Computational Fracture Mechanics - Uni Kiel

Computational Fracture Mechanics Students Project 2015 Final report

Wolfgang Brocks Silvio Rabbolini

May 2015

Dipartimento di Meccanica Politecnico di Milano report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 1 -

Part 1: Geometry:

Material

Determination of a true stress-strain curve as ABAQUS input from tensile test data flat tensile bar (rectangular cross section) width W0 = 8.01 mm, thickness B0 = 2.9 mm measuring length L0 = 30 mm Al 5083 Young’s modulus E = 70300 MPa Poisson’s ratio ν = 0.3 0.2% proof stress Rp0.2 = RF(εp=0.002) = 242 MPa

F [kN]

10

Al 5083 8

6

4

test 2.1.21

2

test 2.1.16

0 0

1

2

3

4

5

∆L [mm]

SCHEIDER, I.; SCHÖDEL, M.; BROCKS, W.; SCHÖNFELD, W.: Engng. Fract. Mech. 73 (2006), 252–263 (1)

Evaluate the test data provided in “tensile-test_Al5083.xls” Note that

•

ABAQUS requires true (equivalent) stresses in dependence on logarithmic plastic strains as flow curve input, , for an elastoplastic analysis,

•

A curve beyond uniform elongation of the tensile bar is needed for the fracture mechanics analysis, which necessitates an extrapolation of the stress-strain curve up to ≈ 3 by a power law = where =

•

, are normalisation values, commonly taken as yield limit = = 0 and = ⁄ .

Number of measured data has to be significantly reduced: no more than 30 points for the curve.

report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 2 -

(2)

Check the stress-strain curve by numerical simulation of the tensile test beyond uniform elongation (maximum force) and compare simulation with test data in terms of force, F, vs. elongation, ∆L . • • • •

Analysis of tensile test

(3)

plane stress and 3D model account for symmetries, displacement controlled, geometrically nonlinear (“large deformations”).

Derive a correction factor, , for the evaluation of the equivalent stress, , from measured force, F, and specimen width, W, beyond the onset of necking according to the procedure of SCHEIDER et al. [2004]. see below, and plot it in dependence on .

Stress and strain measures 1. Homogeneous uniaxial stress state (uniform elongation) =

Linear strain Logarithmic (“true”) strain

ε

Nominal (“engineering”) stress

σ

"

∆

≤

= ln

*

%

%

!

(εu = uniform strain)

= ln&1 + ε )

=+ =,

≤ R* =

-

/01

+

(Rm = tensile strength) σ2

“True” (Cauchy) stress

!3

4

4

= 5 = 67

Assumption of isochoric deformation 8 9 = 89 , σ2 VON MISES equivalent

stress =

Yield condition

!3

4

= σ2

!3

ε =

,

4

=5=5 "

=

3

2 2: "

−

2 2:

=

=

+

* &1

+

≈

)

σ? @

is the (uniaxial) “flow curve” required as input for ABAQUS 2. Triaxial principal stress state (beyond uniform elongation = onset of necking) VON MISES equivalent

stress

B

= AC D&

B

−

C)

C

+&

C

−

E)

C

+&

E

−

B)

CF

SCHEIDER, I.; BROCKS, W.; CORNEC, A.: “Procedure for the determination of true stress–strain curves from tensile tests with rectangular cross section specimens”, J. Eng. Mater. Techn. 126 (2004) nominal area equivalent stress

7 8G = H C 6 4

= 5I

eq. (17) eq. (18)


1.1. Evaluation of test data

250

σnom [MPa]

σnom [MPa]

In a first step, the force elongation curves, F(∆L), measured in the tests are converted into engineering stress-strain curves, σ * &ε ). A magnification of the initial part σ ≤ 240 MPa exhibits measuring offsets of the test data, see Fig. 1.1, which result in wrong plastic strains, = − ⁄ . After offset corrections of ∆ε = +0.0002 for test 2.1.21 and ∆ε = -0.000475 for test 2.1.16 the elastic parts of the test data coincide,1 see Fig. 1.2 .

200

250

200

150

150

100

100 test 2.1.21 test 2.1.16

test 2.1.21

50

50

E*eps

test 2.1.16 E*eps

0

0 0

0,002

0,004

0

0,006 εlin [-]

Fig. 1.1: Engineering stress-strain curves for σ ≤ 240 MPa showing measuring offseta of the test data

0,006

Fig. 1.2: Offset corrected test data

= − ⁄ , Fig. 1.3, provides

• yield strength, = = 0 = 200 MPa , and * • 0.2% proof stress (or “offset strength”) .C = *

2

0,004

εlin [-]

Plotting (nominal) stresses2 vs. (linear) plastic strain,

1

0,002

= 0.002 = 242 MPa

Except the very first strain values below σ = 50 MPa of test 2.1.21 The difference between nominal and true stresses is small, σtrue/σnom ≤ 1.0055%, for εp ≤ 0.002. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 4 -

σnom [MPa]

250

225

200 test 2.1.21 test 2.1.16 ABAQUS

175

150

Fig. 1.3: Nominal stresses vs. plastic strains: determination of yield strength3

0

0,0005

0,001

0,0015 p

ε

0,002 lin

[-]

True stresses result from 2 !3

4

4

=5=5

=

* &1

+

)

(1.1)

under the assumption of isochoric deformation since no necking occurred in the tests. The true stress-strain curves are shown in Fig. 1.4. 500 test 2.1.16

stress [MPa]

stress [MPa]

500

400

300

200

test 2.1.21 400

300

200

100

100

nom stress

nom stress

true stress

true stress

0 0

0,05

0,1

0,15

strain [-]

0 0

0,05

0,1

0,15

strain [-]

Fig. 1.4: Conversion of engineering stress-strain curves to true stress-strain curves (a) test 2.1.16 (b) test 2.1.21 Extrapolation beyond the test range, 0.12 ≤ εp ≤ 3.0 is required for the later analyses of a fracture mechanics specimen, as respective plastic deformations will occur at the crack tip. This is commonly done by a power law, P

P

3

=

RS

QR T ,

(1.2)

The square dots denote the input data for ABAQUS in the range R0 ≤ σ ≤ Rp0.2, see Fig. 1.7 report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 5 -

which at the same time provides a fit curve of the scattering stress data and allows for data reduction. The stress-strain curves obviously do not allow for a unique power-law fit over the whole range of εp. Therefore, only the final part is used to define the power law.4 Normalisation is recommended in order to obtain a dimensionless equation so that α does not depend on the specific dimension used for the stresses. The normalising values can be arbitrarily selected, commonly taken as the yield-limit values, =

= 200 MPa and

2,1

2,1 y=

σtrue/σ0 [-]

σtrue/σ0 [-]

⁄ = 0.00284.

