121169010 3 pavement stresses (1) u

8/13/2019 121169010 3 Pavement Stresses (1) u

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PavementPavement

ResponseResponse

Under LoadUnder Load

20122012

1



StressesStresses

DistributionDistribution

Wheel Load

Load Distribution in Pavements

150psi

3psi

Wearing C.

Base

Sub-base

Sub-grade

2



PavementStructure

Not Drawnto Scale

Weel !oad

Vertical Stress

"S

"B

"SG

"SB

Subgrade

Weel !oad

PavementStructure

Not Drawn

to Scale

Horizontal Stress

Subgrade

StressesStresses

DistributionDistribution3



#

PressPress

ureure p $ %ire Pressure $ &ontactPressure

%irePressure

P

a $ P'( p

) %*pe o+ !oad , Static and D*namic.

) %angential /orces , cceleration and Braing.

p

ntact Areantact Area is circle with radius =

Loads andLoads and

Contact AreaContact Area%andem%ridem

Single

Dual



5

Axles TypesAxles Types



The stresses, strains, and deflectionare the pavement response to theapplied load. Stress is a force load

per unit area, and strain is thechange in dimension. Pavementstresses, strains, and deflection arecaused by traffic loading, daily or

seasonal temperature and moisturevariations and by any change in the

1 41 1

2 42 2

3 43 3

n 4n n

P

6nter+ace 1

6nter+ace 2

6nter+ace 3"7

"r

"t

d7

dr

ulti!Layerulti!Layer

"lastic T#eory"lastic T#eory8



9

LoadLoad

RepresentatRepresentationion



Single LoadSingle Load Location is de$ined by % &r

:1:2

P

a

2

1

p

;< Stress σ ! σ"

;< de+lection # ! #"

:

P

a

2 1

p

;< Stress σ ! σ"

;< de+lection # ! #"

r

Dual LoadDual Load

:

P

s

1

p

r

P

r

p

1 2

σ $σ due to load%σ due to load"

# $# due to load%# due to load"

LoadLoad

RepresentatRepresentat

ionion

=



&andem&andemde'ined b( z)a and r)a

σ $ *σ due to loadi

# $ *# due to loadi

P PPP

S1S2

St

1

Ho+ to Calculate Stress , De'lectionHo+ to Calculate Stress , De'lection

-ne La(er &heor(-ne La(er &heor(

Stress

De+lection

/or an* s*stem o+ loads

Stress

De+lection

&+o La(ers &heor(&+o La(ers &heor( >nder single load

LoadLoad

RepresentatRepresentat

ionion

10



11

41 $4p$ ?

42 , Subgrade.

P

ap

No de+lection in te pavement structure@<<all de+lection is in te subgrade

So te la*ered s*stem is composed o+onl* AN4 !4C ,Subgrade.

%is la*er +ollows te same assumptiono+ te 4lastic ultila*er %eor*

Vertical Stress Vertical Stress σσ/ , Vertical De'lection / , Vertical De'lection ##//

41 $4p$ ?

42 , Subgrade.

P

ap

1. >sing ,7'a r'a. /igure ,1. m$ "'p "

2. >sing ,7'a r'a. /igure ,2. / E$ paF/' 42

Single LoadSingle Load

'ne Layer

T#eoryussinsq Theory)



12

Dual LoadDual Load

:

P

s

p

P

p12

σ total $ σ % σ "

"1 ,7'a r'a.@@<m1 $ " 1 ' p

" 2 ,7'a r'a.@@<m2 $ " 2 ' p

# total $ # % # "

E 1 ,7'a r'a.@@/1@< E1 $ pa/1'4subgrade

E 2 ,7'a r'a.@@/2@< E2 $ pa/2'4subgrade

Same procedure +or &andem Load @@@@< # times

'ne Layer




13

01am2le01a m2le

:$1# inc

42$5000 psi

P

a$8 p$50psi

r$8

12

71'a $ 2 r1'a $ 0 @@@<m1 $ 30G

m1$ "1 ' p @@< "1 $ 0<3F50 $ 15 psi

72'a $ 2 r2'a $ 1<0 @@@m2 $ 20G

m2$ "2 ' p @@< "2 $ 0<2F50 $ 10 psi

71'a $ 2 r1'a $ 0 @@@<+1 $ 0<8

E1 $ pa/1'4sub$ 50F8F0<8'5000 $ 0<0#=

72'a $ 2 r2'a $ 1 @@@<+2 $ 0<58

E2 $ pa/2'4sub $ 50F8F0<58'5000 $0<0#

'ne Layer




1#

(ertical Stress(ertical Stress

))σσ**m)+*



15

(ertical(ertical

De$lection )De$lection )##**



1

4 pav< $ 41

4 Subgrade $ 42

P

ap

- Same assumption as in the

!ultilayer theory

"- #n this theory, pavementdeflection is considered$- %eflection at depth & in a t'olayer system given by Δ , 1-1.pa$/"sub (igid Pav.)

