The AASHTO Flexible-Pavement Design Procedure
*Serviceability Concept
Present Serviceability Index (PSI) = pavement performance,
at any point in time.
Terminal Serviceability Index (TSI) = a point in time when
pavement can no longer perform in a serviceable manner.
AASHO Road Thickness Index:
Thickness Index = 0.44D1 + 0.14D2 + 0.11D3
Where:
D1 = surfacing thickness, inches (2 in. minimum)
D2 = base thickness, inches (3 in. minimum)
D3 = subbase thickness, inches
“thickness index is the same as the structural number”
Flexible-Pavement Design Equation
log10 W18 = Z R S O + 9.36[log10 ( SN + 1)] − 0.20 +
log10 [PSI /(2.7)]
+ 2.32 log10 M R − 8.07
5.19
0.40 + [1094 /(SN + 1) ]
Where:
W18 = is the 18-kip equivalent single-axle load.
ZR = reliability (z-statistic from the standard normal curve.
SO = is the overall standard deviation of traffic.
SN = is the structural number.
ΔPSI = is the loss in serviceability from when the pavement is
new until it reaches its TSI.
MR = is the soil resilient modulus of the subgrade in psi.
W18: Automobiles and truck traffic provide a wide range of
vehicle axle types and axle loads. The problem of handling
mixed traffic loading is solved with the adoption of a
standard 18-kip (80.1-kN)-equivalent single-axle load
(ESAL).
ZR: Defined as the probability that serviceability will be
maintained at adequate levels from a user’s point of view
throughout the design life of the facility.
SO: The overall standard deviation takes into account the
designers’ inability to accurately estimate the variation in
future 18-kip-equivalent axle loads, and the statistical error
in the equations resulting from variability in materials and
construction practices. Typical values are in order of 0.30 to
0.50.
MR: The soil resilient modulus is used to reflect the
engineering properties of the subgrade (the soil).
Measurement of the resilient modulus is not performed by all
transportation agencies; therefore, a relationship between MR
and the California Bearing Ratio, CBR, has been determined.
The CBR is the ratio of the load-bearing capacity of the soil to
the load-bearing capacity of a high-quality aggregate,
multiplied by 100.
MR = 1500 x CBR
SN: The structural number represents the overall structural requirement
needed to sustain the design’s traffic loadings.
SN = a1D1 + a2D2M2 + a3D3M3
Where:
a1, a2, and a3 = are structural-layer coefficients of the wearing surface,
base, and subbase layers, respectively.
D1, D2, and D3 = are the thicknesses of the wearing surface, base, and
subbase layers (in inches), respectively.
M2 and M3 = are drainage coefficients for the base and subbase,
respectively.
Example Problem No. 1
A pavement is to be designed to last 10 years. The initial PSI is 4.2 and
the TSI (the final PSI) is determined to be 2.5. The subgrade has a soil
resilient modulus of 15,000 psi (103.43 MPa). Reliability is 95% with an
overall standard deviation of 0.4. For design, the daily car, pickup truck,
and light van traffic is 30,000, and the daily truck traffic consists of 1000
passes of single-unit trucks with two single axles and 350 passes of
tractor semi-trailer trucks with single, tandem, and triple axles. The axle
weights are:
cars, pickups, light vans = two 2,000-lb single axles
single-unit truck = 8,000-lb steering, single axle
= 22,000-lb drive, single axle
Tractor semi-trailer truck = 10,000-lb steering, single axle
= 16,000-lb drive, tandem axle
= 44,000-lb trailer, triple axle
M2 and M3 are equal to 1.0 for the materials in the pavement
structure. Four inches of hot mix asphalt is to be used as the wearing
surface and 10 inches of crushed stone as the sub-base. Determine
the thickness required for the base if soil cement is the material to
be used.