=

1,0274x 0,1755

2 1,9

y = 1,03x 0,1769

2 1,9

1,8

1,8

1,7

1,7 test 2.1.21

test 2.1.16

1,6

1,6

Pot.(test 2.1.16)

1,5 10

20

30

40

50

Pot.(test 2.1.21)

1,5 10

20

30

40

50 εp /ε0 [-]

εp/ε0 [-]

Fig. 1.5: Power law fit of true stresses vs. logarithmic plastic strains (a)

test 2.1.16

(b)

test 2.1.21

The extrapolation is finally executed for α = 1.03 and n = 1.76, see Fig. 1.6.5 An extrapolation over a range of 25 times the last strain value of the tensile test may be considered as improper. There is no other choice, however, and the analyses of the M(T) specimen will have to legitimate it. The data points of the flow curve for ABQUS, Fig. 1.7, are established as

4

5

=

•

first point:

•

8 points including

•

13 data points σ ε law fit.

=0 ,

from the tensile test data up to = 0.06 = 0.002 as fourth point (see Fig. 1.3, above), .C =

in the range 0,12 ≤ ε ≤ 3.0 from the extrapolation by power

The fit parameters will in any case depend on the chosen interval. Varying this interval is advisable. Extrapolating a flow curve up to 25 times the final test point of plastic deformation seems rather disputably, but there is actually no other choice, and the simulation results of the M(T) specimen will have to justify it. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 6 -

700

RF [MPa]

σ [MPa]

700

600

600 500 400

500 300

test 2.1.21 extrapol

200

400

ABAQUS

100 300

0 0

0,5

1

1,5

2

2,5

3

0

1

1,5

2

2,5

3

εp [-]

εp [-]

Fig. 1.6: Extrapolation of the true stress vs. logarithmic plastic strain curve.

0,5

Fig. 1.7: Flow curve as input for ABAQUS

1.2. Numerical simulation of the tensile test Due to a two- or threefold symmetry in a 2D or 3D analysis, the tensile bar can be represented as a quarter or eighth model, respectively, Fig. 1.8. This is realised by applying boundary conditions along the respective line or surfaces, ZC &[B , 0, [E ) = 0, ZB &0, [C , [E ) = 0, ZE &[B , [C , 0) = 0. The symmetry to the x1-axis guarantees that necking occurs at x2 = 0. In order to realise a homogeneous stress and strain state over the measuring length, L0, up to uniform elongation, the fixation of the specimen with a larger width at the upper end has to be incorporated in the model. Fig. 1.9. In a panel of constant width, necking might occur in the section where the prescribed displacement is applied instead of the centre section. The correct location of necking has to be checked, see below. A finer mesh is applied in the centre segment of the bar. The specimen is loaded in displacement control, as the axial force will decrease under still increasing elongation in the simulation due to local necking beyond uniform elongation.6 A prescribed displacement is applied at the upper cross section of the model by connecting all respective nodes to some reference point using the coupling constraint option in ABAQUS. This point moves in the x2-direction with displacement u2. The corresponding external force is obtained as reaction force at the reference point.

6

This does not occur in the tests as the specimens broke before. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 7 -

Fig. 1.8: Quarter model (2D) of tensile bar

Fig. 1.9: 3D mesh esh of tensile bar (1/8 model)

The deformed mesh at ∆L ∆ = 10mm with iso-contours contours of transverse and longitudinal displacements, u1, u2, respectively, is shown in Fig. 1.10. Necking has occurred in the centre section, whereas in some distance, the deformation is homogeneous, again. Plastic deformation is revealed in Fig. 1.11. 1.1

Fig. 1.10: Deformed 3D mesh showing localisation of deformation (necking) at ∆L = 10mm (a) transverse displacement, displace u1 (b) longitudinal displacement, displace u2 Fig 1.12 finally shows a comparison between the numerical results of a 2D plane stress and a 3D model with the test data in terms of force, F, versus elongation, ∆L. ∆ There is little difference between the 2D and the 3D simulation. Due to the symmetry conditions, the t total external force results from the reaction force at the reference point as F = 2 FRP in the 2D model and F = 4 FRP in the 3D model, and the elongation is ∆L = 2 u2(L0/2) for both models. Both simulations coincide perfectly with each other up to uniform elongation and with the test data. This means that the plane stress assumption is legitimate.

report_Brocks-Rabbolini, wbrocks, wbrocks 24 June 2015, - 8 -

F [kN]

10

8

6 test 2.1.21 test 2.1.16

4

FE: 2D FE: 3D

2

0 0

2

4

6

8

10

∆L [mm]

Fig. 1.11: Deformed mesh with iso-contours of equivalent plastic strain, strain εeq, at ∆L = 10mm

Fig. 1.12: Comparison of test data with 2D and 3D numerical simulations simulation

Maximum force, ultimate (linear) stress and uniform elongation7 are: Fmax (kN)

σu (MPa)

(∆L)u (mm)

εu (-)

2D (plane stress)

8.24

354.8

5.62

0.187

3D

8.26

355.6

5.65

0.188

In a geometrically nonlinear analysis, analysis the he external force decreases due to local necking of the bar in the mid-section. Fig. 1.10. 1.1 . The corresponding value of uniform elongation is affected by any small imperfection of the bar, e.g. a reduction of diameter by 1% or a mesh refinement in the mid-section section as in Fig. 1.9. The plane stress and the 3D simulation start to differ after uniform elongation.

7

which were ere not reached in the tests: tests εf = 0.14 report_Brocks-Rabbolini, wbrocks, wbrocks 24 June 2015, - 9 -

1.3. Relation between true effective stress and applied force beyond uniform elongation Flow curves which are required as input for finite element analyses can generally not be directly determined from measured force versus elongation curves beyond the onset of necking, as the strain field becomes inhomogeneous and the stress state triaxial. An analytical solution derived by BRIDGMAN8 is used for round tensile bars to determine the effective stress in the necked section. It requires the continuous measurement of the reduction of diameter and the necking radius during the test, which necessitates advanced testing techniques. The BRIDGMAN correction does not work for specimens with rectangular cross sections, however, as the respective solution assumes an axisymmetric stress and strain field with constant plastic strain in the smallest necking section. The strain gradients in the cross section increase with the aspect ratio of rectangular bars. Difficulties also arise with the experimental determination of the cross section area as the rectangular cross section adopts the shape of a cushion. ZHANG et al.9 proposed a method using the local thickness reduction in the necking area to determine the actual cross section area, but the triaxiality of the stress state after occurrence of local necking is not taken into account. SCHEIDER et al.10 developed a procedure based on in-situ measurement of the deformation field on the specimen surface and derived an approximate formula relating true effective stresses and applied force which is validated by numerical parameter studies. Since the actual area of the necking section cannot be measured any more after the onset of necking due to its cushion like shape, they define a ”nominal” area, 7 8G = H C 6 ,

(1.3)

where the width, W, is a measurable quantity. The true effective stress results from the measured force by means of a correction factor, fcorr, 4

= 5I

,

(1.4)

which does not only include the conversion of the nominal area, 8 * , to the actual area, A, but also the effect of triaxiality of the stress state. A finite element parameter study has been conducted for determining the correction factor for various materials. The quantities F and 8G result from the finite element calculation and is the employed flow curve. The equivalent plastic strain, εp, serves as a scalar quantity related to the load step of the simulation. In testing, it is approximately calculated from the measured deformation field in the central area of the specimen, ]

≈ ̅ = AE &

C ^

+

^ ^^

+

C ^^ )

.

