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MTO OVERLAY DESIGN The following presentation contains references to Figures 6.02 and 6.03, Tables 6.01 and 6.05, “Overlay Design Model” and “Overlay Example Problem”, all of which are posted under subsection 2.6 of the course notes on the instructor’s website. Viewer discretion is advised as some scenes contain material of a graphic nature. The basic problem is to test the pavement’s strength and compare it to the strength required to carry an estimated number of standard axle load repetitions Instead of coring surveys and guesstimates of strength, its more economical to measure deflections under standard loads as with… THE BENKELMAN BEAM or THE DYNAFLECT Deflection Tests The Dynaflect is more efficient for mass inventory (many tests) but the original strengthening technology is based on Benkelman Beam Rebounds (BBR’s) The Dynflect Sensor 1, (DYN) readings must first be converted into BBR’s in Ontario this conversion is made using the following equation: BB 0.0024 0.019DYN 0.0095DYN 2 Statistical Criteria 1 i n m ean X X i n i 1 X 2sx 0+210 0+180 Chainage 0+150 0+060 X sx 0+030 0+000 X 0+120 X sx X 2sx 0+090 Deflection Plotting deflections along the road alignment gives a deflection profile: i n Standard Deviation = s x X i 1 2 i nX n 1 2 Design Deflection, DD Assuming the variation in deflections is random and therefore normally distributed, using the mean plus two standard deviations will ensure that the maximum deflection is taken into consideration 95% of the time Engineers always like to consider the worst case scenario in design The pavement is at its weakest in the early spring, when the frost has just come out of the ground... TABLE 6.05 Design Deflection, DD (Continued) SEASONAL VARIATION OF STRENGTH RATIO (After: Chong and Stott, Dept. of T ransportation and Communication, Ontario, Report No. IR42, October, 1971) of Type of Spring/Fall The testsType are usually taken some time later, Soil Construction Ratio Remarks quite often inClay theStandard autumn when the pavement Heavy gravel or crushed When estimating the maximum Silt rock base with maximum 3.5" spring deflection for heavy clay structure has stabilized asphalt concrete surface 2.0 of high plasticity in standard An adjustment is therefore made to of 2.0 reflect to 2.5 should be used Light to For all standard construction the pavement’s strength in the early spring 1.8 Medium Clay including heavy clay and silt construction, a spring/fall ratio To Summarize: Design Deflection isgenerally occurs during the spring-thaw period (April). Rockthe mean plus two1.5standard calculated as Deep strength construction Peak deflection generally All Types 1.4 deviations of thewithBenkelman Beam Rebounds thin granular base layer occurs in June or July rather Full depth construction - more than in April. Peak-to-peak on a pavement to spring by than 7" of asphalt concrete curvature, unlike standard All Typesadjusted 1.4 conditions directly on subsoil construction, is gradual. factoring by a spring/fall Soil cement-stabilized baseratio All Types 1.4 Sand and Gravel 1.6 with asphalt concrete surface above, the peak deflection Maximum DD = 0.067” Tolerable Deflection, MTD DTN DTN==48.9 125 If the MTD was how much traffic Say, The Studies for MTD example, in for Ontario a0.067” DTN that (and of itthen is 125 elsewhere) desired is 0.05287”. to know have could it carry? whether correlated or the not design a pavement deflection with aof DD a pavement of 0.067” withwill itsbe ability able to carry another traffic loads 10 years as Since DD > MTD, this pavement can’t handle It could carry a DTN of 48.9. of traffic characterized as reflected by bya the DTN DTN. of 125. this much traffic. For the portion of this relationship for DTN > 10, the following equation can be used instead of the graph: 0.1787907 MTD DTN 0.2523450 0.10" Required Strengthening Through extensive research, the Asphalt Institute and MTO have developed design curves relating Design Deflection to Strengthening Requirements The EachStrengthening curve is for a different is in termsMTD of inches value of which(new GBE range Granular from 0.02” A) up to 0.10” Interpolation What happens when the MTD is between the curves? Example: DD = 0.045” & MTD = 0.033” Proportion 0.033"0.030" 0.3 (0.040"0.030") T0.033" 6.913"(6.913"2.026") (0.3) 5.447" T0.033” = 5.447” T0.03” = 6.913” T0.04” = 2.026” DD = 0.045” Overlay Thickness Equations Each curve on these graphs has an equation with