CT4860 STRUCTURAL DESIGN OF PAVEMENTS

(reinforced concrete pavement) from equalizing the equations 21 and 22b.
The determination of ∆tl is an iterative calculation.
It is however not necessary to calculate the limit temperature gradient ∆tl. The
actual temperature gradient stress σt in the centre of the longitudinal edge is
the smallest value resulting from the equations 21 and 22a, and similarly the
actual temperature gradient stress σt in the centre of the transverse edge or
crack is the smallest value resulting from the equations 21 and 22b.
To obtain the temperature gradient stresses along an edge or crack of the
concrete slab equation 20 can be used.
The temperature gradient stresses calculated by means of this Dutch method
are in good agreement with finite element calculation results (12,13,14).
Example
In this calculation example the temperature gradient stresses according to the
Dutch method are calculated in a plain concrete road pavement at 2 possibly
critical positions of the concrete slab, i.e. the center of the longitudinal edge
and the transverse edge in the wheeltrack.
The starting points for the calculation are:
- modulus of substructure reaction k = 0.105 N/mm 3
- concrete slabs with dimensions L = 4.5 m, W = 3.75 m and h = 210 mm
- concrete quality C28/35 (B35):
• Young’s modulus of elasticity E = 31000 N/mm 2
• Poisson’s ratio υ = 0.15
• coefficient of linear thermal expansion α = 10 -5 °C -1
- positive temperature gradient ∆t = 0.06, 0.05, 0.04, 0.03, 0.02, 0.01 and
0.0025 °C/mm
- the truck width is equal to 2.25 m, which means that a wheel track is at a
distance of 0.75 m from the longitudinal edges of the concrete slabs.
The calculated temperature gradient stresses are shown in table 9.
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