APSEd

# Rigid Pavement Design by Westergard's Method with IRC Recommendations

Rigid pavements are pavements that are made up of cement and concrete materials. By virtue of the materials, rigid pavements offer flexural strength. There are several components of rigid pavement like contraction joints, dowel bars, etc., in order to design a rigid pavement all of which are explained further.

## Rigid Pavement Characteristics

Before getting into rigid pavement design it is important to understand certain characteristics of rigid pavement. As said, rigid pavements are made of concrete, i.e., plain cement concrete, reinforced cement concrete, or prestressed concrete.

### Rigid Pavement Layers

The concrete slab can be used as both wearing surface and base course. But in general, a rigid pavement consists of a concrete slab below which a granular base or sub-base is provided. The provision of a base course improves the life of the rigid pavement.

### Load Transfer in a Rigid Pavement

Unlike flexible pavement where the load is transferred by grain to grain action, the load in the rigid pavement is transferred by slab action.

The main difference between rigid pavement and flexible pavement is that the critical condition of stress in the rigid pavement is the maximum flexural stress occurring in the slab due to wheel load and temperature changes whereas in flexible pavement it is the distribution of compressive stresses.

## Rigid Pavement Design

Westergard is considered the pioneer in the analysis of rigid pavements. IRC has also adopted the Westergard method for the design of rigid pavements with slight modifications and recommendations.

Before getting into details of Westergard's design it is important to understand certain terms related to concrete in pavement design.

The concrete grade to be used for rigid pavement should be

*M40*According to IS456:2000, the flexural strength of concrete is given by the relation

where fck is the characteristic compressive strength of concrete.*fcr = 0.7*(fck^0.5),*It is also called the modulus of rupture or bending tension.According to IRC, the modulus of elasticity of concrete of rigid pavement shall be,

(i.e., 5000*(fck^0.5)) for M40 concrete.*Ec = 3 * 10^4 MPa*The poison ratio of concrete shall be

*0.15*

Now let us discuss some important concepts that Westergard had theorized for rigid pavement design.

### Westergard's Modulus of Sub-Grade Reaction

The design of rigid pavement is dependent on the modulus of subgrade reaction. It is defined as the ratio of pressure applied on a circular plate to the settlement of the plate. It is represented by the letter k. Its unit is kgf/cm^3.

*k = p/Δ*

where,

k - modulus of subgrade reaction

p - pressure applied

Δ - settlement of the plate

According to IRC, the settlement of a 75 cm circular plate shall be 1.25 mm. Therefore,

*k = p/0.125*

The maximum value of subgrade reaction for rigid pavement is 5.5 kgf/cm^3.

### Relative Stiffness of Slab to Subgrade

The subgrade offers a certain amount of resistance to the slab deflection. This is dependent on the properties of both the subgrade material and slab.

The resultant deflection of the slab which is also the deformation of the subgrade is a direct measure of the magnitude of subgrade pressure. The pressure deformation characteristic of rigid pavement is thus a function of the relative stiffness of the slab to that of the subgrade.

Westergard coined this term "Radius of relative stiffness (l)"

*l = ((Eh^3/12k(1-(µ^2)))^(1/4)*

where,

l - radius of relative stiffness (cm)

E - modulus of elasticity of cement concrete (kg/cm^2)

µ - Poisson's ratio of concrete

h - slab thickness (cm)

k - modulus of subgrade reaction (kg/cm^3)

### Equivalent Area of Resisting Section

As the name suggests, this gives the equivalent area of the slab that is effective in resisting the bending moment applied to it. Westergard approximates this by the radius of wheel load distribution and slab thickness.

**b = (1.6a^2 + h^2)^(1/2) - 0.675h**

where,

b - equivalent radius of resisting section (cm)

a - radius of wheel load distribution (cm)

h - slab thickness (cm)

For a better understanding both the formulas are given in a pictorial format below.

## Stresses on a Rigid Pavement

### Critical Load Positions

There are three locations namely the interior, edge, and corner where the differing conditions of slab continuity exist.

**Interior Loading:** Load is applied in the interior of the slab surface at any place remote from the edges

**Edge Loading: **Load is applied at the edges of the slab at any place remote from a corner

**Corner Loading:** Load is applied at the corner that is formed by the intersection of two edges.

In most cases, critical stresses caused by these loads will be given directly. We should just find the most critical combinations out of the given loads and use that load as the design load for calculation. But still, it is important to understand how these stresses are calculated from the given load. Westergard provided a simple method to find the stresses for all conditions as given below.

### Westergard's Stress Equation for Wheel Loads

According to Mr. Westergard, the flexural stress developed in concrete due to wheel load is inversely proportional to the square of the thickness of the rigid pavement.

