Contents 1 Introduction 1 2 Stresses and Strains in Flexible Pavements 2 3 KENLAYER Computer Program 11 4 Stresses and Deflections in Rigid Pavements 19 5 KENSLABS Computer Program 32 6 Traffic Loading and Volume 42 7 Material Characterization 49 8 Drainage Design 58 9 Pavement Performance 63 10 Reliability 70 11 Flexible Pavement Design 84 12 Rigid Pavement Design 91 13 Design of Overlays 102 Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com Chapter 1 Introduction 1-1. The wheel configuration fdr a dual-tandem axle is as follows : ' 8 [3 s "ns ! 9 - 4 48in. ~> 1 The 40 Kip load is applied over 8 tires. Thus, each tire bears 5000 Lbs (22.2 kN) load with 100 Psi (690 kPa) tire pressure. The contact area of each tire, Ac= Load of each tire/tire pressure = 5000/100 = 50 in.2 (8.2 x 10° mm<), The dimension of the contact area is : From Eq.1.7. (Lecture Text) : J L = Jhc/o. 522 = y50/0.5227 = 9.78 in. (248 mm). / . width = 0.6 L = 5.87 in (149 mm.) The most realistic ‘contact area consisting a rectangle and two semi-circles as shown "in following figure 918i, a) CD “5 = H 1 Ki . 3d Each tire imprint,if considered as a rectangular area, should have l a length Of 0.8712 L, or 8.52 in. and a width of 0.6 L, or 5.87 in. -— «Sin. JE — 1-2. Freezing Index = (32 - 24) x 30 +(32 + 3) x 31 + (32 - 14) x 31 + (32 - 16) x 28 + (32 - 22) x 31 + (32 -25) x 30 = 2851 degree days. Yes, this value is likely to be different because the last few days in October and the first few days in May may have mean daily temperatures lower than 32°F, so the degree days for these two months may not be zero. + StudyXY Chapter 2 Stresses and Strains in Flexible Pavements 2-1. Jo bition * FE rd AT 2 =2 - Rom Fg. 2.20 £ : C5) inert = 18 yas i Go = 018% t ke ' % Fron Fo 2.2 R lL 3 oo = 4.0 ee (= = oodect “= ' i oaa, Ge? Ron Fs 2.9 £860, £327 9%) 100 / = ot Ge =o0.0uYy Tr yooh = Bo FRom Fr. 2.40 (5 Jee A 4 © Tre = 0.6089. He prineipsl thorers Gon be aloo baited By Flowing phemuih: Trt 6 Vor 5 (5-0) Grocrol Soon: U7, 0% = DY 4 Nag + (BT) [1 07, 02 = Gur Oc += (ETE (REY | Tar _ 0.80 + 0.0405 PTT ee ) = ESA PEIN ELE ag) 1 0 = 029 4_ T= 0001 4 - Te —_— A. AY (e-r plane). Thre ia so Sheer SARebs in Ft plane FAeefort Uz = (Oz = oo vg © principal Seed L. ee Vig Acol Hvp lacorat : Fon Fp. 2b, pm wil & K:2 > F:oo.5] fw owe Baro: asf A ye Soins , Zo: FLO x (Gr GY] = Lo: ~ 0.0 (0.004 + 0.010) ] . $= 0.212 YE / TT Ee = 2/¢ i-p (a+) = 2/[ Coos ~or(0.22 40.0040] Lo: ovo We J <r: 20h -p(Ge®)] = [ 0.0041 - 0.5 (p22 4 0.011)] Gx comd Ue J co UTE ee y 02:001% , G3: 0.00419 LF 02 Y fy imp yor Ble; Fg x 01nd UE PVR A 9.57 