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Chapter 3

Pavement Design

14 teaching hours 16 marks Transportation Engineering II

3.1 Introduction, types and structural components of pavement208220802073

Past Exam Questions on this Topic

  1. Describe the functions of pavement layers. What are the different types of joints used in the rigid pavement? Explain in brief: [4+4] (2082 Bhadra, Q5)
    Solution

    (a) Functions of the pavement layers (top → bottom):

    LayerFunction
    Wearing / surface courseProvides a smooth, skid-resistant, dust-free riding surface; resists abrasion and the direct wheel load; waterproofs the layers below.
    Binder courseLower part of the bituminous surfacing; takes the high shear from the wheel load and hands it to the base with a coarser, cheaper mix.
    Base courseThe main load-spreading layer — distributes the wheel load over a wider area so that the stress reaching the sub-base is small.
    Sub-baseSpreads the load further, provides drainage, prevents the subgrade from intruding into the base, and acts as a frost/capillary cut-off and a working platform.
    SubgradeCompacted natural soil (top 300-500 mm) that finally carries the load; its CBR / MR governs the total crust thickness.

    (b) Joints in rigid pavement

    • Expansion joint — full-depth transverse gap (20-25 mm) with a compressible filler and sealant; allows the slab to expand in summer. Load transfer by dowel bars (one end free to slide). IRC spacing ≈ 140 m on rough sub-base / 90 m on smooth sub-base.
    • Contraction joint — a groove cut to ≈ ¼ of the depth so that shrinkage/contraction cracks form at a chosen line; spacing ≤ 4.5 m for plain slabs. Load transfer by aggregate interlock (and dowels for thick slabs).
    • Construction joint — provided wherever concreting stops for the day or is interrupted; made to coincide with a contraction joint wherever possible.
    • Warping (hinge) joint — provided to relieve the stresses caused by warping; the slabs are held together by tie bars, so no movement is permitted.
    • Longitudinal joint — provided when the carriageway is wider than 4.5 m; it takes care of longitudinal cracking due to warping and differential settlement. Tie bars are used (deformed bars, fully bonded at both ends).

    Layer functions and joint types & spacings as given in Manual.pdf, Highway Pavement (Ch.2, book pp.113-115) and IRC joint recommendations, book p.168.

  2. Describe four types of rigid pavement with simple sketch: Discuss the importance of standard axle load in flexible pavement design; [5+3] (2080 Bhadra, Q6)
    Solution

    (a) Four types of rigid (cement-concrete) pavement

    TypeDescription / sketch
    JPCP — Jointed Plain Concrete PavementPlain concrete slabs, no temperature steel; closely spaced contraction joints (≤ 4.5 m) with dowel bars for load transfer. Sketch: slab – joint – slab, dowels across the joint.
    JRCP — Jointed Reinforced Concrete PavementLight mesh reinforcement (2-4 kg/m²) holds the intermediate cracks tight so that joints can be spaced wider (10-15 m); dowels at the joints. Sketch: longer slab with mesh at mid-depth.
    CRCP — Continuously Reinforced Concrete Pavement0.6-0.7% continuous longitudinal steel; no transverse contraction joints — fine transverse cracks at 0.6-2 m are held tightly closed by the steel. Sketch: continuous slab with hairline cracks.
    PCP — Prestressed Concrete PavementPre-compressed slab, so the flexural tensile stress under load is largely cancelled; allows a much thinner slab and very long panels (100 m+). Sketch: slab with prestressing tendons.

    (b) Importance of the standard axle load

    • Highway traffic is a mixture of axles of very different magnitude. Pavement damage is not proportional to load — it follows the fourth-power law, so a single heavy axle can do the damage of thousands of car axles.
    • A common yardstick is therefore fixed: the standard axle = 80 kN (8.16 t) single axle with dual wheels. Every other axle is converted with the equivalent-axle-load factor EF = (axle load / 80)4, and per vehicle VDF = Σ(axle load/80)4.
    • The whole mixed traffic then becomes one number — the cumulative standard axles CSA = 365·A·[((1+r)n−1)/r]·LDF·VDF (msa) — which is the single traffic input for the IRC-37/DoR catalogues and for the fatigue and rutting criteria.
    • It also gives a rational basis for legal axle-load limits, for overloading penalties and for comparing the damage caused by different vehicle classes.

