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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.
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.
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.
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)
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