72 J - 2-2. : Given. _ 12 in , f 00 psi | | Pol | —_— vv XN C Woz? Y= 100 pet Ke =0b | P= 035 in zm 0-6 in : E=26c0 °° z=12" Ia / Bouszinesg's stress dighi bution ie valid 526: tO 0c v V2 (1% 26) etl mn - 12 3 = -— 1 wo [1 Griz = 28.444 psi - _% _2UrWE Zz? Or 0% Tz [ 1+27 (2+) + aa” | wo 2(1+e3c712 123 = = | +2703 - L827 4 Zo] zt TORE Tren 7 (ay? = 00294 ps5 _ i : 60 G = 28.406 + 2700294 + — Zo (142404) = 2003 ps T = 2000 §%°F = 19489.34 px Ww, = {Ee 2 (1-9) = FL Ae : — 2( 11-0282; 10D 7 £ ] - / h = gaeeid 2954 7 10%e vad ¥ - 2-3: 10600 Lb 4 Pg Ade ’ or Pla of lochs 0.2 in, gigid plate — L Ae 00s, 2. plate © = j2in => Ae - 7 ew / Abprode E27 7 7 Ae " Leftrcen § plac Cam be Oulonlabed by: We Blip) 2a Wezozin [az fini Mio. ZE, PN Y/Y ea 7 lr -o0Yx930x 6 2 Ae eZ X 0.2 LT CE 3310 pre. J 21200 £6 4. irre oni LL, zsz200 Le bin 7: Prati yar 2 18]5pac. al £2 N A= 10in. Wp = OXI Juoh gree a=6m Fut 331070 Tiro-lafer Lyrdem —> Fommicdn AR rs pe7 fr Bp plade Me 2178 3:2 72 | Me Ee 0223300 Foi Tiga EN LOA ’ G = |. . : a 167 Fe c08, oom Fp 27, Sofee fon be cbdonian Ef = 4 — F =lF200 FI / + StudyXY e/. Soom #6 + P. Looss 100 PY Ao: T° Tree = 500m" - A = /R4 17 . We zo.in ; 4 % wre £ (3230 pei. hist foal el ] Woe LEXUL £3 23310pA° Ko Ex 0.2 x33/0 fa = 7 Fas TE ga T hrs sox Fo = 0.35 Som Ap 2.07, Sor #2 28.35 wad A = 4 A LANCE Zr: S54: 55x 12672 in. wo Ez :33/0 p00; £3200 pa; Ap 2 TO in. v4 2-4. . 10,000 1b Griven | S0' Pst [T1i0 ee E,= 200000 psi [| WoT gin v= 05 : | C0 E, = 16000 psi w=? ) %=0.5 | es? ho=g'n . ! 4:80 psi 10000 _ ' Tat go - |Jdo®oo in a Tego 6.31 I) & Wes LEE FE (Bg. =20, hh 1265) ES . F.=033 ( Eig 247 . ME + Fo +63) § 79000 ¥ 0.33 = co2s in / we ES op ( 8c =20 "a-126,%0 E/E = 10 F c0495 _ Po #63) Zot, se! . Temes CoE pei B/E, =25 Frod425 ‘ E/g, 20 [:0483 = 0.0z44 Vn / | B/E, <0 KZ “oz 9s = 0.153 Toderpokte Ep os Of 0425 (Faz18y E/E, =25 Oc/n =0.153 Oc = ous3 8 sized ps J %/p = 0.785 i Mole The linear joderpolatio ic used get the cgficiont Value between @ chart gic valges The noliice. jderpataric fas beer also Seied ane no Ly differen vite Shown sng Linens interpo Jad ic 9/8 s0'153 se 2 ponds. Wolipeas inderpolat io bese vous wie 3 points oAL-V7 ' . git 7 000 ® oer fo) Corfoc? area A 30 Les : ~~. . AzST0 int . % TA - | a = fet. 6 in. 3 PEN | Sd = 28 in ~ ~ = fe = 4o in. St obo in V/A) FE Ete Y pose 14 = goer Ht ger ‘ E = 1590000 PIO £3 _ eer WTF 7 DAC — £, = pp 000 PIS Hezor Hee oti hort Gre bused on [ol 2e2ef in. tHer, Sar Sol 22d A,X, Se! met be modified Als Je 2 = ox #- sr. in. Ce arid . . a’: 2a = 0.8 in" A= HAL: & FE in. I~ StudyXY Determinsher LomvseSio Sache