Effect of Material Non-Linearity on the Flexural Response of Fiber Reinforced Concrete Chote Soranakom, Barzin Mobasher, Saurabh Bansal Dept. of Civil & Environmental Engineering Arizona State University 8th International Symposium on Brittle Matrix Composites (BMC8) in Warsaw, October 22-25, 2006

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Need for unified representation of fiber reinforced cement based composites 

Strain softening behavior (mostly discrete fiber systems) – SFRC, PFRC, GFRC, SIMCON, SIFCON



Strain hardening behavior (mostly continuous fiber systems) – Textile reinforced cement (TRC), Ferrocement (FRC), ECC,



Similarity in material constitutive models – Tensile strength << Compressive strength – Compressive response ~ some ductility – Tensile response ~ various levels of ductility

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Typical Material Response

s

SIFCON

TRC

s ECC

TRC

SFRC

SIFCON Paste

ECC SFRC

Paste Compression Model

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Tension Model

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e

Class of Strain Hardening Cement Based Composites20 AR Glass Fabric

Stress, MPa

16 E-Glass Fabric

GFRC Vf =5%

12 8

PE Fabric

4 ECC- PVA 0

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0

0.01

0.02 0.03 Strain, mm/mm

0.04

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0.05

Reinforcing Mechanisms

Bond for Polyethylene Mesh

Micromechanical Modeling

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Various stages of cracking

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Stiffness degradation and Crack Spacing Relationship 1000

Tangent Stiffness, MPa

Glass Fabrics 100

10

1 80

Polyethylene Fabric

60 40 20 Crack Spacing, mm

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0

Flexural Response of Cement Composites

500

0

Displacement, in 0.4 0.8

1.2

400 300

Load, lbs

Load, N

80

200

40

Actuator Displacement LVDT Displacement

100 0

0

10

20

30

0

Displacement, mm

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Comparison of Tensile and Flexural Results for AR Glass Fabric Composites

Equivalent tensile strength obtained from flexural results significantly over-estimates the actual tensile data

Flexural Strength, MPa

18

16

14

12

10

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0

2

4

Tensile Strength, MPa

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6

Parametric Strain hardening Material Case Et0  t0



Et1 Et0

1 

 t1  t0

Stress Et1 etu = arbitrary terminating tensile strain

Et0 ec0 et0

et1

e

Strain

Ec0 ,  ecu = arbitrary terminating tensile strain

 t2  t0 E   c0 Et0    tbot  t0

2 

f c’

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c = Ec0( c  c2 )

Stress and Strain distribution across the specimen thickness  z  d  kd   t 0 z  f c1  Et 0 t 0 ( k  1 )2 d 2 ectop

fc(y) Fc1

kd I

et0 II

et1 III

et2

0 

yc

ht1 ht2 ht3

yt1

ft1

yt2 ft2

yt3

Ft1 Ft2 Ft3

IV

etbot

b

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 1      ft   1    1     2  Et0 t 0  1   2  0

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0   1 1    1

1     2 2    

Moment-Curvature Diagram    

Incrementally impose 0 < t < tu (or 0 <  < tu) S F = 0, determine Neutral axis, k kd M = S Ciyci+ S Tiyti and f=c/kd Fc1  b  f c1  y dy 0 Normalization M’=M/M0 and f’=f/f0 kd

yc1 

b Fc1



0

f c1  y  ydy

ec k

0 < et < etu f

C1

M

T1 d

T2 T3

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f

Closed Form Solutions- strain Hardening material

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fi  fi 'f0

f0 

2 t 0 d

M i =M i'M 0

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1 M 0  bd 2 Et 0 t 0 6

Load Deflection Response  

Moment area method Crack localization rules

S

Moment

P/2

P/2

S

Non-Localized Zone

S

M0

Localized Zone

fj-1,Mj-1)

L M

Mmax

M0 fj,Mj)

Mfail Non-Localized Zone

P Localized Zone

f f0

Loading Unloading

cS S

Axis of Symmetry

S/2 Curvature

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Simulation of Glass Fabric Reinforced Cement 

Compare – Uniaxial tension test (10 x 65 x 200 mm) – Four point bending test (10 x 75 x 356 mm) Span= 254 mm

 

Cement paste (w/c = 0.4) Reinforcement – AR-glass (Saint-Gobain Technical Fabrics) – Grid size 25.4x25.4 mm, 2 yarns in longitudinal and transverse directions, VLong = VTrans = 0.70 % – One yarn contains 1579 filaments of 19 mm in diameter



