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dissertation entitled

TRUCK TIRE/PAVEMENT INTERACTION ANALYSIS BY THE
FINITE ELEMENT METHOD

presented by
Lan Meng

has been accepted towards fulfillment
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6/01 cJCIRC/DateDuepas-p. 1 5

TRUCK TIRE/PAVEMENT INTERACTION ANALYSIS BY THE F INITE ELEMENT
METHOD

By

Lan Meng

DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Materials Science and Mechanics

2002

ABSTRACT

TRUCK TIRE/PAVEMENT INTERACTION ANALYSIS BY THE FINITE ELEMENT
METHOD

By

Lan Meng

A simulation is developed for study of interactions between a rolling truck tire and a
rigid pavement structure. A three-dimensional finite element method is required to solve
this nonlinear dynamic contact problem. The ABAQUS finite element code is employed.
A Goodyear 295/75R22.5 Unisteel G167A Low Profile Radial smooth tire is modeled to
roll over the rigid pavement. Fiber reinforced composite model and rubber material model
is utilized to simulate the real tire structure. The Coulomb friction law is used to study the
contact between the truck tire and pavements. Different tire pressure levels, load levels,
moving speeds, slip angles, braking or traction, and friction are applied in the simulation.
Results for a static loaded tire against a rigid flat plate are compared with experimental
data from the literature. Contact stress distributions and contact areas for the rolling tire
over the rigid pavement at various conditions are presented.

Results from this investigation lead to the following observations and conclusions:

1. For a free rolling tire model, normal stress contours of the free rolling tire model are

shifted forward; high stress areas are moved toward to two sidewalls; high stress areas

are increased in fore-back direction; normal maximum stresses at higher tire pressure
(100 psi and 110 psi) are higher than for the static loaded tire.

For a free rolling tire model at a deflection of 50 mm and a speed of 65 mph, the ratio
between the maximum stress and the average static average is above 2.0 at 100 psi
and 110 psi.

. For traction or braking tire models, normal stress contours for the traction models are
shifted forward more than for the free rolling model; normal stress contours for the
braking models are shifted backward.

For traction or braking tire models, contact areas are larger and contact stress are
higher than for the free rolling model.

. For the slip-rolling tire model, high stress areas are moved toward the sidewalls in the
slip direction. The contact areas are smaller than for the free rolling model. The
maximum normal stresses are higher than for free rolling models and traction or
braking models.

. For the slip-rolling tire model, a larger slip angle produces higher normal stress and
shear stress.

. The maximum ratios of max/av. stress vary from 1.98 for static case to 1.97 for the
free rolling case, and 3.22 for the traction case to 3.29 for the slip rolling case. These
findings suggest that the average stress used in the AASHTO seriously underestimates
the maximum stresses served by pavements, with the result that highways can be

seriously underdesigned.

To My Family

iv

ACKNOWLEDGMENTS

Many people have been very helpful to me throughout my graduate studies. I
specially appreciate my advisor Gary Lee Cloud for his encouragement, guidance,
and support throughout my study at the Michigan State University.

Thank you to my Ph.D. committee members, Ronald C. Averill, Larry
Segerlind, and Zhengfang Zhou for their advice, comment, and help during this
project.

Thank you also to Bill Kenis at the Federal Highway Administration for his
support and interest and for providing the opportunity for me to do this project.

I thank the Goodyear Company for providing the truck tire information.

Thank you to the Federal Highway Administration for granting me the Dwight
David Eisenhower Transportation Fellowship, Michigan State University for
granting me the Graduate School Summer Acceleration Fellowship, Cybernet
Systems Corp., the Composite Materials and Structure Center (MSU), and the
Materials Science and Mechanics Department (MSU) for their financial support.

Last, but not the least, a special thanks go to my husband, Binshan Ye, and my

daughter, Mengyuan Ye, for their love, support, and understanding.

TABLE OF CONTENTS

LIST OF TABLE ............................................................................................................. viii

LIST OF FIGURE .............................................................................................................. ix

CHAPTER 1

INTRODUCTION .............................................................................................................. 1
1.1 INTRODUCTION .................................................................................................. 1

CHAPTER 2

BACKGROUND ................................................................................................................ 4
2.1 DYNAMIC ANALYSIS OF PAVEMENT-TIRE INTERACTION ..................... 4
2.2 TIRE CONTACT PATCH ANALYSIS ................................................................ 7
2.3 RESEARCH OBJECTIVES ................................................................................ 12

CHAPTER 3

APPLICATION OF THE FINITE ELEMENT ANALYSIS ........................................... 13
3.1 INTRODUCTION OF ABAQUS/STANDARD CODE ..................................... 13
3.2 GOVERNING EQUATIONS .............................................................................. 14
3.3 STEADY —STATE TRANSPORT ANALYSIS .................................................. 18
3.4 REINFORCING ELEMENTS IN THE ANALYSIS .......................................... 24
3.5 MAJOR MATERIALS USED IN THE TIRE MODEL ...................................... 26

CHAPTER 4

SIMULATION OF TIRE/PAVEMENT INTERACTION AND STATIC ANALYSIS . 30
4.1 DESCRIPTION OF TIRE/PAVEMENT MODELING ....................................... 30
4.2 THE FINITE ELEMENT MODEL ...................................................................... 30
4.3 VALIDATION OF THE SIMULATED TIRE MODEL ..................................... 36
4.4 STATIC RESULTS FOR THE CONTACT STRESS ......................................... 44

CHAPTER 5

ROLLING TRUCK TIRE/PAVEMENT INTERACTIONS ............................................ 65
5.1 TIRE MODEL ...................................................................................................... 65
5.2 SIMULATION RESULTS ................................................................................... 66
5.3 REFERENCE DATA FOR A SLOW ROLLING TIRE TEST ......................... 125

CHAPTER 6

DISCUSSION ................................................................................................................. 128
6.1 FREE ROLLING TIRE MODEL ...................................................................... 128
6.2 TRACTION OR BRAKING TIRE MODEL ..................................................... 131
6.3 SLIP-ROLLING TIRE MODEL ........................................................................ 132

CHAPTER 7

CONLUSIONS ............................................................................................................... 135

vi

CHAPTER 8
FUTURE WORK AND RECOMMENDATIONS ........................................................ 140
BIBLIOGRAPHY ........................................................................................................... 142

vii

LIST OF TABLES

Table 4-1 Ply and belts reinforcement properties of tire model ....................................... 33
Table 4-2 Bead and flipper properties of tire model ......................................................... 35
Table 4-3 Rubber properties of tire model ........................................................................ 35
Table 4-4 Results of normal stress for tire pressure of 90 psi .......................................... 45
Table 4-5 Results of normal stress for tire pressure of 100 psi ........................................ 46
Table 4-6 Results of normal stress for tire pressure of 110 psi ........................................ 47
Table 5-1 Detail information for tire of 90 psi ................................................................. 67
Table 5-2 Detail information for tire of 100 psi ............................................................... 68
Table 5-3 Detail information for tire of l 10 psi ............................................................... 69
Table 5-4 Results for Free Rolling .................................................................................... 70
Table 5-5 Results for Traction and Braking ..................................................................... 71
Table 5-6 results for Slip Rolling ...................................................................................... 72

viii

LIST OF FIGURES

Figure 3.1 Constant cornering motion .............................................................................. 19
Figure 3.2 Rebar in a solid or two-dimensional element .................................................. 25
Figure 4.1 Section of a typical tire structure ..................................................................... 31
Figure 4.2 Cross-section of the tire model ........................................................................ 32
Figure 4.3 Truck ire/pavement model ............................................................................... 34

Figure 4.4 Load-deflection curves for the experimental analysis and the numerical

analysis (90 psi) .......................................................................................................... 37
Figure 4.5 Load-deflection curves for the experimental analysis and the numerical

analysis (100 psi) ........................................................................................................ 37
Figure 4.6 Load-deflection curves for the experimental analysis and the numerical

analysis (110 psi) ........................................................................................................ 38
Figure 4.7 Deformation contour at pressure of 100 psi and deflection of 3.9 mm ........... 39
Figure 4.8 Deformation contour at pressure of 100 psi and deflection of 16.4 mm ......... 41
Figure 4.9 Deformation contour at pressure of 100 psi and deflection of 27.7 mm ......... 40
Figure 4.10 Deformation contour at pressure of 100 psi and deflection of 44.6 mm ....... 42

Figure 4.11 Deformation contour at pressure of 100 psi and deflection of 48.9 mm ....... 43

Figure 4.12 Normal stress contour at pressure of 90 psi and deflection of 4.6 mm ......... 48
Figure 4.13 Normal stress contour at pressure of 90 psi and deflection of 9.6 mm ......... 49
Figure 4.14 Normal stress contour at pressure of 90 psi and deflection of 17.6 mm ....... 50
Figure 4.15 Normal stress contour at pressure of 90 psi and deflection of 28.3 mm ....... 51
Figure 4.16 Normal stress contour at pressure of 90 psi and deflection of 45.2 mm ....... 52

Figure 4.17 Normal stress contour at pressure of 90 psi and deflection of 49.2 mm ....... 53
Figure 4.18 Normal stress contour at pressure of 100 psi and deflection of 3.9 mm ....... 54

Figure 4.19 Normal stress contour at pressure of 100 psi and deflection of 8.4 mm ....... 55

ix

Figure 4.20 Normal stress contour at pressure of 100 psi and deflection of 16.4 mm ..... 56
Figure 4.21 Normal stress contour at pressure of 100 psi and deflection of 27.7 mm ..... 57
Figure 4.22 Normal stress contour at pressure of 100 psi and deflection of 45.6 mm ..... 58
Figure 4.23 Normal stress contour at pressure of 100 psi and deflection of 48.9 mm ..... 59
Figure 4.24 Normal stress contour at pressure of 110 psi and deflection of 6.8 mm ....... 60
Figure 4.25 Normal stress contour at pressure of 110 psi and deflection of 14.6 mm ..... 61
Figure 4.26 Normal stress contour at pressure of 110 psi and deflection of 26.3 mm ..... 62
Figure 4.27 Normal stress contour at pressure of 110 psi and deflection of 43.8 mm ..... 63
Figure 4.28 Normal stress contour at pressure of 110 psi and deflection of 48.4 mm ..... 64

Figure 5.1a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
22.2 mm, and velocity of 5 mph ................................................................................. 74

Figure 5. lb Shear stress contour (xz) for a free rolling at pressure of 90 psi, deflection of
22.2 mm, and velocity of 5 mph ................................................................................. 75

Figure 5.2a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
42.2 mm, and velocity of 5 mph ................................................................................. 76

Figure 5.2b Shear stress contour (xz) for a free rolling at pressure of 90 psi, deflection of
42.2 mm, and velocity of 5 mph ................................................................................. 77

Figure 5.3a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, and velocity of 65 mph ............................................................................... 78

Figure 5.3b Shear stress contour (xz) for a free rolling at pressure of 90 psi, deflection of
49.6 mm, and velocity of 65 mph ............................................................................... 79

Figure 5.4a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
29 mm, and velocity of 65 mph .................................................................................. 80

Figure 5.4b Shear stress contour (xz) for a free rolling at pressure of 100 psi, deflection of
29 mm, and velocity of 65 mph .................................................................................. 81

Figure 5.53 Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, and velocity of 45 mph ............................................................................... 82

Figure 5.5b Shear stress contour (xz) for a free rolling at pressure of 100 psi, deflection of
48.9 mm, and velocity of 45 mph ............................................................................... 83

Figure 5.6a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, and velocity of 65 mph ............................................................................... 84

Figure 5.6b Shear stress contour (xz) for a free rolling at pressure of 100 psi, deflection of
48.9 mm, and velocity of 65 mph ............................................................................... 85

