Legal and standard axle load; rigid stresses; joint spacing

Legal axle load and standard axle load

• Legal axle load is the maximum axle load permitted by law on public roads; it is the legal limit for a single axle (about 10.2 t for a dual-tyre single axle in India/Nepal, as per Nepal Road Standards / transport rules) and for tandem axles. Loads above it are overloads and cause rapid damage.
• Standard axle load is the reference axle used in design: a single axle of 80 kN (8.16 t, 18 kip) with dual wheels. All other axle loads are converted to it by the fourth power law $(P/80)^4$ to get cumulative standard axles (msa) in IRC:37 and AASHTO ESAL.

Stresses induced in a rigid pavement slab

1. Load stresses (flexural, from wheel loads) - interior, edge and corner stresses by Westergaard and Kelly. The edge stress is usually the largest bending stress, the corner load gives the largest deflection and cracks at corners.
2. Warping stress - from the temperature difference between top and bottom of the slab (tension at bottom by day, at top at night). Bradbury: $\sigma_{te}=\dfrac{CE\alpha\Delta T}{2}$.
3. Frictional stress - from restraint to the slab movement by the sub-base: $\sigma_f=\dfrac{WLf}{2\times10^4}$, tension when the slab contracts.
4. Stress due to expansion (compression) when joints are blocked, and shrinkage stress in early age.
5. Subgrade-related stresses - settlement and pumping lead to loss of support and cracking.

Spacing between contraction joints in jointed plain concrete pavement

Contraction joints are placed so that the tensile stress caused by sub-base friction at the middle of the slab does not exceed the allowable tensile stress of the (young) concrete.

$$\sigma_f=\frac{W\,L\,f}{2\times10^{4}}\le S_c\ \Rightarrow\ L_c=\frac{2\times10^{4}\,S_c}{W\,f}$$

where $W$ is the unit weight of concrete (2400 kg/m3), $f$ the coefficient of friction (1.5) and $S_c$ the allowable tensile stress in concrete (about 0.8 kg/cm2 for young concrete). So $L_c=\dfrac{2\times10^4\times0.8}{2400\times1.5}=4.4$ m, which gives the practical spacing of 4.5 m (IRC:58). For reinforced slabs, $L_c=\dfrac{2S_sA_s}{W h f}$ with $S_s$ the allowable steel stress and $A_s$ the steel area per metre width. The spacing also keeps warping stresses (through $L/l$) acceptable.