(1.5)

The curves fcorr(εp) by SCHEIDER et al. have a typical shape: they start at a value of one which is kept until maximum force is reached at the respective limit value, εu, of uniform elongation, and decrease for εp > εp(εu) in the beginning but raise to a value of one again for 8 9

10

BRIDGMAN, P.W.: Studies in large plastic flow, McGraw Hill, New York, 1952. ZHANG, Z.L.; HAUGE, M.; ODEGARD, J.; THAULOW, C.: Determining material true stress-strain curve from tensile specimens with rectangular cross section, Int. J. Solids Struct. 36 (1999), 3497–3516 SCHEIDER, I.; BROCKS, W.; CORNEC, A.: “Procedure for the determination of true stress–strain curves from tensile tests with rectangular cross section specimens”, J. Eng. Mater. Techn. 126 (2004) report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 10 -

εp ≈ 0.8. For higher values of εp, the curves spread significantly, depending on the hardening exponent of the flow curve. Fig 1.13 shows the nominal reduction of the mid-section area as calculated by the assumption of isochoric deformation, 8⁄8 = 9 ⁄9, and by eq. (1.3), 8G⁄8 = &H ⁄H )C . Both approaches coincide below uniform elongation, εp ≤ εp(εu)11. Due to the localisation of plastic deformation, the area reduction in the centre section increases much faster than the elongation, L0/L, which is better matched by the definition of the nominal area, 8G⁄8 , according to eq. (1.3)12. Fig 1.14 finally presents the factor relating external force and true effective stress, fcorr(εp), according to eq. (1.4) for the present material, which additionally includes the effect of triaxiality of the stress state. The initial part corresponds with the results of SCHEIDER et al., i.e. fcorr = 1 for εp ≤ εp(εu) and fcorr < 1 for εp > εp(εu), but a value of fcorr = 1 is reached again not until εp ≈ 1.2. 1,2

fcorr [-]

A / A0 [-]

1

0,8

1,1 0,6

0,4

L0/L

1

(W/W0)^2

0,2 0,9

0 0

0,5

1

1,5

2

εp [-]

Fig. 1.13: Nominal reduction of the midsection area as calculated by the assumption of isochoric deformation and by eq. (1.3)

0

0,5

1

1,5

2

εp [-]

Fig. 1.14: Correction factor, fcorr(εp), for the relation between true effective stress and applied force according to eq. (1.4) red vertical line indicates point of uniform elongation, εp(εu)

11 12

εp(εu) = εp(0.188) = 0.171 Note, that also eq. (1.3) provides only an approximation of the „real“ cross section area, which cannot be calculated from B and W due to the cushion like shape of the cross section. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 11 -

Part 2:

Numerical analysis of a centre cracked panel M(T) with stationary crack

Geometry:

width

2W = 300 mm,

initial crack length 2a = 60 mm,

Material:

thickness

B = 2.9 mm

length

2L = 1.5·2 W = 450 mm

Al 5083 as above for Part1 YOUNG’s modulus E = 70300 MPa POISSON’s ratio

ν = 0.3

0.2% proof stress

Rp0.2 = 242 MPa =

yield limit

=0

F [kN]

200

load vs elongation 150

VL = ∆(2L) = “load point” displacement, elongation 100

50 M(T)150 #1.1.3

Test results in “MT_150_Al5083_test.xls”

M(T)150 #2.1.8

0 0

1

2

3

4

VL [mm]

5

SCHEIDER, I.; SCHÖDEL, M.; BROCKS, W.; SCHÖNFELD, W., Engng. Fract. Mech. 73 (2006), 252–263

(1) Compare FE simulation with test data in terms of force vs. elongation, F(VL), up to approx. maximum force, i.e. _% ≈ 2`` Note:

•

Even below maximum load, the test results include significant crack extension (up to ab ≈ 6``), which is not considered in the present FE analysis! For simulation of crack extension see Part 3!

•

There is also an apparent measuring offset for F → 0.

(2) Evaluate J-integral by domain integral in ABAQUS and plot J(VL); compare with analytical evaluation of J as “energy release rate” from F(VL) curve, see below. (3) Investigate path dependence of J.


2.1.

Analysis of test data

The test data show some measuring offset for VL → 0 which becomes visible if the line of elastic compliance is plotted into the diagramme, Fig. 2.1. The elastic load point displacement results from a superposition, _%3 = _%

: c

+ _%

: c

= d%

: c

+ d%

: c

F = d% f ,

(2.1)

of the elongation of an uncracked panel, d%

: c

=

C

@5

=

C

C@76

=

@76

,

(2.2)

and a contribution which is due to the crack, d%

: c

=

g g _ Q T @6 C 6

,

(2.3)

where the function V2 is given by TADA et al.13. g

g C

g

g E

g ]

_C Q T = −1.071 + 0.250 Q T − 0.357 Q T + 0,121 Q T − 0,047 Q T + 6

g j

0,008 Q6T −

6 B, kB ln Q1 :⁄,

g

6

6

− 6T .

6

(2.4)

According to TADA, eq. (2.4) holds for 9⁄H ≥ 3 but obviously gives a good approximation of the elastic compliance for 9⁄H = 1.5 as in the present case. The contribution of d% : c is only 5.8% of the total compliance, CL, however. 200

F [kN]

F [kN]

200

150

150

100

100

M(T)150 #1.1.3

50

M(T)150 #1.1.3

50

M(T)150 #2.1.8

M(T)150 #2.1.8

elastic

elastic

0

0 0

0,5

1

1,5

2

0

VL [mm]

Fig. 2.1:

0,5

1

1,5

2

VL [mm]

Test data exhibiting a measuring Fig. 2.2: Test data after applying an offset offset for VL → 0 correction

The offset corrections are obtained as _%

13

23m2 = _%23m2 + ∆_% = _%23m2 + d% f&B) − _%&B) .

(2.5)

TADA, M.; PARIS, P.C.; IRWIN, G.R.: The Stress Analysis of Cracks Handbook, Del Research Corp., Pennsylvania [1973] report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 13 -

After the respective shifts of the displacements by ∆VL = 0.0485 mm for test #1.1.3 and ∆VL = 0.0930 mm for test #2.1.8, the measured curves coincide in the elastic range, Fig. 2.2. The offset correction is particularly important for the calculation of plastic deformations, _% in Part 3.

The (corrected) test data, F(VL), can be used to evaluate the J-integral according to its interpretation as an energy release rate. To begin with, crack growth, ∆a, is not considered. Fig. 2.3:

Schematic of an F(VL) curve at constant crack length with definitions F

n = o fp_% = n 3 + n , B

B

U*

n 3 = C f_%3 = C f _ −_% , n = o fp_% , B

B

n ∗ = n − C f_% = n − C f_% .

