Report

PRESENTATION ON ROAD PAVEMENT DESIGN BY Engr. Ejaz Ahmad Khan CEE 320 Winter 2006 Deputy Director Pakhtunkhwa Highways Authority CEE 320 Winter 2006 CEE 320 Steve Muench OUTLINE Section - 1 • Pavement Structure Section - 2 • Design of Pavement Structure Section - 3 • Flexible Pavement Design • How to Design • Practical Example Section - 4 CEE 320 Winter 2006 Section - 5 Section - 1 CEE 320 Winter 2006 Pavement Structure PAVEMENT : Combination of various layers between road top surface / Finished Road Level (FRL) and the subgrade is known as pavement structure. CEE 320 Winter 2006 Pavement Structure: PAVEMENT PURPOSE CEE 320 Winter 2006 • • • • • • Load support Skid Resistance Good ride Less VOC Time Saving Drainage CHAPPAR - DARBAND ROAD (30 KM) PHASE-I CEE 320 Winter 2006 PHILOSOPHY OF PAVEMENTS • Pavements are subjected to moving traffic loads that are repetitive in nature. • Each traffic load repetition causes a certain amount of damage to the pavement structure that gradually accumulates over time and eventually leads to the pavement failure. • Thus, pavements are designed to perform for a certain life span before reaching an unacceptable degree of deterioration. • In other words, pavements are designed to fail. Hence, they have a certain design life. CEE 320 Winter 2006 PAVEMENT TYPES • Flexible Pavement – Hot mix asphalt (HMA) pavements – Called "flexible" since the total pavement structure bends (or flexes) to accommodate traffic loads – The load transmit to the subgrade through particle to particle contact. • Rigid Pavement – Portland cement concrete (PCC) pavements – Called “rigid” since PCC’s high modulus of elasticity does not allow them to flex appreciably – The load transmit to subgrade through beam action. FLEXIBLE PAVEMENT CEE 320 Winter 2006 • Structure – Surface course – Base course – Subbase course – Subgrade RIGID PAVEMENTS In rigid pavements the stress is transmitted to the sub-grade through beam/slab effect. Rigid pavements contains sufficient beam strength to be able to bridge over localized sub-grade failures and areas of inadequate support. CEE 320 Winter 2006 Thus in contrast with flexible pavements the depressions which occur beneath the rigid pavement are not reflected in their running surfaces. RIGID PAVEMENT CEE 320 Winter 2006 • Structure – Surface course – Base course – Subbase course – Subgrade Section - 2 CEE 320 Winter 2006 Design of Pavement Structure PAVEMENT THICKNESS DESIGN Pavement Thickness Design is the determination of required thickness of various pavement layers to protect a given soil condition for a given wheel load. Given Wheel Load 150 Psi Asphalt Concrete Thickness? CEE 320 Winter 2006 3 Psi Given In Situ Soil Conditions DESIGN PARAMETERS CEE 320 Winter 2006 • Subgrade • Loads • Environment SUBGRADE CEE 320 Winter 2006 • Characterized by strength and/or stiffness – California Bearing Ratio (CBR) • Measures shearing resistance • Units: percent • Typical values: 0 to 20 – Resilient Modulus (MR) • Measures stress-strain relationship • Units: psi or MPa • Typical values: 3,000 to 40,000 psi Picture from University of Tokyo Geotechnical Engineering Lab SUBGRADE CEE 320 Winter 2006 Some Typical Values Classification CBR MR (psi) Typical Description Good ≥ 10 20,000 Gravels, crushed stone and sandy soils. Fair 5–9 10,000 Clayey gravel and clayey sand, fine silt soils. Poor 3–5 5,000 Fine silty