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# TTE 4811 TTE 4811

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###### Class Notes

##### HIST 1200 (History, Angela Bell, Survey of American History Since 1865)

###### SC_Megan Buckallew

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###### Class Notes

##### HIST 1200 (History, Angela Bell, Survey of American History Since 1865)

###### SC_Megan Buckallew

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This 98 page Class Notes was uploaded by Leonor Langosh on Friday September 18, 2015. The Class Notes belongs to TTE 4811 at University of Florida taught by Reynaldo Roque in Fall. Since its upload, it has received 18 views. For similar materials see /class/206881/tte-4811-university-of-florida in Engineering and Tech at University of Florida.

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Date Created: 09/18/15

TTE4811 Topic 9 Summary Fall 2005 Distress modes in PCC Pavements Blowup Expansion problem in hot weather insufficient width of j oint for expansion because of in ltration of incompressive material into the joint space Measured by counting number that exists Corner Break Crack that intersect the joint at a distance less than 6ft Due to load repetitions loss of support poor load transfer thermal curling Faulting on Transverse Joints and Cracks Different of elevation across the joint or crack Cause by loose material under the trailing slab near the joint or crack plus the depression of the leading slab Loose material caused by pumping due to heavy loading Joint Load Transfer System Associated Deterioration Transverse crack happens at a short distance from a transverse joint usually at the end of joint load transfer dowels Due to extensive corrosion or misalignment of dowel system Lane Shoulder Drop off 0r Heave Where there is a difference in elevation between traffic lane and shoulder Outside shoulder settles due to consolidation settlement or pumping of underlying layers Heave of the shoulder may be caused by frost action or swelling soil Dropoff of granular or soil shoulder is generally caused from bowling away of shoulder material from passing trucks Longitudinal Cracks Occur parallel to the centerline of the pavement Due to heavy load loss of foundation support curling and wrapping stresses improper construction Longitudinal Joint Faulting Difference in elevation at the longitudinal joint between two traffic lanes Pumping and Water Bleeding Ejection of material by water through joints or cracks Caused by de ection of slab under moving load TTE4811 Topic 9 Summary Fall 2005 Spalling Transverse and Longitudinal Joint 0r Crack Cracking breaking or chipping of the slab edges within 2ft of the joint or crack Caused by the in ltration of incompressible materials and subsequent expansion or traf c loading May be caused by disintegration of concrete or weak concrete at the joint SpallingCorner Raveling or breakdown of the slab within 2ft of the corner Caused by freezethaw condition D cracking Swell Upward bulge on the pavement surface Usually caused by a frost action or by swelling of soil Transverse and Diagonal Cracks Cracks that are caused by combination of heavy load repetitions and stresses due to temperature gradient moisture gradient and drying and shrinkage Edge Punch out Major structure distress of CRCP Due to loss of aggregate interlock underneath A short longitudinal crack form between two transverse cracks Flexible Pavement Distress Dr Christos Drakos University of Florida 3 gs nk G A T 0 R 66 Topic 2 Flexible Pavement Distress 1 Alligator or Fatigue Cracking Series of interconnecting cracks caused by the fatigue failure of asphalt surface or stabilized base under repeated traffic loading Initiates at Cracks propagate to Occurs only in areas Topic 2 Flexible Pavement Distress 1 Alligator or Fatigue Cracking cont Q m v ft v v s w Moderate w High Traffic g Law ixv N k S 0 zzymm C Ean Stripe Topic 2 Flexible Pavement Distress 1 Alliqator or Fatique Crackinq cont q Topic 2 Flexible Pavement Distress 2 Block Cracking Thermal Crackingl Block cracks divide the asphalt surface into approximately rectangular pieces Blocks range from 1 to 100 ft2 in area 0 Caused mainly 0 Not 0 Caused by Topic 2 Flexible Pavement Distress 2 Block Cracking cont Mauevaxe Adjacent Low Seventy Cracking High Adlacem Moderate Seveiiw Cracking Topic 2 Flexible Pavement Distress 3 Joint Reflection Cracking from Concrete Slab Occur on pavements that have an asphalt surface over a jointed concrete slab Cracks occur over transverse and longitudinal joints were pavement was widened Why Caused by the i I Generally Topic 2 Flexible Pavement Distress 4 LaneZShoulder Dropoff or Heave Difference in elevation between the traffic lane and the shoulder Main Causes Dropoff due to Heave Soil shoulder dropoff due to Topic 2 Flexible Pavement Distress 5 Longitudinal Cracking Longitudinal cracks are running parallel to the pavement centreline while transverse cracks extend across the centreline Main Causes Asphalt Reflective Poor construction of paving lane joint Usually Topic 2 Flexible Pavement Distress 5 Longitudinal Cacking Top Down Cracking naming Topic 2 Flexible Pavement Distress 6 Water Bleeding amp Pumping Water bleeding occurs when water seeps out of joints or cracks or through an excessively porous HMA layer Pumping occurs when water and fine material is ejected from underlying layers through cracks in the HMA layer under moving loads Main Causes 0 Porous pavement as a result of 0 Hi h 0 Poor drainage Topic 2 Flegtltible Pavement Distress 7 Rutti g Rutting is characterized by depressions that form in the wheel paths Main Causes 0 High 0 Structural pmblen39s o Asphalt cement grade Topic 2 Flexible Pavement Distress 71 Consolidation Rutting The result of excessive consolidation of the pavement along the wheel path due to either reduction of the air voids in the asphalt concrete layer or the permanent deformation of the base or subgrade subgrade defamation weak subgrade or underlying layer F Topic 2 Flexible Pavement Distress 71 Consolidation Rutting Topic 2 Flexible Pavement Distress 71 Consolidation Ruttin Topic 2 Flexible Pavement Distress 72 Instability Rutting Failure is attributed strictly to the asphalt mixture properties and usually occurs within the top 2 inches of the asphalt concrete layer original pro le Topic 2 Flexible Pavement Distress 72 Instability Rutting F quotg 1 Topic 2 Flexible Pavement Distress 72 Instability Rutting Topic 2 Flexible Pavement Distress 8 Bleeding Bleeding is characterized by excess asphalt binder on the surface of the pavement Main Causes 0 Exoess o Exoess o Draindown o Contamination with diesel or similar Topic 2 Flexible Pavement Distress 9 Slippage Slippage is characterized by crescent or halfmoon shaped cracks generally having two ends pointed into the direction of traffic Main Causes Thin Layers Tracking of tack coat with equipment Contamination Rolling procss Topic 2 Flexible Pavement Distress 9 slippage cont Topic 2 