two coefficients, a and b in the a( DD MTD) form… T WARNING Only the MTD values tabulated can be used in this equation!!!!!! ( DD MTD) b MTD a b MTD a b 0.020 33.880888 0.0250955 0.070 14.948930 0.0892606 0.030 26.179140 0.0418036 0.075 11.960133 0.0723573 0.040 19.330437 0.0427002 0.080 11.927564 0.0832181 0.050 14.308789 0.0414468 0.090 9.600747 0.0728982 0.060 14.342670 0.0740099 0.100 12.565574 0.1565992 OVERLAY EXAMPLE PROBLEM The following Dynaflect Sensor 1 readings were recorded along a 4-lane section of highway in October. If the present AADT is 16000 vpd with 20% heavy vehicles and an annual growth rate of 2.0%/year, design an asphalt concrete overlay for a design period of 12 years if the subgrade in the area is a light clay. Then Finally, First calculate Next step calculate multiply is tothe mean convert by spring/fall mean andDynaflect plus standard two ratio standard Sensor to find1 TABLE 6.05 STA Sensor 1 BBR deviations: Design readings deviation: Deflection: to Benkelman Beam rebounds: SEASONAL VARIATION OF STRENGTH RATIO 0+030 0.013964 0.65 (After: Chong and Stott, Dept. of Transportation and Communication, Ontario, Report No. IR42, October, 1971) Type of 0.60 Spring/Fall0.012420 Soil Construction Ratio Remarks Heavy Clay Standard gravel or crushed When estimating the maximum 0.013032 0+090 0.62 Silt rock base with maximum 3.5" spring deflection for heavy clay 0+120asphalt concrete surface 0.57 2.0 0.011517 of high plasticity in standard Type of 0+060 0.013341 of 2.0 to 2.5 should be used 0+150 0.63 Light to 0+180 Medium Clay Sand and 0+210 Gravel 0.61 1.8 0.69 1.6 above, the peak deflection 0.015233 0.82 1.5 0.019568 All Types with thin granular base layer 0+270 0.75 1.4 0.017194 occurs in June or July rather than in April. Peak-to-peak curvature, unlike standard construction, is gradual. Rock 0+240 Deep strength construction Full depth construction - more than 7" of asphalt concrete directly on subsoil Soil cement-stabilized base with asphalt concrete surface construction, a spring/fall ratio 0.012725 including heavy clay and silt For all standard construction generally occurs during the spring-thaw period (April). Peak deflection generally All Types 0+300 0.68 1.4 0.014913 All Types 0+330 0.70 1.4 0.015555 in 0.0024 0.019DYN 0.0095DYN 2 1 BB 1 2 0.159462 XBB X 0.0144965 i 0.019 0.0024 0.65 0.00950. 65 0.013964 60 60 012420 n i1 11 in sx X i1 i 2 nX n 1 2 0.00236657- 110.0144965 11 - 1 2 0.0023439 0.019184 x 2s 0.0144965 20.0023439 Therefore, DD = 0.034531” x DD 0.019184 1.8 0.034531 For a light clay, S/F = 1.8 according to Table 6.05. OVERLAY EXAMPLE (Continued) TABLE 6.01 MTO LANE DISTRIBUTION FACTORS TFi = 0.90 TFf = 0.96 Truck Factor, TF Now to find the DTN from the traffic data given: 12 20,292 16,000 Highway Type 2.0 /lane 5073 AADTfif/lane 1 4000 AADT 16,000 20291.8687 2 lanes 44 100 Since AADTi =16,000, LDFi = 0.75 Since AADTf =20,292, LDFf = 0.75 Ti = 0.20; Tf = 0.20 4 lanes (After: "Pavement Design and Rehabilitation Manual", 1990) AADT (vpd) Lane Distribution Factor, LDF All 1.00 < 5000 0.85 5000 - 15000 0.80 15000 - 25000 0.75 > 25000 0.70 40% H 1.5 30% H ESALi/day 16,000 0.5 0.20 0.75 0.90 1080 < 15000 20% H 10% H 0.60 ESALf/day 20,291.868 7 0.5 0.20 0.7515000 0.96 1461.0145 0.96 - 25000 0.55 1.0 0.90 6 lanes 1080 1461.0145 > 40000 0.45 DTN 1270.5073 25000 - 40000 Then find the DTN: and finally the MTD… 0.50 2 0.5 0.1787907 MTD 0.02945" 0.2523450 1270.5073 0.0 2000 3000 4000 5000 6000 7000 Traffic/Lane (= AADT/4 for 4-Lane Highways) FIGURE 6.03b - M TO TRUCK FACTORS: 4-LANE HIGHWAYS 8000 OVERLAY EXAMPLE (Continued) Since MTD is between the 0.02” and 0.03” curves, evaluate T0.02” and T0.03” and interpolate For T0.02”: For T0.03”: a = 33.880888 b = 0.0250955 a = 26.179140 b = 0.0418036 T0.02" 33.880888(0.034531"0.02") 12.424" (0.034531"0.02") 0.0250955 T0.03" 26.179140(0.034531"0.03") 2.560" (0.034531"0.03") 0.0418036 The proportion that 0.02945” 0.02945"0.020" 0.945 divides the interval from 0.02” Proportion (0.030"0.020") to 0.03”: The interpolated thickness of extra GBE required: MTD a b MTD a b T0.02945" 12.424" (12.424" 2.560" ) (0.945) 33.880888 0.0250955 14.948930 3.103" 0.0892606 0.020 0.070 And finally,the overlay thickness in millimetres of new AC: 0.030 26.179140 0.0418036 0.075 11.960133 0.0723573 3.103" 25.4 0.080 mm/" 11.927564 0.0832181 0.0427002 0.040 T(mm AC)19.330437 39.4 40mm 2.0 14.308789 0.0414468 9.600747 0.0728982 0.050 0.090 0.060 14.342670 0.0740099 0.100 12.565574 0.1565992