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Rigid Pavement Stress Analysis

Dr. Antonis Michael

Frederick University

Notes Courtesy of Dr. Christos Drakos

University of Florida

Topic 8 Rigid Pavement Stress Analysis

Curling

Load

Friction

Cause of Stresses in Rigid Pavements

Where is the tension zone?

1. Curling Stresses

Topic 8 Rigid Pavement Stress Analysis

1.1 Curling Because of Temperature

Topic 8 Rigid Pavement Stress Analysis

1.3 Curling Because of Shrinkage

1.2 Curling Because of Moisture

Topic 8 Rigid Pavement Stress Analysis

X due to curling in X-direction:

X due to curling in Y-direction:

1.4 Curling Stress of Infinite Plate

2

T tYX

==

Assume linear t = coefficient of thermal expansion

T

T+T

Topic 8 Rigid Pavement Stress Analysis

1.5 Bending Stress of Finite Slab

LX

LYX

Y

)2(1

TEC

)2(1

TEC

2Y

2X

X

tt

+

=

)C(C)2(1

T YX2X

t +

=

Topic 8 Rigid Pavement Stress Analysis

Correction Factor Chart

Topic 8 Rigid Pavement Stress Analysis

Maximum Interior Stress @ Center of Slab

)C(C)2(1

TE

)C(C)2(1

TE

XY2Y

YX2X

t

t

+

=

+

=

Edge Stress @ Midspan

C2

TE t

=

1.5 Bending Stress of Finite Slab (cont)

may be x or y, depending on whether C is taken as Cx or Cy

Topic 8 Rigid Pavement Stress Analysis

1.6 Temperature Curling Example

12

25

8k=200 pcit=5x10

-6 /oF

t=20oFEc=4,000,000 psi=0.15

Calculate Stresses

i. Radius of Relative Stiffness:

1/4

2

3

)k(112

Eh

=

l

X

Y

Topic 8 Rigid Pavement Stress Analysis

)C(C)2(1

TE

)C(C)2(1

TE

XY2Y

YX2X

t

t

+

=

+

=

ii. Maximum Interior Stress @ Center of Slab

Topic 8 Rigid Pavement Stress Analysis

)C(C)2(1

TE YX2Xint

t +

=

)C(C)2(1

TE XY2Yint

t +

=

1.6 Temperature Curling Example (cont)

Topic 8 Rigid Pavement Stress Analysis

iii. Edge Stress @ Midspan

XX C2

TE t

=

1.6 Temperature Curling Example (cont)

Topic 8 Rigid Pavement Stress Analysis

1.7 Combined Stresses

Joints and steel relieve and take care of curling stresses (as long as the cracks are held together by reinforcement and are still able to transfer load they will not affect performance)

Curling stresses add to load stresses during the day and subtract to load stresses during the night

Fatigue principle is based on # of repetitions; curling effect limited compared to load repetitions

Curling stresses are high, but usually not considered in the thickness design for the following reasons:

Topic 8 Rigid Pavement Stress Analysis

2. Loading Stresses

Three ways of determining & : Closed form solutions (Westergaard single-wheel) Influence charts (Picket & Ray, 1951 multiple-wheel) Finite Element (FE) solutions

2.1 Closed-form solutions Westergaard theory

2.1.1 Assumptions

All forces on the surface of the plate are perpendicular to the surface

Slab has uniform cross-section and constant thickness Slab length Slab placed

Topic 8 Rigid Pavement Stress Analysis

2.1.2 Limitations

Only corner loading/edge loading or mid-slab deformation and stresses can be calculated

No discontinuities or voids beneath the slab Developed for single wheel loads

Topic 8 Rigid Pavement Stress Analysis

2.1.3 Corner Loading

=

=

ll

l

2a0.881.1

k

P

2a1

h

3P

2c

0.6

2c

Where:k = modulus of subgrade reaction

l = radius of relative stiffness

a = load contact radiusP = load

2.1.4 Interior Loading

+=

+

=

2

2i

2i

a0.673

2

a

2

11

8k

P

1.069b

4h

0.316P

lll

l

ln

log 0.675hh1.6ab

ab

22 +=

= when a1.724h

when a

Topic 8 Rigid Pavement Stress Analysis

2.1.5 Edge Loading

=

+

=

ll

l

l

a0.821

k

0.431P

0.034a

0.666a

4h

0.803P

2e

2elog

Topic 8 Rigid Pavement Stress Analysis

2.1.6 Dual Tires

Assume that:

Then, area of the equivalent circle:

0.5227qL d

P

( )1/2

ddd

d22

0.5227q

PS

q

P0.8521a

L0.6LS0.5227L2a

+=

+=

Topic 8 Rigid Pavement Stress Analysis

2.1.7 Dual Tire Example

14

P=10000 lbq=88.42 psik=100pciSd=14Ec=4,000,000 psih=10

Calculate stresses.

Topic 8 Rigid Pavement Stress Analysis

iii. Corner Stress:

=

0.6

2c

2a1

h

3P

l

iv. Interior Stress:

+

= 1.069b

4h

0.316P

2i

llog

2.1.7 Dual Tire Example (cont)

Topic 8 Rigid Pavement Stress Analysis

v. Edge Stress:

+

= 0.034a0.666a

4h

0.803P

2el

llog

2.1.7 Dual Tire Example (cont)

Topic 8 Rigid Pavement Stress Analysis

3. Friction Stresses

L

L/2

h

Friction between concrete slab and its foundations induces internal tensile stresses in the concrete. If the slab is reinforced, these stresses are eventually carried by the steel reinforcement.

What happens to PCC w/ T?

Where:

c=Unit weight of PCC

fa=Average friction between slab & foundation

Topic 8 Rigid Pavement Stress Analysis

Steel Stresses: Reinforcing steel Tie bars Dowels

Wire fabric or Do Increase

L/2

ht

f

Where:As = Area of required steel per unit widthfs = Allowable stress in steel

3.1 Reinforcement

Topic 8 Rigid Pavement Stress Analysis

3.1.1 Welded Wire Fabric

What does (6 x 12 W8 x W6) mean?

Transverse

Longitudinal

Orientation

Minimum wires W4 or D4 (because wires are subjected to bending and tension)

Minimum spacing 4in (allow for PCC placement and vibration) Maximum 12x24

Wire fabric should have end and side laps: Longitudinal: 30*Diam. but no less than 12 Transverse: 20*Diam. but no less than 6

Fabric should extend to about 2in but no more than 6in from the slab edges

Wire Reinforcement Institute Guidelines:

Topic 8 Rigid Pavement Stress Analysis

Topic 8 Rigid Pavement Stress Analysis

3.2 Tie Bars

s

'ca

sf

hLfA =

L = distance from the longitudinal joint to

L

L

L

Length of tie bars

= allowable bond stressd = bar diameter

Many Agencies use

Placed along the

Spacing of tie bars

Topic 8 Rigid Pavement Stress Analysis

4. Joint Opening

Where: = Joint openingt = Coefficient of thermal contraction = Drying shrinkage coefficientL = Slab lengthC = adjustment factor for subgrade friction

Stabilized = Granular =