Ansys Fracture Tutorial

8/13/2019 Ansys Fracture Tutorial

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Universitat de Girona

FRACTURE MECHANICS

Computer lab sessions

D. Trias

October 2012This document can be found at: ftp://amade.udg.edu/amade/mme/MecFrac/MecFrac.pdf


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Contents

1 Singular stresses 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Effect of element type, size and shape . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.2 Finite element discretisation of stresses at a crack tip . . . . . . . . . . . . . . . . . . 4

1.2 Quarter point / crack tip elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Creating quarter mid-nodes at crack tip with ANSYS . . . . . . . . . . . . . . . . . . . . . . 7

1.3.1 Meshing with usual tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.2 Meshing with special tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Summary and conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.5 Suggested problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.6 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Computational Fracture Mechanics I: Computation of G 172.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Finite Crack Extension Method (FCEM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Crack Closure Method (CCM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 Virtual Crack Closure Technique (VCCT) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.5 Suggested exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.6 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Computational Fracture Mechanics II: Computation of K 27

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.2 The stress intensity factor (K) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.2.1 Numerical estimation of the stresses at the crack tip . . . . . . . . . . . . . . . . . . 28

3.2.2 Computation of K by stress extrapolation . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2.3 Computation of K by displacement extrapolation. . . . . . . . . . . . . . . . . . . . . 30

3.2.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.3 Displacement extrapolation with quarter node elements . . . . . . . . . . . . . . . . . . . . . 32

3.3.1 Formulae for the stress intensity factor . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.4 ANSYS commands for the computation of K . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.4.1 Crack opening displacement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.4.2 KCALC command . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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3.5 Proposed exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.6 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Computational Fracture Mechanics III: Computation of the J-integral 37

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2 The J integral with ANSYS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38


4.4 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5 Computational Fracture Mechanics IV: Cohesive zone modeling 41

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.2 Cohesive laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.2.1 Bilinear law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5.2.2 Exponential law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.3 Cohesive elements in ANSYS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.3.1 Cohesive zone modeling with interface elements . . . . . . . . . . . . . . . . . . . . . 44

5.3.2 Cohesive zone modeling with contact elements . . . . . . . . . . . . . . . . . . . . . . 48

5.4 Some remarks on element size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51


5.6 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5.7 Aknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51


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Chapter 1

Singular stresses

1.1 Introduction

Linear Elastic Fracture Mechanics deals with cracked solids. This means the stress and strain fields within

this loaded solids are assumed to be influenced somehow by the presence of cracks. In order to analyze

which is the efect of these cracks let us model a simple cracked plate.

Example 1.1 (Analysis of a cracked plate). Let us assume we consider the analysis of the cracked plate

component of Figure1.1 by the Finite Element Method.

Figure 1.1: Plate with a central sharp crack.

For the sake of simplicity let us assume the material of the plate is steel and the applied pressure is 1

MPa.

Solution to Example 1.1. The ANSYST M command sequence for this example is listed below. You can either

type these commands on the command window, or you can type them on a file, then, on the command window enter

/input, file, ext or just use copy and paste.

FINISH/CLEAR

/TITLE Stress singularity

1


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2 Fracture Mechanics

!PRE-PROCESSOR *******************************

/PREP7

!Geometrical parameters

L = 50 ! Half length of component

a = 10 ! Crack half length

b = 0 ! Crack width

th = 30 ! Component thickness

el_len = 5 ! Element length (mesh)

el_shape=0 ! 0: quad, 1: triangle

ET,1,PLANE42 !4 node solid element

!ET,1,PLANE82 !8 node solid element

!Material properties

MP,EX,1,210000 !Young modulus

MP,PRXY,1,0.3 !Poissons coef.

KEYOPT,1,3,3 !Keyoption for thickness activation

R,1,th !Thickness for element type #1

!Geometry definition by keypoints

K,1,a,0

K,2,L,0

K,3,L,L

K,4,0,L

K,5,0,b

!Lines from keypoints

L,1,2,(L-a)/el_len

L,2,3,L/el_len

L,3,4,L/el_len

L,4,5,(L-b)/el_len

L,5,1,a/el_len

!Areas from lines

AL,1,2,3,4,5

!Mesh

LCCAT,5,1

MSHAPE, el_shape, 2D

MSHKEY,2

AMESH,ALL

FINISH

!SOLUTION ***********************************

/SOLU


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Chapter 1. Singular stresses 3

DL,1,1,SYMM

DL,4,1,SYMM

SFL,3,PRES,-1 !Pressure on top tip

SOLVE

FINISH

!POST-PROCESSOR ****************************

/POST1

PLDISP,1 !Deformed shape

PLNSOL,S,Y ! Stresses Y direction

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

! Path plot of stress components at theta=0

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!PATH,0DEG,2,10,50

!PPATH,1,,a,0,0

!PPATH,2,,3*a,0,0

!PDEF,SX,S,X,NOAV

!PDEF,SY,S,Y,NOAV

!PDEF,UY,U,Y,NOAV

!PDEF,SY2,S,Y,AVG

!PLPATH,SY2

!PLPATH,SX

!PLPATH,UY

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

! Path plot of stress components at theta=45 deg

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

PATH,45DEG,2,10,50

PPATH,1,,a,0,0

PPATH,2,,3*a,3*a-a,0

PDEF,SX,S,X,NOAV

PDEF,SY,S,Y,NOAV

PDEF,UY,U,Y,NOAV

PDEF,SY2,S,Y,AVG

PLPATH,UY

PLPATH,SX

PLPATH,SY

PRPATH,SY

FINISH

This file can be found at:

ftp://amade.udg.edu/mme/FEmet/1_stress_sing.dat


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1.1.1 Effect of element type, size and shape

Let us consider now the effect of element size, type and shape in the stress field of the analyzed plate. To

do so, let us fill in the following table:

Element size Triangle Triangle Quad Quad

(el_l en) PLANE42 PLANE82 PLANE42 PLANE42

1 mm

2 mm

3 mm

4 mm

1.1.2 Finite element discretisation of stresses at a crack tip

Try to explain what happens when regularelements are employed to approximate the stress field at a crack

tip. You may employ Fig1.2 by drawing on it the stress field, different discretization sizes and the approxi-

mation obtained by each one. You may start thinking what happens when constant strain elements are used

and then try to figure out which is the solution given by linear elements.

Figure 1.2: Finite element discretization of the stresses near the crack tip

As you may have noticed, traditional engineering maximum stress-based failure analysis become senseless

when stress singularities appear, so other approaches must be used. These are basically stress intensity factor

(K) based approaches and energy release rate (G) based approaches. While G-based and K-based approachesare the main topic of next chapters, the next section gives an introduction of quarter node tip elements,

which are used to capture efficiently the stress singularity in a finite element discretization.


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1.2 Quarter point / crack tip elements

The aim of these elements is to introduce in the element formulation a stress singularity of the 1/

r type.

This is useful for stress related approaches, such as the numerical computation of the stress intensity factor.

To do so we may start by recalling the isoparametric formulation of a 1D quadratic Lagrangian element:

Figure 1.3: 8 node isoparametric quadrilateral element

N1 = 1

2(1 ) (1.1)

N2 =

(12) (1.2)

N3 =1

2(1 +) (1.3)

Since for an isoparametric element the same approximation for the geometry and for the displacements

is used, the geometry of the 1-3 edge may be expressed:

x=n=3

i

Nixi= 1

2(1 )x1 + (1 2)x2 +

1

2(1 +)x3 (1.4)

Figure 1.4: 8 node isoparametric element with quarter-side located mid-nodes


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Then if we locate the origin of a quad element of side length Lat node 1 and locate the mid-node (node

#2) at x2 = L/4, as shown in Figure 1.4:

x

=

1

2

(1

+)L

+(1

2)

L

4

(1.5)

And solving for :

= 1 + 2

x

L(1.6)

As you remember, the displacement approximation is given by:

u=n=3

i

Niui= 1

2(1 )u1 + (1 2)u2 +

1

2(1 +)u3 (1.7)

whereu1, u2and u3are the displacements at nodes 1,2 and 3. Using equation1.6in the former equation

we obtain the expression of the displacements as a function of the geometry (x):

u= 12

1 + 2

x

L

2 2

x

L

u1 + 4

x

L x

L

u2 +

1

2

1 + 2

x

L

2

x

L

u3 (1.8)

If we now compute the strain in the x direction:

x=u

x= x

u

= 1

2

3

xL 4

L

u1 +

2

xL 4

L

u1 +

1

2

1

xL+ 4

L

u3 (1.9)

We may easily verify that the former expression is a function of (1/

x) and, consequently, the strain

field presents a singularity.

