2 Traffic Loading
-
Upload
kwasi-agyeman-boakye -
Category
Documents
-
view
227 -
download
0
Transcript of 2 Traffic Loading
-
8/13/2019 2 Traffic Loading
1/39
KAAF UNIVERSITY COLLEGE
Civil Engineering DepartmentCollege of Engineering
__________________________________
Highway Engineering IICIV 467
Lecture 2_ Traffic Loading
Kwasi AgyemanBoakye ( [email protected])
1
-
8/13/2019 2 Traffic Loading
2/39
KAAF UNIVERSITY COLLEGE
Overview of the Course ( CIV 467)
Time Period
* 7:00am10:00am
Days
* Saturdays , 21 Sept, 28 Sept , 05 Oct, 12 Oct , 19 Oct , 26 Oct , 02 Nov, 09 Nov, 16 Nov, 23Nov, 30 Nov, Dec 07.
Mode of Assessment
* Coursework and 2 Test - 30%; Exams - 70%
Recommended Reading
oHighway; The location, Design Construction and the Maintenance of Road Pavement, C. A. OFlarherty, Fourth Edition (2002).
oTraffic and Highway Engineering, Garber N.J and Hoel A.L, Bill Stenquist (2002).
oPavement Design Manual, Ghana, First Edition (1998).
oStandard Specification for Roads and Bridges, Ghana , 2007
2
-
8/13/2019 2 Traffic Loading
3/39
KAAF UNIVERSITY COLLEGE
Vehicle Categories
In evaluating the traffic condition of a facility and for assessing the geometric design
requirements, it is necessary to consider all types of vehicles using (or expected to use) the
facility. In Ghana the following categories of vehicles are used for design; Axle Load
Configuration.pdf
But for pavement design, only the vehicles that carry significantly heavy loads are important
commercial vehicles. Thus the from the Light Trucks to the Extra Large Truck & Others category.
3
Cars Heav
y Truck
Taxis Semi-trailer (Light)
Pick Up/Vans/4 WD Veh Semi-trailer (Heavy)
Small Bus Truck Trailer
Medium Bus/Mummy Wagon Extra Large Trucks & Others
Large Bus
Light Truck
Medium Truck
http://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/Axle%20Load%20Configuration.pdf -
8/13/2019 2 Traffic Loading
4/39
KAAF UNIVERSITY COLLEGE
Traffic Loads
Vehicle loads can be categorized into various categories such as Gross Load, Axle Load, Wheel
Load for different purposes. Also the axles have various categorisations.
4
200kN
80kN 120kN
-
8/13/2019 2 Traffic Loading
5/39
KAAF UNIVERSITY COLLEGE
Load Transfer through Wheels
Load transfer is done through the wheels of vehicles. These wheels are pneumatic tyres inflated
with air. Three parameters are considered in application of loads through wheels;
Total wheel load
Shape of contact area
Distribution of pressure over the contact area
5
-
8/13/2019 2 Traffic Loading
6/39
KAAF UNIVERSITY COLLEGE
Load Transfer through Wheels
Contract Pressure and Tyre Pressure
Rigid Factor = Contact Pressure / Tyre Pressure
RF=1, when average tyre pressure = 0.7 MPa
RF>1, when average tyre pressure < 0.7 MPa
RF 0.7 Mpa
Tyre Pressure = Inflation
Contact Pressure = Wheel Load/(area of imprint)
6
For high inflation pressuresWall of tyre
in tension, Contact pressure is less than
tyre pressure
For low inflation pressures
Wall of tyre incompression, Contact pressure is greater
than tyre pressure.
-
8/13/2019 2 Traffic Loading
7/39
KAAF UNIVERSITY COLLEGE
Load Transfer through Wheels
Stresses
Vertical Stresses
Unidirectional surface shear stresses (breaking and acceleration)
Centripetal shear stresses
The pressure distribution (vertical, centripetal or unidirectional) is not normally uniform. Normallyonly uniformly distributed vertical surface stress equal to tyre pressure is considered for
analysis.
