IRC Rigid Pavement Design

Atul Narayan, S. P.

IIT Madras

September 17, 2015

Outline

Introduction

Thickness design

Joints, Reinforcement and Tie bar design

Dowel bar design

Subordinate learning objectives

To analyze the stress-strain distribution in pavements for givenloading conditions. 3 To estimate pavement distresses based on stresses andstrains in pavement structure. 3 To explain the effect of mechanical properties on pavementbehavior and performance. 3 To analyze the stresses and distresses caused by vehicleloading. 3 To estimate the expected volume of traffic in design life. 3

General

The IRC design method is given in IRC 58: 2002 This design method is very similar to the design method ofPortland Cement Association. Recommended design life is 30 years.

Pavement Analysis Method

Analysis is essentially conducted using the finite elementprogram - IITRIGID Stresses may also be obtained using Westergaards solutions Recommended critical loading condition is edge loading fordoweled pavements and corner loading for undoweledpavements IRC recommends a contact pressure of 0.8 MPa IRC does not specify shape of contact area

Material PropertiesSubgrade

Modulus of subgrade reaction should be determined as per IS9214 Standard plate diameter for determining k is 75 cm Smaller plate diameter of 30 cm may be used. Conversion equation:

k75 = 0.5k30 (1) k-value can also be determined from CBR through tables

provided in IRC 58.

Material PropertiesConcrete

Modulus of rupture should be at leastS = Sc + Z (2)

where Sc is the characterisitic strength used in the design, Zis the single-tail normal deviate for reliability and is thestandard deviation. For design, Sc of 4.5 MPa can be used. Recommended elastic modulus and Poissons ratio are 30GPa and 0.15 respectively Recommended coefficient of thermal expansion is10 106 /0C

Traffic

Just like in IRC 37, IRC 58 recommends conducting axle-loadsurvey An Axle-load distribution table is constructed from the data IRC recommends multiplying axle loads by a Load SafetyFactor (LSF). Recommended value of LSF is 1.2. Cumuluative number of axle-loads is given by

C = 365A{(1 + r)n 1}r

(3)

There are no directional and lane distribution factor in use.

Traffic (cont.)

Not all axle loads occur at the edge of the slab. There isvehicle wander.

For two-lane two-way highways, IRC conservativelyrecommends 25% of the total axles (in both directions) beused for determining fatigue life.

That is, 25% of the total axle load occurs at the edge of thepavement. (PCA design method uses 6%)

For highways with multiple lanes each way, IRC recommends25% of the total axle load in one direction.

Climate

Effect of moisture onsubgrade propertiesare not considered IRC recommendstemperaturedifferential for curlingstresses based onlocation

Pavement DistressesFatigue Equation

For SR 0.55 logNf = 0.9718 SR0.0828 (4a)For 0.45 SR 0.55 Nf = ( 4.2577SR 0.4325)3.268 (4b)For SR 0.45 Nf = + (4c)

Pavement DistressesErosion and pumping

Erosion is not considered in design method The reason is that erosion is severe only for tandem axles andtandem axles are rare in India Erosion being smaller for single-axle loads does not meanthey are negligible.

Outline

Introduction

Thickness design

Joints, Reinforcement and Tie bar design

Dowel bar design

Procedure1. Assume a thickness for the slab.2. Divide the axle load distribution into axle groups Create an axle group for every 2-ton interval for single axles Create an axle group for every 4-ton interval for tandem axles3. Take the mean axle load of the group and multiply it by LSF.4. Determine the critical stress for each axle group using the

charts provided in IRC 58.5. Determine Nfi , the number of repetitions to failure, for the

critical stresses.6. Determine the expected number of axles in each axle group.

25% of these axles are expected to occur at the edge of thepavement. (ni )

7. Determine the damage caused by each axle group asniNfi

8. Determine if the total damage niNfi

is less than 1.

9. Determine design thickness by trial and error.

IRC Chart for critical stresses

Sample calculations

Sample calculations (cont.)