    Manual.pdf, Highway Pavement — loads and factors controlling pavement design (book pp.116-121) and the CSA/VDF definition used with the IRC catalogues.

  3. What are the advantages of rigid pavement over flexible pavement? Explain the concept of cumulative standard axle load. (2073 Shrawan, Q5)
    Solution

    (a) Advantages of rigid pavement over flexible pavement

    • Slab (beam) action distributes the wheel load over a very wide area, so a weak subgrade is much less critical — the subgrade stress is only a fraction of that under a flexible pavement.
    • Longer design life (30-40 yr against 15-20 yr) and much lower maintenance.
    • High flexural strength; no rutting, shoving, corrugation or bleeding.
    • Unaffected by oil, fuel and chemical spillage — important at junctions, bus bays, toll plazas and parking areas.
    • Light coloured surface → better night visibility and lower lighting cost.
    • Lower rolling resistance → better fuel economy for heavy vehicles; lower life-cycle cost.
    • Not softened by high temperature and not damaged by standing water.

    (b) Concept of cumulative standard axle load (CSA)

    Mixed traffic cannot be used directly for design, so every axle is first converted to the number of standard 80 kN axles that would cause the same damage:

    EF = (axle load / 80 kN)⁴ VDF = Σ EF summed over all the axles of one vehicle

    The daily commercial traffic is then projected to the year of completion, distributed to the design lane and accumulated over the design life with the growth series:

    A = P (1 + r)ˣ x = years from the count to completion CSA = 365 × A × [((1 + r)ⁿ − 1)/r] × LDF × VDF

    where P = commercial vehicles per day at the last count, r = growth rate, n = design life, LDF = lane distribution factor (single lane 1.0, two-lane single carriageway 0.75, four-lane single carriageway 0.40, dual two-lane 0.75 of each direction) and VDF = vehicle damage factor.

    CSA, expressed in million standard axles (msa), is the traffic input with which the pavement composition is picked from the IRC-37 / DoR design catalogue, and against which the fatigue-cracking and rutting criteria are checked.

    Lane-distribution factors and the design-traffic equation are taken verbatim from Manual.pdf book p.122 (§2.4 — Lane distribution factor) and §2.4.1.

Definition

Structural Components of Pavement

Pavement Layers — Indian practice (left) and American practice (right):

Surface course

Base and Subbase course

Subgrade

Binder course

Types of Pavement

Depending on structural behavior:

Flexible Pavement

Rigid Pavement

Semi-Rigid Pavement

Composite Pavement

Difference between Flexible and Rigid Pavement

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3.2 Factors controlling pavement design20812078

Past Exam Questions on this Topic

  1. Explain the various factors affecting the pavement design. [8] (2081 Bhadra, Q5)
    Solution

    The design of a pavement is controlled by four groups of factors.

    1. Traffic and loading factors

    • Axle load — the wheel load is transmitted to the surface through the tyre; the heavier the wheel load the thicker the pavement. Design is governed by the commercial vehicles.
    • Wheel configuration — single / dual / tandem wheels change the stress distribution; a dual wheel set is converted to an Equivalent Single Wheel Load (ESWL).
    • Axle configuration — single, tandem, tridem axles carry more load with less damage per tonne; converted by the fourth-power law.
    • Number of load repetitions — accumulated as CSA (= 365·A·[((1+r)n−1)/r]·LDF·VDF).
    • Contact area / tyre pressure — fixes the radius of the loaded area a; for thin pavements the tyre pressure governs, for thick pavements the total load governs.
    • Vehicle speed — slow (creep) traffic loads the layer far longer, lowering the resilient modulus of the bituminous layers and increasing rutting; critical on gradients and at junctions.