Load Contact Area
Shape of contact area depends on;
-Inflation
- Tyre age
- Pavement Surface
7
-
8/13/2019 2 Traffic Loading
8/39
KAAF UNIVERSITY COLLEGE
Contact Area Shapes
Different contact shape areas are considered for analyses, such as;
Circular
Rectangular
Rectangular with semi-circular ends
More exact shapes for rigorous analysis
Circular Contact Area
Area (A) = Wheel Load (P)/ contact pressure(p)
For circular contact area, radius of contact area is obtained as;
a = (1/p)(P/p)0.5
Example: For 20kN load (P) transmitted through a pressure (p) of 0.7MPa, the radius of contact
area is given by
a2= 20,000/0.7
a = 95.365mm
8
-
8/13/2019 2 Traffic Loading
9/39
KAAF UNIVERSITY COLLEGE
Contact Area Shapes
Rectangle with Semi Circular Ends
Contact Area (A) = 0.522L2
In An Example;
For a P = 20kN and p= 0.7MPa determinethe length of contact area if the contact area
is assumed to be a rectangle with a semi
Circular ends.
Ans L= 233.8mm
Note: Where the rectangle with semi circular ends is converted to an equivalent rectangle the
dimensions take this form;
9
0.8712L
0.6LA= 0.522L2
-
8/13/2019 2 Traffic Loading
10/39
KAAF UNIVERSITY COLLEGE
Design Loading Considerations
Traffic loads applied over several years ( design life period) traffic volume increases each
year.
Vehicles on a given road carry different loads
Vehicles on different facilities carry different loads
The wheel loads are carried over different portions of the pavement and not at a single location.
The manner of the transmission of the load to the pavement depends on the speed of the
vehicle.
Pavement is designed to carry traffic load over a specific period (design life)
Thus it is essential to have a good estimate of the total number of vehicles expected to use the
facility during the design life period
10
-
8/13/2019 2 Traffic Loading
11/39
KAAF UNIVERSITY COLLEGE
Traffic Forecast
It will also help in design if the traffic volumes during different periods (even on a yearly basis) of
the design life can be estimated. These estimates can be done from the traffic volumes
prevailing in a base year and by selecting appropriate growth factors and projection techniques.
Projection of cumulative commercial traffic over design life is given by;
N = 365 x A [ (1+r)n1]
r
Where A= Initial design traffic in the year of completed construction (com.veh/day)
r = annual growth rate of commercial vehicles expressed as a fraction
n= Design period (years)
Note: This cumulative volume will be used to determine the Cumulative Equivalent Standard
Axle (CESA).
11
-
8/13/2019 2 Traffic Loading
12/39
KAAF UNIVERSITY COLLEGE
Traffic Forecast
The traffic in the year of completion of construction is estimated using the expression;
A = P (1+r)x
Where P = Number of commercial vehicles/day as per last count
x = No. of years between the last count and the year of completion of construction
The cumulative traffic (commercial vehicles) for the design period N, would have to be adjusted
for the following to get the design traffic;
Directional distribution of traffic
Lateral placement characteristics of wheels on pavement
Load spectrum
12
-
8/13/2019 2 Traffic Loading
13/39
KAAF UNIVERSITY COLLEGE
Traffic Forecast
Try this
Determine the cumulative volume of
commercial vehicles for the design of the
pavement for construction of a new bypass with
the following data:
1. Two lane carriage way2. Initial traffic in the year of completion of
construction = 400 CVPD (sum of both
directions)
3. Traffic growth rate = 7.5 %
4. Design life = 17 years
5. Construction is expected to occur over a
period of 4 years.
Ans. 4.7x106cvpd.
Example
The average two-way traffic per day on an
existing 2 lane highway counted in 2010
was 4000 commercial vehicles.
Determine the cumulative volume ofcommercial vehicles over the design life of
the road if;
The annual growth rate of commercial
vehicles , r=7%
N= design period is 15 years
And construction of road is expected to be
completed in 2013Solution
Determine A;
A = 4000(1+0.07)2= 4580cvpd
N= 365x 4580x[ (1+0.07)151] =42x106cvpd
0.07
13
-
8/13/2019 2 Traffic Loading
14/39
KAAF UNIVERSITY COLLEGE
Lateral distribution of wheel loads
All the commercial vehicles do not take the same lateral position on highway. Depending on the
type of facility (two-way, multi lane), number of lanes, etc the paths that the wheels of
commercial vehicles tread differ. As a result all the wheels of all the commercial vehicles utilizing
the pavement during the design period do not stress the same point on the pavement. Each part
of the pavement get different repetitions of loads.
