3 rd Stage, 1 st semester, 2020-2021 - PowerPoint Presentation

1. 3rd Stage, 1st semester, 2020-2021Lecture 7. Two Layer System: Critical Tensile Strain, Single Wheel , dual wheels and dual-tandem wheels Lecturers Dr. Abeer K. JameelPavement Analysis University of Mustansiriyah, College of Engineering, Highway and Transportation Engineering Department 1

2. Critical Tensile Strain The tensile strains at the bottom of asphalt layer have been used as a design criterion to prevent fatigue cracking. Two types of principal strains could be considered:The overall principal strain based on all six components o f normal and shear stresses. the horizontal principal strain based on the horizontal normal and shear stresses only. The overall principal strain is slightly greater than the horizontal principal strain, so the use of overall principal strain is on the safe side.2

3. Critical Tensile Strain The critical tensile strain is the overall strain and can be determined frome= ………2.17e = is the critical tensile strain Fe = the strain factor, which can be determined from the charts The charts are developed according to the number of wheels  3

4. Critical Tensile Strain Single Wheel Figure 2.21 presents the strain factor for a two-layer system under a circular loaded area. 42-21Step 1:Enter h1/a (for example h1/a=1.5)Enter E1/E2 (for example =10)Use Eq 2.17 to find e

5. Critical Tensile Strain Single Wheel In most cases, the critical tensile strain occurs under the center of the loaded area, where the shear stress is zero 5

6. Critical Tensile Strain Single Wheel However, when both h1/a and E1/E2 are small, the critical tensile strain occurs at some distance from the center, as the predominant effect of the shear stress. Under such situations, the principal tensile strains at the radial distances 0, 0.5a, a, and 1.5a from the center were computedthe critical value was obtained and plotted in Figure 2.21 .62.21

7. Critical Tensile Strain Example 2.8 :Figure 2.22 shows a full-depth asphalt pavement 8 in.(203 mm) thick subjected to a single-wheel load of 9000 lb (40 kN) having contact pressure 67.7 psi (467 kPa). If the elastic modulus of the asphalt layer is 150,000 psi (1 .04 GPa) and that of the subgrade is 15,000 psi (104 MPa), determine the critical tensile strain in the asphalt layer.Yoder; E. J. and M. W. Witczak, “Principles of Pavement Design”, A Wiley- Interscience Publication, John Wiley & Sons Inc., U.S.A., 1975.7

8. Critical Tensile Strain Solution: Given: a = =6.5 in.(165 mm), h1/a = 8/6.5 = 1.23, E1/E2 = 150,000/15,000 = 10from Figure 2.21, Fe = 0.72 . From Eq. 2.17, the critical tensile straine= ………2.17e = 67.7 × 0 .72/150,000 = 3.25 × 10-4 82.21

9. 3rd Stage, 1st semester, 2020-2021Two Layer System: Critical Tensile Strain, Dual Wheels Lecturer Dr. Rana Amer , Dr. Abeer K. JameelPavement Analysis University of Mustansiriyah, College of Engineering, Highway and Transportation Engineering Department 9

10. Critical Tensile Strain: Dual Wheels To determine the strain factor for dual wheels, the following variables are considered:The contact radius a Layer thickness h1Elastic modules E1 and E2The dual spacing Sd Thus the strain factor depends on Sd/a , E1/E2 and h1/a, tYoder; E. J. and M. W. Witczak, “Principles of Pavement Design”, A Wiley- Interscience Publication, John Wiley & Sons Inc., U.S.A., 1975.10

11. Critical Tensile Strain : Dual Wheels However, this approach requires a series of charts, and the interpolation could be quite time-consuming . To avoid these difficulties, a unique method was developed that requires only one chart, as shown in Figure 2.23 Figure 2.23Yoder; E. J. and M. W. Witczak, “Principles of Pavement Design”, A Wiley- Interscience Publication, John Wiley & Sons Inc., U.S.A., 1975.11

12. Critical Tensile Strain : Dual Wheels The two-layer theory indicates that the strain factor for dual wheels depends on h1/a, Sd/a, and Ei/E2 . As long as the ratios hi/a and Sd/a remain the same, the strain factor will be the same, no matter how large or small the contact radius a may be .Consider a set of dual wheels with Sd = 24 in. (610 mm) and a = 3 in . (76 mm). The strain factors for various values of h1 and E1/E2 were calculated and the conversion factors were obtained and plotted as a set of curves on the upper part of Figure 2.23. Another set of curves based on the same Sd but with a = 8 in.(203 mm) is plotted at the bottom. Yoder; E. J. and M. W. Witczak, “Principles of Pavement Design”, A Wiley- Interscience Publication, John Wiley & Sons Inc., U.S.A., 1975.12

