(rigid pavement)
TRANSCRIPT
,v i n istry ., ,.fi::iiililTj};:':,d Transport
Department of Roads
Planning Monitoring and Evaluation Unit
PAVEMENT DESIGN GUIDELINES
(Rigid Pavement)
$urface r mo othn ecs
Lmg frudtnal Jolnt, Tr*n*uersa joinr
Sur{aca Texture
Contrete maertds
Dosl barsTlebar*Suhgrrde
Suhhaee or b*e
Submitted by:
@,T
December, 2016
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Shb Thicknes*
GRID CONSULT PVT. LTD.
Anamnagar, Kathmandu
: o ie of Contents
LIST OF FIGURES
LIST OF TABLES
ACKNOWLEDGEIVENT
ABBREVIATIONS
3IIAPTERl INTRODUCTION
1.1 lntroduction.,
1.2 Sc0pe..........
1.3 Pavementdesignapproach......
1.4 Types of concrete pavements................
1.4.1 Joint Plain Cement Concrete Pavement (JPCP)...............
1.4.2 Continuous Reinforced Concrete Pavement (CRCP)........
1.4.3 Precast Panel Concrete Pavement (PPCP)
OHAPTER 2 DESIGN FACTORS....
2.1 Axle Load Characteristics...................
2.2 DesignPeriod..........
2.3 Traffic Considerations........
2.3.1 Designlane..............
2,3.2 Design traffic............
2.4 Temperature Differential....
2.5 Characteristics of Subgrade Soil and Subbase......
2.5.1 Embankment
2.5.2 Subgrade.....
2.5.3 Subbase.....
2.6 Separation Layer between subbase and pavement.............
2.7 Drainagelayer.............
2.8 Characteristics of concrete
2.8.1 Design strength
2.8.2 [Modulus of Elasticity and Poisson's Ratio of Concrete .....
2.8.3 Coefficient of thermal expansion
2.8.4 Fatigue behavior of cement concrete
CHAPTER 3 DESIGN OF SLAB THICKNESS
3.'l Criticalstressconditions...............
3.2 Calculation of flexural stress...,.,.....
3.3 Cumulative Fatigue DamageAnalysis.......
3.4 Drainage 1ayer............
3.5 Tied Concrete Shoulder and Widened Outer Lane...
3.6 BondedRigidPavement..................
3.7 Recommended procedure for Slab design
CHAPTER 4 DESIGN OF JOINTS
4.1 Spacing and layout....
4.2 Load Transfer at Transverse Joints........
4.3 Dowel group acti0n...........
4.4 Design of Tie Bars for Longitudinal Joints
4.5 Reinforcement in Cement Concrete Slab to Control Cracking..............
]HAPTER 5 COUNTINUOUSLY REINFORCED CONCRETE PAVEMENT.
5.1 lntroduction..
5.2 Design considerations...................
CHAPTER 6 PRECAST PANEL CONCRETE PAVEENT
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5
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6.1 lntroduction..
5.2 Precast Panel Concrete pavement System.........6.3 Design Considerations of ppCp....
APPENDIX I:VALIDITY OF FATIGUE EQUATIONS
APPENDIX ll: EQUAToNS FOR FLEXURAL STRESS tN CONCRETE s1AB.......... . .
APPENDIX III: EXATMPLE FOR THE DESIGN oF DRAINAGE LAYER
APPENDIX IV: EXAMPLE FOR THE DESIGN oF SLAB THICKNESS
APPENDIX V: DESIGN OF DOWEL BARS.
APPENDIX Vl: DESTGN OF TtE BARS..... .....
APPENDIX Vll: TYPICAL DESTGNS rN THE CASE oF LtMTED TRAFFC DATA............
APPENDIX Vlll: SAMPLE DESTGN TEMPLATE oF DEPTH oF CONCRETE s1A8 ........
APPENDIX IX: SOME FIGURS FoR CONSTRUCTION TECHNOLOGY oF PCCP SLAB
APPENDlXX:STRESSCHARISFoRBoTToMUPCRACKANALYSIS...
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.39
.40
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LIST OF FIGURES
' , = ' ypicalCross Section of Concrete pavement_1 ........ ...... ...........10' ; =  ypicalCross Section of Concrete pavement _2..... . . ....10: , .  : : l.css Sectron of Concrete pavement3..,....,
...10'.','_:::::" of EffectiveCBRof Subgrade ........,.............14: _ _ = , : a:i I ar tne lvliddle of the Slab during lvlid_day
.. .. .... .. .. .. .l g" :
':=.=: :: Axres for ft/aximum Edge Flexural Stress at the bottom of the slab without Tied Concrete
: ';racementof TwoAxlesofaCommercialVehicleonaSlabCurledDuringNightHours............ . ..1g: ' : I ferent Axle Load Positions causing Stress at the Top Fiber of the slab with Tied concrete shoulder.....20" , . 1  Scncept used for Obtaining Combined Flexural Stiffness .....,....,...23
: ' _cngitudinaljoint with tie rod between two 1anes............ , ......26
x
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v
LIST OF TABLES
: : ' 3"nperature differential values for various slab thickness ,"....... ......
.. .  ::proxrmate kvalue corresponding to CBR values homogenous soilsubgrade.. ,' =  r,value for Granular and Cement Treated Subbases.. .
 :. :  {values for DLC subbase
, . = : , alues of Standard Normal Variate, Za ...............
, : = : =xpected values of Standard Deviation
 : . Stress ratio and allowable repetition for the concrete slab .. . . ..
 ,  = : Recommended Dimensions of Dowel bars. ...... .'...
 ,: = 3 Recommendations for Tie Bars for Longitudinal Joint of TwoLane Rigid Pavements .
13
13
14
14
15
16
17
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ACKNOWLEDGEMENT
".,:^i was prepared for the draft stage to finalize the 'Guidelines for Rigid Pavement Design'. GRID',r;t Ltdwasentrustedtopreparethisguidelines.Thequalityforprofessionalservicewastakenasthe
,  ::: Juring the preparation engagement in this assignment.
'  :: ,rould ltketo appreciate Design Branch, Planning lVonitoring and Evaluation Unit, Deparlmentof Roads
:  : ai on to prepare the guidelines. I would like to extend my sincere thanks to DDG lVr. Daya Kant Jha,: :  : ^ ef SDE Mr LaxmiDatta Bhatta, and lt/r. Ramesh Kumar Singh for their support and trust in doing this
.' :'ai assignment. Similarly, consultant would like to express thanks to engineers Mr. Pradeep Niraula and' 's'rna Bahadur Kunwar for their support during the study. Similarly, I would like to thank [Vlr. Prabhat Kumar
: '' Shiva Adhikariand lVr. BinodSapkota fortheircontribution on finalizing the guidelines afterthe workshop
. radma Bahadur Shahi
.r r CONSULT Pvt. Ltd.
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ABBREVIATIONS
 . j ^ r mber of axles per day in the year when the road is opened to traffic
 :: s section al area of one tie bar in mm2
.'a2 ct steel in mm2
,^3'ilidth in m
l:,'re diameter, mm
:e'mrssible bond stress of concrete in It/Pa:actor for transverse joint efficiency in topdown cracking
3ottomup cracking
lumulative number of axles during the design period
Spacing of transverse joints m
Calrfornia Bearing Ratio in%
Cumulative Fatigue Damage
Diameter of tie bar in mm
Dry Lean Concrete
[/odulus of elasticity of concrete in I\4Pa
Allowable bearing stress in lVPa
Coefficient of friction
lMaximum bearing stress in lVPa
Characteristic flexural strength at 28 days in IMPa
Target average flexural strength at 28 days in MPa
Finite element method
Thickness of slab, m
Crack inflltration rate m3/day/m
Total number of load groups
lModulus of Subgrade Reaction in MPa/m
ltlodulus of dowel support in MPa/m
Rate of infiltration through uncracked pavement surface (m:/day/m)
lModulus of subgrade reaction (MPaim) with plate diameter rD
N/odulus of subgrade reaction (MPa/m) with plate diameter 750 mm(k)
Length of tie bar in mm
Radius of relative stiffness in m
Load Transfer Efficiency, in %
Fatigue life
Rate of Cone Penetration in mm/blow
Design period in years
Number of predicted repetitions for the /t load group
Number of longitudinal joints/cracks
Number of allowable repetitions for the fl,, load group
Single/tandem axle load in kN
Perimeter of tie bar in mm
Pound per cubic inches
Pavement Quality Concrete
pcir.
\Pdc
IR
r
Sts
SR
rcsTD
TT'C
w
fl,we
Lo
p
Y
F
oAI
lnfiltration rate per unit area,m3/day/m2
Flexural stiffness in tMNm
Annual rate of growth of commercial traffic volume (expressed as decimal)
Allowable working stress of steel in fMPa
Flexural stress in slab in MPa
Skess Ratio
Tied Concrete Shoulder
Temperatu re Differential
Topdown cracking
Weight of slab in kN/m2
Length of the transverse cracks or joints, m
Width of pavement subjected to infiltration, m
A factor corresponding to the desired confidence level, which is 1.96 for 5 % Confidence Level
Coefficient of thermal expansion in/'C
Relative stiffness of dowel bar embedded in concrete in MPa/m
Unit weight of concrete in kN/cum
Poisson's ratio of concrete
Standard deviation of field test samples in MPa
Temperature differential in'C
I)
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CHAPTER 1 INTRODUCTION
:,.sntdesignguidelinesforflexiblepavementtypewaspublishedin20l4.Theshiftfromtheflexibletypeinto
, , d has been realized by many practicing engineers. The need for the uniformity in design approach and
: : . ,0 utilization of resources are the rationale behind the making of such guidelines.
':se guidelines, general conditions for the selection of pavement type, traffic calculations, subgrade
: : )'.0n, structuraldesign approach and design illustrations have been elaborated.
: llcument is the guidelines forthe practicing engineers, and has the status of departmental approvalforthe
: ':^: c application for the design,
,l.2 Scope
e guidelines cover the design of plain jointed cement concrete pavement and brief descriptions for the reinforced
ad prestressed cement concrete pavement. The guidelines are applicable for roads having a daily commercial
ffic (vehicle laden weight exceeding 3 T) of over450. They are not applicable to low volume rural roads.
1 3 Pavement design approach
= early approach to design of rigid of rigid pavements was based on'Westergaard's analysis. However,
. ; s:ntly this approach has been advanced by the application of more analytic techniques which may incorporate
= :ffect of present day vehicle fleet, considering cumulative fatigue damage due to the combined effect of load
:  : pavement temperature variations. The guidelines include procedure for design of pavements with widened
::' lane, tied concrete shoulder, pavements bonded to cemented subbase, design of longitudinal joints,
: , : ansion and contraction joint. Main features of the guidelines are listed as:
. Design of pavements considering the combined flexural stress under the simultaneous action of load and
temperature gradient for different categories of axles,
Design for bottomup fatigue cracking caused by single and tandem axle load repetitions,
Design for topdown fatigue cracking caused by single, tandem and tridem axle load applications,
Design guidelines for pavements without concrete shoulders and with tied concrete shoulders,
Consideration of Concrete slabs with unbounded as well as bonded cement bound subbase and
Design of pavements with widened outer lanes.
1.4 Types of concrete pavements
1.4.1 Joint Plain Cement Concrete Pavement (JPCP)rlP is the most common type of rigid pavement, JPCP is constructed with longitudinal and transverse joints to
:^trol where cracking occurs in the slabs. Tie bars and dowel bars are provided for the transfer of wheel load
rn one to the neighboring concrete slabs. Dowel bars are placed along the longitudinaldirection and across the
:nsverse joints. Similarly, tie bars are placed along the transverse direction and across the longitudinal joints.
!
may be distinguished as below
II
ntroduction
:, :.:^. s one of the important components of road transport infrastructure. lt has substantial role for the
: :'^i efficiency of the traffic movement. The wheel load distribution to the ground beneath and frictional
 : :''r with the tyres and repetition or cyclic loading natures are the engineering features of the pavement
. = Magnitude of wheel load and the subgrade soil strength are prime factors for the structural design.
,:' i.) 7 pavements are designed as the flexible type However, in the case of weak subgrade or poor drainage
, ... and extremely overloaded vehicular movement are the basic criteria for the option of rigid pavement ,":t on,
l
4v
125 N'licron thicksheet
PQC
DL(/ Cement Treatcd Sub base
tiSts as drainage lay..r
liiltcr Scparation L,ayer S ubgrade
Subgrade bJ Debonding Iayer of .10 r'nm BC over Centent
treated Subbase laverebonding layer of Polythene shcct ovcrCernent trearted Subbase laver
, . 'e l Typical Cross Section of Concrete Pavement 1 Figure 2 Typical Cross Section of Concrete Pavement 2
Subgrade
cl PQC over 100 mm DBM and Granular Subbase
Figure 3 Typical Cross Section of Concrete Pavement 3
]QC  Pavement Quality Concrete; DLC  Dry Lean Concrete; BC  Bituminous Concrete; DBM  Dense
3rtuminous Macadam; GSB  Granular Subbase;
1.4.2 Continuous Reinforced Concrete Pavement (CRCP)
 :1ough this type of pavement is limited in construction. However, CRCP is still a relatively new concept and
.ferred for construction in High Mountain and High Desefi climate regions. Since CRCP uses reinforcing steel'::rer than weakened plane joints for crack control, saw cutting of transverse joints is not required for CRCP. The
::rtinuous reinforcement in the pavement holds the cracks tightly together. CRCP typically costs more initially'an JPCP due to the added cost of the reinforcement. However, CRCP is typically more costeffective over the life
'the pavement on high volume routes due to improved longterm performance and reduced maintenance. ln
RCP, there are no transverse joins, properly built CRCP should have better ride quality and less it has less
aintenance than JPCP. Detail description of the CRCP is given in Chapter 5.
1.4.3 Precast Panel Concrete Pavement (PPCP)
PPCPs use panels that are precast offsite instead of castinplace. The precast panels can be linked together with
dowel bars and tie bars or can be posttensioned after placement. PPCP offers the advantages of:
. lmproved concrete mixing and curing in a precast yard,
. Reduced pavement thicknesses, which is beneficial when there are profile grade restrictions such as
vertical clearances and. Shorter lane closure times, which is beneficial when there are short construction duration.
PQC
Bituminous Concrete
Cement Treated Granular Subbase
GSB as drainage layer
Filter Separation Layer
PQC
DBM
Granular Subbase
GSB as drainage layer
Filter Separatlon Layer
I expl and design criteria for the PPCP is given in Chapter 6.
II
{
CHAPTER 2 DESIGN FACTORS
. : ': joverning design consideration are single, tandem axle and tridemloads, their repetition, tire presume
: :'  ire ffient characteristic of commercial vehicles.
: ' lrle Load Characteristics
 .: ,.sLble axle loads in Nepal is taken as 10.2tonnes, lgtonnesand 24 tonnesfor single axles, tandem axles
: axles respectively However, most of freight vehicles plying along the National Highways as well as
:  ='. r:ads carry much more than these legal axle load limits.
 , :::'r of axle load distribution of commercial vehicles is necessary for the purpose of computation of number
: , : . I rs of single and tandem axles of different weights expected during the design period. Axle load survey is
 :!::i to conductfor the contlnuous for 48 hours, covering minimum sample size of ten percent in both
' :: :"s of commercialvehicles for exceeding 6000 per day (CPVD), fifteen percentfor 3000 to 6000 CVPD and
: ' : " nercent for /ess than 3000 CVPD, lf the spacing of consecutive axles is more lhan 2.4 m each axle may be
, :=':d as single axle. For the purpose of fatigue damage analysis axle load group is considered as below.
Single axle
Tandem Axle
Tridem axle
10 kN
20 kN
30 kN
::: ^j between successive axles of commercial vehicles is necessary to identify the proportion of axles that' :  : be considered for estimating topdown fatigue cracking caused by axle loads during night period when the
,   as the tendency of curling up due to negative temperature differential. Data on the spacing of axles may be
. :ed during the traffic survey. As discussed in subsequent sections of these guidelines, if the spacing between
:' :: r of axles is less than the spacing of transverse joints, such axles need to be considered in the design traffic
:rputing topdown fatigue cracking damage. Wheel bases of trucks of different models generally range from
:  to more than 5.0 m whereas the commonly used spacing of transverse joints is 4.5 m. Thus, axles with
.: ^g of more than 4.5 m are not expected to contribute to topdown fatigue cracking. However, if the actual
: : :  'g of transverse joints is different from 4.5 m, design traffic for estimation of top down cracking damage may
:; .: ected appropriately. The percentage of commercial vehicles with spacing between the front and the first rear
: : ess than the proposed spacing of the transverse joints in the concrete slab should be established from axle
.:l SUrVey.