14
-
8/13/2019 2 Traffic Loading
15/39
KAAF UNIVERSITY COLLEGE
Lateral distribution of wheel loadsSome recommended load distributions
Single lane roadsDesign is based on total number of commercial vehicles in both direction
Two-lane Single carriageway road
Design is based on 75% of the total number of commercial vehicles in both direction
Four lane Single Carriageway Road
Design is based on 40% of the total number of commercial vehicles in both directions.
Dual Carriage Roads
Design of dual two lane carriageway roads should be based on 75% of the number of
commercial vehicles in each direction. For dual three lane carriageway and dual four lane
carriageway the distribution factor will be 60% and 45% respectively.
15
-
8/13/2019 2 Traffic Loading
16/39
KAAF UNIVERSITY COLLEGE
Traffic Load Considerations in Design
Pavements of highways and airports carry different types of vehicles. And
also vehicles carry different magnitude of loads and occur repeatedly.
The question is; which vehicle and how many repetitions are considered in
design?
Load Considerations in Design
Three different approaches occur, namely;
Fixed Traffic
Fixed Vehicle
Variable Traffic and Vehicle
16
-
8/13/2019 2 Traffic Loading
17/39
KAAF UNIVERSITY COLLEGE
Traffic Load Considerations in Design
In the Fixed Traffic Design Approach heaviest anticipated vehicle is the main concern for
design. The number of repetitions is not considered. Pavements are designed for a single wheel
load. Multiple wheel loads are converted to an Equivalent Single Wheel Load (ESWL) Eg.
Airport pavements and highways that would carry very heavy loads. Not commonly used.
For Fixed Vehicle Design Approach the design is governed by the number of repetitions of a
standard vehicle or axle. 80KN single axle is considered as the standard axle. Axles that are not
either single or not equivalent to 80kN are converted into equivalent standard axle loads using
Equivalent Standard Axle Load Factor. Multiplying the repetitions of a given axle load by the
EALF gives the equivalent number of 80kN axle load repetitions.
Sum of the equivalent repetitions obtained for all the axle loads anticipated (during the design
period) is used as a design parameter. And most design approaches use this method.
In the Variable Vehicle and Traffic Approach variations in loads and repetitions of each
individual load are considered as important for design. There is no need to deal with traffic in
terms of ESWL or ESAL. It is normally used for procedures adopting a cumulative damage
approach.
17
-
8/13/2019 2 Traffic Loading
18/39
KAAF UNIVERSITY COLLEGE
Equivalent Standard Wheel Loads
(ESWL)
This is defined as the load on a single tyre that will cause an equal magnitude of pre selected
parameter (stress, strain, deflection or distress) at a given location within a specific pavement
system to that resulting from a multiple wheel load at the same location within the pavement
structure.
Here the equivalency is in terms of a selected parameter ( for a selected pavement and a
selected location). To determine ESWL the following parameters must be considered;
Equal vertical stress
Equal vertical deflection
Equal tensile strain
Equal contact pressureEqual Contact radius
These design parameters can be theoretically calculated or experimentally determined as
specified in a given design methodology.
18
-
8/13/2019 2 Traffic Loading
19/39
KAAF UNIVERSITY COLLEGE
Determining ESWL - Using Equal Vertical
Stress Concept
This concept depends on the fact that there is an equal maximum vertical stress on thesubgrade. This is based on approximation of stress distribution in a one layer system.
For a pavement thickness less than d/2, no stress overlaps. Hence ESWL will be Pd. At adepth of approximately 2Sd, the effect of overlap is such that it is equivalent to stress causedby 2Pd.
For an intermediate depth it is assumed that a linear relationship exists between load andthickness plotted on a log-log scale.
