Topic 8 Complete- Rigid Pavement Stress Analysis

Rigid Pavement Stress Analysis

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


1.1 Curling Because of Temperature


1.3 Curling Because of Shrinkage

1.2 Curling Because of Moisture


• σX due to curling in X-direction:

• σX due to curling in Y-direction:

1.4 Curling Stress of Infinite Plate

2∆Tεε t

YXα

==

)2(1∆TE

σ 2X ναt−⋅⋅

=

)2(1∆TEν

σ 2X ναt−⋅⋅⋅

=

Bending occurs in both X & Y directions, solution must be superimposed

Sum of both directions:

)2(1∆TEσX ν

α t−⋅⋅

=

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

T

T+∆T

ε

ε


1.5 Bending Stress of Finite Slab

LX

LY

X

Y

)2(1∆TEC

)2(1∆TEC

σ 2Y

2X

X ναν

να tt

−+

−=

)C(C)2(1

∆TΕσ YX2X νν

αt +−

=

CX & CY = correction factors for finite slab

Correction factors depend on LX/l & LY/l

l = radius of relative stiffnessh = slab thicknessk = modulus of subgrade reaction

1/4

2

3

)k(112Eh

⎥⎦

⎤⎢⎣

⎡−⋅

=ν

l


Correction Factor Chart


• 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∆TE

σ 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


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(112Eh

⎥⎦

⎤⎢⎣

⎡−⋅

=ν

l in 30.57)(200)(0.15)(112

)810(41/4

2

36

=⎥⎦

⎤⎢⎣

⎡−⋅×

=l

σXσY


)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

Have to calculate CX & CY first!

4.7130.57144L

9.8130.57300L

Y

X

==

==

l

l

CX=1.07

CY=0.63

LX/lLY/l


)C(C)2(1

∆TEσ YX2Xint νν

α t +−

=

238.26psi)0.15(0.63)(1.07)0.152(1

)2010)(510(4σ 2

66

Xint =+−

××=

−

)C(C)2(1

∆TEσ XY2Yint νν

α t +−

=

161.74psi)0.15(1.07)(0.63)0.152(1

)2010)(510(4σ 2

66

Yint =+−

××=

−

σXint > σYint Was that expected? Why?

1.6 Temperature Curling Example (cont)


Concrete Tensile Strength≈10% fc’ ≈ 400 psi

iii. Edge Stress @ Midspan Since the critical stress is associated with the longest dimension, omit σY

XX C2∆TEσ tα=

214psi1.072

)2010)(510(4σ

66

X =××

=−

So, a wheel load stress of only 186 psi (400-214) will exceed the tensile stress when thermal and load stresses are combined

25’ Slabs are almost always reinforced

1.6 Temperature Curling Example (cont)


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:


2. Loading StressesThree 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• Shear deformations are small• Slab length is infinite• Slab placed on subgrade consist of discrete springs


2.1.2 Limitations

Interior EdgeCorner

Interior & Edge: Tension @ the bottom

Corner: Tension @ the top

• 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

Where is the tension zone?


2.1.3 Corner Loading

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ll

l

2a0.881.1

kP

δ

2a1

h3P

σ

2c

0.6

2c

Where:k = modulus of subgrade reactionl = radius of relative stiffnessa = load contact radiusP = load

2.1.4 Interior Loading

⎪⎩

⎪⎨⎧

⎪⎭

⎪⎬⎫

⎟⎠⎞

⎜⎝⎛⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛+=

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=

2

2i

2i

a0.673

2a

21

18kP

δ

1.069b

4h

0.316Pσ

lll

l

lnπ

log 0.675hh1.6ab

ab22 −+=

= when a≥1.724h

when a<1.724h


2.1.5 Edge Loading

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

⎥⎦

⎤⎢⎣

⎡−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

ll

ll

a0.821

k0.431P

δ

0.034a

0.666a

4h

0.803Pσ

2e

2e log


2.1.6 Dual Tires

Assume that:

Then, area of the equivalent circle:

Pd = single tire loadPd = P/2

0.5227qL dP≈

( )1/2

ddd

d22

0.5227qPS

qP0.8521

a

L0.6LS0.5227L2a

⎟⎟⎠

⎞⎜⎜⎝

⎛+

×=

−+×=

ππ

π


2.1.7 Dual Tire Example

14”

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

Calculate stresses.

i. Effective Radius:

7.85ina

42)0.5227(88.(5000)(14)

(88.42)0)0.8521(500a

1/2

=

⎟⎠⎞

⎜⎝⎛+=

ππ

ii. Radius of Relative Stiffness:

What do we need first?

