<p>Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>UNITUNIT 14 14</p><p>FLEXIBLE PAVEMENT DESIGN</p><p>OBJECTIVES General Objective </p><p>To know the methods and procedures in designing the flexible pavement for roads in Malaysia.</p><p>Specific Objectives</p><p>At the end of the unit you should be able to :- describe the basic layers of road design calculate the design using the required formula and figure. design the basic flexible pavement using JKR method. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>INPUT </p><p>FLEXIBLE PAVEMENT DESIGN</p><p>14.0 INTRODUCTION</p><p>A typical section through a pavement is shown in the following sketch ( not to scale )</p><p>Surfacing Wearing course Base course</p><p>Upper Road-base Lower</p><p>Sub-base</p><p>Capping</p><p>Sub-grade</p><p>In some type of construction some layers maybe combined.</p><p>14.0.1 Foundation</p><p>Foundation provides uniform support to the pavement through its life so that maintenance operation is confined to the upper level of the pavement. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p> i. Sub-grade – either natural soil or material placed to form embankment.</p><p> ii. Capping – on sub-grade with a low CBR, a capping layer is required provide working. Platform on which sub-base construction can proceed with minimum intersection from wet whether a minimize effect at weak sub-grade on road performance.</p><p>14.0.2 Sub-base</p><p>This layer forms the upper of the pavement foundation and provides a regulated working platform at a consistent strength on which to transport, place and compact the bound layers of pavement.</p><p>14.0.3 Road Base</p><p>Road base is a main structural element that purposes to spread induced by repeated wheel loads over the foundation and to with stand internal stresses without excessive deformation.</p><p>14.0.4 Surfacing</p><p>Surfacing is done in order to provide acceptable running surface of adequate skid resistance and to reduce water penetration to underlying layer. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>14.1 FACTORS FOR DESIGN 14.1.1 Failure Criterion</p><p>The definition from Croney, failure as transformation shape or deflection of 20mm tire’s lane not speeding that measuring from ground level.</p><p>Slow lane Fast lane</p><p>20 mm</p><p>14.1.2 Traffic Loading</p><p>Protection of the sub-grade from the loading imposed by traffic is one of the primary functions of a pavement structure. The designer must provide a pavement that can withstand a large number of repeated applications of a variable-magnitude loading. The primary loading factors that are important in flexible pavement design are 1. Magnitude of axle (and wheel) loads. 2. Volume and composition of axle loads. 3. Tire pressure and contact area.</p><p>The magnitude of maximum loading is commonly controlled by legal load limits. Traffic survey and loadometer studies are often used to establish the relative magnitude and occurrence of the various loading to which a pavement during its design life is a very difficult but obviously important task. Most design Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN procedures provide for an increase in traffic volume on the basis of experience by using some estimate growth rate.</p><p>14.1.3 Climate or Environment</p><p>The climate or environment in which a flexible pavement is to be established has an important influence on the behavior and performance of the various materials in the pavement and sub-grade. Probably the two climatic factors of major significance are temperature and moisture.</p><p>The magnitude of temperature and its fluctuation affect the properties of certain materials. For example, high temperatures cause asphaltic concrete to lose stability whereas at low temperatures asphaltic concrete becomes very hard and stiff. Low temperature and temperature fluctuations are also associated with frost heave and freeze-thaw damage. Granular materials, if not properly graded, can experience frost heave. Likewise, the sub-grade can exhibit extensive loss in strength if it becomes frozen. Certain stabilized materials can suffer substantial if large numbers of freeze-thaw cycles occur in the material.</p><p>Moisture also has an important influence on the behavior and performance of many materials. Sub-grade soils and other paving materials weaken appreciably when saturated, and certain clayey soils exhibit substantial moisture- induced volume change.