design of rigid pavement.pdf

Design of Rigid Pavement

Elements of a Typical Rigid Pavement

A typical rigid pavement has three elements :i) Subgrade; ii) Sub‐base; iii) Concrete slab

Subgrade is the in situ soil over which the pavement structure is supported.Stiffness of the subgrade is measured by modulus of subgrade reaction (K).K is determined with the assumption that the slab is resting on dense fluid and thus the reactive pressure of soil on pavement is linearly proportional to the deflection of the slab. Value of K is widely dependant upon the soil type, soil density, and moisture content.K is determined by plate bearing test.

Elements of a Typical Rigid Pavement (contd)

Sub‐base is the layer of selected granular materials placed on the subgrade soil and immediately below the concrete pavementIt is provided for the following purposesTo provide an uniform and reasonable firm pavement support.To prevent mud pumping.To provide levelling course on undulated and distorted subgrade.To act as capillary cut off.It is not a part of the rigid pavement structure as it is not provided to impart strength to the pavement structure.


Construction of sub‐base is generally done byGranular material like natural gravel, crushed slag, crushed concrete, brick metal, laterite, soil aggregate etc. Granular construction like WBM or WMMStabilized soilSemi rigid material like Lime clay Puzzolana Concrete, Lime Flyash Concrete, Dry Lean Concrete .

Concrete Slab is designed on the basis of flexural strength of concrete.Due to repeated application of flexural stresses by the traffic loads, progressive fatigue damage takes place in the cement concrete slab in the form of gradual development of micro‐cracks.

The ration between flexural stress due to the load and the flexural strength of concrete is termed as the Stress Ratio (SR). If SR < 0.45 the concrete is expected to sustain infinite number of repetition.Various properties of concrete as recommended for use as rigid pavement are

Flexural strength: 45 kg/cm2; Modulus of Elasticity: 3 x 105 kg/cm2;Poisson's ratio: 0.15;Coefficient of thermal expansion: 10 x 10‐6 /⁰C.


Types of Rigid PavementThere are four types of Rigid PavementJointed and unreinforced concrete pavementJointed and reinforced concrete pavementContinuously reinforced concrete pavementPrestressed Concrete Pavement

Most of the rigid pavements in India are jointed and unreinforced concrete pavement. The necessary IRC design guidelines areIRC: 58– 2002 (Guideline for the Design of Plain Jointed Rigid Pavements for Highways)IRC: 15‐ 2002 (Standard Specifications and Code of Practice for Construction of Concrete Roads)IRC: SP: 62‐ 2007

Stresses in Rigid Pavement Stresses in concrete pavement are produced due to following reasonsApplied wheel loadChanges in temperatureChanges in moisture contentVolumetric changes in soil subgrade

In design of rigid pavement stresses due to applied wheel load and changes in temperature are considered.

Since the nature of the stresses due to changes in moisture content is reverse that of stresses due to changes in temperature, it is not considered in thickness design.

Stresses due to volumetric changes of subgrade soil is taken care by properly selected sub‐base course.

Westergaard AnalysisH. M. Westergaard is considered to be pioneer person in rigid pavement design.

The basic assumptions in Westergaard (1925) analysis for computation of stresses are i. Concrete slab acts as a homogenous, isotropic, and

elastic solid in equilibrium.ii. The reaction of subgrade are vertical only and they are

proportional to the deflection of the slab. This reaction of subgrade per unit area at any given point is equal to a constant K multiplied by the deflection at that point.

iii. The thickness of the slab is uniform.iv. The load at the interior and at the corner of the slab is

distributed uniformly over a circular area of contact.

Westergaard Analysis (contd)v. For corner loading the circumference of the area of

contact is tangential to the edge of the slab.vi. For the load at the edge of the slab is uniformly

distributed over a semi‐circular area contact. The diameter of the semi‐circle is with the edge of the slab.