Figure 5.73 Normal stress contour for a free rolling at pressure of 110 psi, deflection of
48.4 mm, and velocity of 65 mph ............................................................................... 86

Figure 5.7b Shear stress contour (xz) for a free rolling at pressure of 110 psi, deflection of
48.4 mm, and velocity of 65 mph ............................................................................... 87

Figure 5.8a Normal stress contour for a traction at pressure of 90 psi, deflection of 49.6
m, V, = 65 mph, v, = 70 mph ................................................................................... 88

Figure 5.8b Shear stress contour (xz) for a traction at pressure of 90 psi, deflection of
49.6 m, V, = 65 mph, v, = 70 mph ........................................................................... 89

Figure 5.9a Normal stress contour for a traction at pressure of 90 psi, deflection of 49.6
m, V, = 65 mph, v, = 75 mph ................................................................................... 90

Figure 5% Shear stress contour (xz) for a traction at pressure of 90 psi, deflection of
49.6 m, V, = 65 mph, v, = 75 mph ........................................................................... 91

Figure 5.10a Normal stress contour for a braking at pressure of 90 psi, deflection of 49.6
mm, v._ = 65 mph, v,_ = 60 mph ................................................................................... 92

Figure 5.10b Shear stress contour (xz) for a braking at pressure of 90 psi, deflection of
49.6 mm, v,_ = 65 mph, v,_ = 60 mph ........................................................................... 93

Figure 5.11a Normal stress contour for a braking at pressure of 90 psi, deflection of 49.6
m, V, = 65 mph, v,_ = 55 mph ................................................................................... 94

Figure 5.11b Shear stress contour (x2) for a braking at pressure of 90 psi, deflection of
49.6 m, V, = 65 mph, v,_ = 55 mph ........................................................................... 95

Figure 5.12a Normal stress contour for a traction at pressure of 100 psi, deflection of 29
m, V, = 65 mph, v, = 68.5 mph ................................................................................ 96

Figure 5.12b Shear stress contour (xz) for a traction at pressure of 100 psi, deflection of
29 mm, Vt. = 65 mph, v, = 68.5 mph ........................................................................... 97

Figure 5.13a Normal stress contour for a traction at pressure of 100 psi, deflection of 29
m, V, = 65 mph, v, = 74.1 mph ................................................................................ 98

Figure 5.13b Shear stress contour (xz) for a traction at pressure of 100 psi, deflection of
29 m, V, = 65 mph, v, = 74.1 mph ........................................................................... 99

xi

Figure 5.14a Normal stress contour for a traction at pressure of 100 psi, deflection of 48.9
mm, vi, = 65 mph, v,_ = 70 mph ................................................................................. 100

Figure 5.14b Shear stress contour (xz) for a traction at pressure of 100 psi, deflection of
48.9 m, V, = 65 mph, v,_ = 70 mph ......................................................................... 101

Figure 5.15a Normal stress contour for a traction at pressure of 100 psi, deflection of 48.9
mm, v., = 65 mph, v,_ = 75 mph ................................................................................. 102

Figure 5.15b Shear stress contour (xz) for a traction at pressure of 100 psi, deflection of
48.9 mm, v1, = 65 mph, v, = 75 mph ......................................................................... 103

Figure 5.16a Normal stress contour for a traction at pressure of 110 psi, deflection of 48.4
mm, v,, = 65 mph, v, = 70 mph ................................................................................. 104

Figure 5.16b Shear stress contour (xz) for a traction at pressure of 110 psi, deflection of
48.4 mm, vt, = 65 mph, v, = 70 mph ......................................................................... 105

Figure 5173 Normal stress contour for a traction at pressure of 110 psi, deflection of 48.4
mm, v., = 65 mph, v, = 75 mph ................................................................................. 106

Figure 5.17b Shear stress contour (xz) for a traction at pressure of 110 psi, deflection of
48.4 m, V, = 65 mph, v,_ = 75 mph ......................................................................... 107

Figure 5.18a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, velocity of 65 mph, f= 1.0, and slip angle 1.50 ........................................ 108

Figure 51% Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, velocity of 65 mph, f= 1.0, and slip angle 2.40 ........................................ 108

Figure 5.20a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, velocity of 65 mph, f= 1.0, and slip angle 3.00 ........................................ 109

Figure 5.21a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, velocity of 65 mph, f = 0.9, and slip angle 1.50 ........................................ 110

Figure 5.22a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, velocity of 65 mph, f = 0.9, and slip angle 2.40 ........................................ 111

Figure 5.23a Normal stress contour for a free rolling at pressure of 90 psi, deflection of
49.6 mm, velocity of 65 mph, f = 0.95, and slip angle 3.00 ...................................... 112

Figure 5.24a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
29 mm, velocity of 65 mph, f = 0.95, and slip angle 0.90 ......................................... 113

Figure 5.25a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
29 mm, velocity of 65 mph, f = 0.95, and slip angle 1.80 ......................................... 114

xii

Figure 5.26a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
29 mm, velocity of 65 mph, f = 0.95, and slip angle 3.00 ......................................... 115

Figure 5.27a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, velocity of 45 mph, f= 0.95, and slip angle 1.50 ...................................... 116

Figure 5.28a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, velocity of 45 mph, f= 0.95, and slip angle 3.00 ...................................... 117

Figure 52% Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, velocity of 65 mph, f = 0.95, and slip angle 0.90 ...................................... 1 18

Figure 5.30a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, velocity of 65 mph, f = 0.95, and slip angle 1.80 ...................................... 119

Figure 5.31a Normal stress contour for a free rolling at pressure of 100 psi, deflection of
48.9 mm, velocity of 65 mph, f= 0.95, and slip angle 3.00 ...................................... 120

Figure 5.32a Normal stress contour for a free rolling at pressure of 110 psi, deflection of
48.4 mm, velocity of 65 mph, f = 0.95, and slip angle 0.90 ...................................... 121

Figure 5.333 Normal stress contour for a free rolling at pressure of 110 psi, deflection of
48.4 mm, velocity of 65 mph, f = 0.95, and slip angle 1.80 ...................................... 122

Figure 5.34a Normal stress contour for a free rolling at pressure of 110 psi, deflection of
48.4 mm, velocity of 65 mph, f = 0.95, and slip angle 3.00 ...................................... 123

Figure 5.35 Measured vertical stress distribution of 40000. N load at pressure of 90 psi
and speed of 5 mph ................................................................................................... 124

Figure 5.36 Measured vertical stress distribution of 80000. N load at pressure of 90 psi
and speed of 5 mph ................................................................................................... 125

Figure 5.37 Measured longitudinal stress distribution of 40000. N load at pressure of 90
psi and speed of 5 mph .............................................................................................. 126

Figure 5.38 Measured longitudinal stress distribution of 80000. N load at pressure of 90
psi and speed of 5 mph .............................................................................................. 127

xiii

Chapter 1

INTRODUCTION

1.1 Introduction

The effect of traffic loads on pavement performance is a fundamental concern in
highway engineering, and it represents one of the most visible aspects of road
transportation to the motoring public. Truck performance has become a critical element of
highway safety in recent years for two reasons. First, the percentage of trucks in the
highway traffic stream has increased dramatically, up to 30 percent on some roadways,
according to the American Association of State Highway and Transportation Officials
(AASHTO, 1993). Second, trucks are now equipped with very powerful engines that
allow them to carry larger loads at higher speeds. The increase in speed and gross weight
produces heavy loads on roadways, thus, high quality performance of pavement has
become an important requirement.

The safety and efficiency of a truck depend heavily upon the few square inches of tire
that are in contact with the ground. This contact area, not much larger than a 3-by-5 inch
postcard, is responsible for the load transfer between the truck and the ground during any
maneuvering behavior (accelerating, cornering, braking, and so on), on any kind of

surface (asphalt, concrete, and so on) in any kind of weather (dry, wet, icy, and so on).

The contact load acts repeatedly and produces damage to the pavement, which eventually
leads to the deterioration of the pavement. Clearly, roadbuilders need more accurate data
on truck-pavement interaction because it is crucial in determining the impact of trucks on
highway safety. This process is very complicated because it involves many analytical
aspects that include three mechanical parts. First, pavement is a composite material with
several layers. Material properties of each layer can be complex, for they might be elastic,
plastic, visco-elastic, visco-plastic, bi-modular, anisotropic and so on. Second, when a
load is applied to the pavement, the complexity increases. Vehicles with various
suspension and tire structures pass over highways at different speeds or duration of
staying on its surface. Vehicle loadings are randomly repeated on pavements. Third,
pavement and vehicles impose forces on each other. These applied loads deform the
pavement, especially at the surface; this leads to excitation of axle suspensions resulting
in ever-greater dynamic load on the pavement structure. The process involves
acceleration which causes variations in pavement conditions and vehicle load to reinforce
each other through time, and to become more significant as the pavement deteriorates
further.

Most of the widely used pavement design methods are based on the principle of
“equivalent loads” which involves a traffic axle-load spectrum represented by a static
single equivalent design load for the pavement. The equivalent loadings are predicted
using the AASHTO method of equivalency factors that was derived from the American
Association of State Highway and Transportation Officials Road Test. The test,
performed in the late 1950’s and early 1960’s, may not accurately quantify the effects of

many vehicle loading parameters, according to recent studies. Vehicle characteristics such

k)

as tire pressure, suspension types, axle configurations, axle loads and gross vehicle
weights have changed significantly since the AASHTO Road Test. This may have the
general effect of causing more damage to the pavement than the axle loads and overloads
used in the AASHTO Road Test. The equivalency factors developed from the road test
tend to underestimate the amount of damage caused to the pavement by modern vehicles.

The most serious limitation of this approach is that it provides no understanding of
the actual mechanisms of dynamic interaction between tire load and pavement structure.
This lack of insight makes it difficult to adapt the AASHTO concepts to change vehicle
or pavement technology. In pavement design, AASHTO uses a static load rather than
dynamic load, and uniform tire contact stress distribution is used to simulate
tire/pavement contact vertical stresses as input towards the design and evaluation of
pavements.

Recently some researchers have started analyzing pavement response under a moving
load. In these analyses, a point load or a uniform stress is used for investigating the
pavement response. Researchers in both the automotive industry and the highway system
are also analyzing tire/pavement contact. But too little data are available for a tire rolling
over pavement, so an actual interface stress distribution when a vehicle passes over
pavement can not be provided.

The purpose of this study is to investigate the contact stresses caused by a real truck
tire rolling over pavement and to better understand the actual interaction between the tire
load and the pavement structure. The results will provide data on real contact stresses at
the tire/pavement interface. These data should be useful for designing longer lasting, safer

pavement.

Chapter 2

BACKGROUND

2.1 Dynamic Analysis of Pavement-tire interaction

For investigating dynamic effects of vehicles on pavements, the two major methods
are numerical analysis and experimental analysis. The problem is so complicated that
classical theoretical analyses are not efficient.

White, Zaghloul, Anderson, and Smith (1995) analyzed the pavement response under
a moving aircraft load by using a three-dimensional finite element model. They modeled
pavement as a layered structure with three groups of materials: asphalt mixture, granular
material and cohesive soils. The asphalt mixture was modeled as a visco-elastic material,
and the granular and fine-grained layers were modeled as elastic-plastic materials. The
simulation consisted of a single-wheel load moving repetitively back and forth over the
pavement model at a speed of 6.7 kph (4 mph). They approximated the load contact area
in the analysis by using a rectangular area and applying average stress on the contact area.

