Highway Design C E 427
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This 16 page Class Notes was uploaded by Charley Wintheiser DVM on Monday October 5, 2015. The Class Notes belongs to C E 427 at California State University - Long Beach taught by Shadi Saadeh in Fall. Since its upload, it has received 19 views. For similar materials see /class/218763/c-e-427-california-state-university-long-beach in Civil Engineering at California State University - Long Beach.
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Date Created: 10/05/15
Design of Rigid Pavement Materials Used in Rigid Pavements Portland Cement Coarse Aggregate Fine Aggregate Water Reinforcing Steel Joints in Concrete Pavements Expansion Joints Associated with expansion occurring with increasing temperature Contraction Joints Associated with contraction occurring with decreasing temperature Hinge Joints To reduce cracking along the center of the highway pavement Construction Joints Placed transversely across the pavement width to provide suitable transition between concrete laid at different times Types of Rigid Highway Pavements Plain Concrete Pavement No temperature steel or dowels for load transfer joints at short distances 10 20 ft Simply Reinforced Concrete Pavement Have dowels for the transfer of traffic loads across joints with joints spaced at large distances 30 100 ft Continuously Reinforced Concrete Pavement Have no transverse joints except construction joints or expansion joints when necessary Pumping of Rigid Pavements The discharge of water and subgrade materials through joints cracks and along the pavement edges Major Design Considerations Reduction of elimination of expansion joints Replace soils that are susceptible to pumping with a nominal thickness of granualr or sandy soils or improve them by stabilization Stabilize the susceptible soil with asphalt or Portland Cement Thickness design of Rigid pavement AASHTO Method Design Considerations Pavement Performance gt Initial Serviceability Index PSIi 45 gtTerminal Serviceability Index PSIt 20 25 Subbase strength Thickness is not less than 6 in and should extend 1 to 3 ft outside the pavement Traffic Similar to flexible pavements ESALS Concrete properties Flexural Strength modulus of rupture at 28 days Drainage Drainage Factor Cd Table 209 PP 1106 Reliability Similar to flexible pavement Thickness design of Rigid pavement AASHTO Method Design Considerations Subgrade Strength Westergaard modulus of the subgrade reaction k Fig 208 PP 1095 textbook The type of the material used in the subbase is important for determination of the effective modulus of the subgrade reaction Fig 209 PP 1097 o k also depend on the erosion of the subbase material loss of support factor Table 205 and Fig 2010 PP 1098 o The presence of bedrock with 10 ft of the subgrade may increase k Fig 2011 PP 1099 Example on calculating k effective modulus of the subgrade reaction Example 212 Computing Effective Modulus ofSu39 grade Reaction for 3 Rigid Pavement Using AASHTCI Method A ti in layer ofcemenbtreated granular material is to he used 15 5uhha5e for a rigid pavement The monthlyr values flquot the roradbed soil resilient modulus and the suhhase elastic reeilientli modulus are given in columns 2 and 3 of l ahle 216 if the melt depth is located 5 ft below the subgrade surface and the projeeted shah thickness is 9 in estirmte the eft ective modulus ef39subgrade reaction uSing the SIITO method Note that this is the example given in the 1993 MSHTO guide Also note that the V l 39 for the modulus of the roadhed and suhhase materialza should be determined as dlStTUSSEd in Chapter 2U and the corresponding values shown in columns 2 and 3 of Table 2L6 should he for the same seasonal period Example on calculating k effective modulus of the subgrade reaction Table 216 Data for and