Upl Uel vL

Assuming “deformation theory of plasticity”14, J is split into an elastic and a plastic part, where the elastic part is calculated from the stress intensity factor, KI, and the plastic part from the area under the f&_%% ) curve, Fig. 2.3, r = r3 + r =

where and

z^ = ∞

=

stu @

g ∞ √|b } Q6T 4

C76

vw S

− QC7 vgT

xy

,

(2.6)

,

(2.7)

.

(2.8)

The geometry function for an M(T) specimen is15 g

•g

} Q6T = Asec QC6T ,

(2.9)

and the plastic J can be calculated from16 r =

w∗

7&6‚g)

.

(2.10)

The result of this evaluation is shown in Fig. 2.4 for the specimen #1.1.3. Up to VL = 1 mm only the elastic part, J e, contributes to the total J. As the evaluation does not account for crack growth, J e decreases beyond maximum load and J p becomes more and more dominant.

14

15 16

Or actually hyperelastic behaviour of the material requiring that the strain energy density represents a potential, ƒ„ = …†⁄… ƒ„ . FEDDERSEN, R.E., ASTM STP 410 (1966), 77-79 RICE, J.R.; PARIS, P.C.; MERKLE, J.G., ASTM STP 536 (1973), 231-245. Note that eq. (2.10) holds for an M(T) only! report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 14 -

1400

J [N/mm]

J [N/mm]

200 J_total J_e

150

J_p

J_total

1200

J_e J_p

1000 800

100 600 400

50

200 0

0 0

0,5

1

1,5

2

0

VL [mm]

1

2

3

4

5

VL [mm]

Fig. 2.4: J-integral for test #1.1.3 evaluated from F(VL) according to eqs. (2.6)-(2.10) neglecting crack growth (a) VL ≤ 2 mm, F ≤ Fmax (b) total test range 2.2.

FE model

The FE model disregards all technical details of a real M(T) specimen, in particular • •

clampings for application of the force17, the machined notch used as starter for fatigue cracking

and accounts for the twofold symmetry with respect to x1 and x2. It is thus reduced to a rectangle of size W×L, Fig. 2.1. The mesh is refined towards the ligament, b ≤ [B ≤ H, Fig. 2.2, reaching a minimum element size of 0.125×0.125 mm2. The regular arrangement of square elements in the ligament is specific for simulations of crack extension (see Part 3) but will yield strongly distorted elements at the crack tip and is hence inappropriate for calculating reliable stress and strain values at a stationary crack. Since the present analysis is focused on global values, namely elongation and J-integral, the meshing at the crack tip is uncritical. The twofold symmetry is realised by fixing the vertical displacements in the ligament, ZC &b ≤ [B ≤ H, 0) = 0, and the horizontal displacements of the vertical centre line, ZB &0, [C ) = 0. The upper edge is kept straight, ZC &[B , 9) = ZC &0, 9). The specimen is loaded in displacement control mode by applying a vertical displacement u2 to a reference point as in exercise #1, which is connected to the upper edge of the specimen. The respective force is picked up as reaction force at the reference point. The elongation of the specimen equals the displacement of the reference point, a&29) = _% = 2ZC &0, 9)

17

This is different from the tensile bar, Fig. 1.8, where the fixation of the specimen had to be incorporated into the model in order to prevent necking in the section where the prescribed displacement is applied. Due to the crack at x2 = 0 this is precluded here. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 15 -

Fig. 2.5: FE mesh of the M(T) specimen Fig. 2.6: Refined FE mesh in the ligament (quarter model) with boundary conditions

200

F [kN]

F [kN]

The analysis is performedd as plane stress with the specimen’s thickness B. Fig 1.13 has shown that the assumption of a plane stress state is admissible. Figs. 2.7 shows show comparisons of the FEM results with test data in terms of force, F,, versus load point displacement VL, where in Fig. 2.7(b) the offset correction of eq. (2.5) has been applied to the test data as in Fig. 2.2. The elastic compliance of the FE model is obviously lower than in the analytical solution of TADA, eqs. (2.1)--(2.5), d% ‡ˆ = 0.91d%Š+‹+ , and for the test data. data The higher stiffness of FE models is generally due du to the finite number of degreess of freedom. Fig. 2.7(b) also contains the FE results of exercise #3 (see below). The cohesive elements obviously reduce the elastic stiffness of the model. 200

150

150

100

100 M(T)150 #1.1.3 M(T)150 #2.1.8 FEM ex#2 FEM ex#3 Tada

M(T)150 #1.1.3

50

50

M(T)150 #2.1.8 FEM ex#2 Tada

0

0 0

0,5

1

1,5

VL [mm]

2

0

0,5

1

1,5

2

VL [mm]

Fig. 2.7: Comparison of the numerical results with test data up to maximum load (a) original test data (b) offset correction, eq. (2.5) Any offset correction is negligible in a plot over the full range of VL, see Fig. 2.8. The FEM curve starts deviating from the experimental data already below maximum force reached in the tests,, as no crack extension is considered in the analysis. The force continues to increase in the simulation whereas it decreases in the tests. Different from the force vs. elongation curve of the tensile bar, Fig. 1.13,, where the load drop was due to plastic necking, it is due to crack extension, here.


F [kN]

200

150

100 M(T)150 #1.1.3 M(T)150 #2.1.8

50

Fig. 2.8: Comparison of the numerical results with test data over the whole range of VL

FEM

0 0

1

2

3

4

5

VL [mm]

2.3. J-integral evaluation The J-integral introduced by RICE18 and CHEREPANOV19 is defined as an integral over a closed contour, Γ, around the crack tip, r = ∮• †p[C −

ƒ„ •„ Zƒ,B pŽ

,

(2.11)

where w is the strain energy density, σij the stress tensor and nj the outer normal to the contour. It is equal to the energy release rate in a plane body of thickness, B, due to a crack extension, ∆a, in mode I, ‘w

r = − •7‘g’ ,

(2.12)

xy

which is exploited in the virtual crack extension (vce) method to calculate J numerically as well as in the experimental determination of J. Applying the divergence theorem, any contour integral can be converted into a domain integral over a finite region, i.e. an area in two dimensions or a volume in three dimensions, surrounding the crack. ABAQUS/Standard uses this domain integral method to evaluate the J-integral and automatically finds the elements that form each ring from the regions defined as the crack tip or crack line. Each “contour” provides an evaluation of the respective domain integral. The number, n, of contours must be specified in the history output request. 20 *CONTOUR INTEGRAL, CONTOURS=n, TYPE=J By definition, J is a “path-independent” integral, claiming that each contour should yield the same J-value. However, the derivation of its path-independence requires (1) 18

19

20

time independent processes,

RICE, J.R.: “A path independent integral and the approximate analysis of strain concentrations by notches and cracks”, J. Appl. Mech. 35 (1968), 379-386. CHEREPANOV, C.P.: “Crack propagation in continuous media" Appl. Math. Mech. 31 (1967), 476-488. Note that the largest domain must not touch the outer surface of the specimen, as the energy release rate is calculated in a ring of elements around this domain. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 17 -

(2) (3) (4)

absence of volume forces and surface forces on the crack faces, small strains (geometrically linear), homogeneous hyper-elastic material (deformation theory of plasticity).

