ce 481 aashto rigid
TRANSCRIPT
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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