However, for a 2D element we might apply the same method for the 8-node serendipity element of Figure

1.4. This way, the singularity would be present only along the edge 1-3, and r would not be singular. We

may obtain a radial singular stress field by collapsing nodes 4, 5 and 6 of the quad of Figure 1.4, ash shown

in Figure1.5.

Figure 1.5: Quarter tip element from collapsed 8 node quadrilateral


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1.3 Creating quarter mid-nodes at crack tip with ANSYS

As seen in the former sections, if we have to deal with the stresses when stress singularities are present it

is useful to use some kind ofspecialelements. We also showed how, actually, these specialelements may

be constructed from regular elements just by placing the mid-side nodes at a quarter of the element sidelength of the crack tip. We usually may do this by our means or by using some special commands provided

by the commercial finite element software.

1.3.1 Meshing with usual tools

We may construct this kind of mesh at a crack tip manually, that is, placing the corresponding mid-side

nodes at a distance of a quarter of the side from the crack tip. The following ANSYS log file analyzes the

same problem of Figure1.1, using N, NGEN and E commands to build the mesh.

Example 1.2. Model the cracked plate of Figure1.1using the isoparametric quarter node elements presented

in Section1.2. Locate the quarter nodes by using ANSYS common node and element generation commands.




FINISH

/CLEAR

/TITLE Stress singularity - Mesh #2

!===================================

!PRE-PROCESSOR

!-----------------------------------

/PREP7

! Geometrical parameters in mm

L = 5 ! Plate length

a = 1 ! Crack length

b = 0 ! Crack heigth

th = 1 ! Plate thickness

el_len = 0.2 ! Element length

pi= 3.1415926535897932384626433832795

fx = a/10

rdiv = 4 ! radial divisions at crack tip (must be 4)

tdiv = 12 ! angular divisions at crack tip

rdiv2 = 10

ET,1,PLANE42


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MP,PRXY,1,0.3 !Poissons ratio

KEYOPT,1,3,3 !Keyoption to introduce thickness in element 42

R,1,th ! Thickness

!Keypoints which define the geometry

K,1,a,0

K,2,L,0

K,3,L,L

K,4,0,L

K,5,0,b

K,6,a,L

K,7,2*a,0

K,8,2*a,a

K,9,a,a

K,10,0,a

K,11,a+2*fx,0

K,12,a,2*fx

K,13,a-2*fx,0

! Lines

L,1,11,100,1000 !1

L,11,7,rdiv2 !2

L,7,2,10 !3

L,2,3,tdiv/4 !4

L,3,6,tdiv/4 !5

L,4,10,10 !6

L,10,5,tdiv/4 !7

L,10,9,tdiv/4 !8

L,9,8,tdiv/4 !9

L,7,8,tdiv/4 !10

L,5,13,rdiv2 !11

L,9,12,rdiv2 !12

!L,1,13,85,20 !13

L,1,13,100,1000 !13

LARC, 11, 12, 1, 2*fx !14

LARC, 12, 13, 1, 2*fx !15

LESIZE,14,,,tdiv/2

LESIZE,15,,,tdiv/2

L,1,12,100,1000 !16

L,6,4,tdiv/4 !17

L,9,6,10 !18

! Nodes

CLOCAL,11,1,a,0,0

CSYS,11


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N,1,0,0,0

N,rdiv,2*fx,0,0

FILL,1,,1,2,1,,,1

NGEN,tdiv*2+1,10,2,rdiv,1,0,90/tdiv,0,0

csys,1

! Generate 1/4 nodes

*DO,I,1,tdiv+1

FILL,1,2+20*(I-1),1,1000+I,,,,3

*ENDDO

*DO,I,1,tdiv+1

FILL,2+20*(I-1),4+20*(I-1),1,2000+20*(I-1),,,,1

*ENDDO

csys,1

ET,2,PLANE82

KEYOPT,2,3,3 !Keyoption to introduce thickness in element 82

R,2,th !thickness

TYPE,2

! Elements

csys,11

*DO,I,1,tdiv

E,2+20*(I-1),2+20*I,1,1,12+(I-1)*20,1001+I,1,1000+I

*ENDDO

E,2,4,24,22,2000,14,2020,12

EGEN,tdiv,20,tdiv+1

AL,2,10,9,12,14 ! Area created from lines

AL,11,15,12,8,7

AL,3,4,5,18,9,10

AL,18,17,6,8

AL,1,14,16

AL,16,15,13

ASBA,1,5,,DELETE,DELETE

ASBA,2,6,,DELETE,DELETE

LCCAT,4,5 !19

LCCAT,9,10 !20

LCCAT,7,8 !21

TYPE,2

!MSHKEY,2

AMESH,1,4 !Mesh

csys,11


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NSEL,S,LOC,X,2*fx-fx/50,2*fx+fx/50

NUMMRG,NODE,0.001

theta=pi/(2*tdiv)

!*DO,I,1,tdiv

!NMODIF,12+20*(I-1),fx*cos(theta)

!NMODIF,14+20*(I-1),2*fx*cos(theta)

!*ENDDO

/PSYMB,ESYS,1

csys,0

ALLSEL

FINISH

/SOLU

ANTYPE,0

D,1,UY,0

D,1,ROTZ,0

D,2,UY,0

D,2,ROTZ,0

DL,1,5,SYMM

DL,2,1,SYMM

DL,3,3,SYMM

DL,6,4,SYMM

DL,7,2,SYMM

SFL,5,PRES,-1 !Pressure applied to upper line

SFL,17,PRES,-1

ALLSEL

SOLVE

FINISH

!===================================

!POST-PROCESSOR

!-----------------------------------

/POST1

PLDISP,1 !Deformed shape

PLNSOL,S,Y ! Stresses Y direction

angle=pi/6

PATH,30DEG,2,10,100

PPATH,1,,a,0,0

PPATH,2,,a+cos(angle),sin(angle),0

PDEF,SX,S,X,NOAV


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PDEF,SY,S,Y

PDEF,UY,U,Y,NOAV

PLPATH,SY

PLPATH,SX

PRPATH

FINISH


ftp://amade.udg.edu/mme/FEmet/1_quarter_mid_nodes.dat

Figure1.6 shows the different path plots for different elements sizes when using ANSYS PLANE42 and

PLANE82 elements compared to the solution obtained with 2 mm Quarter mid-side node elements which

were constructed the way described above. Results show as these elements are able to provide a solution

similar to that obtained by much smaller elements. Since these elements include the 1/

rsingularity, they

do not provide a value for r= 0, which is the singular point.

0 2 4 6 8 101

2

3

4

5

6

r0(mm)

y

y(MPa)

0.5 mm PLANE42 TRI

2 mm PLANE42 TRI5 mm PLANE42 TRIQuarter midside nodes

0 2 4 6 8 101

2

3

4

5

6

7

r0(mm)

y

y(MPa)

1 mm PLANE82 TRI2 mm PLANE82 TRI5 mm PLANE82 TRIQuarter midside nodes

0 2 4 6 8 101

2

3

4

5

6

r0(mm)

yy

(MPa)

0.5 mm PLANE42 QUAD

2 mm PLANE42 QUAD5 mm PLANE42 QUADQuarter midside nodes

0 2 4 6 8 101

2

3

4

5

6

r0(mm)

yy

(MPa)

1 mm PLANE82 QUAD

2 mm PLANE82 QUAD5 mm PLANE82 QUADQuarter midside nodes

Figure 1.6: y yat the crack tip region for the analyzed plate, as obtained with different element type/size

and with Quarter mid-side node elements (45o path)

1.3.2 Meshing with special tools

Since doing this may be tricky, commercial finite element software usually have special tools for meshing

the crack tip. The next box shows ANSYS command for crack tip meshing.