Sample calculations (cont.)

Sample calculations (cont.)

Sample calculations (cont.)

Sample calculations (cont.)

Class Problem

A particular highway section requires a new concrete pavement.The daily truck traffic at the start of the design period is 1100trucks and is expected to grow at 5% per annum. Traffic iscompletely composed of single axles and composition is as shownin the table below:

Load (tons) Percentage7-9 60.59-11 19.811-13 11.713-15 6.315-17 1.217-19 0.4719-21 0.03

The modulus of rupture of the concrete is 45 kg/cm2. The modulusof subgrade reaction of the subgrade is 10 kg/cm2. Use a load

Class Problem (cont.)

safety factor of 1.2. Design the thickness of the concretepavement.

Class Problem (cont.)

24 tons22 tons20 tons18 tons16 tons14 tons12 tons10 tons8 tons6 tons

Flexuralstress

(kg/

cm2)

Slab thickness (cm)15 20 25 30 35

0

10

20

30

40

50

60

70

Outline

Introduction

Thickness design

Joints, Reinforcement and Tie bar design

Dowel bar design

Reinforcements

Reinforcement is provided to the concrete slab to take tensilestresses due to contraction. Reinforcements are not for increasing structural capacity ofslab. (They do not assist in flexure) They are provided at mid-depth of the slab. Contraction causes tensile stresses in concrete due to frictionin the concrete-subgrade interface. Total tensile force per unit width due to contraction is

cLhfa2

,

where c is the unit weight of concrete, L and h are the lengthand thickness of the slab, respectively, and fa is the coefficientof friction.

Reinforcements (cont.) Assuming the diameter of the steel bars (usually in the range

of 6mm to 12 mm), the number of bars required per unitlength is calculated. If the cross-sectional area of one bar is As, the number ofsteel bars required per unit length is:

nbars = cLhfa2Asfs (5)where fs is the allowable stress in steel The number of bars required along the width of the slab perunit length is

cWhfa2Asfs

, where W is the width of the slab.

fa is usually taken as 1.5 and allowable stress is usually 66%of yield strength. If reinforcements are not provided, as with JPCP, concreteshould resist tensile stress.

Reinforcements (cont.)

Tensile stress in concrete, in the absence of reinforcement isc = cLfa2 (6)

Tie-bars

Tie-bars are provided along the longitudinal joint to hold theslabs together. Forces trying to separate the slabs are again the forces due tocontraction The number of steel bars required per unit width if thecross-sectional area of one bar is Ast is

Ast = cWhfafs (7)since total force due to contraction actin on the tie bars iscWhfa.

Tie-bars (cont.)

Tie-bars should be of adequate length to ensure sufficientbonding between the tie-bar and concrete. (To preventpull-out failure) Length of tie-bar is

t = 2( At fsO

) = fsd2

(8)

where At is the cross-sectional area of one tie-bar, is theallowable bond stress, O is the bar perimeter and d is thediameter of the bar. Allowable bond stress is usually taken as 2.4MPa

IRC recommendations for tie-bars

JointsTypes of Joints

Contraction joint Relieves tensile stresses caused by contraction

Expansion joint Leaves room for concrete slabs to expand

Longitudinal joint Separates slabs along the transverse direction

Construction joint Point of separation between old and new slabconstructions

Contraction Joint Spacing

Joint spacing depends on Tensile stresses due to contraction Extent of increase in joint width Presence of reinforcements Change in joint width depends on slab length as well as

thermal and shrinkage properties:

L = L(eT + ) (9)where e is the coefficient of thermal expansion, T is thetemperature decrease and is the shrinkage strain.

Contraction Joint Spacing (cont.)

Tensile force due to contraction is cLhfa2

; it depends on the

length of the slab

Tensile force should be resisted by either the concrete or thereinforcement

Spacing for JPCP is about 3 to 10 m and for JRCP, it is about10 to 30 m.