C, rd Figs, 2.26 a 2.27 CAR Sein; Epc 2 gL 286 Ca: p10 KL Gz p08 Ley roz2(a"Yxlc-¢) = 1168 CBR Ser Juin Toco; Bs 404 [SAE RP A: sez 40.2 L108-312~112) 3 fd Se: fain = a = ..4 Si STAIN LT fon abe lade VARZ ET — Zz 84 Jf ~ 1768 Z pr sels fin a pdf = (G55) 7-1-9) EY 3 Seain Fachr(Fe): ’ Pl a XY fm Aor x 2087 oat £ = VE = 52 s Fe: or Sods feat Fe = Fez Ex Fe 2 LAR XSF Fo. i” ) Ce Ez lerReant Aroste SAhain © € = I, Te = _s00 3 v4 roo am ez ot 0x10" (Git paver Al. = 4 erent Hf loko oy dur Fh ce V4 Arbproct Uncles JH Bmer of one Vi 2224 ( aay , Pont 2). sting Fig. 2.17 cbftohom Fotre Fo fen be dedermineol | Loa» mié | Lonprné LomDING lop snla (1). Soe: dBecls Lon figu- i Tom!" > @> ator | E/E so So | so so Afa__ | 0.635 | 0.635 | 0.625 | 0.625 alm) | a4 2.6 | 12.4 2.6 rl) | 0 1 40 66.2 | 28 52” 22: 0.20 Jos |o.31 Feonl = 1.2¢ a _ so0x R6 i . we BEF 2 BET 2.0583 = / Sree + Cf tos ale oss in. 2-6 4opo0 |b & ren 160 pi NEN BR ] E,=4o0000 psi i E=° 2,205 57& un BRA SN Sa E.= 20600 pe noes 23 i'n | | C98... A : E;=técoo ps : Vy 70% vo _ [Fuers in a=, arose OE! EL £E (Forze K E- ze 3 =) koe = 2 ES I, 2 A ha. 23 04 ian) —- H = ho ETC pon Co a . ye PN 23 = TC 0o007s (tension) pi-zEl Fr 2.31¢) 2 If use Tale 23 221-RrR] = 286779 = _@ (Rpi-2z1) ; I Er =z Ne big difference betes: _— iso Ep Using Table from Wing Foooro | 2 Frgieres . = —bcoo yrs (ler sien > : whew P= o0&. Table 2:3 G2: ~ Grr = $ (222 BL.) (zz:-pe2z vm0wlsq = qgoxo digg 22 Ga. = 0us © © -o0) fk, = 212% /2 = 10.61 PSI Eps Cex bux x _ clog2 J TTY ooo TE. enc Vs 277. : : Given, Ey = - 0000725 I . Ez = o000l062 ~Gooo lb A=921 . “octo 1b bo psi | 152 psi E=<fs0000 _& | 575 in E=7 : - Lo2g7s E,=0 | E,=l0,000 Pe i Cs | =F Er E, e EE, 0.000725 1 4060600 p= Lob 0000725: d0be00 S 7 = 1933 / h/a = 0.624 / . TE, 2 20 Lom Bg. 207 = Bi /fop~ 400660 / 0, = Zoo \ / FT rer TY em Zn ee Ey = @o0/ol From Ee 2.2/, £ =oLE,= go00f3 By Fria) and. error | ;F was Sound E, = FLooo pI because when &, = 3S sso £264 Fo = LIA ITE cn s2g } 5 28.7% _ Gerd keith == S2 = 3.72 i rem FoF z.z/ Z - 35 or E, = 3S x /oose = ER P54 2 PE 2-8 = y Lor Fer the Meaxwel] model DCéd= z= 7 z woo ZC-PORR | zRRTIXFOG oF . zE /e os = 60574 C+ ort) (A) For #riangular Joao. ze ~ d=. Let) = +! Hor Ft=o Ee = Z Since Fhe First Ser of Wy 18 Constant 4 N Independent of ¢ So BoHaminns Superpostan Pra<pk need. anly fe AYphed Jo Fhe Second form Sz o W, = 00879 + 087% x 2 So tat = 00574 + G00t]8 27) o _ pj)o09 Mm. oo z l-5 by jor haversine Jeadmi, = from Eps 2.57 and 2.58 W,= 2°87¢ + cos 7p Ere Sin(ZE>¢ 44 - 5 + ]° = a0§7) = 0.0075 | Zr Sin (028) cates] = oos7p rms (57) = 0.423 Chapter 3 