Module 4: Pavement Design

Lecture 28 IRC method of design of flexible pavements

1

2

3

4

5

6

7

8

9

10

11

12

13

Solutions

1. Distribution factor = 0.75

2. $$\begin{eqnarray*} N&=&\frac{365\times{\left[(1+0.05)^{15}-1)\right]}}{0.05}\times{300}\times{0.75}\times{2.5}\\ &=&4430348.837\\ &=&4.4~msa \end{eqnarray*}$$
   
   

3. Total pavement thickness for CBR 4% and traffic 4.4 msa from IRC:37 2001 chart1 = 580 mm
4. Pavement composition can be obtained by interpolation from Pavement Design Catalogue (IRC:37 2001).
   1. Bituminous surfacing = 20 mm PC + 50 mm BM
   2. Road-base = 250 mm Granular base
   3. sub-base = 280 mm granular material.