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ANSYS Command: Meshing the crack tip

KSCON, NPT, DELR, KCTIP, NTHET, RRAT

NPT number of the node located at the crack tip

DELR Radius of first row of elements about crack tip

KCTIP Crack tip singularity key. For our purposes its value should be 1

NTHET Number of elements in circumferential direction. Default is one per 30o.

RRAT Ratio of 2nd row element size to DELR. Default is 0.5

The following ANSYS log file performs the same analysis but now the command KSCON is employed

to mesh the crack tip.

Example 1.3. Model the plate of Figure1.1using quarter node elements for the crack tip obtained with

the KSCON command.




FINISH

/CLEAR

/TITLE Stress singularities - KSCON

/PREP7

! Geometrical parameters in mm

L = 50 ! Plate length

a = 10 ! Crack length

b = 0 ! Crack heigth

th = 30 ! Thickness

el_len = 2 ! Element length

ET,1,PLANE82


MP,EX,1,210000 ! Young modulusMP,PRXY,1,0.3 ! Poissons ratio

KEYOPT,1,3,3 ! Keyoption to introduce thickness


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R,1,th ! Thickness

K,1,a,0

K,2,L,0

K,3,L,L

K,4,0,L

K,5,0,b

K,6,2*a,0

K,7,a,a

K,8,0,a

K,9,a+sqrt(2)*a/2,sqrt(2)*a/2

K,10,a,L

L,1,6,(2*a)/el_len

L,6,2,(L-2*a)/el_len

L,2,3,L/el_len

L,3,10,(L-a)/el_len

L,10,4,a/el_len

L,4,8,(L-a)/el_len

L,8,5,a/el_len

L,5,1,a/el_len

L,9,3,sqrt((L-a)**2+L*L)/el_len

L,1,7,a/el_len

L,7,8,a/el_len

LARC,7,9,1,a,6

LARC,9,6,1,a,6

L,7,10,(L-a)/el_len

KSCON,1,0.15,1,12,0.25

AL,2,3,9,13

AL,12,9,4,14

AL,1,13,12,10

AL,8,10,11,7

AL,11,14,5,6

AMESH,ALL

FINISH

/SOLU

DL,6,5,SYMM

DL,7,4,SYMM

DL,2,1,SYMM

DL,1,3,SYMM

SFL,4,PRES,-1

SFL,5,PRES,-1

SOLVE

FINISH

/POST1


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PLDISP,1

PLNSOL,S,Y

FINISH


ftp://amade.udg.edu/mme/FEmet/1_stress_sing_KSCON.dat

Figure1.7shows pathplots for y yas obtained with quarter mid-side node elements. Two of the plots

correspond to elements obtained with the KSCON command. The third one corresponds to the solution

given by the elements obtained in section1.3.1.

0 0.5 1 1.5 2 2.5 30

5

10

15

20

25

30

r0(mm)

yy

(MPa)

KSCON1

KSCON2modified mesh

Figure 1.7: y yat the crack tip region for the analyzed plate, as obtained with with Quarter mid-side node

elements: several mesh refinement with KSCON and by-hand mesh modification (45o path)

1.4 Summary and conclusions

When stress singularities are present in some kind of problem, the finite element analysis and specially

the mesh must be carefully considered.

If regular elements are employed, a very fine discretization is needed

Special crack tip elements may be used to estimate adequately the stress singularity with a lower

number of elements.

However the stress solution should still be used carefully, knowing that any solution with a finer mesh

at the crack tip will lead to a higher value of the maximum value of the stress (which analytical value

is)

Energy release rate (G) - based analysis or stress intensity factor (K) - based analysis are a much

better option for linear elastic fracture mechanics regime.


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1.5 Suggested problems

Problem 1.1. Let us assume we want to model the bi-metal shown in Figure 1.1. Consider that the

bi-metal thickness is 30 mm. Apply the displacement on the right side of the steel block. Material data:

Esteel = 210000 MPa, steel = 0.3, EAl= 70000 MPa, Al= 0.3.

Solve the problem for three different meshes and analyze the stress in the vertical direction.

Identify the zone which reaches higher stresses. What happens with the maximum value (absolute

value) of the stress when remeshing? Why?

Construct an adequate mesh using KSCON command and check that, at least for a range of some

element size at the crack tip, solution is not mesh dependent.

Figure 1.8: Aluminium-Steel Bi-metal for Suggested Problem1.1

1.6 Further reading

Henshell R.D and Shaw K.G. (1975) Crack tip elements are unnecessary. International Journal for Numerical

Methods in Engineering 9(3): 495-507.

Saouma V.E. and Schwemmer D. (1984) Numerical evaluation of the quarter-point crack tip element. In-

ternational Journal for Numerical Methods in Engineering20(9): 1629-1641.

Gray L.J., Phan AV. et al (2003) Improved quarter point crack tip element.Engineering Fracture Mechanics

70:269-283


20/55


21/55

Chapter 2

Computational Fracture Mechanics I:

Computation of G

2.1 Introduction

As seen in the former chapter, the traditional-materials strength stress analysis of a cracked component

may be hardly tackled. Although the stress discretisation may be improved by using crack tip elements, the

meaningful analysis is generally that performed using the energy release rate (G).

The energy release rate may be defined as the rate at which energy is dissipated () when a crack grows,

under constant boundary conditions:

G= A

constant B.C

(2.1)

As seen, the crack opening is measured in terms of the created area (A). This is possible because G is

a state function which means it only depends on the updated geometry and the geometry but not on how

they change in the fracture process. So no matter if the crack grows we can compute G using different

crack lengths.

In the following sections, differents ways for the numerical computation of G, according to its definitionof equation2.1will be summarized. We will apply these methods to the classical example of a cracked plate

shown in figure3.2.1, using an initial crack length a=10mm, L=50mm, unit pressure and a plate thickness

of 30 mm.

Although the crack length should be much more longer to satisfy the main assumptions involved, you

may compare the results we will obtain with the Classical Beam Theory approximation, in order to get a

rough idea of the magnitude:

GC BT=

P2a2

B E I =12P2a2

B2h3E

(2.2)

17


22/55


Figure 2.1: Plate with a side sharp crack.

GCBT = N/mm

Recall that equation 2.2 is an approximation and the analyzed geometry does not satisfy the basic

assumpttions of the Classic Beam Theory, so the result for sure includes large errors. You may use it only

to check the order of magnitude of the computed energy release rate in the following sections.

2.2 Finite Crack Extension Method (FCEM)

In Linear Elastic Fracture Mechanics (LFEM), in quasi-static conditions (which means the kinetic energy

involved in the process may be neglected), the elastic stored energy () equals the difference between the

strain energy (U) and the external work (W):

= UW (2.3)

where the strain energy U:

U=

1

2 i ji j (2.4)

According to the definition of G, it may be computed:

G= (a+a) (a)A

(2.5)

where, ifB is the thickness of the component, A= a B. By these assumption, Gmay be computedusing two different finite element models with two different crack lengths and the same boundary conditions:

Finite Element Model 1. BC and crack length a

Finite Element Model 2. Same BC but crack length a+a


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Chapter 2. Computational Fracture Mechanics I: Computation of G 19

ANSYS Command: Computation of Strain Energy

1. Using User GUI:

General Post-processor Element Table Define Table Add General Post-processor Element Table Sum of each item

2. Using ANSYS commands:

AVPRIN

ETABLE

SSUM

Example 2.1. Compute the energy release rate (G) for the side-cracked plate of Figure3.2.1by means of

the Finite Crack Extension Method. You may use the parametrized ANSYS model given below.