</p><p>14.2 FLEXIBLE PAVEMENT DESIGN METHOD</p><p>14.2.1 JKR Method</p><p>This method is a combination of two methods above using a formula and figures from the result of the testing. A complete guideline for pavement design can be found in “Arahan Teknik (Jalan) 5/85”. The thickness of the Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p> pavement depends on the CBR value and the Total Cumulative of Standard Axle ( JBGP ). </p><p>Some data need to be collected before starting any design. They are; i. Design life. ii. Road hierarchy base of JKR classification. iii. Average daily traffic volume. iv. Percentage of commercial vehicle. v. Yearly rate of traffic growth. vi. CBR value for sub-grade. vii. Topography condition.</p><p>14.2.3.1 Design Life</p><p>The design life on JKR Design Method is suggested for 10 years. The design life begins from the road starts in use for traffic until the maintenance is required.</p><p>14.2.3.2 Road Hierarchy Base Of JKR Classification.</p><p> a. Road Classification and its Construction Material.</p><p>CLASS TYPES OF CONSTRUCTION A1 Concrete Surfacing A2 Hard Bituminous Metalled B Hard Waterbound Metalled C Hard Bituminous Sealed D Gravelled Waterbound E Soil Surfacing</p><p> b. Category and its road width. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>CATEGORY WIDTH OF ROAD RESAVE W (m) R (cm) 01 4.5 20.0 02 5.0 30.0 03 6.0 30.0 04 7.0 40.0 05 7.5 40.0 06 3.5 per lane 40.0 or more Notes : 01 – 03 – Village Roads. 04 – 06 – Urban Roads. 14.2.3.3 Classification by JKR Standard</p><p>Road hierarchy Description</p><p>The lowest of hierarchy and geometry design level. Traffic for one R1U & R1(A)U1(A) way. This road hierarchy is same like R3 type. Geometry design R2U 2 U3 R3 level is lowest from U3 type. The lowest hierarchy for single carriageway.</p><p>This road is design for local traffic. Geometry design level – low R3U3 and non inflow traffic control.</p><p>Another road is allowed to intersect in the same level. Geometry R4U 4 design level is intermediate. Allowed maximum velocity – intermediate. </p><p>R5U5 Inflow control degree – half. Distance – quite far. Geometry design level – high. R U 6 6 Inflow control degree – fully. Distance – far. Geometry design level – high </p><p>Notes : R – Rural U – Urban</p><p>14.2.3.3 Traffic Estimation Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>To design using this method, commercial vehicles without loading weight more than 1500 kg are to be taken. </p><p>The formula is include:</p><p> 1 Pc Vo = PLH x365x 2 100 </p><p>Where :</p><p>Vo = Total of Yearly Commercial Vehicle for one direction.( JBKP ) PLH = Average Daily Traffic Ratio for two directions.</p><p>Pc = Commercial Vehicle Percentage.</p><p>To determine the Total of Yearly Commercial Vehicle ( JBKP ) for the one direction for ever lasting Design Life, we have to apply following formula;</p><p> x Vo 1 r 1 Vc = r</p><p>Where;</p><p>Vc = JBKP at one direction for ‘x’ year. r = Rate of Traffic Growth. x = Road pavement Design Life ( in year’s unit ) Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>To determine the Total Cumulative of Standard Axle ( JBGP ) for traffic mixture, equivalent axle load concept is used. </p><p>JBGP = JBKP x equivalent factor</p><p>Where ;</p><p>JBKP = Vc Equivalent Factor = use the data in Table 10.7 = e</p><p>Thus,</p><p>JBKP = Vc x e</p><p>Using the JKR Method, the traffic volume checklist is used by comparing the maximum traffic volume. The formula is;</p><p>X Vx = V1 ( 1 + r )</p><p>Where;</p><p>Vx = The total traffic volume (commercial and non commercial) at the end of pavement design life of one direction after ‘x’ year.</p><p>V1 = The Daily Traffic Volume of one direction r = Rate of Yearly Traffic Growth. x = Life design (in year’s unit) Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p> c = I x R x T Where; c = Maxima Traffic Loading per hour for one way. I = Absolute Traffic Loading per hour - ( Refer Table 10.8 ) R = Road Decreasing Factor – ( Refer Table 10.9 ) T = Traffic Decreasing Factor – ( Refer Table 10.10 )</p><p>In JKR Standard, Traffic Loading for an hour is assumed that equal with 10 % daily loading, as:</p><p>C = 10 x c Where; C = Daily Traffic Loading ( 24 hours traffic loading at one direction) c = Traffic Loading per hour.</p><p>With comparison ‘C’ value and ‘Vx’, we can conclude that;</p><p> a). C > Vx - Road still obtain to support the Traffic Volume at the end of the life design for ‘x’ years. b). C < Vx - Road cannot obtain to support the Traffic volume at the end of the life design for ‘x’ years.</p><p>For (b) condition, this formula are used, Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>C = V ( 1 + r )n</p><p>Where;</p><p> log(C /V ) n = log(1 r)</p><p>C = Daily Traffic Loading for one way ( 24 hours ) V = Daily Traffic Volume for one way that assumed loaded for road. r = Rate of Yearly Traffic Growth. n = Life Design.