Module 4: Pavement Design

Lecture 29 Rigid pavement design

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2.Given h=20 cm, b=7/2=3.5m, $S_s=1750~kg/cm^2$ $W=2400~kg/cm^2$ $f=1.5$ $S_b=24.6~kg/cm^2$.  
 
Step 1: diameter and spacing:

$$\begin{eqnarray*} A_s=\frac{3.5\times{}20\times{}2400\times{}1.5}{100\times{}1750}=1.44~cm^2/m \end{eqnarray*}$$

Assume $\phi=1~cm,~\Rightarrow~A=0.785~cm^2$. Therefore spacing is $\frac{100\times{}0.785}{1.44}=54.57~cm$, say $55~cm$  
 
Step 2: Length of the bar:

$$\begin{eqnarray*} L_t=\frac{1\times{}1750}{2\times24.6}~=~36.0~cm \end{eqnarray*}$$

[Ans] Use $1~cm~\phi$ tie bars of length of $36~cm~@~55~cm~c/c$