Development of Design Guides for Strain Hardening and Strain Softening Fiber Reinforced Concrete Materials Chote Soranakom, Masoud Yekani-Fard and Barzin Mobasher Dept. of Civil & Environmental Engineering Arizona State University Session Honoring Antoine E. Naaman ACI Conventions Spring 2008 March 31, 2008, Los Angeles, USA
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Outline
Overview of fiber reinforced concrete (FRC) Areas of applications Strain softening FRC – Material models – Design guideline – Example
Strain hardening FRC – Material models – Design guideline – Example
Conclusions
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Precast panels
FRC can be used in precast industries – – –
Better fracture toughness Higher impact resistance Lighter panel
Elevated Slab made of SFRC and minimum steel in column strip to prevent progressive collapse
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Drainage trough and cover
Glass fiber reinforced concrete (GFRC) can be used to create small structural applications such as light weight drainage trough and covers
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Material models (Strain softening)
Compression model
Tension model
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Material Model for strain softening
Two material parameters (ecr ,E) and four normalized parameters (w, m, lcu, btu), independent variable l.
sc
st E
scr = ecrE E
sp = mecrE
ec ecy = wecr
ec = lecr
ecu = lcuecr
Compression model
et ecr
etu=btuecr
Tension model 1 M cr = bd 2 E cr 6
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2 cr cr = d
Stress and Strain Distribution ec=lecr
ec=lec
ec=lecr
r
kd
wecr
kd ecr
d
d
kd d
ecr et
et
sc 1
sc1
yc1
st1
0
sc1 Fc2
Fc1 yc1
Fc1
yt1 Ft1
et
yt2 st1 st2
yt1
y yc1 Fc1 c2 Ft1 yt1 yt2 Ft2
st1
Ft1 Ft2
1
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st2
w
Moment Curvature Diagram
Incrementally impose 0 < et < etu Strain Distribution Stress Distribution SF = 0, determine k M = SCiyci+ STiyti and f=ec/kd strain
Fc1 = b
kd
f c1 y dy
0
b yc1 = Fc1
stress
kd
0
f c1 y ydy
Moment curvature diagram
ec C1
k f
M
M
C2 T1
d
T2 T3
0 < et < etu You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)
f
Model for strain softening Stage
k
M’=M/Mcr
1
1 2
2k
2 2 2 ( 1) 1
(2 3 3 2 3 2)k 2 3 (2k 1) 2
2 2 2 ( ) 2 1
(3 2 3 3 2 3 2)k 2 3 (2k 1) 2
0
2 1
w l < lcu
1 M = M cr M ' = bd 2 E cr M ' 6
M cr =
1 bd 2 E cr 6
f’=f/fcr
2k
cr =
2 cr d
Soranakom, C., and Mobasher, B., “Closed-Form Solutions for Flexural Response of Fiber-Reinforced Concrete Beams,” Journal of Engineering Mechanics, Vol. 133, No. 8, August 2007, pp. 933-941 You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)
Effect of Softening Region Tensile strength, m 3
w = 10 ecu = 0.004 etu = 0.015
0.4 0.3 0.2
m=1.00
0.1 m=0.01
m=0.35 m=0.18
m=0.68
0
0 4 8 12 16 Normalized top compressive strain, l
M '( ) = 3 +
m=1.00
M’=2.727 Normalized Moment, M'
Neutral axis depth ratio, k
0.5
2
m=0.68
M’= 1.910
M’=1.0145
m=0.35
1
M’= 0.530 M’=0.03 0
0
m=0.18 m=0.01
20 40 Normalized Cuvature, f'
1 M cr = bd 2 E cr 6
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cr =
60
2 cr d
Calculation Example 1 2 M ult = 3 bd E cr + 6
What is the moment capacity of a fiber reinforced concrete beam? Given that: – b=4 in, d=4 in – E = 3x106 psi, scr = 300 psi, sp = 150 psi – fc’ = 4500 psi, scy ~ 0.8fc’
Calculations – m = sp/scr = 0.50 – w = scy/scr = 12 – M’∞ = 3mw/(w+m) = 1.44 (no unit) – Mcr = 1/6bd2scr = 3,200 lb-in – M∞ = M’∞Mcr = 4,600 lb-in – Mu = 0.90M∞ = 4,150 lb-in Moment capacity
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Normalized Moment at Infinity
3 Minf/Mcr
Development of Design Guides Deflection Softening-Deflection Hardening Transition
crit =
2 1
0 15 1 10 omega
3 1
0.5 mu
5 0
m = 0 – 1 and w = 5 – 15
– For w = 10, mcrit=0.355
Normalized Post Crack Tensile Strength
3 =
M ' = 3 M '
M cr =
1 bd 2 E cr 6
1 mu (required)
M '
0.5
0 15 3 10 omega
1 5 0
2 Minf/Mcr
M’∞ = 0.1 – 2.5 and w = 5 - 15
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Probability based model Frequency Q
R
Frequency
Load failure VR= Coefficient of variation of R VQ= Coefficient of variation of Q
failure
=
ln(R/Q) VR2 VQ2