</p><p>14.2.3.4 CBR-Sub-grade Value</p><p>To determine the CBR value, 1.0 m sub-grade soil must be taken from the hard rock level. To get CBR for sub-grade design, this formula must be applied.</p><p>1/3 1/ 3 1/3 3 (NGC ) h1 (NGC ) h2 ...... (NGC ) hn NGC (%) = 1 2 n 100 </p><p>Where;</p><p>NGC1, NGC2 , = CBR value for layer 1,2 …. h1, h2 = soil deepness from form level for sample 1,2 …. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>14.2.3.5 Pavement Thickness Design</p><p>Figure 10.8 shows the Nomograph Thickness Design that is used to design pavement thickness. An early process of design, the CBR (Sub-grade) and JBGP value must be determined first. To use the Nomograph, the following steps are normally applied.</p><p>Line A value is fixed to 3% and line B is the required JBGP. Draw a straight crossed line. Then determine C values i.e the thickness Equivalent, (TA).</p><p> a. Insert CBR design value at the line A and draw a line by using previous C value until crossing line D, Determine D value i.e the Equivalent Thickness Interval (TA’), if all the entire pavement is constructed from the wearing course or road base.</p><p>To determine the thickness of each pavement layer, table 10.11 and table 10.12 is used with the following formula below i.e to</p><p> determine D1, D2 and D3 value of the surface layers, base and road sub-base.</p><p>SN = a1D1 + a2D2 + a3D3</p><p>Where,</p><p> a1, a2, a3 = Determine from the table 10.11 based on the types of pavement requirement at the certain layers. = Layer Structure Coefficient. </p><p>D1, D2, D3 = Approximate thickness design of the certain Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p> layer ( minimum thickness value according to table 10.12 ) Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>ACTIVITY 14</p><p>TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT Question </p><p>1. State the factors of design that will give impact on to the designing of flexible pavement.</p><p>2. What is the meaning of R2U 2 in road hierarchy outlined by JKR standard. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>FEEDBACK ON ACTIVITY 14</p><p>Answer </p><p>1. a. Failure Criterion b. Traffic Loading c. Climate or Environment d. Moisture</p><p>2. R2U 2 - R3 This road hierarchy is same like U3 type. Geometry design level is R3 lowest from U3 type. The lowest hierarchy for single carriageway. R is for Rural and U is for urban, Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>Question </p><p>A road with hierarchy of 05 has a surface width of 7.0 m and road reserve of 40.0m is to be built as a main road in a residential area. It has a initial average daily traffic of 7000cv/day in both directions. The rate of traffic growth is 7%. Percentage of commercial vehicle is 25%. Design a flexible pavement for the road which needs a design life of 10 years. The CBR for sub-grade of the road is 5%. ( Employ the JKR Malaysia Design Method ).</p><p>Note: Requirement of pavement layers: i. Wearing Course = Asphalt Concrete. ii. Road-Base Course = Broken Aggregate. iii. Sub-Base Course = Broken Aggregate. Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>Answer</p><p> 1 Pc Vo = PLH x365x 2 100 </p><p> 1 25 Vo = 6800 x365x 2 100 = 310250 </p><p> x Vo 1 r 1 Vc = r 10 310250 1 0.07 1 Vc = 0.07</p><p>= 4286552.98</p><p>JBGP = JBKP X EQUAVALENT FACTOR</p><p>JBKP = Vc x e Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>JBKP = 4.29 x 106 x3.0</p><p>= 12.87 x 10 6</p><p> x Vx = V1 ( 1 + r )</p><p>10 Vx = 6800/2( 1 + 0.07 )</p><p>Vx = 6689</p><p> c = I x R x T</p><p>I = Absolute Traffic Loading per hour - ( Refer Table 10.8 ) = 2000/2 = 1000</p><p>R = Road Decreasing Factor – ( Refer Table 10.9 ) = 1.0</p><p>T = Traffic Decreasing Factor – ( Refer Table 10.10 ) Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>= 100/(100 + 25 ) = 0.8</p><p> c = I x R x T = 1000 x 1.00 x 0.8 = 800 vec/hr/lane</p><p>C = 100c</p><p>= 100(800 vec/hr/lane )</p><p>= 8000 vec/day/lane</p><p>From Nomograph Diagram;</p><p>D = 43 cm</p><p>From table10.11 and table 10.12</p><p> a1 = 1.00</p><p> a2 = 0.32</p><p> a3 = 0.25 Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>From table 10.12</p><p>Wearing Course = 4 + 5 </p><p>= 9 cm</p><p>Base Course = 10 cm</p><p>Sub Base = 10 cm granular</p><p>SN = a1D1 + a2D2 + a3D3</p><p>SN = 1D1 + 0.32D2 + 0.25D3 = 43 cm</p><p>Try and Error</p><p>1. D1 = 9 D2 = 10 D3 = 10</p><p>SN = 1(9) + 0.32(10) + 0.25(10) = 14.7 cm < 43 cm </p><p>2. D1 = 20 D2 = 40 D3 = 50</p><p>SN = 1(20) + 0.32(40) + 0.25(50) = 46 cm > 43 cm </p><p>3. D1 = 18 Highway Engineering C3010 / UNIT 14 FLEXIBLE PAVEMENT DESIGN</p><p>D2 = 40 D3 = 50</p><p>SN = 1(18) + 0.32(40) + 0.25(50) = 43.3 cm < 43 cm OK</p>
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