AASHTO Rigid Pavement Design

Transcription

1 AASHTO Rigid Pavement Design Dr. Antonis Michael Frederick University Notes Courtesy of Dr. Christos Drakos University of Florida 1. Introduction Empirical design based on the AASHO road test: Over 200 test sections JPCP (15 spacing) and JRPC (40 spacing) Range of slab thickness: 2.5 to 12.5 inches Subbase type: untreated gravel/sand with plastic fines Subbase thickness; 0 to 9 inches Subgrade soil: siltyclay (A6) Monitored PSI w/ load applications developed regression eqn s Number of load applications: 1,114,000

2 2. General Design Variables Design Period Traffic what changes? ( Reliability Based on functional classification Overall standard deviation (S 0 = ) Performance criteria PSI = PSI 0 PSI t 3. Material Properties 3.1 Effective Modulus of Subgrade Reaction (k) Need to convert subgrade M R to k: Pavement Without Subbase If there is no Subbase, AASHTO suggests: Correlation based on 30in plateload tests k value becomes too high because k=fnc(1/a) More accurate k if plate test was run w/ bigger plates; too expensive & impractical Pavement With Subbase k PCC SUBBASE SUBGRADE If subbase exists, need to determine the BEDROCK

3 3.1.2 Pavement With Subbase (cont.) Example: Subbase thickness=10 Subbase modulus= psi Subgrade M R = psi To get k: Figure Rigid Shallow Depth If bedrock is within 10ft, it will Figure Example: Rigid depth=5 From prev. page: Subgrade M R = psi k = 600 pci

4 3.1.4 Effective Modulus of Subgrade Reaction K eff Equivalent modulus that would result in the same damage if seasonal variations were taken into account (similar to flexible design) u r = ( D 0.39k ) Month k u r _ u r = n u r n xxx yy.y K eff Example Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Subgrade Modulus, M R (psi) 12,000 12,000 7,000 7,000 Subbase Modulus, E SB (psi) Figure Figure Composite k value, k (pci) Rigid Foundation, k (pci) Relative Damage, u r Σu r

5 3.1.5 K eff Example (cont.) _ u r u = n D = 8in r = = Figure Loss of Support (LS) Reduction of k eff by a factor LS to account for Best case scenario,

6 3.1.7 Table for Estimating K eff 20, ,000 Fig Fig EQUATION 15, , Portland Cement Concrete (PCC) Elastic Modulus of Concrete (E c ): Correlated with compressive strength Modulus of Rupture (S c ): Thirdpoint 28 days

7 3.3 Pavement Structure Characteristics Drainage Coefficient (C d ): Quality of drainage & percent time exposed to moisture (Table 12.20) Load Transfer Coefficient (J): Ability to transfer loads across joints and cracks (Table 12.19) Lower J 4. Thickness Design 4.1 Input Variables Modulus of Subgrade Reaction, k eff =70 pci Traffic, W 18 =5 million Design Reliability, R =95% Overall Standard Deviation, S 0 =0.30 PSI =1.7 Elastic Modulus, E c =5,000,000 psi Modulus of Rupture, S c =650 psi Load Transfer Coefficient, J =3.3 Drainage Coefficient, C d =1.0 Use Nomograph (Figures 12.17a&b) or solve equation

8 4.2 Nomograph 4.2 Nomograph

9 4.3 Equation W 18 := Z R := S 0 := 0.3 PSI := 1.7 k := 70 S c := 650 J := 3.3 C d := 1.0 p t := 2.8 E c := D := 4.5 Given ( ) ( Z R S 0 ) 7.35 log D + 1 log W 18 Find( D) = 9.9 PSI log ( ) p t 1 + ( D + 1) 8.46 ( ) log ( ) S c C d D J D E c k 5. Other Design Features 5.1 Slab Length Jointed Plain Concrete Pavement (JPCP) Governed by

10 5.1.2 Jointed Reinforced Concrete Pavement (JRPC) Always doweled Use same typical values from before: L L = 6 C α T + ε = L 75ft 4 ( ) ( ) t δ = 894 Guideline JRCP Reinforcement If (when) concrete cracks, steel picks up stress faγclh A s = 2f s Where: A s = Area of required steel per unit width f s = Allowable stress in steel f a = Average friction coefficient between slab and foundation Example tie bars γ c = pci h = 10 in f a = 1.5 f s = 43,000

11 5.1.3 JRCP Reinforcement (cont) 5. Design Example Given the following information: Roadbed soil M R : 20,000 psi (December January) 8,000 psi (February March) 15,000 psi (April November) Subbase Information: Loss of Support =0.5 Friction factor =1.5 Thickness =6 inches Elastic Modulus =100,000 psi Design Factors: Design Reliability, R =90% Overall Standard Deviation, S 0 =0.40 PSI =1.5 Traffic =37.9 million ESAL Drainage coefficient =1.0 Shoulders =10ft wide PCC Temperature drop =55 o F

12 5. Design Example (cont) PCC: Elastic Modulus, E c = 4,500,000 psi Modulus of Rupture, S c = 725 psi Limestone rock Indirect Tensile Strength = 500 psi Design a JPCP (w/o dowels) and a JRCP (35ft, w/ dowels). For each pavement determine the slab thickness, joint spacing (for the JCPC), and reinforcement (mesh designation for the JRCP) 5.1 Effective modulus of subgrade reaction Next page