    2. Material properties (including the subgrade)

    • Subgrade strength expressed as soaked CBR or resilient modulus MR (MR = 10.3×CBR MPa); the design value is the 87.5 / 90 percentile of the tested samples.
    • Elastic / resilient moduli and Poisson's ratios of the bituminous, granular and cemented layers; for repeated loading the resilient modulus is the correct parameter.
    • Quality, grading and durability of aggregates; binder type and content.

    3. Environmental factors

    • Temperature — changes the modulus of the bituminous layer (softening in summer, low-temperature cracking in winter); in rigid pavements it produces the warping and frictional stresses.
    • Precipitation / drainage — infiltration raises the subgrade moisture, lowers its CBR and causes pumping and loss of support; frost action in cold regions.
    • Road geometry (gradient, curvature) has a small but real effect through speed and drainage.

    4. Failure criteria — the design must satisfy the distress models used by the method:

    • Fatigue cracking — limits the horizontal tensile strain at the bottom of the bituminous layer.
    • Rutting — limits the vertical compressive strain on top of the subgrade (or the rut depth).
    • Thermal cracking; and for rigid pavements, fatigue of the concrete under the combined load + warping stress against the modulus of rupture.

    The four-group classification is exactly the one used in Lecture 14 (§3.2 of these notes) and in Manual.pdf §2.3 — "Loads and other factors controlling pavement design", book pp.116-121.

  2. What are the factors that affect in pavement design? How are they considered in AASHTO method? [1+7] (2081 Baishakh, Q5)
    Solution

    (a) Factors affecting pavement design — traffic and loading (axle load, wheel/axle configuration, repetitions, contact pressure, speed), subgrade and material properties, environment (temperature, moisture, drainage, frost) and the adopted failure criteria.

    (b) How AASHTO-1993 accounts for them

    All of the above enter the single empirical design equation obtained from the AASHO Road Test:

    log₁₀W₁₈ = ZRS₀ + 9.36 log₁₀(SN+1) − 0.20 + log₁₀[ΔPSI/(4.2−1.5)] / [0.40 + 1094/(SN+1)5.19] + 2.32 log₁₀MR − 8.07
    FactorHow it enters the AASHTO method
    Traffic & loadingW18 = cumulative 80 kN (18-kip) ESAL in the design lane, obtained from the traffic count × truck factor × growth factor × directional factor × lane-distribution factor.
    ReliabilityZR (standard normal deviate for the chosen reliability R) and S0 (overall standard deviation, 0.40-0.50 for flexible pavements) shift the design to the safe side.
    Serviceability (riding quality)ΔPSI = pi − pt (initial minus terminal present serviceability index, e.g. 4.2 − 2.5).
    SubgradeEffective roadbed resilient modulus MR (psi), obtained from the soaked CBR (MR ≈ 1500×CBR psi = 10.3×CBR MPa) with a seasonal correction.
    MaterialsStructural layer coefficients a1, a2, a3 correlated with the layer moduli, e.g. a1 = 0.323·log10E1 − 1.42, a2 = 0.249·log10E2 − 0.977, a3 = 0.227·log10E3 − 0.839.
    Environment / drainageDrainage coefficients m2, m3 (0.40-1.40) applied to the unbound layers; the seasonal variation of MR is averaged over the year.

    The equation is solved for the required structural number SN, and the thicknesses follow from

    SN = a₁D₁ + a₂D₂m₂ + a₃D₃m₃

    worked layer by layer — SN1 is computed with the MR of the base, SN2 with that of the sub-base and SN3 with that of the subgrade, so that each layer is protected by the layers above it.

    Manual.pdf §2.4.4 — AASHTO method of flexible pavement design, book pp.144-149.