Flegtltible Pavement Distress 9 Other types of Distress Swell corrugation Depression Potholes Ravelling and weathering Traffic Characterization Dr Christos Drakos University of Florida 3 SC r nk G A T o R 66 Topic 4 Traf c Characterization 1 Introduction Traf c is the most important factor in pavement design thickness is based on the number of load repetitions traffic 11 Traf c Characterization Procedures a Fixed Traf c Thickness governed by singlewheel load use the highest anticipated load for design Used for heaw load low volume ie airfields b Fixed Vehicle Traffic Thickness governed by of repetitions of a standard vehicle or axle load Convert all traffic to 18kip single axle loads c Variable Traffic and Vehicle Loads are divided into groups load spectra and the corresponding stresses and strains are used for design More appropriate for mechanistic design methods l 12pm m u Topic 4 Traffic Characterization Figtlted Vehicle Traffic 0 Design is based on the total number of passes of the standard agtltle load 18 kip Equivalent Single Axle Load ESAL during the design period Covert all traffic to the standard agtltle load ESAL Nd 1365 X 109 ac W77 N N Load cycles to failure Nf 00796st393291E1 0854 d l ESAL Basic premise Must determine how many 18kip single agtltle loads would cause the same damage as one Xkip load How many ESAL does a 24kip agtltle amount to Topic 4 Traffic Characterization 12 Eguivalent Agtltle Load Factor EALF aka LEF De nes the damage per pass to pavement by the axle in question relative to the damage per pass of a standard 18 k agtltle Load Eguivalence Factor LEF Depends on 0 Type of pavement Thickness structural capacity Terminal conditions definition of failure 20 of lane arm with fatigue cracking 2 inch rutting Theoretical analysis Nf18Nfx Based on experience AASHO Road Test Table 64 exible pavements Table 67 rigid pavemeniS Topic 4 Traffic Characterization 13 Theoretical Analysis to get LEF m 24kip m 18kip AC AC r the same structure BASE BASE apply 24 amp 18kip load SUBGRADE SUBGRADE 0 KENLAYER 18 Kipel 20005 gt Nf18 1612000 24Kip a 26705 gt N424 623000 0 So we can get an equivalent damage factor Nf18Nf24 259 o It would take 259 18kip load single agtltles to cause the same damage as one 24kip agtltle Topic 4 Traffic Characterization 13 Theoretical Analysis to get LEF I 39 39 39 I LEF No of 18k Single Axle Load to cause specific damage No of k Single Axle Load to cause specific damage Issues with theoretical analysis 0 Does the LEF change if we modify structural configuration thickness modulus etc 0 Which one is more critical fatigue cracking or rutting analysis Due to the many factors that influence the LEF it is almost impossible to select an appropriate a single value that applies to all situations For a truly mechanistic design method each load group should be analyzed separately Topic 4 Traffic Characterization 13 AASHTO E uivalent Factors Em irica 479 x WK LMLZS 10G szrn EALF WE Lx l sz IOAIX WX Where WX axle applications inverse of equivalency factors W18 No of 18kip single axle loads LX axle load being evaluated kips L18 18 standard axle load in kips LB code for standard axle 1 single axle L2 code for axle load being evaluated 2X 1 for single axle 2X 2 for tandem axle 2X 3 for triple axle added in the 1986 AASHTO Guide lll Topic 4 Traffic Characterization 13 AASHTO E uivalent Factors Em irica w L L W 10 w X 18 is G LZXFH EALFJ WE Llezx 10 Bix WX Where pt quotterminalquot serviceability index point at which the pavement is considered to be at the end of its useful life 42 function of the ratio of loss in servioeabilit at time t to G Log 2 P J v 5 4 1 5 the potential loss taken at a point where pt 1 7 0 4 0081LX LZX 3 23 function which determines the relationship B 7 39 l 519 3 23 between serviceability and axle load applications SN 1 LZX Topic 4 Traffic Characterization 1 AASHTO Eguivalent Factors Example 1 3 mm ievmmai sewiceabiiiiy iSZ 5 18 2 3945 vv1E predictednumbevuii rkipsmgieaxieiuadappiicaiiuns 7m 25 WX piEdidEd numbevui rkip singie axie iuad appiicaiiuns LX 39 Pl 39 LX L an3n 17Xi SN 2 9X 1SingiEaXiE 323 323 DEIle 5 mm X Lfm g 4245 sin 3945 smn 14x i m i w wx L g x LALF 8 g m Mm WX i in i quot L L5Y ix EALF M m Pix EALF7935 Lx i n m J Topic 4 Traffic Characterization 1 5 AASHTO Eguivalent Factors Example 2 Caicuiaie the LEFiuv aAEI EIEIEI ii iandemaxie iuad The simciuvai number SN i5 equaiiu m 5 and iheievmnai sewiceabiiny is 2 5 Liam 15 W1E predictednunbevuii rkipsingieaxieiuadappiicaiiuns 725 WX predictednunbevuiAElrkipiandemaxieinadappiicaiiuns PV Lx L 40 squot iw 2iand2maxie 323 323 mm um L Gquot Am mn4 X I ixu4 LX 2 L a 45 smnm 115323 smnm qu i w wx L 5W ix Em 18 18 1 in My Wx W12 wax i m 3 y i quot L SW ix W amp 11133 mm LN L mm Topic 4 Traffic Characterization 2 Computation of Design ESALs ESAL Equivalent Single Agtltle Load ESALs Cumulative ESALs for a IEICES over the entire design period we can also calculate ESAL for specific vehicle type ESALs ADT0TTFGYDL365 0 Where ADTEl Initial Average Daily Traffic T Percent Tmcks decimal TF Tmck Factor decimal G Growth Factor Y Dsign Period L Lane Distribution Factor decimal D Directional Distribution Factor decimal Topic 4 Traffic Characterization 2 Computation of Design ESALs Design ESALs ESALS EESALG m vehicle types 3 53225 f ESAli ADToXTXTiTFiXGXYXDl365 New terms T Distribution ofspeci c type of truck within all trucks decimal TF Truck factor for the speci c truck type decimal 21 Average Daily Traffic ADT Unless otherwise stated ADT is in all lanes amp both directions Also ADT includes 0 Cars 0 Singleunit trucks amp buses o Multipleunit trucks Topic 4 Traffic Characterization 22 Average Daily Truck Traf c ADTT or T 0 Minimum traffic information required for pavement design everything else can be found in tables 0 Very important in pavement design Effort to collect actual data Table 69 guide to distribution of truck types among total amount of trucks 221 Example 0 4000 ADT 20 Trucks Rural SystemPrincipal Find the of 2agtltle 4tire trucks 2axle 4 tire trucks 365 ADT T Topic 4 Traffic Characterization mu 55 rm mom 7 m min M rm dim W m hinciznl Auuhl w Minor in lrmmu I mm l39uriuwl m a um i u gtw H u m n w t m m m xi mm 3 ll n m m x i u u n u u m x l v i r i i 1 x i4 n M m Lil m x xi w w rm i v 1 i m u ll m A M m u u i 4 i x I I i u H i i l i ll ui mm m n 7 i4 n u m u in u i lt u m My im im iili im im mu m m mu w m n in quotmm H s huhml mum minnmmvn bmwm in W m Topic 4 Traffic Characterization 22 Average Daily Truck Traffic ADTT or T 0 Minimum traffic information required for pavement design everything else can be found in Tables 0 Very important in pavement design Effort to collect actual data Table 69 guide to distribution of truck types among total amount of trucks 221 Example 0 4000 ADT 20 Trucks Rural SystemPrincipal Find the of 2agtltle 4tire trucks 2axle 4 tire trucks 365175000year ADT T Table 69 Topic 4 Traffic Characterization 23 Truck Factor TF Sum of ESALs divided by the number of trucks weighed count of trucks not