2 Traffic Loading

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KAAF UNIVERSITY COLLEGE

Civil Engineering DepartmentCollege of Engineering

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Highway Engineering IICIV 467

Lecture 2_ Traffic Loading

Kwasi AgyemanBoakye ( [email protected])

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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

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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.

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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
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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.

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200kN

80kN 120kN


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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

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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)

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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.


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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

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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

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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;

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0.8712L

0.6LA= 0.522L2


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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

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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).

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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

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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

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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.

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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.

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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

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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.

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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.

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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.

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Pd PdSd

d

2Sd

d/2


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Stress Concept

Thus,

Where log (ESWL) = log Pd + 0.301 log (2 Z/d)

log (4Sd/d )

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Z=d/2

2Pd

Z Z=2Sd

P

ESWL(logscale)

Pd

Depth Z (log scale)


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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


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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 )

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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

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Sd

P P

h

a

E1, m1

E2, m2

Max. Deflection h

ESWL

M2 =0.5

E1 = E2

M1=0.5

E2


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Deflection Concept

F is a deflection factor , a function of rand z. A chart of it is shown below;

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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


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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.

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Determining ESWL for 2 layer System

Charts for Determining Factors

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Determining ESWL for 2 layer System

Charts for Determining Factors

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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


Total F = 0.5+0.28 = 0.78

Point 2

Load A r/a = 100/100 = 1


Load B r/a = 200/100 = 2


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

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m1=0.5

m2=0.5


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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?

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200mm

ESWL

m2 =0.5

E1 = E2

M1=0.5

E23


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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.

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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

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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.

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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.

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AASHTO EALF

.

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AASHTO EALF

.

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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.

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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


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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%.

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2 Traffic Loading