Pavement Design

Topics: OSI model, Elasticity, Young's modulus Pages: 24 (1160 words) Published: March 12, 2014
Chapter 2
Stresses and Strains in
Flexible Pavements

1

Layers of Flexible Pavements

Surface (wearing) Course
Binder Course
Base Course
Subbase Course
Subgrade

2

Contents





Single layer analysis
Two-layer system
Three-layer system
Viscoelastic solution

3

Single Layer Analysis

r

t

z
4

Boussinesq Theory (1885)






Homogeneous elastic half-space
A concentrated load is applied
Stresses, strains, and deflections
are calculated by integrating
concentrated loads on the surface
Further developed by Burmister
(1943)
5

Vertical Normal Stress (σz)

R

2R

Material Properties are more
important in higher layers

3R

4R

R is the radius of circular loading
6

Analysis Methods



Charts (Foster and Ahlvin)
Closed form solutions





Flexible plate
Rigid plate

Computer programs





Kenlayer (provided in textbook)
BISAR
ILLI-Layer
others
7

Charts developed by
Foster and Ahlvin


Circular load with a radius a and an
intensity q




q= P/A = P/(πa2)

Assumed incompressible half-space
(ν = 0.5)
 lateral
Poisson’s ratio (ν) = 



 longitudinal

Bulk modulus




ν = 1/3  K=E
ν = 0  K=E/3
ν = ½  K=∞

K

E
3(1  2 )
8

Vertical Stresses Due to
Circular Loading
a

z/a=3
r/a=2
Also, find radial stress (σr), tangential stress (σt), shear stress (τrz), and vertical deflection (w) from the consecutive graphs 9

r

t
10

 rz

Vertical
displacement
(w)

11

Strains
1
 z   z   ( r   t )
E
1
 r   r   ( t   z )
E
1
 t   t   ( z   r )
E
12

Principal Stresses
τmax

 1, 2 

 z r
2

2

  r 
2
  rz
  z

 2 

σ1

σ2

2

 z  r 
2
  rz

 2 

 max  
3  t

Mohr’s Circle
13

Example 2.1
q=

z=

a=5”
14

Flexible Plate v.s. Rigid Plate

15

Closed Form Solutions for
Flexible Plate
• Limitation: Stresses, strains, and deflection at the
center of the circular area on the axis of symmetry
due to a single circular load


z3
 z  q 1  2 2 1.5 
 (a  z ) 

r 

2a =


q
z
2(1   ) z
 2
1  2  2
  t
2
(a  z 2 ) 0.5 (a  z 2 )1.5 
3

q=


z3
(1   )q 
2z
z 
 2
1  2  2
2 0.5
2 1.5 
E 
(a  z )
(a  z ) 

Z=


z3
(1   )q 
2(1   ) z
 2
r 
1  2  2


2E 
(a  z 2 ) 0.5 (a  z 2 )1.5 
w


a
(1   )qa 
1  2 2

(a  z 2 ) 0.5  z 
 2
2 0.5
E
a
 (a  z )






 rz  0
16

Rigid Plate Loading





Deflection is the same at all points
Pressure distribution under the plate is
not uniform
Pressure distribution (Ullidtz, 1987)
qa
q(r ) 
2(a 2  r 2 ) 0.5



Plate deflection
w0 

 (1   2 )qa

The smallest pressure is at
the center and equal to
one-half of the average
pressure.
The pressure at the edge
of the plate is infinity

2E
17

Vertical Surface Deflection
Flexible v.s. Rigid Plate



ν = 0.5, z = 0
Flexible plate
1.5qa
w0 
E



Rigid plate
w0 

 (1  2 )qa

2E
1.18qa
w0 
E



 (1  0.52 )qa
2E

vertical surface deflection under a rigid plate is only 79% of that under the center of a uniformly distributed load in flexible plate 18
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