19
Pd PdSd
d
2Sd
d/2
-
8/13/2019 2 Traffic Loading
20/39
KAAF UNIVERSITY COLLEGE
Determining ESWL - Using Equal Vertical
Stress Concept
Thus,
Where log (ESWL) = log Pd + 0.301 log (2 Z/d)
log (4Sd/d )
20
Z=d/2
2Pd
Z Z=2Sd
P
ESWL(logscale)
Pd
Depth Z (log scale)
-
8/13/2019 2 Traffic Loading
21/39
KAAF UNIVERSITY COLLEGE
Example Using Equal Vertical
Stress Concept
SolutionExample
Determine the ESWL for the pavement
loading below;
300mm
20kN 20kN
200mmE1 = E2
100mm
Subgrade (E2)
Pavement (E1)
300mm
20kN 20kN
200mm
E1 = E2
100mm
Pavement (E1)
Subgrade (E2) 600mm
50mm
50
40
Z 600
?
ESWL(logscale)
20
Depth Z (log scale) 21
-
8/13/2019 2 Traffic Loading
22/39
KAAF UNIVERSITY COLLEGE
Example Using Equal Vertical
Stress Concept
Thus ESWL =29.44kN
Try
Calculate ESWL by equal stress criteria for a
dual wheel assembly carrying 2044 kg each for
a pavement
thickness of 5, 15, 20, 25 and 30 cms. The
distance between walls of the tyre is 11 cm.Use either graphical or
functional methods.
(Hint: Pd=2044kg, Sd=27cm, d=11cm). [Ans:
2044, 2760, 3000, 3230 and 4088]
Therefore
log (ESWL) = log Pd + 0.301 log (2 Z/d)
log (4Sd/d )
= log 20 + 0.301 log (2 x200/100)
log (4x300/100 )
22
-
8/13/2019 2 Traffic Loading
23/39
KAAF UNIVERSITY COLLEGE
Determining ESWL - Using Equal Vertical
Deflection Concept
For single layer system deflection at any depth and radial distance is given by;
D= p x a x F 1
(E)
Where D is the deflection at depth zand radial distance (measured from the centre of theload) rand E is the elastic modulus of the pavement (subgrade modulus in case of a two
layer system) and
23
Sd
P P
h
a
E1, m1
E2, m2
Max. Deflection h
ESWL
M2 =0.5
E1 = E2
M1=0.5
E2
-
8/13/2019 2 Traffic Loading
24/39
KAAF UNIVERSITY COLLEGE
Determining ESWL - Using Equal Vertical
Deflection Concept
F is a deflection factor , a function of rand z. A chart of it is shown below;
24
-
8/13/2019 2 Traffic Loading
25/39
KAAF UNIVERSITY COLLEGE
Calculating ESWL - Using Equal Vertical
Deflection Concept
The deflection for ESWL is given by;
D ESWL= pESWLx a x FESWL ..2
(ESUB)
For the multiple wheel the deflection is also given by
DMultiple= pMultiplex a x Fmax 3
(ESUB)
Thus according to the Equal vertical deflection concept , the deflection should be equal;
D ESWL= Dmultiple
pESWLx FESWL = pMultiplex Fmax .4
pMultiple is known and given
Fmax is a function of wheel configuration rand depth z
FESWL is a function of h(and r=o)
Thus pESWL and ESWL can be determined.25
-
8/13/2019 2 Traffic Loading
26/39
KAAF UNIVERSITY COLLEGE
Determining ESWL for 2 layer System -
Using Equal Vertical Deflection Concept
For the two layer system having modulus E1 and E2, the deflection at the interface of the two
layers is most important. As such it is given by;
D= p x a x F where F is the interface deflection factor, a function of radial distance rand
(Esubgrade) and pavement thickness h.
26
-
8/13/2019 2 Traffic Loading
27/39
KAAF UNIVERSITY COLLEGE
Determining ESWL for 2 layer System
Charts for Determining Factors
27
-
8/13/2019 2 Traffic Loading
28/39
KAAF UNIVERSITY COLLEGE
Determining ESWL for 2 layer System
Charts for Determining Factors
28
-
8/13/2019 2 Traffic Loading
29/39
KAAF UNIVERSITY COLLEGE
Determining ESWL for 2 layer
System - Using Equal Vertical
Deflection Concept
Trial Question
Where m is Poisson distribution values.
