1. Subgrade Models for Rigid Pavements

2. Development of theories for analyzing rigid pavements include the choice of a subgrade model. When the chosen model closely represents the actual situation, the resulting analyses are useful tools for design engineers. The model for analyzing a concrete pavement consists generally of an elastic plate that rests on a foundation model. Whereas the equations for plates that are subjected to mechanical load and thermal effects are well established, the modeling of the supporting base is not. The simplest and most popular model for the foundation base is the one proposed by Winkler (1867), consisting of closely spaced independent springs. For static loads, the response of this base model is the same as the response of a liquid base. An early solution for the infinite plate on a Winkler base subjected to a vertical load was presented by Hertz (1884). Solutions for a semi-infinite plate with a “free” boundary - for interior, edge, and corner loads - were presented by Westergaard (1926, 1939, 1947). These solutions have been at the

3. heart of slab design since the 1920’s. According to Iounnides et al. (1985), several investigators have noted repeatedly that although the Westergaard solution agreed fairly well with their observation for the interior condition, it failed to give even a close estimate of the response for edge and corner loads. It appears that this failure has less to do with the accuracy of the Westergaard solutions than with the inadequacy of the Winkler foundation model, which consists of closely spaced independent springs. From an interior load, the interaction between these springs is provided by the pavement, whereas along the “free” edge, these interactions - which take place in an actual subgrade - are missing. They are included in the Pasternak and Kerr subgrade models.

4. k w P w k G P (I) Winkler Model(II) Pasternak Model P G klkl kaka (III) Kerr Model

5. 0 1.0 0.6 0.2 1.4 q=3472 lb/in 2 Displacemets, in -0.4-0.2 0 0.2 0.4 x/L Kerr Model Pasternak Model Finite Element Winkler Model L/2 q H (E,v)

6. -0.4-0.2 0 0.2 0.4 0 1.0 0.6 0.2 1.4 1.8 2.2 Pasternak Model Kerr Model Winkler Model Finite Element lb/in H (E,v) L/2 FF F x/L Displacements, in

7. -0.4-0.20 0.2 0.4 0 200 600 1000 Pasternak Model Winkler Model Kerr Model Finite Element x/L Moments, Kips in/in

8. -5000 -4000 -3000 -2000 -1000 0 -0.4-0.2 0 0.2 0.4 x/L Moments, Kips in/in Pasternak Model Kerr Model Finite Element Winkler Model

9. P Most Foundation models are in between these two. Winkler Foundation P(x,y)=k w(x,y) w Figure 1 Elastic Foundation P (a) Figure 2

10. Semi-infinite continuum (isotropic solid) more difficult mathematically. Does not represent all foundation materials {relative to deflection away from loaded region}. Improvement of foundation modeling is approached on the basis of assuming some kind of interaction between spring elements. P Figure 2 Continued (b)

11. Filonenko-Borodich Foundation Top end of springs are connected to a stretched elastic membrane of constant- tension T Equilibrium in the z direction of a membrane element yields a load displacement relation: Laplace operator in the x and y direction Also, Het’enyl Foundation D= Flexural rigidity of a plate Spring interaction provided by an embedded plate

12. TT stretched membrane, plate in bonding, or shear layer Spring interaction is a function of T Figure 3

13. Assumes existence of shear interaction between spring elements. Connection of springs by incompressible vertical elements which deform by transverse shear. Pasternak Foundation w x x z Figure 4 Shear Intaction

14. y y+dy y x x NyNy P NxNx qsqs Vertical Equilibrium Figure 4 Continued

15. Pasternak Foundation Valasov Foundation Uses a ‘continuum’ approach by means of a variational method. Certain restrictions are imposed on possible deformations of the elastic layer. A load-deflection relation is obtained which can be re-dined to match the Pasternak expression. Plate Element Element of Shear Layer qsqs y y+dy y x x+ax x qsqs m sy m sx dx dy Figure 12

16. Assuming homogenous & isotropic in the x-y plane Shear force per unit length of shear layer Equilibrium Equation Effect of shear interaction of vertical element

17. Reissner Foundation Based on equations of a continuum. Assuming in-plane stresses throughout the foundation layer to be small: and that horizontal displacements are zero at the upper and lower surfaces of the foundation layer yields: where E,G = elastic soil constants H = soil layer thickness If c 1 = k & c 2 =G then the pasternak foundation is matched for a constant p 0

18. A consequence of the above assumption is that the shear stresses  zx and  xy are independent of z and thus are constant throughout the depth of the foundation for a given surface point (x,y); physically a nonrealistic result particularly for relatively thick foundation layers. However, in view of the fact that foundation models are introduced to study the response of the foundation surface to loads and not the stresses caused within the foundation, this deficiency may in general be of no serious consequence.

19. Deflection profiles for typical highway pavement under a point load

20. Figure 2a

23. Deflection profiles for typical highway pavement under a uniformly distributed load

24. Figure 2b

27. Shear Layer Plate in Bending Figure 7 p q q Reaction Distribution Bending Moment Figure 8

28. Plate R Foundation Region F t n Figure 9 D c G k q Plate in Bending Shear Layer Figure 10 (a)

29. q p Figure 10 (b)

30. q r a a Figure 11