Design spacing for a dowel bar system for following data:
Design wheel load = 8000 kg
Percentage of load transfer = 40 %
Slab thickness = 30 cm
Joint width = 2 cm
Radius of relative stiffness = 105.45 cm
Characteristic compressive strength = 400 kg/cm²

The detailed procedure to find the spacing of the dowel bar system is explained in Appendix-3 of IRC 58:2002. The first step is to find the bearing stress in concrete as below,
\(F_b=\frac{(10.16-b) f_{ck}}{9.525}\)
The diameter of dowel bar, b is assumed to be 3.2 cm, so,
\(F_b=\frac{(10.16-3.2)×400}{9.525}\)=292 kg/cm²
The spacing between dowel bars is assumed a 30 cm and that the first dowel bar is placed at 15 cm from the edge of the pavement. The length of the dowel bar is assumed to be 50 mm. These are taken as per IRC 58:2002.
Only the number of dowel bars taking part in the load transfer is to be considered. Dowel bars up to a distance equal to the radius of subgrade reaction from the point of application of load are considered to be effective in the load transfer process.
Therefore, the number of dowel bars = 1+\(\frac{l}{spacing}=1+\frac{105.45}{30}=4.52?5\) bars
Next, we have to assume that the load transfer by the first dowel bar is Pt and the load on dowel bar system is at a distance of 1 from the first dowel bar?s load to be zero. This is expressed as below:
\(\left(1+\frac{l-s}{l}+\frac{l-2s}{l}+?\right)P_t\)
i.e.
\(\left(1+\frac{105.45-30}{105.45}+\frac{105.45-60}{105.45}\frac{105.45-60}{105.45}\right)P_t = 2.293 P_t\)
The load carried by the outer dowel bar P_t, must be equated to 40% of design wheel load, i.e.
2.293 P_t=0.40×8000
P_t=1395.5 ≈1396 kg
Next step is to check the above design for bearing stress.
Bearing stress=\((P_t k)\frac{2+?z}{4?^3 EI}\)
Where k is the modulus of dowel/concrete interaction = 41500 kg/cm²/ cm
E is the modulus of elasticity and it is assumed as 2×10⁶ kg/cm²
I is the moment of inertia of dowel bar and can be found out using the equation,
\(I=\frac{?b^4}{64}=\frac{?×3.2^4}{64}=5.147 cm^4\)
? is the relative stiffness of bar embedded in concrete which can be found as below:
\(?=\left[\frac{kb}{4EI}\right]^{1/4}=\left[\frac{41500×3.2}{4×2×10^6×5.147}\right]^{1/4}\)=0.238
Z is the joint width = 2 cm
Therefore, the bearing stress is obtained as
Bearing stress=1396×41500×\(\frac{2+0.238×2}{4×0.238^3×2×10^6×5.147}=258.41?258\) kg/cm²
The bearing stress in dowel (258 kg/cm²) is found to be less than the bearing stress in concrete (292 kg/cm²), therefore the design is correct. The assumed spacing and diameter of the dowel bar are safe.