AASHTO Rigid Pavemen Design Dr. Chrisos Drakos 1. Inroducion Empirical design based on he AASHO road es: Over 00 es secions JPCP (15 spacg) and JRPC (40 spacg) Range of slab hickness:.5 o 1.5 ches Subbase ype: unreaed gravel/sand wih plasic fes Subbase hickness; 0 o 9 ches Subgrade soil: silyclay (A6) Moniored PSI w/ load applicaions developed regression eqn s Number of load applicaions: 1,114,000 Universiy of Florida. General Design Variables Design Period Traffic wha changes? (EALF Table 6.7) Reliabiliy Based on funcional classificaion Overall sandard deviaion (S 0 5 35) Performance crieria PSI PSI 0 PSI 3. Maerial Properies 3.1 Effecive Modulus of Subgrade Reacion (k) Need o conver subgrade M R o k: 1. Wihou Subbase. Wih Subbase 3. Shallow bedrock 3.1.1 Pavemen Wihou Subbase If here is no Subbase, AASHTO suggess: M k R 18.8 Correlaion based on 30 plaeload ess k value becomes oo high because kfnc(1/a) More accurae k if plae es was run w/ bigger plaes; oo expensive & impracical 3.1. Pavemen Wih Subbase k PCC SUBBASE SUBGRADE BEDROCK If subbase exiss, need o deerme he composie modulus of subgrade reacion (k) 3.1. Pavemen Wih Subbase (con.) 3.1.3 Rigid Foundaion @ Shallow Deph Example: Subbase hickness10 Subbase modulus psi Subgrade M R psi To ge k: Cu across he E SB Cu across M R o he TL Verically mee oher le Read kvalue k600pci Figure 1.19 If bedrock is wih 10, i will confe he maerial (subgrade) and will produce a higher k. Example: Rigid deph5 From prev. page: Subgrade M R psi k 600 pci k 800pci Figure 1.18
3.1.4 Effecive Modulus of Subgrade Reacion K eff Equivalen modulus ha would resul he same damage if seasonal variaions were aken o accoun (similar o flexible design) 3.1.5 K eff Example Monh Jan Subgrade Modulus, M R (psi) 1,000 Subbase Modulus, E SB (psi) Figure 1.18 Figure 1.19 Rigid Composie k Foundaion, value, k (pci) k (pci) 700 Relaive Damage, u r 31.84 u r 75 5 ( D 39k ) 3. 4 Feb Mar Apr 1,000 7,000 7,000 700 400 400 31.84 43.45 43.45 Monh k u r 1 500 33.5 450 3.5...... _ u r u n r May Jun Jul Aug Sep Oc n xxx yy.y Nov Dec Σu r 444.41 3.1.5 K eff Example (con.) _ ur 444.41 u r 37.03 n 1 D 8 3.1.6 Loss of Suppor (LS) Reducion of k eff by a facor LS o accoun for erosion and/or differenial soil movemen Bes case scenario, LS0 (Slab is full conac wih subbase) Figure 1.0 37 k eff 540pci LS 1.0 Fig1.1 k eff 540pci k acual 170pci 3.1.7 Table for Esimag K eff GRANULAR > 10 * 6 9 5 (*) IF<10 THEN FILL IN (5) Fig. 1.19 EQUATION Fig. 1.18 0,000 100,000 1,300 37 3. Porland Cemen Concree (PCC) Elasic Modulus of Concree (E c ): Correlaed wih compressive srengh ' E c 57,000 fc Modulus of Rupure (S c ): Thirdpo loadg @ 8 days L/3 L/3 L/3 15,000 100,000 1,000 43 534 1 44.5 Fig. 1.1 945 480 534 S c
3.3 Pavemen Srucure Characerisics Draage Coefficien (C d ): Qualiy of draage & percen ime exposed o moisure (Table 1.0) Load Transfer Coefficien (J): Abiliy o ransfer loads across jos and cracks (Table 1.19) Lower J beer performance/less conservaive 4. Thickness Design 4.1 Inpu Variables Modulus of Subgrade Reacion, k eff 70 pci Traffic, W 18 5 million Design Reliabiliy, R 95% Overall Sandard Deviaion, S 0 30 PSI 1.7 Elasic Modulus, E c 5,000,000 psi Modulus of Rupure, S c 650 psi Load Transfer Coefficien, J 3.3 Draage Coefficien, C d 1.0 Use Nomograph (Figures 1.17a&b) or solve equaion 4. Nomograph 4. Nomograph E c 5E6 D9.8 J3.3 C d 1.0 74 S c 650 74 Use D10 W 18 5 millon k70 S 0 3 R95% 4.3 Equaion W18 : 5000000 ZR : 1.645 S0 : 3 PSI : 1.7 k : 70 Sc : 650 J : 3.3 Cd : 1.0 p :.8 Ec : 5000000 D : 4.5 ( ZR S0) 7.35 log( D + 1) log W18 Fd( D) 9.9 + 06 PSI 4.5 1.5 + + 1.64 10 7 4. 