,gher axle loads stimulate high stresses in the pavement and result in the consumption of fatigue resistance of
:srcrete slab. Pavement design should determine the fatigue deterioration caused by the different axle load
loups. Stress due to the wheel load also depends upon the tyre pressure as well as the shape of the contact
reas. ln general, the value of tyre pressure for the commercial vehicles ranges from) 0.7 to 1 MPa. However,
sesses due to variation in the tyre pressure (for the given range of 0.7  1.0 MPa) are not significant for the slab
lrckness of 20 cm and above. ,/r case of limited data on traffic volume count and axle load pattern, typical approach is given in APPENDIX Vl.
g
2.2 Design Period
Generally, cement concrete has the life span of 30 years. When the traffic intensity cannot be predicted accurately
bra long period of time, and for low volume roads, a design period of 20 years may be considered. However, the
Jesign Engineer should use his judgment about the design life taking into consideration the factors, like, traffic
lu , the traffic growth rate, the capacity of the road and possibly of g rowth of capacity
t1
 : : ase of analyzing top.down cracking, the design traffic will be a portion of the design traffic considered for
,up cracking analysis. Commercial vehicles with the spacing between the front axle and the first rear axle
: : :  3 rt the spacing of transverse joints should be considered for topdown cracking analysis, This ration will be
, =: shed from the axle load/ traffic survey. A default value of fifty percent of the design traffic used for
: , up cracking analysis may be considered.
 :ated number of repeated movement of different axle load groups during the design period can be estimated
, : lhe details of commercial traffic volume and expected rate of growth of commercial traffic and the
ation about axle load spectrum and the numberof single, tandem and tridem axles obtained from axle load
 .; Since front axles (steering axle)with single wheels on eitherside cause only negligible bottomup fatigue
, .. iJe it is only the rear axles that may be included in the axle load spectrum.
' ': .ase of new road links, where no traffic count data is available then the traffic data from the roads of similar
: :: lry and imporlance may be used to predict the design traffic intensity.
 . : lmulative number of repetitions of axle during the design period may be computed from the formula.
365xA{(1+r)n1}r/
U:re,
C = cumulative number ofrepetitions of axles during the design period,
r = Annual rate of growth of commercial traffic (expressed in decimals),
A = lnitial number of standard axle per day in the year when the road is open to traffic,
n = Design period in years= design cumulative number of axle load repetitions for fatigue be obtained from the cumulative
of al vehicles as mentioned in the previous parag:W
L
i 3 Traffic Considerations
r1 Designlane': io carrying the maximum number of heavy commercial vehicles ts termed as design lane. Each lane of a
,. rwolane highway and the outer lane of multilane highways can be considered as design lanes.
 :2 Design traffic ,: Jme should normally be analyzed on sevenday 24hour count. Actualgrowth rate for heavy commercial
, .^ould be determined afterthe analysis of previous trafficcount records. However, minimum growth rate
.n :3(en as 5 percent and maximum of T.Spercent depending upon the economic development of the
', ; ^t time and day traffic analysis is important for the analysis of topdown and bottomup cracking analysis
,  ' V
, :  l: flexural stress caused by axle loads for bottom up cracking is the maximum when the tyre imprint of
 :: wheel touches the longitudinal edge. When the tyre position is away even by 150 mm from the
. : al edge, stress in the edge region is reduced substantially. The edge flexural stress is small when the
:: S 3re close to the transverse joints. Generally, only very few commercial vehicles are making their tyre
.  s :angential to the longitudinal edge/joint on two lane twoway roads.Therefore, it is recommended that 25
:;'::"t cf the totaltwolane twoway commercialtraffic may be considered as design traffic fortwolane twoway
,., ':. the analysis of bottom.up cracking.ln the case of fourlane and other multilane divided highways,25
 :=: of the total traffic in the direction of predominant traffic may be considered for design of pavement for
, rp cracking.
  :mperature Differential
' '' ': difference between top and bottom of the concrete pavements causes the concrete slab to warp,: 'se in stresses. The temperature differential depends on the intensity of solar radiation received by the. =' ..]'face at location, losses due to wind velocity etc. and thermal diffusivity of concrete. Therefore, it is: . : ' Jeo$r?phical features of the pavement location As far as possible, values of actually anticipated: :"'. dtfferentials at the location are estimated by using relevant geographical features and material: :' S.:s ln the absence of local data, the values of tentative temperature differentials are given in Table 1.
Table 1 Temperature differentiar values for various slab thickness
:'::Jre differentials are positive when the slab has the tendency to have a convex shape during the day
' 'J negative with a concave shape during the night. The axle load stresses should becomputed forfatigue' ' ' 'vhen the slab is in a warped state due to the temperature differential during day as well as night hours.
 5 Characteristics of Sub.grade Soil and Sub.base
 : i Embankment
: " 2 Bearing Ratio (CBR) of embankment soil placed below the 500 mm selected subgrade should be .' ^ed for estimating the effective cBR of subgrade and its ,k'
value for design.
 a 2 Subgrade' ' ' '=ngth of subgrade is expressed in terms of modulus of subgrade reaction k, which is defined as pressure
'  : jeflection of the foundation as determined by plate bearing tests. As limiting design deflection for cement: '::: pavements is taken as 1 .25 mm, the kvalue is determined from the pressure sustained at the deflection
' :eflection, The kvalue is necessary to correlate with the soaked CBR test value as shown in Table 2.Table
Table 2Approximate kvalue corresponding to cBR values homogenous soil subgrade
Soaked
<value,
Note: 1 00 pci = 2.Ttkglcms =27.2 MPa/m (Source: IRC: 58201 5)
= l3R of the 500 mm thick compacted subgrade is significantly largerthan that of the embankment below it, =::tive CBR of the subgrade can be estimated from Figure 4 (lRC:5820'lS) .A minimum subgrade CBR of g :=^. is recommended fordesign.The insitu CBR of the subgrade soilcan also be determined quickly fromthe  ,  c cone Penetrometer (60" cone) tests using the following relationship: /o
ogroCBR = 2.465  i,.j,Z LogroNoro Equation 1
=': rVrrpRate of cone penetration (mm/blow)
, ':commended to provide filter and drainage layers above the subgrade for drainage of water to prevent (i)=ssrve softening of subgrade and subbase and (ii) erosion of the subgrade and subbase parlicularly, under='se moisture condition and heavy dynamic loads, Synthetic geocomposite layer can also be used at the
=ace of subgrade and granular subbase Iayer for filtration and drainage. lt will not allow migration of finerade soil to the granular drai nage layer above
2
15 cm
Hilly region 12.5 13.1 14.3 15.8Terai region 15.6 16.4 16.6 16.8
2 J 4 E7 10 15 20 50 100
21 28 35 42 48 55 62 AO 140 220
:esof
.o.
S/N Region oClthickness slab
20 cm 25 cm 30 cm
0tto
 , .z tested on the wet condition of subgrade is less than 6.0 kg/cm2/cm, then concrete pavement should
: , : directly over the subgrade. A Dry Lean Concrete (DLC) subbase is generally recommended for
. ,:rrete pavements, pafticularly those with high traffic intensity.
Ctsrl o, enftbankmrlnt $orl bclnw 500 fim of $e,ecl srrbgradel0
u6t t)oro
E:o&aa
U) rrE
l0
7la
5%
1Yo
2.5Y,
2Yo
1.5v"
5
0
0r02010CgR of Compacted Eorroly Msterial
40 50500 mrn Thi6k
60
Figure 4 Chart for Estimation of Effective CBR of Subgrade
: 5 3 Subbase": a n purpose of the subbase is to provide a uniform, stable and permanent support to the concrete slab laid
i' : lt must have sufficient strength so that it is not subjected to disintegration and erosion under heavy traffic
, ': :Cverse environmental conditions such as excessive moisture, freezing and thawing. ln the light of these
ements,subbaseof DryLeanConcrete(DLC) having a7day averagecompressivestrengthof l0MPais..  rmended. lVinimum recommended thickness of DLC for major highways is150 mm.
:=::ve kvalues of different combinations of subgrade and subbase (untreated granular and cement treated
*  1um 7 day compressive strength of 10 MPa) thicknesses laid over granular subbase consisting of filter and
: ": ^age layers can be adopted fromTable 4. The contribution of granular subbase placed below the DLC layer
:. 5e ignored for estimating the effective modulus of subgrade reaction of the foundation. The values given in ,: e 4 are based on theoretical analysis and an upper limit of 300 MPa/m is recommended considering the loss
. .rbgrade support expected to be caused by heavy traffic.
Table 3 Kvalue for Granular and Cement Treated Subbases
28
56
84
::,,en day unconfined compressive strength of cementtreated granularsoilshould be minimum of 2.1 IVIPa. DLC
, :uld have minimum compressive strength of 7 MPa at seven days.
Table 4Kvalues for DLC subbase.,.value
225150 300 100 150 200
JV 44 53 76 108 141
63 75 BB 127 4 7') 225
92 102 119
21 28 42 48 55 62
56 97 '166 208 278 300
97 138 208 277/ 1 300 300
value of effective k shall be taken as 300 MPa/m
lective
,4A
grade, (MPa/m)
Effective k (MPalm) of untreated Granular
layer subbase of thickness in mm
Effective k (MPa/m) of Cement treated sub
base of thickness in mm
mm:ective k
 : Separation Layer between subbase and pavement
:.   ayer below concrete slabs should be smooth to reduce the inter layer friction. A separation membrane
 :htckness of 125 micron polythene is recommended to reduce the friction between concrete slabs and_ _: ase.
  Srainage layer
: ::3 the quick disposal of water that is likely to enter the subgrade, a drainage layer may be provided' ,  .^e pavement throughout road width above subgrade. An example and Specifications for drainage is  ^e Appendix lll of this guidelines.
 : Characteristics of concrete
:' Designstrength
: ::;nof therigidpavementisbasedontheflexural stressesof concrete.Flexural strengthof concretecan ::':'ed after testing the concrete beam as per procedure (lS. 516). Alternatively, it can be derived from the, ,:.:'stic compressive strength using following relationship.
f,, = 01xyl7i Equation 2
f ,, = flexural strength (modulus of rupture), Mga, frt = characteristic compressive cube strength ofconcrete, [MPa
Sr= Characteristic flexural strength at 28 days, Mpa.
S = Target mean flexural strength at 28 days, Mpa.
Za = Tolerance factor for the desired confidence level, known as the standard nonnal variate, which is
1.96 for 5% confidence level.
o = expected standard deviation of field test samples, Mpa
en the target average flexural strength is given as:
s = s1 * Zoo Equation 3
,:e values of Standard Normal Variate zois given in the table below. The values of z odepends on the importance:r the road. For example it is recommended that the value of zomay be adopted as 1.96 and for expressways itay be taken as 2.33.
Table 5 Values of Standard Normal Variate, Z o
depends on the importance of the road.
'l in 15 1E Fair
1in20 t.b3 Good
1in40 196 Very Good
1 in 100 Excellent
^e values of
rt/
Accepted proportion oflow results (tolerance)
Standard Normal Yariate, Z o Quality Level
2.33
,, l:ld: means construction with semimechanized methods and site mixed/semiautomatic batching plant,' :' i bar/dowel bars and joint cutting by manualmethod/Joint cutting by machine (usually for low traffic
: . ' ery Good: means construction with semi mechanized/ fixed form pavingmachines and batch mixed
: : , .' semiautomatic/automatic batching plantinsertion of tie bars and dowel bars by manual method
 edium traffic roads
I ::d to Excellent: means construction with fixed form/slip form paving machinesand batch mixed concrete
,':attc batching plant insefiion of tie bars and dowelbars by manual/automatic dowel/tie bar inse(ion
,  . method usually for heavytraffic roads/expressway
  .1 by modulus of rupture tests underthird point loading. The preferred size of the beam should be 15cm
' , C cm when the size of the aggregate is more than 20 mm. When the maximum size of the aggregate is=^ 20 mm, 10 cm x 10 cm x 50 cm beams may be used.
'   ^3ton of flexural strength by correlating with cube strength (compressive strength)shall not be allowed for, . ' ::.ects as the correlation is not well established.
:::l deviation depends upon the degree of quality control, exercised during production of aggregate and
':'::e mix for major projects using batch type mixing plant with modern aggregates crushing plants, standardi :::''t will be relatively much less as compared to the locations where mix is prepared using semimechanized: : :: on process. The values of standard deviation o used in
:. ..' :^ 3 for major projects shall according be adopted corresponding to the deviation in flexural strength actually
. =l n the field. However, for the initial mix design for major projects value of standard deviation shall be
: : ?S per the table below.
 :: : i Expected values of Standard Deviation
3.0
3.5
4.0
4.5
2 8 2 lVodulus of Elasticity and Poisson's Ratio of Concrete= nodulus of elasticity (E) and the Poisson's ratio (p) of cement concrete are known to vary with concrete*'.:rials and strength. The elastic modulus increases with the increase in strength, and Poisson's ratio decrease
 the increase in modulus of elasticity. While it is desirable that the values of these parameters are ascertained
.,:erimentally forthe concrete mix and materials actually to be used in the construction, this information may not
;,,ays be available at the design stage. Even a 25 percent variation in E and p values does not have any
,:^flcant effect on the flexural stresses in the pavement concrete. lt is suggested that for design purpose, the
: cwing values may be adopted for concrete for the flexural strength of 4.5 MPa.
':dulus of elasticity of the concrete E = experimentally determined value or 30,000 l\tlPa (3x'l0s kg/cm2)
0.38 055
0.35 0.50 0.55
0.32 0.45 0.50
0.29 0.40 0.45
0.26 0.35 0.40
r: sson's ratio p = 0.15
&/
Standard deviation for different degree of control, MPa
Very good Good Fair
0.6
strength, MPa
I cefficient of thermal expansion
:.:l of thermal expansion of concrete (a) varies with the type of aggregate. However, for design: : , ,e of cr=10 x 'l0s per oC may be adopted.
:  :attgue behavior of cement concrete.  :"r : :pplication of wheel loads create repeated flexural stresses, which results in the progressive fatigue =entconcreteslabintheformof gradual developmentof microcracksespeciallywhentheapplied
. *. :f flexural strength of concrete is high The ratio between the flexural stressdue to the load and the::: of concrete is termed as the Stress Ratio (SR). lf the SR is less than 0.45, the concrete is
  , .slain infinite numberof repetitions. As the stress ratio increases, the numberof repetitions required: ':: ' ^g decreases. The ratio between fatigue life (IV/ and stress ratio is given as:
N = unlimited for SR < 0.45
m ]3 'zuu wh.n 0 45 < sR s 0 55Equation 4
I Or 1 0 CD
= ' ForSR>0.550.0828
Equation 5
" :  , :f these fatigue equations is discussed in the Appendix l.
 : :s lf fatigue life for different values of stress ratio are given in the table below. Use of fatigue criteria isI :  e basis of lViner's hypothesis. Fatigue resistance not consumed by repetitions of one load is available ' :' : ns of other loads.
Table 7 Stress ratio and allowable repetition for the concrete slab
045 6.279x107 5.83x10r
0.46 1.4335x'107 0.67 4.41x103
0.47 5.2x106 068 3.34x10e
0.48 2.4x104 0.69 2531
0.49 1.287x106 0.7 1970
0.50 7.62x105 1451
051 4.85x105 1 099
0.52 3.26x10s 0.73 p.1.'.)
0.53 2.29x105 0.74 630
0.54 1.66x105 0.75 477
0.55 1.24x105 0.76 361
0.56 9.41x1 0+ 0.77 274
0.57 7.12x10q 0.78 207
0.58 5.4x10a 079 157
059 4.08x104 0.8 1 '19
3.09x104 0.81
0.61 2.34x10a 0.82 68
0.62 1.77x10a 0.83 52
0.63 1.34x10q 0.84 39
0.64 1.02x104 0.85 <Il
0.65 7.7x103I
rffi
Stress
Ratio
Allowable
RepetitionStress
Ratio
Allowable
Repetition
0.66
0.71
0.72
U.bU 90
q,
CHAPTER 3 DESIGN OF SLAB THICKNESS
. 3 ritical stress conditions
. :3rete pavements are subjected to stresses due to the variety of factors, acting simultaneously. The' .,' rbinations of drfferent factors that induce the maximum stress in the pavement will give the critical
" ,. ard temperature differentials between top and bottom fibers of concrete slab, as the two assumed to
'. : stress at the bottom layerof the concrete slab is the maximum during the day hour; when the axle,   dway on the pavement slab while there is a positive temperature gradient as illustrated in Figure 5 and
. : rrs condition is likely to produce bottomup cracking (BUC).
