2 Traffic Loading

Civil Engineering Department College of Engineering KAAF UNIVERSITY COLLEGE

_________________________________ _____________ _____________________ _ Highway Engineering II CIV 467 Lecture 2_ Traffic Loading Kwasi Agyeman – Boakye ( [email protected])

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Overview of the Course ( CIV 467) KAAF UNIVERSITY COLLEGE

•Time

Period * 7:00am – 10:00am

•Days

* Saturdays , 21 Sept, 28 Sept , 05 Oct, 12 12 Oct , 19 Oct , 26 Oct , 02 Nov, Nov, 09 Nov, Nov, 16 Nov, Nov, 23 Nov, 30 Nov, Dec 07. •

Mode of Assessment * Coursework and 2 Test - 30%; Exams - 70%

Recommended Reading location, Design Construction and the Maintenance of Road Pavement o Highway; The , C. A. O’ Flarherty, Flarherty, Fourth Edition (2002). Traffic and Highway Engineering, Garber N.J and Hoel A.L, Bill Stenquist (2002). oTraffic oPavement Design Manual, Ghana, First Edition (1998). oStandard Specification for Roads and Bridges, Ghana , 2007

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Vehicle Categories KAAF UNIVERSITY COLLEGE

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 Cars

Heavy Truck Heavy

Taxis

Semi-trailer Semi-traile r (Light)

Pick Up/Vans/4 Up/Vans/4 WD Veh

Semi-trailer Semi-traile r (Heavy)

Small Bus

Truck Trailer

Medium Bus/Mummy Wagon

Extra Large Trucks & Others

Large Bus Light Truck Medium Truck

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. category. 3

Traffic Loads KAAF UNIVERSITY COLLEGE

Vehicle loads can be categorized into various categories such as Gross Load, Axle Load, Wheel Load for different purposes. Also the axles have various c ategorisations. 200kN

80kN

120kN

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Load Transfer through Wheels KAAF UNIVERSITY COLLEGE

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

For high inflation pressures – Wall of tyre in tension, Contact pressure is less than tyre pressure

For low inflation pressures – Wall of tyre in compression, Contact pressure is greater than 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<1, when average tyre pressure > 0.7 Mpa Tyre Pressure = Inflation Contact Pressure = Wheel Load/(area of imprint) 6


Stresses Vertical Stresses Unidirectional surface shear stresses (breaking and acceleration) Centripetal shear stresses The pressure distribution (vertical, centripetal or unidirectional) is not normally uniform. Normally only 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 KAAF UNIVERSITY COLLEGE

Different contact shape areas are considered for analyses, such as; •Circular •Rectangular •Rectangular •More

with semi-circular ends

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 Πa2 =

20,000/0.7

a = 95.365mm

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Contact Area Shapes KAAF UNIVERSITY COLLEGE

Rectangle with Semi Circular Ends Contact Area (A) = 0.522L 2 In An Example; For a P = 20kN and p= 0.7MPa determine the 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;

A= 0.522L2

0.6L

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Design Loading Considerations KAAF UNIVERSITY COLLEGE

•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 specifi c 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

Traffic Forecast KAAF UNIVERSITY COLLEGE

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) n – 1] 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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The traffic in the year of completion of construction i s 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 •Lateral •Load

distribution of traffic

placement characteristics of wheels on pavement

spectrum

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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 of commercial 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 2013 Solution Determine A; A = 4000(1+0.07)2 = 4580cvpd

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 way 2. 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. 4.7x106 cvpd.

N= 365x 4580x[ (1+0.07)15 – 1] =42x106 cvpd 0.07 13

Lateral distribution of wheel loads KAAF UNIVERSITY COLLEGE

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 loads Some recommended load distributions KAAF UNIVERSITY COLLEGE

Single lane roads – Design 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 KAAF UNIVERSITY COLLEGE

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 KAAF UNIVERSITY COLLEGE

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

Equivalent Standard Wheel Loads (ESWL) KAAF UNIVERSITY COLLEGE

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 pressure

•Equal

Contact radius

These design parameters can be theoretically calculated or experimentally determined as specified in a given design methodology. 18

Determining ESWL - Using Equal Vertical Stress Concept KAAF UNIVERSITY COLLEGE

This concept depends on the fact that there is an equal maximum vertical stress on the subgrade. This is based on approximation of stress distribution in a one layer system. Pd

Pd

Sd d

d/2 2Sd

For a pavement thickness less than d/2, no stress overlaps. Hence ESWL will be Pd. At a depth of approximately 2Sd, the effect of overlap is such that it is equivalent to stress caused by 2Pd. For an intermediate depth it is assumed that a linear relationship exists between load and thickness plotted on a log-log scale. 19

Determining ESWL - Using Equal Vertical Stress Concept KAAF UNIVERSITY COLLEGE

Thus, 2Pd ) e l a c s g P o l ( L W P S E d

Z=d/2

Z

Z=2Sd

Depth Z (log scale)

Where

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

log (4Sd/d )