agtltles TF What is the importanoe of TF ESALS of Trucks For the same ESAls ifTF increases 9 Less of trucks If less of trucks produoe the same ESALS 9 More severe loads 0 Single TF can be applied to all trucks weighed average or separate for each truck type if the growth rates are different 0 Table 610 Truck Factors If we use all trucks we do not have to calculate ESALs for each truck type unu HM mm Topic 4 Traffic Characterization oili 4m w H mm 0 in Topic 4 Traffic Characterization 24 Directional Distribution D 0 Usually assume D 05 Where could that be diffe 25 Lane Distribution L 0 Function of ADT amp of lanes Outer traf c rent Table 614 We design for Outer Lane but everything is built the same Inner Lane usually under loa e Topic 4 Traffic Characterization 26 Growth Rate Factor G 0 Assuming a yearly rate of growth r Gx11rY Asphalt Institute Table 613 GY combined Topic 4 Traffic Characterization 27 Examgle 1 4lane RuralPrincipal ADT Determine the ESAls for 20 Trucks 2axle 6tire trucks 20year design r4 ESAli ADToXTiXTEXGXYXDL365 0 Distribution of trucks Table 69 T 10 for 2axle 6tire 20 trucks 002 0 Truck factor Table 610 TF 025 0 Growth Table 613 Y 2978 0 Lane Distribution Table 614 39 L 03994 ESAL4000002025298705094365 102175 ESAIsZO years Topic 4 Traffic Characterization 28 Example 2 4Iane RuralPrincipal ADT Determine the total ESALs 20 Trucks 20year design r4 ESAIs ADTOXTXTFXGXYXDL365 0 Trucks T 20 trucks 0 2 0 Truck factor Table 610 139 ESAIs 400002038298705094365 155quot10 ESAIsZO years Rigid Pavement Stress Analysis Dr Christos Drakos University of Florida 3 gs nk G A T 0 R 66 Topic 8 Rigid Pavement Stress Analysis Cause of Stresses in Rigid Pavements o Curling 0 Load Friction 1 Curling Stresses Q Where is the tension zone ugiim mu Topic 8 Rigid Pavement Stress Analysis 11 Curling Because of Temperature T 39 Tb Topic 8 Rigid Pavement Stress Analysis 12 Curling Because of Moisture Trapped High Moisture Water 13 Curling Because of Shrinkage High Shrinkage Low Shrinkage lg Topic 8 Rigid Pavement Stress Analysis 14 Curling Stress of Infinite Plate TAT i l mAT tX tv 7 2 0 Assume linear AT oz1 coefficient of thermal expansion 0X due to curling in Xdirection 0X due to curling in Ydirection Topic 8 Rigid Pavement Stress Analysis 15 Bending Stress of Finite Slab Y a 7 CXEatAT CYantAT L i X 7 217v2 217v2 V L gt x EaLAT c x 2 Wm Y Topic 8 Rigid Pavement Stress Analysis Correction Factor Chart length or of slab of relative stiffness Topic 8 Rigid Pavement Stress Analysis 15 Bending Stress of Finite Slab cont 0 Maximum Interior Stress Center of Slab E AT 0x mx ch EULAT Oy21vzcyvcx Edge Stress Midspan 7 EaLAT C 0 may be oX or 0 depending on a 2 whether C is taken as CX or Cy Topic 8 Rigid Pavement Stress Analysis 16 Temperature Curling Example quot k200 pci I a15gtlt10395 F At20 F Ec4000000 psi v015 Calculate Stresses i Radius of Relative Stiffness 3 14 L 1217v2k Topic 8 Rigid Pavement Stress Analysis ii Maximum Interior Stress Center of Slab EatAT 0x mmx t ch Ea AT crY 2CY VCX 21V or of relative Topic 8 Rigid Pavement Stress Analysis 16 Tem erature Curlin Egtltam e cont EaAT 0m WCX ch EMT 21 V2Cv ch 0Vth Topic 8 Rigid Pavement Stress Analysis 16 Tem erature Curlin Egtltam e cont iii Edge Stress Midspan 0X EaiATCX Topic 8 Rigid Pavement Stress Analysis 17 Combined Stresses Curling stresses are high but usually not considered in the thickness design for the following reasons Joints and steel relieve and take care of curling stresses as long as the cracks are held together by reinforcement and are still able to transfer load they will not affect performance Curling stresses add to load stresses during the day and subtract to load stresses during the ni Fatigue principle is based on of repetitions curling effect limited compared to load repetitions Topic 8 Rigid Pavement Stress Analysis 2 Loading Stresses Three ways of determining o amp 6 Closed form solutions Westergaard singlewheel Influence charts Picket amp Ray 1951 multiplewheel Finite Element FE solutions 21 Closedform solutions 39 theory 211 Assumptions All forces on the surface of the plate are perpendicular to the surface Slab has uniform crosssection and constant thickness Slab length Slab placed Topic 8 Rigid Pavement Stress Analysis 212 Limitations Only corner loadingedge loading or midslab deformation and stresses can be calculate No discontinuities or voids beneath the slab Developed for single wheel loads Topic 8 Rigid Pavement Stress Analysis 213 Corner Loading Where k modulus ofsubgrade reaction u 5 a elrli l t radius of relative stiffness p 3 a load oontact radius 5E W117088 p load 5 214 Interior Loading ba when a21724h 0316P t 2 U h 2 410g E 1069 b11682 h 70675h when alt1724h 5 8 th 1 llng owls Topic 8 Rigid Pavement Stress Analysis 215 Edge Loading ce 01410g 06668 0034 a 0431P a 5 17082 e kgz lummm Topic 8 Rigid Pavement Stress Analysis 216 Dual Tires A tht L 13 ssume a N 05227q Then area of the equivalent circle naz 2 X 05227L2 sd 06LL a 08521xi3dsd Pd 1 2 qr 7 05227q i 3131 Topic 8 Rigid Pavement Stress Analysis 217 Dual Tire Example p1oooo lb Calculate stresses q8842 psi k100pci sd14quot 554000000 psi h I I 10 1439 Topic 8 Rigid Pavement Stress Analysis 217 Dual Tire Egtltam le cont iii Comer Stress 3P a m 0Eh717T iv Interior Stress 0316P t O h 2410gBj1069 Topic 8 Rigid Pavement Stress Analysis 217 Dual Tire Egtltam le cont v Edge Stress cg 4logij0666ijr 0034 h a e Topic 8 Rigid Pavement Stress Analysis 3 Friction Stresses Friction between concrete slab and its foundations induces internal tensile stresses in the concrete If the slab is reinforced these stresses are eventually carried by the steel reinforcement What happens to PCC w AT Where vcUnit weight of PCC faAverage friction between slab ampfoundation Topic 8 Rigid Pavement Stress Analysis Steel Stresses Reinforcing steel 0 Tie bars Dowels 31 Reinforcement Wire fabric or 0 Do 0 Increase LZ tlIh E Where AS Area of required steel per unit width f5 Allowable stress in steel Topic 8 Rigid Pavement Stress Analysis 311 Welded Wire Fabric What does 6 X 12 W8 gtlt W6 mean Orientation Longitudinal Transverse Wire Reinforcement Institute Guidelines 0 Minimum wires W4 or D4 because wires are subjected to bending and tension Minimum spacing 4in allow for PCC placement and vibration Maximum 12x24 Wire fabric should have end and side laps Longitudinal 30 iam ut no less than 12quot ns erse 20Diam but no less than 6quot Fabric should extend to about 2in but no more than 6in from the slab edges Topic 8 Rigid Pavement Stress Analysis um u wziems Am nmmsms or warm wm manic Crmwcumvni men m m n m ccnlnrlwcml r mm mm i Dumclu wmm A himmlli lklmmul mm M 2 x 4 x ml in u 5 I w i m l 24 9x a m ww um um Him i m i u on m n wzx mm a w u i M i ll H w 4 mr n 575 m x i m m 3 m m 4 six in 7 4H m m 4 m a my am as I In m w an m rm 0 1 I n 7 u m 7 um um i M 4x 1 1 in MI I i a at z m Man 7 x a 4 ix ml in m u 1 2 w um m 1 2 m Um inn m m z u um 21 I u m u w u 7 Di in no 5 as n m m cm w z u x u m u m u u m i z 4 2i n ms um w m 1 rm m mm