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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

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Page 1: Rigid Pavement Stress Analysis - FIT Staffwebstaff.fit.ac.cy/eng.ma/acet406notes/Topic 8 - Rigid Pavement Stress... · Topic 8 –Rigid Pavement Stress Analysis • σX due to curling

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

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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

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Topic 8 – Rigid Pavement Stress Analysis

• σX due to curling in X-direction:

• σX due to curling in Y-direction:

1.4 Curling Stress of Infinite Plate

2

∆Tεε tYX

α==

• Assume linear ∆Τ• αt = coefficient of thermal expansion

T

T+∆T

ε

ε

Topic 8 – Rigid Pavement Stress Analysis

1.5 Bending Stress of Finite Slab

LX

LYX

Y

)2(1

∆TEC

)2(1

∆TECσ

2Y

2X

αν

ν

α tt

−+

−=

)C(C)2(1

∆TΕσ YX2X ν

ν

αt +−

=

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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

∆TEσ

)C(C)2(1

∆TEσ

XY2Y

YX2X

νν

α

νν

α

t

t

+−

=

+−

=

• Edge Stress @ Midspan

C2

∆TEσ tα=

1.5 Bending Stress of Finite Slab (cont)

σ may be σx or σy, depending on whether C is taken as Cx or Cy

Page 5: Rigid Pavement Stress Analysis - FIT Staffwebstaff.fit.ac.cy/eng.ma/acet406notes/Topic 8 - Rigid Pavement Stress... · Topic 8 –Rigid Pavement Stress Analysis • σX due to curling

Topic 8 – Rigid Pavement Stress Analysis

1.6 Temperature Curling Example

12’

25’

8”k=200 pciαt=5x10

-6 /oF

∆t=20oFEc=4,000,000 psiν=0.15

Calculate Stresses

i. Radius of Relative Stiffness:

1/4

2

3

)k(112

Eh

−⋅=

νl

σX

σY

Topic 8 – Rigid Pavement Stress Analysis

)C(C)2(1

∆TEσ

)C(C)2(1

∆TEσ

XY2Y

YX2X

νν

α

νν

α

t

t

+−

=

+−

=

ii. Maximum Interior Stress @ Center of Slab

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Topic 8 – Rigid Pavement Stress Analysis

)C(C)2(1

∆TEσ YX2Xint ν

ν

α t +−

=

)C(C)2(1

∆TEσ XY2Yint ν

ν

α t +−

=

1.6 Temperature Curling Example (cont)

Topic 8 – Rigid Pavement Stress Analysis

iii. Edge Stress @ Midspan

XX C2

∆TEσ tα=

1.6 Temperature Curling Example (cont)

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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 σ & δ:– 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

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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

2a1

h

3Pσ

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

1.069b

4h

0.316Pσ

lll

l

lnπ

log 0.675hh1.6ab

ab

22 −+=

= when a≥1.724h

when a<1.724h

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Topic 8 – Rigid Pavement Stress Analysis

2.1.5 Edge Loading

−=

+

=

ll

l

l

a0.821

k

0.431Pδ

0.034a

0.666a

4h

0.803Pσ

2e

2e log

Topic 8 – Rigid Pavement Stress Analysis

2.1.6 Dual Tires

Assume that:

Then, area of the equivalent circle:

0.5227qL dP≈

( )1/2

ddd

d22

0.5227q

PS

q

P0.8521a

L0.6LS0.5227L2a

+×=

−+×=

ππ

π

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Topic 8 – Rigid Pavement Stress Analysis

2.1.7 Dual Tire Example

14”

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

Calculate stresses.

Topic 8 – Rigid Pavement Stress Analysis

iii. Corner Stress:

−=

0.6

2c

2a1

h

3Pσ

l

iv. Interior Stress:

+

= 1.069b

4h

0.316Pσ

2i

llog

2.1.7 Dual Tire Example (cont)

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Topic 8 – Rigid Pavement Stress Analysis

v. Edge Stress:

+

= 0.034a

0.666a

4h

0.803Pσ

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:

• γc=Unit weight of PCC

• fa=Average friction between slab & foundation

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Topic 8 – Rigid Pavement Stress Analysis

Steel Stresses:• Reinforcing steel• Tie bars• Dowels

• Wire fabric or • Do • Increase

L/2

hσt

σf

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:

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Topic 8 – Rigid Pavement Stress Analysis

Topic 8 – Rigid Pavement Stress Analysis

3.2 Tie Bars

s

'ca

sf

hLγfA =

L’ = distance from the longitudinal joint to

L’

L’

L’

Length of tie bars

µ = allowable bond stressd = bar diameter

Many Agencies use

• Placed along the

Spacing of tie bars

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Topic 8 – Rigid Pavement Stress Analysis

4. Joint Opening

δ

Where:δ = Joint openingαt = Coefficient of thermal contractionε = Drying shrinkage coefficientL = Slab lengthC = adjustment factor for subgrade friction

• Stabilized = • Granular =