Δ , 1- pa$/"sub (*le+ible

Pav.)

Δ vertical deflection under ofthe applied load

p tire pressure (psi)

a /P0(p

To Layer

T#eory rmister Theory)



18

2- This theory is applied only to oneoad and r0a 3

4- 5hen &0a 3 * .3 forany 1" 0 1 ratio

6- #f it is required to calculatedeflection in t'o layers system

under dual or dual tandem loadsystem, the thic7ness conversionmethod described earlier can beused to convert the t'o layers to

one layer system.

To Layer




19

La(er 03uivalenc(La(er 03uivalenc(

#t is the conversion of a thic7nessof a layer of material 'ith 7no'nmodulus to an equivalent thic7nessof another material 'ith 7no'nmodulus using the formula

&sub. & pav. $

/101"

4 pav< $ 41

4 Subgrade $ 42

P

ap

4 Subgrade $ 42

P

ap

4 Subgrade $ 42

To Layer




1=

a $ H15000 ' (F100 $ 8<0

&+o la(er&+o la (er

7'a $ 2 42 ' 41 $ 1'100 @<+I $ 0<13

E $ J pa+ '4sub$ 1<5 F100F8F0<13'1000$ 4.56

-ne la(er-ne la (er

7'a $ 2 r'a $ 0 @@@<+1 $ 0<=

E $ pa/1'4sub $ 100F8F0<='1000 $4. 785

ConversionConversion

: $ : pav< 3H41'42 $ 1#3H100000'1000 $5

E $ pa/1'4sub $ 100F8F0<18'1000 $4. "

:$1#

42$1000 psi

P $ 15000lb

p$100psi

41$100000 psi

01am2le01am2le

To Layer




20

(ertical Stress(ertical Stress

))σσ**m,G.



21

(ertical(ertical

De$lection )De$lection )##**

:'a



22

8se of !ultilayer theory in pavementevaluation

- Pavement evalaution 999..to find 1

: 1"

- Since the t'o layer theory ( ormultilayer thoery, in general) lin7sbet'een the pavement deflection and itscharacteristics, i.e. 1 : 1", then if 'e

can measure the deflection, it 'ill beeasy to bac7 calculate 1 : 1".

- Plate Bearing Test is used to measurethe deflection of subgrade under givenload.

P

a 4igid plate

o to 3et "1 & "2 Usin3 t#e Plate4earin3 Test

- Put the Plate on the top of subgradeand measure the deflection ( Δ) at the

top of subgrade ( i.e. ; 3.3)

p P 0 π (4)"

using the t'o layer theory

Δ < p a *= 0 1"

P

a 4 igid platSubgrade

Pavement

"valuation



23

o to 3et "1 & "2 Usin3 t#e Plate

4earin3 Test

"- Put the Plate on the top of the base (or pavement materials) and measurethe deflection ( Δ) at the top of

subgrade using the t'o layer theoryΔ < p a *= 0 1" 99 get *=

<no'n *= and ;0a ...get 1" 0 1 9get 1

&choused

1"

P

1

"xample"xample

? certain fle+ible pavement consists of "-inbituminous surface, @-in crushed stone basecourse, and A-in sandy subbase. ? $3-in rigidplate 'as used to determine the load- deflectioncharacteristics. The follo'ing results 'ereobtained

- subgrade deflection 3.3 in at "3 psi

"- Pavement deflection 3.3 in at A> psi

Solution5

step ()

Δ 3. < p a *= 0 1"3. .> "3 4 0 1"

"in

@inA

inSubgrade

>inPavement

(1)

1"

$3 in

Pavement

"valuation



Civen eq.

S5 !ulti'heel load

P , &

P, &

P, " ma+.

P, E ma+.

& , "

ma+.&, E ma+

"

E

&

&

P

P

hart

hart,

1quationhart

hart,1quation

hart

%ra'nelation (Δ , P)

hart

hart, 1quation

%ra'n elation(" , &)

%ra'n elation(Δ , &)

%ra'n elation

(" , P)%ra'n elation(Δ , P)

2#

step (")

Δ 3. < p a *= 0 1"

3. .> A> 4 *= 0 $423

*= 3."32 ( using hart 'ith ; 0 a >04

." and f= 3."32 )

1" 0 1 0 >3 99 1 >3 $423 ">$"33 psi

&>1"$423

psi

1

Civen &, p, E ma+., 1"

eq. P

- ?ssume P a / P 0 ( p

"- alculate E

$- #f E E ma+. 99..P req. P2- if not repeat $ to 2 times to dra' the sho'n

Pavement

"valuation

121169010 3 pavement stresses (1) u

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