Page 109 - Lecture Notes DCC3113

Design of Layer Thickness

TA = a1D1+ a2D2+ ... + anDn

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

Taking into consideration the minimum thickness requirements, the pavement structure then comprise
of the following layer thicknesses

Wearing - 5 cm
Binder - 10 cm
Base - 20 cm
Subbase - 20 cm

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