Module 4: Pavement Design

Lecture 29 Rigid pavement design

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Solution:

Given, $P=5000~kg$, $l=80~cm$, $h=25~cm$, $\delta=2~cm$, $F_s=1000~kg/cm^2$, $F_f=1400~kg/cm^2$ and $F_b=100~kg/cm^2$; and assume $d=2.5~cm$ diameter.

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

$$\begin{eqnarray*} L_d&=&5\times2.5~\sqrt{\frac{1400}{100}\frac{(L_d+1.5\times{2}... ...imes{2})}}\\ &=&12.5\times~\sqrt{14~\frac{(L_d+3)}{(L_d+17.6)}} \end{eqnarray*}$$

Solve for $L_d$ by trial and error:

put $L_d=45.00~\Rightarrow{}L_d=40.95$
put $L_d=45.95~\Rightarrow{}L_d=40.50$
put $L_d=45.50~\Rightarrow{}L_d=40.50$
Minimum length of the dowel bar is $L_d+\delta~=~40.5+2.0~=~42.5~cm$, So, provide $45~cm$ long and $2.5~cm~\phi$. Therefore $L_d=45-2=43~cm$.

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

$$\begin{eqnarray*} P_s=&0.785\times2.5^2\times1000&=~4906~kg\\ P_f=&\frac{2\time... ...rac{100\times2.5\times43.0^2}{12.5~(43.0+1.5\times{2})}&=~804~kg \end{eqnarray*}$$

Therefore, the required load transfer capacity

$$\begin{eqnarray*} \max\left\{\frac{0.4\times{5000}}{4906},~\frac{0.4\times{5000}... ...000}}{804}\right\}\\ \max\left\{0.41,2.77,2.487\right\}=2.77\\ \end{eqnarray*}$$

Step-3 : Find the required spacing: Effective distance of load transfer $=1.8~l~=~1.8\times{80}~=~144~cm$. Assuming $35~cm$ spacing,

Actual capacity is

$$\begin{eqnarray*} 1+\frac{144-35}{144}+\frac{144-70}{144}+\frac{144-105}{144}+\f... ...144-140}{144}\\ =2.57~<~2.77~~~(\mathrm{the~required~capacity}) \end{eqnarray*}$$

Therefore assume $30~cm$ spacing and now the actual capacity is

$$\begin{eqnarray*} 1+\frac{144-30}{144}+\frac{144-60}{144}+\frac{144-90}{144}+\frac{144-120}{144}\\ =2.92~>~2.77~~~(\mathrm{the~required~capacity}) \end{eqnarray*}$$

Therefore provide $2.5~cm~\phi$ mild steel dowel bars of length $45~cm~@~30~cm~$ center to center.