rigid pavement design course crack pattern development
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Rigid Pavement Design Course
Crack Pattern Development
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
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.
Rigid Pavement Design Course
Free Edge
‘t’
L
Longitudinal Joint
Traverse Crack
CL
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
Rigid Pavement Design Course
Extensive bond slip
Crack
L/2=nSmin
Good bond
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
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
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.
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
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
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
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 )
Rigid Pavement Design Course
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
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
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 )
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
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.
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
Rigid Pavement Design Course
Bond Stress
Relative Slip Between Concrete and Steel
As Modeled in Computer Program
Actual bond Stress-Slip Relationship
(c)
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
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
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
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 &
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
Rigid Pavement Design Course
F230T IF 2p
ffETα t
yss
ssy
t
ETαf2
fp
ssy
t
ETαf2
f
0074.
minpp
ys ff
2yL
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
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
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
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
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
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
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