Critical stress locationsInterior: This is the position within the slab which is at any place remote from all the edges.Edge: This is the position of the slab which is situated in the edge, remote from the corners.Corner: This is the position which is situated at the bisector of the corner angle.

Westergaard Analysis (contd)

Interior CornerEdge

Stresses due to Wheel LoadUnder the wheel load the interior and the edge of the slab behaves like a simple supported beam having tension at the bottom.

Under the action of wheel load corner may behave as a cantilever specially when the slab is casted panel by panel.

The maximum tensile stress may be found at corner as this location is considered as discontinuous from all the directions.

As the edge is discontinuous in one direction this location may encounter lesser stress than the corner.

Loads applied at the longitudinal edge can produce more stress than that at the transverse edge.

Least stress is occurred at the interior as this position of theslab is continuous in all directions.

Stresses due to Wheel Load (contd)

Computation of stress at edge locationThe original equations of Westergaard has been modified by several researchers.As per IRC the stresses due to wheel load may be determined by the software IITRIGID developed at IIT Kharagpur. The stresses at edge may also be computed by the following equation as modified by Teller & Sutherland

]4048.0)b(log)bl(log4[)54.01()

hP529.0( 10102le −++= μσ

σle = Wheel Load Stress at Edge Region (kg/cm2)

P = Design Wheel Load (kg) or ½ of Single Axle Load (kg) or ¼ of Tandem Axle Load (kg)

h = Pavement thickness (cm)

μ = Poisson's Ratio

E = Modulus of Elasticity of Concrete (kg/cm2)

k = Modulus of Subgrade Reaction (kg/cm3)

l = Radius of Relative Stiffness (cm)

b = Equivalent Radius of Resisting Section (cm)

a = Radius of Load Contact Area (cm)


Relative stiffness of slab to sub‐gradeA certain degree of resistance to slab deflection is offered by the sub‐grade. The sub‐grade deformation is same as the slab deflection. Hence the slab deflection is direct measurement of the magnitude of the sub‐grade pressure. The resistance to deformation depends on the stiffness of the supporting medium as well as on the flexural stiffness the slab.This pressure deformation characteristics of rigid pavement lead Westergaard to define the term radius of relative stiffness (l). l in cm is given by

42

3

k)1(12Ehl

μ−=


Equivalent Radius of Resisting Section The wheel load concentrates on a small area of the pavementThe area of the pavement that is effective in resisting the bending moment due to that load may be more than tyreimprint area. The maximum bending moment occurs under the loaded area and acts radial in all directions.The area of the pavement that is effective in resisting the bending moment due to a wheel load is known as Equivalent Radius of Resisting Section or also as Radius of Equivalent Distribution of Pressure.

724.1haforh675.0ha6.1

724.1haforab

22 <−+=

≥=

Computation of Stress at Corner LocationWheel load stress at corner region is obtained as per Westergaard’s analysis modified by Kelley

σlc = Wheel Load Stress at Corner Region (kg/cm2)

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2.1

2lc l2a1

hP3σ


Stresses due to Temperature Variation

Stresses are induced in the slab due to variation of temperature

The temperature variation may be of two typesdaily variation resulting in a temperature gradient across the thickness of the slab, and seasonal variation resulting in uniform change in the slab temperature.

The former results in warping stresses and the later in frictional stresses.

Stresses due to Temperature Variation (contd)

Temperature Warping Stresses Cement concrete pavement undergoes a daily cyclic change of temperature as thermal conductivity of concrete is low.The top surface of the pavement becomes hotter than bottom during day time and cooler during night.In the daytime thus the top surface of the pavement expands more than that in the bottom. This results the slab to warp upwards (top convex). The restraint offered to this warping tendency by self‐weight and the dowel bars of the pavement induces stresses in the pavement. This is known as warping stress.


