Module 4: Pavement Design

Lecture 29 Rigid pavement design

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Solutions

1. Given, $P=4000~kg$, $l=90~cm$, $h=20~cm$, $\delta=3~cm$, $F_s=1000~kg/cm^2$, $F_f=1500~kg/cm^2$ and $F_b=100~kg/cm^2$; and assume $d=2.5~cm$ diameter.

   Step-1: length of the dowel bar $L_d$,

   $$\begin{eqnarray*} L_d&=&5\times2.5~\sqrt{\frac{1500}{100}\frac{(L_d+1.5\times{3}... ...es{3})}}\\ &=&12.5\times~\sqrt{15~\frac{(L_d+4.5)}{(L_d+26.4)}} \end{eqnarray*}$$
   
   

   Solving for $L_d$ by trial and error, it is =39.5cm Minimum length of the dowel bar is $L_d+\delta~=~39.5+3.0~=~42.5~cm$, So, provide $45~cm$ long and $2.5~cm~\phi$. Therefore $L_d=45-3=42~cm$.

   Step 2: Find the load transfer capacity of single dowel bar

   $$\begin{eqnarray*} P_s=&0.785\times2.5^2\times1000&=~4906.25~kg\\ P_f=&\frac{2\t... ...{100\times2.5\times42.0^2}{12.5~(42.0+1.5\times{3})}&=~758.71~kg \end{eqnarray*}$$
   
   
   Therefore, the required load transfer capacity (refer equation)
   $$\begin{eqnarray*} \max\left\{\frac{0.4\times{4000}}{4906.25},~\frac{0.4\times{40... ...{758.71}\right\}\\ \max\left\{0.326,2.335,2.10\right\}=2.335\\ \end{eqnarray*}$$
   
   

   Step-3 : Find the required spacing: Effective distance of load transfer $=1.8\times l~=~1.8\times{90}~=~162~cm$. Assuming $35~cm$ spacing,

   Actual capacity is

   $$\begin{eqnarray*} 1+\frac{162-35}{162}+\frac{162-70}{162}+\frac{162-105}{162}+\frac{162-140}{162}\\ =2.83 \end{eqnarray*}$$
   
   
   Assuming 40cm spacing, capacity is,
   $$\begin{eqnarray*} 1+\frac{162-40}{162}+\frac{162-80}{162}+\frac{162-120}{162}+\frac{162-160}{162}\\ =2.52 \end{eqnarray*}$$
   
   
   So we should consider 2.52>2.335 as it is greater and more near to other value. Therefore provide $2.5~cm~\phi$ mild steel dowel bars of length $45~cm~@~40~cm~$ center to center.