13. Critical Tensile Strain : Dual Wheels It can be seen that, for the same dual spacing, the larger the contact radius, the larger the conversion factor . However, the change in conversion factor due to the change in contact radius is not very large, so a straight-line interpolation should give a fairly accurate conversion factor for any other contact radii . Although Figure 2.23 is based on Sd =24 in. (610 mm), it can be applied to any given Sd by simply changing a and h1 in proportion to the change in Sd , so that the ratios h1/a and Sd/a remain the same. Yoder; E. J. and M. W. Witczak, “Principles of Pavement Design”, A Wiley- Interscience Publication, John Wiley & Sons Inc., U.S.A., 1975.13

14. Critical Tensile Strain : Dual Wheels Then, the method of determination of the critical tensile strain resulted from dual wheels is as follows:1. Replacing the dual wheels by a single wheel with the same contact radius a, so that Figure 2.21 can be used. 2. Finding the conversation factor C to adjust the results of single wheel to dual wheel using Figure 2.23 as follows:(C=the ratio between dual- and single-wheel strain factors)a. From the given Sd, h1 , and a, determine the modified radius a' and the modified thickness h`1:a'= ………(2.18a)h`1= h1 ……….(2.18b) 14

15. Critical Tensile Strain: Dual Wheels b. Using h` and E1/E2 to find conversion factors C1 and C2 from Figure 2.23 15Step 1, enter h1’ (for example h1’=10Step 2: Enter E1/E2 for both the upper and bottom charts (for example E1/E2=5)

16. Critical Tensile Strain: Dual Wheels b. Using h` and E1/E2 to find conversion factors C1 and C2 from Figure 2.23 c. Determine the conversion factor for a' by a straight-line interpolation between 3 and 8 in. (76 and 203 mm), or useC=C1 +0.2(a'—3)(C2 —C1 ) ………… (2.19)3. Multiplying the conversion factor C by the strain factor obtained from Figure 2.21 to yield the strain factor Fe for dual wheels .4. Find the critical tensile strain ee= ………2.17 16

17. Critical Tensile Strain: Dual Wheels Example 2.9:For the same pavement as in Example 2.8, if the 9000-lb (40-kN) load is applied over a set of dual tires with a center-to-center spacing of 11.5 in. (292 mm) and a contact pressure of 67 .7 psi (467kPa), as shown in Figure 2.24, determine the critical tensile strain in the asphalt layer 17

18. Critical Tensile Strain : Dual Wheels Solution: Identify the given information: Given a = Sd = 11.5 in. (292 mm), h1 = 8 in. (203 mm), E1 = 150000, E2 150001. Find Fe for single Wheel using Figure 2.21, Fe for single wheel = 0.47 2. Find Ca. Adjust a and h1 to a` and h1` respectively using Eq. 2.18a and 2.18b, a' = 24 × 4.6/11.5 = 9.6 in. (244 mm) h1` = 24 × 8/11.5 = 16.7 in. (424 mm)  18b. Use Figure 2.23 to find C1 and C2Enter El/E2 = 10 and h1` of 16.7 in. C1= 1.35 and C2 = 1.46 .

19. Critical Tensile Strain : Dual Wheels c. Interpolating C1 and C2 to find C using Eq. 2.19, C = 1.35 + 0.2 (9.6 –3)(1.46 – 1.35) =1.50. 3. Find the Fe for dual wheels Fe for dual wheels =1.50×0.47=0.7054. Find the critical tensile straine =67.7 ×0.705/150,000 = 3.18×10-419

20. 3rd Stage, 1st semester, 2020-2021Two Layer System: Critical Tensile Strain, Dual-Tandem Wheels Lecturer Dr. Rana Amer , Dr. Abeer K. JameelPavement Analysis University of Mustansiriyah, College of Engineering, Highway and Transportation Engineering Department 20

21. Critical Tensile Strain: Dual-Tandem Wheels To determine the strain factor for dual Tandem wheels, the following variables are considered:The contact radius a Layer thickness h1Elastic modules E1, E2The dual spacing Sd The tandem spacing St21

22. Critical Tensile Strain: Dual-Tandem Wheels Charts similar to Figure 2.23 with dual spacing Sd of 24 in. (610 mm) and tandem spacing St of 24 in. (610 mm), 48 in. (1220 mm), and 72 in.(1830 mm) were developed for determining the conversion factor due to dual-tandem wheels, as shown in Figures 2.25, 2.26, and 2.27. 22Figures 2.25 Figure 2.26 Figure 2.27