Flexural tensile stress will be generated at the bottom surface during day time. Conversely, in the night the slab warp downward (top concave). Flexural tensile stress will be generated at the top surface. As the restraint offered to warping at any section of the slab is a function of weight of the slab upto the section, the corner has very little of such restraint for slabs without dowel bars and is free to warp. Thus warping stress is negligible. The interior can offer maximum restraint to warp and has maximum warping stress.The equations for warping stresses are available due to Westergaard.


The critical combination of stress indicates that most critical location is the edge.

The equations for the warping stress at the edge as recommended by IRC is obtained as per Westergaard’sanalysis using Bradbury’s coefficient.

σtwe = Temperature Warping Stress at Edge Region (kg/cm2)

E = Modulus of Elasticity of Concrete (kg/cm2)

α = Coefficient of thermal expansion of concrete

t = temperature difference between top and bottom of slab

C = Bradbury’s coefficient depends on L/l or W/l of slab

C2

tEtwe

ασ =

Stresses due to Temperature Variation (contd)Temperature Friction Stresses

Uniform seasonal temperature variation cause the slab expands and contracts in the longitudinal direction. This expansion and contraction of the slab is prevented by the friction between the slab and the subgrade. Stresses are thus set up in the slab.

σcAc

L/2

100002WLf

fW100

h2LB100Bh

c

c

×=∴

××××=×××

σ

σ


B = Slab width (m)

h = slab thickness (m)

L = Length of the slab (m)

σtfe = Temperature friction stress in concrete (kg/cm2)

W = Unit weight of concrete in (kg/cm2)

f = Coefficient of friction between concrete and subgradeThe temperature friction stress is taken care in rigid pavement by providing joints in plain jointed pavement or by reinforcement in reinforced concrete pavement

Critical Combination of Stresses

Combination of flexural stresses due to wheel load and that to temperature warping provides the critical stress for design of rigid pavement.Maximum combined stress at the three critical locations will occur when these two stresses are additive.Warping stresses at three locations decrease in the order of interior, edge and corner whereas the wheel load stresses decrease in the order of corner, edge and interior.Therefore, critical stress condition is reached at edge location where neither wheel load stress nor the warping stress is minimum.Since at night due to warping the corner may behave as cantilever it is recommended to check the wheel load stress at corner.

Joints in Rigid Pavement The rigid pavement slab is deliberately divided into blocks of appropriate sizes in order to take care the effects of temperature friction stress or stresses due to moisture variation.

These deliberate planes of weaknesses in the slab are known as joints. A good joint should have the following functional requirements:Must be waterproof [proper sealing to be provided]Riding quality should not be deterioratedShould not make any structural weakness [for example staggered joints should be avoided]

Joints in Rigid Pavement (contd)Classification of the joints according to location in the pavementLongitudinal JointsTransverse Joints

Longitudinal JointsTransverse Joints

Joints in Rigid Pavement (contd)

Classification of Joints according to FormsDummy JointButt JointTongue and Groove JointJoints with Clear Gap

Classification of Joints according to FunctionExpansion JointContraction JointLongitudinal JointConstruction Joint


t

t+12

75mm t+6

SealerFillerDowel Bar [Fully Bonded part]

Dowel Bar [Bitumen Painted part]

Expansion Cap with Cotton Waste at the Back

Schematic Drawing of Expansion Joint with Dowel Bar

Expansion Joint


The pavement slab tends to expand when the temperature rises above that at which the pavement was laid.

Expansion of the slab is prevented by friction between the slab and the subgrade.

Compressive Stress is thus set up and this may try to buckle or blow up the slab .

In order to prevent this stress, Expansion Joints in the transverse direction of the pavement are provided to allow space for expansion of the slab.

The joint is formed by maintaining a gap of about 20 to 25 mm between two slabs.

The gap is filled up by a non‐extruding compressive filler material.


A sealing compound is provided on the top of the filler material to prevent entry of water and dust.