Chatti (1992) developed a dynamic model for analyzing rigid discontinuous
pavements subjected to moving transient loads in the time domain and in the frequency
domain. He modeled the pavement slabs using a four-noded, twelve-degrees-of-freedom,

medium thick-plate bending element. Load transfer mechanisms used a one-dimensional

bar element or a vertical spring element, respectively. The foundation support was
represented by either a damped Winkler model (which consists of vertical independent
springs) with uniformly distributed frequency-dependent springs and dashpots, or a
system of semi-infinite horizontal layers resting on a rigid base or a semi-infinite
halfspace. In this analysis the moving load was represented by using a direct shape
function method in which the uniformly distributed global moving load vector was
composed of zero entries except at the nodes of the elements where the distributed loads
are positioned.

Hendrick, Marlow, and Brademeyer (1992) developed methodology to investigate the
interaction between vehicle and pavements that involved two parallel components. One
component employed computer simulations of various vehicle loads, plus modeling axle
configuration, suspension type and behavior, and tire structure and flexing, to predict
dynamic loads along the length of the pavement over time. The second component
adapted computer simulations of asphalt pavements and portland cement concrete
pavements to predict the response of these pavements to moving dynamic loads and to
estimate the accumulation of different manifestations of damage over time. Furthermore,
the surface profile of the simulated pavement at specified times was used as input to the
vehicle model to predict dynamic loads, which in turn were used as inputs to the
pavement model to predict subsequent response and deterioration. In the vehicle model
simulation, the first mathematical models for each component were developed, then these
sub-systems were put together in a simulation program. Rolling was neglected, but
vertical motion and pitch were included. The developed models were fully nonlinear,

time domain models. In the pavement simulation part, both mathematical and finite

element models were established for flexible and rigid pavement. These parametric
studies show that consideration of dynamic loading is necessary and that considerable
variation of dynamic loads is produced by alternative tandem axle configurations.

Researchers have developed another major analysis method to evaluate primary
response of pavement under traffic loads. In 1990 the Federal Highway Administration
(FHWA) constructed a test section on the access road to the Turner Fairbank Research
Facility in McLean, VA. They used instrumented pavement test sections in conjunction
with instrumented vehicles to measure primary pavement response to the dynamic wheel
loads. The test sections consisted of thick and thin sections of asphalt concrete, 178 mm
and 89 mm, respectively. Each test section was 30.4m long, separated by a 7.6m
transition zone. The road was equipped with longitudinal strain gages and deflectometers.
The F HWA vehicle was instrumented with LVDT’s between the axles and chassis to
measure chassis-axle displacements on both ends of the front and rear axles. The vehicle
was also equipped with accelerometers to measure the vertical accelerations, plus strain
gages in conjunction with the accelerometers at the ends of each axle to calculate
dynamic wheel forces. Tests were conducted at four check speeds including 15 mph, 25
mph, 35 mph, and 45 mph. Three vehicle loads, empty, intermediate, and full were used.
This experimental work provides the primary response of the pavement as well as

simultaneous displacements of axles and wheel forces under different loads and speeds.

2.2 Tire Contact Patch Analysis

Stress, not force, damages pavement. The contact patch (footprint) is a critical factor
in pavement design and analysis. Though engineers have researched contact patches for
many years, they are still important problems for highway engineering and the automotive
industry. Due to complicated tire structure and geometry, the problem is inherently
nonlinear. The most general deformation of a pneumatic tire is in reality a coupled-
mechanical-thermodynamic process. Because the material properties of tires are
nonhomogenous and anisotropic, composite material theory needs to be applied. In
addition to these concerns, both the contact area and the contact load intensities are not
known in advance, and they vary during the loading history.

For a moving truck, especially one moving on a highway with a very high speed,
getting a footprint is very difficult and expensive. Much experimental analysis has been
done in laboratories. Pottinger (1992) discussed the three-dimensional contact patch
stress field of solid and pneumatic tires. He based his discussion on experimental studies
of a smooth-treaded urethane solid tire and a low-pull-force smooth-treaded radial, to
represent solid and pneumatic tires. The results were in static and free rolling states of
both tire types. All experiments were performed on the BF Goodrich Flat Surface Tire
Dynamic Machine. He discussed footprint size and shape, normal stress, lateral stress and
longitudinal stress for solid and pneumatic tires.

de (1994) discussed an experimental vehicle-road surface pressure transducer array
(vrspta) system. He designed this system to simultaneously measure vertical, longitudinal,
and lateral loads; and load distributions transferred to the road surface by a passing wheel.

He based his research on the concept of instrumented and calibrated pins buried in the

surface of the pavement designed to measure loads and load distributions over the
complete tire contact area. Strain-gauged load cell pins were installed in a row, at right
angles to the direction of travel of the passing wheel. He used a heavy vehicle simulator
for different wheel load ranges, different wheel velocities, and one tire pressure in the
test. He gathered data on vertical stress distribution, transverse stress distribution and
longitudinal stress distributions at the given speeds, loads and tire pressures. The wheel
speeds were very low: 4 km/h, 6k m/h and 8 km/h.

The development of nonlinear finite element techniques has enabled researchers to
develop comprehensive mathematical tire models. Roberts, Tielking, Middleton, Lytton,
and Tseng (1985) developed a finite element model of a free-rolling tire for analyzing the
effect of tire inflation pressure and load on pavement contact stresses. In this model an
assembly of axisymmetric shell elements was positioned along the carcass mid-ply
surface. They used a 10-ply 10.00-20 truck tire. The ply structure in each finite element
was specified separately to define the laminated cord-rubber tire carcass. Specifying the
lamination of each element separately allowed tire construction features such as belts,
sidewall reinforcements, and bead turn-ups to be included in the tire model. The finite
elements were homogeneous orthotropic, with moduli determined by the laminated
carcass. The elements were connected at the node points. The tire model was loaded first
by specifying the inflation pressure. They obtained an axisymmetric solution for the
inflated shape. Prior to applying the axle load they calculated structural stiffness of the
inflated tire. They brought the model into contact with the pavement by specifying the
axle height, which was equivalent to specifying the tire deflection relative to the

pavement. The computer tire model was deflected against a frictionless, flat rigid surface.

DeEskinazi, Yang, and Soedel (1978) analyzed a steel belted radial tire in contact
with a flat surface by the use of the finite element method. They expressed their cross-
sectional profile by two fifth-degree polynomials in x and y. Because of symmetry, only
one octant of the tire was modeled. They modeled the tire configuration by using flat
triangular plate elements. They divided the radial profile into four zones; each zone was
assumed to be homogeneous. Laminate properties were included in the formulation of
the stiffness matrix of the triangular plate element. They considered inflation of the tire
and vertical loading in the analysis, and they discussed the deflections of the tire and the
effect of bending material properties.

Trinko (1984) modeled a steel-belted radial tire subjected to inflation pressure and
footprint loading using laminated, geometrically nonlinear finite elements of the
MSC/NASTRAN program. He used a one-quarter model to take advantage of symmetry
with respect to the meridional plane and a plane perpendicular to the axis of rotation. The
analyzed tire had two steel belts and two polyester radial plies. He divided the 1800
segment into 11 circumferential sections, and smaller subsections were located in and
near the eventual footprint. He modeled rubber portions using solid eight-noded
hexahedrons and a limited number of six-noded pentahedrons; and he modeled the cord-
reinforced layers using four-noded quadrilateral membrane elements. Two belts were
described by one layer of membrane plate element.

Ridha, Satyamurthy, Hirschfelt, and Holle (1985a) discussed the contact loading of a
rubber disk by using three-dimensional isoparametric elements. They analyzed the contact
of a deformable body loaded against rigid surfaces; the body was a 25 mm-thick rubber

disk modeled by lZ-noded-three-dimensional isoparametric elements. The inflation

pressure and footprint loads were applied on the disk model. In the contact area, both
frictionless and no-slip results were discussed. The computed load-deflection responses
show good agreement with measured results. The shape and size of the footprint contact
were also presented and exhibited good agreement between the computed and measured
results. Contact pressures were calculated from the contact force by using the principle of
static equivalency of the two force systems. Ridha, Satyamurthy, Hirschfelt, and Holle
(1985b) used the same approach to model a homogeneous pneumatic tire subjected to
inflation pressures and footprint loads. The computed shapes and sizes of the footprint
area at different levels of rim deflection are shown to be in good agreement with
experimental results.

Tanner (1996) modeled the Space Shuttle nose-gear tire using a two-dimensional
laminated anisotropic shell theory, in which the effects of variation in material and
geometric parameters, transverse shear deformation, and geometric nonlinearities were
included. In this model he developed a contact algorithm with static friction effects that
predict tire response to combined inflation-pressure and static vertical-loading conditions.
He also carried out experimental measurements on the Space Shuttle orbiter nose-gear
tire to define its response to combined inflation-pressure and static vertical-loading
conditions. The predicted normal and tangential contact load intensity distributions were
in fair to good agreement with experimental measurements. This author notes that
additional experimental and analytical studies are needed.

Faria, et al. (1989, 1992) and Oden, et a1. (1988) derived a steady-state formulation
for the rolling contact problem with friction. For steady motion the explicit time

dependency was removed from the problem. In this analysis the motion of the structure

10

was decomposed into two successive motions, a point in the structure underwent first a
rigid rotation and then moved to the point occupied by the material point at time t. In the
contact area, a nonpenetration condition was applied. They used Coulomb fiction for
stick/slip conditions and a variational formulation for the equilibrium and all boundary
conditions. F aria analyzed a viscoelastic spinning cylinder and a passenger tire. In the tire
model, the cross section of the tire was represented by a combination of layered
anisotropic thick-shell elements in the carcass and belts and isotropic bricks in the tread
region. For the contact area (footprint) he used a friction coefficient of 0.5 1. In Oden’s
study, a viscoelastic rubber cylinder was analyzed as a numerical example.

Mousseau and Clark (1994) used a ring on an elastic foundation model as a pneumatic
tire analog for two-dimensional contact problems. They applied two-node beam elements
on the ring model. The problem considered was one of large-deformation, quasi-static
contact. The tire was first inflated by incrementally increasing the pressure loading. In
this step, the nonlinear solver in the finite element program calculated a series of
equilibrium solutions until the specified inflation pressure was reached. While this
process took place, position and rotation constraints on the hub prevented the tire from
moving. Once inflated, the tire was then pushed onto a flat surface by incrementally
decreasing the wheel center height. Finally, the hub rotational constraint was released
and the longitudinal position of the ground surface was moved incrementally backward.
This motion, along with the friction between the ring elements and ground allowed the
tire to roll over surface. They completed these steps by using the ABAQUS nonlinear
finite element program. Experimental hub force measurements verified the simulation

results, and the experimental results agree with the simulation.

11

2.3 Research Objectives

Much work has been done in tire/pavement interaction analysis, but very little data are
available for a real tire structure rolling over pavement. In this study, actual tire/pavement
contact stresses for steady-state moving conditions are analyzed. To complete this task,
the finite element method was used to model a real truck tire rolling over a rigid
pavement. A three-dimensional tire model with real tire pressure simulated the interaction
between a truck tire and a rigid pavement. Different steady-state moving conditions of
trucks, tire pressure levels, different tire load levels, and different friction models are
considered. The purpose of this work is to find out the effects of these factors on truck
tire/pavements interaction. These analyses provide fundamental data and establish
mechanisms for understanding truck tire/pavement interaction.

Given the complexity of truck tire/pavement interaction analysis, the average static
normal stress of vehicle loads is still used in pavement design. The results of this analysis
provide contact information for a steady—state rolling truck tire. The accuracy of the
method for considering vehicle loads in pavement design can be evaluated, and

improvements to the design procedure can be implemented.

12

Chapter 3

APPLICATION OF THE FINITE ELEMENT ANALYSIS

3.1 Introduction of ABAQUS/Standard code

Based on the characteristics of truck-pavement interaction, the ABAQUS/Standard
finite element code (Hibbitt, Karlsson & Sorensenm Inc. 1998) is chosen for this analysis.
ABAQUS is a general-purpose finite element program.