Solution
E1/E2=250/50=5
Z/a =200/100 = 2
Explore points 1, 2 and 3 for maximum deflection
Point 1
Load A r/a = 0/100 = 0
(r/a = 0 , h/a = 2) thus F=0.5 ( from chart slide 27)
Load B r/a = 300/100 = 3
(r/a = 3 , h/a = 2) thus F=0.28 ( from chart slide 27)
Total F = 0.5+0.28 = 0.78
Point 2
Load A r/a = 100/100 = 1
(r/a = 1 , h/a = 2) thus F=0.4 ( from chart slide 24)
Load B r/a = 200/100 = 2
(r/a = 2 , h/a = 2) thus F=0.35 ( from chart slide 24)
Total F = 0.4+0.35 = 0.75
Point 3
Load A r/a = 150/100 = 1.5
(r/a = 1.5 , h/a = 2) thus F=0.38 ( from chart slide 24)
Load B r/a = 150/100 = 1.5
(r/a = 1.5 , h/a = 2) thus F=0.38 ( from chart slide 24)
Total F = 0.38+0.38 = 0.76
Thus maximum deflection, Fmax = 0.78 occurs at Point 1.
300mm
20kN 20kN
200mm
100mm
E1=250
E2=50
BA
1 2 3
29
m1=0.5
m2=0.5
-
8/13/2019 2 Traffic Loading
30/39
KAAF UNIVERSITY COLLEGE
Determining ESWL for 2 layer
System - Using Equal Vertical
Deflection Concept
Thus for an ESWL acting at point 3,
(r/a = 0 , z/a = 2) thus F=0.5 ( from chart slide 27)
Thus from equation 4
pESWLx FESWL = pMultiplex Fmax
Where pMultiple= 20,000
x 1002
pMultiple = 0.6366MPa
Thus
pESWL= pMultiplex FMax
FESWL= 0.6366x 1
0.7
= 0.9930MPa
ESWL = pESWL
x contact area
= 0.909 x x 1002
= 31.2kN
Try thisCalculate the surface deflection under the centre of a tyre (a =
152 mm, p = 552 kPa) for a 305 mm pavement having a 345
MPa modulus and subgrade modulus of 69 MPa from two-layer
theory. Also calculate the interface deflection.
a. A circular load with a radius of 152 mm and a uniform
pressure of 552 kPa is applied on a two-layer system. The
subgrade has an elastic modulus of 35 kPa and can support a
maximum vertical stress of 55 kPa. What is the required
thickness of full depth AC pavement, if AC has an elastic
modulus of 3.45 GPa.
b.Instead of a full depth AC pavement, if a thin surface
treatment is applied on a granular base (with elastic modulus of173 MPa), what is the thickness ofbase course required?
30
200mm
ESWL
m2 =0.5
E1 = E2
M1=0.5
E23
-
8/13/2019 2 Traffic Loading
31/39
KAAF UNIVERSITY COLLEGE
ESWL Other Criteria
Other Criteria exist for determining ESWL apart from the vertical Stress and Deflection
methods. These include;
Equal Tensile StrainFor pavements with bituminous layers
Equal Tensile StressConcrete Pavements.
31
-
8/13/2019 2 Traffic Loading
32/39
KAAF UNIVERSITY COLLEGE
Equivalent Axle Load Factors (EALF)
In some design methods pavements are designed for a selected number of repetitions of a
standard load (Standard axle load80kN). As such EALFs are used to convert different axle
loads into equivalent repetitions of a standard axle.
EALFs defines the damage caused to the pavement by one application of the axle load under
consideration relative to the damage caused by a single application of a standard axle.
Design is based on the total number of applications of standard axle load during the design
period which is known as the Equivalent Standard Axle Load (ESAL).
Where m = number of axle load groups
Fi= EALF for the ith-axle load group
ni= number of applications of the ith group during the design period
32
-
8/13/2019 2 Traffic Loading
33/39
KAAF UNIVERSITY COLLEGE
Equivalent Axle Load Factors (EALF)
For the design of pavements, it is important to have information on EALFs and the expected
axle load spectrum for the design period. Axle load spectrum describes the number of
passes of axles for different groups of axle loads ( 05 kN, 5-10 kN, etc).