Location of lVaximum tensile
Stress
Concrete Slab
Foundation
Figure 5 Axle Load Placed at the Middle of the Stab during Mid.day
  : : 's of points of maximum flexural stress at the bottom of the pavement slab without tied concrete shoulder:  I e tandem and tridem axles are shown in Figure 6. The tyre imprints are tangential to the longitudinal edge.: . '=l concrete shoulders also, the maximum stress occurs at the same locations. Single axles cause highest
..='ollowed by tandem and tridem axles respectively. Spacing between rndividualaxles fortandem and tridem, =. ;aries from 1 .30 m to about 1 .40 m. There is practically no difference in stresses for axle spacing between
'   and 1 .40 m. A spacing of 1 .30 m has been used in these guidelines for stress computation. ./o
   ^; the night time, the top surface is cooler than the bottom surface and the ends of the slab curl up resulting in. ,, :f support for the slab as shown in Figure 7. Due to the restraint provided by the selfweight of concrete and: e dowel connections, temperature tensile stresses are caused at the top. Figure Tshows the placement of, = cads close to transverse joints when there is negative temperature gradient during night period causing high
=''al stresses in the top layer leading to topdown cracking. Positioning of axles of different configurations on the,.: rvrth successive axles placed close to the transverse joints is shown in Figure 8. These axle positions can
':te topdown cracking (TDC) during the night hours when the pavement has the tendency to cud up. Builtin =
*anent curl induced during the curing of the concrete slab further worsens the problem.
I Wheel Load
X
Transveise Joint
fl[
Tridem Axle
fr[
Joint
fr[flf
Tandem Axle
'3 rsverse Joint
B!;: ngle Axle
a of lvlaximum Tensile Stress at bottom of slab tied shoulders
' ; 'e 6 Placement of Axles for Maximum Edge Flexural Stress at the bottom of the slabwithout Tied ConcreteShoulders
F
Wheel + I Wheel Load
Location of maximum tensile
StreSS
Foundation
Figure 7 Placement of Two Axles of a Commercial Vehicle on a Slab Curled During Night Hours
3.1 Calculation of flexural stress
. :: the loads causing failure of pavements are mostly applied by single,tandem, tridem and other multiple axles,: ':sses should be determined for theconditions illustrated in Figure 5, Figure 6, Figure 7 and Figure 8. Various,:cutational techniques have been developed based on the Picket and Ray (1951)'s work on computation of:":sses in infinite slabs. lt is used for the computation of load stress in the edge region of pavements without tied. :  3rete shoulders if there is no temperature gradient in the slab. : following combinations of pavements and loading were considered for the analysis of bottomup and topdown.'a:king. For bottomup cracking case, the combination of load and positive nonlinear temperature differential:
3ure 5) has been considered whereas for topdown cracking analysis, the combination of load and negative:ar temperature differential (Figure 7) has been taken. For bottomup cracking analysis, srngle/tandem axles
,re been placed on the slab in the positions indicated in Figure 6. ln bottomup cracking case, single axle load
:JSes the largest edge stress followed by tandem and tridem axles. Since the stresses due to tridem axles are
,rall, they were not considered for stress analysis for bottomup cracking case. For topdown cracking analysis,q load poEition considered for analysis is as shown in Figure 8. As indicated in the figure, only one axle of> ' 'r
/
Concrete Slab
Transverse Joint
=
o
Bfl
Single Axle
:.::* tridem axle units has been considered for analysis in combination with front axle. Front axle weight
 'ssumed to be 50 percent of the weight of one axle of the rear axle unit (single/tandem/tridem) for
' ;  'e 8 Different Axle Load Positions Causing Stress at the Top Fiber of the Slab with Tied Concrete Shoulder
ng cases shall be considered:
Bottomup cracking
. Pavement with tied concrete shoulders for single rear axle. Pavement without concrete shoulders for single rear axle
. Pavement with tied concrete shoulders for tandem rear axle. Pavement without concrete shoulders for tandem rear axle
: Topdown cracking:.
'.nts with and without dowel bars having front steering axle with single tyres and the first axle of the rear
=  ^ t (single/tandem/tridem) placed on the same panel as depicted in Figure 8.
:.. :avy traffic conditions, dowel bars are usually provided across transverse joints for load transfer. Tied.  ':ie shoulders are also necessary for high volume roads. However, for smaller traffic volumes smaller than
. :rmmercial vehicles per day tied concrete shoulders and dowel bars are not generally warranted. Finite. : . 3rt analysis has been carried out for pavements with and without
dowelled transverse joints and
tied concrete shoulders.
:'ninal load transfer efficiencies (LTE) for dowelled transverse joints and tied joints between the slab and
" :'ete shoulder shall be taken as 50 percent and 40 percent respectively.
= results of finite element analysis of a large number of concrete pavements with different pavement
:'gurations subjected to various combinations of axle loads and temperature differentials have been presented ^e form of Charts in AppendixVlll. The charts can be used to obtain the edge flexural stress caused by a,=:ified magnitude of single/tandem axle (positioned as indicated inFigure 6in combination with a specified
 : s tive temperature differential for a given pavement structure. Linear interpolation can be done for obtaining
: 3sses for intermediate loads and temperatures from the charts given in Appendix.Vlll.
^e finite element analysis results have also been used to develop regression equations for estimation of the':x11ral tensile stress for bottomup as well as topdown cracking cases. Whlle a single regression equation was
\ , '."e '
I
Joint
=
a
Tandem Axle
Transverse Joint
E" "fl
Tridem Axle
EO
' :e adequate for estimation of flexural tensile stress in the slab for topdown case, separate equations'= =cped for the bottomup case for different pavement types and foundation strengths. The regression
' ': are presented in Appendix'll.Designers can develop their own excel sheets for analysis and design : :: .egression equations.
. i Cumulative Fatigue Damage Analysis
: : 'en slab thickness and other design parameters, the pavement willbe checked for cumulative bottomup':iown fatigue damage. For bottomupcracking, the flexuralstress at the edge due to the combined action. ortandemrearaxle load and positive temperature differential cycles is considered. This stresscan either '==:ted from the stress charts given in AppendixVlll or by using theregression equations. Charts explains
 i ' :1e interplay ofthickness, modulus of subgrade reactions, axle loads and temperature differentials.Similarly,
"sessing the topdown fatigue damage caused by repeated cycles ofaxle loads and negative temperature.'=^: al, flexural stress can be estimatedusing the regression equation.
 '' 'sxllralstress is divided by the design flexuralstrength (modulus of rupture) of the cement concrete to obtain
"s ratio (SR). lf the stress ratio (SR) is less than 0.45 then allowable number of cycles of axle load is infinity.r '" ::'ess ratio values greater than 0.45, allowable cycle of loading (axle load + temperature) can be estimated' : :quation 4 and Equation 5. The concrete slab undergoes fatigue damage through crack growth induced by
'=;:ed cycles of loading. The cumulative fatigue damage caused to the slab during its service life should be: :  a I to or less than one. : ''sis indicates that contribution to cumulative fatigue damage (CFD) for bottomup cracking is significant only
^g day time (say 10 AtM to 4 PM) because of higher stresses due to simultaneous action of wheel load and
 ': : ve temperature gradient. Thus, the day hour traffic during the six hour (10 AM to 4 ptM) is to be considered for:':l'Trup cracking analysis. Forthe topdown cracking analysis, only the CFD caused during the period between
 !1 and 6 Afi/ is imporlant. Hence, the six hour night time traffic (0 AM and 6 Alv) only is to be taken for,:*cuting CFD fortopdown cracking analysis, lf the exact proporlionsof traffic expected during the specified six"r periods are not available, it may be assumed that the total night timetraffic is equally distributed among the'':]ve night hours. Simitarly, the total day time traffic may be assumed to be distributed uniformly among the :lve day hours.The Cumulative fatigue damage (CFD) expressions for bottomup and topdown cracking cases.: given by
:quation 6 and Equation 7 shows respectively (Source IRC: 582015), The times indicated in the equations will'ary depending on the geographical location of the project site but the duration of each period will practically':main the same.
CFD(BUC) =
CFD(TDC) =
jIijIi
(tr
@,
(10 AM to 4 PM)
(0 AM to 6 AM)
Equation 6
Equation 7
{
IV; = allowable number of load and temperature differential cycles for the wheel load group during thespecified sixhour period;
n, = predicted number of load and temperature differential cycles for the Ihsixhour period
Where,
j = total number of load groups
group during the specified
. 3 Drainage Layer
': 'rle loads are very common for major highways in Nepal and therefore, it should be ensured that the' : layers do not undergo unacceptable permanent deformation under repeated loading. Entrapped water in ':'ade and granular subbase may cause erosion of the foundation material since pore water pressure".' =J by tandem and tridem axle loads may be substantially high. lt may be mentioned that pavement ::'l"r due to heavy tandem and tridem axles can be as high ffi t.o mm which may result in the formation of
" : :: ow the pavement due to the permanent deformation of the foundation material. presence of excessI'': accumulated in the unbound foundation layers due to infiltration or due to thawing in snowbound regions.  :,crve for deveropment of permanent deformation in these rayers.
' :ate the quick disposal of water that is likely to enter the subgrade, a drainage layer together with a;' :::aration layer may be provided beneath the subbase throughout the road width. The filter/separation layer= =:s fines from pumping up from the subgrade to the drainige layer. lt also provides a platform for the"' :tton of the drainage layer. The amount of water infiltrating into ttre pavement should be assessed and a='':e layer having the required permeability needs to be designed. ln no case should the coefficient of' =:brlityof drainagelayerbelessthan30m/dayevenforlowrainfaltarea.Therequirementofthecoefficient:
= eability can be as high as 300 m/day or more in some cases and it is essential to design the drainage layer: ' '.^' ately for major highways in areas having annual rainfall in excess of 1000 mm. The diainage layer can be
'='= trith 2 percentcementor2.5 percent bitumen emulsion so as to permitthe construction traffic withoutany:= ':v displacement and/or shoving of the aggregates. lf granular rayers are not needed because of high'::r subgrade, synthetic geocomposite with reduced thickness of granular layer can be used over the_ ::ie to function both as a filter as well as a drainage layer. ='ample of the design of a drainage rayer is given in the Appendix lil.
3 4 Tied Concrete Shoulder and Widened Outer Lane .  ::ment concrete shoulders are recommended to protect the edge of high volume highway pavements. These" : nes provide for design of concrete pavements with tied concrete shoulders. wroening'of outer lanes of' 
":te pavement by 0'5 m to 0 6 m can be adopted for twolane twoway roads to reduce the flexural stresses in' "reel
path region Analysis of typical concrete pavements shows that provision of a widened outer lane ::ning as a monolithic concrete shoulder reduces the edge flexuralstress by 20 to 30 percent. This will result'=:lction of pavement thickness. The total quantity of concrete is likely to be nearly the same as that without  cer Rough texture, if provided to the widened part, will bring in additional safety to vehicles particularly during:: rours Thicknesses of pavements with widened outer lane as well as tied concrete shoulder are almost the
=  = An example of designing concrete pavements with widened outer lanes has been included in Appendix.lv.
3,5 Bonded Rigid pavement
 ::"crete pavement layer may be laid directly over the DLC layer. ln this case pQC and DLC act as monolithic'' :" between layers can be achieved and thickness of the concrete slab can be reduced. DLC surface is made':r by using wire brush Bounding is achieved by pouring the waterand cementslurry before laying the pec.::':avements constructed with full bonding between the slab and DLC layer, transverse joints may be formed in= ILC layer by cutting grooves to 'l /3rd of its depth at exactly the same rocations where transverse joints are to
' = :'cvtded in the upper PQC layer in order to prevent random reflection cracking of upper layer due to cracks in= .rnjointed DLC layer' rhe 7day compressive strength of the DLC layer in bonded rigid pavement should not: :ss than 10 Mpa.
 :'anular subbase of 200 mm to 250 mm thickness may be provided below the DLC layer for bonded concrete: ' ' ement for filtration and drainage. The effective modulus of subgrade reaction of the subgradegranular subbase:rbination can be estimated from Table 3. Totalslab thicknesrlr,;or.rthe granular layer is worked out forthe ':n traffic and other design parameters. A part of the PQC of thickness 'h' is to be replaced with 150 mm of DLC,i:rat!>combined flexuralstiffness of the pavementslab layer(thickness of h,) and I) f.tavar/rhintrnocc nrh \
, ,'greaterthan the flexuralstiffness of the slab of thickness'h'overthe granular layer. Flexuralstiffness' ' : ckness,h. is given as:
EI Eh3 Equation 8
7  tt' 12(1 1r2)
'  I p, l,hare the modulus of elasticity, Poisson's ratio, moment of inertia and thickness of the slab
  . =i to provide a DLC layer of thickness 'hz" bonded to the concrete slab of thickness "h1", the thickness of ::e slab (h7) can be obtained by equating the flexural stiffness of the design slab to the combined flexural.:: l'theDLClayerandconcreteslab.Figuregshowsthebondedsectionwithneutral axisinthepavement
1m
h1/2d
f> h2/2
E2/E1
1
Figure 9 Concept used for Obtaining Combined Flexural Stiffness
e reutral axis depth 'd' can be calculated using the
, _ o.s(htr) . (fr) hz(h,  0.sh1) Equation e
h + c;)h,'= 'exural stiffness of the two layers can be determined using the equations below.
,,({f) +h{doshl)'z) Equation 10Flexural sttf fness(7,PQC) = 1 p?
Equation 11
FLexural stif fness(Z,DLC) = (t  p3)
ne combined flexural stiffness of two layers should be equal to or more than the requirement of design flexuralstiffness given by the Equation 8. 28day strength of DLC may be taken as 13.6 MPa. fhe'E' of DLC layer may betaken as 13600 MPa. Poisson's ratio of DLC is taken as 0.2.
3.6 Recommended procedure for Slab design
The Concrete Pavement slab is recommended to adopt the following steps. Example of design is given in the
,,\ry. (ft) ttzut, +?  ot'
Appendix lV
G'.l
rm
Step 1: Specify design values for the various parameters related to the traffic calculation, subgradem,Erqfi and temperature variations at night and day{ime and etc.
Step 2: Select a trial design thickness of pavement slab
Step 3: Compute the repetitions of axle loads of different magnitudes and different Categories during thedesign life
Step 4: Find the proportions of axle load repetitions operating during the day and Night periods
Step 5: Estimate the axle load repetitions in the specified sixhourperiod during the day time. Themaximum temperature differential is assumed to remain constant during the 6 hours for analysisof bottom.up cracking
Step 6: Estimate the axle load repetitions in the specified sixhour period during the night time. Themaximum negative temperature differential during night is taken as half of day{ime maximumtemperature differential. Built in negative temperature differential of 50oC developed during thesetting of the concrete is to be added to the temperature differential for the analysis of top.downcracking' Only those vehicles with spacing between the front (steering) axle and the first rearaxle less than the transverse joint spacing need to be considered for topdown cracking analysis.
Step 7: Compute the flexural stresses at the edge due to the single and tandem axle loads for thecombined effect of axle loads and positive temperature differential during the day time.Determine the stress ratio (Flexural stress/ Modulus of Rupture) and evaluate the cumulativefatigue damage (CFD) for single and tandem axle loads. Sum of the two CFDs should be lessthan '1.0 for the slab to be safe against bottomup cracking.
Step 8: Compute the flexural stress in the central area of the pavement slab with the front axle near theapproaching transverse joint and the rear axle close to the following joint in the same panelunder negative temperature differential. Determine the stress ratio and evaluate the CFD fordifferent axle loads forthe analysis of topdown cracking. CFD should be less than 1.0 fortopdown cracking design
N
v
CHAPIER 4 DESIGN OF JOINTS
_' : :  +
I n the cement concrete pavement ry crucial item of work because theseis VC
I I ^ a'ii g S lg nifica nt effect 0n pavement performance he oints also need to beT;:1 well Vario US types of joints rn Cement Co ncrete P'ave me nts
n
are shown in Figure
mffi ,*[:=,;d]*,*irffi :#ffi tr*#rf?r ;1*ff ,iffi #:x.,,;l#l
M'#:,:xilH,r,,:ff ,:*,'i:ffi ff T,,[ff $iJ.Ji,Tfi :#:fr ::r*l.,il:fi :hffim ra'srrl tocking of the joints and preventing expansion oi concrete slabs. They are, therefore, no longertmtm".r'=r, dl. p"rn.nt structure like bridges and culverts.ri
hMImM olnB should, as far as possible, be placed at the location of contraction joints except in case of@''*T tnErl a key joint may be used.Longitudinaljoints are required in pavements of width greater than 4.5 mhdlhr rtrrr,_sverse contraction and warping
m* meis of contraction joint, longitudinatlointand expansion jointareshown in Figure 11, Figure 12, Figure
:: :raction Joints
MS round Dowel bar
Figure 10 Types of CC pavements
Top of the groove is widened forsealing purpose
d
Plastic sheathing
Joints in CC Pavement
1/3 d
 _,",gf Guidelines (Riqid)
Figure 'll joint with Dowel bar
Transverse Joints Longitudinal Joints
Construction Joints Expansion Joints
"^m
Top of the groove is widened forsealing purpose
d
Tie bar150 mm length painted with
bituminous paint
Figure 12 Longitudinal joint with tie rod between two lanes
Compressible Dowel cap filled with compressivematerials
d
l\ils round Dowel bar Plastic sheathing
Figure 13 Expansion joint with dowel bar
4.2 Load Transfer at Transverse Joints
oad transfer to relieve part of the load stresses in edge and corner regionsof pavement slab at transverse joints is
3rovided by means of mild steel round dowelbars. ln the areas of high rainfall, coated/corrosion resistant dowel
tars are oftenused to provide long term load transfer. The coating may be zinc or lead based paint orepoxy
:oating. Dowel bars enable good riding quality to be maintained by preventingfaulting at the joints.