FINISH

/CLEAR

/TITLE Computation of G - Crack length: a

/PREP7 !PRE-PROCESSOR

L = 50 !Length of component (mm)

a = 10 ! Crack length (mm)

b = 0 ! Crack heigth (mm)

th = 30 !Thickness (mm)


ET,1,PLANE42KEYOPT,1,3,3 !Keyoption to activate thickness

R,1,th !Thickness assignment



K,1,a,0

K,2,L,0

K,3,L,L

K,4,0,L

K,5,0,b

L,1,2,(L-a)/el_len

L,2,3,L/el_len

L,3,4,L/el_len


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L,4,5,(L-b)/el_len

L,5,1,a/el_len

AL,1,2,3,4,5

LCCAT,5,1

ARSYM,Y,1

LCCAT,7,8

MSHKEY,2

AMESH,ALL

ALLSEL

NSEL,S,LOC,X,a,L ! All nodes at y=0, but those of the crack

NSEL,R,LOC,Y,0 ! are selected

CPINTF,ALL

FINISH

/SOLU !SOLUTION

DL,10,2,ALL,0

SFL,3,PRES,-1 !Pressure

nsel,s,loc,y,L

CP,1,UY,ALL

allsel

SOLVE

FINISH

/POST1 !GENERAL POST-PROCESSOR

PLDISP,1

PLNSOL,S,Y

FINISH


ftp://amade.udg.edu/mme/FEmet/2_comp_g.dat

You may fill in the following table to compute the energy release rate according to the Finite Crack

Extension Method:

Magnitude FE model 1 (a) FE model 2 (a+a)Strain energy

Displacement

Force

W

GFCEMI

= N/mm


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2.3 Crack Closure Method (CCM)

The Crack Closure Method assumes that the energy which is dissipated when a crack grows some a is the

same energy needed to open the crack some a. So if, as in the former method, two different finite element

models are employed, the forces and displacements needed to close the crack by some amay be computedas follows.

Finite Element Model 1 (Crack length = a). Used to obtain the force acting at the crack tip. Since

this force is actually a reaction, some rigid link orMulti Point Constraint (MPC)should be employed

at the crack tip to obtain this force value.

Finite Element Model 2 (Crack length = a+a). The longer crack length is achieved by removingthe rigid link at the former crack tip and keeping a rigid link at the new one. This model will be used

to read the displacements needed to close the crack by a.

The Energy Release Rate (G) may be computed:

G=

F(1)x (2)x + F(1)y (2)y 1

2 a B (2.6)

where the first term of the addition corresponds to GI Iand the second term to GI, that is:

GI=1

2

F(1)y (2)ya B (2.7)

GI I

=

1

2

F(1)x (2)x

a B(2.8)

and (2)x = u(2)x u(2)x , (2)y = u(2)y u(2)y and superscripts (1) and (2) denote the model where the variableis taken.

Figure 2.2: Closure Crack Method

ANSYS Command: Removing CPs

CPINTF

NUMMRG


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ANSYS Command: Reading displacement and force results

PRNLD

PRRSOL

You may fill in the following table to compute G by using the nodal values of force and displacement,

according to the CCM:

Magnitude FE model 1 (a) FE model 2 (a+a)

Displacement (top) -

Displacement (bottom) -

y -

Force (Fy) -

GCCMI

= N/mm

2.4 Virtual Crack Closure Technique (VCCT)

The Virtual Crack Closure Technique makes a further assumption: the crack grows in a self-similar manner.

This means that if we only look at the nearby of the crack tip, from one growth step to the next one, we would

see about the same crack shape -the same displacements- and about the same forces acting at the crack tip.

Consequently, instead of using two different models to get the forces at the crack tip and the displacement

needed to close the crack by some a, we may use the same model, so the computational efforts are reduced.

Figure 2.3: Virtual Crack Closure Technique

So, now G may be computed as:


27/55


G=

Fx x+ Fy y 1

2 a B (2.9)

where, again, the first term of the addition corresponds to GIand the second term to GI I, that is:

GI= 12

Fxxa B (2.10)

GI I=1

2

Fy ya B (2.11)

where and x= ux ux, y= uy uy.Finally, you may compute G through the formula derived by the Virtual Crack Closure Technique:

Magnitude FE model (a+a)Displacement (top)

Displacement (bottom)

y

Force (Fy)

GVCCTI

= N/mm

2.5 Suggested exercises

Problem 2.1. Model the CT specimen of Figure2.4using ANSYS with W=5mm, A=25 mm, B (thick-

ness)=10 mm, C=50 mm, D = 62.5 mm, E=5mm and F=60 mm.

Figure 2.4: CT specimen

With this model you are going to analyze G as a function of the crack length (a), using different methodsand for two different load cases: constant force (P=1000 N) and constant displacement (0.053mm). To

make this you have to use in the ANSYS model Constraint Equations in the line where the crack will grow.


28/55


The crack should grow between 2 mm and 16mm, so a element size of 2 mm may be a good choice. You

may use the provided fileCT.dat.

In each model you have to keep the following data:

External load or applied displacement

In the constant force loadcase, the displacement of the node where the force is being applied.

In the constant displacement loadcase, the reaction at the node where the displacement is being

applied.

Force at the crack tip, in the vertical direction.

Displacement of the nodes closer to the crack tip, in the vertical direction.

Strain energy

Compute:

1. For both loadcases, get the compliance curve for the specimen as a function of the crack length

(C= f(a)). Derive numerically the obtained curve and use the computed derivative to computeG= P22B

dCda

2. G(a) for both loadcase using the FCEM, CCM, VCCT methods. Plot in the same graph the obtained

curves together with the curve in the former question.

(The four methods should give similar results, for the following, use only the curve obtained with

FCEM.)

3. Assume that the material R-curve is given by:

R=

Gc

1 (1 a/cf)3

for a cfGc for a> cf

wherecf= 1.25 mm andGc= 0.3N/mm. Find what loadPproduces instability.

4. Assume you are performing a laboratory test using a CT speciment, trying to measure the R-curve of

a material. Which loadcase would you use? Why?

5. Analyze the effect of the element size on FCEM, CCM, VCCT methods. Obtain G for three different

meshes with different element sizes using the three methods.

FINISH

/CLEAR

/TITLE Computation of G - Crack length: a

/PREP7 !PRE-PROCESSOR

a0=8


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A= 25

W= 5

B=10

C=50

D=62.5

E=5

F=60


ET,1,PLANE42

KEYOPT,1,3,3 !Keyoption to activate thickness

R,1,B !Thickness assignment



K,1,0,0

K,2,C-A,0

K,3,C-A+E,w/2

K,4,D,w/2

K,5,D,F/3

K,6,D,F/2

K,7,C,F/2

K,8,0,F/2

K,9,C,F/3

L,1,2,(D-(D-C)-A)/el_len

L,2,3,sqrt(E*E+W*W/4)/el_len

L,3,4,(D-C+A-E)/el_len

L,4,5,(F/3-W/2)/el_len

L,5,6,(F/2-F/3)/el_len

L,6,7,(D-C)/el_len

L,7,8,C/el_len

L,8,1,(F/2)/el_len

L,5,9,(D-C)/el_len

L,9,7,(F/2-F/3)/el_len

AL,1,2,3,4,9,10,7,8

AL,9,5,6,10

LSYMM,Y,ALL

AL,11,12,13,14,19,20,17,18

AL,19,15,16,20

ACCAT,1,2

ACCAT,3,4

MSHKEY,2

AMESH,5

AMESH,6

ALLSEL

NSEL,S,LOC,X,0,C-A-a0 ! All nodes at y=0, but those of the crack


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NSEL,R,LOC,Y,0 ! are selected

CPINTF,ALL

FINISH

/SOLU !SOLUTION

KD,9,UX,0

KF,9,FY,1000 ! Comment for displacement loadcase

!KD,9,UY,10 ! Uncomment for displacement loadcase

KD,18,ALL

ALLSEL

SBCTRAN

SOLVE

FINISH

/POST1 !GENERAL POST-PROCESSOR

PLDISP,1

PLNSOL,S,Y

FINISH


ftp: //amade.udg. edu/ mme/FEmet/ CT.dat

2.6 Further reading

Krger R. (2002) The Virtual Crack Closure Technique: History, Approach and Applications. NASA/CR-

2002-211628. ICASE. Report No. 2002-10.


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Chapter 3

Computational Fracture Mechanics II:

Computation of K

3.1 Introduction

Although the use of the energy release rate is normally preferred in advanced analysis and in crack propaga-

tion simulation, the stress intensity factor K is widely used for design and verification of structures. While

G is a energy-based magnitude, Kis a stress related value and so, any computational method used to com-

pute it will have to deal somehow with the stress singularity and its related issues we introduced in Chapter 1.

As we saw in the first chapter, the stress discretisation in a finite element mesh may be improved byusing crack tip elements, to avoid the strong mesh dependence produced by the stress singularity.