m u i m v21 u w u ix m n m An N m nu um 21 m m um w um u i m m Topic 8 Rigid Pavement Stress Analysis 32 Tie Bars Placed along the A fay L h s f L39 distance from the longitudinal joint 0 Length of tie bars p allowable bond stress d bar diameter Spacing of tie bars Topic 8 Rigid Pavement Stress Analysis 4 Joint Opening Where Joint open n clt Coef cient of thermal contraction e Drying shrinkage coef cient L Slab length C adjustment factor for subgracle friction o Stabilized o Granular on Pavement Design Principles Dr Christos Drakos University of Florida 3 ga nk G A T 0 R 66 Topic 1 Pavement Design Principles What is Design 0 Conceive amp develop plans for something to serve a speci c function 0 Must de ne function prior to design What is the Function of a Pavement 0 To provide vehicle access between two points Is this specific enough for N you to proceed w design 0 More specific function for pavements Access 9 Under all climatic conditions drain Durable 9 Sufficient structure for loads Safe 9 Adequate friction Smooth 9 Good ride quality level i ugnm mu Topic 1 Pavement Design Principles How many Elements into the Design ofa Highway 0 Route 9 Geometric design 0 Materials 9 Mix design Pavements 9Thickness design What information do you need as an Engineer Topography Route Access Drainage Hydrology Cut Fill Cost Egtltisting soils Must carry loads Determine structural requirements Affects drainage and drainage requirements Topic 1 Pavement Design Principles What information do you need as an Engineer cont 0 Weather Rainfall Temp variation Drainage eave Durability of AsphaltPCC Roads Rutting Bleeding HOT Low temperature cracking COLD Traffic Load levels Structural requirements 0 Use Primary or secondary facility Acceptable quality High or low maintenance 0 Design life High initial cost 9 Low maintenance cost high access Low initial cost 9 High maintenance cost low access Topic 1 Pavement Design Principles What information do you need as an Engineer cont 0 Available materials Contractors Type of structure AC or PCC Stabilization requirements MANY ASPECTS TO DESIGN OTHER THAN SIZING COMPONENT Topic 1 Pavement Design Principles What will this course cover Pavement types Flexible Asphalt Concrete Rigid Portland Cement Concrete 0 Failure modescauses Ruttlng racking Stresses in the pavement structure Distribution of stresses and strains Multilayer analysis 0 Traffic Characterization Predict traf c loads over a certain period ofyears Material Characterization De ne material properties 0 Design Procedures AASHTO Topic 1 Pavement Design Principles Pavement Histom Major Developments 0 Wheel 3500 BC ASIA Roads begun movement to Egypt 0 First Longdistance Highway 3500 to 323 BC Persia 1755 miles threemonth trip 0 First Engineered Road 300 BC Romans Appius Claudius Built 53000 miles of road Via Appia 360 miles 14ft wide 35 ft thick hand placed Lasted 2000 years Recognized two essentials Dry Road bed Impervious surface Topic 1 Pavement Design Principles First Engineered Road Concrete Stone amp other SUMMUM DaRsuM f u 393 a o a h 39 Qi ift fsiaflequot 1quot PAVIMENTUM Squared Stones Fine Dry Soil Wellcompacted Topic 1 Pavement Design Principles First Modern Roads 0 1764 France Tresaguet 0 Labor costs too high smaller stones thinner sections 0 Maintained two essentials mentioned above 10 years design life 3390F SMALL STONFS 5 m 70 LARGER STONES H 5125 OF WALNUT 10 in xze39v t i we a r FOUND1 TIEV STONES Topic 1 Pavement Design Principles Use of Tar and Asphalt 0 1830 s USA England McAdam o Impervious surface asphalttar mixed hot sand added to ll voids IHREE LA YERS 0F 21 5T0NES 1 M KKG IUI39 y s IMPERVUUS Topic 1 Pavement Design Principles Pavement Histo Other Develo ments 0 Portland Cement Concrete PCC 1850 Austria rst PCC roads Rubbertired Motor Cars 900 USA caused dust amp pollution problen39s Generated need for binders Highers eed requiring more smoothness Highway Research Board Currently TRB 1920 USA Resarch efforts to improve pavement design Looked at better materials ampconstruction methods Initiated rapid development in pavement technology 20th Century Pavements Better understanding ofstress distribution Use stifferstronger materials near the surface Topic 1 Pavement Design Principles 1 Introduction 11 What is Pavement Structure ll 12 Pur ose of Pavement Structure Topic 1 Pavement Design Principles 13 Stress Distribution under Wheel Load Two ways to reduce oz Pavement III39 um Subgrade Topic 1 Pavement Design Principles 14 What do we need to compute Subgrade Need 0 amp 8 within 4oz Ac 0 Max 02 BASE 0 02 decreases 0 As the 0 decreases SUBBASE quot SU BGRADE Topic 1 Pavement Design Principles 2 Pavement Types 21 Two Concepts 0 Thicker section of lower stiffness materials L Traffic Lane gt AC Surface lt Shoulder gti A Pavement Structure Granular Base Granular Subbase Su rad 39 gtg Irgmmm Topic 1 Pavement Design Principles 21 Two Concepts 0 Thinner section of higher stiffness materials L Traffic Lane gt 4 Shouler gt T Structure PCC Slab r Granular Subbase V Sub d m2 Topic 1 Pavement Design Principles 22 Typical Flegtltible Pavement Structures V 39 l a O V 39 39 v r 39 L39o 39y III f NW 39quot l39quot7 7 Pre ared Sub rade 39 w ilii p g Topic 1 Pavement Design Principles 22 T ical Flegtltible Pavement Structures 39 39 Asphaltaggregate mixture Base Granular material sometimes oement treated Subbase Granular material or selected soil Normally not treated Prepared Subgrade Topic 1 Pavement Design Principles 22 Typical Flegtltible Pavement Structures Asphalt Concrete Surfaoe Base Portland Cement Concrete 397 l Prepared Subgrade W Z z 7s Topic 1 Pavement Design Principles 23 T ical Ri id Pavement Structures Jointed Plain Concrete Pavement JPCP mnmm Joints ml or Purpose of Joints mm a s 5 Q15 5 important l is 0 an n xs m m n Topic 1 Pavement Design Principles 23 T ical Ri id Pavement Structures 0 Jointed Reinforced Concrete Pavement JRCP Purpose of Reinforcement Transverse Johns Hilli Dowels i I I laVZD Longitudinal Join a 9 with Tie Bars y y I I I I I l I I u I I Hm Fabn 0 an to 100 t Purpose of Dowels Load 114 to 112 2 z 18 long Topic 1 Pavement Design Principles 23 T ical Ri id Pavement Structures 0 Continuously Reinforced Concrete Pavement CRCP Cracks spaced 38 ft No Joints 0 Prestressed Concrete Pavement PCP Wire Strands Slab Length 300 to 700 ft Topic 5 Material Characterization Dr Christos Drakos University of Florida 3 SC r nk G A T o R 66 Topic 5 Material Characterization 1 Introduction Pavement design inputs covered in previous topics 0 Predicted stresses and strains load magnitude 0 Traffic load cycles What material properties have we used up to now 0 Most paving materials are not elastic but experience some permanent deformation after each load application 0 If the load is small compared to the strength of the material and is repeated for a large number of times the deformation is almost recoverable and can be considered elastic l 12pm m u Topic 5 Material Characterization Resilient Modulus MR Type and duration of loading 1 is