uiuc cee 575 final project

7/25/2019 UIUC CEE 575 Final Project

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CEE 575 Fracture Mechanics

Final Project

Travis FillmoreMunci InonuMaozhe Gong

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Figure 1: A semi-innite strip with crack

Question 1

Recall the expression of the energy release rate G and of the mode I stressintensity factor K I as a function of the prescribed displacement , the mid-height h, the Youngs modulus E and the Poissons ratio . Assume Plane strain

conditions.For 2D plane strain, zz = 0 and

yy = E

(1 + )(1 2 ) [(1 )yy + xx ]

zz = (xx + yy ).

For this problem, xx = xy = 0 and xx = xy = 0. Substituted into the above equations this gives

yy = E

(1 + )(1 2 ) [(1 )yy ]

zz = (yy ).

Strain energy U = 12 = 12 yy yy . Substituting in yy (E, , yy ) and yy =

2 h yields

U = 18

E (1 )2

(1 + )(1 2 )h2.

Strain energy U is related to energy release rate G by 2Uhbda = Gbda which yields G = 2Uh.Substituting in values for U produces

G( ,h ,E, ) = 2

4E (1 )

(1 + )(1 2 )h.

The energy release rate G is related to the rst mode stress intensity factor K I by G = K 2I

E =K 2I

E1 2

. This means that the stress intensity factor can be expressed in terms of known variables by

K I ( ,h ,E, ) = 2

E (1 + ) (1 2 )h

.

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Question 2

Deformed conguration

The deformed congurations are analyzed by Abaqus CAE using the CAE le provided, the resultsare shown in Figure 2 and Figure 3.

Figure 2: Deformed conguration

Figure 3: Deformed conguration close to the crack tip

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Vertical reaction force on the top surface

The result is 38031 .8N . This result is a summation of all nodal reaction forces on the top surface,the nodal reaction forces are calculated by Abaqus CAE.

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Question 3

Calculate the stress intensity factor in mode I , K I , from the displacements of the node close to the crack tip. Compare with the theoretical value.

The approximation of K I is

K I C d 2u B

12

u C n (1)with

C = E 28 1 2

(2)

C = 3000MP a 2

8 1 0.32 (3)

C = 1032.95MP a (4)

From the CAE results,

u B = 3 .5981 102 mm (5)

u C = 7 .0977 102 mm (6)

d = 0 .543mm/ 4 = 0.1358mm (7)

Substitute the calculated parameters to equation (1) to get K I ,

K I = 1032.95 0.1358 23.5981 10

2

12 7.0977 10

2 MP a m (8)

K I = 198.36 MP a mm (9)To nd the theoretical value, K I solution from question 1 is used,

K I = 2

E (1 + ) (1 2 )h

(10)

K I = 0.4

23000

(1 + 0 .3) (1 2 0.3)10MP a m (11)

K I = 230.77 M P a mm (12)The theoretical value for K I is 14% higher than the computational value from ABAQUS.

This dicrepancy likely comes from discretization error in the model, which leads to too low a valueu B compared to u A . We expect that as the mesh is rened the computational value for K I will increase and approach the theoretical K I .

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Question 4

Evaluate the energy release rate G using the J -integral

The energy release rate G using the J -integral method is 13 .03MN/mm , the result is calculatedby Abaqus CAE.

Compare with the Griffith-Irwin estimate

The Griffith-Irwin estimate isK 2I E

= K 2I

E 1

2

.

The numerical value is= 11 .94 MN/mm .

Compare with the theoretical solution provided in Question 1.

In question 1, G is derived in this equation,

G = 2

4E (1 )

(1 + )(1 2 )h (13)

then,

G = 0.42

43000(1 0.3)

(1 + 0 .3)(1 2 0.3)10 (14)

G = 16.15 MN/mm (15)

The Griffith-Irwin estimate is 26% lower than the theoretical solution while the J-integralis 19% lower than the theoretical solution. In this problem J-integral is a better estimate of thetheoretical energy release rate G. This is probably because the path independence property of theJ-integral relies less on the renement of the mesh than the point-displacement method of calcu-lating K I in question 3.