To ensure transfer of load between the two slabs on each side of the joint dowel bars are provided.Dowel bars are usually mild steel round bars of short length.Half length is bonded into concrete on one side of the joint and the other half is painted by bitumen in order to prevent bonding with concrete.A metal cap with cotton waste at the back is provided at the painted half end of the dowel bar. This ensures free movement of the slab during expansion.


Dowel bars not only permits the expansion of the slabs but also holds the slab ends on each side of the joint as nearly as possible.

Deflection of one slab under load is resisted by the other slab which, in turn is caused to deflect and thus carry a portion of the load imposed upon the first slab.

The spacing of expansion joint may vary from twenty meters to a few hundred meters.


Contraction Joint with Butt Joint

Contraction Joint with Dummy Joint

Contraction Joint

Dowel Bar

Sealer

Sealer


Stresses are also generated in the concrete pavement slab due to contraction of concrete when the temperature is reduced with respect to that during laying.

Contraction Joints are thus provided to reduce tensile stress due to contraction or shrinkage of concrete.

There are two types of contraction joint:Dummy Joint: In this type no joint is made in reality. Only a small groove is cut on top of the slab for a depth of ¼ to 1/3 of the thickness of the slab. If stress becomes more than that the slab can withstand, a crack may develop at the location of the grove as this is the weakest plane in the slab. Simple dummy joint may not contain any dowel bar. If dowel bar is not provided the load transfer is ensured by particle interlocking.


Butt Joint: In case of a butt joint two slabs abut each other. Therefore, a clear plane of separation will exist in this joint. Dowel bars may or may not be provided.

In case of dummy or butt joint good sealing material is provided at the top of the joint in order to prevent entry of water and dust inside the joint.

Spacing of joints varies with thickness of the slab and also with the existence of reinforcement. For slab of thickness upto 250mm joints maximum spacing may be 4.5m whereas upto 350mm thick pavement, maximum spacing will be 5.0m.

Longitudinal JointLongitudinal Joints are necessary in the concrete slab for the pavement having more than 4.5m wide. Longitudinal joint prevents longitudinal cracking.Mild steel bars known as Tie bars are provided across the longitudinal joint to hold the joint tightly closer and to keep both the slabs at the same level. Tie bars are not provided to act as load transfer device.Both the ends of the tie bars are fully bonded in the concrete.Longitudinal joints may be butt type or keyed [Tongue and Groove] type.


Construction Joint

Construction Joints are provided in the transverse direction whenever the placing of concrete is suspended for more than 30 minutes.

As far as possible construction joint should coincide with either expansion joint or contraction joint.

If a construction joint is provided at the location of any contraction joint it should be of butt type with dowel bar.

If a separate construction joint is needed, it should be provided within the middle third of two contraction joints.


Arrangement of Joints

Staggered JointWhen transverse joints are staggered with respect to the longitudinal joint, sympathetic cracks may occur.These cracks often occur in the line with the joint in the other side of the transverse crack.


Sympathetic cracks

Skew and Acute Angle JointUse of skew joint increases the risk of cracking at the acute angle corners.At the acute angles the stresses become very excessive.At the time of warping the acute angles become completely unsupported and cause more stresses than that would occur in right angle corner.


Design of Rigid PavementsStep 1: Stipulate design values for various parameters

Step 2 : Decide type and spacing between joints.

Step 3 : Select a trial design thickness of the pavement slab.

Step 4 : Compute the repetitions of axle loads of different magnitudes during design period

Step 5 : Calculate stresses due to single and tandem axle loads and determine cumulative fatigue damage (CFD).

Step 6 : If CFD is more than 1.0, select a higher thickness and repeat the procedure from step 4.

Design of Rigid Pavements (contd)

Step 7 : Compute the temperature stress at edge. If sum of the temperature stress and the flexural stress due to highest wheel load is greater than modulus of rapture select a higher thickness and repeat the procedure from step 4.

Step 8 : Design the pavement thickness on the basis of corner stress, if no dowel bar is provided and no load transfer is possible due to lack of aggregate interlocking.


