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MTO EMPIRICAL PAVEMENT DESIGN The following presentation contains references to Tables 6.02 and 6.03, and “MTO Empirical Design Examples”, all of which.

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Presentation on theme: "MTO EMPIRICAL PAVEMENT DESIGN The following presentation contains references to Tables 6.02 and 6.03, and “MTO Empirical Design Examples”, all of which."— Presentation transcript:

1 MTO EMPIRICAL PAVEMENT DESIGN The following presentation contains references to Tables 6.02 and 6.03, and “MTO Empirical Design Examples”, 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 tabular nature.

2 ANY QUESTIONS?

3 There are? OK, well here’s a little more… You may have noticed that there were two design charts: a)T he first is for “Kings” Highways and Expressways b)T he second is for secondary highways SSSSo, a) has 5 levels (rows) of traffic (AADT) which are different than the 6 for b) TTTThe 6 columns for different subgrade types are the same for both EEEEach cell has a conventional pavement design AADT>200 vpd Subgrades range from: STRONGEST to weakest AADT<3000 vpd Same old subgrades

4 Example 1 A 2-lane county road is expected to have an initial AADT of 1750 vpd and is to built over a silty sand subgrade with 30% Passing the No. 200 sieve. a)D etermine a conventional flexible pavement design. b)A couple of subsections require the use of a deep strength design using cement treated base and no subbase. Give the appropriate layer thicknesses.

5 for AADT of 1750 use second row for silty sand with 30% passing no. 200 use second column And the winning design is: HM: 50 mm B: 150 mm SB: 250 mm GBE: 415 mm For a county road with an AADT of 1750 vpd, try Table 6.03b

6 Layer Layer Thickness (mm) Hot Mix 50 Base150 Subbase250 Equivalency Factor Granular Base Equivalency (mm) Granular Base Equivalency (mm) Total Granular Base Equivalency (mm): Lets calculate the GBE for this design:  T T T The tabulated GBE was 415.  T T T The tabulated GBE’s have been rounded to the nearest 5 mm.  N N N Now on to part b)

7 Example 1 b) Part b) calls for a deep strength design This would require that the base and subbase be replaced by a cement treated base layer (CTB) The strength of the base and subbase is = mm of new Granular A The Equivalency Factor for CTB is… 1.80

8 If the required thickness of CTB is TCTB, then the GBE of the CTB is 1.80TCTB = mm Solving, TCTB = ≈ So the deep strength design would be: Layer Layer Thickness (mm) Hot Mix 50 CTB180 Example 1 b) mm Equivalency Factor Granular Base Equivalency (mm) Total Granular Base Equivalency (mm): 424

9 Example 2 A 12 km stretch of Highway 99 has 75 mm of hot mix over 180 mm of granular base over 300 mm of granular subbase. If the AADT has grown to 2500 vpd, the sandy subgrade has 22% passing the No. 200 sieve and the hot mix has lost 65% of its strength, what minimum thickness of hot mix overlay will restore the pavement to its required strength?

10 for AADT of 2500 use third row for silty sand with 22% passing no. 200 use second column The required design is: HM: 90 mm B: 150 mm SB: 300 mm GBE: 530 mm For a highway with an AADT of 2500 vpd, try Table 6.03a

11 Layer Layer Thickness (mm) Hot Mix 90 Base150 Subbase300 Equivalency Factor Granular Base Equivalency (mm) Granular Base Equivalency (mm) Total Granular Base Equivalency (mm): 531 The recommended design:Layer Layer Thickness (mm) Hot Mix 75 Base180 Subbase300 Equivalency Factor Granular Base Equivalency (mm) Granular Base Equivalency (mm) Existing GBE (mm): The existing design: EF of HM = 2-(2-1)*0.65 = % of HM strength lost

12 Example 2 Extra Strength Required = 531 – = mm of new Granular A Since overlay will be with new Hot Mix, only half of this thickness of new hot mix will be needed. Overlay Thickness Required = /2 = ≈ mm of new Hot Mix


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