ce 481 aashto rigid

8/18/2019 Ce 481 Aashto Rigid

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1993 AASHTO Rigid Pavement Design

CE481A

LECTURE 10

AASHTO Rigid Pavement Design GuideDevelopment

Based on AASHTO road test - Data collected onrigid pavement sections

Interim guide published in 1972

Revised in 1986 & 1993

Equations are in the same form as for flexiblepavements, with different values for regressionconstants

Design Parameters

Traffic expressed in ESAL’s oStructural number (flexible) replaced with Slab thickness

in rigid pavements

Reliability of design accounted for

Performance measured in terms of PresentServiceability Index (PSI)oTerminal serviceability 2.5 (1.5 for flexible pavement)


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Serviceability

Present Serviceability Index (PSI)oMeasuring physical attributes and relating them to driver

ratings (PSR)

oPhysical attributes

Surface deterioration (Slope variance)

Surface deformation (Cracking, Patching)

Result is usually a numerical scale

Present Serviceability Index (PSI)

Initial serviceability◦ Fn (pavement type, construction quality)

◦ Flexible pavements – 4.2

◦ Rigid pavements – 4.5

Terminal serviceability◦ Lowest index that will be

tolerated before rehabilitation,resurfacing and reconstruction

0 “Road

closed”

5.0 Just constructed PSI

4.5 Initial PSI ( pi)

Terminal PSI ( pt )PSI

Performance

Where

wt – Axle load application at time t

ρ – Expected number of load repetitions to terminal serviceability (pt)

β – function of design and load variables that influencesthe shape of ρ vs wt curve


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Modified AASHTO Design Equation forRigid Pavements

Where

oW18 – Number of 18 kip single axle load applications (from traffic projections)

oZR – Standard normal deviate (user defined reliability)

oS0 – Combined standard error of traffic and performance predictions (0.35 – 0.4 assumed)

25.075.0

75.0

46.8718

)//(42.1863.215

)132.1(log)32.022.4(

)1/(10624.11

)5.15.4/(log06.0)1(log35.7log

k E D J

DC S p

D X

PSI DS Z W

c

d ct

o R

Design Equation Cont’d…

D – Slab thickens in inches

ΔPSI – Allowable drop in Present Serviceability Index (user defined)

k – Effective modulus of subgrade reaction (known)

Pt – Terminal serviceability index

Sc – Modulus of rupture of concrete

Cd – Drainage coefficient

J – Load transfer coefficient

Ec – Modulus of elasticity of concrete

Drainage Coefficients (Cd)


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Load Transfer Coefficients

Modulus of Concrete

Elastic modulus of concrete

Concrete modulus of rupture

oThree point bending test (28 days)

concreteof strengthecompressiv f Where

f E

c

cc

'

5.0')(000,57

Modulus of Subgrade Reaction

Without sub-baseoFrom plate load test[OR]

oInferred from resilient modulus valuesk = MR/19.4

With sub-base layeroComposite modulus of subgrade reaction can be estimated

based on AASHTO chartsFunction of subgrade k value, sub-base layer tk., quality of sub-base

oAssumption - subgrade extends to infinite depth


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Example Problem

Determine composite modulus of subgradereaction using AASHTO Charts

oD = 6 inches

oSub-base resilient modulus Esb =20,000 psi

oSubgrade resilient modulus = 7,000 psi

SHTO chart for composite modulus

Rigid Foundation at Shallow Depth

Rigid foundation at depth (Dsg) less than 10 feet(3m)

oBed rock

Modification charts for (k) developed by AASHTO

Applicable to slabs with or without sub-base


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Sample Problem

Determine the modulus of subgrade reactionusing AASHTO charts if

oSoil Resilient Modulus Mr = 4,000 psi

oDsg = 5 feet

oSoil composite modulus K∞ = 230 psi

Climatic Effects on Subgrade Modulus

Effective modulus of subgrade reactionoEquivalent modulus that will result in the same amount of

damage if seasonal modulus for the entire year wereused

oWhereD = slab thickness

ki = seasonal moduli of subgrade reaction

k = effective modulus of subgrade reaction

n

i

ik Dn

k D1

42.325.075.042.325.075.0)39.0(

1)39.0(


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Climatic Effects Cont’d… Estimating relative damageto rigid pavements

AASHTO charts are alsoavailable

42.325.075.0 )39.0( k Dur

Sample Problem

Determine the effectivemodulus of subgradereaction for the givenseasonal modulus values

oD = 9 inch

Sample Problem

Determine k valuefor each month

k = MR/19.4

Calculate therelative damage Urusing equation orfrom chart

42.325.075.0 )39.0( k Dur


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Loss of Subgrade Support

Loss of foundation support (LS)

oErosion

oDifferential vertical soil movement

Effective modulus of subgrade reaction reduced by factor LS

Loss of Subgrade Support

Joints in PCC Pavements


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Effect of Volume Change on Concrete

Caused by variation in temperature or moisture

oInduces tensile stresses and causes concrete to crack

oCauses joints to open and decreases LTE

Joint spacing in JPCPoFriction-induced stresses does not induce wide enough cracks

in PCC causing reduction in LTE to intolerable levels

JRCPoReinforcing steel or wire keeps cracks that form between joints

small enough to effect good LTE

Reinforcements in Rigid Pavement

Frictional force on the slab balanced by tensile force in theconcrete which is ultimately carried by steel

Area of steel,

hhLf

cac

2

steel in stresses Allowable f

inthickness slabh

subgradeand slabbetween frictionof coeff f

in slabof length L

concretein stressesWhere

s

a

c

)(

.