supposed to simulate that g occurring in the eld Time 5 E U 2 MR d o 5 Topic 5 Material Characterization Typical Flegtltible Pavement Structure AC Decreasing BASE COURSE stiffness SUBBASE counse Optional I if SUBGRADE SOIL 1 Surface Course 0 La er that Domes to oontact with traf c normally contains the highest quality material 2 Base 0 Provides additional load distribution and contributes to drainage and frost res stance 3 Subbase optional 0 Functions primarily as structural support but it can also help with drainage and frost action 4 Subgrade 0 Native soil the only nonengineered material in the pvt structure Topic 5 Material Characterization Subgrade performance generally depends on Load bearing capacity subgrade must be able to support loads passed on from the pavement structure 2 Moisture content it tends to affect a number of subgrade properties including load bearing capacity 3 Shrinkage andor swelling some soils shrink or swell depending upon their moisture content Remedies for poor subgrade conditions 1 Remove and replace subgrade soil can be removed and replaced with higher quality fill 2 Stabilization adding an appropriate binder such as lime portland cement or emulsified asphalt can increase subgrade stiffness andor reduce swelling tendencies 3 Additional base layers include a subbase or increase thickness of base Topic 5 Material Characterization Subgrade Seasonal Variations MR 50000 psi F Normal MR Thaw MR Fleexe may Time Time Time Time Topic 5 Material Characterization 2 Design Resilient Modulus Base ampSubgrade 21 Correlations Maybe there is information already available Also MR1500CBR MR1155555R Topic 5 Material Characterization 211 California Bearing Ratio CBR Basically a penetration test Piston penetrates soil at constant rate 005 inmin Pressure is recorded 0 Take the ratio to the bearing capacity of a standard rock 0 Range 0 worst 100 best Pressure to cause 01quot penetration to the sample CBR Pressure to cause 01quot penetration for standard rock Topic 5 Material Characterization 211 California Bearin Ratio CBR De ection dial loading Surcharge con nement De ection dial swelling Proctor 4 sample 45 6 0 Topic 5 Material Characterization 211 California Bearing Ratio CBR Coarsegrained soils Finegrained soils Topic 5 Material Characterization 212 Stabilometer Rvalue Resistance value of a soil determined by stabilometer Closedsystem Triagtltial test Apply vertical pressure 0 Measure horizontal pressure induced in the fluid 100 25 D2 pi pv amp p Vertical and horizontal pressure respectively D2 displacement of Stabilometer uid to increase 100 psi measured in revolutions ofa calibrated pump handle R1007 Topic 5 Material Characterization 212 Stabilometer Rvalue 160 psi Pressure Gauge Apply the vertical horizontal pressure Fluid under p ressure R 100 i 100 Topic 5 Material Characterization 22 Insitu Testin Plate Loadin Test 0 Circular plate 30 in Z series of plates used to minimize nding Apply load at constant rate to reach 10 psi 0 Pressure held constant until the deflection increases no more t an 0001 inmin forthree consecutive minu Use average dial reading to determine the deflection k E A k Modulus ofsubgrade reaction p pressure on the plate 10 psi A de ection of the plate Topic 5 Material Characterization 22 Insitu Testin Plate Loadin Test Reaction Steel Beam Figure 736 Correlation of k with MR Topic 5 Material Characterization 23 Laborato Testin Triagtltial Test mu quot Topic 5 Material Characterization Triagtltial Test 23 La borato Testin lllll G In Triaxial cell Sample cylinder 0203 i U 1 a Hill gt Confining Pressure 6263 gt Deviator Stress Gd Deviator Stress Axial stress in excess of the State of con nement con ning pressure in Triaxial cell de ned by the rst invariant What can affect the results L9 Level of oon nement 0110170393 6039103920393 Topic 5 Material Characterization 231 Triagtltia Test Granular Material For Granular subgrade k MR function confinement MR k1 X19 2 k1 amp k2 experimentally determined values Run Triagtltia test at certain levels of confining pressure and vary the deviator stress Example 72 Huang E 2 k2 3 k1 log 6 psi Topic 5 Material Characterization 231 Triagtltia Test Granular Material Sample Conditioning Set the confining pressure to 5 psi and apply a deviator stress of 5 psi and then 10 psi each for 200 repetitions Set the confining pressure to 10 psi and apply a deviator stress of 10 psi and then 15 psi each for 200 repetitions Set the confining pressure to 15 psi and apply a deviator stress of 15 psi and then 20 psi each for 200 repetitions N w Resilient Modulus Test After sample conditioning the resilient modulus test follows a constant confining pressureincreasing deviator sequence and the results are recorded at the 200 11 repetition of each deviator stress Topic 5 Material Characterization 231 Triagtltia Test Granular Material Test Procedure Set the confining pressure to 20 psi and apply a deviator stress of 1 2 5 10 15 and 20 psi Reduce the confining pressure to 15 psi and apply a deviator stress of 1 2 5 10 15 and 20 psi Reduce the confining pressure to 10 psi and apply a deviator stress of 1 2 5 10 and 15 psi Reduce the confining pressure to 5 psi and apply a deviator stress of 1 2 5 10 and 15 psi Reduce the confining pressure to 1 psi and apply a deviator stress of 1 2 5 75 and 10 psi stop the test after 200 repetitions of the last deviator stress level or when the specimen fails Iquot E 5 S Topic 5 Material Characterization 231 TrIagtltIal Test Granular Material 052an 555 5 55 5555555 5555 5555 5555 5 s 5 55m 555 2 555555 5 5 5 x x 5 55 55555 5555 525 555255 52 m 5 555 55 55555 5 55 2255 555 555555 55 552 555 555525 55 5525 5555 555555 55 5255 55 555555 55 I512 122 15625E 47 Ex RecoverableDeforrnation lZEIEI 325 153846 SH 5 2 555 525 555555 55 Gage Distance 4 555 555 555555 55 4572 5545 574555 55 M 7 675 5525 55 525555 55 5 5 I 672 168 WAX 32 1 55 5555 55 555555 55 5555 555 555555 45 Scd303 5552 555 555555 55 5555 525 55555 55 5555 25 55552 55 5 2225 555 55525 25 5 5555 555 552555 25 5 5555 5542 55 22 55 5555 555 52555 5 5555 225 55555 5 5 2 555 5 55 5255 555 52525 555 5 555 5555 55555 55 Topic 5 Material Characterization 231 Triagtltia Test Granular Material Wmdmmmxa y 368mm mumman MR 3686 x 0351 Topic 5 Material Characterization 232 Triagtltia Test Cohesive Material For Finegrained cohesive subgrade MR function deviator stress 0d k1 k2 k3 amp K4 experimentally determined values Run Triagtltia tests at certain values of deviator stress and vary the confining pressure Example 73 Huang MR kITk3k270 d Vad ltk2 MR k1k4a39d7k2 V0 gtk2 MR psi Topic 5 Material Characterization 233 Asphalt Institute Subgrade Characterization 0 To determine subgrade MR Asphalt Institute suggests Con ning stress 03 02 2 psi Deviator stress od 01 03 6 psi 0 For granular material For example use data from Fig 78 o For finegrained material For example use data from Fig 79 ivymm Topic 5 Material Characterization 234 Granular material example MR k1 6quot2 3960 42035 94495 sues Invzriani p5 