Module 4: Pavement Design

Lecture 29 Rigid pavement design

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

Tie bars

In contrast to dowel bars, tie bars are not load transfer devices, but serve as a means to tie two slabs. Hence tie bars must be deformed or hooked and must be firmly anchored into the concrete to function properly. They are smaller than dowel bars and placed at large intervals. They are provided across longitudinal joints.

Step 

Diameter and spacing: The diameter and the spacing is first found out by equating the total sub-grade friction tot he total tensile stress for a unit length (one meter). Hence the area of steel per one meter in $cm^2$ is given by:

$$A_s\times{}S_s=b\times{}h\times{}W\times{}f$$

$$A_s=\frac{bhWf}{100S_s}$$ (1)

where, $b$ is the width of the pavement panel in $m$, $h$ is the depth of the pavement in $cm$, $W$ is the unit weight of the concrete (assume $2400~kg/cm^2$), $f$ is the coefficient of friction (assume $1.5$), and $S_s$ is the allowable working tensile stress in steel (assume $1750~kg/cm^2$). Assume $0.8$ to $1.5~cm~\phi$ bars for the design.

Step 

Length of the tie bar: Length of the tie bar is twice the length needed to develop bond stress equal to the working tensile stress and is given by:

$$L_t=\frac{d~S_s}{2~S_b}$$ (2)

where, $d$ is the diameter of the bar, $S_s$ is the allowable tensile stress in $kg/cm^2$, and $S_b$ is the allowable bond stress and can be assumed for plain and deformed bars respectively as $17.5$ and $24.6~kg/cm^2$.