The point-displacement method relies on the models ability to accurately model the 1 r de-pendence, which relies on the renement of the mesh near the crack tip. However, the J-integralcan be calculated far away from the crack tip, implying that it relies less on the accuracy extremelyclose to the crack tip and the renement there.

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Question 5

Run several simulations corresponding to different values of between 0 and0.6mm . Plot G = f () .

The results are shown in Figure 4.

Figure 4: G = f ()

Comment on the scaling of the energy release rate and compare to the analyticalprediction

According to the results, the analytical solution is larger than the FEM solution, with an almostconstant error of 20%. This error must come from something that does not change signicantlyfor changing . Possible reasons include the analytical solutions assumption that the plate isinnite is not met and the FEM discretization error.

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Question 6

The LEFM local solution predicts a scaling of the stresses in 1 r . Plot 22 ( = 0, r )as a function of the distance r to the crack tip.

Plot 22 ( = 0, r ) as a function of the distance r to the crack tip

In order to obtain the 22 results from ABAQUS, a feature called path was used. A path wascreated along = 0 starting at the crack tip and extending until the end of the domain. Then inXYData manager the path option was used to plot the 22 versus radius r as shown in Figure 5.

Figure 5: 22 versus r for the FEM and theoretical solutions

Very close to the crack tip there is some large error over the domain in which the quarternode quadrilateral element is used. This oscillation is considered discretization error and has nobearing on the K-dominance region. As the radius r increases, the theoretical solution decreaseswhile the FEM solution remains fairly constant (as one would expect in a plate with a patch-test

type problem such as this one). Depending on the tolerance = Theory22

F EM 22

Theory22, the K-dominant

zone can be determined.

Estimate the region of the K-dominance.

The error between the theoretical and the FEM solutions for 22 is shown in Figure 5.

The error is chosen to be = 0 .5. The radius r = 3 mm at that tolerance. Thus theK-dominant zone is estimated to be a circle with the crack tip at its center and a radius of 3 mm .

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Figure 6: Error versus r

Question 7

The analytical solution found in Question 1, applies when hL


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Figure 7: hL versus G for constant aL = 0 .4

Figure 8: aL versus G for constant hL = 0 .1

values of a or very high values.

As with the lack of theoretical dependency G(a), the fem solution also lacks any dependencyexcept for possible discretization error when the crack is very short or very long.

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Question 8

We call 2 D = {2 D11 (r, ), 2 D22 (r, ), 2 D12 (r, )} the LEFM solution given in slide 24 onChapter 2. We call =

{11 (r, ), 22 (r, ), 12 (r, )

}T the actual full eld solution

provided by the nite element simulation in Abaqus. We dene the error as:

e = | 2 D11 11

11 |2 + | 2 D22 22

22 |2 + | 2 D12 12

12 |2 . Show the contour of the error e for theinitial conguration and determine the zone of K-dominance.

Since this question deals with the error between the FEM solution found in ABAQUS and theK-eld analytical solution in the notes, a qualitative comparison may give some insight as to whatto expect. The contour plots of the analytical solutions were made using Matlab and screenshotswere taken of the Abaqus solutions. The scaling in each gure is approximately the same. The S11solution is shown in Figure 9.

(a) Exact (b) ABAQUS

Figure 9: 11 solution from analytical and the fem results

The analytical K-eld shape takes a dumbell shape; however, abaqus only supplies the righthalf of the dumbell well. The left half rounds off near the crack. Thus, large error in s11 willprobably be found near = . The S12 solution is shown in Figure 10.

The two plots bear some resemblance. The region to the left of the crack tip has an increasedmagnitude in the same manner as the analytical solution. However, the other regions little incommon, with the abaqus solution failing to reach the comparable magnitudes. It is expected that

this discrepancy in behavior will result in large errors close to the crack tip. Finally Figure 11 isinspected.

