Rigid Pavement Stress Analysis

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1 Rigid Pavement Stress Analysis Dr. Antonis Michael Frederick University Notes Courtesy of Dr. Christos Drakos University of Florida Cause of Stresses in Rigid Pavements Curling Load Friction 1. Curling Stresses Where is the tension zone?

2 1.1 Curling Because of Temperature 1. Curling Because of Moisture 1.3 Curling Because of Shrinkage

3 1.4 Curling Stress of Infinite Plate T+ T ε ε = ε αt T = T ε Assume linear Τ α t = coefficient of thermal expansion due to curling in -direction: due to curling in -direction: 1.5 Bending Stress of Finite Slab L L CEαt T CνEα t T = + (1 ν (1 ν Εα T = t (C νc + (1 ν

4 Correction Factor Chart 1.5 Bending Stress of Finite Slab (cont Maximum Interior Center of Slab Eαt T = (C (1 ν Eα t T = (C (1 ν + νc + νc Edge Midspan Eαt T = C may be x or y, depending on whether C is taken as C x or C y

5 1.6 Temperature Curling Example 5 8 k=00 pci α t =5x10-6 / o F t=0 o F E c =4,000,000 psi ν= Calculate Stresses i. Radius of Relative Stiffness: 3 Eh l = 1 (1 ν k 1/4 ii. Maximum Interior Center of Slab Eαt T = (C (1 ν Eαt T = (C (1 ν + νc + νc

6 1.6 Temperature Curling Example (cont int Eα T = t (C νc + (1 ν int Eα T = t (C νc + (1 ν 1.6 Temperature Curling Example (cont iii. Edge Midspan Eα T = t C

7 1.7 Combined Stresses Curling stresses are high, but usually not considered in the thickness design for the following reasons: 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. Loading Stresses Three ways of determining & δ: Closed form solutions (Westergaard single-wheel Influence charts (Picket & Ray, 1951 multiple-wheel Finite Element (FE solutions.1 Closed-form solutions Westergaard theory.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

8 .1. 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.1.3 Corner Loading δ c c 3P a = 1 h l P = kl 0.6 a l Where: k = modulus of subgrade reaction l = radius of relative stiffness a = load contact radius P = load.1.4 Interior Loading b = a when a 1.74h 0.316P l i = 4log h b = 1.6a + h 0.675h when a<1.74h b P δi = 8kl 1 + a ln π l 1 a l

9 .1.5 Edge Loading δ e e 0.803P l a = 4log h a l 0.431P a = kl l Dual Tires Assume that: L Pd 0.57q Then, area of the equivalent circle: πa a = = 0.57L P qπ d Sd + π ( S 0.6L d L Pd 0.57q 1/

10 .1.7 Dual Tire Example 14 P=10000 lb q=88.4 psi k=100pci S d =14 E c =4,000,000 psi h=10 Calculate stresses..1.7 Dual Tire Example (cont iii. Corner Stress: c 3P a = 1 h l 0.6 iv. Interior Stress: 0.316P l = 4log h b i

11 .1.7 Dual Tire Example (cont v. Edge Stress: 0.803P l a = 4log h a l e 3. Friction Stresses 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. L L/ h What happens to PCC w/ T? Where: γ c =Unit weight of PCC f a =Average friction between slab & foundation

12 Steel Stresses: Reinforcing steel Tie bars Dowels 3.1 Reinforcement Wire fabric or Do Increase L/ t h f Where: A s = Area of required steel per unit width f s = Allowable stress in steel Welded Wire Fabric What does (6 x 1 W8 x W6 mean? Longitudinal Orientation Transverse Wire Reinforcement Institute Guidelines: Minimum wires W4 or D4 (because wires are subjected to bending and tension Minimum spacing 4in (allow for PCC placement and vibration Maximum 1x4 Wire fabric should have end and side laps: Longitudinal: 30*Diam. but no less than 1 Transverse: 0*Diam. but no less than 6 Fabric should extend to about in but no more than 6in from the slab edges

13 3. Tie Bars Placed along the L L L ' faγclh A s = fs L = distance from the longitudinal joint to Length of tie bars Spacing of tie bars µ = allowable bond stress d = bar diameter Many Agencies use

14 4. Joint Opening δ Where: δ = Joint opening α t = Coefficient of thermal contraction ε = Drying shrinkage coefficient L = Slab length C = adjustment factor for subgrade friction Stabilized = Granular =

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