ACP Inspector Certification Homework (oops, not homework) Solutions January 2007 ACP Inspector Certification Homework (oops, not homework) Solutions

PG 64-22 or PG 64-28 may be substituted for the grade of asphalt cement specified in the Contract for guardrail flares, mailbox turnouts, and road approaches. 2.Pavement legends may be left in place in areas where an overlay greater than 2” is being placed. True – 00745.11 (a) – Asphalt Cement, Additives, and Aggregate Treatment, Page 658. False – 00745.42 – Preparation of Underlying Surfaces, Page 669.

3.When leveling irregular surfaces, the compacted thickness of intermittent areas of 1,000 sf or less should not be more than 4 inches. 4.Compaction of thin pavements to a specified density is not required for leveling, patches or where the dense graded pavement is less than 2 inches. True – 00745.46 (c) – Hauling, Depositing and Placing, Page 670. True – 00745.49 (c) – Compaction QC (Thin Pavement), Page 674.

5.For a mix designated as 2.0” Level 2, ½” Dense Graded HMAC, the Level 2 applies to the smoothness requirements for the project. False – See Student Manual or 00745.02 Definitions page 653

6. Using the paving example presented in class, determine the Post Bid Mass calculation for a pavement thickness of 3.0”.

Post Bid V olume Calculations For the following calculations, use the cross section and mix design provided. Calculate the volume for one side of the section and then multiply it by two to determine the total HMAC volume. 2.6 52 Length (miles) = _________ Width (ft) = ____________________ Length (ft) = Length (miles) x 5280 ft/mile= ___________________ Surface Area (sq ft) = L x W = _____________________________ Lift Thickness of HMAC (in) = _________________________________ Volume (cu ft) = SA x T/12 = _____________________________ 13,728 13,728 x 52.0 = 713,856 3.0 713,856 x 3/12 = 178,464

Post Bid Mass Calculations 2.417 Gmm (Rice value) from JMF = _____________________ Min Compaction per Specification = _________________ Expected Compaction = Min Compaction + 1% = ______ Mix Unit Wt. (lbs/cu ft) = Gmm x 62.4 lbs/cu ft x Expected Compaction/100 = ______________________ Est. Mix Quantity (tons) = V x Mix Wt./2000 = _____________________________________________ 92% 93% 2.417 x 62.4 x 93/100 = 140.3 178,464 x 140.3/2000 = 12,519.25

7. Tack – Understanding Dilution 1:1 As Manufactured Add Water Diluted 1:1 1/3 Asphalt 2/3 Water Residual AC = 0.67/2 = 0.33 1/3 Water 1 Water + = 2/3 Asphalt Residual AC = 0.67

7. Tack – Understanding Dilution 1:1 – Example Using Volume As Manufactured Add Water Diluted 1:1 100 gal Asphalt 200 gal Water Residual AC = 100/300 = 0.33 50 gal Water + = 150 gal Water 100 gal Asphalt Residual AC = 100/150 = 0.67

7. Tack – Understanding Dilution 1:1 Contractor is applying diluted tack at 1:1. What should the application rate be to wind up with a residual asphalt application of 0.05 gal/sq yd? Application rate: Residual AC Needed/Residual AC in Tack Application rate: 0.05/0.67 = 0.07 gal/sy

8. Determine the Expected Yield (Mass) for the pavement section below: Actual Gmm = 2.559 Actual % Compaction = 92.0 % Plan Lift Thickness = 3.0 in Paved Width = See below Section Length = See below

AREA CALCULATION AREA 1 = (3435-2100) * 12 = 16,020 sq. ft. TOTAL AREA = 1335 + 920 = 16,940 sq. ft.

Expected Yield (Mass) Calculations Area (sq. ft.) = __________________ Lift Thickness for HMAC (in) = ___________________________________ Volume (cu ft) = L x W x T/12 = _________________________________ MAMD (lbs/cu ft) (Gmm x 62.4 lbs/pcf) = __________________________ %Compaction from CDT = ______________________________________ Mix Unit Wt. (lbs/cu ft) = MAMD x %C/100 = ______________________ Expected Yield (tons) = V x Mix Wt./2000 = _______________________ 16,940 3 16,940 x 3/12 = 4235.00 2.559 x 62.4 = 159.682 92.0 159.682 x 92.0/100 =146.907 311.08

I will use an average width= Total Area/Total Length 9. Complete the attached Material Delivery and Yield Check Sheet which is based on the pavement section shown in Problem 8. Does the Load Yield calculated meet the tolerance required when compared to the theoretical yield? I will use an average width= Total Area/Total Length 16,940 sqft / 1,335 ft = 12.69 ft average width

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