KENLAYER Computer Program 1 (1) NPROB Problem 3.1 2 0 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 2 1 80 1 0 (5) NL NZ ICL NSTD NUNIT 12 (6) TH 0.35 0.35 (7) ER o (8) zc 1 (9) NBOND 3000 3000 (11) E [] (13) LoAD 6 100 (14) Cr cp 1 (16) MR 0 (mn Re 2 2 (25) NOLAY ITENOL 1 0 2 0 (26) LAYNO NCIAY 12 12 (27) zcwonL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 1 (29) RELAX 145 0 (30) eam 0.55 0.6 (31) K2 KO . 0.55 0.6 (31) K2 KO . 9 0 3000 (33) PHI K1 answer : 0.05339 ) 0 3000 (33) PHI K1 f 1 (1) Nero PROBLEM 3.2 1 0 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 2 2 80 5 0 (5) NL NZ ICL NSTD NUNIT 8 (6) TH 0.5 0.5 (7) PR [4] 8 (8) zc 2 N 1 (9) NBOND jer: a / ”, 200000 10000 (11) E dns 7 a zs . 0 (13) LoAD 0. 0Z3%] 7 6.308 80 (14) CR CP N . 1 (16) MR 17. 39% Pse o an r 1 (1) NPROB PROBLEM 3.3 1 0 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 2 1 80 9 0 (5) NL NZ ICL NSTD NUNIT 8 (6) TH 0.5 0.5 (7) PR 8 (8) zC wer: — 2o =-o HS ans 20685 E-o4 1500000 30000 (11) E ~ 2 (13) Loap 0.08722 1, 12.616 100 (14) cr cp 1 (19) NT 60 28 0 0 (20) XW YW XPT + StudyXY 1 (1) NPROB PROBLEM 3.4 1 0 1 1 (3) MATL NDAMA NPY NIG 0.001 (4) DEL 3 2 80 9 0 (5) NL NZ ICL NSTD NUNIT 5.75 23 (6) TH - 0.5 0.5 0.5 (7) PR AASWE? 5.75 28.7501 (8) zC 1 (9) NBOND —-7.259 E-o4 400000 20000 10000 (11) E 0 (13) roap / obs E-03 9.213 150 (14) CR CP 1 (16) Wm 0 (17) mc 2 (1) NPROB PROBLEM 3.5 (a) 2 0 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 4 2 80 9 0 (5) NL NZ ICL NSTD NUNIT 8 2 2 (6) TH 0.45 0.3 0.3 0.4 (7) ER 8 12.0001 (8) zC 1 (9) NBOND 500000 30000 15000 5000 (11) E 0 (13) LoaD 4.5 75 (14) CR CP 1 (16 Ww 0 (17! Rc 3 15 (25) NOLAY ITENOL 2 0 3 0 4 1 (26) LAYNO NCIAY 9 11 13 (27) zCNoL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX R 145 135 135 130 (30) GAM gnswer 0.5 0.6 (31) K2 KO _ 0.5 0.6 (31) K2 KO ~~). oI E-o4 6.2 1110 178 0.8 (31) K2 K3 K4 KO 0 10000 (33) PHI KL 2. 845E —o 4 0 10000 (33) PHI KL 1827 7682 3020 (33) EMIN EMAX K1 PROBLEM 3.5 (b) 2 0 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 3 2 80 9 0 (5) NL NZ ICL NSTD NUNIT 8 4 (6) TH 0.45 0.3 0.4 (7) PR 8 12.0001 (8) zc 1 (9) NBOND 500000 20000 5000 (11) E : 0 (13) LOAD . 4.5 75 (14) CR cP answer. 1 (16) WR = 0 (17 RC — 1 040E-o4 2 15 (25) NOLAY ITENOL — 2 0 3 1 (26) LAYNO NCIAY 2 883 E-04 9 13 (27) zZCNoL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 145 135 130 (30) GAM 0.5 0.6 (31) K2 KO 6.2 1110 178 0.8 (31) K2 K3 K4 KO 10000 10000 (33) PHI KL 1827 7682 3020 (33) EMIN EMAX K1 + StudyXY 1 (1) NPROB PROBLEM 3.6 2 1 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 3 0 80 9 0 (5) NL NZ ICL NSTD NUNIT 4 8 (6) TH 0.4 0.3 0.45 (7) IR 1 (9) NBOND 400000 10000 10000 (11) E 1 (13) LOAD 4.5 75 (14) CR cP 3 (19) NPT 0 13.5 0 0 0 4.5 0 6.75 (20) XW YW XPT 1 15 (25) NOLAY ITENOL 2 [J] (26) LAYNO NCLAY 6 (27) zcNoL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX i 145 135 0 (30) GAM 0.5 0.6 (31) K2 KO 8000 8000 (33) PHI Ki answer: $5.49 Years 1 1 (46) NLBT NLIC 1 (47) LNBT 3 (48) INTC 36500 (49) TNIR 0.0796 3.291 0.854 (50) FTL FT2 FT3 1.365E-09 4.477 (51) FT4 FTS 1 (1) NPROB PROBLEM 3.7 2 1 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 6 0 80 9 0 (5 NL NZ ICL NSTD NUNIT 4 2 2 2 2 (om 0.4 0.3 0.3 0.3 0.3 0.45 (7) ER 1 (9) NBOND 100000 50000 40000 30000 20000 10000 (11) E 1 (13) LOAD 4.5 15 (14) CR CP 3 (19) NPT 0 13.5 0 0 ©0 4.5 0 6.75 (20) XW YW XPT 4 15 (25) NOLAY ITENOL 2 0 3 0 4 [J 5 [2] (26) LAYNO NCLAY 5 7 9 11 (27) zCNoOL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 145 135 135 135 135 130 (30) GaM 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO . 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0 8000 (33) PHI Ki R 0 8000 (33) PHI KI answer: $5.08 Gears 0 8000 (33) PHI KI o 8000 (33) PHI K1 1 1 (46) NLBT NLTC 1 (47) NBT 6 (48) INIC 36500 (49) TNIR .0796 3.291 .854 (50) FT1 FT2 FT3 1.365E-09 4.477 (51) FT4 ETS + StudyXY 1 (1) NPROB PROBLEM 3.8 2 0 1 1 (3) MATL NDAMA NPY NIG 0.001 (4) DEL 3 1 8 9 0 (5 NL Nz ICL NSTD NUNIT 4 8 (6) TH 0.4 0.3 0.45 (7) PR 6 (8 zc > . Lolo) amon gnswer: 2956 E+ od fie 400000 8000 10000 (11) E . 1 (13) LoAD |G e023 Psa 4.5 75 (14) CR CP N 1 (19) wer [0113 Pew [2] 13.5 o 6.75 (20) XW YW XPT ~ 1 15 (25) NOLAY ITENOL —3 083 Pso 2 0 (26) LAYNO NCLAY * 6 (27) zCNoL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 145 135 0 (30) GAM 0.5 0.6 (31) K2 KO 8000 8000 (33) PHI Kl 1 (1) NeroB PROBLEM 3.9 3 0 1 1 (3) MATL NDAMA NPY NIG 0.001 (4) DEL 2 1 80 1 0 (5 NL Nz ICL NSTD NUNIT 8 (6) TH RN - 0.5 0.5 (7) PR GosWEY : 0.0156) m. oo1bS5] mM. , 0 (8) zc - . . 1 (9) nBoND 0.0195) 7. 5 co208é 1m, 0 10000 (11) E - . [J (13) LOAD ; 2020 743, 6 15 (14) CR cP 0.0208] m. eZ >z : 1 (16) MR 0 (an Rc 0 (36) DUR 11 (37) NVL LNV 6 (38) NTRME 0 0.01 0.1 1 10 100 (39) TYME 0.113 70 (41) BETA TEMPREF 0.000002 2.37E-06 3.81E-06 4.38E-06 4.57E-06 0.0000046 (42) CREEP 70 (44) TEMP 1 (1) NPROB PROBLEM 3.10 . 3 1 1 1 (3) MATL NDAMA NPY NIG 0.001 (4) DEL 2 0 80 9 0 (5) NL NZ ICL NSTD NUNIT 8 (6) TH 0.5 0.5 (7) PR 1 (9) NBOND 0 10000 (11) E - o Va 7S 0 as toms gnswey: Z/) oD YE 6 75 (14) CR CP 1 (16) NR 0 (17) RC 0.1 (36) DUR 11 (37) NVL INV 11 (38) NTRME 0.001 0.003 0.01 0.03 0.1 0.3 1 3 10 30 100 (39) TYME + StudyXY