Highway Engineering -II

Chapter 4

Design of Rigid Pavements

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Learning Outcome

� design road pavement structure using various design methods and computer software applications.

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Factors Affecting Design of Rigid Pavements

� Loading

� Wheel load and its repetitions

� Area of contact of wheel

� Location of load with respect to slab

� Properties of subgrade

� Subgrade strength and properties

� Provision of subbase

� Properties of concrete

� Strength

� Modulus of elasticity

� Poisson’s ratio

� Shrinkage properties

� Fatigue behavior

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Factors Affecting Design of R -Pavements

� External Conditions� Temperature changes

� Friction between slab and subgrade

� Joints� Arrangement of joints

� Joints Sealing

� Reinforcement� Quantity of reinforcement

� Continuous reinforcement

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Critical Rigid Pavement Responses

� Critical responses occur at specific locations within the pavement structure � Slab edge

� Slab corner

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Key Rigid Pavement Distresses

� Fatigue cracking (bottom-up)� Fatigue cracking (top-down)� Joint faulting (undoweled and doweled JPCP)� Punchouts (CRCP)

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Fatigue Cracking (Bottom -Up)

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Fatigue Cracking (Bottom -Up)� Critical response is the tensile stress at the bottom of the PCC slab� Location of critical stress is usually at mid-slab location at the bottom of

the slab� Traffic and climatic forces contribute to critical stresses

Plan view

Long. X-sectional view

Outer wheel path is more critical

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Fatigue Cracking (Top -Down)

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Fatigue Cracking (Top -Down)

� Critical response is tensile stress at the top of the PCC slab� Critical location varies with axle configuration� Traffic and climatic forces contribute to this critical response

Critical stress location

Shoulder

Traffic

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Joint Faulting

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Joint Faulting (Undoweled)

� Critical responses are deflections of loaded and unloaded slab

� Critical locations are at slab corners

� Traffic loading, foundation erosion, and presence of moisture contribute to this critical response

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Joint Faulting (Doweled)

� Critical responses and responses locations same as for undoweled slabs

� Dowel-Concrete bearing stresses are used by some researchers

Foundation: Base and Subgrade

Traffic

P

Critical Response Location

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Punchouts (CRCP)

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Punchouts (CRCP)� Critical slab structural response is tensile stress� Critical location is at the top of the slab between two adjacent cracks � Crack spacing, material properties, subgrade friction, and external loads

affect this response

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Sources of Slab Stresses

� Traffic Loads� Thermal Curling� Moisture Warping� Shrinkage from Curing� Contraction and Expansion from Temperature

Changes

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Traffic Induced Stresses and Deflections

� Major source of stresses in pavements� Traffic load creates a bending stress (tensile stress at the

bottom of the slab)� Repeated applications can result in fatigue cracking� Critical location for traffic loading is generally along outside

slab edge. As a rule of thumb, the stress under edge loading isabout twice that of the stress under interior loading.

� The highest deflections in a slab occur when the load isplaced at the corner slab location. Approximately 2.5 to 3times higher than that produced from loading the slab at aninterior location

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Stresses and Strains due to Thermal Loading� Stresses due to daily temperature reversals during the

service life of the pavement (curling stresses),� Stresses due to the temperature conditions at the time

of construction (built-in temperature gradient that is locked into the slab).

Where is tension zone?

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Temperature Induced Curling Stresses� Temperature differentials are usually expressed in terms of a linear

temperature gradient (e.g. 2.0 oF/in) .� Positive (daytime) temp. gradients curl the slab down at the corners. � Negative (nighttime) temp. gradients curl the slab up at the corners

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Temperature Induced Curling Stresses

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Temperature Induced Stresses and Deflections

� Positive gradients produce tensile stresses at the top of the pavement slab� Negative gradients produce tensile stresses at the bottom of the pavement

slab� Magnitude depends on slab properties,

support conditions, and thermal gradient� Temperature differentials are usually

expressed linear temperature gradients� Field studies have shown that

temperature gradients are non-linear� Typically range from 0.5 to 3.0 oF/in

(0.01 to 0.07 oC/mm). � Tests have shown that maximum positive temperature differentials occur

during the daytime in the spring and summer months.

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Moisture and Shrinkage Induced Stresses

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Warping Stresses

� Caused by differences in moisture content between the top and bottom of the slab (moisture causes the slab to expand)

� Greater moisture at top of slab results in downward warping, and vice versa� Moisture contents through slabs in:

� Wet climates - fairly constant

� Dry climates - top is drier

than the bottom

� Difficult to measure strains due to moisture

� Equivalent temperature gradients due to moisture differences are much smaller than those due to temperature and are often neglected

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Drying Shrinkage Stresses� Fresh concrete contains more water than required for hydration� Loss of moisture as concrete cures leads to shrinkage of slab� Shrinkage resisted by friction of the base, which induces the stress

development � Introduction of joints in slab reduces magnitude of shrinkage stresses� CRCP have joint spacing of 3-8 ft, results in punchouts� Factors responsible for shrinkage stresses;

� Water content.