In this chapter we will show how to use quarter mid-side node elements to discretize the stress field and

then, some method to compute K.

3.2 The stress intensity factor (K)

The plane stress field in the nearby of a crack tip of a crack loaded in mode I can be approximated by the

following expressions:

Ix =KI2r

cos

2

1 sin

2

sin

3

2

(3.1)

Iy =KI2r

cos

2

1 + sin

2

sin

3

2

(3.2)

Ix y =KI2r

cos

2

sin

2

cos

3

2

(3.3)

were superscriptIdenotes mode I and and rare the polar coordinates (angle and distance, respectively)in a polar coordinate system with center at the crack tip.

Analogously, the stress field ahead the crack tip in a Mode II situation is given by:

27


32/55


I Ix = KI I2r

sin

2

2 +cos

2

cos

3

2

(3.4)

I Iy =

KI I

2rsin

2cos

2cos

3

2 (3.5)

I Ix y =KI I2r

cos

2

1 sin

2

sin

3

2

(3.6)

The stress at the crack tip might be seen as the limit:

limr0

Ii j=KI2r

fIi j()=c

(3.7)

where cdenotes a constant. So if we are able to somehow know the stress field at the nearby of the

crack tip we are able to compute KI:

KI= limr0

Ii j

2rfI

i j()=c

(3.8)

since fIi j

() are known trigonometrical functions. Analogously for KI I:

KI I= limr0

I Ii j

2r

fI Ii j

()=c

(3.9)

3.2.1 Numerical estimation of the stresses at the crack tip

Let us recall the cracked plate of Figure that we analyzed in the former chapter, using the file 1_quarter_mid_nodes.Using the ANSYS commands of the next box, you may obtain the stresses at a given path (radius).

ANSYS Command: Path Plots

Command Comments

PATH,NAME,nPTS,nSETS,nDIV Defines geometrically a path by nPTS

PPATH,POINT,NODE,X,Y,Z,CS Defines one of the points of the path.

POINT is the ID of the point. NODE is

a node number if the point is located in a

NODE. X,Y,Z may be used to define the lo-

cation of the point

PDEF,LABEL,ITEM,COMP,AVGLAB Defines the ITEM (for instance STRESS)

and COMP (for instance X) to plot and give

it a LABEL

PLPATH,NAME Plots the path labelled with NAME


33/55

Chapter 3. Computational Fracture Mechanics II: Computation of K 29

Example 3.1. Obtain a plot foryfor the side-cracked plate, using a path plot.




FINISH

/POST1

! Path plot of stresses

PATH,0DEG,2,6,100

PPATH,1,,a,0,0

PPATH,2,,a+5,0,0

PDEF,SX,S,X,NOAV

PDEF,SY,S,Y,NOAVPLPATH,SY

PRPATH,SX,SY !Path results in a text file


ftp://amade.udg.edu/mme/FEmet/3_path.dat

A plot similar to that in Figure3.1 should be obtained.

Figure 3.1: yat the nearby of the crack for = 0o.


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3.2.2 Computation of K by stress extrapolation

Since we know how to obtain the stresses in the nearby of the crack tip we are able to obtain KI. If the

stresses when r 0would not tend to we could compute KIwith the stresses when r= 0, but since theydo tend to

, we have to somehow compute numerically the limit of Eq. 3.8. To do so we compute K for

each value of the stress i jusing equations as a function ofrand plot the pairs (K,r). Since the values of

the stress for small rare affected by the stress singularity we will neglect them and fit the linear variation

ofi j(r). The extrapolation for r= 0gives a good approximation ofKI, as shown in Figure 3.2

0 5 10 15 20 256

7

8

9

10

11

12

r (mm)

KI

(MPa

mm

1/2)

y = 0.183*x + 6.5

Figure 3.2: Computation ofKIby stress extrapolation (from y

3.2.3 Computation of K by displacement extrapolation

The former procedure is strongly affected by the stress singularity. In a finite element procedure, stresses

are obtained from the displacements and so may contain larger errors, specially in cases like this one where

large stress gradients are present. For this reason, a more precise option is to use the displacement solution

for the computation of the stress intensity factor, K. In this case for the region near the crack tip, the

relations between the displacement field and KI and KI I are:

KI

cos2 ( cos)sin2 ( cos)

= 2G

2

r

uI

vI

(3.10)

KI I

sin 2

(2 ++ cos)cos2 (2 cos)

= 2G

2

r

uI I

vI I

(3.11)

whereG is the shear modulus:

G

=

E

2(1 +)(3.12)

and is a parameter which allows the simultaneous consideration of plane stress and plane strain cases,

with:


35/55


= 3 1 + for plane stress (3.13)

= 3 4 for plane strain (3.14)

Since in a finite element solution, the displacement field is generally a better solution than the stress

field, the value ofKIobtained in this manner (see Figure3.4) should provide a better approximation.

0 0.5 1 1.5 2 2.5 3 3.5 46

6.5

7

7.5

8

8.5

9

r (mm)

KI

(MPa

mm

1/2)

y = 0.449*x + 6.78

Figure 3.3: Displacement extrapolation technique for the computation ofKI (using uy).

You may compare the numerical result with the analytical one, which may be obtained using the hand-

book formula of Figure3.4.

Figure 3.4: Handbook expression for the analyzed case.

3.2.4 Remarks

The reviewed techniques for the computation of K are first approximations. Further developments

exist and are still object of current research.

Both stress and displacement extrapolation need of fine meshes to converge to the correct value of

K. Crack tip elements are strongly recommended for the stress extrapolation method.


36/55


3.3 Displacement extrapolation with quarter node elements

When the stress singularity is very well discretized, which practically means when quarter-point isoparametric

elements are used, some simple formulae can be applied with surprisingly accurate results. This formuale

are obtained making = in expressions3.10and3.11, since for this angle the error is minimum.

If you recall the quarter node elements shown in Section 1.2, the approximation of the displacement

along edge 1-3 is given by:

u= u1 + [4u2 u3 3u1]

r

L+ [2u3 +2u1 4u2]

r

L(3.15)

and the same expression is valid for the vertical displacement v.

If we substitute this approximation of the displacement field in expressions 3.10and 3.11we obtain:

KI

cos2 ( cos)sin2 ( cos)

= 4G

2

L

4u2 u3 3u14v2 v3 3v1

(3.16)

and for mode II:

KI I

sin2

(2 ++ cos)cos2 (2 cos)

= 4G

2

r

4u2 u3 3u14v2 v3 3v1

(3.17)

So we can compute KI and KI Isubstituting any value of angle and using the displacements at nodes

1,2 and 3. If we particularize the former expressions for =and since v

1 =0:

KI=2G

+ 1

2

L(4v2 v3) (3.18)

KI I=2G

+ 1

2

L(4u2 u3) (3.19)

If we denote node 2 with an A and node 3 with a B and make L= the former expressions may be

written in the form given by Guinea et al. (2000):

KI= E4

2

(4vA vB) (3.20)

where E is the effective elastic modulus defined as equal to E for plane stress and E/(1 2) for planestrain. vAis the vertical displacement of the quarter mid-side node and vBthe vertical displacement of the

outer vertex node (See Figure3.5) .

3.3.1 Formulae for the stress intensity factor

With some similar approach the following expressions may also be obtained to compute KI (Guinea et al,

2000):

KI=E

2

2

vA (3.21)


37/55


Figure 3.5: Quarter-point singular elements and coordinates for near crack-tip field description. Source:

Guinea et al, 2000

KI=E

12

2

(8vA vB) (3.22)

These methods provide objective ways of computing K. Similar expressions can be obtained for KI I,

using3.11. Again in mixed mode situations the superposition principle may be applied

3.4 ANSYS commands for the computation of K

3.4.1 Crack opening displacement

ANSYS offers a built in method for the computation of the Stress intensity factor (K). Although this method

is related to displacement extrapolation it is actually based on the concept of Crack Opening Displacement

(COD) and uses the formula obtained by Paris and Sih which describes the crack opening near the crack

tip for linear elastic-plastic materials:

Vr

= KI2G

1 +2

(3.23)

This expression can be easily obtained from3.10for =.Since the crack opening displacement V can be obtained from the displacement solution at the nodes

which define the crack face, the parameters A and B can be obtained by a simple linear fit.