3 p log Sc Cd D 75 1.13 75 18.4 15.63 J D 5 Ec k 5. Oher Design Feaures 5.1 Slab Lengh Wha does his (lengh) depend on? 5.1.1 Joed Pla Concree Pavemen (JPCP) Governed by jo openg C L α T + ε Where: Jo openg α Coefficien of hermal conracion ε Dryg shrkage coefficien L Slab lengh C adjusmen facor for subgrade fricion For NO dowels, deerme L for 05 L 05 L C( α T 65( 5. 5 10 60 + 1. 0 10 4 L 15 Remember (?): If 05 hen USE dowels 179" ) IF SLAB > 15 USE DOWELS
5.1. Joed Reforced Concree Pavemen (JRPC) Always doweled Remember (?): 5 o LIMIT bearg sress Use same ypical values from before: L 5 L 4 C α T + ε 65 5. 5 10 60 + 1. 0 10 L 75 Guidele 894 Lenghs ypically beween 30 100 40 slab lengh has been found o be OPTIMAL for JRCP 5.1.3 JRCP Reforcemen If (when) concree cracks, seel picks up sress faγ clh Where: A s f A s s Area of required seel per uni widh f s Allowable sress seel f a Average fricion coefficien beween slab and foundaion Example 4 40 ie bars γ c 0868 pci h 10 f a 1.5 f s 43,000 Longiudal: 0868 480 10 1.5 0077 087 Transverse: 053 Table 4.3 selec welded wire fabric 5.1.3 JRCP Reforcemen (con) Longiudal: 087 Transverse: 053 Selec: 6x1 W4.5 x W5.5 5. Design Example he followg formaion: Roadbed soil M R : 0,000 psi (December January) 8,000 psi (February March) 15,000 psi (April November) Subbase Informaion: Loss of Suppor 5 Fricion facor 1.5 Thickness 6 ches Elasic Modulus 100,000 psi Design Facors: Design Reliabiliy, R 90% Overall Sandard Deviaion, S 0 40 PSI 1.5 Traffic 37.9 million ESAL Draage coefficien 1.0 Shoulders 10 wide PCC Temperaure drop 55 o F 5. Design Example (con) PCC: Elasic Modulus, E c 4,500,000 psi Modulus of Rupure, S c 75 psi Limesone rock Indirec Tensile Srengh 500 psi Design a JPCP (w/o dowels) and a JRCP (35, w/ dowels). For each pavemen deerme he slab hickness, jo spacg (for he JCPC), and reforcemen (mesh designaion for he JRCP) 5.1 Effecive modulus of subgrade reacion Nex page 5. JPCP Design 5..1 Slab Lengh for no Dowels (<05 ) C L( α T L C ( α T + ε ) 05 α 3.8 x 10 6 / o F (Table 1.3, Limesone) ε 00045 (Table 1., Indirec Tensile Srengh 500 psi) C 65 (Cemen Treaed) L 05 116.7" 4 C( α T 65(3.8 10 55 + 4.5 10 ) L 9.7 Use SLAB LENGTH 9
5.. Slab Thickness Declare he variables: W 18 : 37900000 C d : 1.0 p : 3 E c : 4500000 Z R : 1.8 PSI : 1.5 S 0 : 4 k : 480 S c : 75 5.3 JRCP Design 5.3.1 Slab Thickness Declare he variables: W 18 : 37900000 E c : 4500000 PSI : 1.5 k : 480 S c : 75 J : 3.9 Table 1.19, No Dowels C d : 1.0 p : 3.0 Z R : 1.8 S 0 : 4 Give an iial esimae: D : 4.5 Solver ieraion: ( Z R S 0 ) 7.35 log( D + 1) log W 18 Fd( D) 13.195 PSI 4.5 1.5 + 06 + + 1.64 10 7 4. 3 p log S c C d ( D 75 1.13) 15.63 J D 75 18.4 5 E c k J :.8 Give an iial esimae: D : 4.5 Solver ieraion: ( Z R S 0 ) 7.35 log( D + 1) log W 18 Fd( D) 11.11 (Table 1.19, Wih Dowels) PSI 4.5 1.5 + 06 + 1.64 10 7 + ( 4. 3 p ) log S c C d D 75 1.13 75 18.4 15.63 J D 5 E c k 5.3. Reforcemen Longiudal: (0868)(11)(35 1)(1.5) 0070 0839 Transverse: 1 (lane)+1 (lane)+10 (shoulder) (0868)(11)(34 1)(1.5) 0068 0816 Fabric: 6 x 1 W4.5 x W8.5