The bearing stress in the concrete that is responsible for the performance of dowel bars at the joints. High concrete
bearing stress can fracture the concrete surrounding the dowel bars, leading to the looseness of the dowel bar and
tre deterioration of the load transfer system with eventual faulting of the slab. Larger diameter dowel bars are
bund to provide better performance. Maximum bearing stress (F6,,J between the concrete and dowel bar is
cbtained fromEquation 1 2.
Equation 12 kmarPt(Z + pZ), Dmax 4B3EI
, here,
4 Kmd.sbd
E= Relative stiffness of the bar embedded in concrete, mm = 4EI
kmds= lvlodulus of dowel support, MPa/m
ba diameter of the dowel bar, mm
Z = Joint width (5 mm for contraction joint and 20 mm
E = Modulus of elasticity of the dowel bar, Ji/Pa
I = Moment of inertia of the dowel bar, mma
transfer by designed dowel bar, kN
S/
1t3 d
joint), in mm
100 mm
:
f* = characteristic compressive strength of the concrete, Mpa (For M 35 concrete, f"1 = 35 Mpa (2gdays); = 42\Apa(90 days)
srce tre initial load transfer efficiency (LTE) at the transverse joint is almost 100 percent and it takes a long timeiT fie LTE to decrease with traffic repetitions, 90 day compressive strength can safely be used for thertprtation of allowable bearing stress.
=r lrcavy traffic' greater than 450 cvPD, dowels are to be provided at thecontraction ioints since aggregate intermi cannot be relied upon to effect loadhansfer across the joint to prevent faulting due to the repeated loading of=avy axles Joint widths of 5 mm and 20 mm may be taken for stress computation in dowel bar atcontraction and:rlansion joint respectively. Recommended diameter and length of dowel bar is given in the table below.
. '  s of dowelsupport ranges from 80,000 to 415,000 tvPa/m. A typicalvalue of 415,000 lvpa/m may be  '' lesign since only the fourth root of lhe kvatue affects the computation ofp.
'"'' bar should be designed for the maximum load being transferredby it for the allowable bearing' 3ased on the expression givenby the American concrete lnstitute (ACl) committee, Equation 13may be: '.' ..:ulationof the allOwable bearing stress on concrete.
(701.6  b)frt Equation 13' =  9825
Fr = allowable bearing stress, tt/pa
0a = dowel diameter, mm
Table I Recommended Dimensions of Dowel bars
:te Dowel bars for the slabs of thickness less than 200 mm is not recommended to provide.The recommended:3 e is suitable for concrete grade M40. lf the concrete used for slab is of grade less than M40, then dowel bars,^all be provided only afterdesign as in the Appendix V.
4.3 Dowelgroup action
hen loads are appried at a joint, a portion of the road is transferred to the other side of the slab through the dowelbars' lf the load is near the joint of a pavement srab tied to a conc rete shoulder, a part of the load is transferred tore shoulder also. The dowel bar immedi ately below the wheer road carries maximum amount of roadand other:cwel bars transfer progressively smaller magnitudes of loads. Repeated loading causes some looseness between:re dowel bars and the concrete slab and recent studies indicate that the dowel bars within a distance of one radiusof relativestiffness (7.0 /)from the point of load application participate in load transfer. Assumi nga linear variation of:he load by different dowel bars within 7.01 the maximumload carried by a dowelgiven in Appendix V,
a
Diameter, mm Length, mm25 360 300
230 30 400 300250 32 450 300280 JO 450 300300 38 500 300350 3B 500 300
of the design of dowel bars is
nbar can be computed
Slab thickness bar
mm200
II
I
r .: Jesign of Tie Bars for Longitudinal Joints
' ::ldinal joint is expected to open up during the service period (in case of heavy traffic, expansive' ':=s etc). The area of steel required per meter length of joint may be computed using the following.: '
Abfw
t sr,Equation 14
::As = dre? of steel in mm2, required permeter length of joint;b = lane width in meter;
f = coefficient of friction between pavement and the subbase/base (usually taken as 1.5)l',7 = weight of slab in kN/mz and
S.r allowable working stress of steel in Mpa
 ' =^::h of any tie bar should be at least twice that required to develop a bond strength equal to the working" ' :' :he steel. The formula for estimating the length of the tie bar is given as in the Equation below.
, _z srtArs Equation 15''  B pprb" : :
[ = length of tie bar (mm)
S,r = allowable working stress in steel (tMpa)
Acs = ctossSectional area of one tie bar (mmz)
Ppo= porimeter of tie bar (mm), and
B = permissible bond stress of concrete; (for deformed tie bars = 2.46 r,Apa.,for prain tie bars 1. z5 Mpa.):  '' 'ted Cement Concrete needs to be provided in pavement panels in curued porlions of radius tess than 45 m' . ' tnderpasses, on steep gradients, and for stabs having manhole cover slab having uB (tength to breadth). .  tre than 1 .5 and in other simitar situations.:
='mit warping at the joint, the maximum diameter of tie bars may be limited to 16 mm, and to avoid' '^:ration of tensile stress they should not be spaced more than 750 mm apart. The calculated length, L, may
: ' :'eased by 50 to 80 mm to account for any inaccuracy that may occur in the placement during construction. = , arple of design of tie bar is given in Appendix Vl.
150 x'10
330
520530
830440510
480560
200 10
12390
3bu620900
5'10
580560
640250 12 450 720 580 640JUU 12
16
370bbu
600'1060
580
720640
800350 12
16
320570
510
910580
720640
800
\
V
.r
Slab thickness barDiameter (d) mm Spacing, mmMaximum Minimum length, mm
Plain Deformed Plain :formed
  : recommended details are based on the following values of different design parameters: S = 125 tr/pa for,'s 200 lVPa for deformed bars; bond stress for plain bars = 1.75 lVpa, for deformed bars = 2.46 Mpa as:  58r2015.
: : Reinforcement in cement concrete srab to contror cracking: 'lement in concrete pavements, is intended to hold the cracked faces tighfly together, so as to prevent
= l of the cracks and to maintain aggregate interlock required for load transfer. lt does not increase the.  .,
strength of unbroken slab when used in quantities which are considered economical.
:' ':'lement in concrete slabs, when provided, is designed to counteract the tensile stresses caused by'aJe and contraction due to temperature or moisture changes. The maximum tension in the steel across the
''' :quals the force required to overcome friction between the pavement and its foundation, from the crack to
' =arest joint or free edge. This force is the greatest in the middle of the slab where the cracks occur first.:' ':'3ement is designed for this critical location. However, for practical reasons reinforcement is kept uniform"
 'rout the length for short slabs.'' '"nount of longitudinal and transverse steel required per meter width or length of slab is computed by the. : ,, 19 formula:
ASLafw Equation 16
2Srt
:'e,
4s = area of steel in mm2 required per m width or length of slab;fu=distance (m) between free transverse joints (for longitudinal steel) or freelongitudinaljoints (fortransverse steel)
f = coefficient of friction between pavement and subbase/base (usually takenas 1.5)lll = weight of the slab in kN/m2 andS.r= allowable working stress in steel in tMPa (usually taken as 50 to 60 percentof the minimum yieldstress of steel.
Since reinforcement in the concrete slabs is not intended to contribute towards its flexural strength, its position*ihin the slab is not important except that it should be adequately protected from corrosion. Since cracks startingTom the top surface are more critical because of ingress of water when they open up, the general preference is forfie placing of reinforcement about 50 to 60 mm below the top surface. Reinforcement is often continued acrossongitudinal joints to serve the same purpose as tie bars, but it is kept at least 50 mm away from the face of theransverse joints and edge. ln special cases, the steel reinforcement shall be provided in acute curve portions,under passes, steep gradients and slabs having manhole coveE and slabs having length to breadth ratio moreftan 1.5 and at acute angled corners.
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CHAPTER5 COUNTINUOUSLYREINFORCEDCONCRETEPAVEMENT
5.1 lntroduction
The technique of Continuously Reinforced Concrete Pavement (CRCP) does not require the expansion andcontraction joints, thus permitting very long slab lengths with improved riding comfort and reduced maintenance ascompared to plain concrete pavements. Conventional CRCP requires relatively high percentage of steel rangingfrom 0'7 to 1'0 percent of concrete crosssection. The technique of CRCP construction with elastic joints (cRCpEJ) facilitates significant reduction in quantity of steel required (0.4  0.5 percent) and also eliminates the randomcracks which occur in conventional continuously reinforced concrete pavements.
The provision of continuous reinforcement in CRCP of the conventional type results in the formation of transversecracks in the pavement which are held tightly closed by the steel without impairment of structural strength. Theclosely held cracks ensure load kansfer across the cracks through aggregate interlocking and also prevent theingress of water and grit into the cracks. The width and spacing of sucrr cracks are dependent on the amount ofsteel reinforcement provided. The greater the amount of steel, the closer is the spacing of the cracks and thesmaller is their opening. An optimum amount of longitudinal reinforcement is called for so that the cracks areneither too widely spaced with resulting overstressing of steel, loss in load transfer provided by aggregateinterlock and accelerated corrosion of steel; nor too closely spaced so as to cause disintegration of the slab.
The elastic joints consist of dummy contraction joints with the reinforcement continuous through them. Thereinforcement is painted with a bondbreaking medium over a specified design length on either side of the jointgroove to provide adequate gauge length for limiting the steel strains due to joint movement.
The use of elastic joints, apart from resulting in reduction of steel stresses by about 50 per cent and enabling theuse of less quantity of steel, also preclude the random cracking associated with conventional construction, sincefie weakened plane provided at such joints localizes the cracking. The usual spacing of such joints works out toabout4to5m.
5.2 Design
calculation for steel Percentage and stresses in steel and Concrete due to continuity at Elastic JointsThe continuity of steel at elastic joints leads to restraint in the slab movement due to shrinkage and temperaturechange, and thus induces stresses in both steel and concrete. However, if steel is provided at middepth of theslab, as is the usual practice, no stress wiil deverop in it due to wheer road and warping.
The stresses (due to continuity of steel at elastic joints) in steel, (os), and concrete, (oc), in the vicinity of elasticjoints may be calculated from Ersenann equations which are given below;
os L00(a. Ar.h. Eo. Es)kg/cmzf'E'(7  1) + (100. h.Ec.1)
and
a.A7.h.Es.E"o, =;ugfiffi,tt kg/cm2
,vhere
o = Coefficient of thermal expansion of concrete per oC,
4r = Difference between the mean temperatures of the slab at the time of construction and the coldest period in"C,
At = Maximum tem re differential between top and bottom of the slab as given in Table 'l of this guidelines,
rature stress at the edge, At is different from lr which is used in these equationsfinding the
difference between the minimum of minimrrms anruho moan famnarafi rrn at *[.^
I,/
*: cf constructicn lr ts not a function of slab thickness whereas At the temperature differential depends,  :'e thickness of slab.
.: :^ ckness in cm
, . r  us of elasticity of concrete in kg/cm2
. : " ls of elasticity of steel in kg/cm2
, , s section of steel in 1 m width of the slab in cm2
: =  :f free, unbonded length of the steel to the slab length between two consecutive elastic joints.
  3 tudinal reinforcement'16rm dia. @26 cm clc
 :  s, erse reinforcement
  Cia. @ 41 cm ctc
Figure 144 Details of Elastic foint Section
' '::ses developed in steel and concrete per 0C of AT for steel percentage range of 0.1  0.6 and for ratio, A
' ' I10.4. of are shown in the charts below. The steel stresses so determined should not exceed the
' ' : e value of 1400 kg/cm2. The concrete stresses are additive to the load and temperature warping stresses' ' '=quired to be taken into account while designing the pavement. The transverse steel may be taken as 25 ' :' longitudinal steel.
' " '
ston of steel enables some increase in the effective slab thickness and its continuity at elastic joints:= ' addttional load transfer over and above that provided by conventional dummy contraction joint. At the' .: the percentage of steel is small enough not to induce any restraint to bending of the slab at elastic
  = to effect of wheel load and temperature warping.
  IRCP without elastic joints the permissible stress in steel in 2800 kg/cm2 (i.e. the steet is allowed to be' :: JO to the yield point), the permissible value in CRCP with elastic joints is restricted to 1400 kg/cm2 only:*al working stress in steel used in conventionalstructures). The lower permissible stress in steel in CRCp
' ":: ioints enables taking advantage of the effective increase in concrete slab thickness due to provision of e the permissible yield stress limit in steel in the case of CRCP without elastic joint prevents such::r
Longitudinal reinforcement coated withbitumen to break bond with concrete over
this length of 150 mm
filled with bitumen coated strip
Depth 5 cm: Elastic Dummy Groove
Mild steel chairs to supportthe longitud inal reinforcement
T*^I
_l
N!
ec
I roocm l
\
D esio n G r id elinec / P in irt)
{
F
ooo0)
Eoo
;oo)
a
200
'180
160
140
120
100
80
70
bU
50
/
, 0.10/o
o 20k
0.34
(
0.59
0.6Yo
.4%
\\\
\ N0.0 0.15 0.20 0.25 0.30 0.3S
A = Ratio of unbonded length of steel to spacing of elastic joints
Figure 15: Design Charts for calculation ofstresses in Steel
(
\
I
ffi
09
0.8
0.7
Y =0.6%
=0.5
Y=0
\Y =0.3%
Y =02o/o
Y =0.10/o
0.6
F
ooo
oo.
c\,1
oo)j
ooooo,;oo
U)
05
0.4
03
0.2
01
0
0.0 0.15 0.24 0.25 0.30 0.35
A = Ratio of unbonded length of steel to spacing of elastic joints
,fl
Figure 16 Design charts for calculation ofstresses in concrete
 ;: 3 n of Slab Thickness
': thickness of cement concrete pavement should be worked out considering PCCP. While working
ruw fE frickness, the ad rete tensile stress should be accounted for as indicated in Figure 15 and
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fhe effective increase in slab thickness due to provision of steel reinforcement may be worked out by usingMallinge/s chart given in Fig. 17. The equivalent CRCP slab thickness may then be calculated by reducing thethickness calculated above, by the amount of the effective increase in slab thickness using Figure 17 the averageof steel in longitudinal and transverse direction may be taken.
Figure 17 Malinger's chart showing the effect of reinforcement on rigid pavements
5.4 Outline of Design procedure
Design of continuous reinforced concrete pavement with elastic joints shall be done according to the followingsteps.
Step l: Assume a thickness and examine wheel load and temperature stresses as per this guideline,
Step ll: For a proposed value, choose a steel percentage Figure 15 such that the steel stress is within thepermissible value. Calculate the concrete stress from Figure 16. ln calculating these stresses the ordinates are tobe multiplied by the corresponding value of ( as applicable to the particular location.
Step lll: Add the concrete stress calculated in Step ll to the value obtained in Step I and the final total stressesshould be within the flexural strength of concrete. The trials may be repeated till the assumed thickness in Step Imeets the requirement.
Step lV: Calculate the effective increase in slab thickness due to provision of reinforcement as per 17 and reducethe thickness obtained in Step lll to account for the increase.
5.5 Cement Goncrete Mix DesignThe mix should be designed on the basis of absolute volume method as per IRC:44  "Tentative Guidelines forcement concrete Mix Design". The flexural strength of concrete at 28 days in the field should not be less 40kg/cm2.
5.5.1 Materials
Cement: Should conform to lS: 26g or lS: 8112.
Coarse and Fine Aggregate: Should conform to lS: 3g3
Steel: The diameter of steel bars should be so chosen as to keep the spacing, between bars around 25 to 35 cmSteel should conform to tS: 432 (part l) tvtild Steet.
,'/ater: Water for both mixing and curing should be clean and free from injurious amount of deleterious matterS:456. Potable water is generally considered satisfactory
n
E
=a
a.o.>EljR
.=6
10
0040
2A 30 4A
008
0.06
050
040
030
020
0.10
,  C should&
I)
ifr
5.6 Construction Details
The construction details are the same as in the case of plain cement concrete pavements (vide IRC: 15) except the
following.