Aging – 0, 14 and 28 days

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Ec 0 

23 mL3 54 bh3



Material Parameters

Tension material parameters – Fit the trilinear tension model to the uniaxial tensile response. – Parameters : et0 ,Et0, h, a1, a2 and btu – Ec0 is back calculated from the initial load deflection response of the bending test; then g = Ec0/Et0 – Assume fc’ = 45 MPa; then ec0=2fc’/Ec0 and =1/(2ec0) – Assume ecu = 0.012, and lcu = ecu/et0 – Obtain all compression parameters : g,  and lcu

23 mL3 Ec 0  54 bh3

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Typical cement based material composite model

10 0 -0.005

0

0.005 -10

-30 -40 -50

0.015

Strain (mm/mm) Stress (MPa)

-20

0.01

et0=0.001, Et0=5000 MPa, g=5, =200, a1=10, a2=50 h=0.5

-60

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Prediction of flexural response using tensile data

Equivalent Stress (MPa)

15

12

9

6 0 day (exp.) 0 day (model) 28 days (exp.) 28 days (model)

3

0

0

5

10

15

20

25

Deflection (mm)

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30

Parametric Study







Direct use of uniaxial tensile response in simulation under-predicts flexural response. To simulate size effect, the tension model is scaled up such that the predicted response matches the experimental response. Scaling effects – Increase first cracking strain, et0 -> increase M and f – Increase Young modulus, Et0 -> increase only M

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Effect of Scaling First Crack Strain, t0 

Both M and f in addition to load and deflection are affected

20000

Moment, N-mm

Tensile Stress, MPa

Load, N

0.001 0.002 0.003 Curvature, 1/mm

0 0

Moment curvature diagram

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07

0 0

5

0.004 0.008 0.012 0.016 0.02 Tensile Strain, mm/mm

6 .00 =0

100

4000 0 0

200

.00 =0 e t0

8000

e t0

7

12000

300

8

08

8 .0 0 =0 7 e t0 .0 0 =0 e t0 06 .0 =0 5 e t0 .0 0 =0 e t0

16000

.0 =0 e t0

.0 =0 e t0

.00 =0 e t0

06 0.0 5 .00 =0 e t0

2

e t0=

4

0 .0 =0 e t0

400

6

M = M’M0

M0=1/6(bd2Eet0)

f = f’f0

f0 =2et0/d

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4 8 12 Deflection, mm

16

Effect of Scaling Tensile Modulus, Et0 

Only M, and Load are affected 20000 16000

2

300 12000 8000

Et0=8000 Et0=7000 Et0=6000 Et0=5000

4000

0 0

0.004 0.008 0.012 0.016 0.02 Tensile Strain, mm/mm

0 0

Load, N

Moment, N-mm

4

400

t0

=8 E 000 =7 E 000 =6 E t 0 00 0 =5 E t0 t0

Tensile Stress, MPa

6

0 00

Et0=8000 Et0=7000 Et0=6000

200

Et0=5000

100

0 0.001 0.002 0.003 Curvature, 1/mm

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0

4 8 12 Deflection, mm

16

212cccfE

'00

Glass Fabric composites (Unaged)

400

s=24.5N s=35 .1N s=52 .4N

300 Load, N

Tensile Stress, MPa

6

4

2

Experiment Fitted model Modified 1.15et0

200

100

Modified 1.53et0&0.77Et0

0 0

0.02 0.04 0.06 0.08 Tensile Strain, mm/mm

Experiment Fitted model Modified 1.15et0 Modified 1.53et0&0.77Et0

0.1

0 0

10 20 30 Mid Span Deflection, mm n

 y  y  i

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s

pred

i 1

n

2

Glass Fabric composites (Aged 14 days)

Experiment Fitted model Modified 1.34et0

400

4

2

0 0

s= 20

Load, N

300

200

100

0.02 0.04 0.06 0.08 Tensile Strain, mm/mm

0.1

. 5N

Modified 1.27et0&1.17Et0

Modified 1.27et0&1.17Et0

0 0

s=

. 31

5N

.5N 83 s=

Tensile Stress, MPa

6

Experiment Fitted model Modified 1.34et0

10 20 30 Mid Span Deflection, mm

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Glass Fabric composites (Aged 28 days)

Experiment Fitted model Modified 1.45et0 Modified 1.43et0&1.02Et0

Load, N

4

Modified 1.43et0&1.02Et0

300

200

.5 73

100

. 5N 21 s= .9N 21 s=

2

s=

Tensile Stress, MPa

6

Experiment Fitted model Modified 1.45et0

400

N

0 0

0.02 0.04 0.06 0.08 Tensile Strain, mm/mm

0.1

0 0

10 20 30 Mid Span Deflection, mm

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Material Model for Strain Softening Response Intrinsic parameters (ecr and E) Non-dimensional parameters (w,m,lcu,btu)