*fload = (P/h^2) * Q*

where,

P - wheel load

h - thickness of the pavement

Q - stress coefficient, *Q = l/b*

l - radius of the relative stiffness of concrete

b - equivalent radius of contact of the wheel

According to IRC the wheel load on a rigid pavement is 5100 kgf i.e., 51 kN. And the thickness of the rigid pavement should not be less than 15cm.

Note that this formula cannot be directly used to compute stresses for interior, edge, and corner loading and requires some more components to calculate the stresses. But still, if the stress coefficient for a loading type, load, and thickness are given one can use this formula to calculate the critical stress.

### Temperature Stresses

Apart from wheel load stresses a rigid pavement is also subjected to stresses due to temperature change. The variation in temperature across the depth of the slab is caused by daily variation in temperature whereas the seasonal variation in temperature causes an overall increase or decrease in slab temperature. Accordingly, there are two types of stresses namely,

Warping stresses

Frictional stresses

**Warping Stress: **Whenever the top and bottom surfaces of a concrete pavement simultaneously possess different temperatures, the slab tends to warp upward or downward. This is caused due to daily variation in the temperature.

According to IRC, the warping stress shall be determined by Mr. Bradbury's formula which is based on Hooke's law.

*f = α * Δt * E*

where

α - coefficient of thermal expansion (10 * 10^-6 /°C)

Δt - change in temperature

E - young's modulus of concrete

**Frictional Stress: **This is caused due to uniform rise and fall of temperature in the slab which results in the overall expansion and contraction of the slab. As the slab is in contact with the subgrade, the subgrade tries to restrain this expansion/contraction by friction thereby inducing the frictional stresses.

*Sf = WLf / (2*10^4)*

where

Sf - unit stress developed (kg/cm^2)

W - the unit weight of concrete, kg/cm^3

f - coefficient of subgrade restraint

L - slab length (m)

B - slab width (m)

### Combination of Stresses

Combinations of the above-mentioned stresses need to be considered while calculating the critical combination of stresses. The most critical combination occurs during daytime only as the wheel load stress and warping stress cause bending tension in the bottom fiber. Whereas, during night thought the bending tension due to wheel load remains the same, warping stress causes compression in the bottom-most fiber.

The following conditions are considered to provide the critical combinations.

**During Summer: **Critical combination of stresses = wheel load stress + warping stress - frictional stress, at the edge region

**During Winter: **Critical combination of stresses = wheel load stress + warping stress + frictional stress, at the edge region

**At Corner: **Critical combination of stresses = wheel load stress + warping stress

The greater value of resultant stress determined at the edge and corner shall not exceed the modulus of rupture of concrete.

*f(wheel) + f(warping) + f(friction) <= fcr = 0.7*(fck^(1/2))*

Apart from designing the thickness of the slab expansion joints and contraction joints also need to be designed.

## Joints in a Rigid Pavement

### Expansion Joints

The expansion joint is an assembly designed to hold the structure together while safely absorbing temperature-induced expansion. According to IRC, the expansion gap in concrete should be 25mm, and the net expansion gap is 25/2 = 12.5mm.

Spacing between expansion joints (L)

Δl = L * α * Δt

25/2 = L * 10 * 10^-6 * 28

**L = 44.64 m**

**Dowel Bars**

Dowel bars are large diameter bars placed along the traffic direction i.e., longitudinal direction. The design strength of dowel bars is taken as 40% of wheel load. The design of dowel bars is dependent on the following parameters,

Shearing strength of dowel bars

Bending strength of dowel bars

Bearing strength of the concrete near dowel bars

### Contraction Joints

Contraction joints are provided to form a weak plane in the pavement surface such that cracks will occur along the plane rather than any other undesired locations. The length of the contraction joint is given by two different expressions based on the presence of dowel bars.

**Case 1: Dowel bars are not present**

*Lc = (2*fct) / (γcc*f)*

where,

fct - allowable tension in concrete = 0.8 kgf/cm^2

γcc - unit weight of the concrete = 2400 kgf/m^3

f - coefficient of friction = 1.5

Substituting all the values in the above equation, we get, the length of the contraction joint as 4.5m.

**Case 2: Dowel bars are present at contraction joint**

*Lc = (2*σst*Ast) / (f*γcc*B*h)*

where,

σst - allowable tensile stress in steel = 140 N/mm^2 for Fe250

Ast - area of tensile reinforcement (mm^2)

f - coefficient of friction = 1.5

γcc - unit weight of the concrete = 2400 kgf/m^3

B - width of the pavement (m)

h - depth of the pavement (m)

#### Tie Bars

Tie bars are smaller diameter bars placed in the perpendicular direction to the traffic flow. It is used to connect two panels. The design is based on friction and not the wheel load. The length of the tie bars is found as the sum of the development lengths of bars in both panels.

## Practise Problem

With this, we hope we had covered all the important topics related to the design of rigid pavements. Want us to cover a topic of your interest in a simple manner? Let us know in the comments below.