� Type and gradation of aggregate.

� Type and quantity of admixtures.

� Wind conditions.

� Temperature.

� Moisture conditions.

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Temperature Shrinkage Stresses

� Daily and seasonal temperature changes cause PCC slab to expand/contract

� Frictional force between slab and base creates stresses in slab

� Magnitude of stress estimated by subgrade drag formula:

2σC =

cL f γ

where:σc = tensile stress in the concrete

f = coefficient of frictional resistanceL = slab lengthγc = unit weight of PCC

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Typical Coefficient of Friction Values

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Combined Load and Curling Stresses

� Stresses result from traffic loading and climatic forces� Combined stress state determined by superimposing

environmentally related stresses on load-associated stresses

� Load and thermal stresses are usually considered only since it is difficult to assess the stresses from other sources at this time.

� Critical condition results when curling/warping and loading stresses are additive.

Calculating Responses In PCC Pavement

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Complexity of Analysis

� Rigid slab on relatively weaker foundation (high modular ratio),

� Varying slab sizes and the presence of pavement discontinuities in terms of longitudinal and transverse joints,

� Load transfer mechanisms (dowel bars, tie-bars, aggregate interlock),

� Built-in temperature gradients,� Loss of support due to erosion of fine materials,� Full range of expected loading conditions in terms of

mechanical and environmental loading.

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Methods for Structural Analysis

� Closed-form equations� Westergaard’s slab on Winkler or dense liquid

foundation

� Slab on elastic solid foundation

� Finite Element Methods (FEMs)

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Westergaard’s Solutions

� Stress and deflection equations for three loading conditions

� Interior

� Edge

� Corner

� Curling stresses at edge and interior locations� Westergaard’s Assumptions

� Slab is homogeneous, isotropic elastic solid (Fully characterized by E and µ)� Infinite or semi-infinite extent (depending on the loading situation) - No load

transfer

� Subgrade characterized by a set of independent springs known as a Winkler or a dense liquid foundation

� Circular contact area for interior and corner; semicircular or circular contact area for edge

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Westergaard’s Loading Conditions

Corner loadingEdge loading

Interior loading

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Westergaard’s Winkler Idealization� Subgade is represented using a series of independent springs� Modulus of subgrade reaction or k value is used to represent subgrade� k value is obtained from plate load test or from Falling Weight

Deflectometer (FWD)

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Falling Weight Deflectometer

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Radius of Relative Stiffness, l

� Radius of relative stiffness was introduced to measure the stiffness of the slab relative to the subgrade

where,E = PCC modulus of elasticityh = Slab thickness

µ = PCC Poisson’s ratiok = Modulus of subgrade reaction (from plate load test)

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Modulus of Subgrade Reaction

K = p/s

Deflection diagram and soil pressure contour.

Soil pressure contour

K = modulus of subgrade reactionp = contact pressure intensityS = soil settlement

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Interior Stress Westergaard Equations

Equation for Interior stress due to circular load (radius = a)

Equation for interior deflection due to circular load

γ : Euler's constant (0.577215)

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Edge Stress (Semi Circular) Westergaard Equations

Equation for edge stress due to semi-circular load

Equation for edge deflection due to semi-circular load

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Edge Stress (Circular) Westergaard Equations

Equation for edge stress due to circular load

Equation for edge deflection due to circular load

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Corner Stress Westergaard Equations

Equation for Corner stress due to circular load (radius = a)

Equation for Corner deflection due to circular load

CURLING STRESSES

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Westergaard Curling Stress

Equation for curling stress at slab edge

where,

L, W = Length and Width of PCC SlabC1, C2 = coefficients for interior stress for the desired direction∆T = Temperature differential in degree FαT = Coefficient of thermal expansion

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Bradbury’s Chart

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Solutions to Westergaard Equations

� Manual Methods� Influence Charts (Pickett and Ray)� Computer Programs (WESTY, WESTER)

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Limitations of Westergaard Theory

� Only interior, edge, and corner stresses anddeformations can be calculated

� Shear and frictional forces on slab surface may not benegligible

� Winkler foundation extends only to slab edge (nosupport of base layer beyond slab dimensions)

� Assumes slab is fully supported� Load transfer between joints and cracks is not

considered.