  3. Explain Vehicle Damage Factor (VDF) used in design of flexible pavements. An existing two-lane single carriageway highway is proposed to be widened to a new 4-lane divided carriageway. Design the flexible pavement for the proposed highway using the information provided in the design catalogues: i) 4-lane divided carriageway; ii) The expected traffic in the year of completion of construction is 4000 commercial vehicles per day in both directions with 30% heavy trucks of 100 kN axle load, 50% light trucks of 80 kN axle load and 20% tractors of 70 kN axle load; iii) Design life = 15 years; iv) Percentage CBR values obtained from seven different locations along the stretch of the highway are 9, 7, 8, 5, 4, 5 and 6.5 respectively; v) Traffic growth rate = 8%. (Design catalogues for CBR 4% and CBR 5% are printed on the question paper.) [2+6] (2078 Bhadra, Q1)
    Design catalogue — subgrade CBR 4%
    Cumulative traffic (msa)Total pavement thickness (mm)BC (mm)DBM (mm)Granular base and sub-base (mm)
    107004080base 250
    sub-base 330
    2073040110
    3075040130
    5078040160
    10080050170
    15082050190
    Design catalogue — subgrade CBR 5%
    Cumulative traffic (msa)Total pavement thickness (mm)BC (mm)DBM (mm)Granular base and sub-base (mm)
    106604070base 250
    sub-base 300
    2069040100
    3071040120
    5073040140
    10075050150
    15077050170
    Solution

    (a) Vehicle Damage Factor (VDF) [2]

    VDF is the number of standard 80 kN axles that cause the same pavement damage as one pass of the commercial vehicle in question:

    VDF = Σall axles (axle load / 80 kN)⁴

    It converts a mixed axle-load spectrum into a single equivalent traffic figure. VDF is obtained from an axle-load survey; in its absence IRC-37 gives indicative values by terrain and by initial traffic volume (e.g. 3.5 for plain/rolling terrain with 150-1500 cvpd). Cumulative standard axles then follow as CSA = 365·A·[((1+r)n−1)/r]·LDF·VDF.

    (b) Design of the widened 4-lane divided carriageway [6]

    Traffic in the year of completion = 4000 cvpd in both directions — 30% heavy trucks @ 100 kN axle, 50% light trucks @ 80 kN axle, 20% tractors @ 70 kN axle · design life n = 15 yr · growth r = 8% · CBR at seven locations = 9, 7, 8, 5, 4, 5, 6.5%

    Vehicle damage factor

    Vehicle classShareAxle loadEF = (L/80)4Share × EF
    Heavy truck30%100 kN(100/80)4 = 2.44140.7324
    Light truck50%80 kN(80/80)4 = 1.00000.5000
    Tractor20%70 kN(70/80)4 = 0.58620.1172
    VDF1.3497

    Traffic in the design lane

    A four-lane divided highway is a dual two-lane carriageway; the Manual (book p.122) gives LDF = 75% of the commercial vehicles in each direction.

    traffic per direction = 4000 / 2 = 2000 cvpd Alane = 0.75 × 2000 = 1500 cvpd

    Cumulative standard axles

    The traffic is already given at the year of completion, so no further projection is needed.

    growth factor = [(1.08)15 − 1]/0.08 = (3.17217 − 1)/0.08 = 27.152 CSA = 365 × 1500 × 27.152 × 1.3497 = 2.01 × 10⁷ ≈ 20 msa

    Design subgrade CBR (87.5 percentile)

    Arrange the seven values in descending order and take percentile = (rank/N)×100:

    CBR (%)9876.5554
    Rank1234567
    Percentile14.328.642.957.171.485.7100
    interpolate between (85.7 %, CBR 5) and (100 %, CBR 4): CBR87.5 = 5 + (87.5 − 85.7)/(100 − 85.7) × (4 − 5) = 4.87 % ≈ 5 %

    Being marginally below 5%, the CBR 4% catalogue is adopted (safe side).

    Read the design catalogue supplied with the question (CBR 4%, 20 msa)

    Selected pavement composition
    LayerThickness
    Bituminous concrete (BC)40 mm
    Dense bituminous macadam (DBM)110 mm
    Granular base (WMM)250 mm
    Granular sub-base (GSB)330 mm
    Total crust730 mm
    Provide 40 mm BC + 110 mm DBM + 250 mm granular base + 330 mm granular sub-base = 730 mm total crust.

