Module 4: Pavement Design

Lecture 28 IRC method of design of flexible pavements

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Solution

1. Distribution factor = 0.75

2. $$\begin{eqnarray*} N&=&\frac{365\times{\left[(1+0.075)^{15}-1)\right]}}{0.075}\times{400}\times{0.75}\times{2.5}\\ &=&7200000\\ &=&7.2~msa \end{eqnarray*}$$
   
   

3. Total pavement thickness for CBR 4% and traffic 7.2 msa from IRC:37 2001 chart1 = 660 mm
4. Pavement composition can be obtained by interpolation from Pavement Design Catalogue (IRC:37 2001).
   1. Bituminous surfacing = 25 mm SDBC + 70 mm DBM
   2. Road-base = 250 mm WBM
   3. sub-base = 315 mm granular material of CBR not less than 30 %