ABAQUS/Standard provides a specialized analysis capability to model the steady-
state behavior of a cylindrical deformable body rolling along a flat rigid surface. The
capability uses a reference frame that removes the explicit time dependence from the
problem so that a purely spatially dependent analysis can be performed. For an
axisymmetric body traveling at a constant ground velocity and constant angular rolling
velocity, a steady state is possible in a frame that moves at the speed of the ground
velocity but does not spin with the body in the rolling motion. This choice of reference
frame allows the finite element mesh to remain stationary so that only the part of the body
in the contact zone requires fine meshing.

The kinematics of the rolling problem is described in terms of a coordinate frame that
moves along with the ground motion of the body. In this moving frame the rigid body

rotation is described in a spatial or Eulerian manner and the deformation in a material or

13

Lagrangian manner. It is this kinematic description that converts the steady moving
contact field problem into a purely spatially dependent simulation.

In the following sections of Chapter 3, a brief review of ABAQUS/Standard theories
related to this analysis is presented. The information is extracted from the

ABAQUS/ Standard manual (Hibbitt, Karlsson & Sorensenm Inc. 1998.)

3.2 Governing Equations

Let V denote a volume occupied by a part of the body in the current configuration,
and let S be the surface bounding this volume. Let the surface traction at any point be t
(force per unit of current area), and let the body force at any point within the volume of
material under consideration be f (per unit of current volume.) Force equilibrium for the

volume is then

jtdS+ jrdv =0
3 v (3-1)

The Cauchy stress matrix 0' at a point of S is defined by
t = n -o (3-2)
where n is the unit outward normal to S at the point. Using this definition, the equation of

(3-1) becomes

[n-odS+ jidV=0 (33)
S V

Applying the Gauss theorem to the surface integral in the equilibrium equation gives

14

a
' quz _ ' dV 3—4
In a ,(ax) a ( )

3'
Since the volume is arbitrary, this equation must apply pointwise in the body, thus

providing the differential equation of translational equilibrium:
3 - o + f = 0 (3-5)
6x

Here the test function can be imagined to be a “virtual“ velocity field, 6v, which is
completely arbitrary except that it must obey any prescribed kinematic constraints and

have sufficient continuity. The “virtual” work rate is

,
j[§'o+f]-6vdV=0 (3-6)

V

a a aov
—- o dV= —. .5 — :
Jléix 6] v Juan”6 v) “(ax

 

)]dV (3-7)

6 66v 33v
Jlgx-p].5vdV = In .o.5vdS— ”63(30611/ -_— JtfivdS— In : (EMV (3-8)

8' P V

Thus, the virtual work statement can be written

jt-ovdS+ jr-ovdsz jo:(§§l)dv (3-9)
S V V ax

The virtual velocity gradient in the current configuration is defined as:

80v

(3x

 

= 5L (3-10)

The gradient can be decomposed in to a symmetric and an antisymmetric part:

15

éL=6D+§Q (3-11)

where

8D 2 sym(5L) = ; (5L + 5LT)

1 (3-12)
59 = asym(8L) = 7 (8L — 5LT)
Since 0 is symmetric,
o : 89 = 0 (3-13)
The virtual work equation in the classical form is obtained as following:
[o : ondv = I8v-tdS+ jov-rdv (3-14)
V S V
where
- 1 .65v 05v
OD = ( +[ P)
2 6x 811

In the equilibrium and virtual work equation, the internal virtual work rate term is
replaced with the integral over the reference volume of the virtual work rate per reference

volume, the equation is written as following:

[a :6ch0 = jtT ~8vdS+ jiT ~6vdV (3-15)
S V

vi.
where dV0 is the natural reference volume, “EC and 8 are any conjugate pairing of material
stress and strain measures, and the superscript T means transpose.
The finite element interpolator can be written in general as
uzNNbvN (3-16)

where NN are interpolation functions that depend on some material coordinate system.

UN are nodal variables.

16

The virtual field, 8v, must be compatible with all kinematic constrains.
5v=NN8vN (3'17)
Now 88 is the virtual rate of material strain associated with 8v. Hence, the interpolation
assumption gives
as = ppm” (3-18)
where [SM is a matrix. It can be written as BN=BN(x, NN) . The equilibrium equation is

approximated as

av” [pp ”211/“ =av~"‘[jl~lf,l,~ -tdS+ IN: -r./V] (3-19)
1*” s l"

srnce the SV are 1ndependent varlables, a system of nonllnear equilibrium equatrons

can be written as:

[13,, may" =jN'; .tdS+ INT, -de (340)
S V

VI]

By taking the variation of the equilibrium, equations can be written:

I(d‘r° :88 + r‘ :d58)dV0 — IdtTBvdS- ItTSVdA' fli' dS— IdfTbvdV — IfT5v1;dS (3'21)
v“ s s v v

Here I = |dV/dV0|, the volume change between the reference and the current volume
occupied by a piece of the structure; Ar = ldS/dSOI is the surface area ratio between the

reference and the current configuration. Using the constitutive theory,

d‘r" = Hsde + g (3 22)

The complete Jacobian matrix is obtained as following:

KMN = [3M 11:0,,th0 + jt° :aNfiMdV” — [NL 0163— [NL -Q’;dv (3-23)
v” v" S V

17

Here

a, za/auN (3-24)
0.8- 6‘32 =0. (3-25)
an”; aNJ=Q§q (3-26)
an”; aNJ=Q§ (3—27)

ABAQUS/Standard generally uses Newton’s method as a numerical technique for solving

the nonlinear equilibrium equations

3.3 Steady —State Transport Analysis
In Figure 3.1 the body is rotating with a constant angular rolling velocity (0 around a
rigid axle T at X0, which in turn rotates with constant angular velocity Q around the fixed

cornering axis 11 through point X,. Then the motion of a particle X at time t consists of a

rigid rolling rotation to position Y,

18

 

 

Figure 3.1 Constant cornering motion

Y = R, - (X - X0) + Xo (3-28)
followed by a deformation to point x, and a subsequent cornering rotation (or precession)
around to position Y. So that

Y =Rs-(x-Xc)+Xc (3-29)
where, R5 is the spinning rotation matrix defined as R5 = exp((bt) andtb is the skew-
symmetric matrix associated with the rotation vector 00 = (0T. The velocity of the particle
then becomes

v=y =Rc-(x-Xc)+ch-ir (3-30)
where Rc the cornering rotation given by Rc = exp(f2t) and fl is the skew-symmetric

matrix associated with the rotation vector . To describe the deformation of the body, we

define a map x(Y,t), which gives the position of point X at time t as a function of its

location Y at time so that x = x(Y,t). It follows that

i=fl€§£+fl (3-31)
6Y 6t 6t
where
%=RS.(x—xo)=mrx(v—x0) (3-32)
Noting that
R, = to - R5 = (0T - Rs (333)
and introducing the circumferential direction
S = T x (Y — X0 )/R (3-34)
where R = [Y - X0] is the radius of a point on the reference body. So that
xszfl-S+%=wR—0x—+g (3-35)
av at as at

where S = S - Y is the distance-measuring coordinate along the streamline. Using this

result. Together with
R, 42.11, =Qri-Rs (3-36)

the velocity of the particle can be written as
5X 0X
v=an -X, +a)R.- +R.-—— 3-37
(y ‘) i 68 ‘ 6t ( )

The acceleration is obtained by a second differentiation and some manipulation:

20

a=(22(nn-I)-(y-Xc)+2(1)o)(2anc~ax+2f2anc-&x
68 at (3-38)

‘7 7

.. .. qZ
+m2R2Rc-a X+20)0)Rc-a X +Rc.° x
as: asat at=

To obtain expression for the velocity and acceleration in the reference frame tired to the

body, using the transformations

v, = R] -y (339)
ar 2 RCT -a (3-40)
It is obtained that
V,=an(x-Xc)+(0Ra—X+Rc-ax (3-41)
aS' 6:

ar =Qz(nn-I)-(x-Xc)+Za)QRnx:§+2§2nx(3x

at (3 49)
+a)2R2 a'x +260R 57X + 5-x
65 3 aSBt 6t 3
For steady-state conditions these expressions reduce to
_ 6X
Vr — an(x-Xc)+toR (3-43)
68
ar —_-Qz(nn-I).(llr-Xc)+201(2Rnx:38C +0)2R2 :8? (3-44)

The first term of right hand side of (3-44) can be identified as the acceleration that

gives rise to centrifugal forces resulting from rotation about 11. Noting that 00R fix is a

measure of velocity, the second term can be identified as the acceleration that gives rise to
Coriolis forces. The last term combines the acceleration that gives rise to Coriolis and

centrifugal forces resulting from rotation about T. When the deformation is uniform

along the circumferential direction, this Coriolis effect vanishes so that the acceleration
gives rise to centrifugal forces only.

The virtual work contribution from the d’Alembert forces is

— jp-ovdv (3-45)
V

élI=—pQZ [(x-Xc)-(nn-I)-5vdV—

6x (35‘, (3-46)

51
2 (OQR nx -5vdV+ a) 22R dV
p Ia p Ias as

and the rate of virtual work becomes

elm =—pQ'-’ Idx-(nn-I)-5vdV—

(3-47)

prQRJnxadx-fivdV-i-pt 022R jadx Md
,, as

05' an

For points on the surface of the deformable body
=an(x-Xc)+60Rax+Rc-ax (3-48)
‘ 08 at

where n is the cornering axis (which must be normal to the rigid surface) and Q is the

cornering angular velocity around n. Assuming that the velocity of a point on the

foundation (or rigid surface) is vR, the relative motion becomes
Vo‘VR=an(x‘xc)+wR:§+Rc':f—VR (3-49)

where r = x-Xc. This equation can be split into normal and tangential components. The

rate of penetration is

h=—n-v=-n-vR —0)Rn-

ax — n- ax (3-50)
6 at

For any point in contact

.axzo
(38

Hence,

61
at

h=-—n.v=-n-vR —n-

which in incremental form reduces to the standard contact condition
Ahzn-(AxR —AxD)

For steady-state conditions

n-AxR =0
AxD =0
Similarly, the rate of slip is
3"=t(1«v=Qt(ll -(nxr)+0)Rta _5X_ta -6x+ta -vR
as at

(3-51)

(3-52)

(3-53)

(3-54)

(355)

where to, (0t=l ,2) are two orthogonal unit vector tangent to the contact surface so that

n = t1 x t2 . For steady-state conditions

6”:0
at

So
y=tu -v=Qtu t(nxr)+(0Rtu-a

Variations in y, yield

(3-56)

(3-5 7)

5y=t -v=§2t(I -(nx6r)+(0Rtu a6681+tu ~6vR (3-58)

(1

To complete the formulation, a relationship between frictional stress and slip velocity
must be developed. A Coulomb friction law is provided for steady-state rolling. The law

assumes that slip occurs if the fiictional stress,

2

t = t, +1: (3-59)
is equal to the critical stress, rem = up , where t, and 12 are shear stresses along ta , p. is

the friction coefficient, and p is the contact pressure. On the other hand, when rm (1

crit ’
no relative motion occurs. The condition of no relative motion is approximated in
ABAQUS by stiff viscous behavior

Ta = xsy; (3-60)
where y; is the tangential slip velocity and K5 is the "stick viscosity," which follows
from the relation

KS =L (3-61)
Ycrit

The allowable viscous slip velocity is defined as a fraction of the circumferential velocity
7...... = ZFMR (3-62)
where Ff is a user-defined slip tolerance.

