Indiacon Civil

Subjects › Pavement Design › Intermediate

Rigid pavement design

Intermediate Pavement Design Reviewed 24 Oct 2026 2 min read

A rigid pavement is a slab of paving quality concrete, usually 250 to 350 mm thick on highways, resting on a dry lean concrete or cement-treated sub-base over a prepared subgrade. Because concrete is stiff, the slab spreads load over a wide area, so stresses on the subgrade are low. The governing stresses are in the slab itself: flexural tension from wheel loads at critical positions, combined with curling stresses as the top and bottom of the slab are at different temperatures.

Westergaard's analysis of a plate on a dense liquid foundation introduced the radius of relative stiffness, which depends on slab thickness, concrete modulus and the modulus of subgrade reaction k. It identified interior, edge and corner as critical load positions. Modern Indian design, in IRC 58, checks cumulative fatigue damage under the expected axle load spectrum: for each axle load group, the flexural stress ratio to the concrete's flexural strength sets an allowable number of repetitions, and the sum of expected over allowable must stay below one. Bottom-up cracking (edge loading with daytime curling) and top-down cracking (with night-time curling and axles straddling a transverse joint) are both checked.

Diagram: Indiacon original, not to scale.

Joints control cracking: contraction joints at regular spacing, longitudinal joints between lanes, and expansion or isolation joints at structures. Dowel bars transfer load across transverse joints; tie bars hold lanes together. Concrete grade, flexural strength (often around 4.5 MPa) and good curing are central.

Tip

Test flexural beams, not only cubes, for paving concrete; the design is based on flexural strength.

Key formulas
  • Radius of relative stiffness: l = [E h³ / (12 (1 − μ²) k)]^0.25
  • Stress ratio = flexural stress / modulus of rupture
  • Cumulative fatigue damage: Σ (nᵢ / Nᵢ) ≤ 1
  • Curling stress (Bradbury): σt = C E α Δt / 2
Solved example

Problem. Find the radius of relative stiffness for a 300 mm slab with E = 30 000 MPa, μ = 0.15 and k = 0.08 N/mm³.

  1. E h³ = 30 000 × 300³ = 8.1 × 10¹¹.
  2. 12 (1 − 0.0225) k = 12 × 0.9775 × 0.08 = 0.9384.
  3. Ratio = 8.63 × 10¹¹; fourth root ≈ 964 mm.
  4. So l ≈ 0.96 m, roughly three times the slab thickness.

Answer. l ≈ 964 mm

Quick quiz

1Which stresses govern rigid pavement design?

Flexural stresses in the slab from wheel loads and curling.

2What is checked with Miner's rule?

Cumulative fatigue damage from all axle load groups.

3What do dowel bars do?

Transfer load across transverse joints.

Code note
  • IRC:58 (India): Guidelines for the design of plain jointed rigid pavements for highways.

Clauses and documents to read

  • IRC:58-2015 (guidelines for the design of plain jointed rigid pavements)
  • IRC:15-2017 (construction of concrete roads)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • E. J. Yoder, M. W. Witczak, Principles of Pavement Design, Wiley, 2nd edition.
  • Y. H. Huang, Pavement Analysis and Design, Pearson, 2nd edition.

Free and official sources

All references for this subject →

Found an error? Write to indiacongroupofengineers [at] gmail [dot] com. For corrections and feedback. We never ask for payment or personal details.