sands, clays, silts, organic soils. TRAFFIC LOADS CHARACTERIZATION CEE 320 Winter 2006 Cars Pickups Buses Trucks Trailers EQUIVALENCY FACTOR Equivalent Standard 18000 - Ibs ESAL (8.2 tons) Damage per Pass = 1 Axle Load • Axle loads bigger than 8.2 tons cause damage greater than one per pass CEE 320 Winter 2006 • Axle loads smaller than 8.2 tons cause damage less than one per pass • Load Equivalency Factor (L.E.F) = (? Tons/8.2 tons)4 EXAMPLE Consider two single axles A and B where: A-Axle = 16.4 tons Damage caused per pass by A -Axle = (16.4/8.2)4 = 16 This means that A-Axle causes same amount of damage per pass as caused by 16 passes of standard 8.2 tons axle i.e., CEE 320 Winter 2006 = 16.4 Tons Axle 8.2 Tons Axle DAMAGE PER PASS 80 70 60 50 40 30 20 10 0 1.0 1.1 2.3 3.3 4.7 6.5 8.7 11.5 14.9 18.9 23.8 29.5 36.3 44.1 53.1 63.4 75.2 AXLE LOAD & RELATIVE DAMAGE CEE 320 Winter 2006 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 SINGLE AXLE LOAD (Tons) SERVICEABILITY CONCEPT OF PAVEMENT • • CEE 320 Winter 2006 • Serviceability Serviceability is the ability of a pavement to serve the commuters for the desired results for which it has been constructed within the designed life and without falling the Terminal level (TSI). Present Serviceability Index (PSI) Present Serviceability is defined as the adequacy of a section of pavement in its existing condition to serve its intended use. Terminal Serviceability Index (TSI) It is defined as that stage of the pavement condition after which it is not acceptable for the road users. SERVICEABILITY CONCEPT OF PAVEMENT • Defined by users (drivers) • Develop methods to relate physical attributes to driver ratings • Result is usually a numerical scale CEE 320 Winter 2006 From the AASHO Road Test (1956 – 1961) Present Serviceability Index (PSI) • Values from 0 through 5 • Calculated value to match PSR PSI 5.411.80log 1 SV 0.9 C P SV = mean of the slope variance in the two wheel paths (measured with the profile meter) C, P = measures of cracking and patching in the pavement surface CEE 320 Winter 2006 C = total linear feet of Class 3 and Class 4 cracks per 1000 ft2 of pavement area. A Class 3 crack is defined as opened or spilled (at the surface) to a width of 0.25 in. or more over a distance equal to at least one-half the crack length. A Class 4 is defined as any crack which has been sealed. P = expressed in terms of ft2 per 1000 ft2 of pavement surfacing. CEE 320 Winter 2006 Serviceability (PSI) PSI vs. Time p0 p0 - pt pt Time PAVEMENT THICKNESS DESIGN Comprehensive Definition CEE 320 Winter 2006 Pavement Thickness Design is the determination of thickness of various pavement layers (various paving materials) for a given soil condition and the predicted design traffic in terms of equivalent standard axle load that will provide the desired structural and functional performance over the selected pavement design life i.e. the serviceability may not fall below the TSI. Section - 3 CEE 320 Winter 2006 Flexible Pavement Design Flexible Pavements CEE 320 Winter 2006 A flexible pavement absorbs the stresses by distributing the traffic wheel loads over much larger area, through the individual layers, until the stress at the subgrade is at an acceptably low level. The traffic loads are transmitted to the subgrade by aggregate to aggregate particle contact. A cone of distributed loads reduces and spreads the stresses to the subgrade. TYPICAL