A geometrically nonlinear elasto-plastic analysis applying incremental theory of plasticity violates the two last-mentioned conditions, so that the values obtained for the various contours will differ and the J-integral becomes path-dependant, more precisely, J increases with increasing size of the contour.21 Contour (or domain) dependence thus is physically significant due to the dissipative character of plastic deformation. What is needed in a fracture mechanics analysis is a saturation value or “far-field” value of J corresponding to that evaluated from the mechanical work done by the external force at the load-line displacement, eq. (2.12). For further details see BROCKS & SCHEIDER22. The user must define the crack front and the virtual crack extension in ABAQUS to request contour integral output; this can be done as follows: •

Defining the crack front

The user must specify the crack front, i.e., the region that defines the first contour. ABAQUS/Standard uses this region and one layer of elements surrounding it to compute the first contour integral. An additional layer of elements is used to compute each subsequent contour. The crack front can be equal to the crack tip in two dimensions or it can be a larger region surrounding the crack tip, in which case it must include the crack tip. By default ABAQUS defines the crack tip as the node specified for the crack front in the so called CRACK TIP NODES option of the contour integral command. *CONTOUR INTEGRAL, CONTOURS=n, CRACK TIP NODES Specify the crack front node set name and the crack tip node number or node set name.

Alternatively, a user-defined node set which include the crack tip can be provided to ABAQUS manually by omitting the CRACK TIP NODES option: *CONTOUR INTEGRAL, CONTOURS=n Specify the crack front node set name and the crack tip node number or node set name. •

Specifying the virtual crack extension direction

The direction of virtual crack extension must be specified at each crack tip in two dimensions or at each node along the crack line in three dimensions by specifying either the normal to the crack plane, n, or the virtual crack extension direction, q, see Fig. 2.9.

21

22

BROCKS, W.; YUAN, H.: On the J-integral concept for elastic-plastic crack extension, Nuclear Engineering and Design 131 (1991), 157-3 BROCKS, W.; SCHEIDER, I.: Reliable J-values - Numerical aspects of the path-dependence of the J-integral in incremental plasticity, Materialprüfung 45 (2003), 264-275. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 18 -

Fig. 2.9: Crack front tangent, t, normal to the crack plane, n, and virtual crack extension direction, q, at a curved crack The normal to the crack plane, n, can be defined in ABAQUS using the so called NORMAL option23 in the contour integral command: *CONTOUR INTEGRAL, CONTOURS=n, NORMAL nx-direction cosine, ny-direction cosine, nz-direction cosine(or blank), crack front node set name (2-D) or names (3-D) ABAQUS will calculate a virtual crack extension direction, q, that is orthogonal to the crack front tangent, t, and the normal, n, see Fig. 2.9. Alternatively, the NORMAL option can be omitted and the virtual crack extension direction, q, can be defined directly: *CONTOUR INTEGRAL, CONTOURS=n crack front node set name, qx-direction cosine, qy-direction cosine, qz-direction cosine (or blank) Note, that the symmetry option (SYMM) must be used to indicate that the crack front is defined on a symmetry plane: *CONTOUR INTEGRAL, CONTOURS=n, SYMM The change in potential energy calculated from the virtual crack front advance is doubled to compute the correct contour integral values. Different combination of the above mentioned options can be used to request contour integral within ABAQUS, more details about this could be found in the ABAQUS documentation. Two options of defining the contours are used in the present study with a number of contours24, n = 20, for both: • The CRACK TIP NODES option with the first contour being the crack tip, hereafter called “automatic option”, Fig. 2.9; • A first contour around the crack tip of finite size is defined via a crack tip node set, hereafter called “manual option”, Fig. 2.10.

23 24

The normal option can only be used if the crack plane is flat Note that the number of contours is limited by the symmetry line of the specimen. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 19 -

Fig. 2.9: Contour (domain) definition25 of J-integral calculation by “automatic option”: option” st (a) 1 domain = crack tip (b) last domain #20 20

Fig. 2.10: Contour (domain) definition for J-integral calculation by “manual option”: option” st (a) 1 domain = „node node set“ set (b) last domain #20 20 Fig. 2.11 shows two comparisons between the domain sizes and the plastic zones. A lower limit of the equivalent quivalent plastic strain, strain εp,=0.002, corresponding to = .C C , has been defined, so that the grey area approximately represents the elastic region. region In order to obtain meaningful far-field values of J, the corresponding contour should imbed the plastic zone emanating from the crack tip. This would be possible for VL = 1.5 mm, Fig. 2.11(a),, but require a different definition of the initial node set, accounting for the specific shape of the plastic zone, since the number of contours is limited to n = 20 by the symmetry line of the specimen, Fig. 2.10(b). It is not possible any more when the plastic zone extends from the crack tip to the free surface of the specimen, specimen Fig. 2.11(b).

Fig. 2.11: Equivalent plastic lastic strain, εp, and domains for J evaluation (manual option) (a) VL = 1.5 mm,, domain #1 (b) VL = 4 mm,, domain #20 Test and simulation results of r&_% ) are compared in Fig. 2.12. The J-integral J has been evaluated from F(VL) for both, test and FE results according to eq. (2.6) and also as domain 25

The terms “contour” and “domain” “ are used more or less synonymously, here. More precisely spoken, the contour is the first fi ring of elements around the he respective domain. report_Brocks-Rabbolini, wbrocks, wbrocks 24 June 2015, - 20 -

integral, eq. (2.11), for contour #20 (manual option), Fig. 2.10, in ABAQUS. The test data coincide satisfactorily with the FE results even a bit beyond maximum load, VL = 2 mm, though the F(VL) curves start differing significantly, Fig. 2.8, since the opposing effects of decreasing force and increasing ∆a cancel each other. Remember that the evaluation of J p from the F(VL) curve according to eq. (2.10) does not consider crack extension. The underlying F(VL) data include crack extension, ∆a, in the tests resulting in the load drop, but do not in the simulation. Also note that J-results for VL > 2 mm have only numerical significance. The “far-field” value of the domain-integral definition is supposed to lie close to the J-definition as energy release rate, J_F(VL). 1000 M(T)150 #1.1.3 J_20 (manual) J_F(VL)

J [N/mm]

J [N/mm]

200

150

M(T)150 #1.1.3 J_20 (manual) J_F(VL)

800

600 100 400 50

200

0

0 0

0,5

1

1,5

2

0

1

2

VL [mm]

3

4

5

VL [mm]

Fig. 2.12: Comparison of FE results with test data #1.1.3 (a) VL ≤ 2 mm, F ≤ Fmax,test (b) total range of VL The first contours, particularly in the automatic option, lie more or less in a near-field around the crack tip where large plastic strains with corresponding rearrangement of stresses occur, Fig. 2.11, which violates the requirements of “deformation theory of plasticity”. This causes the “path dependence” of J, Fig 2.13. It is still not pronounced for VL ≤ 2 mm (maximum force in the tests), Fig 2.13(a). Both, automatic and manual option yield a far-field value for the domain #20 which coincides with the J-value calculated from F(VL), and even J_01 (node set) of the manual option does so, though the respective domain covers a large part of the plastic zone, Fig. 2.11(a). In the automatic option, J_01 = Jcrack-tip is significantly lower, of course. Due to increasing plastic deformations for VL > 2 mm, significant and growing path dependence occurs, Figs. 2.13(b) and 2.14. For domain #1 of the automatic option, J_01 = Jcrack-tip even becomes physically meaningless as J must not decrease with increasing global deformation. J_20 (manual) differs slightly from J_F(VL) for VL > 3 mm, which calls for even larger contours, see Fig. 2.12(b).