These expressions contribute to the standard virtual work contribution for slip,

on = jjtuoyudA (3-63)
A

and rate of virtual work for slip,

em = “(drubya + tudbyu )dA (3-64)
A

3.4 Reinforcing Elements in the Analysis

Reinforcing in elements is defined by using REBAR option. Rebars are defined
unaxial reinforcement in solid, membrane, shell, and beam elements. Rebars have
material properties that are distinct from those of the underlying element Rebars can be
defined as individual reinforcing bars in beam and solid elements. Rebars can also be
defined as layers of uniformly spaced reinforcing bars in shell, membrane, and solid
elements. Rebars do not contribute to the mass of the model.

Let g,, i=1, 2, be the element's usual isoparametric coordinates. Let r be an
isoparametric coordinate along the line where the face of the element intersects the plane
of reinforcement, with — l S r $1 in an element. The plane of reinforcement is always

perpendiclar to the element face. The detailed information is shown in Figure 3.2

4
4\t
P

g2

Figure 3.2 Rebar in a solid, two—dimensional element.

 

g1

The volume of integration(AV), and rebar strain (,8) are calculated as

_ Ar (
S Dr

I'

91%

AV 3;)2 tow, (3-65)

 

where
to = 21tx,
A0 = rebar cross-sectional area
So = spacing of rebar
WN = Gauss weight associated with the integration point along the (r) line,

x = x(gi) is position.

3).
6r 6gi 6r

Strain is

1 til2
8 = 1n ~ 3-67

2 ( dlfi ) ( )
where d1 and dlo measure length along the rebar in the current and initial configurations.

For the deformation allowed in these elements,

d12

dlz )2 = cos2 (112, + sin2 alcz, (3-68)
0

(

where or is the orientation for the rebar from the plane of the model,

, 8x 2x ,
Mina” “.6“ (3-69)
Or at 6r (it
is the squared stretch ratio in the r-direction, and A, (A1 = x 1/ x 0 for axisymmetric

elements) is the stretch ratio in the t-direction.

3.5 Major Materials Used in the Tire Model
3.5.] Incompressible Mooney-Rivlin Rubber

It is known that the tread part of a truck tire is made of rubber material. In this
analysis the Mooney-Rivlin model is applied on the tread part. The strain energy density

function is defined in terms of the input constant A, B and Poission‘s ratio, v, as :

W(Ii,12,13) = A(1,-3)+ Bu.-3)+C( 1.-1)+ 003-1)" (3-70)

h

where

D 2 A(5v-2)+ Bun/-5)

2(1-2v) (3'71)

C = 0. 5A + B (3-73)

G = 2 (A + B) = shear modulus of linear of elasticity

I], 12, I3 = strain invariant in terms of the principal stretches

1, = if + if + is" (3-73)
12 = if if + if A." + 232 if (3-74)
13 = xii/lit." (.3-75)

The principal components of Cauchy stress, 0, are given

J0, : /l, 0w
8,1,

(3-76)
The derivation of the constants C and D is completed by considering uniform dilation. For
uniform dilation

xii : xi: 2 43 Z A (3-77)

thus the pressure p, is obtained

all 812 613 (3-78)
The relative volume, V can be defined in terms of the stretches as:

V : ’13 : VOIumem/w

(3-79)

volumeom
For small volumetric deformations, the bulk modulus K, can be defined as the ratio of the

pressure over the volumetric strain as the relative volume approaches unity:

. p
K = , 3-80
1123(V — I) ( )

In the limit, as the stretch approaches 1 the pressure must approach zero. So equation
(3-78) must approach zero as the stretch approaches one. With this condition and

equation (3-80) the constants C and D can be solved.

3.5.2 Orthotropic Composite Material

The material law that relates stresses to stains is defined as:
C = 1""(7, T (3-81)
where T is a transformation matrix, and CL is the constitutive matrix defined in terms of

the material constants of the orthogonal material axes, a, b and c. The inverse of CL is

 

 

defined as
t‘ 1 Vin: _ ca 0 0 0 .
Ea Eh E“
_ vuh l _ 13,, 0 0 0
E. E, E,
_ V... _ ”~- 1 , 0 0 0
CE, : Eu Eh EC 1 (3'82)
0 0 0 g 0 0
(Job
0 0 0 0 1 0
(1,“.
0 0 0 0 0 G1
Note that
Vuh : Vbu , Vat : Va" . VF” : V " (3-83)

E. Et’E. E.’ E. E6

29

Chapter 4

SIMULATION OF TIRE/PAVEMENT INTERACTION AND STATIC ANALYSIS

4.1 Description of Tire/Pavement Modeling

In this analysis, a Goodyear 295/75R22.5 Unisteel G167A Low Profile Radial
smooth tire is modeled. The design maximum load of the tire is 6175 lbs (27479 N) at
l 10 psi for a single wheel and 5675 lbs (25254 N) at 100 psi for each wheel of a dual
wheel. The tread of the tire has 5 plies of steel cord; the sidewall has 1 ply of steel cord.
The overall diameter of the tire is 40.77 inches (inflated); A typical tire structure of this
kind of tire is shown in Figure 4.1. The tire information is provided by Smithers
Scientific Services, Inc. (Akron, Ohio). The rim data required for this tire is obtained
from the Tire and Rim Handbook (1993).

A rigid pavement was considered in this analysis.

4.2 The Finite Element Model

The tire model consists of the following parts: ply, belts, tread cap, sidewall, bead,
and flipper. The axisymmetric finite element meshes of the cross—section of the body are
used as a starting point. The model is discretized with axisymmetric elements. The cross-

section of the tire model is shown in Figure 4.2. The ply and belts are modeled by

30

MOLD TREAD CAP
TREAD BASE

 
 
 
  
 
  

    

NOD'SKID TREAD CUSHION/BELT EDGE
#4 BELT OVERLAY GUMSTRIP
#3 BELT BELT EDGE WEDGE
#2 BELT

# l BELT

BELT EDGE
UNDERLAY GS

SHOULDER WEDGE

    
  
 
   
  

 

C/L INSERT
SIDEWALL

SCUFF RIB
GOODYEAR TIRE & RUBBER CO.
#1 APEX

#3 APEX

PLY

GUM CHAFER
PLY/CHIPPER GS
FLIPPER
LINER CHIPPER
FABRIC CHAFER
BARRIER

BEAD

Figure 4.1 Typical tire structure

31

 

 

335 out 05 mo noncommmeo N4. onE

2.»;

:1..:r_.: :..:<.

 

32

defining rebar in the continuum elements. They are constructed from fiber reinforced
rubber composites. Bead and flipper are modeled as a linear elastic material. The tread
and sidewall are made of rubber. The rubber is modeled as an incompressible hyperelastic
material, and the fiber reinforcement is model as a linear elastic material. To save
computing time, the rim was modeled as a rigid body. For a real truck tire, the rim is
made of alloy. Its modulus is much higher than that of rubber material. It is often modeled
as a rigid body in tire analysis.

A three-dimensional model is created by revolving the cross-section about the
symmetry axis. The axisymmetric elements are converted to three-dimensional continuum
elements during the model generation. Fifty divisions make up the circumferential
direction of the tire. To refine the mesh in the contact zone, a higher density of elements
is chosen in this area. The road is defined as a rigid surface in the three-dimensional
model. The truck tire/pavement model, shown in Figure 4.3, has a total of 16600
elements.

The material properties for the tire model were provided by the Goodyear Company

(Akron, Ohio). They are listed in the Table 4.1, Table 4.2, and Table 4.3.

Table 4.1 Ply and belts reinforcement properties of tire model.

 

 

 

 

Part E, Mpa D
Bar 1 1515800 0.3
Bar 2 1102400 0.3

 

 

 

 

33

L002: 30503805 x02... «iv 0BwE

 

34

Table 4.2 Bead and flipper properties of tire model.

 

 

 

 

Part E, Mpa v
Bead 1515800 0.3
Flipper 1 10240.0 0.3

 

 

 

 

Table 4.3 Rubber properties of tire model.

 

Part

 

 

 

 

 

A, Mpa B, Mpa v
tread 1.1 0.0 0.495
rubber at bead area 5.52 0.0 0.495
Rubber at side wall 3.0 0.0 0.495
Rubber at belts 1.1 0.0 0.495

 

 

 

 

In this analysis a rotating tire is simulated by three steps: (1) applying tire pressure;

(2) applying vertical deflection which is equivalent to load; and (3) rotating the tire with

different speeds and accelerations.

For the static part only the first two steps are applied. In the first step, the pressure is
applied on the inner surface of the tire model. In this analysis three pressure levels are
considered: 90 psi, 100 psi, and 110 psi. In the second step an upward vertical

displacement was applied on the rigid pavement to contact with the tire model for a given

pressure.

35

 

4.3 Validation of the Simulated Tire Model

In this analysis the load-deflection ratio of the tire was used for validating the tire
model. The load-deflection curve, widely used for both numerical and experimental
analysis, is a very important factor in tire/pavement interaction analysis. The deflection of
the tire is defined as the displacement of the tire axis after applying load on it. In static
analysis three tire pressure levels are used: 90 psi, 100 psi and 110 psi. Load-deflection
curves for three pressure levels are presented in Figure 4.4, Figure 4.5, and Figure 4.6.
Experimental results prepared by Goodyear (Akron, Ohio) are presented also. The
differences between FEA results and relative experimental results are increased as
deflection increased. The differences are 21%, 15.7%, and 16.8% at a deflection of 50.8
mm for pressure of 90 psi, 100 psi, and 110 psi. The mass of the tire is 119.38 lbs. The
mass of FEA tire model is 115.79 lbs. Contour maps showing the total deformation at

inflation pressure of 100 psi and different deflections are presented in Figure 4.7 to Figure

4.11.

36

50000

400001

30000 ‘

Load, N

   

200001

10000 p + 90psi FEA

+ 90psi test 1

O 10 20 30 40 50 60
Deflection, mm

Figure 4.4 Load-deflection curves for the experimental analysis and the numerical

analysis (90psi)

50000

30000

Load, N

20000:

  

I0000 . V—o— lOOpsi FEA!
. + lOOpsi test
0 ‘ 7 j::,,
0 10 20 30 40 50 60
Deflection, mm

Figure 4.5 Load-deflection curves for the experimental analysis and the numerical

analysis (lOOpsi)

37

60000 ~

50000

40000 ..

Load, N
i

   

+ l lOpsi FEA

+ l lOpsi test

0 10 20 30 40 50 60
Deflection, mm

Figure 4.6 Load-deflection curves for the experimental analysis and the numerical

analysis (110psi)

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43

4.4 Static Results for the Contact Stress

For the three pressure levels used, summaries of these results are listed in Table 4.4,
Table 4.5 and Table 4.6 for pressures of 90psi, lOOpsi, and llOpsi respectively.
Representative contact stress distributions are presented in Figures 4.12 to Figure 4.28.

For a pressure of 90psi, the absolute maximum normal stress level varies from 0.343
Mpa (defl=4.6 mm) to 1.433 Mpa (defl=49.6 mm). For a pressure of 100 psi, the absolute
maximum normal stress level varies from 0.2938 Mpa (defl=3.9mm) to 1.540 Mpa
(defl=48.9 mm), For a pressure of l 10 psi, the absolute maximum normal stress level is
from 0.0392 Mpa (defl=l .04 mm) to 1.584 Mpa (defl=48.4 mm). For the same pressure
level the following results are found: (1) When deflection increases, the maximum stress
level increases. (2) Contact shapes and sizes are different when deflections are different;
contact areas increase as deflections increase. (3) The maximum normal stress is located
at the center part of the contact area when deflection is small, but the maximum normal
stress occurs in two separate patches which move toward the tire sidewalls when
deflection is large. For the same deflection, a higher pressure gives higher maximum
normal stress level. As mentioned earlier, normal average stress is used in pavement
design. From Tables 4.4 to 4.6, one sees that some average normal stresses are about half
of the corresponding maximum normal stresses. If pressure or deflection is higher, the
ratio of Max/Av. is higher. The data show that error is high at high pressure and

deflection if the average normal stress is used in pavement design.