)(

,

s

ac s

f

hLf A

2


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Design of Tie Bars

Placed along longitudinal joint to tie slabs together

steel in stresses Allowable f

thickness slabh

subgradeand slabbetween frictionof coeff f

exist barstienowhereend freetheto joal longitudinthe fromlength L

Where

s

a

.

int'

, s

ac s

f f hL Arequired steel of Area ',

d f t bar tieof length s

2

1,

stressbond allowable

bar theof diameter d

steel in stressallowable f Where s

,

Increase length ‘t’ by 3” to account for misalignment

Sample Problem

Design reinforcements and tie bars for a

o2 lane concrete pavement (8”) thick

oLength – 60’ – 720 in

oWidth - 24’ – 288 in

oγc – 0.0868 psi (23.6 KN/m3)

ofs – 43,000 psi

ofa – 1.5

Sample Problem

Longitudinal steel

Transverse steel

For four lane pavement with all four slabs tied together,the transverse reinforcement for the inside 2 lanes shouldbe doubled, length L = 48’ instead of 24’

)/222(/105.043000*2

5.1*720*8*0868.0 22 mmm ft in A s

)/9.88(/00349.043000*2

5.1*288*8*0868.0 22 mmm ft in A s


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Sample Problem

For the same pavement determine diameter,spacing and length of tie bar

Fs = 27,000psi, µ = 350 psi

L’ – 12’ (144’’)

Using 0.5in bars (Area – 0.2in2)

Spacing = 0.2/0.00556 = 36 in. (3’)

s

ac s

f

f hL Arequired steel of Area

',

inin /00556.0000,27

5.1*144*8*0868.0 2

Sample Problem

t = (27000*0.5)/(2*350) = 19.3”

Correction for misalignment = 19.3 + 3 = 23.3 ≈ 24”

2 feet long, 0.5 inch diameter tie bars to be providedat a spacing of 3 feet

Design of Dowels

Size of dowels to be used depends on slabthickness

oTypically 1/8th of slab thickness

oPCA recommendation

1.25 in dia dowel for slab tk. below 10”

1.25 to 1.5 in dia dowel for slab tk. above10”


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Design of Dowels

Size and spacing determined by bearing stress between dowel andconcreteoAllowable bearing stress

f b = [(4 – d)/3]f’cwhere, d – dowel diameter (in)

f’c – compressive strength of concrete

oBearing stress inducedb = Kyo = [KPt (2 + z)]/(4

3EdId)

= [(Kd/4EdId)]0.25

Where, K – modulus of dowel support (3x105 to 1.5x106 psi/81.5 to 409GN/m3)

Pt – load on one dowel bar

Design of Dowels

Id – moment of inertia of dowel

β – relative stiffness of dowel embedded in concrete

Ed – young’s modulus of dowel

yo – maximum deformation of concrete under the dowel

Z – Joint width

4

64

1d

4

4d d

I E

Kd

Design of Dowels

Based on westergaards analysiso maximum negative moment for interior and edge loading

occurs at “1.8 l” from load,

where,

reaction subgradeof ulusk

ratio poissonsv

thickness slabh

concreteof elasticityof ulus E Where

k v

Ehl stiffnessrelativeof radius

mod

mod

)1(12,

25.0

2

3


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Sample Problem

Figure below shows a 9.5”slab resting on afoundation with k=50 psi. 12 dowels at 12 in

spacing are placed at the joint on the 12 feet lane.Two 9000 lb loads are applied at points A & B.Determine maximum load on one dowel

Assume a load factor of 1 for dowel A

Determine load factors of other dowels using similar triangles

Sum of these factors = 4.18 effective dowels

Load carried by dowel at A = 4500/4.18 = 1077 lb

Determine load carried by others by proportions

Sample Problem

inl

inl stiffnessrelativeof radius

888.1

17.4950)15.01(12

)5.9(10*4,

25.0

2

36

Similarly find load carried by dowels due to load at point B

Sum of load factors = 7.08

Determine load carried by dowel at B = 636 lb

Now, determine load carried by other dowels due to load B(proportions)

Combined effect on dowels due to both loads is given below


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Sample Problem

Load carried by edge dowel is most critical andshould be used for design purposes

Direct determination of load carried by the dowel

)3.5(119118.0/4500*18.018.4/4500 kN lb Pt

ce 481 aashto rigid

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