J [N/mm]

J [N/mm]

200 J_01 (automatic) J_01 (manual) J_20 (automatic)

150

J_20 (manual)

800 J_01 (automatic) J_01 (manual) J_20 (automatic)

600

J_20 (manual)

J_F(VL)

J_F(VL)

100

400

50

200

0

0 0

0,5

1

1,5

2

0

1

2

3

VL [mm]

4

5

VL [mm]

Fig. 2.13: Contour dependence of J: 1st contour = node set (Fig. 2.9) (a) VL ≤ 2 mm, F ≤ Fmax,test (b) total range of VL

800

800

J [N/mm]

J [N/mm]

J increases with increasing size of the domain, Fig. 2.15, see BROCKS & YUAN 21. The finite slope at the end of the J(contour) curve for VL = 4.5 mm indicates that no saturation value, Jfar-field, has yet been reached. J_01 J_02 J_03 J_05 J_10 J_15 J_20 J_F(VL)

600

400

600

400

200

200

VL=3mm VL=4mm VL=4.5mm

0

0 0

1

2

3

4

VL [mm]

Fig. 2.14: Contour dependence of J for manual option of contour definition, Fig. 2.10

5

0

5

10

15

20

contour

Fig. 2.14: Contour dependence of J (manual option)


Part 3:

Geometry:

Simulation of ductile crack extension in a centre cracked panel, M(T) width 2W = 300 mm, initial crack length 2a = 60 mm, thickness B = 2.9 mm length 2L = 1.5·2 W = 450 mm

Material

Al 5083 as above for exercises #1, #2 Young’s modulus E = 70300 MPa Poisson’s ratio ν = 0.3 0.2% proof stress Rp0.2 = 242 MPa =

yield limit

=0

F [kN]

δ5 [mm]

200

5

4

150 3

100 2 M(T)150 #1.1.3 M(T)150 #2.1.8

50 1

M(T)150 #1.1.3 M(T)150 #2.1.8

0

0 0

1

2

3

4 VL [mm]

5

0

20

40

60

80

∆a [mm]

Test results in “MT_150_Al5083_test.xls” (1) Perform a numerical simulation of crack extension applying cohesive elements (see SCHEIDER, I.: “The Cohesive Model - Foundations and Implementation“, Geesthacht [2011]) (2) Compare simulation with test data in terms of force vs. load-point displacement, F(VL), and CTOD R-curve, δ5 (∆a) (3) Evaluate J by the domain integral method and compare to the “far-field” J evaluated from F(VL,∆a) (see formulas given in exercise #2 and below). (4) Investigate the effect of cohesive parameters σc, Γc


Cohesive law (traction-separation law) suggested shape parameters: 1

δ1 = 0.05 δc 0,8

σ / σc

δ2 = 0.5 δc 0,6

“effective” cohesive parameters (plane-stress model)

0,4 0,5

σc = 2.3 Rp0.2 = 560 MPa

0,2 0 0

0,2

0,4

0,6

0,8

1

δ / δc

Γc = 10 kJ/m2 δc = 0.024 mm

Evaluation of J from F(VL, ∆a) for extending crack r&ƒ) = r&ƒ‚B)

f&ƒ‚B) _%&ƒ) − f&ƒ) _%&ƒ‚B) “&ƒ) + “&ƒ‚B) 2”“&ƒ‚B)

“ = H − b ; _% = _% − _%3 = _% − d% f see also

• formulas in exercise #2 • BROCKS, W.; ANUSCHEWSKI, P.; SCHEIDER, I.: “Ductile Tearing Resistance of Metal Sheets”, Engineering Failure Analysis 17 (2010), 607-616.


3.1.

Analysis of test data including crack extension

Whereas the F(VL) curves of the two tests #1.1.3 and #2.1.8 differ significantly beyond VL > 3 mm, Fig. 3.1, the δ5(∆a) curves nearly coincide, Fig. 3.2. The reason of this discrepancy will be discussed and analysed below. In any case, it is another indication, that local quantities like δ5 and ∆a are better suited for calibrating cohesive parameters by comparison of test and simulation data than global quantities like F and VL. 5

F [kN]

δ5 [mm]

200

150

4

3 100 2

50

M(T)150 #1.1.3

1

elast

M(T)150 #2.1.8

M(T)150 #1.1.3 M(T)150 #2.1.8

0 0

0 0

2

4

VL [mm]

Fig. 3.1: Force vs. elongation (offset corrected)

20

40

60

6

80

∆a [mm]

Fig. 3.2: CTOD R-curves

Contrary to the unique CTOD R-curves, δ5(VL) and ∆a(VL) differ between the two tests, Figs. 3.3 and 3.4. The source of this inconsistency of the test data can evidently be found in the kinematics between (local) CTOD and the (global) elongation of the specimens. 80

4

∆a [mm]

δ5 [mm]

5

M(T)150 #1.1.3 M(T)150 #2.1.8

M(T)150 #1.1.3 M(T)150 #2.1.8

60

3 40 2 20 1

0

0 0

1

2

3

4

0

5

CTOD in dependence on VL

2

3

4

5

VL [mm]

VL [mm]

Fig. 3.3:

1

Fig. 3.4:

Crack extension in dependence on VL


The maximum crack extension of more than 60% of the initial ligament, Fig 3.4 is quite remarkable. The test data, F(VL, ∆a), can be used to evaluate the J-integral according to its interpretation as an energy release rate like in section 2.1, eqs. (2.6) to (2.9). The evaluation of the plastic J now has to accounts for crack extension, ∆a. Since (a/W) is not constant anymore and F changes with both VL and ∆a, a step-wise procedure is needed, r& ) = r& ‚B)

–&—)

–&—˜™)

“ = H − b.

where

Eq. (3.1) requires a f _% and a plastic part by

+

S

S

4&—˜™) xy&—) ‚4&—) xy&—˜™) C7–&—˜™)

,

(3.1) (3.2)

curve. The total load point displacement is split into an elastic

_% = _% − _%3 = _% − d% f ,

(3.3)

where the compliance, CL(a/W). results from eqs. (2.2) and (2.3). Beyond maximum load, _%3 stays constant whereas _% increases linearly with VL, Fig. 3.5. 1200 a=a0

J [N/mm]

VLe, VLp [mm]

5 Ve=CL*F

4

Vp=V-Ve V_L

3

a=a0+Da

1000 800 600

2 400 1 200 0 0

1

2

3

4

5

VL [mm]

0 0

1

2

3

4

5

VL [mm]

Fig. 3.5: Elastic and plastic parts of the total load point displacement according to eq. (3.3) for specimen # 1.1.3