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Problem 1 – Problem Definition Given

Slab thickness = 10 in.PCC modulus of elasticity = 4,000,000 lbf/in 2

Poisson's ratio of PCC = 0.15Wheel load = 9,000 lb; Tire pressure = 80 lbf/ in2

Temperature differential = 25 oFModulus of subgrade reaction = 100 lbf/in 2/inSlab length =15 ft; Slab width = 12 ftPCC thermal coefficient of expansion = 5.5 x 10 -6 /oF

Determine

Stress and deflection responses due to loading and Curling using Westergaard's equations

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Problem 1 – Solution

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AASHTO Rigid Pavement Design Equation

W18 = predicted number of 80 kN (18,000 lb.) ESALs

ZR = standard normal deviateSo= combined standard error of the traffic prediction

and performance predictionD = slab depth (inches)pt = terminal serviceability index

∆PSI = difference between the initial designserviceability index, po, and the design terminal serviceability index, pt

= modulus of rupture of PCC (flexural strength)

Cd = drainage coefficientJ = load transfercoefficient (value depends upon the

load transfer efficiency)Ec = Elastic modulus of PCCk = modulus of subgrade reaction

= PCC compressive strength

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Example� Total ESAL = 5 million� Reliability = 90% (ZR = -1.282)� S0 = 0.40� Pi = 4.5� Pt = 2.5� ∆PSI = 2.0� Elastic modulus (PCC) = 4,000,000 psi� Modulus of rupture = 700 psi� Drainage factor = 1.0� Load transfer coefficient = 3.2� Modulus of subgrade reaction = 200 pci

� Calculated slab thickness = 8.9 in = 9.0 in

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Reinforcement Design

� The maximum tension in the steel across the crack equals the force required to overcome friction between the pavement and the foundation, from the crack to the nearest joint or free edge

� The force is greatest when the crack occurs at the middle of the slab. Thus for greater joint spacing, the more is the reinforcement is required.

� In order to obtain the greatest advantage from a given weight of reinforcement, a major part of this weight should be in longitudinal direction which has the greatest dimensions in the slab. The longitudinal bars should thus be heavier and more closely spaced than the cross bars.

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Reinforcement Design � Design Equation for longitudinal and transverse steel.

� A = area of steel in cm2 required per m width or length of slab� L = distance in m between transverse joints (for longitudinal steel) or free

longitudinal joints (for transverse steel)� W = weight of slab in kg/m2

� f = coefficient of friction� Ss = allowable working stress in steel (kg/cm2) (50-60% of yield strength)

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Example

� Design the reinforcement of a cement concrete slab 200 mm thick, assuming the following.

� Density of concrete = 2.3 gm/cc� f = 1.5� Transverse joint spacing = 15 m� Pavement width = 3.75 m � Working stress in steel = 140 MN/m2 = 1400 kg/cm2

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Solution to Example

� W = weight of slab =

� Area of longitudinal steel =

� Provide 13 mm dia bars at 33 cm centers giving an area of 3.9 cm2

� Area of transverse steel =� Provide 10 mm dia bars at 50 cm spacing = 1.42 cm2

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Spacing of Joints

� Spacing is governed by many factors� Temperature

� Thickness of slab

� Amount of reinforcement provided

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Design of Dowel Bars

� The load transfer capacity of a single dowel bar is given by the following equations.

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Design of Dowel Bars

� For balanced design, for equal capacity in bending and bearing, the length of embedment of the dowel bar is first obtained.

� Knowing z, d, and r, the load transfer capacity of a single dowel bar is determined from equations

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Design of Dowel Bars

� It is assumed that dowel bars up to 1.8l (where l is radius of relative stiffness) in both the directions participate in the load transfer, and the load carried by the dowel bars varies linearly.

� The load on single dowel bar should not exceed its capacity.� This form the basis of finding the interval of placing of the

dowel bars.

n = load transfer capacity required for the dowel system / load transfer capacity of a single dowel bar

n = capacity factor (used to calculate the spacing of dowel bars)

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Example� Design a dowel bar system for a cement concrete slab for the following

conditions.

� Design wheel load = 4100 kg

� Design load transfer = 40 %

� Slab thickness = h = 25 cm

� Joint width = z = 2.0 cm

� Permissible flexural stress in dowel bar = 1400 kg / cm2

� Permissible shear stress in dowel bar = 1000 kg / cm2

� Permissible bearing stress in concrete = 100 kg / cm2

� Modulus of sub-grade reaction = k = 8 kg / cm2/cm

� E = 3 x 105 kg / cm2

� µ = 0.15`

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Solution to Example

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Design of Tie Bars� The area of the tie bar is obtained from the following formula.

� As = area of steel in cm2 per m length of joint� b = distance between joint in question and nearest free edge in m� h = thickness of slab� W = weight of slab in kg/m2

� f = coefficient of friction = 1.5� Ss = allowable working stress in steel (kg/cm2) (50-60% of yield strength)

� L t = length of tie bar (cm)� d = diameter of the bar, cm� Sb = permissible bond stress, (kg/cm2)

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Example