Solution to Examplezl 2 42 4 5 6 V Rondberl a Camp 02k Amwayzquot regammmgz 1 010400 My Suth Mollyu Value lbm mm Fomxdnlmn 0 Manll IllI712 Ewanm Fw21 r J1 F 2112 J1nunry 43000 50000 Hot 0 5 rgbmm 70000 50mm 1100 0 55 Marle 2500 15000 100 036 Apnl 4000 1quot 000 10 07gt Maw 4000 5000 30 0 73 June 7 000 20000 100 000 11 000 2 000 400 000 lug n 7000 20000 400 000 SEpwmbev 7 000 20000 400 050 0mm 7000 20000 400 000 m rmlzcr 4000 15000 230 o 73 Dmmbn 20000 50000 l mo Tau Suhhwv pa Granular Avengt 7 60 u 12 Effective madulus ofsubgmdc rumor k lbJ Corrccmd rm hm ofsuppon la lbm 7n Example on calculating k effective modulus of the sqggrade reaction 1 I L L l periods conszdered A Ins sng ht arm om cmstcncc ofhgdruck within Sal ution Tam 7 l o in mtc Cpth The composite subgmdc I is then dmmmd for cash seasonal perlnd usmg dlc chart In Figure 219 The proccdmc tnvolves he follow mg scqu Step 1 Esumaw L For mmplc in Snpmmbcr renalde modulus MZ 7900 lbln 2and subbzse modulusEsu 20000 Ilaquot12 r 1 Enwrdxechanm Fig auubbasckhlcknzsauf mand draw 2 Erucal line to mxcrscct Lhc subbas elastic mpdulus graph of 20000 lbm atA and um mam n 7000 lbfm m5 2 D m 1 m ng nea C nc upward from L m mm sta the horizonu n dmm m A 135th This point ofimcncction D l thucompas modulus ofsubgmdr rexllnu 44m 117 m Step 2 Adjan urpmmc nf39rnrilvctlwithm IU nfsubgrlldvmnhcc This m7 39 739 16quot 39 r rgt71lquotquot v 39 39 r J d Emu FIguIch llxtmudhcd sml I39L stlxcnrmodlllusluk arms lbin mm t vm llincminmrtcrtthc glan mmpnndiugm tlwdcplhol um ruckl ml bum dxuaubgradc summ 1n luscglsc n y s n The mur 5mm palnt K From Al dr w l mwonul lmu m uncrscu mt mppmpnuu L Ilnu 1 am use me mmnccmn pmm s B u L From 15 dnw I vcmml lmc downwer m dcmnmne llIc ltdjnslud nmdulm ul39mbgmdc mama I Thu u obtaincd lb 50 Ilrm 1 Entur xln 1e values in cnlnmn 5 Example on calculating k effective modulus of the subgrade reaction Step 1 Determine mu drama modulus ufzubgmdz mmon bydcm mming the mgc mmng 5 The stcps Involved am la a c av l 1 Lllosc for cxlhlt va m mm hm m lm msc mm 2112 15 used obtained m 3ch 2 hm l 500 lbm l meal the graph rtprcscnnng m d slab xlmkness at m at pomtA 2 From A Prujc 1 horizontal Ime m dcmmmc he pperl Ntc u 111 ths case n an percent mom Tllcsr valucs an retarded m ml mm 6 afTal Vle 21 r 3 DetarminathcrncnnuasshowlillTlelc 21 6 mus l n 6 4 Using ll mem ll 0106 ablam the cffculve modulus of cubgmda reacnnn From Flgurc 2112 3 500 Ilauni l1 81 H5 1 Sup 4 Adjust 01 Emma modulus nl subgmdc rcllcunn dctcrmmcd in VIC 3 m mmn m he pottlmal log nfsubl asesul pun dun m Emamn usingFlg me 21w Tm LS Cantor given m Table 215 ls mud Slnu he mums camm s of mummama granular mammal Ls 1 and k on lbmF Tll mmcmd modulus ofmhgradc mam is 170 Illm 3 g Ems Luzl my 55 115 an um E Example 213 Dcmgymg a ngm Pavemevn Um the MEMO Mcthud Tm use afrhe hang n dsmmmmml ch m cxaxup c gwcn H mm 21 13 In nus ms Inputvalues For cgmcm 1 mm chart 4mm 2 13 1er H 39cuvc nmdulus of ibg z ndc ruuuon X 1 Wm Mean mnam mndi lus oFvaturc I osu nm 3 Lou ransth coef cmvIJ 32 rnmagc mcmcmu M n The m ucs m Lmd to dcmmmc a wluc um 21 3 1mm hmABCnEF I um on h much lmc u shown m F r l pmpammttels hrwgmentllt c2l u ofdlr Mmh Imc mm dcmmincd m 1gy7ut w 74 mmbn 39lusg PS 4577 my Rquot rcm u 9 pi n 0 mm swam devuuml Ag Cnmnmm Is k1 ESAI a x 10quot 4129 Design Example Figure 2113 Desugn Chan mv Rvgm Pavements Bastd on 05mg Mean mm for Eanh 1mm vamame segmcm n Design Example M Mm mm 214 He Iquot k m for ma HvLmuzs am an Usmg um Va u S M 2m mm v n r m 21 Design Example nmum mu mm mm HMer 5 AM Figure 21 Vaviame 529mm I 13 Damn Chan w ngm Pavcmfnts mm on Using Mean Vaiuzs for E Valanzhuqmm 2 m n hr 0 MAAAJ mm 2m Has an cm 01 Mn mums um an Hung Nun mm on Fem mm Homework Due on Wednesday Dec 9 Solve problem 2013 from your textbook
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