Crack Pattern Development
description
Transcript of Crack Pattern Development
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Rigid Pavement Design Course
Crack Pattern Development
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Rigid Pavement Design Course
CRC Pavement
Vetter, C.P. 1933
• Reinforced Concrete
Drying Shrinkage
Temperature Drop
Consider a unit Length (L) between cracks
a. is restrained by the reinforcement
b. Causes tension in concrete & compression in the steel.
c. Bond stress between steel & concrete and the concrete & subgrade
shrε
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Rigid Pavement Design Course
(1) Bond stress in the vicinity of crack
(2) Compression in steel and tension in the concrete increases until steel = concrete. In this region there
is no bond slip or stress.
d. Subsequent crack form in concrete when bond stress exceeds the concrete tensile strength.
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Free Edge
‘t’
L
Longitudinal Joint
Traverse Crack
CL
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Rigid Pavement Design Course
Crack
Asfs Asfsc
Acft
Section XX Section YY
Forces Acting on CRC Pavement Section
Probable Strain Distribution Adjacent to a Crack
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Rigid Pavement Design Course
Extensive bond slip
Crack
L/2=nSmin
Good bond
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Rigid Pavement Design Course
Unrestrained shrinkage strain
Concrete strain
Crack width: equation 2.10a
Steel strain
Smin or (L-2x)/2
Co
mp
ress
ion
Ten
sio
nS
trai
n
Crack width: equation 2.10b
rε
cε
sε
sΔε
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Rigid Pavement Design Course
Concrete Stress Steel stress
Stressed, full restraint
c.g. of bond
h
b/2cCondition of no stress
L
cw
Stresses and Strains in Fully Restrained, Cracked Reinforced Concrete for Decreasing Temperature
ss Aφ
tφcfAssφA
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Rigid Pavement Design Course
Assumptions of Vetter Analysis
1. Volumetric ‘s are uniformly distributed.
2. Compatibility exists in bonded region.
3. Total bond force=Total Tensile Force=
Total change in the steel stress
4. Total length of steel will remain unchanged. Total elongation = Total shortening
5. Equilibrium exists between forces at crack & the forces in the fully bonded region.
• In partially bonded region; compatibility of deformation does not exist.
• Crack width results from relative displacement
between the steel and the concrete.
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Rigid Pavement Design Course
Stress Distribution Between Cracks Subject to Shrinkage
LC of Crack
u
xL
Bond StressBond Stress
Tension
b) Concrete Stresses
a) Steel Stresses
Compression
TensionTension
x1
ftz
fsz
fsz
c) Bond Stresses
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Rigid Pavement Design Course
cs εε
c
tz
s
sz
E
fz
E
f
tzssz nfzEf
szszstzco ffAfAu(x)
sz
szsz
f
ffxL
(1) Center of crack spacing
(2) Bond Force = Concrete tensile force = Change in steel force
(3) Total length of steel bars remain unchanged total shortening= total elongation
s
sz
szszs
sz
szszs
sz
E
f
ff2
x
E
f
ff2
x
E
fx
2
L 22
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Rigid Pavement Design Course
Total Shortening = Total Elongation
Note
szf
szf
szf
2
x
sEsz
f
szfszfsz
f
2
x
sEsz
f
szfx
2
L
sE
1
sEsz
fx
2
Ldx
1x
0 sE
f(x)Δs
x1x
szff(x)
szfszfszf
1x
1sz
1
sz x2
f
2
x
x
ff(x)
x1
0
2
s
sz
s
sz1
s
sz
E
fx
E
f
2
L
2
x
E
fΔs
szsz
sz
s
szsz
s ff
f
2
x
E
ffx
2
L
E
1
2xLfff
ffx sz
szsz
szsz22
szszszsz f2xfLfxxf
Lf
ffx
sz
szsz
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Rigid Pavement Design Course
uQ
f
pu
fA
u
fAff
u
Ax tztzstzc
szszs
ooo
sc Ap
AVol. Conc.
