Design of rigid pavement

1. FLEXIBLE PAVEMENT
DESIGN
Rohaya Binti Rasmin

2. INTRODUCTION
Pavement consists of several layers made of appropriate material selection, the task of
disseminating the vehicle load (stress and strain) on the basis of the road (subgrade),
provides a surface that has adequate skid resistance and have a long lifespan without
need lots of maintenance work.

3. Factors that are
considered in
designing the
thickness of
flexible
pavement :
Failure criteria
Traffic
loadings
Traffic
decaying
power
Environmental
effect

4. DESIGN FACTORS OF
THICKNESS OF PAVEMENT
 a. Traffic load
Imposed load by private vehicles does not affect much on road
damage. Only commercial vehicle unlade weight exceeding 1500 kg
only be taken into account for the design. Total Cumulative can be
obtained after the design life and traffic growth rates determined.
Total cumulative value of commercial vehicles must be converted into
a number of incremental axle standard.
b. Design life
Design life selected based on the type of road and utility. The design
life of 10 years is recommended because the rate of traffic growth
and trade in developing countries is not only high but difficult to
estimate accurately.

5. c. Sub-grade condition
This is a key factor in determining the thickness of the pavement.
Subgrade strength (from an estimated value California Bearing Ratio,
CBR) depends strongly on the type and moisture content of the soil.
When the pavement (assuming impermeable water) built on the
formation, subgrade beneath backing up gradually increased to its
ultimate level. CBR value at the highest moisture content must be
identified for the purpose of design.
d. Drainage
Try to avoid the ground water level from rising to a level of 600 mm
from the end of the road. This can be done by draining underground
or elevated levels by the end of the road or embankment reclamation.

6. FLEXIBLE PAVEMENT
DESIGN
METHOD
PWD (Arahan Teknik 5/85)
AASTHO (American Association of State Highway and Transportation
Official)
AI (Asphalt Institute)

7. PWD METHOD
STEP
Initial annual commercial traffic for one way,
Vo
Total number of commercial vehicles one
direction for x years, Vc
Total Equivalence Standard Axles, ESAL
Maximum hourly one way traffic flow, c
One way traffic capacity (24 hours), C
Volume of daily traffic after x year in one
lane, Vx

8. PWD METHOD
EXAMPLE
The following conditions are given:-
Class of Road JKR
05
Initial daily traffic volume (ADT)
6,600 vph
Percentage of commercial vehicles, Pc
15%
Annual growth rate , r
7%
Equivalence factor , e
2.0

9. PWD METHOD
STEP 1 : Initial annual commercial traffic for one way, Vo
Vo = ADT x 0.5 x 365 x Pc/100
Pc = Percentage of commercial vehicles
Vo = 6600 x 0.5 x 365 x 15/100
= 180,675 vph

10. PWD METHOD
STEP 2 : Total number of commercial vehicles one direction for x
years, Vc
Vc = Vo [(1+r)x – 1] / r
= 180,675 [(1+0.07)10 – 1] / 0.07
= 180,675 (0.967)/ 0.07
= 180,675 (0.967)/ 0.07
= 2.5 x 106 vph

11. PWD METHOD
STEP 3 : Total Equivalence Standard Axles, ESAL
ESA = Vc x e
e = Equivalence factor or table 3.1
ESA = 2.5 x 106 x 2
= 5 x 106 vph

12. PWD METHOD

13. PWD METHOD
STEP 4 :Maximum hourly one way traffic flow, c
c = I x R x T
I = The ideal hourly capacity (Table 3.2)
R = The roadway factor (Table 3.3)
T = The traffic reduction factor (Table 3.4)
Assume that road 2 lanes and carriageway width is 7.5m with 2m
shoulder width
c = 1000 x 1 x 0.769
= 769 vph

14. PWD METHOD

15. PWD METHOD

16. PWD METHOD

17. PWD METHOD
STEP 5 : One way traffic capacity (24 hrs), C
C = 10 x c
C = 10 x 769
= 7690 v/d/ln

18. PWD METHOD
STEP 6 : Volume of daily traffic after x year in one lane, Vx
Vx = Vl (( 1 + r )x / direction
Vl = initial daily traffic in one direction
Vx = 6600 (1 + 0.07)10 / 2
= 6492 v/d/ln

19. PWD METHOD
STEP 7 : Equivalent Thickness, TA
From figure 2, with ESAL = 5 x 106 vph and Subgrade CBR is 5%,
required TA is 26 cm
TA = a1D1 + a2D2 + ........+ a3D3
a1,a2...an = Structural coefficients of
each layer (Table 3.5)
D1,D2.....Dn = Thickness of each layer
(Table 3.6)

20. 5 %
5 x 106
26 cm
3 % (*From the
bottom of
this
nomograph)
TA’= 26 cm

21. PWD METHOD

22. PWD METHOD

23. PWD METHOD

24. PWD METHOD

25. TRY & ERROR
1st trial
Nominate D1 = 12.5 cm
D2 = 18 cm
D3 = 20 cm
TA = 1 (12.5) + 0.32 (18) + 0.23 (20)
= 12.5 + 5.76 + 4.6
= 22.65 cm < TA

26. TRIAL & ERROR
2rd trial
Nominate D1 = 15 cm
D2 = 20 cm
D3 = 23 cm
TA = 1 (15) + 0.32 (20) + 0.23 (23)
= 15 + 6.4 + 5.29
= 26.69 cm > TA OK
Taking into consideration the minimum thickness
requirements, the pavement structure than comprise of the
following layer thicknesses:
Wearing Course - 5 cm
Binder Course - 10 cm
Base Course - 20 cm