LOAD & STRESS DISTRIBUTION IN FLEXIBLE PAVEMENTS. Wheel Load Bituminous Layer Vertical stress CEE 320 Winter 2006 Foundation stress Sub-grade CEE 320 Winter 2006 EFFECT OF PAVEMENT THICKNESS ON STRESS DISTRIBUTION BASIC EQUATION OF AASHTO PROCEDURE FOR FLEXIBLE PAVEMENT DESIGN. PSI log10 4.5 1.5 log10 W18 Z R So 9.36 log10 SN 1 0.20 2.32 log10 M R 8.07 1094 0.40 SN 15.19 The various terms/parameters which are used in the basic equation of AASHTO Procedure for the Design of flexible pavements are: CEE 320 Winter 2006 i). W18 (ESAL): It is the accumulated traffic load converted to 18kips or 8.2 tons. This is also known as 18-kips Equivalent Standard Axle Load (ESAL). That the pavement will experience over its design life. ii). Standard Deviation (S0): Standard deviation accounts for standard variation in materials and construction, the probable variation in traffic prediction and variation in pavement performances for a given design traffic application. The recommended value of S0 for flexible pavement is 0.4 to 0.5. iii) Reliability (R): CEE 320 Winter 2006 Design Reliability refers to the degree of certainty that a given pavement section will last for the entire design period with the traffic & environmental condition. The recommended level of reliability for freeways in rural areas varies from 80% to 95%. A high reliability value will increase the thickness of pavement layer and will result in expensive construction. TABLE FOR REALABLILITY Recommended Level of Reliability ( R ) Reliability (%) Functional Classification Urban Rural 85-99.90 80-99.90 Principal Arterial 80-99 75-95 Feeders 80-95 75-95 Local 50-80 50-80 CEE 320 Winter 2006 Interstate and Other Freeways CEE 320 Winter 2006 iv). Standard Normal Deviate (ZR): It is defined as the probability that serviceability will be maintained at adequate levels from a user’s point of view throughout the design life of the facility. This factor estimates the probability that the pavement will perform at or above the TSI level during the design period and it accounts for the inherent uncertainty in design. The relationship of reliabilities with ZR is given in the table: Value of (ZR) Reliability R (%) 50 60 70 75 80 85 90 91 92 93 94 95 96 97 98 99 99.9 99.99 Standard Normal Deviate (ZR) 0.000 -0.253 -0.524 -0.674 -0.841 -1.037 -1.282 -1.340 -1.405 -1.476 -1.555 -1.645 -1.751 -1.881 -2.054 -2.327 -3.090 -3.750 v). Structural No (SN): Structural No is the total structural strength value required to cater for the cumulative equivalent standard axles load (CESAL) during design life so that the serviceability may not fall below the Terminal serviceability Index (TSI) Definition of Structural Number CEE 320 Winter 2006 Subgrade Structural Coefficient (a): a = fnc (MR) SN = SN1 + SN2 + SN3 vi) Loss of Serviceability Index ∆ PSI. Serviceability (PSI) ∆ PSI = Initial Serviceability Index – Terminal Serviceability Index The recommended value for initial serviceability index is 4.2 while for terminal serviceability index it is to 2 to 2.5. ∆ PSI = 4.2 – 2.5 = 1.7 p0 p0 - pt CEE 320 Winter 2006 pt Time vii). Resilient Modulus (MR): CEE 320 Winter 2006 It is defined as repetitive or cyclic stress divided by recoverable strain. Resilient Modulus is a measure of stiffness of the soil. MR = Repetitive stress / recoverable strain MR can be determined from the resilient modules test in the laboratory or from the