Solution to Problem No. 1
Given:
𝑛 = 10 𝑦𝑒𝑎𝑟𝑠
𝑃𝑆𝐼 = 4.2
𝑇𝑆𝐼 = 2.5
∆𝑃𝑆𝐼 = 4.2 − 2.5 = 1.7
𝑀𝑅 = 15,000 𝑝𝑠𝑖
𝑅 = 95%; 𝑍𝑅 = −1.645
𝑆0 = 0.4
𝑀2 = 𝑀3 = 1.0
𝐷1 = 4 𝑖𝑛. ; 𝑎1 = 0.44
𝐷2 =? ; 𝑎2 = 0.20
𝐷3 = 10 𝑖𝑛. , 𝑎3 = 0.11
Determine the 18−kip equivalence of vehicle (Assume: SN = 4)
For cars, pick−ups, light vans:
2 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.0002 2 = 0.0004
For single−unit trucks:
8 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.041
22 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 2.090
𝑇𝑜𝑡𝑎𝑙 = 2.131
For tractor semi−trailer trucks:
10 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.102
16 − 𝑘𝑖𝑝 𝑡𝑎𝑛𝑑𝑒𝑚 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.057
44 − 𝑘𝑖𝑝 𝑡𝑟𝑖𝑝𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.769
𝑇𝑜𝑡𝑎𝑙 = 0.928
𝐷𝑎𝑖𝑙𝑦 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 0.0004 30,000 + 2.131 1,000 + 0.928 350
𝐷𝑎𝑖𝑙𝑦 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 2,467.8
𝑇𝑜𝑡𝑎𝑙 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 2,467.8 365 10
𝑇𝑜𝑡𝑎𝑙 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 9,007,470
log10 (9,007,470)
log10 [ 1.7/2.7]
= −1.645(0.4) + 9.36[log10 ( 𝑆𝑁 + 1)] − 0.20 +
0.40 + [1094/(𝑆𝑁 + 1)5.19 ]
+ 2.32 log10 (15,000) − 8.07
𝑆𝑁 = 3.94
𝑆𝑁 = 𝑎1 𝐷1 + 𝑎2 𝐷2 𝑀2 + 𝑎3 𝐷3 𝑀3
3.94 = 0.44 4 + 0.20 𝐷2 1.0 + 0.11 10 1.0
𝐷2 = 5.4 𝑖𝑛𝑐ℎ𝑒𝑠; 𝑈𝑠𝑒 𝐷2 = 5.5 𝑖𝑛𝑐ℎ𝑒𝑠
Activity
A flexible pavement is constructed with 4 in. (10.16 cm) of hot mix asphalt
wearing surface, 8 in. (20.32 cm) of emulsion/aggregate-bituminous base, and 8
in. (20.32 cm) of crushed stone subbase. The subgrade has a soil resilient modulus
of 10,000 psi (68.95 MPa), and M2 and M3 are equal to 1.0 for the materials in the
pavement structure. The overall standard deviation is 0.5, the initial PSI is 4.5,
and the TSI is 2.5. The daily traffic has 1080 20-kip (89.0-kN) single axles, 400 24kip (106.8-kN) single axles, and 680 40-kip (177.9-kN) tandem axles. How many
years would you estimate this pavement would last (i.e., how long before its PSI
drops below a TSI of 2.5) if you wanted to be 99% confident that your estimate
was not too high, and if you wanted to be 95% confident that your estimate was
not too high?
The AASHTO Rigid-Pavement Design Procedure
log10 PSI /(3.0)
log10 W18 = Z R S O + 7.35log10 ( D + 1) − 0.06 +
1 + 1.624x107 /( D + 1) 8.46




Sc' Cd D 0.75 − 1.132
+ (4.22 − 0.32TSI)log 10 
0.75
0.25
215
.
63
J
D
−
[
18
.
42
/(
E
/
k
)
]}
c







Where: W18 = is the 18-kip-equivalent single-axle load is the same
concept as that discussed in the flexible-pavement design procedure.
ZR = is the reliability (z-statistic from the standard normal curve.)
SO = is the overall standard deviation of the traffic.
D = is the slab thickness in inches.
TSI = is the pavement’s terminal serviceability index.
ΔPSI = is the loss in serviceability from when the pavement is new until
it reaches its TSI.
S’c = is the concrete modulus of rupture in psi.
Cd = is a drainage coefficient.
J = is the load transfer coefficient.
Ec = is the concrete elastic modulus in psi
k = is the modulus of subgrade reaction.
Sc’: The concrete modulus of rupture is a measure of the
tensile strength of the concrete and is determined by loading
a beam specimen, at the third points, to failure.
Cd: The drainage coefficient is slightly different from those
used in flexible-pavement design. In rigid-pavement design,
it accounts for the drainage characteristics of the subgrade. A
value of 1.0 for a drainage coefficient represents a material
with good drainage characteristics.
J: The load transfer coefficient is a factor that is used to
account for the ability of pavement to transfer a load from
one PCC slab to another across the slab joints.
Ec: The concrete modulus of elasticity is derived from the
stress-strain curve as taken in the elastic region.
k: The modulus of subgrade reaction depends upon several
different factors including the moisture content and density
of the soil.
Example Problem No. 3:
A rigid pavement is to be designed to provide a service life of 20
years and has an initial PSI of 4.4 and a TSI of 2.5. The modulus of
subgrade reaction is determined to be 300 pci (81.3 N/cm3). For
design, the daily car, pickup truck, and light van traffic is 20,000 and
the daily truck traffic consists of 200 passes of a single-unit truck with
a single and tandem axle, and 410 passes of a tractor semi-trailer
truck with a single, tandem, and triple axle. The weights are:
Cars, pickups, light vans = two 2-kip single axles
Single-unit truck = 10-kip steering, single axle
= 22-kip drive, tandem axle
Tractor semi-trailer truck = 12-kip steering, single axle
= 18-kip drive, tandem axle
= 50-kip trailer, triple axle
Reliability is 95%, overall standard deviation is 0.45, the concrete’s
modulus of elasticity is 4.5 million psi, the concrete’s modulus of
rupture is 900 psi, the load transfer coefficient is 3.2. The drainage
coefficient is 1.0. Determine the required slab thickness.