V

r=A+ Br (3.24)

Then, since:

limr

0

V

r= A (3.25)

K can be computed from3.23:

KI=

22G A

1 + (3.26)


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3.4.2 KCALC command

The main steps needed to perform the computation of K in a two-dimensional model are:

1. Define a path with three nodes. Where NODE1 must be the crack tip and NODE2 and NODE3 two

nodes in the same crack face. If quadratic elements are used, a choice which gives good results is to

use the three nodes of the crack tip element.

2. Define a cartesian local coordinate system with origin at the crack tip.

3. Execute the KCALC command

ANSYS Command: KCALC, KPLAN, MAT, KCSYM, KLOCPR

KPLAN Key to convert plane stress results into plane strain stress intensity factors:

0 - Plane strain and axisymmetric cases (default)

1 - Plane stress

MAT Material number used in the extrapolation (defaults to 1).

KCSYM Symmetry key:

0 or 1 - Half-crack model with symmetry boundary conditions in the crack-tip coordinate system.

KII = KIII = 0. Three nodes are required on the path.

2 - Like 1 except with antisymmetric boundary conditions (KI = 0).

3 - Full-crack model (both faces). Five nodes are required on the path (one at the tip and two on

each face).

KLOCPR Local displacements print key:

0 - Do not print local crack-tip displacements.

1 - Print local displacements used in the extrapolation technique.

Example 3.2. Compute the Stress Intensity Factor for the plate of Figure 3.2.1 using ANSYS KCALC

command.





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SOLVE

FINISH

CS,12,0,1,4,124 !Define local coordinate system at crack tip

CSYS,12 ! Activate local coordinate system

PATH,K1,3,10,50 ! Define 3-node path

PPATH,1,1

PPATH,2,1013

PPATH,3,242

KCALC,0,1,0,1 !Execute KCALC


ftp://amade.udg.edu/mme/FEmet/3_kcalc.dat

3.5 Proposed exercises

Problem 3.1. Numerical validation of Irwins hypothesis

Irwins hypothesis may be used when plastic strains appear in the region near the crack tip. It is based on

defining an equivalent case in the elastic regime, with an equivalent crack length. Let us keep working with

the model1_stress_sing_KSCON.datwe used in Chapter 1.

1. Introduce a perfect plasticity model as the material model. You can do this by adding the following

lines after the material properties definition:

TB,BKIN,1,1

TBDATA,1,270,0

where the value 270 MPa is the yield stress and0 the hardening modulus.

2. Now increase the applied stress to a value that ensures that plastic strains appear near the crack tip

(representative results are obtained for about 40 MPa).

3. Obtain a curve ofSyfor r between 0 and 0.3 mm, approximately.

4. Compute the equivalent crack length according to the Irwins hypothesis. In the former plot you can

obtain the crack length forSy=270 MPa (the yield strength). Compare both values of the equivalent

crack length.

5. Using the analytical expression of the stress field in the nearby of a singularity, plot theSy curve for

the equivalent crack length of the former point. Compare this curve with the one of the question 3

6. Observe the results and comment on about the validity of Irwins hypothesis.


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Problem 3.2. Consider again the side-cracked plate of Figure3.2.1. Compute the mode I stress intensity

factor using equations3.21, 3.22, 3.20. Compare the results with those obtained with the other methods

seen in this chapter. Comment on the results.

Problem 3.3. Superposition principle

Proof the principle of superposition can be used as schematized in Figure3.6.

Figure 3.6: Proposed exercise # 2.

Consider a cracked plate submitted to an stress(A). Consider the same plate with the same stress

but also closing stresses which make the crack remain closed (B). Consider the plate submitted only to the

closing stresses but in the oposite direction (C).

1. Considering that the superposition principle is applicable for a single opening mode, discuss how could

you computeK(C)I

.

2. Proof thatK(A)I

= K(B)I

+ K(C)I

3.6 Further reading

Guinea G.V, Planas J. and Elices M. (2000) KI evaluation by the displacement extrapolation technique.

Engineering Fracture Mechanics 66:243-255.

Tada H., Paris P.C., and Irwin G.R. (2000) The Stress Analysis of Cracks Handbook. ASME Press. 3rd

Edition


41/55

Chapter 4

Computational Fracture Mechanics III:

Computation of the J-integral

4.1 Introduction

To complete the review of computational analysis for Linear Elastic Fracture Mechanics we will summarize

the concept of J-integral and we will use ANSYS to compute it.

Let us consider a line integral going around the crack tip and starting in one side of the crack and ending

at the other side of the crack, as shown in figure 4.1.

Figure 4.1: J integral.

It can be shown that the following integral is independent of the path for any curve which satisfies the

former conditions:

J

= Udy

t

u

x ds (4.1)

were U is the strain energy density (U= 12 : ), tis the traction vector defined by the external normaln, u is the displacement field and ds is an infinitesimal in the direction of the curve.

37


42/55


The integral is actually an equilibrium, for any path not including the crack, that is starting and ending

at the same point, J=0, so if the curve starts at one side of the crack and ends at the other side, its value

equals the energy inverted on the crack. The J-integral is also useful in non-linear fracture mechanics but,

since in LEFM its value equals the energy release rate G,

4.2 The J integral with ANSYS

As usual, ANSYS help describes properly the procedure to compute the J-integral. Here we summarize this

procedure for bidimensional cracks:

1. Start the new computation of the J-integral with: CINT,NEW,ID where ID is an integer identifying

the path, for instance 1.

2. Define the node at the crack tip and the crack plane normal with:

CINT,CTNC,CMNAME

where CMNAME is the name of a node component1

CINT,NORMAL,par1,par2

where par1 is a coordinate system identifier and par2 is an axis of the coordinate system

3. Specify the number of contours n to compute with the command:

CINT,NCONTOUR,n

4. Activate the option for symmetry conditions, if present:

CINT,SYMM,ON

5. Specify the output controls:

OUTRES,ALL

or

OUTRES, CINT

6. Finally, the results for the value of the J integral may be listed or plotted:

PRCINT,ID

PLCINT,PATH,ID

where ID is the crack identifier.

Example 4.1. Compute the J-integral for the cracked plate of Figure3.2.1, by using the ANSYS built-inmethod.

1The command CM,CMNAME,NODE stores the selected nodes under a node component of name CMNAME.


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Chapter 4. Computational Fracture Mechanics III: Computation of the J-integral 39

Solution to Example 4.1. The following commands may be used in any of the parametrized models we

used before, with the crack tip located at (a,0) to define the J-integral computation.

The ANSYST M command sequence for this example is listed below. You can either type these commands on the

command window, or you can type them on a file, then, on the command window enter /input, file, ext or just use

copy and paste.

FINISH

\PREP7

CSYS,1

NSEL,S,LOC,X,a !Select the crack tip node, located at (a,0)

NSEL,R,LOC,Y,0

CM,CRACK,NODE

NSEL,ALL

CINT,NEW,1

CINT,CTNC,CRACK

CINT,NORMAL,0,2

CINT,NCONTOUR,20

CINT,SYMM,ON

OUTRES,CINT


ftp://amade.udg.edu/mme/FEmet/4_j_int.dat

After solving the model and in the \POST1 module, results ofr the J integral may be obtained with

the commands PRCINT,1 or PLCINT,PATH,1. It is important to set a sufficient number of contours in thecommand CINT,NCONTOUR,n, so the integral converges to a value.


Problem 4.1. Compute the J-integral for the model1_stress_sing_KSCON.dat we used in Chapter 1.

Compare the value of J, with that of G and K, obtained in the corresponding examples.

4.4 Further reading Rigby R.H. and Aliabadi M.H. Decomposition of the mixed-mode J-integral - Revisited. International

Journal of Solids and Structures 35(1):2073-2099, 1998.


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Chapter 5

Computational Fracture Mechanics IV:

Cohesive zone modeling

5.1 Introduction

Whilst Linear Elastic Fracture Mechanics assumes the presence of a crack in a perfectly elastic brittle or

quasi-brittle material this is an idealization. Generally, in the nearby of the crack tip there exists a zone

where the material is damaged due the presence of microcracks. When the number of microcracks grow a

larger crack is formed and crack growth takes place. This region of the material is called Failure Process

Zone or, in the case of crack growth modeling, cohesive zone.

