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

2. PG or PG 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 – (a) – Asphalt Cement, Additives, and Aggregate Treatment, Page 658.
False – – Preparation of Underlying Surfaces, Page 669.

3. 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 – (c) – Hauling, Depositing and Placing, Page 670.
True – (c) – Compaction QC (Thin Pavement), Page 674.

4. 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 Definitions page 653

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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

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

13. 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 =
2.559 x 62.4 =
92.0
x 92.0/100 =
311.08

14. 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
= ft average width

16. QUESTIONS/COMMENTS