Design Example

Design a cement concrete pavement for a two lane two way National Highway in Karnataka State. The initial total two way traffic is 3000 commercial vehicles per day. The other deign parameters are: Flexural strength of cement concrete: 45 kg/cm2.Effective modulus of subgrade reaction of the DLC sub‐base: 8 kg/cm3.Spacing of contraction joints: 4.5 m.Width of slab: 3.5mTyre pressure: 8 kg/cm2.Rate of traffic increase: 7.5%.

Axle load spectrum obtained from axle load survey is given below


Single Axle Load Tandem Axle Load

Axle load (tonnes)

% of axle loads

Axle load (tonnes)

% of axle loads

19‐21 0.6 34‐38 0.3

17‐19 1.5 30‐34 0.3

15‐17 4.8 26‐30 0.6

13‐15 10.8 22‐26 1.8

11‐13 22.0 18‐22 1.5

09‐11 23.3 14‐18 0.5

Less than 9 30.0 Less than 14 2.0

Total 93.0 Total 7.0


Design Traffic

Present traffic = 3000 cvpd; Design life (assumed) = 20 years;

Cumulative repetition in 20 years = 47, 418, 626 cvN = 365 A [(1+r)n – 1]/r where A is the initial number of axles in the year when road is operational ; r is the rate of annual growth of traffic; n is the design life.

Design traffic = 25% of repetition of commercial vehicles

= 11,854, 657 cv

Total repetitions of single axles and tandem axles are


Single Axle Load Tandem Axle Load

Load (tonnes) Expected repitions

Load (tonnes) Expected repitions

20 71127 34‐38 35564

18 177820 30‐34 35564

16 569023 26‐30 71128

14 1280303 22‐26 213384

12 2608024 18‐22 177820

10 2762135 14‐18 59273

Less than 10 3556397 Less than 14 237093


Thickness Design

Trial thickness = 32 cm; Load safety factor =1.2To take care unpredicted heavy truck loads the magnitude of axle loads should be multiplied by load safety factor (LSF).National highways and other roads where there will be uninterrupted traffic flow and high volumes of truck traffic: 1.2Lesser important roads with lesser incidence of truck traffic : 1.1Residential and other local streets : 1.0


Axle load tonnes(AL)

AL x 1.2 Stress (kg/cm2)

Stress Ratio

Expected Repetition (n)

Fatigue life (N)

Fatigue life consumed (n/N)

Single axle

20 24.0 25.19 0.56 71127 94.1 x 103 0.76

18 21.6 22.98 0.51 177820 4.85 x 103 0.37

16 19.2 20.73 0.46 569023 14.33 x 104 0.04

14 16.8 18.45 0.41 128030 Infinity 0.00

Tandem axle

36 43.2 20.07 0.45 35560 62.8x 106 0.0006

32 38.4 18.4 0.40 35560 Infinity 0.00

Cumulative fatigue life consumed 1.1705 > 1


Relation between fatigue life (N) and Stress ratio (SR)N = unlimited for SR, 0.45N= [4.2577/ (SR ‐0.4325)]1.324 0.45≤SR≥0.55Log10 N = (0.9718 – SR) / 0.0.828 for SR > 0.55

Since CFD fro thickness of 32 cm>1, increase thickness

Take next trial thickness = 33cm

Repeat the steps from 4

Cumulative fatigue life consumed for thickness of 33 cm = 0.47

Highest stress load stress for thickness of 33 cm = 24.10 kg/ cm2


Temperature Warping stress

For E= 3 x 105 kg/cm2, α = 10 x 10‐6 /⁰C, t= 21⁰ [slab thickness 33 cm in Karnataka state], K= kg/cm3, L= 4.5m,], Temperature Warping stress = 17.3 kg/ cm2

Total of temperature warping stress and the highest axle load stress = 17.3 + 24.1 = 41.4 kg/ cm2 < 45 kg/ cm2

design of rigid pavement.pdf

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