    If the CSA is rounded up to the next catalogue step of 30 msa the total becomes 750 mm with 130 mm DBM.

    LDF = 0.75 per direction for a dual two-lane carriageway and the 87.5-percentile tabulation are from Manual.pdf book pp.122 and 178-179; the catalogue is the one printed on the question paper (2078 Bhadra, page 1).

Factors Controlling Pavement Design

  1. Traffic and loading factors
  2. Material properties
  3. Environment factor
  4. Failure criteria

1. Traffic and Loading Factors

Axle load

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Flexible pavement: Wheels on only one side is considered

Rigid pavement: Wheels on both sides are considered

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Number of Load Repetitions

Contact Area

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Vehicle Speed

2. Material Properties

3. Environment Factor

Temperature

Both temperature and precipitation affect elastic moduli of pavement layers.

Flexible pavement:

Rigid pavement:

Precipitation

4. Failure Criteria

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Rigid pavement:

Practice Tasks from the Lecture

3.3 Historical development of pavement structures (Flexible pavement design): Theoretical (Boussinesq and Burmister approach); Semi-empirical (Tri-axial approach); and empirical methods

Flexible Pavement Design: Historical Development

  1. Theoretical / mechanistic methods
  2. Semi-empirical method
  3. Empirical methods
  4. Mechanistic-empirical method

Theoretical/Mechanistic Design Method

Boussinesq’s Theory

Assumptions The soil mass is:

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Drawbacks

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Burmister’s Theory

Assumptions

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Displacement factor(𝑭𝒘)

Burmister’s Theory

Burmister’s Theory

Problem Plate bearing tests were conducted with a 75cm diameter plate on soil subgrade and a granular base. The stress was 0.02MN/m2 when the deflection on the subgrade soil was 0.25cm. The same plate yielded 0.25cm deflection on the base course of 15cm under a stress of 0.05MN/m2. Design the pavement for an allowable deflection of 0.5cm under a wheel load of 40KN and tire pressure of 0.5MN/m2.

Semi-Empirical Method

Triaxial Method

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Empirical Method

Mechanistic-Empirical Methods

Practice Tasks from the Lecture

3.4 Design of flexible pavement by IRC, DoR guidelines, Road Note 31, AASHTO method20822080207620752074207220712068

Past Exam Questions on this Topic

  1. A new four lane divided highway is to be designed for the ADT of 8000 truck traffic based on the last count. The directional split of traffic is 55:45. Vehicle damage factor based on axle load survey was 4.0. Design a suitable pavement section for a design life of 15 years. The last traffic count was taken 1 year back and the project would be completed in 2 years from now. Growth rate of traffic = 7%. The results of subgrade soil CBR at seven locations obtained in a certain stretch of road are given in the table below. [8] (2082 Bhadra, Q6)
    Subgrade CBR test results along the stretch
    S.No.ChainageTest CBR result (%)
    10+5012
    20+3508
    30+5007
    40+6508
    50+8006
    60+9505
    71+5005
    Solution
    New four-lane divided highway · last count ADT = 8000 commercial vehicles/day (both directions) · directional split 55 : 45 · VDF = 4.0 · design life n = 15 yr · count taken 1 year ago, construction finishes 2 years from now · growth r = 7% · subgrade CBR at seven chainages = 12, 8, 7, 8, 6, 5, 5%

    Traffic in the predominant direction at the year of completion

    time from the count to opening, x = 1 + 2 = 3 years heavier-direction traffic now = 0.55 × 8000 = 4400 cvpd A = 4400 (1 + 0.07)³ = 4400 × 1.225043 = 5390 cvpd

    Lane distribution factor

    A four-lane divided highway is a dual two-lane carriageway → design on 75% of the commercial vehicles in each direction (Manual p.122).