area bondq oo
bd
4p pqQ
p
ff
A
Aff tz
tzs
cszsz
)nf(zEup
fA
p
f
f
1
u
AL
tzs
tzstz
sz
s
o2
2
2
2
o )fuqn(zEp
f
tzc
tz2
2
c
s
E
En
;q ;n ;p ;u asL zf tz ;
For temp. drop
Both
2t
2c t
fL
p uqn(αtE f )
2t
2c c t
fL
p uqn(αtE zE f )
2t
2c tφ
fL
p uqn(αtE f )
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a) Steel Stresses
b) Concrete Stresses
Bond Stress Bond Stressy
L
c) Bond Stresses
u
Tension
C of CrackL
Stress Distribution Between Cracks Subject to Temperature Drop
sφsφ
tφf
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Rigid Pavement Design Course
cs εε
tcc
tφms
s
s tαE
ftα
E
φ
msstcstφs tαEtαEnfφ
(1) Center of crack spacing
(2) Bond Force = concrete tensile force = change in steel force
)φ(φAfAu(y) ssstφco
(3) Total length of steel bars remain unchanged total shortening= total elongation
s
s
s
ssms E
φy
2
L
2E
φφytα
2
L
s s
s s m s
φ φL y
E α t φ
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Rigid Pavement Design Course
uQ
f
pu
fA
u
fAff
u
Ax tztzstzc
szszs
ooo
sc Ap
AVol. Conc.
area bondq oo
bd
4p pqQ
p
ff
A
Aff tz
tzs
cszsz
)nf(zEup
fA
p
f
f
1
u
AL
tzs
tzstz
sz
s
o2
2
2
2
o )fuqn(zEp
f
tzc
tz2
2
c
s
E
En
;q ;n ;p ;u asL zf tz ;
For temp. drop
Both
2t
2c t
fL
p uqn(αtE f )
2t
2c c t
fL
p uqn(αtE zE f )
2t
2c tφ
fL
p uqn(αtE f )
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Rigid Pavement Design Course
Ave
rag
e C
r ack
Sp
a cin
g (
ft)
Ratio of Steel bond Area to Concrete Volume x 10-2 (in.2/in.3)
Relationship Between Steel Bond Area and Crack Spacing
2 3 4 5 6 7
20
18
16
12
8
4
0
Pavements Placed During Winter = Summer =
0 b
bc b
πd p 4pdA dπ4
q
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Rigid Pavement Design Course
Development Length
Allowable Bond Stress
ss Af Design Strength
of the Bar
ACI Definition of Development Length
(a)
Assumed and Actual Bond Stress-Slip Relationships.
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Rigid Pavement Design Course
Actual Bond Stress
Development Length
Vetter
Allowable Bond
StressForce in Bar Under Working Stress
Condition
Stress Transfer Length
(b)dx
dεEAu
o
ss
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Rigid Pavement Design Course
Bond Stress
Relative Slip Between Concrete and Steel
As Modeled in Computer Program
Actual bond Stress-Slip Relationship
(c)
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Rigid Pavement Design Course
concrete tension specimen
steel bar
P(a)
(b)
(c)
a b
sε
cε
x
Sb
sΔ
cΔ
eS
ssEA
P
dx
dεslope s
cf1.50)Slip3100(1.43xu • x-Displ.• Slip• cf
Determination of Slip from Strain Functions
b
c
d
f9.5u
ACI
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Rigid Pavement Design Course
If L =∞ (i.e., no cracking)0fzEthen tzc
Shrinkage Limiting is E
fz
c
t
SZ yBut f <f
)f(fAfA szszstc
sztsszszt fnfzEff
p
f
z)tα(tαEnp
1fzEn
p
1ff csststsz
minthe min p p to prevent yielding of steel
tmin
y s t
f shrinkage
f zE nfp
drop temp.nff
f
ty
t
cs αα :Note
AASHTO multiplier-A on steel %
1.0A 1.5 if 0.2-1.3A 1.0A 3.0 if 0.1- 1.3Aprefer
tzssz nfzEf
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Rigid Pavement Design Course
L/2x IF
sz
t
f
1
p
f
2
LL
Lf
ffx
sz
szsz
tsszt nfzEf
2p
f
ctt znEnf
2p
f
12pn
1
E
fz
c
t
y
t
f
fp ngSubstituti
y y tt
c c
f f 2nf1z f 1000
E 2n 2nE
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Rigid Pavement Design Course
Relative to temperature drop
Max. drop to cause L = 2x
substitute for in equation tαs z
tss
t
cs
t
o nfEαt2
fp
Eα
12pn
1f
to
ty
tmin nff
fpp setting
tty
cst
t
csf
2n
nff
Eα
1f
2pn
f
Eα
1t
y t
s s
f nf with steel yielding
2α E
minpt
For a greater temp. drop t2……only if syssys φfEα t,ff 2
y s minbut f φ then p p
otherwise
np
1fφ ts
2o ttT &
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Rigid Pavement Design Course
np
1ffEαt tyss2
n
EαEα ss
cs
12pn
1nfEαtEαt tsscs oo
n2p
1f t
np
1ffn
2p
1fETα tyzss
2p
ff t
y
ss
ty
ss
ty
Eα2p
ff
E2pα
f2pfT
0.0078(400)60ksi
400pmin
F219.4
36
18.0072
1400
αE
12pn
1f
tc
t
1
min
y tp
s c
f nf 60 8 400t 219.4 F
2nα E 2 8 6 3
t
y
s s
400f 60f2 .0072p
T 219.4 Fα E 6 8 3
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F230T IF 2p
ffETα t
yss
ssy
t
ETαf2
fp
ssy
t
ETαf2
f
0074.