5.6.1 Construction of Joints
Elastic joints
These are dummy type joints which should be induced at intervals similar to that for dummy contraction joints. Thejoint grooves may be formed as in the case of conventional dummy joint, and filled with sealing compound.
Alternatively, bitumencoated plywood strips of 50 mm width and 3 mm thickness may be inserted therein. On
either side of the elastic joint, steel should be coated with bitumen for a length of 113  114 joint spacing in order to
break the bond of steel with concrete and to provide greater length for elongation of the steel due to joint opening
for reducing the stress in the reinforcing steel.
Expansion joints:
The expansion joints are provided only at the ends of the CRCPEJ sections and there is no need of providing
these inbetween. The width of such expansion joints is kept upto double that of conventional concrete pavement
to accommodate the greater end movements. Details of these joints should be as shown in IRC: 15.
5.7. Reinforcement
The steel mats, assembled at site, are placed over suitable chairs at mid depth of the slab before concreting is
done. The bars should be continuous across elastic joints and any construction joints. Where overlap of bars isrequired, a minimum overlap of 30 diameters should be provided. Such overlaps should be staggered. lt shouldalso be that no overlap of steel bars is provided at the location of elastic joints.
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CHAPTER6 PRECAST PANEL CONCRETE PAVEENT
6.1 lntroduction
Generally, precast pavement technology comprises new and innovative construction methods that can be used torneet the need for rapid pavement repair and construction. Precast pavement components are fabricated orassembled offsite, transporled to the project site, and installed on a prepared foundation (existing pavement or regraded foundation). The system components require minimal field curing time to achieve strength before openingto traffic' These systems are primarily used for rapid repair, rehabilitation, and reconstruction of both asphalt and:ement concrete pavements in highvolumetraffic roadways. The precast technology can be used for intermittentepairs or fullscale, continuous rehabilitation. ln intermittent repair of concrete pavement, isolated fulldepthrepairs at joints and cracks or fullpanel replacements are conducted using precast concrete panels. The repairsare typically fulllane width. The process is similar for fulldepth repairs and fullpanel replacement. One technologydeveloped for continuous applications is precast prestressed concrete pavement (ppcp). Use of precast concretepavements'for reconstruction and rehabilitatlon is a very viable alternative to conventional castinplace concretepavement construction, especially in situations where high traffic volumes and consideration of the delay costs to'lsers due to lane closures favor reconstruction and rehabilitation solutions that allow expedited opening to traffic.frecastconcrete also offers the advantage of being "factory made" in a more controlled environment than cast inplace construction and thus is potentially more durable and less susceptible to construction and material variability.
6.2 Precast Panel Concrete pavement System
The basic precast prestressed pavement concept consists of a series of individual precast panels that are post:ensioned together in the longitudinal direction after installation on site. Each panel can be pretensioned in theransverse direction (long axis of the panel) during fabrication, and ducts for longitudinal posltensioning are castrto each of the panels. The basic features (typical) of the ppcp system are as follows:
' Panel size: up to 1 1.6 m wide, typically 3 m long, and 178 mm to 203 mm thick (or as per design).' Panel types: Base, joint, and central stressing panels (as originally developed).. Tongueandgroove transverse epoxied joints
' Posltensioning details: a. 15 mm diameter 7wire monostrand tendons, typically spaced at 600 mm. b.Tendon load: 75 percent of ultimate tendon load, typically. c. Prestress force: sufficient to ensure about1'0 to 1'4 MPa residual prestress at the midpoint of each series of prestressed panels. D. Grouted posltensioning ducts.
. Expansion joint spacing: about 76 m, typically.
' Base type: a' Hotmix asphalt concrete base with polyethylene sheet over base or asphalt concreteinterlayer in the case of an overlay application. b. Permeable base (as used in the Missouri demonstrationproject). c. Lean concrete base, d. Aggregate base
' Injection of bedding grout to firmly seat panels (after posttensioning)
6.3 Design Considerations of ppCp
Ihe following factors need to be considered:
The PPCP systems require placement of the panels on a smooth base or interlayer to ensure that thefriction between panel and base is as low as possible. othenrise, a larger portion of the prestressingforce is consumed in overcoming this friction.The PPCP base and the foundation need to be of high quality and stiff to minimize slab deflections at theexpansion joint.
The PPCP can be designed to achieve a minimum residual prestress of about 1.0 to 1.4 Mpa at midof the series of posttensioned slabs. This residual prestress adds to the concrete,s flexural
and allows use of PPCp systems that are about 75 to 100 mm ress in thickness than
a.
b
concrete pave ments for the same traffic loading and environmental conditions
6
ffi d
6
The PPCP systems can be designed to incorporate expansion joints at 50 to 75 m. The longer jointspacing requires use of more prestressing tendons (more prestressing force) to balance the higher prestress losses due to longer prestressing lengths involved.The prestressing tendon size (diameter) and spacing should be selected to achieve the desired stresslevelat slab ends and at midlength of the posttensioned panels. Generally it is kept as 15 mm diameter,Grade 270 7wire stressrerieved tendons for highway apprications.The AASHT0 (2004) mechanisticempirical design procedure can be used to determine the requiredthickness of the PPCP using only the "fatigue cracking" distress criteria. The flexuralstrength used shouldbe the concrete's flexural strength plus the residual prestress at midlength of the posttensioned seriesof the panels.
The expansion joint should be designed to allow for large slab end movements (typically 51 to 76 mml,depending on environmental conditions) and to provide desired load transfer r.rogs th. wider joints.
i.
f
3
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q.
REFERENCES
AASHTO (1993), 'AASHTO Guide for Design of Pavement Structures', American Association of StateHighway and Transport Officials, Washington, D.C.AUSTRoADS (2004), 'A Guide to the Structural Design of Road pavements,ARRB Transport Research Ltd,Australia
NCHRP (2004), '2002 Design Guide Design of New and Rehabilitated structures' Draft Final Report, NCHR,study 137A, Nationarcooperative Highway Research program, washington, DCPea (1980) 'Joint Design for concrete Highways and skeet Pavements', porfland cement Association, USA.MoRTH (2001) 'specifications for Road and Bridge Works', published bylndian Road Congress, New Delhi,2001
6' A'T Papagiannakis, E.A' Masad (2008), 'Pavement Design Materials,JohnWiley & Sons, lnc.7' Ridgeway, H'H'(1976), 'lnfiltration of water through Pavement surface',Transportation Research Record,Transportation Research Board, 616, pp gg_100.
! cedergren H'R' (1974), 'Drainage of
.Highway
and Airfield pavements, Johnwiley and Sons,.9. IRC:sP:42, 'Guiderines on of Road Drainage, rndian Roads congress.10' IRC:sP:52, 'Guiderines on Urban Drainage', rndian Roads congless.11' ls 4332 (Part lv)  1968 'Methods of Test for stabilized Soits part 4 wettingand Drying, and Freezing andThawing Tests for Compacted Soil_CementMixtures,
" ;;:I::t' G, and G'K' Ray(1951). 'lnfluence charts for concrete pavements,,Transactions,
ASCE, vot. 116, pp.
13' croney' D' and croney, P. (1992),'The Design and Performance of Roadpavements, second Edition, 1992,McGraw Hills lnternational Series in CivilEngineering.14' JamshidArmaghani and Roger schmitt(2006), 'LongLife concrete pavements: TheFlorida perspective,
lnternational conference on LongLife concrete pavements chicago,lllinois, USA.15' v'v'Subramanian(1964),'lnvestigation on Temperature and Fricjion stressesin Bonded cement concretePavement' Ph' D thesis rransportationEngineering section, civil Engineering Department, llT Kharagpur.16' Kohn's'D' and Tayabji,s'D.(2003), 'Best Practice for Airportportaland cement concrete pavementconstruction',lPRF01Gu02l,washington,DC:lnnovative pavement Research Foundation.17' Neville, A.M, and Brooks, J. i. (2010)'concrete Technorogy,, prentice Hail18' IRC;582015; 'Guidelines for the design of plain jointed ngiJ pavements for Highways,, Third Revision19' IRC:SP:62, 'Guidelines for the Design and Construction of Cement Concretepavement for Rural Roads,,Indian Roads Congress.l
ry
fl
APPENDIX l: VALIDITY OF FATIGUE EQUATIONS
The fatigue equation adopted in the lVechanistic Empirical Pavement Design Guide (MEPDG) recently developed
USA (NCHRP , 2004) for a reliability of 90 percent gives fatigue lives that are practically the same compared tothose obtained by using the fatigue equations adopted in the present guidelines. The design thicknesses obtained
for a typical concrete pavement using the fatigue criteria adopted by IIIEPDG (NCHRP, 2004) and the fatigue
cnteria adopted in these guidelines are practically the same. The IVEPDG fatigue equation for concrete is
SRrto'z+{Logro (1  P)}0.277 Equation 17
Log6N, = 0.07L2
Where,
Nr fatigue life at load level r
SR; StfeSS ratiO at lOad leVelr  stresscausedatload"tevetr
mo dulusruptur e o f c oncr ete
p = probability of 10 percent fatigue cracking failure at the end of designperiod (in decimal)
ReliabilitY = 'l P
Suggested reliability values for different types of roads are:
Village roads 60%
District Roads 70%Feeder Roads 80%
National Highways 90%
# Type of Roads Reliability
q/
APPENDTx il: EeuATrONS FoR FLEXURAL srREss rN coNcRETE SLAB
Regressions equations are given in this appendix for estimation of the maximum tensile stress in the slab in theedge region due to the combined effect of axle loads and temperature differential. The equations are given forbottomup cracking and for topdown case. Flexural stress for bottomup case has been computed for nonlinearpositive temperature differential occurring in the slab during day time. The stress for topdown cracking case is forthe combination of axle loads and linear negative temperature differential in the slab occurring during night time.
For the computation of stress for bottomup cracking analysis, only the rear axles (single as well as tandem) withtwo wheels (dual wheel sets) on either side of each axle have been considered as the front axles do not contributeto any significant fatigue damage. For topdown cracking, rear axle is considered at one end and the front axle atthe other end' only one axle of the tandem and tridem axles is assumed to be placed on the slab underconsideration' Thus, for a tandem axle, 50 percentof the tandem axle weight is considered for analysis. For atridem axle, 33 percent of the tridem axle weight may be taken for analysis. The corresponding front axle is takenas 50 percenl of the rear axle, (25 percent of rear tandem axle and one sixthof rear tridem axle).
1) Expressions for maximum tensile stress at the bottom of the slab (for bottomup cracking case)l. singre axre' pavement with tied concrete shourders
a)K < g0 trrlpa/m
s = o ooa  6 72Wl. r rrl#1+ o 0266^r
lt.
s = o a4z + 3.26Wl. r.rrf#l + o oszz^r
Single axle . Pavement without concrete shouldersa)k s 80 MPaim
s = 0 14e  z 6owl. 3.Bl#l+ o ozsz^r
b)k > 80 MPa/m, k s 150 Mpa
.s = 011e  zssffil. r.rll#1+ o o4s6^r,
b)K > 80 MPa/m, k < 150 Mpa
c)K> 150 MPa
.s = o oB  s 6sful+ z.osl#1. o.o4os^r
c)k> '150 lVPa
.s = 0 238 + T o2W1. r ^rl#)+
o osas^r
lll. Tandem axle . Pavement with tied concrete shoulders
a)k < 80fr/Pa/m
.S=0.188+0.93
b)k > 80 tt/Pa/m, k < 150 MPa
c)k> 150 MPa
a)k < 80 MPa/m
S=0.223+2.73
s = o tr4 + 1..21W1. o.Brl#)+ 0s364^r
Yr,,l. Lo,sl#l+ o ozo,^r
Wl.rcsl#)+oozzs^r
s = 0.210 + 3.BB W). o.nl#l+ 0.0s06^r
lV. Tandem axle . Pavement without tied concrete shoulders
b)k > 80 MPa/m, k < 150 MPa
W)PhS=0.276+5.78 + 1.74
equations have the following meaning:
t>^ /
+ 0.0404A7kl+
c) k> 150 MPa
2) Expression for maximum tensile stress at the top of the slab (for topdown cracking case)
fopdown cracking analysis is performed only with the consideration of rear axle load. Front axle load is assumedbe 50% of the rear axle load (tandem/tridem).
.s = 0.3 + e.BB W).0.es6 l#1.0.0s43^r
4 s = o.zle + 1.686f#* rcB.4ll,/l
fl
flil
rI
'{fl
The sym In
+ 0.108947
lJ
S = flexural stress in slab, MPa
AT= maximum temperature differehtial in "c during day time for bottomupcracking(Sum of the maximum night time negative temperature differential and builtin negative temperaturedifferential in 'C for topdown cracking)
h = thickness of slab, m
/r = effective modulus of subgrade reaction of foundation, Mpa/m
I = radius of relative stiffness,m
Eh3
nkoaE = elastic modulus of concrete, MPa
1r= Poisson's ratio of concrete
y = unit weight of concrete (24 kN/m3)
P = For Bottomup cracking analysis: single/tandem rear axle load (kN). Nofatigue damage estimated forfront (steering) axles for bottomup crackingcase
P = For Topdown cracking analysis: 100 percent of rear single axle, 50 percent of rear tandem axle, 33percent of rear tridem axle. No front axle weight is required to be given as input for topdown cracking. 50percent of rear single axle, 25 percent of rear tandem axle, 16.5 percent of rear tridem axle, have beenconsidered in the finite element analysis as the front axle weights for single, tandem and tridem rear axlesrespectively.
B = 0.66 for transverse joint with dowel bars (load transfer efficiency was taken as 50 percent)= 0.90 for transverse joint without dowel bars (load transfer efficiency was taken as 10 percent)
1t
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v
AppENDrx ilr: EXAMpLE FoR THE DEsrcN oF DRATNAGE LAYER
A fourlane divided cement concrete pavement will be constructed for a high kaffic volume road in an area havingan annual rainfallof 1500 mmlyear. The width of each carriageway will be 7.0 m. 2.5 m wide (1.5 m concrete, .1.0m unpaved) shoulders will be provided' Transverce ioint spacing wiil be 4.5 m. The highway has a longitudinalgradient of 3 percent and a cambe r of 2.5 percent. side slopes oiembankme nl are 2:1(horizontal to vertical). Thepavement has a 300 mm thick concrete slab placed over 150 mm thick DLC layer. Estimate the requirement ofpermeability of the drainage layer material to be used if the layer thickness is 150 mm.
Solution:
combined thickness of the srab and DLC rayer = 300 + ,150 = 450 mm
The drainage layer will be provided below the DLC layer at a depth of 4s0 mm from pavement surfaceThe drainage layer will be extended to the full width of embankment.Width of drainage layer= 7 m (pavement)+ 2.5 m (shoulder)+ 2x0.4Sm = 10.4 mThe figure below indicates the direction of flow of water arong AD (diagonar)ln this figure, AB = 10.4 m; AC = 10.4 x (0.03/0.0251 = 12.4g m;
Drop of elevation along AC = 12.4gx 0.03 = 0.37 4 m;Elevation drop alongcD = 10.4 x 0.025 = 0.26 m;Elevation drop along AD = 0.37 4 + 0.260 = 0.634 mGradient along AD = r = drop arong AD/rength pf AD= 0.634116.24.= 0.039
AD= AC2 + CD2 = 10.42+1.2.482=16.24
Longitudinal Slope, 3 %A
c
su?Nt'ollEE(J
DB
The infiltration rate per unit area gi can be estimated using following. The equation is reproduced here forconvenience.
et=+(#,.#)
rate = 0.223 m3/day/mnumber of longitudinal joints/cracks 3 (joints between lanes, between lane and shoulder and paved
shoulder edge)
[llp= width of pavement subjected to infilhation= 1.0 + 2.5= 9.5 mlile =length of the transverse cracks or joints = T .0 + 1 .5 = g.5 mCs =spacing of transverse joints= 4.5 m
TK,
', here;
l"=IVc
Direction of flow ofwater
I
[;Jiil,#;;J#[ush uncracked pavement sudace = 0 mitdaytmwhich is armost nesiisibre ror
. Thus, the rate of infiltration of water into pavemen l, qi= 0.115 m!daylm. Ai1fl;['i11T!ff:i::,,ff"'rensth nowins ,.* in. path AD or the drainase rayer,. Rate of flow through drainage layer,. e = KlA. Since /= 0.039, t<R=t.g+glO 039 = 49.92;A=1x0.15=0.15m2' Hence' the required coefficient of permeabili ty , K = 4g.g7 10. 15 = 333 m/day
 : ^sidering the coefficient of permeability values adopted by AASHTo (1gg3) for different aggregate gradations it''^ be expected that the grading r, ri anJ rrr of the coars. grrd.o granurar subbase materiar wi, have::;::flll,Y;fl:',',:ffi:ii:H'.',Jjf,i,'n' AASHTo soi,o,n, and the correspondins coerricient or:=':ent ror ensurins sood permeabirity rr il. ;i:i::'fhmffi ;il :::,Hff::i:il:nffil*r;J:T:X.H#:iJ;:li::i:iflli;,:'* nll* sranurar subbase materiar is recommendld berow the coarse
20 100 100 100 100 10012.5 10085 B4 83 81.5 79.59.5 7577.5 76 74 72.5 69.54.76 6358.3 56 52.2 49 43.52.36 3242.534 29.5 222.00 5.839 35 JU 25 tt0.84 0to.5 22 15.5 oa
00.42 018.2 13.3 6.3 0 00,25 013.0 7.5 0 0 00.1 05 06.0 0 0 0 00.075 00 0 0 0 0 0ofmlday
,)35 100 350 850 950
44 I Q . * ^
Pavement Desigi Guidetines (Rig@
% passin
Grading 1 Grading 2 3
@
b)
a)
2
3
c
!