    Alane = 0.75 × 5390 = 4043 cvpd

    Cumulative standard axles over the design life

    growth factor = [(1.07)15 − 1]/0.07 = (2.759032 − 1)/0.07 = 25.129 CSA = 365 × Alane × 25.129 × VDF = 365 × 4043 × 25.129 × 4.0 = 1.483 × 10⁸ = ≈ 148 msa

    Design subgrade CBR (87.5 percentile)

    Arrange in descending order; percentile = (rank/N) × 100 with N = 7:

    CBR (%)12887655
    Rank1234567
    Percentile14.328.642.957.171.485.7100

    At 87.5% (and equally at the 90 percentile used by IRC-37 for major roads) the value read off is design CBR = 5%, because the two weakest samples are both 5%.

    Pick the section from the DoR-2021 / IRC-37 catalogue

    Effective CBR 5% and 150 msa (the next catalogue step above 148 msa):

    Design section — CBR 5%, 150 msa
    LayerThickness
    Bituminous concrete (BC)90 mm
    Dense bituminous macadam (DBM)150 mm
    Granular base — WMM300 mm
    Granular sub-base — GSB300 mm
    Total crust840 mm
    Provide 90 mm BC + 150 mm DBM + 300 mm WMM base + 300 mm GSB = 840 mm over a subgrade compacted to CBR 5% (top 500 mm), with the drainage layer and 2.5% cross-fall.

    Design-traffic equation and LDF from Manual.pdf book p.122; percentile tabulation from book pp.178-179; catalogue thicknesses from the DoR-2021 / IRC-37 "Effective CBR 5%" table reproduced in §3.4 of these notes (Lecture 18-19).

  2. The flexible pavement was designed for a two lane two way road as shown in the figure. The drainage coefficient for the base layer is 1, the layer coefficients for the surface and base layers are a1 = 0.323 log10(E1) - 1.42 and a2 = 0.249 log10(E2) - 0.977, where E1 and E2 are in psi. Determine the initial commercial vehicles per day the road was designed for. Traffic growth rate = 6%, construction period = 2 yrs, design life 10 yrs, VDF = 2.5. [Figure: HMA surface E1 = 365 ksi, 5.3 in thick; granular base E2 = 30 ksi, 8.6 in thick; subgrade E3 = 9 ksi.] [8] (2080 Bhadra, Q5)
    Solution
    From the figure — HMA surface E1 = 365 ksi, D1 = 5.3 in · granular base E2 = 30 ksi, D2 = 8.6 in, m2 = 1.0 · subgrade E3 = MR = 9 ksi · two-lane two-way road · r = 6% · construction 2 yr · design life n = 10 yr · VDF = 2.5

    This is the AASHTO design run backwards: section → SN → W18 → initial commercial vehicles per day.

    Structural layer coefficients (E in psi)

    a₁ = 0.323 log₁₀(365 000) − 1.42 = 0.323 × 5.5623 − 1.42 = 0.3766 a₂ = 0.249 log₁₀(30 000) − 0.977 = 0.249 × 4.4771 − 0.977 = 0.1378

    Structural number of the existing section

    SN = a₁D₁ + a₂D₂m₂ = 0.3766 × 5.3 + 0.1378 × 8.6 × 1.0 = 1.996 + 1.185 = 3.18

    Allowable ESAL from the AASHTO-1993 equation

    The paper does not restate the reliability data, so the values printed with this same figure in the companion paper are adopted: R = 90% (ZR = −1.282), S0 = 0.5, ΔPSI = 2.0, MR = 9000 psi.

    ZRS₀ = −1.282 × 0.5 = −0.641 9.36 log₁₀(SN+1) = 9.36 × 0.62126 = +5.815 (SN+1)5.19 = 4.1815.19 = 1676 → 0.40 + 1094/1676 = 1.0527 log₁₀(2.0/2.7)/1.0527 = −0.13033 / 1.0527 = −0.124 2.32 log₁₀(9000) = 2.32 × 3.95424 = +9.174 ——————————————————————————————————————— log₁₀W₁₈ = −0.641 + 5.815 − 0.20 − 0.124 + 9.174 − 8.07 = 5.954 W₁₈ = 105.954 = 9.0 × 10⁵ ESAL

    Back-calculate the initial commercial traffic

    For a two-lane two-way road LDF = 0.75 (Manual p.122); the traffic must first grow through the 2-year construction period:

    W₁₈ = 365 × P(1+r)² × [((1+r)ⁿ−1)/r] × LDF × VDF [(1.06)10 − 1]/0.06 = 13.181 (1.06)² = 1.1236 365 × 1.1236 × 13.181 × 0.75 × 2.5 = 10 136 P = 9.0 × 10⁵ / 10 136 = ≈ 89 cvpd
    The road was designed for an initial traffic of ≈ 89 commercial vehicles per day (about 100 cvpd at the time the road is opened).

