12a) or Equation (6.12b), which in turn is used to estimate Keff.

• A new r1 and consequently a new aeff are computed from the last Keff

estimate, and the process is repeated until successive Keff estimates converge. Typically, no more than three or four iterations are required for reasonable convergence.


ሻKeff = σ πa F( Τa eff b


• In certain cases, this iterative procedure is unnecessary because a closed-form solution is possible.

• For example, the effective Mode I stress intensity factor for a through crack in an infinite plate in plane stress is given by

for plane stress



Conditions for the LEFM

• An important restriction to the use of LEFM is that plastic zone size at the crack tip must be small relative to the crack length as well as the geometrical dimensions of the specimen or part:

• A definite limiting condition for LEFM is that net (nominal) stresses in the crack plane must be less than 0.8 σy (80% of yield strength).

• Under monotonic loading r1 ≤ (1/8) a.• Other restrictions include r1 ≤ 1/8 of B and (w-a) where B

(sometimes referred as t) is the thickness and (w-a) is the uncracked ligament along the plane of the crack.

• Otherwise, a plasticity correction (given by eq (6.13)) is required for the stress intensity factor, K, or elastic-plastic fracture mechanics using J-integral shall be used, which will be explained later in this course.


Any Questions?


Definition of the J-integral



Reminder: The linear elastic energy release rate GI, was derived and shown to be the area under the load-displacement curve for a given body between two load states.

G


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.

J• In the same manner as GI, 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 GIC.



• In practice, the nonlinear energy release rate J, is calculated using the J-integral, written as,

where w is the strain energy, defined by,

and Ti are the boundary tractions, given by,

The summation convention applies here with i = x and j = y.

Rice, J. R., Mathematical Analysis in the Mechanics of Fracture, in FRACTURE An Advanced Treatise, Vol. II, Chapter 3, edited by H. Liebowitz, Academic Press, New York, NY 1968.



• The contour over which the integral is evaluated is shown in Figure 6.11.

• The origin of the x-y axes is taken 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.

• The choice of Γ is arbitrary.

• The traction vector T, unit normal vector to the contour n and the increment along the contour ds, are also shown in Figure 6.11.

• As mentioned above, the integral expression of eq. (6.24) defining J is equivalent to the area between the loading and unloading curves in Figure 6.10, i.e. the nonlinear elastic energy release rate.



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



• It can be shown that

• Thus, the J-integral defined in eq. (6.24) is equivalent to the nonlinear energy release rate.

• When the material is linear elastic, the ERR GI is equal to J.

• Note that, for a linear elastic materials, or for small scale yielding (case I in Figure 6.2), the J-integral, ERR, SIF and the CTOD are equivalent parameters. For mode I:



• 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 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.



Elastic-plastic fracture: Properties of J-integral

Proof of property 2• Consider the two arbitrary contours

Γ1 and Γ2, shown in Figure 6.13(a).We want to show that J, evaluated oneach of the two contours, is

equivalent, i.e. J|Γ1 = J|Γ2 . For this wedefine the closed contour തΓ, shown inFigure 6.13(b). The contour iscomposed of four individual contoursegments തΓ= Γ1∪ Γ2

*∪ Γ3∪ Γ4. Usingthe property 1, one can write

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



• Note that,

as the direction of the contour has been reversed to form the closed contour തΓ. Using the last equality and that when the contour is closed we can write,

• Considering that the crack faces are traction free, for the contributions of the contours Γ3 and Γ4 one has,

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



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

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


Example 1


Figure 6.16 A symmetric contour for the calculation of J.

Example 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 boarder 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 contour can be chosen along the boundaries of the specimen or in it interior.



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

• As defined before, w is the strain energy density, σij is the stress tensor, n is the outward normal to the element of ds of the contour and ui is the displacement. The stress vector T, and the outward normal are related by the following relation,

• 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,



• Using the stress-strain relations for plane stress (eqs 4.15), W is expressed as,

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

• 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),



or,

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

Along AB or B´A´ dx=0 and dy≠0

Along BC or C´B´ dx≠0 and dy=0



Along CD or DC´ dx=0 and dy≠0

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,

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


Example 2


Example 2: Double cantilever beam under constant displacement

The double cantilever beam of unit thickness with a>>h is loaded under constant displacement uy, along the two horizontal boarders as shown in Figure 6.17. Determine the J-integral along the contour OABCDEFO. Note that contour contains the crack surfaces but not the crack tip.