0
2
3
le4
)
c
7
8
CBR ofcompacted subgrade, % o Given value
bgrade ( 2)IModu ofUS Su reaction PalmM Table 44.18 lnterpolatedof granular subbase (provided), mmThickness
150 Assumed
thicknessThickness of DLC subiaseMPa, mm
m 7 day compressive strength of 10with a minimu 150 Assumed
thicknessVEEffecti odulusm SUof reaction c0of binmbgrade founed ationd of +Aradsubg
ranu suar bbase and LD SU bbase Tr 4ble M Palm(from238.55 Interpolated
ValuePolythene sheet of 1 25 ness between DLC and concrete slabmicron thick Yes Adopted provision
y compressive strength of cement concrete, Mpa28da'35 Standard Value
2Bday Flexu ral strength of cement concrete, Mpa 4.14 Formulastrength of cement concrete, Mpa90day Flexural 4.56 10 % more than
28day strengthModulus of concrete, E , MpaElastic
30000Poisson's ratio of concrete, ;r = 9.1 5 0.'1 5
weight of concrete, y = 11713Unit t+Max. day{ime Temperature Differe bottomup cracking), 'C forntial in slab (forplain area
16.8
(for topd cracking)jN ht Ttimeg rature renDiffe intialempe slab 0wn d mti Aay 13.4
Design period , years 30.00 Given Value
traffic (expressed in decimat)Annual rate of growth of commercial 0.065 Assumed as 6.5 %
per day (in the year of completion ofTwoway commercial traffic volume
c0n
in direction percent
2,000.00 Tow way atraffic
0.50 Percentage of Traffic in
Design lane
APPENDIX IV: EXAMPLE FoR THE DESIGN oF SLAB THICKNESS
A cement concrete pavement is to be designed for a fourlane divided National Highway with two lanes in eachdirection in plain terrain. Design the pavement for a period of 30 years. Lane width is 3.5 m, transverse jointspacing is 4.5 m.
It is expected that the road will carry in the year of completion of construction, about 2000 commercial vehiclesper dayin both direction. Traffic growth rate is assumed as 6.5 %per annum. Axle load survey of commercialvehicles indicated that the percentages of front single (steering) axle, rear single axle, rear tandem axle and reartridem axle are 31 percent, 9 percent, 2Spercent and 32 percent respectively. The percentage of commercialvehicles with spacing between the front axle and the first rear axle less than 4.5 m lb ii percent. Traffic countindicates that 30 percent of the commercial vehicles travel during night hours (6 plM to 6 AM). The percentage ofday time traffic for six hours (10 AM  4 AM) is assumed as75 o/oof the total day time traffic. Details of axle loadspectrum of rear single, tandem and tridem axles are given in the table below. Front (steering) axles are notincluded. The average number of axles per commercial vehicle is 2.5 (due to the presence of multiaxle vehicles).Effective CBR of subgrade is 6%.
Solution:
1
I
7
11
Selection of modulus Values Explanation
Traffic Calculation for design period
19
20
21
22
23
1
r3
tb
27
:8
2)
il
3'l
:2
JJ
X
Trraffi, tnc apredomin dnt commerciirection, ^tdt haffic volume daper v 1,000.00 Design trafficmberNUAverage of standard axles (steering/si ngle/tandem/tridem per 2.50 Data from axle load
survey and analysisNumber of standard axies tn predominant direction the(in ofyear letion ofc0mpconstruction 2,500
Growth Factoroo Growth factor formulaCumulative
direction
number of standard axles during design period in predominant 78,817,063 ofGrowth factorTotal
Totaltime 2day hour(1 traffic % of traffic n inanpredom directionofnumber standard 70.0Y0
TotalAxle)time Tday raffic n ofumber standard
55,171,944Six hour timeday traffic 0 AM(1 4Pt\4)
4't,378,958n T rafficDesig (for Bottomup ofYoCracking) dardstan inaxle nanpredom
fordirection hsix 0ur tirmeday 0.25 Stan dard percentage
value of Design traffic
for BUCmNUDesign ofber sta ndard axles of(2s percent antpredomin direction trafficfourfor lane divided h igh ways10,344,740
NUn mberDesig of standard axles axle repetitions for Bottom up Crackingan alysis (K) 10,344,740
Total Night
(number of ,
time (12hour)
standard axle)Totaltraffic of0/
/o traffic in dpredomin dint rection 30.0%
Total Night tiaxle)
me (12hour) traffic in predominant direction (number of standard 23,645,119
6hour night time traffic in predominant direction (number of standard axles) 11,822,560 Assumption 50% oftotal night time traffichourSix timenigh traffic h a wheelung base thanIess m4.5 EEO/oJ /0 of srx houn traffictimeight m ofber(nu standard axle) (J)
6,502,409
Trial thickn ess of the slab, m0.32 Trrial ThicknessRadi US of relative stiffness m0.77 CalculatedLoad transfer forfactor Dowel bars thethrough transverse tSjoin066 Standard given value
q
tars across transverse jointsconcrete with
pavement
axle)
commercial vehicle
Values exptanations
Front single axel (Nl )
Rear single axle (N2)
Tandem axle (N3)
Tridem axle (N4)
Total
Front single (steering) Axles not considered
Rear single Axles = K* N2
TandemAxles=K*N3
TridemAxles=K*N4
rront single (steering) Axles not considered
RearsingleAxles=J N2
TandemAxles=J*N3
Tridem Axles = J. N4
'lq 10E
51 85
rf, I /J
:l l0c
35 145
) LJ
_l 15
.:105
:: 95
<85
ll
w
n[
fl
f,
V
0.31
009
0.28
0.32
1.00
0
931,027
2,896,527
3,3'10,317
0
585,217
1,920,674
2,080,770
190 000 380  400 390 000 530560 545 0
180 000 360  380 370 000 500530 515 0
170 200 340  360 350 000 470500 485 0
160 300 320  340 330 000 440470 455 0
150 10 00 300  320 310 0.00 410440 42s 0
140 12.00 280  300 290 100 3804 1 0 10(1
130 10.00 260  280 270 6.00 350380 365 4
120 20 00 240 260 2s0 10 00 320350 5
110 16 00 220 240 ,?n 28 00 290320 305 to100 10 00 200 220 210 25.00 260290 nlELI J 30
90 700 180  200 190 18 00 230260 245 16
80 10 00 < '190 170 12.00 < 230 18
100.00 100.00 100.00
*
Ratio of total
Design Axle Load Repetitions for Fatigue Analysis
For Bottomup Cracking Analysis
For TopDown Cracking Analysis
Axle
Rear
Group
Rear Tandem Axle Rear Tridem Axle
Load Group
Frequency
('/r)
Load
Group
(kN)
MidPoint of
Load Group
(kN)
Frequency
(%)
Load
Group
(kN)
MidPoint
of Load
Group (kN)
Frequency
(%)
Fatigue damage Analysis of concrete slab
0
0
18621
27931
.'11n2
'111723
33103
1 86205
'48964
i31 03
,5172
331
931 027
2.615 0.574 63554 0.000 0 2.221 0 488 1 483099 008 0 000
2.523 0 554 1 1 1458 0 000 0 2.1473 4.471 4623131 042 0 000
2.431 0 534 202879 0 092 0 2.0736 0.455 26823814.01 0 000
, ??o UC]J 420010 0 067 0 1.9999 0 439 infinite 0.000
2.247 0 493 1072570 0 087 0 1.9262 0.423 infinite 0 000
2.155 0.473 4014459 0 028 28965 '1 8525 a 407 infinite 0 000
2 063 0 453 3807331 6 0 002 173792 1.7788 0.390 infinite 0.0001.971 0.433 infinite 0 000 289653 1.7051 0.374 infinite 0 0001 879 0.412 infinite 0.000 811028 1 6315 0 358 infinite 0 000
1.787 0 392 infinite 0.000 724132 1 5578 0.342 infinite 0 000
1.695 0.372 i nfin ite 0 000 521375 1.4841 0 326 infinite 0 000
1 603 0 352 infinite 0 000 347583 1.4104 0 310 infinite 0 000
Fat Dam from Single Axles = 0.28 2896527 Fat Dam from Tandem Axles = 0.000
N
Flex
Stress
MPa
Stress
Ratio
(SR)
Repetitions
(Ni)
Fatigue
Damage
(ni/Ni)
Expected
Repetitions
(ni)
Flex
Stress
MPa
Stress
Ratio
(SR)
Allowable
Repetitions
(Ni)
Fatigue
Damage
(ni/Ni)
Repetitions
(ni)
Total Bottom.up Fatigue Damage due to Single andTandem axle loads =
6
0.28 + 0.00 0.28
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APPENDIX V: DESIGN OF DOWEL BARSInput data for Design of Dowel
',lodulus of Elasticity of concrete, Ec, Mpa
rc sson's Ration, p
: ab Thickness h m
: nt width, z mm (contraction joint)
',todulus of Subgrade reaction, k, Mpa/m
iadius of relative stiffness, mm; I
Solution:
':dulus of Elasticity, E for Dowel bar, lVPa
'odulus of Dowel support, Knas, MPalm
.'axrmum srngle axle load, kN
','aximum Single wheel load , kN
Assume a load hansfer of 30 % at terminal stage to the tied concrete shoulder or to the side slab.lf no concrete shoulders are provided, no load transfer to shoulder may be assumed
, heel Load to be considered for Dowel bar design (95x0 7), kN
:ssume the percentage of load kansfer through dowel bar is 50%
rermissible bearing stress in concrete is calculated as; Fo
,he
Fck  Compressive strength of Concrete, Mpa (for M35 Grade)
bo diameter of dowel bars assume. mm
:,  Permissible bearing stress in concrete, [Mpa
:cacing ofdowel bars, mm (assume)
:irst dowel bar is placed at a distance of '150 mm from the pavement edge
:ssume the length of dowel bar, mm
l:wel bars up
: ad transferto a distance of 1.0 x radius of relative stiffness (l) from the point of load application are effective in
Number of dowel bars participating in load transfer when the wheel load is just over the dowel bar close to the edgeof the slab = 1+l/spacing = l+(963.88/300) = 4 nos.
ssuming.^+)t
that the load transferred by the first dowel is Pt and that the load on dowel bar at a distance of I from thebar is zero, the total load transferred by the dowel bar system is p
(0 600 l
DttDt 1* , *
L
*tCoefficient to the Pt
oad carried by the outer dowel bar, pt
lheck for Stress
Relative stiffness of dowel bar embedded in the concrete, B
'.loment of inertia of dowel, I, mm4
Bearing stress in dowel bar Forr,,
Searing stress developed in dowel bar < permissible stress concrete
30000.00
0.15
030
20.00
80.00
963 BB
l
200000
41 5000
180
90
63.0
31.5(101.6  b)f,t
Fu=95.25
?^
Jd
23.37
300
500
2.13
14.77
0.0210 B:4K*arba
4EI
102301 985 . rc (b)n64
19.17
Fbma*rFb
(Pt*k^a,)*(2+Bz)E' Dmax
483 EI
'lence, the dowel bar and diameter assumed are safe,
I
t*l t *n.
l1r(1  tN
APPENDIX Vl: DESIGN OF TIE BARS
ln
Slab thickness, m
ane width, b, m
ieight of the slab in kN/m2 W
Coefficient of friction, f
)ensity of concrete, kN/m3
 lowable tensile stress in plain bars, Srt, IMPa
: Iowable tensile stress in deformed bars, MPa
 owable bond stress for plain tie bars, B lMPa
 cwable bond stress for deformed tie bars, B, MPa
 ssumed diameter of tie bar dr, mm
:a of plain steel bar required per meter width of joint to resist the frictional
,:e at slab bottom, As, mm2/m
,:ss sectional area of the tie bar A mm2
: = meter of tie bar, Ppto, tTlrn
: ::cing of tte bar, mm (lr/As)
r: rrde spacing of 340 mm c/c
.gth of the bar, L, mm
 'ease length by 100 mm for ross of bond due to painting and another 50  for tolerance in placement. Therefore, the required rength of tie barx
0.3
?E
7.92
1.5
24
laEI L\)
200
I aE.t J
2.46
12
332.64 b*f*Wit
H" S,,
1 13.04
rd2A4
37.68 Ppta = Ttd
339.83
428.57 . 2,, Srt * A
B x pptb
578 mm adopt 600 mm
of Plain Tie bars
q
Assumed diameter of tie bar dr, mrn 12
Area of plain steel bar required per meter width of joint to resist the frictlonal force at slab
bottom, As, mm2/m207.9
Cross sectional area of the tie bar A. mm2 11304
Perimeter of tie bar, Ppto, rnffi 37.68
543.72
487.80
637 8
Design of Deformed Tie Bars
Spacing of tie bar, mm (A/As)
Provide spacing of 550 mm c/c
Length of the tie bar, L, mm
lncrease length by 100 mm for loss of bond due to painting and another 50 mm for tolerance in
placement.
Calculated thickness (487+100+50 = 637 mm); adopt required length of tie bar 637 mm and adopt 640 mm
Design parameters:
1. Slab Thickness. 32 cm of concrete M35 Grade
2. Dowel bars. 32 mm Dia, 50 Cm long, 30 cm c/c spacing, first bar 15 cm from the edge of the slab3. Tie bars: 12 mm Dia, 60 cm long, 35 cm c/c spacing, first bar 15 cm from the edge of the slab,4. Layout is given in the Fig.
375m 3.75 m 3.75 m
Eq
Eo+
Eo$
\
Figure 15 Typical layout of the dowel and tie bars
APPENDTx vr: TyprcAL DESTGNS rN THE cAsE oF LrMrrED TRAFFTC DATAThe design of Jointed Plain concrete Pavement requires traffic data in terms of its vorume as we, as axre roadingspectrum' However, to get such data is resource consuming. ln the case of limited traffic and axle load datafollowing assumptions could be helpful for design of concrete slab.
b
c,
a. Consideronly the commercial vehicles from the 'Traffic count data available at DoR, The total number ofAADI commercial vehicles in both direction shall be takenDetermine the design traffic volume for the predominant traffic directionFor conversion of average number of axles for commercial vehicle take the multiply by
Busy highways 3.2Feeder road 2.1
The percentage of front singlerespectively.
, rear single , tandem and hidem axle may be taken as 31, 9, 2g and32
The load spectrum can be taken as:
Stress due to axle
calculated using the
loading may be determined from the chafts given in this guidelines or may beregression equations.
d.
e.
f.
Rear Single Axle Rear Tandem Axle Rear Tridem AxleLoad Group
(kN)
MidPointofLoad
Group (kN)
Frequency
(%)Load Group
(kN) Load Group
ofFrequency
f/")Load Group
(kN)
MidPointof
Load GroupFrequency
%t185195 190 100 380  400 390 0.00 530560 545 0175185 180 1.00 360  380 3i0 0.00 500530 515 0'165 1 75 2.00 340  360 350 1.20 470500 485155165 05160 500 320  340 330 150 440470 455 1
'145155150 '10.00 300  320 310 150 410440 425135145 4140 12.00 280  300 290 2.00 38041 0 ?o5125135 5130 10.00 260  280 270 12.00 350380 365 10115125 120 20.00 240  260 250 14.80 320350 7)E
11'105 1 15 110 10.00 220  240 230 20.00 290320 305951 05 16100 10 00 200 220 210 22 00 260290 a1E
8595 18on 7.00 180  200 190 15 00 230260 245<85 2080 12 00 <190 170 '10.00 < 230 215 145100.00100.00
100
#
170
g
ill
APPENDIx Vilt: SAMpLE DESTGN TEMPLATE oF DEprH oF coNcRETE SLAB
1' Design of depth of cement concrete slab with the given design conditions in the Appendix lv are givenin the table below:
2' other design trials can be generated with the help of Excel sheet developed during the preparation ofthis design guidelines.