Figure 73 Example 72 1 pal 59 km ll mum Topic 5 Material Characterization 235 Finegrained material example 0d gtk2 MRk1k4039diz 56003886752 5910 Resilient Modulus lax psi Deviatax Stress psi F igum 79 Example 7 3 1 psi 69 kPn7 Topic 5 Material Characterization 3 HotMix Asphalt 31 Structural Layer Coefficient Used in AASHTO design procedure Describes the quality of the material 0 Function of MR amp position in the pavement Why 32 Marshall Test 0 Used in the Marshall Mix Design procedure Performed on cylindrical specimens 0 Measure Stability fail load and Flow deformation 33 Cohesiometer Test 0 Used to measure the cohesion of cemented materials 0 Apply load at controlled rate until failure Topic 5 Material Characterization 3 HotMix Asphalt cont MI 05 33 o 80 1D quot3 3 as 3 wquot 50 E A 2395 g E m 3 a 5 E n 73 gigs 1390 ozux E g D 1m V175 i aquot a 39 LS g 5 mac g 3quot 3 i E g mquot Eus 25 a i w m an on saw me u m W 02 L0 3 Surface Course bi Base Coursc Topic 5 Material Characterization 4 Bases 41 Untreated Granular Base 40 F 1W IS 20 i 70 so 39 2 50 w 7 g m g 70 a 30 5 S so u a 5 2 g 5quot w 35 a 1 2 5 5 10 Topic 5 Material Characterization 42 Stabilized Granular Base 040 030 m E j a mo m 0 tut mm 5 90 030 m u a 151 30 E 022 5 5 ao mg 2 W 53 25 39R 020 5 50 3 7390 5 w 2 20 quota 0m g 5 020 5 90 gt E 63955 wo g 50 E 700 L5 g ou 2 g 22 042 20 g 50 m 200 I m 010 5 010 mo g E L Bituminous Treated Cement Treated i Topic 5 Material Characterization 5 SubBases ozo 3 5 100 a a 014 7o 2 9 5 2quot 20 so T 80 5 01239 40 In 70 I 339 J 300 so gt 390 14 0403 20 5 3 123 nos3 10 4v 3 006 g 0 10 so quot 539 25 5 0L L l L Topic 5 Material Characterization 6 Drainage Important to keep water away from the pavement structure 61 Detrimental effects of water Reduces strength of unbound material amp subgrade Causes pumping faulting cracking amp shoulder deterioration Porepressure increase 9 Pumping of fines 9 Loss of support Load goes over saturated base amp soil Porepressure increases Water is incompressible moves up Causes heaving swelling of soils Porepressure within the AC layer causes stripping durability cracking Frost action in Northern climates Topic 5 Material Characterization 62 Sources of water 39 Seepage Raise of water table 39 Infiltration From proximity of water table Capillary moisture held in the pores from surface tension Vapor movement Movement of water associated with fluctuating temperature and pressure 2 E lt Topic 5 Material Characterization 62 Sources of water l imam H 3 Fm EMMAQDW Eird Sebsre W Thrcugh Permeable Surface i l from Edue p 5eevage From i High Ground i Q me WalerlTuble I 1 Vapor Mov I Upward Movemen 2 I i C or WavyTable ems i i i i i i i i m 43 Waler Table Topic 5 Material Characterization 63 Protecting the Pavement Structure A Need to minimize availability of water C be Impervious surfaceshoulders How sealants pz ormd Drainage to remove water quickly 9 Ditch after Drainage layer for subsurface water a construction Design phase Construction Three drainage installations for subsurface water 1 Drainage layer or blanket 2 Longitudinal drain 3 Transverse drain Topic 5 Material Characterization 64 Drainage Deficiencies for Pavements with Ditch 641 Typical Pavement with a Ditch IMPquotd hvuldcr Ennuh am an quotmm 39rlnch nign l a ditch Mung mm Topic 5 Material Characterization 642 Water infiltrating from rutted shoulder Qquot 4 liter Infiltrltinn 643 Water infiltratind due to debris caused Dondina Erm and Debris build nu Iltzr in rut I ll Pending BiIuinoux Caizrat l y I tquot i Inu Infiltrltion Topic 5 Material Characterization 644 Water infiltratina due to differential settlement owning an d of mum dun tn nntrlctiln Ind settleen Bitulinou caneretl Fund 31 C Added 1 r t luter Infiltration Topic 5 Material Characterization 65 Subsurface Drainage 651 Draina e la er Blanket Pavement Surface Base Course as rainage layer Subbase f5 Fifter I V Topic 5 Material Characterization 652 Longitudinal drainage Shoulder 39 IBaseICour5 2 as ij Drainage Blanket Topic 5 Material Characterization 653 Transverse drainage Same concept as longitudinal drainage but in this case the drainage runs across the lanes Where would we use the transverse drain AASHTO Rigid Pavement Design Dr Christos Drakos University of Florida lt I 3 tgw s39 m 1 A T R Engmeenng Topic 10 AASHTO Rigid Pavement Design 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 25 to 125 inches Subbase type untreated gravelsand 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 1114000 Slab Thickness k value ESAL PCC Modulus Statistical Regression Model gt Output PSI finn I mwnwnn Topic 10 AASHTO Rigid Pavement Design 2 General Design Variables Design Period 0 Traffic what changes 0 Reliability Based on functional classi cation Overall standard deviation 50025 035 0 Performance criteria APSI PSI0 P51t 3 Material Properties 31 Effective Modulus of Subgrade Reaction kl Need to convert subgrade MR to k 1 2 3 Topic 10 AASHTO Rigid Pavement Design 311 Pavement Without Subbase If there is no Subbase AASHTO suggests 0 Correlation based on 30in plateload tests 0 k value becomes too high because kfnc1a o More accurate k if plate test was run w bigger plates too expensive amp impractical 312 Pavement With Subbase l quot PCC If subbase egtltists need to suggASE determine the k SUBGRADE n BEDROCK Topic 10 AASHTO Rigid Pavement Design 312 Pavement With Subbase cont Example Subbase thickness10quot Subbase modulus30000 psi Subgrade MR10000 psi To get k T 10 AASHTO Pavement If bedka is within 10ft it will p Exam e Rigid depth539 me prev pa e Subgrade MR10000 psi km 600 pci Munbwsw mum Mnnuiul mum n a emeg mm 1031 ameA sd 3 snmpnw sd munw ammaa unnepunnj A ansndmng aseqqns MW smnpnw pm apmaqng GI39ZI BJT DH SI39ZI ELIquotDH 3 max new u6sea 1U9LU9 9d m m OLHSW 01 3d0L I 4 53925 0517 Z Z 1 n 1 uluow Zv1sr0316E04 9500 quot u gsap BqlgtltBJ01JELUS1UI 10CDE olu usgt121 BJBM SUO12JEAEUOSEBS 1 362pr BLUES sq u 1nsaJ pn0M12qlsnnp0Lu1u32Agnb3 33 uonoeea epeJBqns J0 snnpow BARDBJJQ 7139s u6sea 1U9LU9 9d m m OLHSW 01 3d0L Topic 10 AASHTO Rigid Pavement Design 315 Ee Examgle cont ampwm 12 n D8in Figure 1220 Topic 10 AASHTO Rigid Pavement Design 316 Loss of Su ort LS 0 Reduction of keg by a factor L5 to account for 0 Best case scenario lrr ucuw Morluiu a Suhgrmlu mumquot s nun Topic 10 AASHTO Rigid Pavement Design 317 Table for Estimating gaff TriaI Subhnse Type Thickness inches Loss of Support LS Depth to Rigid Foundation feet Projected Slab Thickness inchest l 2 3 4 5 6 kValue lpci Roadhed Subbnse Composite 39 39 Relative Modulus Modulus n i In l Damae u mm M nun psi i 20000 100000 l Jan I I I I I I I I I v v v Dec l 15 000 100 000 l l J Eu Summation in Average u T Effective Modulus of Subgrade Reaction k pci Canceled for Loss of Support k pci l l mimvnw Topic 10 AASHTO Rigid Pavement Design 32 Portland Cement Concrete PCC Elastic Modulus of Concrete EC Correlated with compressive strength Modulus of Rupture S c Thirdpoint loading 28 days Topic 10 AASHTO Rigid Pavement Design 33 Pavement