 

sc

wecrE

st Tension Model

Compression Model

mecrE

E wecr

E lcuecr

ec

ecr

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btuecr

et

Closed form solutions for Elastic-Quasi Plastic material model 2, 

Variable, top compressive strain 0 < ec < ecu (or 0 < l < lcu)  1 2  2  k  2    2 (  1)  1  2    2  2 (   )  2  1

for 0    1 for 1     for   

   2k  (2 3  3 2  3  2)k 2 M '( , k ,  ,  )    3 (2k  1) 2    (3 2 +3 2  3   3  2)k 2   3 (2k  1) 2  

f '( , k ,  ,  ) 

for

0  1

for

1   

for

 

 2k

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for   0

M = M’M0

M0=1/6(bd2Eet0)

f = f’f0

f0 =2et0/d

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Simplified Closed Form Solutions (fix m and/or w ) Stage 1: Elastic compression & Tension, 0 < l< 1

1 2



 

M’ ’

Stage 2: Elastic compression, Bilinear Tension 1 < l< w Elastic compression, bilinear tension

m=1

m=0.75

m=0.5

m=0.25

m=0.1

2   2 -1+2  2 

2  +2  1

3 2 +3  1

  +1

 2 +  1

 5 +  4

M

(2 3 +2+3 2 -3 ) 2 2 6   3

3  1  +1

3 2 -4 +2 (2  1) 2

15 2 -26 +13 2(5 -4) 2



 2

 2 +2  1 4

2 2 +  1 2

5 2 +  4 2



2

2

6( 3 +2 2 - +1) 3 2 -2 +1 (2  1)(2 2  3  1) 2 2

2 2 +3  1 6

 1 2

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2

2

Simplified Closed Form Solutions (continue) Stage 3: w < l< lcu

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Effect of compression to Tension strength, w 3

0.4

0.3

w=1 b = 100 mm d = 100 mm E = 25000 MPa ecr = 0.00015 ecu = 0.0035 m= 1.00 w=4

0.2

Normalized Moment,M'

Neutral axis depth ratio, k

0.5

w=7 w=4

2

w=10

w=1 1

w=7 w=10 0.1 0 5 10 15 20 25 Normalized top compressive strain, l

0 0

20 40 60 80 Normalized Cuvature, f'

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Effect of Post-Peak Tension strength, m

3

b = 100 mm d = 100 mm E = 25000 MPa ecr = 0.00015 ecu = 0.0035 w = 10

0.4 0.3 0.2

m=1.00 0.1 m=0.01

m=0.33

0

Normalized Moment, M'

Neutral axis depth ratio, k

0.5

m=1.00 2 m=0.67

1 m=0.33

m=0.67

0 5 10 15 20 25 Normalized top compressive strain, l

m=0.01 0 0

20 40 60 80 Normalized Cuvature, f'

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Conclusions

 

 



Closed form solution provides a direct and fastest way to determine moment curvature diagram Moment curvature diagram can be used with area moment method and crack localization rule to predict load deflection response of four point bending test The direct use of uniaxial tensile response under predict the flexural response The discrepancy between the uniaxial tension test and flexural beam test can be addressed by two scaling parameters, first cracking strain et0 and tensile Young modulus, Et0 Simplified model for closed form solutions of strain softening material has direct applications fr design and analysis of FRC structures.

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Acknowledgements



US National Science Foundation, Award # 032466903, Program Manager, P. Balaguru.

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Parametric Study-Homogenized Material fc

ft ecu

ec0

ec et0

for  t   t0 for  t0   t   tu for  t   tu

etu

fc ’

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Tension parameters  t Et 0  Et1 = hEt0  t = f t   tn et1 = a1et0 0  et2 = a2et0   2    Compression  f c'  c   c  parameters  c     c0   c0    Ec0 = gEt0 0 fc = Ec0(ec-kec2)

2

   

Material parameters et0=0.0001, ft =2.0, Et0 = 20 GPa a =10, n=0.2, etu=0.001 fc’=-30 MPa, ec0=-0.002, ecu =-0.003

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for  c   cu for  c >  cu

Comparison of simulation and experiments 16

MOR, MPa

14 12

MOR = 1.70 * sut+ 6.47 (experiment) R2 = 0.78

10 8 6 3

MOR = 2.63*sut - 1.48 R2 = 0.98

(simulation)

4 5 Tensile Strength, MPa

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6

Parameter Extraction for correlation of tensile and flexural response

7

15

6

Stress (MPa)

5

Equivalent Stress (MPa)

0 days (exp.) 0 days (model) 0 day (1.27y*model) 28 days (exp) 28 days (model) 28 day (1.55y*model)

4 3 2

12

9

6 unaged (exp.) unaged (fitted model) unaged (modified 1.25et0) aged (exp.) aged (fitted model) aged (modified 1.50et0)

3

1 0

0

0

0.02

0.04

0.06

Strain (mm/mm)

0.08

0.1

0

5

10

15

20

Deflection (mm)

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25

30