44

 

 

 

 

 

 

 

 

 

 

 

 

 

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Chapter 5

ROLLING TRUCK TIRE/PAVEMENT INTERACTIONS

5.1 Tire Model

A comprehensive analysis for a steady-state rolling truck tire is completed by a series
of simulations. Different pressures, loads or deflections, speeds, slip angles, braking or _
traction, and frictions are discussed. As mentioned in the Chapter 4, a rigid pavement is
simulated. The purpose of this analysis is to find out the efifects of these factors on the
contact patch between the tire and pavement.

In Chapter 4, static stress distributions were presented for given pressures and
deflections. Based on the deflected tire model, a free rolling tire model was obtained by
applying a translational ground velocity and a corresponding spinning angular velocity on
the deflected tire model. Here Vt represents translational ground velocity and V0,
represents spinning angular velocity. The slip angle, 6, is the angle between the direction
of travel and the plane normal to the axle of the tire. A free rolling model with different
slip angle was obtained by gradually changing the velocity angle of the free rolling model.
A traction or braking model was obtained by changing the moving ground speed of the

free rolling model.

65

5.2 Simulation results

As with the static analysis, three tire pressures were considered: 90psi, lOOpsi, and
llOpsi. The detail information of simulations is listed in Table 5.1, Table 5 .2, and Table
5.3 for the three pressure levels, respectively. For each simulation the contact stress
distributions were obtained. Summaries of the results of these simulations are listed in
Table 5.4, Table 5.5, and Table 5.6 for free rolling, traction or braking, and slip angle
respectively. Figures 5.1 to 5.34 show detailed contact stress plots that are keyed to the

simulation cases summarized in Table 5.4 to 5.6.

66

Table 5.1 Detail information for tire of 90psi

 

Simulation case

Detail information

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P90_Free_l Defl. = 22 mm, Friction = 0.95, v=5 mph, free rolling
P90_Free__2 Defl. = 42 mm, Friction = 0.95 , v=5 mph, free rolling
P90_Free_3 Defl. = 50 mm, Friction = 0.95, v=65 mph, free rolling
P90_Slip_l Defl. = 50 mm, Friction = 1.0, v=65 mph, slip angle =1.50
P90_Slip_2 Defl. = 50 mm, Friction = 1.0, v=65 mph, slip angle =2.40
P90_Slip_3 Defl. = 50 mm, Friction = 1.0, v=65 mph, slip angle =3.00
P90_Slip_4 Defl. = 50 mm, Friction = 0.9, v=65 mph, slip angle =1.5°
P90_Slip_5 Defl. = 50 mm, Friction = 0.9, v=65 mph, slip angle =2.4O
P90_Slip_6 Defl. = 50 mm, Friction = 0.9, v=65 mph, slip angle =3.0O
P90_Tr_l Defl. = 50 mm, Friction = 0.95, Vt=65 mph, V,=70 mph, Traction
P90_Tr_2 Defl. = 50 mm, Friction = 0.95, Vt=65 mph, V,=75 mph, Traction
P90_Bk_1 Defl. = 50 mm, Friction = 0.95, Vt=65 mph, V,=6O mph, Braking
P90_Bk_2 Defl. = 50 mm, Friction = 0.95, Vt=65 mph, Vr=55 mph, Braking

 

 

67

 

Table 5.2 Detail information for tire of 100psi

 

Simulation case

Detail information

 

P1 00_Free_l

Defl. = 29 mm, Friction = 0.95, v=65 mph, free rolling

 

P 1 00_Free_2

Defl. = 50 mm, Friction = 0.95, v=45 mph, free rolling

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P100_Free_3 Defl. = 50 mm, Friction = 0.95, v=65 mph, free rolling
P100_Slip_1 Defl. = 29 mm, Friction = 0.95, v=65 mph, slip angle =0.9U
P100_Slip_2 Defl. = 29 mm, Friction = 0.95, v=65 mph, slip angle =1.3°
P100_Slip_3 Defl. = 29 mm, Friction = 0.95, v=65 mph, slip angle =3.0O
P100_Slip_4 Defl. = 50 mm, Friction = 0.95, v=45 mph, slip angle =1.50
P100_Slip_5 Defl. = 50 mm, Friction = 0.95, v=45 mph, slip angle =3.00
P100_Slip_6 Defl. = 50 mm, Friction = 0.95, v=65 mph, slip angle =09“
P100_Slip_7 Defl. = 50 mm, Friction = 0.95, v=65 mph, slip angle =1 .80
P100_Slip_8 Defl. = 50 mm, Friction = 0.95, v=65 mph, slip angle =3.0O
P100_Tr_1 Defl. = 29 mm, Friction = 0.95, Vt=65 mph, V,=68.5 mph,
Traction
P100_Tr_2 Defl. = 29 mm, Friction = 0.95, Vt=65 mph, V,=74.l mph,
Traction
P100_Tr_3 Defl. = 50 mm, Friction = 0.95, Vt=65 mph, Vr=70 mph, Traction
P100_Tr_4 Defl. = 50 mm, Friction = 0.95, Vt=65 mph, V,=75 mph, Traction

 

 

68

 

 

Table 5.3 Detail information for tire of 110psi

 

Simulation case

Detail information

 

 

 

 

 

 

 

P110_Free__1 Defl. = 50 mm, Friction = 0.95, v=65 mph, free rolling
P110_Slip_1 Defl. = 50 mm, Friction = 0.95, v=65 mph, slip angle =0.9O
P110_Slip_2 Defl. = 50 mm, Friction = 0.95, v=65 mph, slip angle =1.8O
P110_Slip_3 Defl. = 50 mm, Friction = 0.95, v=65 mph, slip angle =3.00
P110_Tr_l Defl. = 50 mm, Friction = 0.95, Vt=65 mph, V,=7O mph, Traction
P110_Tr_2 Defl. = 50 mm, Friction = 0.95, Vt=65 mph, V,=75 mph, Traction

 

 

69

 

 

 

 

 

 

 

 

 

 

 

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124

5.3 Reference Data for a Slow Rolling Tire Test

As mentioned in Chapter 2, de (1994) investigated the vehicle-road surface contact
problem by using an experimental method. In his analysis, a smooth Goodyear tire was
used, and the rolling speed was low-only 5 mph (8 km/h.) Different load levels were
applied. Some typical contact vertical and longitudinal stress distributions from de’s work
are shown in Figures 5.35 through 5.38. These data provide very useful information for

comparison with the numerical results, as discussed in Chapter 6.

Vertical contact stress [MPal

    

0.0
front
Longitudinal
rear I 4 lateral

Figure 5.35 Measured vertical stress distribution of 40000. N (8989 lb.) load, tire
pressure 90 psi (0.62 Mpa,) and speed of 5 mph (de 1994).

Vertical contact stress [MPa]

 

Figure 5.36 Measured vertical stress distribution of 80000. N (17978 lb.) load, tire
pressure 90 psi (0.62 Mpa,) and speed of 5 mph (de 1994).

126

0.2

0.0 ‘

    
 

Longitudinal contact stress [MPatl

front

Longitudinal l4 Lateral

rear

Figure 5.37 Measured longitudinal stress distribution of 40000. N (8989 lb.) load,
tire pressure 90 psi (0.62 Mpa,) and speed of 5 mph (de 1994).

//
6'

Zf/fl/

/”

   
 
   
 
  

f. ////II//- ‘

front

Longitudinal contact stress [MPa]

Longitudinal

Figure 5.38 Measured longitudinal stress distribution of 80000. N (17978 lb.)
load, tire pressure 90 psi (0.62 Mpa,) and speed of 5 mph (de 1994).

127

Chapter 6

DISCUSSION

The Goodyear 295/75R22.5 truck tire is analyzed by using the finite element method.
Different pressures, loads or deflections, speeds, slip angles, braking or traction, and
frictions are considered. These data were presented in Chapter 5. As results of a tire
rolling over a rigid pavement, these data are very interesting and useful for tire/pavement

interaction analysis.

6.1 Free Rolling Tire Model

Results for the free rolling tire model were illustrated in Table 5.4. The pressure
levels were the same as for the static loaded tire model: 90psi, lOOpsi, and 110psi. It was
found that there are some differences between the static results and the free rolling
results. First, the normal stress contours of the free rolling tire model are shifted forward.
Because of the deformation in the leading portion of the contact patch, the vertical
pressures are shifted forward. The centroid of the vertical force does not pass through the
spin axis and therefore rolling resistance is generated. Second, the high stress areas are
moved toward the two sidewalls in the free rolling tire model. High stress areas are

increased in fore-back direction. When the deflection of the rolling tire model is

128

 

increased, this phenomenon is more pronounced. Deformations for the rolling tire model
were larger in the contact areas due to frictional forces between the tire and the pavement.
Sidewall areas carry larger load because they have higher stiffness. So the higher stresses
of the free rolling tire model are shifted to the sidewall areas. Third, the contact areas are
found to be smaller for the free rolling tire model. For the rolling tire, the inertial force of
the rolling tire has an effect on the contact area. When the tire is rolling about its axle, the
centrifugal force is distributed in the tire tread. Since the centrifugal force is proportional
to (02 (a) is the angular velocity of the rolling tire) a higher rotating speed produces a
larger centrifugal force. Similar to increasing tire pressure, increasing the centrifugal
force will increase the tire stiffness. Thus the contact area will be decreased at a given
deflection level. Fourth, the normal maximum stress for the free rolling tire model is
higher than that of the static loaded tire at higher tire pressure levels. At a pressure of 90
psi and a deflection of 49.6 mm, the maximum normal stresses are 1.492 Mpa and 1.481
Mpa for the statically loaded tire and the free rolling tire respectively. At a pressure of
100 psi and a deflection of 48.9 mm, the maximum normal stresses are 1.54 Mpa and
1.628 Mpa for the statically loaded tire and the free rolling tire respectively. At a pressure
of 110 psi and a deflection of 48.4 mm, the maximum normal stresses are 1.584 Mpa and
1.694 Mpa for the statically loaded tire and the free rolling tire, respectively.

It was mentioned earlier, the static average stresses are used in pavement design. Even
for the static loaded tire model, normal stresses are underestimated by the static average.
For the free rolling tire at 100 psi and 110 psi the underestimation is worse. When
deflections of the static tire were close to 50 mm, the ratio between the maximum normal

stresses and the average normal stresses is close 2.0. For a free rolling tire at 100 psi and

129

110 psi the ratio is 2.045 and 2.12. It seems reasonable to use a multiplying factor of 2.2
to modify the average normal stress for the approximate actual normal stress found for
the free rolling model at 100 psi and 110 psi. As discussed earlier, the tire/pavement
model is with smooth tread and rigid pavement. In the real world, tires may have different
types of grooves and pavements may be concrete or asphalt. The detailed tire/pavement
information will change the factor value.

For a pressure level of lOOpsi, two deflections and two velocities were shown. It can
be seen that normal stress, shear stresses, and contact area are increased as the deflections
increased. This is similar to the static results. Results of the static loaded model and the
free rolling model prove that the deflection is a key factor in contact stresses. The results
for 45mph and 65mph show that the normal stress, shear stress, and contact area are
increased as the speeds increased, but it seems that the deflection has more effect on
stress level than the speed does.

