Flexible and Rigid Pavement Design
Flexible pavement design with traffic in msa, vehicle damage factor and fourth power law, CBR method, rigid pavement stresses (Westergaard), radius of relative stiffness, temperature stresses and joints.
A flexible pavement (bituminous) spreads load through layers; a rigid pavement (concrete slab) carries it by bending. Both are designed for the traffic they will carry over their life.
Traffic for flexible pavements (IRC:37)
Traffic is expressed in million standard axles (msa) of 80 kN:
N = 365 A [(1 + r)ⁿ − 1]/r × D × F
- A = initial commercial vehicles per day (after growth up to opening), r = growth rate, n = design life (years).
- D = lane distribution factor (for example 0.75 per direction on a two-lane road).
- F = vehicle damage factor (VDF).
- Fourth power law: damage ∝ (axle load/standard axle load)⁴. A 160 kN axle does 16 times the damage of an 80 kN axle.
Flexible pavement design
- CBR method: the older approach read the total thickness from design charts for the subgrade CBR and traffic; IRC:37 now uses mechanistic–empirical design.
- Mechanistic checks: horizontal tensile strain at the bottom of the bituminous layer (fatigue cracking) and vertical compressive strain on top of the subgrade (rutting).
- Equivalent single wheel load (ESWL) for dual wheels.
- Layers: subgrade, granular sub-base, base, binder course, surface course.
Rigid pavement stresses (Westergaard)
- Radius of relative stiffness: l = [E h³/(12(1 − μ²) k)]^(1/4).
- Load stresses are critical at the corner, edge and interior; the edge and corner usually govern.
- Corner stress (Goldbeck, simple): σ = 3P/h².
- Temperature (warping) stresses: σ = (Eα Δt/2) C (Bradbury's coefficient C).
- Frictional stress when a slab contracts: the subgrade friction on half the slab length, σ_f = γ_c L f/2 (γ_c = unit weight of concrete, L = slab length, f = friction coefficient). It limits the spacing of contraction joints.
Joints in concrete pavements
- Contraction joints (control cracking), with dowel bars for load transfer.
- Expansion joints (allow expansion), with gap filler.
- Longitudinal joints, held together by tie bars.
- Spacing of contraction joints depends on friction and slab weight.
Worked example
Initial traffic A = 2000 CVPD, r = 7.5%, n = 15 years, D = 0.75, F = 3.5.
- Growth sum = [(1.075)¹⁵ − 1]/0.075 = 26.12.
- N = 365 × 2000 × 26.12 × 0.75 × 3.5 = 50.05 msa.
Practice
By the fourth power law, how many times more damage does a 120 kN axle do than an 80 kN standard axle? (Give 4 decimal places.)
Show answer and explanation
Answer: 5.0625(120/80)⁴ = 1.5⁴ = 5.0625.
A slab has E = 30,000 MPa, h = 250 mm, μ = 0.15 and k = 0.06 N/mm³. What is the radius of relative stiffness in mm? (Round to the nearest integer.)
Show answer and explanation
Answer: 903l = [30,000 × 15.625 × 10⁶/(12 × 0.9775 × 0.06)]^(1/4) = (6.66 × 10¹¹)^(1/4) ≈ 903 mm.
Using the simple corner formula σ = 3P/h², what is the corner stress (MPa) for a 50 kN wheel load on a 250 mm slab?
Show answer and explanation
Answer: 2.4σ = 3 × 50,000/62,500 = 2.4 MPa.
Dowel bars in a concrete pavement are provided at:
Show answer and explanation
Answer: ADowels transfer wheel loads across transverse joints while letting the slabs move; tie bars hold longitudinal joints together.