    AASHTO-1993 equation, layer-coefficient correlations and LDF as given in Manual.pdf §2.4.4 (book pp.144-149) and p.122. R, S0 and ΔPSI are the values printed with the identical figure in the 2080 Baishakh paper (Q6).

  3. A 4-lane divided highway is to be constructed on a new alignment. Traffic volume forecasts indicate that the AADT in both directions during the first year of operation will be 12,000 with 50% passenger cars (5 kN/axle), 33% two-axle single unit trucks (25 kN/axle) and 17% three-axle single unit trucks (30 kN/axle). The vehicle mix is expected to remain the same throughout the design life of the pavement. If the expected annual traffic growth rate is 4% for all vehicles, design the pavement with AC of thickness 7.5 cm, base and sub base. CBR of subgrade 6.5%, E_sub-base = 275 MPa, E_base = 500 MPa and E_surface = 2400 MPa. [8] (2076 Ashwin, Q5)
    Solution
    New 4-lane divided highway · AADT both directions in the first year = 12 000 — 50% passenger cars (5 kN/axle), 33% two-axle single-unit trucks (25 kN/axle), 17% three-axle single-unit trucks (30 kN/axle) · growth r = 4% · AC surfacing fixed at 7.5 cm · subgrade CBR = 6.5% · Esurface = 2400, Ebase = 500, Esub-base = 275 MPa · design life not stated — take n = 20 yr

    Truck factor of each class (fourth-power law, all axles of the vehicle)

    ClassNo./dayAxlesAxle loadEF per axleTruck factorDaily ESAL
    Passenger car600025 kN(5/80)4 = 0.00001530.00003050.18
    2-axle truck3960225 kN(25/80)4 = 0.0095370.01907375.53
    3-axle truck2040330 kN(30/80)4 = 0.0197750.059326121.03
    Total daily ESAL, both directions196.7

    Design-lane ESAL

    Divided highway → directional distribution factor D = 0.50; dual two-lane carriageway → LDF = 0.75 (Manual p.122).

    growth factor = [(1.04)20 − 1]/0.04 = (2.19112 − 1)/0.04 = 29.778 Ns = 365 × 196.7 × 29.778 × 0.50 × 0.75 = 8.0 × 10⁵ ESAL (for n = 15 yr the growth factor is 20.024 → Ns = 5.4 × 10⁵)

    Design subgrade resilient modulus

    MR = 10.3 × CBR = 10.3 × 6.5 = 67.0 MPa

    Full depth of asphalt concrete

    From the Asphalt Institute design chart (full-depth AC) attached to the paper, for Ns = 8.0 × 105 EAL and MR = 67 MPa → full depth ≈ 210 mm (chart read-off).

    Convert to the real layers

    Manual's thickness-equivalency rule tlower = tupper-equivalent × (Eupper/Elower)1/3:

    AC surfacing fixed at 75 mm → AC still to replace = 210 − 75 = 135 mm base equivalent = 135 × (2400/500)1/3 = 135 × 1.686 = 228 mm provide granular base = 125 mm → remaining = 103 mm sub-base = 103 × (500/275)1/3 = 103 × 1.222 = 126 mm provide = 150 mm (practical minimum)
    Section (top → bottom): 75 mm asphalt concrete / 125 mm granular base (E = 500 MPa) / 150 mm granular sub-base (E = 275 MPa) over the subgrade of CBR 6.5%.