following equations: MR = 1500 * CBR for CBR < 10 % MR = 2555 (CBR)0.63 for any value of CBR viii). Computation of Required Pavement Thicknesses The structure Number (SN) requirement as determined through adoption of design parameters as discussed above is balanced by providing adequate pavement structure. Under AASHTO design procedure the following equation provides for the means for converting the structural number into actual thickness of surfacing, base and sub base materials. SN = a1D1 + a2D2m2 + a3D3m3 CEE 320 Winter 2006 a1, a2, a3 = D1. D2, D3 = m 2, m 3 = _______________ (2) Layer coefficients representative of surface, base and subbase courses respectively. It depends upon the modulus of resilient. Actual thicknesses (in inches) of surface, base and subbase courses respectively. Drainage coefficients for base and subbase layers respectively. This equation does not have a single unique solution. There are many combinations of layer thicknesses that can be adopted to achieve a given structural number. There are, however, several design, construction and cost constraints that may be applied to reduce the number of possible layer thicknesses combinations and to avoid the possibility of constructing an impractical design. According to this approach, minimum thickness of each layer is specified to protect the under lying layers from shear deformation. ix). Recommended Value of Layer Coefficients Asphaltic Wearing Course, a1 = 0.44/inch (0.1732/cm) Asphaltic Base Course, a1 = 0.40/inch (0.1575/cm) Water Bound Macadam, a2 = 0.14/inch (0.0551/cm) Granular Subbase, a3 = 0.11/inch (0.0433/cm) CEE 320 Winter 2006 OR Nomograph can be used to work out SN. CEE 320 Winter 2006 NOMOGRAPH Section - 4 CEE 320 Winter 2006 How to Design How to Design Step 1. Fix the design life of the pavement. Step 2. Work out MR value of the subgrade MR = 1500 CBR for CBR <10% OR MR = 2555 (CBR)0.63 for CBR > 10 OR Work out MR in the laboratory. Step 3. Conduct 7-days traffic count. CEE 320 Winter 2006 Step 4. Classify the traffic and consider the commercial vehicles i.e. Bus, Tractor , Trolley, 2-Axle, 3-Axle, 4-Axle, 5-Axle and 6-Axle Trucks. Step 5. Take Growth rate from the table on the next slide. CEE 320 Winter 2006 Growth Rate S. No Vehicle Class Growth Rate 1 Bus 8.4% 2 Tractor Trolley 7.9% 3 Mini Truck 7.9% 4 2-Axle 7.0% 5 3-Axle (Single) 7.0% 6 3-Axle (Tandem) 7.0% 7 4-Axle 7.0% 8 5-Axle 7.0% 9 6-Axle 7.0% CEE 320 Winter 2006 CONVERT THE TRAFFIC TO EQUIVALENT STANDARD AXLE LOAD. ESAL = TRAFFIC X EQUILLANCY FACTOR , EQUIVALENCY FACTOR FOR VARIOUS CLASSES OF VEHICLES ARE GIVEN IN THE FOLLOWING TABLE. S. No Vehicle Class Equivalency Factor (Empty) Equivalency Factor (Loaded) 1 Bus 0 0.939 2 Tractor Trolley 0.1 1.19 3 Mini Truck 0.0172 2.596 4 2-Axle 0.043 6.49 5 3-Axle (Single) 0.072 16.62 6 3-Axle (Tandem) 0.072 18.48 7 4-Axle 0.206 19.00 8 5-Axle 0.084 19.59 9 6-Axle 0.165 27.96 Calculation of CESAL ESA Factor CEE 320 Winter 2006 Vehicle Type ADT Annual Traffic Growth Rate % Growth Factor Total Traffic for 10 Years 80% 20% Loaded Empty CESAL for 10 Years Buses 20 7300 6 13.18 96,214 0.939 0 72,276 Tractor Trolly 139 50735 6 13.18 668,687 1.19 0.1 649,964 Trucks2XL 500 182500 6 13.18 2,405,350 6.49 0.043 12,509,263 Trucks 3XL 250 91250 6 13.18 1202675 18.48 0.072 17,797,666 Cumulate the future traffic throughout the design life with the help of the selected growth rate. Following