3
il
Bothwaytraffic,cvpd
CBR,
6.00 8.00 10.00 12.00 15.00
500 30 30 30 30 30750 30 30 30 30 30
1000 30 30 30 30 30
1500 32 32 30 30 30
2000 32 32 32 32 322500 32 32 32 32 323000 3Z 32 3Z 32 32
X
s
w
APPENDIx lX: sOME FtcuRS FOR CONSTRUcfloN TEcHNoLocy oF pccp SLAB
Iloxel t'h:rirr Fired int*Sublxre beforc ('oncretrPouring
609,ir Don'el Length ('overerllf ith Plarfic $hetthing rrndDox'rl Chrirr OerignerlProperlr'
Tie Bln Psinred inllidrlle lS{} mnr
Prnperlr' *lignetl Tie BlnPl:rcerl in Sirle t'or.rn
i:
:i
iii
i:
i
ia
i
iI
U
1
e/
il
il
APPENDIX XI: AN ILLUSTRATIVE EXAMPLE OF DESIGN OF SLAB THICKNESS
L Design Parameters:
Design Wheel Load, P : 5100 kg
Present Traffic lntensity : 300 veh/day
Design Tyre Pressure, p . 7.2kglcmz
FoundationStrength,k : 6kgicm2
Concrete FlexuralStrength, fn : 40 kglcm2
Ec = 3.0 x 10 5 kg/cm 2
U = 0'15
o=10x104/'CAt = 14.3 'C against thickness of 25 cm
2. Design Procedure:
Step I : Assume h = 25 cm, spacing of elastic joints = 4.5 m
As per IRC:581988
ole = 18.50 kg/cm2
ote = 15.50 kg/cm2
o Total = 34.00 kg/cm2
Step ll : ForA = 0.33 and Ar = 20'C
r = 0.4 per cent (Steel reinforcement)
From Fig. 2, os = 56 x20  1 1 20 kglcmz
< 1400 kg/cm2
OK
From Fig. 2, oc = 0.235 x20 = 4.7 kg/cm2
Step lll : From Step ltotal o = 34.00 kg/cm2
From Step ll oc = 4.70kolcm2
Total o = 38.70 kg/cm say 39.00 kg/cm 2
The total stress 39,00 kgicm2 is less than the flexural strength of concrete 40 kg/cmz andtherefore the thickness of 25 cm is O.K.
Step lV : Average steel reinforcement (r.,)
fl'fl
H
0.004 + 25o/o of 0.OO4 0.005:2
0.005= 0.25 per cent
2
From Figure 17,lor r a= 0.25o/o, the effective increase in slab thickness is 31%.
Reducing the thickness proportionately
l1,.=;fu =19.08cm
H
Design thickness of pavement slab = 19 cm
Beta value w be 0 66 for dowe ed lo nt and 0 90 lor w thout dowe s case)
4laneilodu!sof subgradereadonof subgrade MPdm (cBRoi6%l
 lF;ffi;, 150
Tilckness ol Dry Lear loicet€ subDase mm 150
Transveree lo nt spac no (ml ?n Efi.Mislbqrade readon orroundator uPa/m 238 55
:El lunn*.miconr"t" irur.' 24
28 day Fl€rua srengih oicementconcrele MPa
Max day tme [email protected] 0 lfereil a q s ab'C (lor bonoqr up t6 8
Design nunrber ol standard axle. axlc repetibons lbrBottonr uD Crackinc aDalvsis l'Kl
 ] F.""Fr,:?"ier.ilac[lnor = dai lme drn/2 + !
134
Six hour nrght tinre haflc having wheel base less thad 4 5 5J02,a08 LtuErrhdre$olconcrele Sab n m
MBnLoadTransfer Etrciency Fz.torlor ToC anayss Beb = 0 66 iDr
090
0.09t lPoEsonsRan..raoncrele L{u
Rear sngle axes = K'N2 931 027
2 496 521
FdTop&m cding tudFi!3,310,317
 
2 A8t l1l
$N) (%){%)
000 380 {0 390 000185t95 190
000 360 380 370
200 3{.360 350
000160 300 320 3{ 330
000150 10 00 300 320 310
280 30C 290 100 t:i: li35 r45 12 00
r30 r0 00 260 280 274 600125 135
l0 00 aa: .115 125 124 20 00 ?4  264 250
220 24 234 2800105 1 15 110 t6 00
10 00 2AA ?24 214 25 00 n
Fronl Single and Rear Tddem illes not @nsidered for b.: 
Opti6
Subsrade CBR i%)=
2 and 3) ([fPa/m) =
W
Cafrageway
03i
Ale caEgary 061
Fd ffim{p C.drg tu.lFi.
A
q/
(%)
0 0 2 6ta I[ 048 I :255132 0 000 0
0 0 ! 0 471 623131 0d? 4537/02 0 000 00 18621 2682381401 1i056201 0 000 00 27931 2 339
0 000 00 93103 2 241
:trrs0 000 0 1
1 111723 2 15, i !: l@ 047 E5:< & :r 0 000 20808
93103 2 a63 IE 0 390 .E lg r[ @ ry:= 0 000 83231
5 186205 1 9/l lr a 314 .E0 000 10{3q
26 14964 1 879 a.: 0 358 G q II :e .E0 000 541000 1
30 93r03 I 78i 0 39: l a 342
TisII es 0 000 6?4231
65172 I 695 t rf 0 000 332923 1 70931.603 0 352 I 310 :aj c 000 3i4539
100 931027 ry :a := 
4 riz. 2080770
nalysis Faligue OanaSe
c 00!
:,: AND TDC< OR EQ,1
6 S ab thickness (m) over DLa  It 65
250 DLC lhickness h2
625
Fn&8tu$!bb r om C
a
I r,
2,000 00
oam rnm u,t" tna *iqinrffir2,500
c",,,G;;;;rc;ffi ri;;
7A.Aah
0.25 st.,,,,r,,,,r r,.^a;tai, ra..r!;;La 344 140
10 341,1 40
30 00/i"".r,123,645,779
11,822,560
rl. Calculation Pavement Designfor
Design Temptarc, Slab thickness, cnr
30
l030 30 30
x
1500 32
323232 3?
32
3: 32
r
v 6
:r 195 190 0.00 380  400 390 0.00 530560 545 0
180 0.00 360  380 370 0.00 500530 515 0
170 2.00 340  360 350 0.00 470500 485 0
160 3.00 320  340 330 0.00 440470 455 0
150 10.00 300  320 310 0.00 41.0440 425 0
140 12.00 280  300 290 1.00 380410 395 1
130 10.00 260  280 270 6.00 350380 365 4
)125 1.20 20.00 240  260 250 10.00 320350 335 5
r 185
:;175, t165,l155
r r 145
_ l135
 l5115 110 16.00 220  240 230 28.00 290320 305 26100 10.00 200  220 210 25.00 260290 275 30
90 7.00 180  200 190 18.00 230260 245 1.6
80 10.00 < 180 170 1.2.00 < 230 215 1B
;105:595<85
fl
e AxleRear Axle
Laad,
ur0up
tkNj
Rear Tandem Axle Rear Tridem Axle
MidPointof LoadGroup
IkN]
Frequency
tVo)
LoadGroup
tkNl
MidPointof LoadGroup
[kN)
Frequency(%o)
LoadGroup
tkN)
MidPointof LoadGroup
tkNl
Frequency(Vo)
w
CHARTS FoR FtExURAL sTRESS: singte Axle loading without concrete shoulder
stress chart for Bottomup cracking Analysis without con*ete shoulders {p= g0 kN, ,{T.0. c)
Slab Thickness, m
2.50
2,00
0.50
000015 02
0.3 0.35 0.4
64,
 1.50
o0E5 l_00xgL
Cracking Analysis Without Concrele Shoulders {p= .120
kN, AT= 0" C)
3,50
3_00
I: 250
ofir*?6
1.50
a
t.00
0.sl
.S ik 300) {S (k 150) xS ik 80) _o_S ik 40)
\H
Slab Thickness, m
1.00
0.00
0.15 0.? 015 0.3 0.35 04
Stress Chart for Bottomup
q/
+sik3c0) :.s(kr50) xs(k80) i.Fs(k40)
025
L
t
{
Stress Chaft for Bottomup Cracking Analysis Without Concrete Shoulders (P 160 kN, 1T= 0o C)
5.00
,150
4.00
3.50GAE soo
n(,ts 2,so6E= 2_00xg
1,50
1.00
0.50
0.000.40.15 02 0.25 0_3 0.35
Slab Thickness, m
\<_,,, \<,_>r<,
Stress Chart for Bottomup Cracking Analysis Without Concrete Shoulders {P= 200 kN, AT= 0' C)
as (k 300) rs ik 150) xs lk 80) as {k 40)7.00
Slab Thickness, m
6.00
1.00
0000.35 0"4025 0.3015
5.00(gq
EB' 1.00l,
tn
E r.ootx{,L
2.00
/L"!
.sik30c) *3(k150) xS(k8C) .Slk,10r
<:
t
v {
Stress Chart for Bottomup Cracking Analysis Without Concrete Shoulders {P= 80 kN, AT= 10" C)
3.00
2.W
I 200
E6ots rso6?xoE 1.1!0
0.50
0.00
015 02 025 0.3 0.35 0.4
Slab Thickness, m
l}.S {k 300) {S rk 150j xS {k 8C) }S ik 40r
\\\\\\.\\
\\>s
_\\\!: *
skess chart for Bottomup cracking Analysis without concrete shoulders (p= 120 kN, .\T 10o c)
3.50
3.el
GtrE 450
oo; zoo
356'.u0
,.00
0.58
.slk300) *s(k150) +s(k8cl as{k40}
0.15
1.00
0.00
02 025 0.3 0.35 0.1
Slab
{
)
t
m
V
Stress Chart for Bottomup Cracking Analysis Without Concrete Shoulders (P= 160 kN, AI= 10, C)
6.00
5.00
lE 4.00
E
IDi r.oo6Ex!,E 200
1.00
r.000.15 0.2 0.25 0.3 035 0.4
Slab Thickness, m
oS {k 3C0l rS ik 150) +S ik 80) aFS ik 40)
<)s{k300) *sik150j xSrk80) }sik40)
\\\\\
Stress Chart for Bottomup Cracking Analysis Without Concrete Shoulders {P= 80 kN, rlT= ,l5o C)
3.00
2.50
0.50
0,000.15 0.2 025 0.3 0.35 04
Slab
g 2.00
E
a,i r,5ooExoa 1.00
s {
\J
t
chart for Bottom'up cracking Analysis tvithout concrete shoulders (p= 120 kH, rr= 15" c)
4.00
3.50
e 3.00
Ea'^,6 2NE6
E 2.oo
r f.il
1.00
050
.S ik 300) *S ik 1501 +S (k 80) aS ik 40t
Slab Thickness, m
,t.50
02 0.25 0.3 0"35 0.4
0.00
Stress
0.15
+Sik300) *S(k1S0) +S{k80) ..Sik40)
Cracking Analysis Without Concrete Shoutders {p= i60 kN, AT= i5o C}
6.00
5_00
1.00
0.000.1s 02
0.35 0.4
Slab
Stress Chart lor Bottomup
* 400
E
a,g 3.oo
6,xoE 2.00
a
L"r
rt
Cracking Analysis ltithout Concrete Shoulders (p g0 kl,,l, jT= 200 C)
.S(k300) {S(k15C) +SikBC) _}Sak40,
^\\\\\ '.x
\=*i<<
Slab Thickness, m
3.50
3.00
0.50
0.00
0.15 02 025 0.3 0_35 a4
Stress Chart for Bottom,up
2506A.Ef 2,00o0E rroxtur
,,_00
Cracking Analysis llYithout Concrete Shoulders {p= 120 kil, AT= 20o C}
4.00
3.50
{ 300
Eei 6 tNo
E ?00
qE 1.50
1.00
0.50
+S ik 300) *Sik 150) +Sik30l +Sik4Cj
0.15
Slab Thickness, m
4.50
02 025 0.3 0.35 0.4
0.00
Stress Chart for BoUomup
ry
fI
Jt
b
StressChartforBottomupCrackingAnalysisWithoutConcrete$houlders {P=160kN,AT=20oC)
6.00
5.00
& 4.80
:o! :ooq
3xoE 200
1.00
0.000.15 02 025 0.3 0_35 0.4
Slab Thickness, m
.s(k300) *s(ki5c) +Sik80i ioSik40)
\\\\\
b
I
L
V
CHARTS FoR FLExURAL STRESS: single Axle loading with concrete shoulder
stress chart due to single Axle Load with concrete shoulder (p=90 kN, tT= 0o c)
rs ik3c0j **s rk i50r *s 1k80) r*s ik;r0l
n li
160
020
03 n lE
Go
=si r e,t,
o(t! ^^^L VOU
orUbU
clc
4)(
Slab Thickness, rn
rs rk 3001 +s ik 1501 *SikBC) +Sika0)
\
\ sr\\: \ \
stress chart due to single fule Load with concrete $houlder {p=120 kN, ar= 0o c}300
200
100
U.JU
6CLEuio
(r>
?ol!
a2 n?4 01n 1{ 025
Slab Thickness, m
U
l ,.\   i N\ j i i j
,tl I
I i I i t\i \\rr*
I i I i I i i NNr
lt{t,ttl I I I T r
litlttll I l I I r*
7
7
L
7\
{
35C
?tn
€ ^_^o l.5u5dP
6:uu6
xo,^LL
100
stress chart due to singleAxle Load with concrete shoulder{p=160 kN, ar= 0o c}
rS ik 3001 rrS ik jaol rts tk80) {s rk40)
nqn
02 030.15 n?q
Slab Thickness, m
chart due to single Axre Load with concrete shourder (p=g0 kN, aT=,r0o c)
r s ik 300) *l rk l{fir * s {k 60) *+ S ik ,l0j
\t\:<1
l
ll1r
IlE1l l
n !(000
2.50
20a
Slab
Stress
6
E 150sigU'G
x lnno"tr
s
J
I
II!
I
ii
i
ii
I
I{
stress chart due to single Axle Load with concrete shoulder (p=120 kN, tr=10" c)
rs 1k 3001 {s ik 150) * s .k 30) r s ik 40i
E[__I
lj*".J"l_11
at 03 0.4015
0.00
6(:E
c</)6
otr
f, )\
Slab Thickness, m
I
.rS rk 3C0) +S,I fnLxS ik 150) rS ikSoj
3.50
6o
=ci cqnett,6 i.^
oTL
t)u
stress chart due to single Axle Load with concrete shoulder (p=160 kN, ar=f0o c)
0.50
c2 c.3 0.4
mSlab
I
I
I
II
v
t
J
t'
to Single Axle Load with Concrete Shoulder (p=g0 kN, AT=15o
*S ik 8Cl r S rk 1Ct+q.! !m,rS ik 3C0j
GaE{;o(n?xstr
Slab Thickness, m
.00
020.3
0.1
c)
n rE
25C
Stress Chart due
*s ik 801 +S ik40)fS ik 3C0l +S 1k.t5Cl
tqn
€aE
a,JE
$t!
1.00
fi7 nr(03 04
Stress Chart due to AxleSingle withLoad Concrete Shoulder f=1AKN,{P=120 5a c)
015
r:
;
7I
t /
q/ c
Stress Cha( due to Single Axle Load with Concrete Shoulder {P=160 kN, 1T=15" C)
r*srk3col **sik 1scr *slk80l eSrk,xct
a2 t? N ?E4.4
J.)U
3$A6o
=.ot$
tnJE
xou
?qn
lnn
050
0.00
0.15 n)6
Slab Thickness, m
ls ik 8cl +s ik.lil} S ik 300r x S (k 150)
t
t_LIfrl 1*.1
1_:ll
chart due to $ingle fule Load with concrete shourder {p=g0 kN, aT=20o c)2nn
,qn
6LE
oc) 'rvG
oTL
4.25 c3 035
m
1.C0
Stress
n 1q0.00
L
e/
I
Irlr
LI
{
stress chart due to single Axle Load with concrete Shoulder (p=80 kN, :\T=1s' c)25C
GaE 1.511
vio
(n
?
o"vtr
nrq
Slab Thickness, m
tJJ it 1i
{sik300l +Sikis) *s1k8ci + q.k ln'
I \\:.\
t.\
\ l\t.
{S ik 30{) *S lk i50) *S ik E0) + S rk 401
,s
\>
*'\ +s:'r
=
stress chart due to single Axle Load with concrete shoulder {p=120 kN, aT=15" c}350
? r1e
c.2 025 03 n ?q 0.4
m
015000
6o
=oatG
or
aL
t
I1rlrl 1I I
I II l II i I
l I
II
II
r \i.I I
I I
I I+ fI
I
I r\ll I I I I t\t
l I I
I II I
I I
I II I
g/ 6
l
i
l
i
i
due to Single Axle Load with Concrete Shoulder {p=120 kN, AT=20o
+S ik40)** s ik 80t+s ik 15cl{S 1k 300)
t
F'']
1nn
i, !5
1an
03 n1( c4
c)Stress Chart
?qn
GoE
ot!,!