Given:
𝑛 = 20 𝑦𝑒𝑎𝑟𝑠
𝑃𝑆𝐼 = 4.4
𝑇𝑆𝐼 = 2.5
∆𝑃𝑆𝐼 = 4.4 − 2.5 = 1.9
𝑘 = 300 𝑝𝑐𝑖
𝑅 = 95%; 𝑍𝑅 = −1.645
𝑆0 = 0.45
𝐸𝑐 = 4.5 𝑥 106 𝑝𝑠𝑖
𝑆𝑐′ = 900 𝑝𝑠𝑖
𝐽 = 3.2
𝐶𝑑 = 1.0
𝐷 =?
Determine the 18−kip equivalence of vehicles
Assume: D = 10 inches
For cars, pick−ups, light vans:
2 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.0002 2 = 0.0004
For single−unit trucks:
10 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.081
22 − 𝑘𝑖𝑝 𝑡𝑎𝑛𝑑𝑒𝑚 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.305
𝑇𝑜𝑡𝑎𝑙 = 0.386
For tractor semi−trailer trucks:
12 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.175
18 − 𝑘𝑖𝑝 𝑡𝑎𝑛𝑑𝑒𝑚 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.132
50 − 𝑘𝑖𝑝 𝑡𝑟𝑖𝑝𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 3.02
𝑇𝑜𝑡𝑎𝑙 = 3.327
𝐷𝑎𝑖𝑙𝑦 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 0.0004 20,000 + 0.386 200 + 3.327 410
𝐷𝑎𝑖𝑙𝑦 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 1,449.27
𝑇𝑜𝑡𝑎𝑙 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 1,449.27 365 20
𝑇𝑜𝑡𝑎𝑙 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 10,579,671
log10 10,579,671 = −1.645 0.45 + 7.35 log10 𝐷 + 1 − 0.06 +
1.9
3.0
+ 4.22 − 0.32 2.5
7
1.624 𝑥 10
1+
𝐷 + 1 8.46
log10
𝐷 = 9.21 𝑖𝑛𝑐ℎ𝑒𝑠
900 1.0 𝐷 0.75 − 1.132
18.42
215.63 3.2 𝐷 0.75 −
4.5 𝑥 106
300
0.25
Recompute the 18−kip equivalence of vehicles
Assume: D = 9 inches
For cars, pick−ups, light vans:
2 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.0002 2 = 0.0004
For single−unit trucks:
10 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.082
22 − 𝑘𝑖𝑝 𝑡𝑎𝑛𝑑𝑒𝑚 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.308
𝑇𝑜𝑡𝑎𝑙 = 0.39
For tractor semi−trailer trucks:
12 − 𝑘𝑖𝑝 𝑠𝑖𝑛𝑔𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.176
18 − 𝑘𝑖𝑝 𝑡𝑎𝑛𝑑𝑒𝑚 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 0.133
50 − 𝑘𝑖𝑝 𝑡𝑟𝑖𝑝𝑙𝑒 𝑎𝑥𝑙𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 2.94
𝑇𝑜𝑡𝑎𝑙 = 3.249
𝐷𝑎𝑖𝑙𝑦 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 0.0004 20,000 + 0.39 200 + 3.249 410
𝐷𝑎𝑖𝑙𝑦 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 1,418.09
𝑇𝑜𝑡𝑎𝑙 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 1,418.09 365 20
𝑇𝑜𝑡𝑎𝑙 18 − 𝑘𝑖𝑝 𝐸𝑆𝐴𝐿 = 10,352,057
Recomputing for 𝐷:
log10 10,352,057 = −1.645 0.45 + 7.35 log10 𝐷 + 1 − 0.06 +
1.9
3.0
+ 4.22 − 0.32 2.5
1.624 𝑥 107
1+
𝐷 + 1 8.46
log10
𝐷 = 9.17 𝑖𝑛𝑐ℎ𝑒𝑠; 𝑈𝑠𝑒: 𝐷 = 9.5 𝑖𝑛𝑐ℎ𝑒𝑠
900 1.0 𝐷 0.75 − 1.132
18.42
215.63 3.2 𝐷 0.75 −
4.5 𝑥 106
300
0.25
Activity
You have been asked to design the pavement for an access
highway to a major truck terminal. The design daily truck
traffic consists of the following:
Daily Count
80 (single axle)
570 (tandem axle)
50 (tandem axle)
80 (triple axle)
Axle Load
22,500 lb
25,000 lb
39,000 lb
48,000 lb
The highway is to be designed with rigid pavement having a
modulus of rupture of 600 psi and a modulus of elasticity of 5 million
psi. The reliability is to be 95%, the overall standard deviation is 0.4,
the drainage coefficient is 0.9, ΔPSI is 1.7 (with a TSI of 2.5), and the
load transfer coefficient is 3.2. The modulus of subgrade reaction is
200 pci. If a 20-year design life is to be used, determine the required
slab thickness.