By comparing results of the three pressure levels it can be seen that the normal stress
and shear stress were increased as the pressure increased. But the contact areas were
decreased as the pressure increased. This is similar to the static results.

de (1994) investigated the interaction between a rolling tire and the pavement by
using an experimental method. Some of his results were shown in Figure 5.33 to Figure
5.36. The speed of the experiment is quite low, only 5mph. The tire pressure is 90 psi.
Two loading levels were used: 40000 N and 80000 N. For comparison purposes, two free
rolling cases of the FEA model were listed: 18871 N and 40000 N at 5 mph and 90 psi
tire pressure. In de’s experimental results, high normal stresses are located at two sidewall

areas. The maximum normal stress is about 0.75 Mpa and 1.5 Mpa for 40000 N and

130

80000 N, respectively. The maximum normal stress is a little shifted forward. These
results are similar to the FEA simulations discussed previously, although de’s tire model
is not same as the F EA model. Maximum normal stress of the FEA model is 1.42 Mpa for
40000 N vertical load. It is almost twice de’s result at 40000 N. Since the tire structures
are different, the two results can not be compared directly, but they have similar normal
stress distribution and shear distribution profiles. These results do provide a reference for
tire/pavement interaction information. Also, since de used pins to measure the normal
load, it is possible that the highest local loads were missed. One wonders also if inertial
and dynamic effects were properly accounted for in the experimental work, particularly

in the choice of strain gages. Unfortunately, de did not provide tire properties in the form

of load deflection-curve so direct quantitative comparison can not be pursued.

6.2 Traction or Braking Tire Model

Results for the traction or braking tire model were illustrated in Table 5.5. The same
pressure levels as for the static loaded tire model were considered: 90psi, lOOpsi, and
1 lOpsi. Normal stress contours for traction models are shifted forward more than they are
for the free rolling model. The reason for the shifting was discussed in the previous
section. Since the rolling resistance is larger for a traction model, the normal stress
contours are shifted more. For the braking model the normal stress contours are shifted
backward. As mentioned earlier, sidewall areas carry larger load. High stresses for
traction models are also located toward sidewalls. But high stresses for braking models

are located at a small area toward sidewalls.

131

For traction models and braking models, the contact areas are larger and contact
stresses are higher than for the free rolling model. High normal stresses for braking
models are located in smaller areas than for traction models, and the maximum normal
stress is higher than for the traction model. For the braking model at a pressure of 90 psi
and a deflection of 49.6 mm, the maximum normal stress is 2.442 Mpa when the
translation speed is 65 mph and the rotating speed is 55 mph. In this case the tire is
slipping. These data show that traction and braking are much more critical in terms of
stress than is the free rolling model. Braking models are more critical than traction
models. For the braking model the ratio between the maximum normal stress and the
static average stress is 3.24.

For the traction model at a pressure of 100 psi two deflections were considered. As
with the other two models, a higher deflection creates a higher stress and a larger contact
area.

By comparing results for the three pressure levels it can be seen that the normal stress
and shear stress increased as the pressure increased. For the static loaded models and free
rolling models, the contact areas decreased as the pressure increased. But for traction or
braking models, the contact areas don’t have the same trend. Deformations in contact
areas are larger for traction models or braking models. They produce the larger contact
areas for traction models or braking models. The effect of tire pressure on the contact area
is not as obvious as for the static loaded models and free rolling models. Results show
that traction models or braking models are more critical at higher pressure than free

rolling models.

132

In comparison with a free rolling model, the interface shear stresses are quite high.
The maximum interface shear stress for the braking model is 1.609 Mpa. The maximum
shear stress ratio between the braking model and the free model is 5.4. The maximum

shear stress is a very important factor in pavement design.

6.3 Slip-Rolling Tire Model

Results for a slip-rolling tire model were illustrated in Table 5.6. The same pressure
levels as for the static loaded tire model were considered: 90psi, lOOpsi, and 110psi. The
high stress areas are moved toward the sidewall in the slip direction. The contact areas are
smaller than were found for the free rolling model. The maximum normal stresses are
higher than for free rolling models and traction or braking models. For the slip-rolling
model at a pressure of 100 psi, 65 mph, and 3-degree slip angle the maximum normal
stress is as high as 2.964 Mpa. It is 3.72 times the static average normal stress.

In this analysis the slip angle was increased from zero to three degrees. Results show
that a larger slip angle produced higher normal stress and shear stress.

For the slip-rolling model at a pressure of 90 psi, two friction coefficients were
considered: 0.9 and 1.0. Results are slightly different between these two cases. When slip
angles was increased, stresses difference was increased. For a 3-degree case the
maximum normal stresses are 2.546 Mpa and 2.631 Mpa for friction coefficients of 0.9
and 1.0, respectively.

For the slip-rolling model at a pressure of 100 psi, two speeds were considered: 45
mph and 65mph. Results show that the high speed created a little higher stress. The

normal stresses at a 3-degree slip are 2.488 Mpa and 2.964 Mpa for speeds of 45 mph and

133

65 mph respectively. The transverse shear stresses at a 3-degree slip are 0.8206 Mpa and
1.058 Mpa respectively.

For the slip-rolling model at a pressure of 100 psi, two deflections were considered:
29 mm and 49 mm. As for other loading cases, a higher deflection created higher stresses.
The normal stresses at a 3-degree slip are 1.74 Mpa and 2.964 Mpa for deflections of 29
mm and 49 mm respectively. The transverse shear stresses at a 3-degree slip are 0.6956
Mpa and 1.058 Mpa respectively.

For the slip-rolling model at a slip angle of 3 degree the normal stresses for 90 psi,
100 psi, and 110 psi are 2.631 Mpa, 2.964 Mpa, and 2.713 Mpa. In previous tire loading
models, a higher pressure created a higher normal stress. But for the slip-rolling model,
results for 90 psi and 110 psi are quite close. At a pressure of 100 psi the normal stress is
the highest one among the three pressure levels. It shows that a little higher pressure than

the design pressure helps support normal load.

134

Chapter 7

CONCLUSIONS

A tire/pavement interaction is simulated using the finite element method for a truck
tire rolling over a rigid pavement. In this analysis the tire is modeled by using two-
dimensional and three-dimensional elements in which geometric nonlinearities and
anisotropic materials are included. Different pressures, loads or deflections, speeds, slip
angles, braking or traction, and friction are considered. The analysis successfully provides
contact interface stress distributions for a tire rolling over a rigid pavement. The results
show that the finite element method is a very useful and powerful tool for tire/pavement
interaction analysis.

Results for a statically loaded tire describe the response of the tire to inflation
pressure loads, contact loads, and normal static loads against a rigid flat plate. Numerical
results of load-deflection curves are quite close to experimental results obtained from the
literature. Contact shapes and contact areas differ at different load levels. The maximum
normal stress is located at the center of the contact area at lower load; the maximum
normal stress is separated into two parts located around the sidewalls at higher load. The
ratio between the maximum normal stress and average normal stress is as high as 1.9817

for the tire model with a pressure of 90 psi and 50 mm deflection. The data shows that the

135

normal stress levels are seriously underestimated if the average normal stress is used in

pavement design.

Results for a rolling tire describe the truck tire subjected to inflation pressure loads,
normal loads, and contact loads against a rigid pavement under moving conditions
including braking, traction, and slip. Results from this investigation lead to the following
observations and conclusions:

1. For a free rolling tire model, normal stress contours of the free rolling tire model are
shifted forward; high stress areas are moved toward to two sidewalls; high stress areas
are increased in fore-back direction; normal maximum stresses at higher tire pressure
(100 psi and 110 psi) are higher than for the static loaded tire. At a deflection of 50
mm and a speed of 65 mph, maximum normal stresses for free rolling models are
increased by 5.7 % and 6.9 % beyond those for the static loaded model for 100 psi

and 110 psi tire pressure respectively.

IN)

For a free rolling tire model at a deflection of 50 mm and a speed of 65 mph, the ratio
between the maximum normal stress and the average static normal stress is above 2.0
for both 100 psi and 110 psi.

3. For a free rolling tire model, increasing deflection, and speed will produce higher
contact stress; at higher tire pressure (lOOpsi and 110 psi) increasing tire pressure will
produce higher contact stress. Among the three factors, deflection has more effect on
contact stresses and contact areas than the other two. For a tire pressure of 100 psi and
a speed of 65 mph, the maximum normal stress of the free rolling model is increased
by 32.6 % when the deflection is increased from 29 mm to 50 m. At a deflection of

50 mm and a speed of 65 mph, the maximtun normal stress is increased by 9.9%

136

when the tire pressure is increased from 90 psi to 100 psi. The maximum normal
stress is increased 10.7 % when the tire pressure is increased from 90 psi to and 110
psi. For a tire pressure of 100 psi and a deflection of 50 mm, the maximum normal
stress is increased by 2.5 % when the speed is increased from 45 mph to 65 mph.

. Comparisons with de’s experimental work suggest that they have similar normal
stress distribution and shear distribution profiles. These results do provide a reference
for tire/pavement interaction information. Also, since de used pins to measure the
normal load, it is possible that the highest local loads were missed. One wonders also
if inertial and dynamic effects were properly accounted for in the experimental work,
particularly by choice of strain gages. Unfortunately, de did not provide tire properties
such as a load deflection-curve, so direct quantitative comparison can not be pursued.
. For traction or braking tire models, normal stress contours for the traction models are
shifted forward more than for the free rolling model; normal stress contours for the
braking models are shifted backward.

. For traction or braking tire models, contact areas are larger and contact stress are
higher than for the free rolling model. For a tire pressure of 100 psi, a deflection of 50
mm, and a speed of 65 mph, the maximum normal stress for the traction model is
28.7 % higher than that for the free rolling model; the contact area for the traction
model is 4.99 % larger than that for the free rolling model. High normal stresses for
the braking models are located in a smaller area than for traction models. The
maximum normal stress for the braking model is higher than that for traction. At a tire
pressure of 90 psi, a deflection of 50 mm, and a speed of 65 mph, the contact area for

the braking model is 2.3 % smaller than that for the traction model; the normal stress

137

 

for the braking model is 34.8 % higher for the braking model than that for the traction
model.

. For traction or braking tire models, increasing deflection will produce higher contact
stress and larger contact area; increasing pressure will produce higher contact stress.
For a tire pressure of 100 psi and traction at 65 mph, the maximum normal stress for
the traction model is increased by 25.6 % when the deflection is increased from 29
mm to 50 mm; and the contact area is increased by 49.4 % when the deflection is
increased from 29 mm to 50 m. At a deflection of 50 mm and traction at 65 mph,
the maximum stress of the traction model is increased by 4.7 % when the tire pressure
is increased from 100 psi to 110psi.

. The maximum interface shear stress for the braking model is higher than that for
static loaded model and free rolling model. At a tire pressure of 90 psi, a deflection of
50 mm, and a speed of 65 mph, the maximum shear stress ratio between the braking
model and the free model is 5.4.

. For the slip-rolling tire model, high stress areas are moved toward the sidewalls in
the slip direction. The contact areas are smaller than for the free rolling model. The
maximum normal stresses are higher than for free rolling models and traction or
braking models. At a pressure of 100 psi, a deflection of 50 mm, and a speed of 65
mph, the contact area for the slip-rolling model with 30 slip angle is 9.2 % less than
that for the free rolling model. At a tire pressure of 100 psi, a deflection of 50 mm,
and a speed of 65 mph, the maximum normal stresses for a slip-rolling model with 30
slip angle is 28.3 % and 45.1% higher than that for free rolling model and traction

model respectively.

138

10.

11.

12.

For the slip-rolling tire‘ model, a larger slip angle produces higher normal stress and
shear stress. For a tire pressure of 110 psi, a deflection of 50 mm, and a speed of 65
mph, the maximum normal stress of the slip-rolling model is increased by 53.5 %
when the slip angle is increased form 0.90 to 30; and the maximum shear stress of the
slip-rolling model increased by 132.5 % when the slip angle is increased from 0.90 to
3".