= if,Jqtr
Slab Thickness, m
+s ik 300) +q,k 1qn. l s ik sc) +q kd,l
350
3CCEo
=o
C)E "^^5oI
15C
1.CC
Slab Thickness, m
AT=20oStress Chart todue Axle LoadSingle with Concrete Shoulder KN,{P:160 c)
rt
rrL
f
CHARTS FOR FLEXURAL STRESS (BUC): Tandem Axle
Stress Chart due to Tandem Axle Load with Concrete Shoulder (P=160 kN, .1T=0" C)
140
d _^^CL i 'U
=qio
3
u
0.40
0.20
000c.2 o r( 03 04
Slab Thickness, m
r,Stk300) .*S1k15C) .S{k80) .SiklC) i
\ I l II l l I
\\ I I I I
I I I l
\\
\l I i I I i\
I i! I I l\
\ \\
rxtlttiltlttXx \ \r f\I I I I
\i t\t 1I I
\
Stress Chart due to Tandem Axle Load with Concrete Shoulder {P=200 kN, lT=0o C)
.Fs ik300)  sik 150) .s lk80) rS,k.ic) 
\\
\:; \
\\ \\ \
xx\ x \+'"I \<i hsi,_*i____i +,"+_+
t
180
2.CC
0.40
0.20
n?
150
nr(
Slab Thickness, m
6oE 1.20
vi6oU'lE:x 030
G
UDU
{
rI i<l t\ T
)C
1
Stress Chart due to Single Axle Load with Concrete Shoulder {P=120 kN, 1T=20" C}
aSak300l +Srk150l *Slk8C; .Sik4Cl
\
Eo
=o
U>
E5 l.)Uxc
050
0.00
02 025
Slab Thickness. m
n1(
+Srk300) +Sik150) a.S(k80) +Sik:tCl
\\
= s\\
N
Stress Chart due to Single Axle Load with Concrete Shoulder {P=160 kN, .lT=20o C}
02 c.4
4.5C
1.00
35C
IJ 5U
6ILEqi tqe
@L
t>E "^^
gIL
1.5C
025
Slab Thickness, m
{
a
i "l
r  i_T
S_I\
S\
I
\\
fi
I
/I
CHARTS FOR FTEXURAL STRESS (BUC): Tandem Axte
stress chart due to randem Axle Load with concrete shoulder {p=160 kN, {T0o c}
,Sik300) * Srk150) +Sik80) a S ik ,tC)
1.2C
o 'i0=q;
o
?
0.40
c.20
1_CU
n ?(
il nn
0.15
Slab Thickness. m
*r,.Sik300)  Sik150) +S 801 + S ik r0)
stress chart due to Tandem Axle Load with concrete shoulder {p=200 kN, aT=0o c)2.4!
18C
16:
lJi
GILE 1.2C
U;aa6Ex nea
c
$lab Thickness, m
c2 03 n ?( U4c 15
rt 7
t )=#*]*__t_t_+._+__+__+__f_*+_
ffiffi
d
Stress Chart due to Tandem Axle Load with Concrete Shoulder (P=240 kN, AT=0o C)2.50
lrsik300l  sik150l +sikacl .slklc)
I
>: \, I
LE 1.50
o
x 1lC
L
0.00
J,: c2 c25 0.3 0,4
Slab Thickness, m
.',Sik30Cr  Sik1501 +S,k80) +Sik,10l
\x l I I
I I II l l I
\\
\
S \ \\ I I\ \I I I I I I I I I i I\ x t\.tltttt\
\ l.\i r\J llil, \ \I'\i l'\
i Fr!l I l
I I
Stress Chart due to Tandem Axle Load with Concret€ Shoulder (P=320 kN, AT=0o C)
Slab Thickness, m
0.50
c.25 03 0.35c15
250
6L
=6
a
= t.e
oL
1.@
I
r 7
Ci I I i 't. i i \i Il i :
a
s s {k 30Cl  s lk 150J +S ik 3Cl $s {k 40)
Slab Thickness, m
4.20.3 UJf
\JC
Shoulder{P=160 kl,l, AT=10" C}1.80
1.60
1.4A
C.,1C
Stress Chart due to Tandem Axle Load with Concrete
:2nG(LE,t , ^
oE cac3
tr
.* s rk 8cl },Sik4C)_.r_s
ik 300).e s (k
a203
0.4
Stress Chart todue Tandem Axle Load with Concrete Shoulder (P=200 T=frl 0oKN, c)
000
oE 150!;
@
=x ItoIt
025
Slab Thickness, m
&/""/
6
I'
stress cha( due to Tandem Axle Load with concrete shoulder {p=240 kN, aT=i0" c}
.*s ik 300)   s ik 1501 .S ik80) rS rk40)
\\ \ \
=_
\\
:
a2 lt1 c1n 1i
?tn
250
rin
icc
0c0
6CL€si0o
(n6
or
Slab Thickness, m
.*Sik300) * Sikt50l r S ak 801 o S ;k .l0l
350
?(n
2.00
stress chart due to Tandem Axle Load with concrete shoulder {p=300 kN, ar=10o c}
025 n ?( 04
Slab
6oEvio
C'']ixou
{
t
I
U i: 02
SlabThickness, m
03 035 0.4
3CC
2.5C
caE
oC''6
xor1.00
 S {3CC,  S,(:50. +!rrii; .S.kicr
Stress Chart due to Tandem Axle Load with Concrete Shoulder {P=320 kN, "{T=10o C)
050
Stress Chart due to Tandem Axle Load with Concrete Shoulder {P=160 kN, fT=15o C),an
18C
160
1.4C
,)n
] .UU
080
lou
c.40
+Sik3&)  Sik150l {Sik80} +S:i(,.0l
\\
*\
$
U i3 02 4.4
6tl=vi
Ear>
6
xoir
c25
Slab Thickness, m
q
r )
\*rirT**r
\ I I I I\\
I
S\
rl,=S!l r!
I Il I I
I I
*
:,,:!..i,+_ ,ri,,,ii__l__._t_'.
"""1+r' :\as],*+']*<+'+' i
{
stress chart due to Tandem Axle Load with concrete shoulder {p=200 kN, ar=i5, c}
:Srk300t  S1k JS0l .s ik80j l* s lk 4cr
l
\ \\ \\
.\ :\a
\ \L
rEi;= I
6oE{io
(t6
xsII
t.,l 03 r1 1q\.).  )
250
fin
nEn
Slab Thickness, m
,*S 1k3m) rS ik 150i . S tk 30) + S rk .l0t
Stress chart due to Tandem Axle Load with concrete shoulder {p=240 kN, trT15o c}
u.t 61 n ?(
1ln
25C
0.00
20c€oEo'o!,rnc)Gxotr
100
n)t
Slab Thickness, m
:
I
'\
C
3i
U
stress chart due to Tandem Axle Load with concrete shoulder {p=300 klrl, ar=f ro c}
, S ik 300) . S rk 150j *S ik BC) +S ik 40)
I
LI
L

l'l
*'ltl
I
ItE::_
 1
4 lEn 1q
00c
lEn
3.00
GoE
oar,6
=)uxotr
'UU
c25
SlabThickness, m
40c
1(n
300
6o 250Elio
Exu
stress chart due to Tandem Axle Load with concrete shoulder {p=320 kN, {T=15o c}
'* S ik 3001 .s ik 1501 {s ik 80) + s rk lci
03
Slab Thickness, m
q/
t
.., s ik 3001 ; S rk 80) r S {k 10)
rtoE6ooaGL
xol.r
00
0.? A)
{P={60 kN, AT=20o C}Stress Chart due to Tandem Axle Load with Concrete Shoulder
 /\
Slab Thickness, m
eS lk 300)  sik150l +q iL An +S rk 40)
.00
c.2 425035
m
Shoulder(P=200 kN, AT=20o C)
c.15
lqn
Stress Chart due to Tandem Axle Load with Concrete
Go
=66e
ctta!
xotr
nno
If
t
 Srk3CCj  Srk15Cl dSik8C) .S1k,i0l
I \ \
\
\
.\
Stress Chart due to Tandem Axle Load with Concrete Shoulder(P=240 kN, lT20" C)
02
30c
250
1na
c.0c
€ELEdo
(r,EL
otr
n14
Slab Thickness, m
UJ t1q
__s,i300 __sk,5c _+_s(3: __r_SxeC
Stress Chart due to Tandem Axle Load with Concrete Shoulder (P=300 kN, _\T=20" C)
015
Slab Thickness, m
050
03c
02 c25 04
250
6(L
=ottt6f 'JU
3t!
1C0
b
'{>'
t , t t t t t t T T T"T_
\I Is
l I I\I
I\
I\ t "'+ i l\ I I j I
I I I I I i I 1 l* I I
l I
I i I! I I
Stress Chart due to Tandem fule Load with Concrete Shoulder (P=320 kN, .1T=20o C)
3.50
_ _S k3JC: _ _S i:S _+_S,kSC: {_Sik jC.
S
\ R *+ FLL\
JJU
Go 250
=,;o
GxO ,in_ IJU
UJU
UUU
025
Slab Thickness, m
n2 .)1( c4
Stress Chart due to Tandem Axle: without Concrete Shoulder (P=160 kN, fT= 0oC)250
a!CL
= t50,t;6o
at)6
x 1nno ,"iT
Slab
U{
L
\\\
\I t \ \
\i \
\ \ \ :\I q
I
'lI
0.00
i, i3 A) 0.35
1n
sI I ls
=
\ I I
t\+. Tl I I itl I
I
I I
I I I
I I
\
{
JUU
,qa
200
1.50
0.50
stress chart due to Tandem Axre: without concrete Shourder (p=200 kN, fT
r. S ik 30Cl  s lk 150) .S tk 301 .* S lk .l0l
rt_=*'i_i*.i
i
h)c3 035
0'c)
G(LEsie
U,
L
olr.
0.00
Slab Thickness, m
stress chart due to Tandem Axre: without concrete shourder {p=240 kN, \T=
e.*S {k 300) {BS (k 150) .+ S ,k 80, + S k +01
05c
000
4.2
0"c)
€TL
=e
(r,G5 I,JUxotr
nfc
Slab Thickness, m
li
Stress Chart due to Tandem Axle: without Concrete Shoulder {P=300 kN, AT= 0oC}
1.CC
G ^^O lf,UE,;qoat '"?xO 1Enlt
1:n
nia
nqn
4.2 t)q
Slab Thickness, m
u.4
,:'S ik 300)  S ik 1501 .S lk80) rS ik40lk I i I l I
I ) i I l I
\ I I ll ) I
r l I I I ) II I I ! I i
I I I I I\ \ 1 t__I I I I 1 II I I I I\I i I 1
I I lk \\
K I I I I I l
\\
\t\lI I I l I
t\.iT' I I I II \ i 'i\.i I 'II I l
x=J sa I l
j Il\1 t\,1Y l
I I
I I{ 1 I l**:i. I I
I \i l=$!I I*
i I Ii .{ i=R\I l I \i! I
lg :*\i 1qI II
I I l l l I I tI l l I I I I i l I I
Stress Chart due to Tandem Axle: without Concrete Shoulder {P=320 kN, 1T= 0oC)
1qn
3C0€(LEvi r<ne(r,
E:oootr
150
100
U3U
0.00
'.*Sik300)  Sik150r +S,kSCl .Sik+3r
\\
\\
\\
\"\\
n 1( 0.3 n?q 04
m
nrq
Slab
1)
f
rx
\ x5i x=
t9
Stress Chart due to Tandem Axle: without Concrete Shoulder {P=160 kN, 1T= 1O"Ci) iir
200
S.35! S,')1. +SiEC +S,r,ir''
\\\t\
r\ \\ \ \*
: :\\I
aa=q;
o.n
6
oTL
1
1
c00
Slab Thickness, m
c3 0.35 0.4
Stress Chart due to Tandem Axle: without Concrete Shoulder{P=200 kN, AT= '10,C)
:S ik 3m)  S lk i50) +S ik 3Cl .S ,k 4C,
\\.
I \
\:\ \{ \ \S
:
o
=u,eoE
tr
035 0.4
m
3.00
2.00
1.50
1,00
050
0.m
1a
t'
li,
(e
Stress Chart due to Tandern Axle: without Concrete Shoulder (P=240 kN, \T= 10oC)
1tn
t)n
6(L
=o
ar1
63Jvx6Ll
100
rS ik 3C0l  Si{15C1 *Slk80) .Sik40l
\
I
I
f"
c403f,)0'i5
Slab Thickness, m
stress chart due to Tandem Axle: without concrete shoulder {P=300 kN, ,lT= 1ooC)
 5 rK JUUI €S (k 150) +*S (k 80) .S (k 40)
4.la
qlq
3.S
3.ffiEo
=lE zsoo,tE z.oo
xoIL
1.5i)
1.{n
Slab Thickness, rn
U
1,4
r
V
,1 50
$tress Chart due to landem Axle: without Concrete Shoulder{p=320 kN, AT= 10oC}
,+S ik 300j aq,k An, r s ak 40j
6aE_ r"^E
tttG ^^^x{,tr
iuu
n 1qiz
Slab Thickness, m
n1c4
I
*tLI
*J
l* S {k 300) .$.S {k 150) + i t n,1, + S ,k 40r
\
Stress Chart due to Tandem Axle: without Concrete Sh oulder (P=1 60 KN, \ I 1 5"C)
6oElhoo
a,G
xor.l
05C
0.00
N'A
Slab
.3
u. tf,
m
c.4
'1
6
Stress Chart due to Tandem Axle: without Concrete Shoulder {P=200 kN, AT= 15oC)
soE{iog(r,G!
xglt
2.50
2AA
1Rn
1.00
c00
,,*s ik 300)  *s ik 15ci +S rkSC) .S ik 401
\! \ \
S \\\
\ \I \\ \
\
= \.
c 15 a2 c25
Slab Thickness, m
c3 n?q
Stress Chart due to Tandem Axle: without Concrete Shoulder {P=240 kN, tT= 15oC)
eSik300j * *Sik1soj eS,x8C) oS,kaC:
=:.
:::
03 n 1€
4.00
3_C)
3$
E zso
=d{,#r*E=xg 1.50ll
1.fi)
c25
Slab Thickness, m
16
S
b
]r
i I
i I I
\
I i
I
I
6
Stress Chart due to Tandem Axle: without Concrete Shoulder {P=300 kN, tT= 15"C}+f,u
1C0
6&=vi t qr.
E<t)
G rrn
otr
1.00
n,0.4
Slab Thickness, m
 S (300.  S.i. 15C .S {8C rS iraor
\\\
\
=\
\\
\s
+s (k4q I*.,S1k300] .Sik1:O) .SikE0)
\\ \
\= \ ==\ *
Stress Chart due to landem Axle: without Concrete Shoulder (P320 kN, ,\T= 15"C)5.00
4.50
4.00
3.50
E6
= 3.00!i*o# zsoE
= 2.00
lL
1,50
J2 n1x1\
Slab Thickness, m
6
)t.''+ 1""i' + +::+ I il.+' I
___1_r,_*__+__r,__i i i:qE__, i _
Stress Chart due to Tandem Ade. ritrrut Concr*e Sftoulder3.00
2fi
2fi6o
=d€o# r.so
G
xctr
+ s tk 80) __._ s {k 40)
+L__Ij
I'1
.i"++*i
$lab Thicknesq m
n) !t5035 0.4
(P=160 kN, fT= 20"C)
0.15
1.00
0.50
0.00
Stress Chart due to Tandem Axle: without Concrete Shoulder
+S rktol +S rk40i+S(t3fi1) +S(k1s0)
03 035
{P=200 kN, "\T= 20,C)3.s0
3.&
2fiac=i 2.fl)ato5 r:oIa
lr
025
Slab Thickness, m
# 6'
*:j
!
,l
FI
a
i
I
LFr*
l
"1_1
l
;.I
'*
)
) iai
)i*
43C
? q,1
?ni
6o 2.50
=cio
;;e,EitL
I,UU
stress chart due to Tandem Axre: without concrete shourde r {p=2[akil, ar=
rs ik 80i rS rk 4c)
050
n2 n ts
20"c)
015JUU
n rq
Slab Thickness, m
stress chart due to Tandem Axre: without concrete shourder (p=300
,+ S lk 300J  rS (k 1501 + S ik 801 + i r 1:r
03
m
kN, \T= 20"C)
350
3Inae=iz:nSoE z.o,xau
,9,
I
s
,b {
1q
l.*