Some modeling techniques treat the material in this more realistic manner: before crack growth the

region at the crack tip follows a failure process.

This chapter summarizes the different possibilities included in ANSYS for the cohesive zone modeling.

5.2 Cohesive laws

Cohesive laws describe mathematically the separation or debonding of two material surfaces. They are usually

presented as - curves. is the stress acting to separate the surfaces and the relative displacement

between them.

The different cohesive laws have some similarities:

Some positive slope region in which, when an increase in implies an increase in .

Some inflexion point m. Once this point is reached, the cohesive material starts the failure/damage

process.

Some negative slope region. Since the material is damaged the stress to achieve larger decreases.

Some m for which = 0, which means the total damage of the material

41


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The behaviour of the cohesive material is sketched in Figure5.1. If the material is loaded with < m,the unload follows the same path since the material is not damaged. On the other hand, if the material is

loaded producing some > m, the material starts to damage and then the unload follows the secant.

Figure 5.1: Cohesive law

This material behaviour can be modeled with different laws. Usually the linear (sometimes called

bilinear), linear-parabolic, exponential and trapezoidal are included in the commercial FE software. They

are sketched in Figure5.2. ANSYS includes only the bilinear and exponential laws.

Figure 5.2: Usual cohesive laws

5.2.1 Bilinear law

This is the cohesive law used for the contact elements so the names of the variables are slightly different

P normal contact stress (tension). Equals

Kn: normal contact stiffness

un: contact gap. Equals


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Chapter 5. Computational Fracture Mechanics IV: Cohesive zone modeling 43

Figure 5.3: Bilinear Cohesive law

un: contact gap at the maximum normal contact stress (tension)

ucn: contact gap at the completion of debonding

dn: debonding (damage) parameter. dn= 0for the virgin material and dn = 1for the totally damagedmaterial

For mode II or mixed mode, additional parameters are required.

5.2.2 Exponential law

This law is the only one available for interface elements.

= expma xnexpn exp2t (5.1)

with:

ma x: stress for which crack opening starts

n: maximum normal displacement

t: maximum tangential displacement

Parameters must be given so:

()d= Gc (5.2)

5.3 Cohesive elements in ANSYS

ANSYS offers two different possibilities for the cohesive zone modelling. A straight forward manner is theuse of interface elements. A second approach, is the use of ANSYS contact elements together with a

cohesive law.


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5.3.1 Cohesive zone modeling with interface elements

Element type

Cohesive elements are referred asinterface elementsin the literature because of their topology. That is, the

element is located in the interface between two solid structural elements to simulate the debonding process

between them.The different interface elements available in ANSYS are shown in the next Table:

Element Characteristics Interface Element Structural Elements

2D, linear INTER202 PLANE42, VISCO106, PLANE182

2D, quadratic INTER203 PLANE2, PLANE82, VISCO88,

PLANE183

3D, quadratic INTER204 SOLID92, SOLID95, SOLID186,

SOLID187

3D, linear INTER205 SOLID45, SOLID46, SOLID64,

SOLID65, SOLID185, SOLIDSH190

Material definition

As mentioned before, when using interface elements, the only material model which can be used is the

exponential. It needs of three parameters:

ma x: maximum stress

n: normal displacement at maximum stress

t: tangential displacement at maximum stress

ANSYS Command: Material definition for interface elements

TB,CZM,MAT,NTEMP,NPTS,EXPO

TBDATA,1,SMAX,DN,DT

where SMAX is ma x, DN is nand DT is t.

Example 5.1. DCB test modeling with interface elements

Since the DCB specimen is controlled to always be in crack-opening situation, interface elements may be

successfully employed in the modeling of this test.

Solution to Example 5.1. The following file reproduces an Example of ANSYS verification manual whichaim is to test its cohesive modeling with a DCB test.


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The ANSYST M command sequence for this example is listed below. You can either type these commands on the

command window, or you can type them on a file, then, on the command window enter /input, file, ext or just use

copy and paste.

FINISH

/CLEAR

/COM,ANSYS MEDIA REL. 11.0 (10/27/2006) REF. VERIF. MANUAL: REL. 11.0

/TITLE, VM248, DELAMINATION OF DOUBLE CANTILEVER BEAM - 2D PLANE STRAIN

/COM, REF: ALFANO, G. AND CRISFIELD, M. A.,

/COM, "FINITE ELEMENT INTERFACE MODELS FOR THE DELAMINATION ANALYSIS

/COM, OF LAMINATED COMPOSITES: MECHANICAL AND COMPUTATIONAL ISSUES"

/COM, INT. J. NUMER. METH. ENGNG 2001, 50:1701-1736.

/PREP7

ET,1,182 !* 2D 4-NODE STRUCTURAL SOLID ELEMENT

KEYOPT,1,1,2 !* ENHANCE STRAIN FORMULATION

KEYOPT,1,3,2 !* PLANE STRAIN

ET,2,182

KEYOPT,2,1,2

KEYOPT,2,3,2

ET,3,202 !* 2D 4-NODE COHESIVE ZONE ELEMENT

KEYOPT,3,3,2 !* PLANE STRAIN

MP,EX,4,1.353E5 !* E11 = 135.3 GPA

MP,EY,4,9.0E3 !* E22 = 9.0 GPA

MP,EZ,4,9.0E3 !* E33 = 9.0 GPA

MP,GXY,4,5.2E3 !* G12 = 5.2 GPA

!MP,GYZ,4,5.2E3

!MP,GXZ,4,3.08E3

MP,PRXY,4,0.24

MP,PRXZ,4,0.24

MP,PRYZ,4,0.46

GMAX = 0.004

TNMAX = 25 !* TENSILE STRENGTH

TB,CZM,5,,,EX PO !* COHESIVE ZO NE MATERIAL

TBDATA,1,TNMAX,GMAX,1000.0

RECTNG,0,100,0,1.5 !* DEFINE AREAS

RECTNG,0,100,0,-1.5

LSEL,S,LINE,,2,8,2 !* DEFINE LINE DIVISION

LESIZE,ALL,0.75

LSEL,INVE

LESIZE,ALL, , ,200

ALLSEL,ALLTYPE,1 !* MESH AREA 2

MAT,4

LOCAL,11,0,0,0,0

ESYS,11

AMESH,2

CSYS,0

TYPE,2 !* MESH AREA 1

ESYS,11

AMESH,1

CSYS,0

NSEL,S,LOC,X,30,100

NUMMRG,NODESESLN

TYPE,3

MAT,5


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CZMESH,,,1,Y,0, !* GENERATE INTERFACE ELEMENTS

ALLSEL,ALL

NSEL,S,LOC,X, 100 !* APPLY CONST RAINTS

D,ALL,ALL

NSEL,ALL

FINISH

/SOLU

ESEL,S,TYPE,,2

NSLE,S

NSEL,R,LOC,X

NSEL,R,LOC,Y,1.5 !* APPLY DISPLACEMENT LOADING ON TOP

D,ALL,UY,10

NSEL,ALL

ESEL,ALL

ESEL,S,TYPE,,1

NSLE,S

NSEL,R,LOC,X

NSEL,R,LOC,Y,-1.5 !* APPLY DISPLACEMENT LOADING ON BOTTOM

D,ALL,UY,-10

NSEL,ALL

ESEL,ALL

NLGEOM,ON

AUTOTS,ON

TIME,1

NSUBST,40,40,40

OUTRES,ALL,ALL

SOLVE !* PERFORM SOLUTION

FINISH

/POST26

NSEL,S,LOC,Y,1.5

NSEL,R,LOC,X,0

*GET,NTOP,NODE,0,NUM,MAX

NSEL,ALL

NSOL,2,NTOP,U,Y,UY

RFORCE,3,NTOP,F,Y,FY

PROD,4,3, , ,RF, , ,20

/TITLE,VM248, DCB: REACTION AT TOP NODE VERSES PRESCRIBED DISPLACEMENT

/AXLAB,X,DISP U (mm)

/AXLAB,Y,REACTION FORCE R (N)