For the slip-rolling tire model, increasing deflection, friction, speed, and slip angle
will produce higher contact stress. Among these factors deflection has more effect
than the other parameters do. For a tire pressure of 100 psi, a deflection of 50 mm,
and a speed of 65 mph with a slip angle of 30, the maximum normal stress of the slip-
rolling model is increased 70.3 % when the deflection is increased from 22 mm to 50
mm.

The maximum ratios of max/av. normal stress vary from 1.98 for static case to 1.97
for the free rolling case, and 3.22 for the traction case to 3.29 for the slip rolling case.
These findings suggest that the average stress used in the AASHTO seriously
underestimates the maximum stresses served by pavements, with the result that

highways can be seriously underdesigned.

139

Chapter 8

FUTURE WORK AND RECOMMENDATIONS

For both the automotive industry and highway system designers, truck tire/pavement
contact analysis, a very complicated procedure, is important. Based on the results
obtained in this study, the following recommendations for fiJture research on the analysis
of rolling tire/pavement contact are offered:

1. Consider different road profiles. It is known that unsmooth surfaces induce heavier
load between pavement and vehicle tire. Investigating a rolling tire over unsmooth
pavements will provide very useful data for both pavement design and tire analysis.

2. Consider dynamic loading due to bouncing. It is known that bouncing accelerates
masses, which may produce loads as high as several times actual load. This is a very
critical loading case, but the analysis is complicated. Experimental bouncing data are
needed as input information for tire/pavement interaction analysis.

3. Consider truck tires with typical grooves and less-idealized pavement models. In
general, grooves of a tire created higher contact stresses than a smooth tire. There are
various types of grooves. It is very useful to analyze tires with typical types of grooves

as they roll over asphalt and concrete pavements.

140

4. Consider temperature influences. The tire/pavement interaction is a thermal-
mechanical coupled problem. In real situations, trucks roll over pavements through
many seasons with temperatures that vary greatly. Material properties for tire and
pavement are temperature-dependent. Investigating the effect of temperature on the
tire/pavement contact is important for the automotive industry and for highway
system designers. First, temperature-related materials properties and temperature
change information related to tires rolling over various kinds of pavements should be
obtained by experimental or numerical methods. Second, researchers need to simulate
the tire rolling over pavements under varying conditions such as different
temperatures and periods of rolling time.

5. More experimental investigation is needed to provide data that will guide

computational further efforts.

141

BIBLIOGRAPHY

AASHTO (1993,) Guide for Design of Pavement Structures, The American Association
of State Highway and Transportation Officials, Washington, DC. 2001.

Aliabadi, M. H., and Brebbia, C. A. edited (1993,) Contact Mechanics, Computational
Techniques, Computational Mechanics Publications, Boston, MA.

Altair Computing, Inc. (1995,) HyperMesh Documentation. Altair Computing Inc.,Troy.
MI 48084.

Ashton, J .E., Halpin, J .C., and Petie, RH. (1969,) Primer on Composite Materials
Analysis, Technomic, Stamford, Conn.

Barbin, W. W. (1994,) “Tire Compound Development,” in Rubber Products
Manufacturing Technology (edited by Bhowmick, Hall, and Benarely), Marcel Dekker,
Inc.

Chatti, K. (1992,) Dynamic Analysis of Jointed Concrete Pavements Subjected to Moving
Transient Loads, Ph. D. Dissertation, The University of California, Transportation
Center.

Chen, SS. (1987,) The Response of Multi-layered Systems of Dynamic Surface Loads,
Ph. D. Dissertation, The University of California-Berkeley.

Clark, S. K. (edited) (1981,) Mechanics of Pneumatic Tires, US. Department of
Transportation, National Highway Traffic Safety Administration, Washington, DC 20590.

Cook, R. D., Malkus, D. S., and Plesha, M. E. (1989,) Concepts and Applications of
Finite Element Analysis, John Wiley & Sons, Inc., New York.

de, B. M. (1994,) Measurement of Tyre/Pavement Interface Stresses under Moving Wheel
Loads, Report, Center for Structural Mechanics: University of Pretoria, South Africa.

DeEskinazi, J ., Yang, T. Y., and Soedel, W. (1978,) “Displacements and Stresses
Resulting from Contact of a Steel Belted Radial Tire with a Flat Surface,” Tire Science
and Technology, TSTCA, Vol. 6, No. 1.

Fancher, P. S., et a1. (1986,) Factbook of The Mechanical Properties of The Components
For Single- Unit and Articulated Heavy Trucks, DOT-HS-807125, University of
Michigan, Transportation Research Institute, Ann Arbor, MI.

Faria, L. 0., Bass, J. M., Oden, J. T., and Becker, E. B. (1989,) “A Three-Dimensional
Rolling Contact Model for a Reinforced Rubber Tire,” Tire Science and Technology,
TSTCA, Vol. 17, No. 3.

Faria, L. 0., Oden, J. T., Yavari, B., Tworzydlo, W. W., Bass, J. M., and Becker E. B.
(1992), “Tire Modeling by Finite Element,” Tire Science and Technology, TSTCA, Vol.
20, No. 1.

Federal Highway Administration (1995,) Pavement Primary Response Testing, Federal
Highway Administration, McLean, VA 22101.

Fleming, R. A. and Livingston, D. I. (edited) (1979,) Tire Reinforcement and Tire
Performance, ASTM STP 694, American Society for Testing and Materials.

Ford, T. F. and Charles, R. S. (1988,) Heavy Duty Truck Tire Engineering, Society of
Automotive Engineers, Inc., Warrendale, PA 15096.

Gillespie T. D. (1985,) Heavy Truck Ride, SP-607, SAE the Engineering Resource for
Advancing Mobility, The Thirty—First L. Ray Buckendale Lecture, Society of Automotive
Engineers, Inc.

Gillespie T. D. (1992,) Fundamentals of Vehicle Dynamics, Society of Automotive
Engineers, Warrendale, PA.

Hendrick, J. K., Marlow, M. J ., and Brademeyer, B. (1992,) The Simulation of Vehicle

Dynamic Effects on Road Pavements, Report, FHWA-RD-90-108, US. Department of
Transportation, Federal Highway Administration, November.

Hibbit, Karlsson & Sorensen, INC. (1998,) Abaqus Theory Manual Hibbit, Karlsson &
Sorensen, INC., Pawtucket, RI 02860-4847.

Hibbit, Karlsson & Sorensen, INC. (1998,) Abaqus/Standard User '3 ManuelLHibbit,
Karlsson & Sorensen, INC., Pawtucket, RI 02860-4847.

Huang, Y. H. (1993,) Pavement Analysis and Design, Prentice Hall, Englewood Cliffs,
New Jersey 07632.

Johnson, K. L. (1985,) Contact Mechanics, Cambridge University Press, New York.

Kenny, T. M. (1983,) “Prediction of Contained Air Temperature from SAE Standard
Rolling Resistance Test Data,” SAE Technical Paper Series, 831796.

Kovac, F. J. (1978,) Tire Technology, The Goodyear Tire & Rubber Company, Fifth
Edition, Akron, Ohio.

143

Kulakowski B. T. (edited) (1994,) Vehicle-Road Interaction, ASTM STP 1225, ASTM
Publication Code Number 04-012250-08, American Society for Testing and Materials.

Kuo, C. M., Hall, K. T. and Darter, M. I. (1995,) “A Three -Dimensional Finite Element
Model for Analysis of Concrete Pavement Support,” The Transportation Research Board
75th Annual Meeting, Washington, DC.

Mase, G. E., and Mase G. T. (1993,) Continuum Mechanics for Engineers, CRC Press,
Inc.

Michelin Tire Corporation, Michelin Truck Industrial & Off T he Road Tire Data Book,
Michelin Tire Corporation,

Moore, D. F. (1975,) The Friction of Pneumatic Tyres, Elsevier Scientific Publishing
Company, Amsterdam, New York.

Mousseau, C. W. and Clark, S. K. (1994,) “An Analytical and Experimental Study of a
Tire Rolling Over a Stepped Obstacle at Low Velocity,” Tire Science and T echnology,
TSTCA, Vol. 22, No. 3.

Oden, J. T., Lin , T. L., and Bass, J. M. (1988), “A Finite Element Analysis of the General
Rolling Contact Problem for a Viscoelastic Rubber Cylinder,” Tire Science and
Technology, TSTCA, Vol. 16, No. 1.

Parhizgar, S., Weissman, E. M., and Chen, (1988), “Determination of Stiffness Properties
of Single-Ply Cord-Rubber Composites,” Tire Science and Technology, TSTCA, Vol. 16,
No. 2.

Pottinger, M. G. (1992,) “The Three-Dimensional Contact Patch Stress Field of Solid and
Pneumatic Tires,” Tire Science and Technology, TSTCA, Vol. 20, No. 1.

Pottinger, M. G. and Yager, T. J. (edited) (1986,) The Tire Pavement Interface, ASTM

Jo.--

Special Technical Publication 929, ASTM Publication Code Number 04-929000-27,
American Society for Testing and Materials.

Purdy, J. F. (1963,) Mathematics Underlying the Design oaneumatic Tires, Edwards
Brothers Inc., Ann Arbor, Michigan.

Raous, M., Jean, M., and Moreau, J. J. edited (1995,) Contact Mechanics, Plenum Press,
New York.

Ridha, R. A. and Curtiss, Walter W. (1994,) “Development in Tire Technology,” in
Rubber Products Manufacturing Technology (edited Bhowmick, A. K. et a1), Marcel
Dekker, Inc.

144

Ridha, R. A., Satyamurthy, K., Hirschfelt, L. R., and Holle, R. E. (1985a,) “Contact
Loading of a Rubber Disk,” Tire Science and Technology, TSTCA, Vol.13, No. 1.

Ridha, R. A., Satyamurthy, K., Hirschfelt, L. R., and Holle, R. E. (1985b,) “Finite
Element Modeling of a Homogeneous Pneumatic Tire Subjected to Footprint Loadings,”
Tire Science and Technology, TSTCA, Vol.13, No. 2.

Roberts, F. L., Tielking, J. T., Middleton, D., Lytton, R. L. and Tseng, K. (1985,) The
Effects of Tire Pressures on Flexible Pavement, US. Department of Transportation
Federal Highway Administration, Report, F HWA-TX-85-3 721.

Sebaaly, Peter E. (1992,) “Summary of Tire Testing Data,” Pavement/Materials Program,
Department of Civil Engineering, University of Nevada, Reno, Nevada 89557.

Structural Dynamics Research Corporation (1993,) I-DEAS FEM User ’s Guide,
Structural Dynamics Research Corporation.

Tanner, J. A. (1996,) Computational Methods for Frictional Contact with Applications to
the Space Shuttle Orbiter Nose-Gear Tire, NASA Technical Paper 3573.

Tinko, M. J. (1984,) “Ply and Rubber Stresses and Contact Forces for Loaded Radial
Tire,” Tire Science and Technology, TSTCA, Vol. 11, No. 1-4, January-December, pp.
20-38.

The Tire and Rim Association, Inc., 1993, The Tire and Rim Handbook.

Tseng, N. T. (1987,) “Finite Element Analysis of Freely Rotating Tires,” Tire Science and
Technology, TSTCA, Vol. 15, No. 2, April-June, pp. 134-158.

White, T. D., Zaghloul, S. M., Anderton, G. L., and Smith D. M. (1995,) “Use of a Three-
Dimensional, Dynamic Finite Element Model to Study the Effect of Unbound Layer
Characteristics on Pavement Response to a Moving Aircraft Load,” Transportation
Research Board, 74th Annual Meeting, January 22-28, 1995, Washington, DC.

Zienkiewicz, O. C., and Taylor, R. L. (1989,) The Finite Element Method, McGraw-Hill
Book Company., New York.

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