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Example Using Equal Vertical Stress Concept KAAF UNIVERSITY COLLEGE

Example Determine the ESWL for the pavement loading below; 20kN

20kN

100mm

300mm

Solution 20kN

20kN

100mm

300mm 50mm

E1 = E2

200mm

Pavement (E1) 600mm

Subgrade (E2) E1 = E2

200mm Pavement (E1) Subgrade (E2) 40 ) e l ? a c s g o 20 l ( L W S E

50

Z Depth Z (log scale)

600 21

Example Using Equal Vertical Stress Concept KAAF UNIVERSITY COLLEGE

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 ) 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]

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Determining ESWL - Using Equal Vertical Deflection Concept KAAF UNIVERSITY COLLEGE

P

P

a

ESWL

Sd

E1, m1

Max. Deflection

h

E1 = E2 M1=0.5

h

E2

E2, m2

M2 =0.5

For single layer system deflection at any depth and radial distance i s given by; D= p x a x F

…………………1

(E) Where D is the deflection at depth ‘z’ and radial distance (measured from the centre of the load) ‘r’ and ‘E’ is the elastic modulus of the pavement (subgrade modulus in case of a two layer system) and

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Determining ESWL - Using Equal Vertical Deflection Concept KAAF UNIVERSITY COLLEGE

F is a deflection factor , a function of ‘r’ and ‘z’. A chart of it is shown below;

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Calculating ESWL - Using Equal Vertical Deflection Concept KAAF UNIVERSITY COLLEGE

The deflection for ESWL is given by; D ESWL= pESWL x a x FESWL ……………………..2 (ESUB) For the multiple wheel the deflection is also given by DMultiple= pMultiple x a x Fmax

…………………

3

(ESUB) Thus according to the Equal vertical deflection concept , the deflection should be equal; D ESWL= Dmultiple pESWL x FESWL = pMultiple x Fmax …………….4 pMultiple is known and given Fmax is a function of wheel configuration FESWL is a function of

‘h’ (and

‘r’ and

depth

‘z’

r=o)

Thus pESWL and ESWL can be determined. 25

Determining ESWL for 2 layer System Using Equal Vertical Deflection Concept KAAF UNIVERSITY COLLEGE

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 (E subgrade)

where F is the interface deflection factor, a function of radial distance and pavement thickness

‘r’ and

‘h’.

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Determining ESWL for 2 layer System – Charts for Determining Factors KAAF UNIVERSITY COLLEGE

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Determining ESWL for 2 layer System – Charts for Determining Factors KAAF UNIVERSITY COLLEGE

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Determining ESWL for 2 layer System - Using Equal Vertical Deflection Concept Trial Question

KAAF UNIVERSITY COLLEGE

Point 2

20kN

20kN 300mm A

100mm

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

B

(r/a = 2 , h/a = 2) thus F=0.35 ( from chart slide 24) Total F = 0.4+0.35 = 0.75

E1=250 m1=0.5

200mm 1

2

3

E2=50 m2=0.5

Where m is Poisson distribution values.

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)

Solution E1/E2=250/50=5

Total F = 0.38+0.38 = 0.76

Z/a =200/100 = 2 Thus maximum deflection, Fmax = 0.78 occurs at Point 1. 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) 29

Determining ESWL for 2 layer System - Using Equal Vertical Deflection Concept ESWL

E1 = E2

200mm

M1=0.5

3

E2 m2 =0.5

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 pESWL x FESWL = pMultiple x Fmax Where pMultiple = 20,000 π x

1002

pMultiple = 0.6366MPa

KAAF UNIVERSITY COLLEGE

Thus pESWL = pMultiple x FMax FESWL = 0.6366x 1 0.7 = 0.9930MPa ESWL = pESWL x contact area = 0.909 x π x 1002 = 31.2kN

Try this Calculate 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 def lection. • 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. W hat 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 30 of

ESWL – Other Criteria KAAF UNIVERSITY COLLEGE

Other Criteria exist for determining ESWL apart from the vertical Stress and Deflection methods. These include; Equal Tensile Strain – For pavements with bituminous layers Equal Tensile Stress – Concrete Pavements.

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Equivalent Axle Load Factors (EALF) KAAF UNIVERSITY COLLEGE

In some design methods pavements are designed for a selected number of repetitions of a standard load (Standard axle load – 80kN). 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

Equivalent Axle Load Factors (EALF) KAAF UNIVERSITY COLLEGE

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 ( 0 – 5 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 be determined 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 KAAF UNIVERSITY COLLEGE

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


.

35


.

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Axle Load Measurement KAAF UNIVERSITY COLLEGE

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. If facility 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 KAAF UNIVERSITY COLLEGE

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 axles measured 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)

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

1000

700

95.0183

5000.595

Total

VDF is used for converting a given traffic volume into equivalent number of standard of standard axles. VDF is a typical value representing the loads carried by the commercial vehicles plying on facility. Which is determined by conducting axle load 38

2 Traffic Loading

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