Stress analysis of rigid pavements

Atul Narayan, S. P.

IIT Madras

September 13, 2015

Outline

Introduction

Westergaard’s solutions

Stresses Due to Curling

KENSLABS

Outline

Introduction

Westergaard’s solutions

Stresses Due to Curling

KENSLABS

Subordinate learning objectives

▸ To analyze the stress-strain distribution in pavements forgiven loading conditions.

▸ To estimate pavement distresses based on stresses and strainsin pavement structure.

▸ To explain the effect of mechanical properties on pavementbehavior and performance.

▸ To analyze the stresses and distresses caused by vehicleloading.

▸ To estimate the expected volume of traffic in design life.

Outline

Introduction

Westergaard’s solutions

Stresses Due to Curling

KENSLABS

Critical Points of Loading

Corner Edge

Interior

Concentrated load at slab corner

▸ Concentrated load P is acting vertically at the corner of theslab.

▸ The subgrade support is neglected.

▸ The slab is assumed to act like a cantilever.

At any distance x from the corner

σc =Px

16(2x)h2

=3P

h2(1)

Approximate solutionsCorner loading

▸ Circular loading at the corner of the slab.

▸ Solution was first obtained by Westergaard.

σc =3P

h2

⎡⎢⎢⎢⎢⎣

1 − (a√

2

l)

0.6⎤⎥⎥⎥⎥⎦

(2)

∆c =P

kl2[1.1 − 0.88(

a√

2

l)] (3)

wherea is the radius of contact area

l is the radius of relative stiffness

h is the thickness of the slab

P is the load

k is the modulus of subgrade reaction

Approximate solutionsCorner loading

▸ The contact area is assumed to be a square.

▸ Empirical solution is obtained by finite element method.

σc =3P

h2[1 − (

c

l)0.6

] (4)

∆c =P

kl2[1.205 − 0.69(

c

l)] (5)

wherec is the side length of the contact area

c = 1.772a

Approximate solutionsInterior loading

▸ Solution was obtained by Westergaard

σi =3(1 + ν)P

2πh2[ln

l

b+ 0.6159] (6)

∆i =P

8kl2[1 +

1

2π{ln(

a

2l) − 0.673}(

a

l)2

] (7)

whereb = a when a ≥ 1.724h

b =√

1.6a2 + h2 − 0.675h when a < 1.724h

Approximate solutionsEdge loading

▸ Solution was obtained by Westergaard

▸ Poisson’s ratio of 0.15 was assumed

Circular loading:

σe =0.803P

h2[4 log (

l

a) + 0.666(

a

l) − 0.034] (8)

∆e =0.431P

kl2[1 − 0.82(

a

l)] (9)

Semi-circular loading:

σe =0.803P

h2[4 log (

l

a) + 0.282(

a

l) + 0.650] (10)

∆e =0.431P

kl2[1 − 0.349(

a

l)] (11)

Contact area for Rigid Pavements

▸ In case of rigid pavements, the Portland Cement Association(PCA) recommend the following shape for the contact area:

0.6L

0.8712L

▸ The odd length measures in the contact area is because ofPCA’s old assumption for the shape of the contact area:

0.6L

L

They wanted the current shape to have the same area interms of L as the old shape of the contact area.

Stresses due to dual tyres and non-circular loading

Sd

0.6L

L

When Pd is the load on a single tyre and q is the contact pressure:

Pd = q [π(0.3L)2 + (0.4L)(0.6L) = 0.5227L2] (12)

⇒ L =

√Pd

0.5227q(13)

Stresses due to dual tyres and non-circular loading (cont.)

The equivalent circular area should include the area between thetwo tyres. The area of the equivalent circular area is

πa2 = 2 × 0.5227L2 + (Sd − 0.6L)L = 0.4454L2 + SdL (14)

⇒ πa2 =0.8521Pd

q+ Sd

√Pd

0.5227q(15)

⇒ a =

¿ÁÁÀ0.8521Pd

qπ+Sdπ

√Pd

0.5227q(16)

Use the equivalent circular contact area in the approximateequations.

Outline

Introduction

Westergaard’s solutions

Stresses Due to Curling

KENSLABS

Curling stresses in finite slab

In finite slabs curling stresses in x and y direction area

σxx =EαT∆T

2(1 − ν2)(Cx + νCy) (17)

σyy =EαT∆T

2(1 − ν2)(Cy + νCx) (18)

where Cx and Cy are stress correction factors.

Stress correction factor for curling stresses

Radius of relative stiffness

For liquid foundation:

l = [Eh3

12(1 − ν2)k]

14

(19)

For elastic foundation:

l = [Eh3(1 − ν2s )

6Es(1 − ν2)]

13

(20)

Curling stress at the edge

▸ At the edge of the slab, curling stresses are influenced only byexpansion along the edge.

▸ Expansion perpendicular to the edge has no influence.

▸ Stress at mid-span is given by:

σ =CEαe∆T

2(21)

Outline

Introduction

Westergaard’s solutions

Stresses Due to Curling

KENSLABS

Introduction to KENSLABS program

▸ It is a finite element program for calculating stresses due toloading and curling

▸ It can handle up to 9 slabs at a time with 12 joints and atotal of 420 nodes

▸ Each slab can have a maximum of 15 nodes in each direction

Major Outputs

▸ Major, Minor stresses and maximum shear stress at each node

▸ σxx , σyy and σxy at each node

▸ σxx and σyy are positive when bottom is in tension.