1/4

2

3

)k(112Eh

⎥⎦

⎤⎢⎣

⎡−⋅

=ν

l 42.97in)(100)(0.15)(112

)1010(41/4

2

36

=⎥⎦

⎤⎢⎣

⎡−⋅×

=l


iii. Corner Stress:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

0.6

2c2a

1h3P

σl

166.82psi42.97

27.851

103(10,000)

σ0.6

2c =⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

iv. Interior Stress:

⎥⎦

⎤⎢⎣

⎡+⎟⎠⎞

⎜⎝⎛= 1.069b

4h

0.316Pσ 2i

llog a<1.724h 0.675hh1.6ab 22 −+=

7.34in0.675(10)101.6(7.85)b 22 =−+=

130.8psi1.0697.3442.974

1000)0.316(10,0σ 2i =⎥⎦

⎤⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛= log

2.1.7 Dual Tire Example (cont)


v. Edge Stress:

⎥⎦

⎤⎢⎣

⎡−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛= 0.034

a0.666

a4

h0.803P

σ 2e lllog

244psiσ

0.03442.977.85

0.6667.8542.97

4log10

00)0.803(10,0σ

e

2e

=

⎥⎦

⎤⎢⎣

⎡−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

2.1.7 Dual Tire Example (cont)


3. Friction Stresses

L

L/2

hσt

σf

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.

Stresses by subgrade reaction:

What happens to PCC w/ ∆T?

1h2L

γf1hσ cat ⋅⋅⋅⋅=⋅⋅

0ΣFX = Nµ1hσ t ⋅=⋅⋅;

2Lγf

σ cat =

Where:• γc=Unit weight of PCC• fa=Average friction

between slab & foundation


Steel Stresses:• Reinforcing steel• Tie bars• Dowels

• Wire fabric or bar mats used to control cracking• Do NOT increase structural capacity• Increase joint spacing (slab length)

L/2

hσt

σf

1h2L

γf1hσ cat ⋅⋅⋅⋅=⋅⋅

sst fA1hσ =⋅⋅ s

cas 2f

LhγfA =

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

3.1 Reinforcement


3.1.1 Welded Wire FabricWhat does (6 x 12 – W8 x W6) mean?

Spacing Size

6” 6” 6”

12”

12”

Transverse

Long

itudi

nal

Orientation

A=0.08 in2

A=0.06 in2

• 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:


Recommended Spacing: 6x12


3.2 Tie Bars

s

'ca

s fhLγf

A =

L’ = distance from the longitudinal joint to the free edge where no tie bar exists

L’

L’

L’Length of tie bars

µdf

21

t s= µ = allowable bond stressd = bar diameter

Many Agencies use the standard tie bar design to simplify construction

Tie Bars: d = 0.5” ; L = 36” ; Spaced @ 30-40” apart

• Placed along the longitudinal joint to tie the 2 slabs together

Spacing of tie bars

)s(A required Steel of Areaentreinforcembar ofArea

pacingS =


4. Joint Opening

δ

δ ≤ 0.05” OR use dowels

ε)∆TCL(αδ t +=

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

• Stabilized = 0.65• Granular = 0.80

δ2

δ2 ≤ 0.25” to reduce potential for bearing failure

moment arm increases w/ increasing δ2


δ

δ ≤ 0.05” OR use dowels

ε)∆TCL(αδ t +=

4.1 Joint Opening Example

0.8C60∆T

40L100.5ε

105α4

6t

==

=×=

×=−

−

( ) ( )( )0.13δ

100.56010512400.8δ 4-6

=×+×××××= −

Dowels required

/oF

oFft

in/in

Topic 8 Complete- Rigid Pavement Stress Analysis

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