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. General Design Variables Design Period Traffic what changes? ( Reliability Based on functional classification Overall standard deviation (S 0 =0.25 0.35) 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: 1. 2. 3. 3.1.1 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 3.1.2 Pavement With Subbase k PCC SUBBASE SUBGRADE If subbase exists, need to determine the BEDROCK
3.1.2 Pavement With Subbase (cont.) Example: Subbase thickness=10 Subbase modulus= psi Subgrade M R = psi To get k: Figure 12.18 3.1.3 Rigid Foundation @ Shallow Depth If bedrock is within 10ft, it will Figure 12.19 Example: Rigid depth=5 From prev. page: Subgrade M R = psi k = 600 pci
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 = 0.75 0.25 ( D 0.39k ) 3. 42 Month k u r 1 500 33.5 2 450 32.5...... _ u r = n u r n xxx yy.y 3.1.5 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 12.18 Figure 12.19 Composite k value, k (pci) 700 700 400 400 Rigid Foundation, k (pci) Relative Damage, u r 31.84 31.84 43.45 43.45 Σu r 444.41
3.1.5 K eff Example (cont.) _ u r u = n D = 8in r 444.41 = = 37.03 12 Figure 12.20 3.1.6 Loss of Support (LS) Reduction of k eff by a factor LS to account for Best case scenario,
3.1.7 Table for Estimating K eff 20,000 100,000 Fig. 12.18 Fig. 12.19 EQUATION 15,000 100,000 3.2 Portland Cement Concrete (PCC) Elastic Modulus of Concrete (E c ): Correlated with compressive strength Modulus of Rupture (S c ): Thirdpoint loading @ 28 days
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
4.2 Nomograph 4.2 Nomograph
4.3 Equation W 18 := 5000000 Z R := 1.645 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 := 5000000 D := 4.5 Given ( ) ( Z R S 0 ) 7.35 log D + 1 log W 18 Find( D) = 9.9 PSI log 4.5 1.5 + ( ) 0.06 + + 1.624 10 7 4.22 0.32 p t 1 + ( D + 1) 8.46 ( ) log ( ) S c C d D 0.75 1.132 215.63 J D 0.75 18.42 0.25 E c k 5. Other Design Features 5.1 Slab Length 5.1.1 Jointed Plain Concrete Pavement (JPCP) Governed by
5.1.2 Jointed Reinforced Concrete Pavement (JRPC) Always doweled Use same typical values from before: L 0. 25 L = 6 C α T + ε 0. 65 5. 5 10 60 + 1. 0 10 = L 75ft 4 ( ) ( ) t δ = 894 Guideline 5.1.3 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 40 24 tie bars γ c = 0.0868 pci h = 10 in f a = 1.5 f s = 43,000
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
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