EALF ( relative damaging effect) is a function of the type of pavement, composition and
strength of pavement and criterion determining performance (damage). EALF can bedetermined from the following;
-Obtained from field observations of performance of pavements carrying different types of axle
loads.
-From AASHTO road test ( which is often used)
-Obtained from theoretical exercise using appropriate mechanistic criteria
Note! EALFs are different for different types of pavements and for different performance
criteria.
33
-
8/13/2019 2 Traffic Loading
34/39
KAAF UNIVERSITY COLLEGE
AASHTO EALF
The AASHTO Road Experiment resulted in ESAL factors as a function of the axle
configuration, load, pavement strength ( structural number or slab thickness), terminal
condition of the pavement.
A simplification of this is the Fourth Power Law given as:
ESAL factor = [ axle load ]4
standard Axle load
Example . The damaging effect caused by 160kN load relative to a standard axle load of 80kN
is given as;
ESAL F= (160/80)4 = 16
Try this; Let number of load repetition expected by 80 KN standard axle is 1000, 160 KN is
100 and 40 KN is 10000. Find the equivalent axle load.
34
-
8/13/2019 2 Traffic Loading
35/39
KAAF UNIVERSITY COLLEGE
AASHTO EALF
.
35
-
8/13/2019 2 Traffic Loading
36/39
KAAF UNIVERSITY COLLEGE
AASHTO EALF
.
36
-
8/13/2019 2 Traffic Loading
37/39
KAAF UNIVERSITY COLLEGE
Axle Load Measurement
To estimate the total projected repetitions of commercial traffic in terms of the repetitions of a
standard load unit (standard axle load) it is necessary to have an estimate of the axle load
spectrum besides the EALFs.
Axle load spectrum is obtained by conducting axle load survey of commercial vehicles. Thus,
measurement of axle loads of a sample of commercial vehicles plying on a given facility. Iffacility is very new, commercial vehicle data is obtained from another facility similar to the one
being designed. This is often conducted with portal weigh pads placed on the site.
37
-
8/13/2019 2 Traffic Loading
38/39
KAAF UNIVERSITY COLLEGE
Axle Load Survey and Determination of
VDF
In conducting an axle load survey
- An adequate number of commercial vehicles are sampled
- Usually only the wheel loads are measured
- Axle load = 2x wheel ( thus, where only one wheel load is measured)
- Commercial vehicles in both directions is considered
An analysis of axle load data from 450 sampled commercial vehicles, totaling 1000 axlesmeasured is as shown,
Thus
1000 axles = 5000.6 standard axles(80kN)
1 axle = 5.0 standard axle
450 commercial vehicles = 5000.6 standard axles (80kN)
1 comm. vehicle = 11.11 stded axles (Vehicle Damage Factor)
VDF is used for converting a given traffic volume into equivalent number of standard of standard axles. VDF is a typical valuerepresenting the loads carried by the commercial vehicles plying on facility. Which is determined by conducting axle load
surveys. 38
Load Group Frequency Mid Pt EALF ESAL
0 - 40 50 20 0.0625 3.13
40 - 80 250 60 0.3164 79.10
80 - 120 400 100 2.4414 976.56
120 - 160 250 140 9.379 2344.75
160 - 200 40 180 25.629 1025.16
200 - 240 10 200 57.19 571.90
Total 1000 700 95.0183 5000.595
-
8/13/2019 2 Traffic Loading
39/39
KAAF UNIVERSITY COLLEGE
Estimation of Design Traffic
The repetitions of standard load (80kN) expected to be applied on the pavement during aspecified period ( design life) is a function of ;
-Initial traffic (commercial) Cumulative traffic over the entire period taking into account projectionsabout rate of growth
-Vehicle Damage FactorLateral placement characteristics of wheel loads
Example
1. Design life =15years2. Traffic growth rate = 8%
3. VDF = 4.5
4. Three lane dual carriage way with 12,000 commercial vehicle per day. (0.6 distribution)
N= 365x[ (1+0.08)151] x 6000 x 0.6 x 4.5 = 160.55msa
0.08
Try this on your own
Estimate the design life traffic for a facility with 2-lane road, ADT = 4000cvpd (two-way),VDF=5.0, Design life = 15years, rate of growth of commercial vehicles =7%.
39