Structure Characteristics Drainage Coef cient Cd Quality of drainage amp percent time exposed to moisture Table 1220 I i I u mmltl mm Jr raw 539 l l 5 l I l illl l l I rl Hi I ll ll l fllrl l l l vl lll l IllI Ill LII l l I rl l l Hlvl 00 l illU HI l M i l n m Immw nwwxn um ltri mm mm m u My 0 u mun m u 39u Load Transfer Coef cient J Ability to transfer loads across joints and cracks Table 1219 LowerJ 9 Tpr of shoulder Axplmli Tied I CC Loud transfer chiccs Yes No Ya No Jl ClJ and JRCT 32 314744 394 l 164 I 39RCP 29 2 NRA Z 3 3 N Topic 10 AASHTO Rigid Pavement Design 4 Thickness Design 41 Input Variables Modulus of Subgrade Reaction keff 70 pci Traffic W18 5 million Design Reliability R 95 Overall Standard Deviation S0 030 APSI 17 Elastic Modulus Ec 5000000 psi Modulus of Rupture S c 650 psi Load Transfer Coefficient J 33 Drainage Coefficient Cd 10 V Use Nomograph Figures 1217aampb or solve equation Topic 10 AASHTO Rigid Pavement Design 42 Nomograph o Cancun 0 TL 20 5 3 3 u me u u w 1 u z 40 a c o z u so gt u g 9 a II n a 5 a u 38 o oi 4 1 4 8 or 60 2 v on E g a o 5 7 3 g a 1 30 f 2 300500 mo 50 I0 g g Q so mum Modulus at Subqlode Reacllon h pm I 3900 ii migng Topic 10 AASHTO Rigid Pavement Design 42 Nomograph Topic 10 AASHTO Rigid Pavement Design 43 Eguation Wm soooooo ZR 45 so 03 APSI 17 k 70 Sc6so 133 cdio p22 Ecsoooooo 945 Given M in M 7 4545 i SccdD 1132 i 7 r 1242 162410 563 1 Dan m on k mew Topic 10 AASHTO Rigid Pavement Design 5 Other Design Features 51 Slab Length 511 Jointed Plain Concrete Pavement JPCP Governed by Topic 10 AASHTO Rigid Pavement Design 512 Jointed Reinforced Concrete Pavement JRPC Always doweled L Use same typical values from before 5 AL 025 e 894 Cmt x AT a 065l55x10 x 60 10 x 10 L 2 75ft A Guideline Topic 10 AASHTO Rigid Pavement Design 513 JRCP Reinforcement If when concrete cracks steel picks up stress Where A faVElh As Area of required steel per unit width 5 2fS f5 Allowable stress in steel f6 Average friction coef cient between slab and foundation Topic 10 AASHTO Rigid Pavement Design 513 JRCP Reinforcement cont um 4 wuwu mm uimmsiom v Wkiiitn WW I ml l mmw mm m w u Topic 10 AASHTO Rigid Pavement Design 5 Design Example Given the following information Roadbed soil MR 20000 psi December January 8000 psi February March 15000 psi April November Subbase Information Loss of Support 05 Friction factor Thicknss 0 24 5390 Elastic Modulus 100000 psi Design Facto Design Reliability R Overall Standard Deviation SEl PSI 90 040 15 ra ic 379 million EAL Drainage coefficient Shoulders 10ft wide PCC Temperature drop 55 DF Topic 10 AASHTO Rigid Pavement Design 5 Design Example cont PCC Elastic Modulus E 4500000 psi odulus of Rupture S39E 725 psi Limestone rock Indirect Tensile Strength 500 psi Design a JPCP wo dowels and a JRCP 35ft w dowels For each pavement determine the sla t ickness joint spacing for the JCPC and reinforcement mesh designation for the JRCP 51 Effective modulus of subgrade reaction Next page Tria Subhase Type M Depth to Rigid Foundation feet L u u Thickness inches 6 Projected Slab Thickness inches L 1055 of Support LS 05 l 2 14 5 6 kValue pci Roadhed Suhhns on Rigid Relative omposite ModulusV Modulus kValue pci Foundation Damage u 33 34 35 Dec S 2 Average r E 1443 ummauon ur II 946 Effective Modulus of Subgradc Reaction k poi Corrected for Loss of Support k pci I o AASHTO Flexible Design Procedure Dr Christos Drakos University of Florida so SC r nk G A T o R 66 Topic 7 AASHTO Flexible Pavement Design 1 Development AMERICAN ASSOCIATION OF STATE HIGHWAY AND TRANSPORTATION OFFICIALS httpwwwaashtoorg 11 AASHO Road Test 0 Late 50 s road test in Illinois 0 Objective was to determine the relationship between the number of load repetitions with the performance of various pavements 0 Provided data for the design criteria 12 Performance Measurements Establishment of performance criteria is critical FunctionalCZI AASHTO VS AI IgtStructura l i21 lk lllll Topic 7 AASHTO Flexible Pavement Design 12 Performance Measurements cont AASHO Road Test performance based on user assessment Dif cult to quantify subjective Highly variable Present Serviceability Rating PSR 01 V Poor 12 Poor 23 Fair A panel of experts drove around In standard 34 Good vehicles and gave a rating for the pavement 45 V Good Measurable characteristics performance indicators sible distress cracking amp rut ing Surface friction Roughness slope varianoe Masure of how much slope varies from horizontal along the direction of traf c Topic 7 AASHTO Flegtltible Pavement Design 13 AASHTO Performance Relations Establish correlation between user assessment ride experience an 39n icators quot 39 39 39 USER ASSBSMENT PERFORMANCE INDICATOIE 01 V Poor Measure of Roughness 12 Poor Measure of Rutting 23 Fair Measure of Cracking 34 Good 45 V Good Present seeroeablll Index PSI How does the tme user performance PSI A0 AlF1 AZF2 AEF3 correlate to the measured performance A3 Regression Coef cients F1 Measure of roug ness F2 Measure of rutting F Measure ofcracking calculated the regrssion coefficienis for the PSI equation Topic 7 AASHTO Flexible Pavement Design 13 AASHTO Design Eguations 131 Perfoa quot 39 amp Design Life PSI scale 1 V Poor 9 5 0 Good PSIU p51 APSI PSI0 PSIt PSIt Terminal PSI known 9 Pvt is no longer functional Time age Design Life AASHTO performance requirement APSI o APSI is such that PSIt is NOT reached before end ofdesign life Topic 7 AASHTO Flegtltible Pavement Design 132 Performance Relation ESAL PERFORMANCE St ct Eff m gmFluency Structural Number SN Raff NW J What are the three factors affecting performance APSI APSI fnc MRem ESAL MRE Aocounis for the environment SN Index relating effectiveness of PVT stmcture known known known Solve for SN Topic 7 AASHTO Flexible Pavement Design 133 Definition of Structural Number Ac D1 a1 SND1xa1 Structural Coefficient a BASE D2 a2 SNZD2X32 SN a fnc E position in PVT m SN SN1 SN2 SN3 W Basic Procedure 0 Determine the traffic ESAL Calculate the effective subgrade modulus MRe Select the performance level APSI Solve forthe required SN needed to protect the subgrade Topic 7 AASHTO Flegtltible Pavement Design 134 Design Notes i Different combination of materials ampthicknesses may result in the same SN ii Yourjob as a designer is to select the most economical combination using available materials and considering the following 0 Geometry requirements CutFill Drainage requirements 0 Frost requirements iiiAASHTO assumes that pavement structural layers will not be overstressed Must check that individual layers meet structural requirements Topic 7 AASHTO Flexible Pavement Design 2 W 21 General Design Variables Design Life Material Properties Traffic Reliability Degree of certainty that the pavement will last the design period Uncertainty in Traffic prediction Performance prediction Materials amp construction Topic 7 AASHTO Flegtltible Pavement Design 22 AASHTO Reliabili Factor FR Adjust traffic for reliability FR fnc R So W18W18gtltFR Reliability level Overall Standard DeVIatIon Where chosen 0 Tra Ic