Stresses and Strains in Pavements prof. dr. ir. André A.A. Molenaar

1. Stresses and Strains in Pavementsprof. dr. ir. André A.A. Molenaar

2. Stresses in a pavement structure

3. Stresses in a half space

4. Example • Calculate stresses at depth z = 150 mm and z = 300 mm. • Load 50 kN, contact pressure 700 kPa. • Draw circles of Mohr. • Determine if failure occurs in dry season if c = 150 kPa and  = 45o. • Determine if failure occurs in wet season if c = 50 kPa and  = 25o.

5. Example dry season wet season

6. Traffic and geostatic stresses • At z= 150 mm geostatic vertical stress is only 2.7 kPa. • This value can be neglected when compared with traffic induced stresses. z  600 mm, depends on contact pressure

7. h1 E1 Em Em Effect of placing a stiffer layer on top of the subgrade

8. A A B B E1 Em Em Odemark’s equivalency theory heq Em heq = n h1 (E1 / Em)0.33

9. How good is theory?

10. Effects of placing a stiffer top layer

11. Two layer pavement system • A stiff top layer is placed on top of the subgrade to reduce the stresses in the subgrade and to avoid excessive permanent deformations. • The stiff top layer is subjected to bending and because of that, horizontal tensile stresses develop at the bottom of the layer. This might lead to cracking.

12. E1, h1 horizontal tensile stress E2 vertical compressive stress Design criteria for pavement with bound top layer and subgrade

13. Horizontal stresses in a two layer system

14. How do we get modulus values • Prof. Jenkins will tell you everything about how they can be estimated for asphalt mixtures. • AAAM will do that for soils, unbound materials and cement treated materials.

15. Vertical and shear stresses in a two layer system

16. Effect of friction on horizontal stresses We don’t know what the friction will be. Most of the time we assume full friction.

17. Effect of friction on vertical stresses

18. Effect of Poisson’s ratio on horizontal stresses

19. Effect of Poisson’s ratio on vertical stresses

20. Horizontal stress at bottom top layer of 2 layer system

21. Variation in stress because of variation in modular ratio and thickness -r / p h/a E1 / E2

22. Conclusion • You THINK you have an exact answer but reality is you have at best a good estimate. • You must always consider the fact that variations occur. • Fancy probabilistic analyses are not always needed. Just make a sensitivity analysis using realistic values.

23. Vertical stress at top of subgrade (2layer system)

24. Maximum deflection at top of two layer system

25. Because of variation in modulus and thickness, deflection can easily vary with 20% 0.4 0.37 0.33

26. Diameter 2a, contact pressure p E1 , h1 E2 , h2 E3 z1 r1 z2 Three layer systems A = a / h2 K1 = E1 / E2 K2 = E2 / E3 H = h1 / h2

27. Chart to predict rr1

28. Chart to predict zz2

29. Example • h1 = 100 mm, E1 = 6000 MPa • h2 = 300 mm, E2 = 300 MPa • E3 = 150 MPa • Loading radius = 150 mm, contact pressure = 700 kPa • K1 = 20, K2 = 2, A = 0.5, H = 0.33

30. Example 3.5 rr1 = 3.5 * 700 = 2450 kPa

31. Some effects on strain conditions

32. Some effects on strain conditions

33. Some effects on strain conditions

34. Conclusion • Magnitude of stresses depends on: - modular ratios, - geometrical ratios, - contact pressure.

35. Stresses due to complex contact stress distribu- tions

36. Multilayer systems • It will be obvious that working with charts for multilayer systems becomes a “p.i.t.n.”. • They even don’t exist for 4+ layer systems. • Using computer programs is a must. • Many are available. • There is a difference in quality. • BISAR is a very stable program from a mathematical point of view.

37. Fvertical = 50 kN Fhorizontal = 20 kN diameter loaded area = 300 mm E1 = 6000 MPa h1 = 250 mm 1 = 0.35 E2 = 100 MPa 1 = 0.35 Example showing effects of shear forces

40. Stress conditions compared

41. Conclusion on effect shear stresses • Taking into account shear stresses is very important in explaining damage that occurs at pavement surface. • Their effect rapidly fades away with depth. • For thickness design they are not so much of an issue.

42. Critical stress and strain locations Load 50 kN,  = 300 mm Asphalt  Unbound or Bound Base  Subbase  Subgrade 1. Tensile strain at pavement surface.  2. Tensile strain at bottom asphalt. 3. Compressive stresses in top unbound base. 4.Tensile strain at bottom bound base 5. Vertical compressive strain at top subbase. 6. Vertical compressive strain at top subgrade.

43. Commonly used mechanistic approach • Use linear elastic multi layer system, so characterise materials with E and . • Assume full adhesion between the layers. • Use static load(s). • Calculate stresses and strains at critical locations. • Use transfer functions (fatigue relations) to calculate pavement life. • Drawbacks: materials are NOT linear elastic. This effect becomes of influence for thin structures, high temperatures and long loading times.

44. Required transfer functions • Fatigue and crack resistance of asphalt concrete. • Resistance to permanent deformation of asphalt concrete. • Resistance to permanent deformation of unbound base materials. • Resistance to permanent deformation of subgrade materials. • Fatigue and crack resistance of bound base and subbase materials.

45. How do we get those transfer functions • Prof. Jenkins will discuss how to do this for asphalt mixtures. • AAAM will discuss this for unbound granular materials, soils and cement treated materials.

46. Design systems • A large number of design systems have been developed. • Analyses have already been made for you. • Results are translated into charts, catalogues, simplified systems etc. • Be aware of the assumptions that have been made to prepare input e.g. load configuration. • 80 kN axle with 525 kPa contact pressure is outdated!

47. Design chart ASCON