minpp
ys ff
2yL
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Rigid Pavement Design Course
Structural Response Models
Uniform Bond Stress Distribution
Vetter: Shrinkage and Temperature Drop
Zuk: Shrinkage
Friberg: Temperature Drop
tsmss
t
nfzEtαEpuQ
fL
2
czcc
c
ctc ff
u
fAL
E
ftαzLcw
o
sts zEnp
1ff
z) p, u, , tf(L,f mt
ctcc
c
c ff u
fAL
E
fzLcw
o
tαEφnptαE
ααtEφ
U
φL
ssscs
cssss
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Rigid Pavement Design Course
Hughes: Shrinkage and temperature Drop (concrete only)
u4E
φdcw
s
sb2
cnpa
αbtEtαEφ cs
sss
tαtαE
φEf cs
s
sct
min
Aulti
Si
Fnp)np(1Lcw 2
22
rmin
Aultist iS
Fnp)(1L
min
Aultisc 2iS
FnpL 2
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Rigid Pavement Design Course
CRCP-2: Computer Model for Shrinkage and Temperature Drop (force equilbrium)
Regression Equations:
1.794.60
5.202.19
1.156.70
1000z1p1
1000
σ1d1
2α
α1
1000
f11.32L w
bc
st
4.554.91
2.206.53
p11000
σ1d1
1000
f1.00932cw w
bt
2.74
0.4943.144.090.425
p1
1000z11000
σ1
1000
f1
100
t147300f t
s
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Rigid Pavement Design Course
Percent Steel (p)
c
tc
m
tb
E
fCtαz
pu
fdCcw 21
mt
1 b
Lu pf
C d
sst
s tαEp
fCf 2
L
xφ f(φ(φ)4C i11/2
01
1/20
φ012 f(φ(φ)dφ8CC
tsy
tmin nfzEf
fp
ss
ty
E2α
nffdrop temp
; prevent yielding
; p=pmin
fs=fy
u
Lx
Um
Non-Uniform Bond Stress Distribution
TTICRCP: Computer model for Shrinkage and Temperature Drop (force equilibrium and energy balance)
Reis: Shrinkage and Temperature Drop
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Rigid Pavement Design Course
2002 AASHTO Guide CRC Design
t 0pcc
m
1 b
2f Cσ 1
hL
u pf2 C d
tσ ff where Strength Tensilef t
.60 xf9.5x c
If n
pccf
K a α
hg α g αpcc
c
c h
m 1u 0.002 K
1469.7f183f107f109K ccc26
1
cf117.2or
LnLcε
nεbaC 2
tot
tot1
K1
K2
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Rigid Pavement Design Course
Crack Width
c
σc
m
σb
E
fCΔtαz
u
fdCcw 21
c
σ
E
fCL 2
totε
H
21cσ
dC
puL
2
fLf o
o
ε
br
mσ
22 L
c
K
baC
r
3rh1Δtα αtot εε
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Rigid Pavement Design Course
Crack Spacing Distribution
L vLn(1 %P)
α
1
L v α Ln(1 %P)
1
maxL v α 10
i i
i i
L u
u i Lprob L L L 100 e e
L v L vα α
2 3 41 3.0626 28.024x 66.374x 64.653x 24.198x
L vα
1Γ 1
Γ 1 1/ (1/ )
1Ln Γ
4 3 21 1 1 1 1
Ln Γ 25.703 61.247 53.0072 21.346 4.0845
v c. spc
%P eα1