Determine the current PCR for this section. Then, assume a new roughness measurement is expected soon. If the PCR falls below 60, the section will be scheduled for rehabilitation. Determine the new IRI value that will trigger the rehabilitation.

Question

An agency uses the following formula to determine pavement condition rating:

PCR = 95.6 – 5.51 (5.0-ROUGH) – 1.59 LNALL – 0.221 AVGOUT – 0.0306 LONG –

0.531 TRAN

where:

ROUGH = roughness measured by present serviceability rating (PSR), scale of 0 to 5

LNALL = natural logarithm of alligator cracking, ft per mile

AVGOUT = outer wheel path rutting (all locations averaged), 0.01 inch

LONG = longitudinal cracking, ft per mile

TRAN = transverse cracking, number of cracks per mile

And, present serviceability rating is determined as:



A pavement section has the following distress data:

IRI = 105 in/mi

Alligator cracking = 110 ft/mi

Average rutting = 0.20 inch

Longitudinal cracks = 130 ft/mi

Transverse cracks = 12/mi



What will be an ideal response?

Answer 1

First, determine PSR, IRI = 105 in/mi

PSR = 2.92

Using the PCR equation as shown above,

PCR = 95.6 – 5.51(5.0 – 2.92) – 1.59Ln(110) – 0.221(20) – 0.0306(130) –

0.531(12)

= 95.6 – 11.46 – 7.47 – 4.42 – 3.98 – 6.37

= 61.9

Therefore, the current PCR for this section is 61.9.

Assume other distress values besides IRI remain unchanged,

PCR = 95.6 – 5.51 (5.0 – PSR) – 7.47 – 4.42 – 3.98 – 6.37

60.0 = 95.6 – 5.51 (5.0 – PSR) – 7.47 – 4.42 – 3.98 – 6.37

PSR = 2.575



IRI = 129.5

Therefore, the IRI of 129.5 in/mi will trigger the rehabilitation.