Calculation Highway 2 PCI

 Example

The table below shows a pavement condition survey for sample unit 07, section JPB-KM2 of  Jalan Parit Botak. Determine the Pavement Condition Index for this sample unit. JALAN PARIT BOTAK (ASPHALT SURFACED ROAD) CONDITION SURVEY DATA SHEET BRANCH: TRAFFIC LABORATORY UTHM 

DATE: 25 AUGUST 2007 

SURVEYED BY: BDD 

SAMPLE UNIT: 07 

SECTION: JPB-KM2  01. Alligator Cracking (m2) 02. Bleeding (m 2) 03. Block Cracking (m2) 04. Bumps and Sags (m) 05. Corrugation (m 2)

06. Depression (m 2) 07. Edge Cracking (m) 08. Joint Reflection Cracking (m) 09. Lane/Shoulder Drop Off (m) 10. Longitudinal & Transverse Cracking (m)

DISTRESS SURVEY

03 M  01 L

SAMPLE AREA: 6 m x 100 m  11. Patching & Utility Cut Patching ((m 2) 12. Polished Aggregate (m 2) 13. Potholes (no.) 14. Railroad Crossing (m2) 15. Rutting (m 2)

QUANTITY

7.8

4.1  

2.5

4.2

TOTAL

16. Shoving (m2) 17. Slippage Cracking (m2) 18. Swell (m 2) 19. Weathering/ Ravelling (m2)

DENSITY %

DEDUCT VALUE

1.8  

6.3 

01 H  11 M  13 M  10 L 10 M 

3.1

0.4  

0.5

0.8  

0.8

1.2

0.3

0.7

1.1

0.4  

0.5  

Solution:

(1) (1)

Dete Determ rmin inee the the Dist Distre ress ss Dens Densit ity y and and Dedu Deduct ct Valu Valuee

DISTRESS SURVEY

03 M  01 L 01 H  11 M  13 M  10 L 10 M 

QUANTITY

7.8

4.1

2.5

4.2

TOTAL

11.9 1.8

8.5

6.3

6.3

3.1

0.4

3.5

0.5

0.8

1.3

0.8

1.2

0.3

0.7

1. 1.1

0. 0.4

0.5

2.8 2.2

DENSITY %

100*(11.9/600    ) = 1.98  100*(8.5/600)    = 1.42  100*(6.3/600)    = 1.05  100*(3.5/600)    = 0.58  100*(1.3/600)    = 0.22  100*(2.8/100)    = 2.80  100*(2.2/100)    = 2.20 

DEDUCT VALUE

6  13  32  7  50  2  6 

(2) (2)

Maxi Maximu mum m allo allowa wabl blee nu numb mber er of dedu deduct cts, s, m

Highest deduct value, HDV = 50 m = 1 + (9/98)(100 – HDV) = 1 + (9/98)(100 – 50) = 5.6 Deduct values in descending order: 50, 32, 13, 7, 6, 6, 2  Number of deduct values = 7 Since the maximum allowable number of deducts is 5.6, therefore only the first 5 and 0.6 of  the 6th deduct value are selected, i.e. 50, 32, 13, 7, 6, and 0.6*6 = 3.6.

(3)

Maximum imum corre rrected de dedu duct ct valu alue, CDV CDV

 Number of deduct values greater than 2, q = 6 Total deduct value = 50 + 32 + 13 + 7 + 6 + 3.6 = 111.6 From Figure B-45, CDV = 58 Reduce the smallest individual deduct value to 2 (q is now 5) and determine the CDV. Repeat until q reaches 1.

No. 1 2 3 4 5 6

Deduct Values

Total

q

CDV

50

32

13

7

6

3.6

111.6

6

58

50

32

13

7

6

2

11

5

58

50

32

13

7

2

2

106

4

61

50

32

13

2

2

2

101

3

65

50

32

2

2

2

2

90

2

63

50

2

2

2

2

2

60

1

60

7 8 9 10

Maximum CDV = 65

(4) (4)

Dete Determ rmin inee the the Pave Paveme ment nt Cond Condit itio ion n Inde Index x

PCI = 100 – CDV max = 100 – 65 = 35 Based on the rating for PCI value of 35, this section of pavement is in very poor condition.