Page 45 - UNIT 5- FLEXIBLE PAVEMENT DESIGN

Civil Engineering Department | DCC3113 : Highway & Traffic Engineering

Example

• Design of Layer Thickness

TA = a D + a D + ... + a D
n
1
2
2
1
n
1st Trial
Nominate D1 = 12.5 cm
D2 = 18.0 cm
D3 = 20.0 cm

Then TA = 1.0 x 12.5 + 0.32 x 18 + 0.23 x 20
= 25.36 cm < TA'

2nd Trial D1 = 15.0 cm
D2 = 20.0 cm
D3 = 20.0 cm

Then TA = 1.0 x 15 + 0.32 x 20 + 0.23 x 20

= 26.0 cm

Make sure TA ≥ TA’, so choose 2 trial
nd

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Prepared by : SITI ZURAIFA BINTI MD SAH, zuraifa@pmu.edu.my