Variation W18 Design ESAl o Performanoe prediction w18 Predicted ESAL ananon Stegs 0 Materials subgrade 1 Define functional class InterstateLocal 2 Select reliability level R Table 1114 3 Select a standard deviation SD 0 Flexible No traf c variation S0035 With traf c variation S0045 0 Rigid No traf c variation S0025 With traf c variation S0035 Topic 7 AASHTO Flexible Pavement Design 23 Performance Criteria Design for serviceability change APSI PSI0 PSIt PSI0 Initial serviceability index 0 Flexible 42 0 Rigid 45 PSIt Terminal serviceability index 0 Major highways gt25 0 Lower volume 20 24 Material Properties 241 Effective Subgrade Resilient Modulus Obtain MR values over entire year 0 Separate year into time intervals 0 Compute the relative damage value uf for each modulus 32 u 118x108 XMR Z Topic 7 AASHTO Flegtltible Pavement Design 241 Effective Sub rade Resilient Modulus cont 0 Compute average uf for entire year 0 Determine effective MR using average uf uf 118x108 XMR m Topic 7 AASHTO Flexible Pavement Design 242 Pavement Structural Layers Layer coefficient ai relative quality as a structural unit 2quot of material with a02 provides the same protection as 1quot material a0 Initially layer coefficients were derived from AASHO road test results have subsequently been related to resilient modulus HotMix Asphalt E AASHTO dos not require test to determine HMA modulus usually assume aHMA0 um mum mm It YMivuns pm Topic 7 AASHTO Flegtltible Pavement Design 242 Pavement Structural La ers cont Untreated and Stabilized Bases Can estimate the base layer coefficient from Figure 715 for Untreated base Bituminoustreated base Oementtrated base 0 For untreated base can also use the following instead of interpolating from the figure 0249X10gE27 0977 Granular Subbases Can estimate the subbase layer coefficient from Figure 716 0 Can also use the following instead of interpolating from the figure a3 0227X10gE37 0839 Topic 7 AASHTO Flexible Pavement Design 25 Drainage 0 AASHTO guide provides means to adjust layer coefficients depending on the effectiveness of the drainage 0 Define quality of drainage of each layer based upon Time required for drainage Percent time moisture levels approach saturation 0 Determine drainage modifying factor m from Table 1120 SN Percenlage uf lime pavemenl slruciure is exposed Quality of drainage to moisture levels approaching saturation Waler removed Less lhan Greater than Raling wilhin l 5 5 25 25 Excellent 2 hours LAO 135 35vl30 130 120 l20 Good I day l35 125 125415 ll5 IUD 00 Fair I week 254I5 115105 LOO030 080 Poor I month ll57I05 LBS 181 080 060 060 Very poor Never drain 105 095 095 075 075040 040 Source After AASHTO 986 M Topic 7 AASHTO Flexible Pavement Design 26 Computation of Reguired Pavement Thickness 261 Basic Approach 0 Determine the required SN for design traffic 0 Identify trial designs that meet required SN 262 Nomograph to Solve for SN x a cm 5mm mm 5 Topic 7 AASHTO Flexible Pavement Design 26 Computation of Reguired Pavement Thickness cont 263 Solving the Equation I l 1 ogkzur 15 1094 04 logw 18 ZR so 936 logSN 1 r 02 232 logM R r 807 SN 1519 Declare the known variables W18 ZR S0 PSI amp MR 0 Give an initial estimate forthe SN 0 Allow the equation solver Matlab Maple Mathcad Egtltcel etc to iterate for the solution Topic 7 AASHTO Flegtltible Pavement Design 264 Pavement Structural Layers SN alD1 aZDzm2 o No Unique Solution Many design configurations will meet the required SN Optimize the design consider the following Design oonstraints drainage minimum thickness available materials Construction oonstraints minimum layer thickness EDDanllS 265 Layered Design Analysis Nomograph determines the SN required to protect the subgrade However each structural layer must be protected against overstressing Procedure developed using the AASHTO design nomograph Determine the SN required to protect each layer by entering the nomograph using the MR of the layer in question Topic 7 AASHTO Flexible Pavement Design First we need to protect the subgrade use the nomograph to get SN needed to provide adequate protection BUT have to protect each layer from overstressing need to get required SN level of protection for each layer 0 Only top AC layer does not need protection For example Base needs SN1 protection BUT SN1 alD1 D 1 a1 E2 32 quot 2 E3 33 m3 Topic 7 AASHTO Flegtltible Pavement Design 266 General Procedure 1 Using E2 as the MR value determine from Figure 1125 the structural number SN1 required to protect the base and oompute the thickness of layer 1 by SN D 1 1 31 2 Using E3 as the MR value determine from Figure 1125 the structural number SN2 required to protect the subbase and compute the thickness of layer 2 by o D ZSNzialDl 2 a2quotquot2 3 Based on the roadbed soil resilient modulus MRem determine from Figure 1125 e total structural number SN3 required and oompute the thickness of layer 3 by o D 2SN3ra1D17a2D2m2 3 a3m3 Topic 7 AASHTO Flexible Pavement Design 27 Other Thickness Considerations 271 AASHTO Suggested Minimums ESAL Asphalt Concrete Aggregate Base 1 4 50000 150000 2quot 4quot 150000 500000 25quot 4quot 500000 2000000 3quot 6quot 2000000 7000000 35quot 6quot gt 7000000 4quot 6quot 271 Construction Stability Layer must be thick enough to act as a unit 0 Thickness gt 2 Maximum Aggregate Size Topic 7 AASHTO Flegtltible Pavement Design 28 Cost Considerations 0 Consider Different combination of materials Cost of materials Cost of excavation cut areas Egtltpress cost as a unit contribution 016X080 312 Magtltimize crushed stone thickness minimize AC thickness Can also stabilize base to use less H A 0 Use gravel only for fill or frost Topic 7 AASHTO Flexible Pavement Design 29 AASHTO Design Example 1 Given 0 Reliability 90 0 Overall Sid Dev 035 0 W18 10 million 0 Design Serviceability Loss 20 WORK EXAMPLE ON THE BOARD Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 Given 0 Reliability 90 0 Design Serviceability Loss 20 Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Construct a material information table Material Layer sqydin m Unit SN Next step is to fill in the information Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Asphalt Concrete structural coefficient a Figure 713 0 mains Idspsi a Surface Course Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Bituminous treated base structural coefficient a Figure 715 ulnaMan ml39 9 E 5mm mm m mumquot mvany onquot b Bituminous Treated Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Cement stabilized base structural coef cient a Figure 715 summi nominalu mm MD7 any mar c Cement Treated iii3m Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Crushed stone base structural coefficient a Figure 715 mm mm in Mmulusdm In a Unirealed Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Crushed stone subbase structural coefficient a Figure 716 Modulus1W psi Topic 7 AASHTO Flexible Pavement Design 210 AASHTO Design Example 2 cont Material Layer lsqvdin Are there any obvious oonclusions Unit SN

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