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=uy/h.

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



Solution• Contour OA

It is free of traction Ti=0 and dy=0, therefore JOA=0.

• Contour OFFor the same reasons as before JOD=0.

• Contours ABIt is traction free Ti=0. 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 JAB=0.

• Contour EFFor the same reasons as before JEF=0



• Contour BCAlong this contour dy=0 and (6.49b) results in,

Since there is no shear stress on this contour σxy=0. Also ∂uy/∂x=0 because the vertical displacement there yuis constant. In addition, dy=0 along the contour and wdy=0. Therefore, JBC=0.

• Contours DEFor the same reasons as before, JDE=0.



• Contour CDIt is free of traction Ti=0. 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,

• 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 Ti=0and dy=0.



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

• Here the stress component given by (4.3b) has been recalled and the Poisson effect has been neglected (i.e., εxx=0). Therefore,


Example 3


Example 3: Double cantilever beam under end load

Consider the double cantilever beam specimen shown in Figure 6.18. The specimen 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,

(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.

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



• 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 uy.

• The other part of the specimen is assumed stress free with w=0.



Solution

(a) Compliance method• The energy release rate G, is given by,

• From beam theory

Del Pedro, M., Th. Gmür & J. Botsis, Introduction à la mécanique des solides et des structures, PPUR, Lausanne, 2001.



• The total compliance (due to both beams),

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



(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,

Where Γ is the contour ABCDEF.

• Contours BC or DETi=0 and dy=0. Thus, JBC=JDE=0.



• Contour CDTi=0 and w=0 due to the assumption that the part of the beam far ahead of the crack tip does not deform. Thus, JCD=0

• Contour AB or EFConsider first the term with w. To evaluate the integral along these contours we proceed as follows

Thus,




To determine the shear strain εxy, one can use the displacement field from the theory of elasticity of a cantilever beam to obtain,

• Therefore, w=0. Now, consider the other term of the intergal, on AB or DE,




• 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.



• Therefore,

or

• 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.


Any Questions?


Summary and Recommendations



• Plastic relaxation and redistribution of the stress field occurs in the plastic zone:• The stress distribution for σy shown below must shift to the right

to accommodate the plastic deformation and satisfy equilibrium conditions.

• As a result, the actual plastic zone size is approximately twice the calculated value.



• Plane strain plastic zone size is usually taken as one-third the plane stress value.

• For a through crack in a thick plate:• The free surfaces with zero

normal and shear stresses are in a plane stress condition.

• The interior region of the plate near the crack tip is closer to plane strain conditions as a result of elastic constraint.

• Thus, the plastic zone size along the crack tip varies similarly to that shown schematically here.



• Under monotonic loading, the plastic zone size, 2 r1, at the crack tip, in the plane of the crack, is:

for plane stress

for plane strain

• In plane strain, yielding is suppressed by the triaxial stress state, and the Irwin plastic zone correction is smaller by a factor of 3.

• Note that the value r1 (referred also as ry or L) is often called the plastic zone radius.

3



• In the same manner as GI, J is used as a fracture criterion, meaning that for a given material at a certain critical value JC, the crack will propagate.

• The two main properties of J-integral:1. J integrated along any closed contour is zero. 2. J is path independent.

• In the case of linear fracture, JC reduces to GC.


References

[1] Schijve J. “Fatigue of Structures and Materials”, 2008: New York: Springer.[2] Anderson T.L. “Fracture Mechanics - Fundamentals and Applications”, 3rd Edition, Taylor & Francis Group, LLC. 2005.[3] Budynas R.G., Nisbett J.K. “Shigley's Mechanical Engineering Design”, 2008, New York: McGraw-Hill.[4] Stephens R.I., Fatemi A., Stephens R.R., Fuchs H.O., “Metal Fatigue in Engineering”, 2nd Edition, John Wiley & Sons, 2000.[5] Socie D., A., University of Illinois, “Fatigue and Fracture”, 2004-2013 (efatigue.com). [6] Botsis J., “Elements of Fracture Mechanics”, Laboratory of Applied Mechanics and Reliability Analysis, EPFL, 2019.

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