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ln(Q/R)
b, number of standard deviations
LRFD Based Design guide for strain softening FRC n Rn
Applied loads < Member Resistance li Qi
Structural resistance = n Rn Theoretical strength, Rn (Mn) resistance over-capacity factor, fn (for flexure n = 0.70) Factor of safety for each load =li Factored loads = li Qi li Qi = Mu= Max(1.4 DL, 1.2 D+1.6 L + 0.5(Lr or S or R))
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ACI Based approach for Flexural Strength of strain softening FRC 3 n M n = p M cr M u
1 2 M n = 3 bd E cr + 6
M ult
f c' = 15.55+2 f c'
cy 0.85 fc' = = = 0.126866 f c' cr 6.7 f ' c
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bd 2 E cr
Design Guide for Strain softening Or, the required normalized post crack tensile strength, µ
M u = 3 p M cr M u
3. The minimum normalized post crack tensile strength For flexure For shrinkage and temperature
min_ flex = 0.40 min_ ST
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140 psi = cr
Design Guide for Strain softening, specimen Variability and Size Effect 4. The mean post crack tensile strength spm from material testing of n samples must be at least p pm = 1.34s kh where, post crack tensile strength size dependent safety factor
p = cr
for h 5in. 1.0 kh = 1.0 0.03(h 5) for 5 in.< h 24 in. 0.4 for h > 24in.
and s is the standard deviation of more than 30 samples
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Design example of strain softening Design four-span floor slab using SFRC fc’=4,000 psi
The problem can be designed as a one way slab per one foot strip
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Design example of strain softening 1. Estimate the thickness from deflection control 2. Estimate the self weight 3. Calculate factored loads
l 8 12 h= = = 4 in. 24 24
4 150 = 50 psi 12 wu = 1.2(50 15) 1.6(75) = 198 psf
wsw =
4. Calculate critical moment at first interior support 1 2 M u = Cm wu ln = 0.198 82 = 1.27 kips-ft/ft 10 ' 5. Calculate cracking tensile strength cr = 6.7 f c = 6.7 4000 = 424 psi
cy 0.85 4000 = = 8.02 cr 424 cr bh 2 424 12 42 1 = = = 1.13 ft-kip 6 6 12,000
6. Calculate normalized compressive yield strain = 7. Calculate cracking moment M cr
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Design example of strain softening 8. Calculate required normalized post crack tensile strength M u 1.27 8.02 = = = 0.573 3 p M cr M u 3 8.02 0.7 1.13 1.27 9. Check with the minimum normalized post crack tensile strength 9.1 For flexure
min_ flex = 0.40
9.2 For shrinkage and temperature
min_ ST =
140 psi 140 = = 0.33 cr 424
10. Thus, µ=0.573 governs the design 11. post crack tensile strength
p = cr = 0.573 424 = 243 psi
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Design example of strain softening 10. Calculate size dependent safety factor from the designed thickness of 4 in. kh = 1.0 for h 5 in.
11. If standard deviation of the material from database (n > 30) is 50 psi, the mean post crack tensile strength spm from (n < 5) samples tested must be at least
pm
p 243 = 1.34 s = 1.34(50) = 310 psi kh 1.0
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Comparison With ASTM 1000
2
Experiment Present Model
Specimen Depth, (in)
Flexural Load, lb
800
600
400
L-056 : 9.5 lb/yd3 FibraShield sample 1 age: 14 days
200
Stress Distribution Softening Zone L056-01
1
0 0
400
800
1200
1600
Stress (psi)
-1 ARS Method, LE material ASU Method, Elastic Softening
0
0
0.01
0.02 0.03 Deflection, in
0.04
0.05
-2
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Conclusions
A design procedures based on the ultimate strength design concept and LRFD Method is proposed. The simplified equation is proposed to calculate the moment capacity of fiber reinforced concrete under flexural loading and tensile failure conditions. The present approach is fundamentally superior to the residual strength method proposed by ASTM.
M ult
f c' = 15.55+2 f c'
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bd 2 E cr
Acknowledgements
The NSF (National Science Foundation), program 0324669-03, Drs. P. Balaguru, and K. Chong. Professor Tony Naaman for being a great teacher, mentor, a role model.
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