Design thickness of a rigid pavement slab for a two-way, two-lane National highway in state of West Bengal. The total two-way traffic is 2500 CVPD at end of construction period. The flexural strength of concrete is 45 kg/cm², modulus of subgrade reaction is 8 kg/cm³, tyre pressure is 8 kg/cm², rate of traffic increase is taken as 7.5 %, spacing of contraction joints is 4.5 m and width of slab is 3.5 m. Assume any or data required. The axle load spectrum obtained from a survey is given below:

Single Axle Load		Tandem Axle Load
Axle load class (tons)	Percentage of axle load	Axle load class (tons)	Percentage of axle load
19-21	0.6	34-38	0.3
17-19	1.5	30-34	0.3
15-17	4.6	26-30	0.6
13-15	11.0	22-26	1.8
11-13	22.3	18-22	1.5
9-11	23.0	14-18	0.5
<9	30.0	<14	2.0

The procedure is followed exactly as mentioned in the Appendix-2 of IRC 58:2002. The first step is to find the design traffic by evaluating the cumulative repetitions of commercial vehicles.
The cumulative repetitions, C=\(\frac{365A[(1+r)^n-1]}{r}\)
Where A is the present traffic = 2500 CVPD
R is the growth rate = 0.075
N is the design life = 20 years
C=\(\frac{365×2500×[(1+0.075)^{20}-1]}{0.075}\)=39515522 commercial vehicles. Now design traffic is to be taken as 25% of C, i.e. the design traffic = 0.25×39515522=9878881 commercial vehicles.
Now the axle load can be taken as the mid-value of the axle load class and the expected repetitions can be found out by multiplying the respective percentages of load by the design traffic. This can be written in tabular format as below:

Now assuming a thickness of 32 cm for the slab and the load factor to be 1.2, the following can be computed and tabulated as below:
The stress from the chart is calculated for slab thickness 32 cm and subgrade reaction modulus of 8 kg/cm³ using the charts in IRC 58:2002 (Refer Book: Guidelines for the Design of Plain Jointed Rigid Pavements Design for Highways, Published by: The Indian Roads Congress).
Stress ratio is obtained as the ratio of flexural stress of concrete i.e. 45 kg/cm² to the flexural stress due to load i.e. stress from charts.
The fatigue life corresponding to stress ratio is available in Table 6 of IRC 58:2002. It is found out until it becomes infinite.

Since the cumulative fatigue life consumed is less than 1, the thickness assumed is safe.
The next step is to check for temperature stress and corner stress.
Temperature stress at the edge is to be checked for using the equation \(St_e=\frac{CEet}{2}\). Finding the radius of relative stiffness as below:
Assuming E = 3×10⁵ kg/cm² and μ as 0.15,
l=\(\left[\frac{Eh^3}{12k(1-?^2)}\right]^{1?4}=\left[\frac{3×10^5×32^3}{12×8×(1-0.15^2)}\right]^{1?4}\)=101.17 cm
L/l ratio for computing C = 450/101.17 = 4.5. Using Figure 2 from IRC 58:2002, the value of C is obtained as 0.58; e is assumed as 10×10^-6 /°C and t for West Bengal can be obtained as 16.8°C from Table 1 given in IRC 58:2002.
\(St_e=\frac{0.58×3×10^5×10×10^{-6}×16.8}{2}\)=14.62 kg/cm²
The stress = temperature stress + maximum load stress = 14.62 + 24 = 38.62 kg/cm², which is less than the flexural stress of concrete (45 kg/cm²), therefore the design is correct.
The equation to find corner stress is S_c=\(\frac{3P}{h^2} \left[1-\left(\frac{a?2}{l}\right)^{1?2}\right]\). The design load P is the 98-percentile axle load = 16 tonnes, so the wheel load is 8 tonnes or 8000 kg. the tyre pressure, p is given as 8 kg/cm². The radius of the contact area, a is found using the relation, \(\frac{?}{4}a^2=\frac{P}{p}\) which can be rewritten as a=\(\sqrt{\frac{4P}{?p}}=\sqrt{\frac{4×8000}{?×8}}\)=35.68 cm.
Therefore, S_c=\(\frac{3×8000}{32^2} \left[1-\left(\frac{35.68?2}{101.17}\right)^{1?2}\right]\) =13.27 kg/cm². The corner stress 13.27 kg/cm² is less than the flexural stress of concrete (45 kg/cm²), therefore the design is correct.
The slab thickness is designed as 32 cm.