23. Critical Tensile Strain: Dual Tandem Wheels The following procedure are used to find the critical tensile strain of dual-tandem wheel1. Modify the tandem spacing , St`=St (24/Sd)2. Modify a and h1 to find a` and h1` using Eq.2.18a and 2.18b3.Find the conversion factors C1 and C2 then interpolate themIf St=24 use Figure 2.25, then interpolate C1 and C2 using Eq 2.19If St=48 use Figure 2.26, then interpolate C1 and C2 using Eq 2.19If St=72 use Figure 2.27, then interpolate C1 and C2 using Eq 2.19If 24<St<48 use Figures 2.25 and 2.26, the interpolate the obtained C as follows: - Find C1 and C2 from Figure 2.25 then interpolate them using Eq 2.19 - Find C1 and C2 from Figure 2.26 then interpolate them using Eq 2.19 interpolate the both above C using Eq 2.20C=C(of smaller St)+[C(of bigger St)-C(of smaller St)][St`-St smaller]/[St bigger-St smaller] ……………..(2.20)23

24. Critical Tensile Strain: Dual Tandem Wheels If 48<St<72 use Figures 2.26 and 2.27 then interpolate the obtained C as follows: - Find C1 and C2 from Figure 2.26, then interpolate them using Eq 2.19 - Find C1 and C2 from Figure 2.27 then interpolate them using Eq 2.19 - Interpolate the both above C using Eq 2.20If 72<St<120 use Figures 2.27 and 2.23 then interpolate the obtained C as follows: - Find C1 and C2 from Figure 2.27, then interpolate them using Eq 2.19 - Find C1 and C2 from Figure 2.23, then interpolate them using Eq 2.19 - Interpolate the both above C using Eq 2.20If St >120 then the wheels configurations is considered Dual wheel , use Figure 2.23 then interpolate C1 and C2 using Eq 2.194. Find the strain factor Fe using Figure 2.215. Multiply the result of step 4 by the C resulted from step 36. Find the critical tensile strain e using equation 2.17 24

25. Critical Tensile Strain: Dual Tandem Wheels Example 2.10:Same as example 2.9, except that an identical set of duals is added to form dual-tandem wheels having the tandem spacing 49 in.(1.25 m), as shown in Figure 2.28 . Figure 2.28 25

26. Critical Tensile Strain: Dual Tandem Wheels Solution: Identify the given InformationGiven a = Sd = 11.5 in , St =49 in, h1=8 in, E1=150000, E2=15000 1. Modify the StSt`=49×24/11.5 = 102.3 in (2.60 m) 2. Modify a and h1 a' = 24 × 4.6/11.5 = 9.6 in. (244 mm) h1` = 24 × 8/11.5 = 16.7 in. (424 mm)  26

27. Critical Tensile Strain: Dual Tandem Wheels Solution: (continue) 3.As St` = 102 , 72<St<120, Then Use Figures 2.27 and 2.23 to find C1 and C2 Enter St = 72 in, a=3 in., h1 =16.7 in Figure 2.27 …..> C1=1.23St=72 in, a=8in , h1=16.7 in Figure 2.27…..> C2=1.3By interpolation of C1 and C2 using Eq 2.19 C = 1.23 + 0.2 (9.6 - 3) (1.30 -1.23) = 1.3227Figure 2.27

28. Critical Tensile Strain: Dual Tandem Wheels Solution (continue)Enter St=120, a=3in., h1= 16.7 in Figure 2.23…….> C1=1.35St=120, a=8in, h1=16.7in Figure 2.23……> C2=1.46Eq 2.19…>C=1.35+0.2(9.6-3)(1.46-1.35)=1.5 Interpolate the C of 1.32 for St= 72 in and 1.50 for St = 120 in using Eq 2.20C=C(of St=72)+[C(of St=120)-C(of St=72)][St`-St smaller]/[St bigger-St smaller]C = 1.32 +(1.50 - 1.32)(102.3-72)/(120-72) =1.43 28Figure 2.23

29. Critical Tensile Strain: Dual Tandem Wheels Solution (continue)4. Find the strain factor due to single wheel Use Figure 2.21, Fe= 0.47 5. Find the strain factor due to dual-tandem wheels through multiplying the Fe of 0.47 by CAdjust Fe= 1.43 ×0.47 = 0.672 . 6. Find the critical tensile strain using Equation 2.17e=67.7×0.672/150,000=3.03×10-4292-21