is the simple relation to project the traffic to any selected year. Pn = (1 + r)n – 1 Where Pn = Projected traffic for nth year r = Growth rate n = year of consideration CEE 320 Winter 2006 Add all the yearly traffic from base year to the last year of the design life. CEE 320 Winter 2006 Step 6. Fix the parameter like R, ZR, So, ∆ PSI etc. The generally taken value of the above parameters is listed below: ∆ PSI = 1.7 R = 90% So = 0.45 ZR = -1.282 Step 7. Put these values in equation 1 and use trial & error method or Nomograph to work out the SN SN = a1D1+a2D2m2+a3D3m3 Step 8. Take the value of m2 and m3 from the table on the next slide. TABLE FOR QULALITY OF DRAINAGE CEE 320 Winter 2006 QUALITY OF DRAINAGE Quality of Drainage Water Removed within Excellent 2 hours Good 1 day Fair 1 week Poor 1 month Very Poor water will not drain Percent of Time Pavement Structure is exposed to Moisture Levels Approaching Saturation Quality of Drainage Excellent Good CEE 320 Winter 2006 Fair Poor Very Poor Less Than Greater Than 1% 1 - 5% 2 - 25% 25% 1.40 - 1.35 1.35 - 1.30 1.30 - 1.20 1.20 1.35 - 1.25 1.25 - 1.15 1.15 - 1.00 1.00 1.25 - 1.15 1.15 - 1.05 1.00 - 0.80 0.85 1.15 - 1.05 1.05 - 0.80 0.80 - 0.60 0.60 1.05 - 0.95 0.95 - 0.75 0.75 - 0.40 0.40 Put the above values in equation at step No. 07, to find out the various combination of thicknesses, keeping in view the minimum thicknesses requirements as mentioned below: – – – – Minimum Asphalt wearing course thickness Minimum asphaltic base course thickness Minimum unbound base course thickness Minimum unbound sub base thickness = 5 Cm = 7.5 Cm = 15 Cm = 15 Cm CEE 320 Winter 2006 Select the most appropriate and economical combination of thicknesses. Section - 5 CEE 320 Winter 2006 Practical Example Practical Example CEE 320 Winter 2006 • Let us work out the thicknesses of various layers for the Pavement of Topi Bypass Road. 1. The ADT is given as follow. Vehicle Type ADT COASTER/ FLYING COACH 250 BUSSES 25 Tractor Trolley 36 Trucks2XL 110 Trucks 3XL 2 Trucks 4XL 5 THE CESAL IS WORKED OUT AS FOLLOW:- Vehicle Type ESA Factor Total Annual Growth Growth Traffic ADT Traffic Rate % Factor for 10 Years 80% 20% Loaded Empty COASTER/FLYING COACH 250 91250 8.4 14.78 1,348,675 0.939 0 1,013,125 BUSSES 25 9125 8.4 14.78 134,868 0.939 0 101,312 Tractor Trolly 36 13140 7.9 14.423 189,518 1.19 0.1 184,212 Trucks2XL 110 40150 7 13.82 554,873 6.49 0.043 2,885,673 Trucks 3XL 2 730 7 13.82 10088.6 18.48 0.072 149,295 Trucks 4XL 5 1825 7 13.82 25221.5 19 0.385 385,309 Total CEE 320 Winter 2006 ESAL for 10 Years 4,718,925 ESAL by taking 100 % of directional factor 4718925.35 = 4.719million ESAL by taking 80 % lane factor 3775140.28 = 3.775million CESAL =3.775 million CEE 320 Winter 2006 • California Bearing Ratio (CBR)= 30 % at 95% MDD • MR= 2555 (CBR)0.63 for CBR > 10 Putting the value CBR , MR= 18000 Psi • Keeping the value of various parameters as follow. R= 90% So=0.45 ∆Psi=1.7 Using Nomograph to work out the SN CEE 320 Winter 2006 Required SN = 3.35 54 • Using the following the equation • SN = a1D1+a2D2m2+a3D3m3 CEE 320 Winter 2006 • Given Data: • a1=0.44, • a2=0.14 , • a3=0.11 and • m2,m3=1 • from Nomograph SN= 3.35 • Putting these values and assuming d1=2 inch, d2=10 inch and d3=10 inch 3.35=0.44*2+0.14*10+0.11*10 3.35≈3.38 Hence the Design thickness are ACWC= 5cm WBM=25cm Granular sub base=25cm CEE 320 Winter 2006 Thank You