/YRANGE,0,60

XVAR,2

PLVAR,4

PRVAR,UY,RF

*GET,TMAX,VARI,4,EXTREM,TMAX !* TIME CORRESPONDING TO MAX RFORCE

FINISH

/POST1

SET, , , , ,TMAX !* RETRIEVE RESULTS AT TMAX

NSEL,S,NODE, ,NTOP !* SELECT NODE NTOP

*GET,RF_NTOP,NODE,NTOP,RF,FY !* FY RFORCE AT NODE NTOP

*GET,UY_NTOP,NODE,NTOP,U,Y !* DISP AT NODE NTOP CORRESPONDING TO RFORCE

RF_MAX = RF_NT OP*20 !* PLANE S TRAIN OPTION AND WIDTH = 20 mm

SET,LAST !* RETRIEVE RESULTS AT LAST SUBSTEP

*GET,RF_END,NODE,NTOP,RF,FY !* FY RFORCE AT NODE NTOP AT LAST SUBSTEP

*GET,UY_END,NODE,NTOP,U,Y !* DISP AT NODE NTOP CORRESPONDING TO RFORCE

RF_END = RF_END*20 !* PLANE STRAIN OPTION AND WIDTH = 20 mm

*DIM,LABEL,CHAR,2,2

*DIM,VALUE,,2,3


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*DIM,VALUE2,,2,3

LABEL(1,1) = RFORCE,DISP

LABEL(1,2) = FY (N),UY (mm)

*VFILL,VALUE(1,1),DATA,60.0,1.0

*VFILL,VALUE(1,2),DATA,RF_MAX,UY_NTOP

*VFILL,VALUE(1,3),DATA,ABS(RF_MAX/60.0),ABS(UY_NTOP/1.0)

*VFILL,VALUE2(1,1),DATA,24,10.0

*VFILL,VALUE2(1,2),DATA,RF_END,UY_END

*VFILL,VALUE2(1,3),DATA,ABS(RF_END/24.0),ABS(UY_END/10.0)

/COM

/OUT,vm248,vrt

/COM,------------------- VM248 RESULTS COMPARISON --------------

/COM,

/COM, | TARGET | ANSYS | RATIO

/COM,

/COM,MAX RFORCE AND CORRESPONDING DISP USING INTER202:

/COM,

*VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3)

(1X,A8,A8, ,F10.3, ,1F10.3, ,1F5.3)

/COM,

/COM,RFORCE CORRESPONDING TO DISP U = 10.0 USING INTER202:

/COM,

*VWRITE,LABEL(1,1),LABEL(1,2),VALUE2(1,1),VALUE2(1,2),VALUE2(1,3)

(1X,A8,A8, ,F10.3, ,1F10.3, ,1F5.3)

/COM,-----------------------------------------------------------

/OUT

FINISH

*LIST,vm248,vrt


ftp://amade.udg.edu/mme/FEmet/5_VM248.dat

Figure5.4 shows the force-displacement curve for the DCB specimen, as obtained from the model using

cohesive elements.

Figure 5.4: Force-displacement curve


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Figure 5.5: Force-displacement curves for different parameters of the cohesive law

5.3.2 Cohesive zone modeling with contact elements

Element type

On the other hand, for complex boundary conditions which may not always tend to open the crack, contact

elements may also be employed. In this case

Element Formulation Usage Target element

CONTA171 linear 2-D 2-Node Surface-to-Surface TARGE169

CONTA172 quadratic 2-D 3-Node Surface-to-Surface TARGE169

CONTA173 linear 3-D 4-Node Surface-to-Surface TARGE170

CONTA174 quadratic 3-D 8-Node Surface-to-Surface TARGE170

Material definition

When using contact elements the only material model which can be employed is the bilinear. This can be

defined in ANSYS in two different ways: by maximum traction and maximum separation (CBDD) or by

maximum traction and critical energy release rate (CBDE).

ANSYS Command: Material definition for cohesive zone modeling through contact elements

TB,CZM,MAT,NTEMP,NPTS,CBDX(changing X by D or E)

TBDATA,1,C1,C2,C3,C4

Example: DCB test modeling with contact elements

The DCB test may also be modelled with contact elements. This option requires higher computational time.


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finish

/clear

/prep7

et,1,182 ! solid 4 node element

keyopt,1,3,2 ! plane strain

et,2,182

keyopt,2,3,2

et,3,169 ! target 2d element

et,4,171 ! 2d contact element

keyopt,4,12,5 ! bonded: cohesi ve law must be defined

et,5, 169 ! target 2d element

et, 6, 171 ! 2d contact element

keyopt,6,4,2 ! Nodal point contact

keyopt,6,2,4 ! Lagrange multiplier method

MP,EX,1,1.353E5 !* E11 = 135.3 GPa

MP,EY,1,9.0E3 !* E22 = 9.0 GPa

MP,EZ,1,9.0E3 !* E33 = 9.0 GPa

MP,GXY,1,5.2E3 !* G12 = 5.2 GPa

MP,GYZ,1,5.2E3

MP,GXZ,1,3.08E3

MP,PRXY,1,0.24

MP,PRXZ,1,0.24

MP,PRYZ,1,0.46

kopen = 1.e6 ! Stiffness contact

smax=25 ! Definition of Cohesive law

gic=0.26

tb,czm,2,1,1,cbde

tbdata,1,smax,gic,smax,gic,1.e-5,1e15

!Geometry

a=35

length=100h=1.5

l=length/2

rectng,0,length,0,h

rectng,0,length,0,-h

e_size=0.5 ! Element size

esize,e_size

type,1

mat,1

local,11,0,0,0,0

esys,11

amesh,2csys,0

type,2

esys,11


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amesh,1

! Contact between specimen two arms

real,3 ! real set of contact between arms

r,3

rmodif,3,3,-kopen ! Normal contact stiffness

rmodif,3,12,-kopen ! Tangential contact stiffness

asel,s,area,,1

nsla,s,1

nsel,r,loc,y,0

type,3

mat,2

esurf

asel,s,area,,2

nsla,s,1

nsel,r,loc,y,0

nsel,r,loc, x, 0, a+e_size/2

real,3

mat,3

type,6

esurf

allsel,all

nsel,all

finish

/solu

dk,6,all

dk,3,uy,30

nsel,all

esel,all

eqslv,front

neqit,200

nropt, unsymm

nlgeom,on

autots,on

time,1

deltime,0.0005,0.000005,0.1

outres,all,all

solve

finish


ftp://amade.udg.edu/mme/FEmet/5_DCB_comp.dat


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5.4 Some remarks on element size

The use of cohesive elements, in any of the available forms, implies that the cohesive zone (failure process

zone) must be meshed with a sufficient number of elements. A rule of the thumb is to use at least three

cohesive elements for the failure process zone. Some estimation for the length of the FPZ should be usedto determine a critical element size. For instance, the length of the fracture process zone for delamination

in a unidirectional test specimen loaded in mode I can be estimated as:

fpz =9

32

E3GI c

(o3)2

(5.3)

whereE3and o3 are respectively the Young modulus and the strength for the direction 3 of the composite

and GI c is the critical energy release rate for mode I. Under mixed-mode loading, the length of the failure

process zone is larger than the obtained using the former expression, so the given estimation is conservative.

For typical CFRP the latter expression leads to element size between 0.1 and 0.5 mm. Obviously this isunsuitable for large structures. Then, some engineering methods may be applied which allow the use of

larger element sizes (Turon et al, 2007).


Perform a mesh-size dependence analysis for the simulation of the DCB test, using interface elements. You

may use the Example given in section5.3.1. Compare the results with equation5.3.

5.6 Further reading

Mi, Y., Crisfield, M. A., Davies, G. A. O. and Hellweg, H. (1998) Progressive delamination using interface

elements. Journal of Composite Materials 32(14):1246 1272.

Alfano, G. and Crisfield, M. A. (2001) Finite element interface models for the delamination analysis of

laminated composites: mechanical and computational issues. International Journal for Numerical Methods

in Engineering 50(7):1701 1736.

Turon A., Dvila C.G., Camanho P.P., Costa J. (2007) An engineering solution for mesh size effects in the

simulation of delamination using cohesive zone models. Engineering Fracture Mechanics74(10):16651682.

5.7 Aknowledgements

Documents

Ansys Fracture Tutorial