Fig. 3.6: J as energy release rate calculated from F(VL, ∆a) neglecting and considering crack extension for specimen # 1.1.3

Fig. 3.6 compares the results of the J evaluations for constant crack length, a = a0, and crack extension, a = a0 + ∆a., where J p is calculated according to eq. (2.10) and eq. (3.1), respectively. For given VL, crack extension reduces J. If J is plotted vs. VL, Fig. 3.7, no differences between the test curves are observed since J has been calculated from the global F(VL) curve which yields consistent results. A crucial indicator, however, is the relation between the J-integral and CTOD, which is supposed to be approximately linear, šj ~ •

œ

S .u

,

(3.4)


as plotted in Fig. 3.8.26 It finally answers the question which of the respective two test curves in Figs.3.1, 3.3 and 3.4 is more reliable. Actually, the test data of #2.1.8 violate the linear relationship and have hence to be regarded as faulty. The problem with testing M(T) specimens is generally the existence of two crack tips for which crack extension may be varying. A local δ5(∆a) curve is not affected, Fig. 3.2, but the global F(VL) is, of course. 1000

J [N/mm]

J [N/mm]

1000 M(T)150 #1.1.3

800

M(T)150 #2.1.8

800

600

600

400

400

200

200

M(T)150 #1.1.3 M(T)150 #2.1.8 d_5*R_p0.2

0 0

1

2

3

4

5

VL [mm]

Fig. 3.7: Relation between the J –integral and elongation VL

0 0

1

2

3

4

5

δ5 [mm]

Fig. 3.8: Relation between the J-integral and CTOD

Plotting J vs. ∆a, Fig. 3.9, yields resistance curves, JR, which are supposed to characterise the material’s resistance to crack extension.27 Since the F(VL) curves of the two tests differ, so the JR curves do, which is inconsistent with the CTOD R-curves, Fig. 3.2, because of the reasons discussed above. There are also some oscillations beyond ∆a > 60 mm which should be regarded as deficient. Finally, the CTOD R-curves, Fig. 3.2, appear as reliable and significant experimental reference data for identifying the cohesive parameters via numerical simulations.

26 27

The coefficient of proportionality appeared to be 1, correspondent to plane stress. JR curves appeared to be geometry dependent, however, a problem generating a countless number of publications. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 27 -

J [N/mm]

1200 M(T)150 #1.1.3

1000

M(T)150 #2.1.8

800 600 400 200

Fig. 3.9:

3.2.

0

JR-curves from F(VL, ∆a), eqs. (2.6) and (3.1)

0

20

40

60

80

∆a [mm]

FE model

The FE model is basically the same as in exercise #2, see Figs. 2.5 and 2.6. Instead of fixing the vertical displacements in the ligament, b ≤ [B ≤ H, [C = 0, cohesive elements28 are applied in the ligament between b ≤ [B ≤ ab*:ž to allow for crack extension. constraint equations &‚)

ZC

&‚)

ZB

&Ÿ)

= −ZC

&Ÿ)

= ZB

cohesive parameters delta_N = δc, traction_N = σc/2

Fig. 3.10: Cohesive elements at symmetry line: Constraint equations and input variables &‚)

&Ÿ)

Symmetry29 is realised by applying linear constraint equations, ZC = −ZC , Fig. 3.1, &Ÿ) &‚) &Ÿ) which guarantee a symmetric opening of the cohesive element, š = ZC − ZC = 2ZC , 30 implying delta_N = δc. These constraint equations induce extra nodal forces , however, which reduce the stresses in the respective element to one half, so that the cohesive strength traction_N = σc/2. 28

29

30

SCHEIDER, I.: The cohesive model – Foundation and implementation, Release 2.0.3, Report, Helmholtz Centre Geesthacht, 2011. BROCKS, W.; ARAFAH, D.; MADIA, M.: Exploiting symmetries of FE models and application to cohesive elements, Report Milano / Kiel, November 2013. http://www.tf.uni-kiel.de/matwis/instmat/departments/brocks/brocks_homepage_en.html “Linear constraint equations introduce constraint forces at all degrees of freedom appearing in the equations. These forces are considered external, but they are not included in reaction force output. Therefore, the totals provided at the end of the reaction force output tables may reflect an incomplete measure of global equilibrium.” - ABAQUS Analysis User's Manual (6.11): 33.2.1 Linear constraint equations. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 28 -

The variables of user elements cannot be displayed by ABAQUS/Viewer. They are hence transferred to the adjacent continuum elements as user output variables by providing an element mapping between cohesive and continuum elements which is given in the .cohes file, *ELEMENT MAP cont_elem_1, coh_elem_1 cont_elem_2, coh_elem_2 … cont_elem_n, coh_elem_n The cohesive law of SCHEIDER26 is applied

( ) ( )

δ δ  2 δn1 − δn1   σ n (δ n ) = σ c ⋅  1  3 δ −δ 1 + 2 δnc −δ 22 − 3

(

)

2

for δ n ≤ δ1 for δ1 < δ n ≤ δ 2 ,

(

δ n −δ 2 δ c −δ 2

)

2

(3.5)

for δ 2 ≤ δ n ≤ δ c

with the cohesive energy δc

1 2



Γ c = ∫ σ n (δ n ) dδ n = σ cδ c 10



2 δ1 δ 2  + . 3 δc δc 

(3.6)

For plane stress structures the thickness change on the ligament must be taken into account31, which is calculated based on the in-plane strain components assuming isochoric deformation, =

&1 +

EE )

=

&1 −

BB

−

CC ).

(3.7)

The keyword *THICKNESS DEPENDENCE assigns the plane stress cohesive elements to be thickness dependent. Since the thickness is determined from the adjacent continuum elements, the keyword *ELEMENT MAP must be used together with *THICKNESS DEPENDENCE. A user defined output variable (UVARM9) in the UEL subroutine indicates completely failed cohesive elements by taking an integer value of -1. The number of failed elements can be counted, and crack extension, ∆a, is calculated by multiplying it with the element length of 0.125mm. A far-field J-integral is calculated as described above in Part #2 for the contour shown in Fig. 3.8.

31

SCHEIDER, I.; BROCKS, W.: Cohesive elements for thin-walled structures. Comp. Mat. Science 37 (2006), 101-109. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 29 -

Fig. 3.11:

3.3.

