Download - Rigid Pavement Design Course Crack Pattern Development

Rigid Pavement Design Course

Crack Pattern Development


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


(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.


Free Edge

‘t’

L

Longitudinal Joint

Traverse Crack

CL


Crack

Asfs Asfsc

Acft

Section XX Section YY

Forces Acting on CRC Pavement Section

Probable Strain Distribution Adjacent to a Crack


Extensive bond slip

Crack

L/2=nSmin

Good bond


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


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


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.


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


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


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


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 )


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


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


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 )


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


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.


Actual Bond Stress

Development Length

Vetter

Allowable Bond

StressForce in Bar Under Working Stress

Condition

Stress Transfer Length

(b)dx

dεEAu

o

ss


Bond Stress

Relative Slip Between Concrete and Steel

As Modeled in Computer Program

Actual bond Stress-Slip Relationship

(c)


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


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


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


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 &


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


F230T IF 2p

ffETα t

yss

ssy

t

ETαf2

fp

ssy

t

ETαf2

f

0074.

minpp

ys ff

2yL


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


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


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


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


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


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


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

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