Rigid Pavement Stress Analysis - FIT 8 - Rigid Pavement Stress... · Topic 8 –Rigid Pavement Stress

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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 =