Joint spacing is decided based on experience

IRC recommendations for joint spacing in JPCP

Contraction Joint Construction

A steel plate is placed upto a certain depth at the joint locationduring construction and removed later. Otherwise, concrete slab is sawed after construction to acertain depth. The gap is then sealed with a bituminous or polymer sealant Cracks form over the rest of the depth due to vehicle loadingresulting in the formation of joint. The cracked surfaces provide aggregate interlock for loadtransfer. Dowels are provided during construction; they cannot preventcracking and thus, cannot prevent joint formation

Contraction Joint before and after cracking

Source:pavementinteractive.org

Expansion Joints

Expansion Joints are provided to leave room for concreteexpansion These joints prevent blow-up failure Expansion joints are no longer used except on bridges A good choice of aggregate, with low coefficient of thermalexpansion, makes expansion joints unnecessary

Outline

Introduction

Thickness design

Joints, Reinforcement and Tie bar design

Dowel bar design

Dowel Bars

Dowels are provided for transfer of load from one slab toanother

In the absence of dowels and aggregate interlock, cornerstress will be higher than edge stress

Dowel bar design is mostly based on experience Primary mode of failure of dowel arrangement is by failure of

concrete below dowel due to excessive bearing stress.

IRC Recommended Dowel Size and Length

Recommended diamater is one-eighth the thickness ofpavement

Allowable Bearing Stress

The allowable bearing stress of concrete is

Fb = (10.16 b9.525 ) fck (10)where b is the diameter of dowel bar in cm and fck is thecharacteristic compressive strength of concrete. Fb and fck bothhave the same units ([NL2])

Actual Bearing Stress due to load transfer by dowel bars

Maximum bearing stress due to a load on a dowel bar wasfound by Friberg. Bearing stress is determined by treating the dowel bar as abeam and the concrete as a liquid (Winkler) foundation.

Pt

d2

y0

Actual Bearing Stress due to load transfer by dowel bars(cont.) Maximum deflection of the concrete below the dowel is at the

edge of the slab. It is

y0 = Pt(2 + z)43Ed Id (11)where Pt is the load on the dowel bar, z is the joint width, Edand Id are the elastic modulus and moment of inertia of thedowel bar. is the relative stiffness of dowel bar, similar to radius ofrelative stiffness:

= 4 Kd4Ed Id

(12)

where K is the modulus of dowel support (similar to modulusof subgrade reaction) The range of K is between 80 to 400 GN/m3

Actual Bearing Stress due to load transfer by dowel bars(cont.)

Maximum bearing stress isb = Ky0 = KPt(2 + z)43Ed Id (13)

Maximum bearing stress depends on Pt , the load on thedowel bar.

Pt depends on the manner of transfer of load by dowels

Dowel Group Action

W

W2

W2

Role of dowels is to transfer axle load from one slab to another Dowels, together, will maximum transfer half the axle load,when they are fully efficent. Usual number is around 40% of the load.

Dowel Group Action (cont.) Friberg found that the dowels over 1.8l distance from the point

of loading act together to transfer the load Heinrichs has suggested a distance of 1.0l instead, which isnow used everywhere.

W 1.0l1.0l

The maximum load on one dowel bar depends on thedistribution

Dowel Group Action (cont.)

Load on dowel bar is assumed to linearly decrease withdistance from load, being zero at 1.0l from the point of load.

If multiple loads are acting at the transverse joint, for eachone, the distribution must be determined.

The total force on a dowel bar is the sum of the forces due toeach load.

Dowel Design Procedure

Pick dowel diameter and length from table. Assume a trial spacing Calculate the maximum force on a dowel bar by consideringthe most critical loading condition The 85 percentile single axle load is taken as the critical load.The critical loading condition is when the single axle is placedflush with the edge of the pavement. Determine maximum bearing stress corresponding to themaximum dowel load If bearing stress exceeds allowable bearing stress, reducespacing If spacing is already too small, increase dowel diameter

IntroductionThickness designJoints, Reinforcement and Tie bar designDowel bar design