Determination Of An Elastic Potential For A Thin Rubber Sheet · 2007-09-20 · Final Project...

EN 227 Advanced ElasticityEN 227 Advanced ElasticityDivision of Engineering, Brown UniversityDivision of Engineering, Brown University

Final Project PresentationFinal Project PresentationDecember 3, 2003December 3, 2003

Determination Of An Elastic Potential For A Thin Rubber Sheet

EN 227 Final ProjectDonald Ward & Brian Burke

12/3/03



ØØOverview Overview

•Introduction•Testing Procedure•Experimental Setup

•Load application•Deformation characterization

•Material Response•Material symmetry•Incompressibility•Elastic Potential



ØØTesting Procedure Testing Procedure •Examine microstructure for clues on material response (e.g. preferential direction, fiber direction, porosity, etc.)

•Uniaxial Test

•Examine material response in several directions looking for Material Symmetry Group

•Biaxial Test

•Measure all three principle stretches

•Known tractions

•Compare results to known elastic potentials



ØØExperimental Setup Experimental Setup

•Self-centering•Pulleys help align the system and ensure that stresses do not contain any shear components•No support for moment loading•Clip and pin minimize slip

Biaxial System

2

1

Pulleys

Load lines

Clip



Computer Imaging of Deformation

• Take digital image of un-deformed configuration.

• Convert image into text file with an intensity value for each pixel.

• Search the values for high intensity sources (black grid lines)

• Store these pixels



95

145

195

245

295

345

70 90 110 130 150 170 190 210 230 25095

145

195

245

295

345

110 130 150 170 190 210

Center(high pixelcount)

5 0 100 150 200 250 300 350

50

100

150

200

250

•From the saved pixels search for those pixels that have a considerable number of neighbors in the 1 and 2 directions•The pixel positions are then averaged to determine the centersfor the corners •This process is done for the deformed and un-deformed images•The images are correlated through the center which contains many more intense pixels than the rest



•Micrographs of material show a randomdistribution of material implying isotropyin the plane•A second observation made is that the material appears somewhat porous this possible implies compressibility but other results show that the material is more

closely represented asincompressible

ØØMicrostructure Microstructure

100 mm

50 mm 20 mm



ØØMaterial Symmetry Group Material Symmetry Group

• Check angular dependence of load response

•Three samples cut from same sheet

•Response measured under uni-axial loading

•Arrows indicate direction of load



ØØMaterial Symmetry Group (cont.)Material Symmetry Group (cont.)

e1

e2 1 2 3

A

B

C

Data examined along horizontal and vertical segments




3 3.5 4 4.52

2.2

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8Sample-1;Uni-Axial Load:100g; y 2 vs. x 2

m m

mm

Line 1Line 2Line 3

•Good agreement in principle directions

•No off-diagnol components in deformation gradient

1.8 2 2.2 2.4 2.6 2.8 31.8

1.9

2

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

Sample-1;Uni-Axial Load:100g; y1 vs. x

1

mm

mm

Line ALine BLine C

3 3.5 4 4.51.8

1.9

2

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

Sample-1;Uni-Axial Load:100g; y1 vs. x

2

m m

mm

Line 1Line 2Line 3

1.8 2 2.2 2.4 2.6 2.8 32

2.2

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8Sample-1;Uni-Axial Load:100g; y2 vs. x 1

m m

mm

Line ALine BLine C

y1 vs. x1 y1 vs. x2

y2 vs. x1 y2 vs. x2




•Response along each line segment averaged for each sample

•The average response for each material direction plotted on same axis

•Slopes matched!

•Material seems isotropic in plane

3 3.2 3.4 3.6 3.8 4 4.2 4.4 4.6 4.8 52

2.5

3

3.5

4

4.5

Average responses of all samples, Load:100; y2 vs. x

2

mm

mm

Sample 1Sample 2Sample 3

λ1

1



ØØBiaxial Experiment Biaxial Experiment –– Typical ResultsTypical Results

2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.22.2

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

4.2Sample-1;Bi-Axial Load:2k; y1 vs. x1

m m

mm

Line ALine BLine CAvg. Res.

3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.42.2

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4


m m

mm

Line 1Line 2Line 3

2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.24.8

5

5.2

5.4

5.6

5.8

6

6.2

6.4


m m

mm

Line ALine BLine C

3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.44.8

5

5.2

5.4

5.6

5.8

6

6.2

6.4


m m

mm

Line 1Line 2Line 3Avg. Res.

y1 vs. x1 y1 vs. x2

y2 vs. x1 y2 vs. x2



ØØOut of Plane StretchOut of Plane Stretch

.769

.803

.803

.934

λ3(Average)

1.1579

1.0493

0.9536

1.0259

det(F)

.0075.0075.008.0075None

.0055

.006

.006

.007

Location 4 (in.)

.0055.0065.0062

.006.0065.0061 ½

.006.0065.0061

.007.0075.007½

Location 3 (in.)

Location 2 (in.)

Location 1 (in.)

Load(kg)

•Thickness measured with modified calipers

•Results summarized below

•Appears Incompressible



Computing Potential•Pick a potential using “educated guess”: incompressibleMooney-Rivlin potential.

•Used 2kg case with Mathematica to solve for unknown constants using data extracted from biaxial test



Checking potential•Once the parameters are determined we can compare the potential with the uniaxial test data•First solve for the nominal stress in the direction of appliedload vs. lambda 1

lambda1 lambda2 lambda3 P11 (Predicted) P22(Predicted) P11(measured) P22(measured) % Error % Error1.115 0 .985 0.934 3.841 2.532 2.190 1.280 42.98% 49.46%1.168 1 .017 1.218 4.073 2.613 3.170 2.260 22.17% 13.50%1.218 1 .073 0.803 4.767 3.358 4.150 3.250 12.95% 3.22%1.284 1 .171 0.769 5.260 4.200 5.260 4.200 0.01% 0.01%



Uniaxial ResponseP 2 2 v s l a m2 O v e r a R a n g e o f l a m 1

-10

- 5

0

5

10

15

20

25

0 0. 5 1 1.5 2 2.5 3

l a m 2

l a m 1 = 0 . 5

l a m 1 = 1 . 0

l a m 1 = 1 . 5

l a m 1 = 2 . 0

l a m 1 = 2 . 5



Concerns About the MethodWhile this method sounded good to start with there are a fewissues that cause a little concern1. The set up does not self-center perfectly2. The method for measuring thickness change has limitations3. The fit parameters are very sensitive to the initial tractions

which are not easily measured(thickness and contact lengthwere not determined perfectly)

4. Some of the forces are disregarded to simplify calculations5. Incompressibility is assumed but fails at large loads



Conclusions:

•The material responds in an isotropic manor in the plane•At low stretch values the material appears incompressiblethis fails with higher applied tractions•A Mooney-Rivlin Potential fits the data pretty well but driftsfrom observations at low stretch values•Over all the data fits well but with more development the experiment would have more success



( ) ( ) ( ) ( )1 2 1 2, 3 1 32

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α α= − + − −

( )( )1 1 (1 )TP pF I F G Fµ α α µ α−= − + + − − −% % % %%

3.486.501.049

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Determination Of An Elastic Potential For A Thin Rubber Sheet · 2007-09-20 · Final Project...

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