Contour (domain) definition for J-integral integral calculation

Simulation Results - Comparison with Test Data. Data

The first set of cohesive parameters suggested above is taken from SCHEIDER et al. [2006]32. Different from the referenced publication, they did not yield a good approximation of the experimental data,, however, see green curves in Figs. 3.12 and 3.13. 3.13 Remember that the F(VL) curve of specimen #1.1.3 is the proper reference. 5

F [kN]

δ5 [mm]

200

150

4

3 100 2 M(T)150 #1.1.3 M(T)150 #2.1.8 FE (par set 1) FE (par set 2) FE (par set 3) FE (par set 4)

50

1

0

0 0

1

2

3

M(T)150 #1.1.3 M(T)150 #2.1.8 FE (par set 1) FE (par set 2) FE (par set 3) FE (par set 4)

4

0

5

20

40

60

80

∆a [mm]

VL [mm]

Fig. 3.12: Force vs. load-point point displacement Fig. 3.13: CTOD δ5 vs. crack extension parameter set

δ1 / δc

δ1 / δc

σc [MPa]

δc [mm]

Γc [kJ/m2]

#1

0.05

0.5

560

0.024

10.0

#2

0.05

0.5

560

0.0365

14.8

#3

0.05

0.5

560

0,043

17.5

#4

0.05

0.5

560

0.049

20.0

Table 3.1:: Cohesive parameters applied in the simulations

32

SCHEIDER, I.; SCHÖDEL, M.; BROCKS, W.; SCHÖNFELD, W.: Crack propagation analyses with CTOA and cohesive model: Comparison and experimental validation. validation Engng. Fract. Mech. Mech 73 (2006), 252–263. report_Brocks-Rabbolini, wbrocks, wbrocks 24 June 2015, - 30 -

The reason for the deviation between test data and the simulation is the substantial overestimation of crack extension, Fig. 3.14. Further simulations were hence run with increased values of the cohesive energy, Γc, as listed in Table 3.1, below. Increasing Γc implies an increase of the critical separation, δc, according to eq. (3.6). Parameter set #4 failed to converge at VL = 3.4 mm (∆a = 33.9 mm). 5


60

δ5 [mm]

∆a [mm]

80


4

3 40 2 20 1 0 0

1

2

3

4

0

5

0

1

2

3

4

5

VL [mm]

VL [mm]

Fig. 3.14: Crack extension vs. load-point displacement

Fig. 3.15: CTOD vs. load-point displacement

The increase of “ductility” in the TSL delays crack extension, Fig. 3.14, and makes the simulations approach the test curves, F(VL) and δ5(∆a). The dependence of the local CTOD on the global deformation is much less affected by the cohesive parameters, Fig. 3.15. Increasing Γc does also affect maximum force and the initiation values of J and CTOD, see Table 3.2. More or less, the initiation values of the simulations lay within the scatter of the measured values, and the overestimations of maximum force is less than 3%. Considering the unavoidable errors in measuring initiation values, the simulations yield quite satisfactory results, after all.

Fmax [kN] ¡šj | ¡r|

.B** DmmF

C .B** DkJ⁄m F

test #1.1.3

test #2.1.8

simul #1

simul #2

simul #3

simul #4

172

172

169

174

175

177

0.16

0.06

0.09

0.13

0.15

0.16

32.3

20.4

21.2

27.8

29.5

39.2

Table 3.2: Effect of cohesive energy on Fmax and initiation values33 of J and CTOD Fig. 3.16 shows the relation between the cohesive parameters δc and Γc and the “initiation” values33 of the CTOD and J resistance curves.

33

“Engineering” definition of initiation at ∆a = 0.1 mm obtained from interpolation of data. report_Brocks-Rabbolini, wbrocks, 24 June 2015, - 31 -

4 J_i / Γ_c

3,5

δ_i / δ_c

3 2,5 2 1,5 1

Fig. 3.16: Ratio of cohesive parameters δc and Γc and the respective initiation values33 of CTOD and J

0,5 0 1

2

3

4

The plots of equivalent plastic strain, εp, at ∆a = 40 mm and 80 mm in Fig. 3.17 indicate that no contour can be found for calculating J which bypasses the plastic zone. zone

Fig. 3.17: Plastic zones at ∆a = 40 mm and 80 mm Due to the dissipation of external work by plastic deformation, the the “far field” J-integral values calculated for the contour shown in Fig. 3.11, have always to be less or equal the limiting values of the (global) energy release rate calculated from the F(VL,∆a) curves, which is confirmed by Fig. 3.18 for all parameter sets. The evaluation of J as energy release rate from the F(VL,∆a)) curves requires the calculation of _% = _% − _%3 according to eq. (3.3) and hence the elastic compliance CL. As was found in Fig. 2.7,, already, the elastic compliance of the FE model is slightly lower, lower i.e. its stiffness is higher, than the value resulting from TADA’s equation. However, the compliance is increased due to the cohesive elements, d% ‡ˆ = 0.98d%Š+‹+ , see Fig. 2.7(b). 2.7(b) Nevertheless, applying TADA’s equation would yield artificial negative plastic deformations in the elastic range. Hence, the compliance for calculating _%3 has been modified, ¡§ ¨©ª « y

d% &b⁄H ) = ¦§ ®¯°¯ &&gy⁄6) ± d%Š+‹+ &b⁄H ) . 6 y

¬ →

(3.8)


1000 M(T)150 #1.1.3

J [N/mm]

J [N/mm]

1000 M(T)150 #2.1.8

800

J_FE (contour) J_FE (en.rel.rate) par set #1

600

M(T)150 #1.1.3 M(T)150 #2.1.8

800


600

400

400

200

200

0

0 0

1

2

3

4

5

0

1

2

3

VL [mm]

5

VL [mm]

(a)

(b)

1000

1000 M(T)150 #1.1.3

J [N/mm]

J [N/mm]

4

M(T)150 #2.1.8

800


600

M(T)150 #1.1.3 M(T)150 #2.1.8

800


600

400

400

200

200

0

0 0

1

2

3

4

5

0

1

2

VL [mm]

(c)

3

4

5

VL [mm]

(d)

Fig. 3.18: J calculated as contour integral (Fig. 3.11) and energy release rate, cohesive parameter sets #1 (a) to #4 (d) The linear relationship between J and δ5 according to eq. (3.4) is confirmed by the simulations independent of the cohesive parameters, Fig. 3.18. Fig. 3.19 finally shows the simulated JR curves. The simulation applying parameter set #4 coincides with the experimental curve of specimen #1.1.3 up to a crack extension of more than 30 mm.


1000

J [N/mm]

J [N/mm]

1000

800

600

800

600

400

400

FE (par set 3) M(T)150 #2.1.8 FE (par set 1) FE (par set 2) FE (par set 3) FE (par set 4)

200


200

0

0 0

1

2

3

4

5

0

20

δ5 [mm]

Fig. 3.18: J-integral vs. CTOD

40

60

∆a [mm]

Fig. 3.19: JR curves

Epilogue and Acknowledgement The report presents the results of an exercise for the students of a PhD course on Computational Fracture Mechanics held at the Dipartimento di Meccanica of the Politecnico di Milano in March and April 2015. Beyond the direct results of the task which have been used as benchmark for the evaluation of the students’ reports, additional information and results are included which have not been part of the definition of the project but may serve as background information. The report can hopefully also give guidance to engineers and scientists working in the field of numerical simulations of fracture problems how to achieve reliable results. Unfortunately, even current contributions to international scientific journals contain erroneous conceptions of the elastic-plastic J-integral and its path dependence. The report also intends to show that not only numerical simulations have to be examined critically but also experimental results have to be checked with respect to consistency. The advantage of a simulation is, at least, that it yields reproducible results. The authors thank MANFRED SCHÖDEL for making the test data available.