Similar to S11, this plot looks pretty good everywhere except to the left of the crack tip. Ithas too small a magnitude in this region.

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In order to compare the analytical results with the fem results, the path feature of abaquswas utilized. Figure 11 shows the mesh lines spreading radially from the crack tip. These lineswere used to dene paths. The entire circular region around the crack tip that uses those linesfor element creation was considered. Paths were created along each radial line and its angle considered. Abaqus calculates solution values at each node along the line at a certain radius r .The XY Data Manager feature is used to calculate 11 , 12 , and 22 for every and r.

These values are compared against analytical values 2 D11 , 2 D12 , and 2 D22 to calculate e asspecied in the question prompt. There are several problems with this analysis however. At cer-tain values = {0, , }, the

2 D solutions are 0. This means that the error |2 D

2 D |2 will be

innite. This makes determining the error in such regions using this method unreliable. In order tocircumvent these shortcomings in these regions, ( 2 D ) is scaled by the next largest magnitudeof value 2 D .

Also, the denition of error e is made difficult by the fem solution values close to the cracktip. At the crack tip, the fem has a nite magnitude. However the analytical solution has aninnite solution (scales with 1 r ). This means that the closer to the crack tip, the higher the error

will be. Also, an unusual phenomenon is observed a little farther away as shown in Figure 12.

Figure 12: Error versus radius r for = 0

The error decreases quickly as expected until at a point (in this case r = 0 .3 mm) the errorincreases quickly. The reason for this increase in error is unknown. It may result from discretiza-tion/modeling error (this jump always occurs within the quarter-point element) in the fem modelor be the actual solution to the problem. After this initial increase in error the error drops againand then gradually increases until K-dominance is lost.

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As a result of the plot above, several contours will be shown. First, the small r value atwhich the error is very small will be shown. Then the error at the terminally increasing zone willbe displayed. The resulting contours for e are shown in Figure 13. The circle outside the errorcontours denotes the area over which the radial elements are dened, while the x and y axis aredened in millimeters.

Figure 13: Contour of several error tolerances including the inner error contour

Above and below the crack tip (located at (0 , 0)) the error is large. To the right of the cracktip the error is at its lowest, while the left of the crack tip has interesting behavior. As one movesfrom above the crack tip to the left the error decreases and then increases dramatically to 0. Partof this dramatic increase comes from the result that all analytical stresses have a magnitude of 0here. This means that according to any of the proposed denitions for error, the error here will bescaled by 0 and innite.

Also, these contours look very similar to the contours of the 11 solution in Figure 9. This

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hearkens back to the qualitative comparison between the analytical and the fem solutions. If anydirections of stress do not match up well, then they will result in signicant error. Thus, since 22matches poorly to the left of the crack tip, error will be high there. While, 12 does not matchthe solution very well close to the crack tip, the error from this coincides with (and may resultfrom) discretization error. However, above and below the crack tip 22 should continue to have anelevated magnitude, which the fem does not. This results in high error between the two above andbelow the crack tip. The unusual branches extending to upper left and lower left from the cracktip in Figure 13 likely result from interaction in the 22 and 11 errors.

In order to estimate the zone of K-dominance one must choose an error tolerance parameter as well as determine whether the jump in error shown in Figure 12 results from discretization

error or is the correct solution for the problem. It will be assumed that the jump results from adiscretization error. The error tolerance will be taken as = 0 .5.

Since the high error above and below the crack tip is due to coarseness of mesh, it will beassumed that the solution in the limit of renement will have a similar error as what is given tothe right of the crack tip. The same will be said for the 0 magnitude solutions to the left of the

crack tip. Thus, the K-dominant zone will be a circle with the same radius as epsilon = 0 .5 when = 0. This is shown in Figure 14 displayed among the fem elements.

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Figure 14: Contour of the K-dominant zone

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uiuc cee 575 final project

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