    Truck-factor tabulation, LDF, MR = 10.3×CBR, the AI design chart and the (E1/E2)1/3 substitution rule are the procedure of Manual.pdf §2.5 — solved problems 2 and 3, book pp.178-182. The full-depth value is a chart read-off, so a 200-225 mm answer is equally acceptable.

  4. Design the flexible pavement for a 4-lane single carriageway road with the following parameters: i) Initial traffic in each direction 2000 CVPD; ii) Design life = 15 years; iii) Construction period = 3 years; iv) Traffic growth rate = 8%; v) Design CBR value 6%; vi) Modulus of elasticity of asphalt concrete surface course = 2500 MPa; vii) Modulus of elasticity of bituminous treated base = 1200 MPa; viii) Modulus of elasticity of granular subbase course = 125 MPa; ix) Axle load distribution of commercial vehicles on the road is given in the table below. [8] (2075 Ashwin, Q6)
    Axle load distribution of commercial vehicles
    Axle load (kN)1030507090110
    No. of axles (%)151520301010
    Solution
    4-lane single carriageway · initial traffic 2000 CVPD in each direction · n = 15 yr · construction 3 yr · r = 8% · design CBR = 6% · EAC = 2500, EBTB = 1200, EGSB = 125 MPa · axle-load spectrum as tabulated with the question

    Vehicle damage factor from the axle-load spectrum

    VDF = Σ fi (Li/80)⁴
    Axle load L (kN)% of axles(L/80)4Contribution
    10150.0002440.000037
    30150.0197750.002966
    50200.1525880.030518
    70300.5861820.175855
    90101.6018070.160181
    110103.5744630.357446
    VDF0.727

    Traffic at the year of completion

    commercial vehicles, both directions = 2 × 2000 = 4000 CVPD A = 4000 (1.08)³ = 4000 × 1.259712 = 5039 CVPD

    Lane distribution

    For a four-lane single carriageway the Manual (book p.122) prescribes design on 40% of the total commercial vehicles in both directions → LDF = 0.40.

    Cumulative standard axles

    growth factor = [(1.08)15 − 1]/0.08 = (3.17217 − 1)/0.08 = 27.152 Ns = 365 × 5039 × 27.152 × 0.40 × 0.727 = 1.45 × 10⁷ ESAL

    Design subgrade resilient modulus

    MR = 10.3 × 6 = 61.8 MPa

    Full depth of asphalt concrete

    From the Asphalt Institute full-depth design chart for 1.45 × 107 EAL and MR = 61.8 MPa → full depth ≈ 400 mm (chart read-off).

    Substitute the real layers

    AC wearing course (minimum) = 50 mm → AC to replace = 400 − 50 = 350 mm bituminous-treated base = 350 × (2500/1200)1/3 = 350 × 1.2772 = 447 mm provide BTB = 125 mm → remaining = 322 mm granular sub-base = 322 × (1200/125)1/3 = 322 × 2.1253 = 684 mm
    Section (top → bottom): 50 mm asphalt concrete (E = 2500 MPa) / 125 mm bituminous-treated base (E = 1200 MPa) / 685 mm granular sub-base (E = 125 MPa) over the subgrade of CBR 6%.

    Identical procedure and layer-substitution formula as Manual.pdf §2.5 solved problem 2 (book pp.178-180); LDF = 0.40 for a four-lane single carriageway, book p.122.

  5. Design a flexible pavement by using the Asphalt Institute method from the following data of a stretch of existing two lane road: a) Current traffic of 80 kN equivalent single axle load = 0.95 x 10^3 EAL/day; b) Traffic growth rate = 7.5%; c) Design period = 15 yrs; d) Construction period = 16 months; e) CBR of sub-grade to be taken = 5%; f) Elastic modulus of asphalt concrete surface course = 2500 MPa; g) Elastic modulus of bituminous treated base = 1200 MPa; h) Elastic modulus of granular sub base course = 125 MPa. Also draw a neat sketch of the pavement layers. [8] (2074 Ashwin, Q6)
    Solution
    Existing two-lane road · current traffic of 80 kN ESAL = 0.95 × 103 EAL/day · growth r = 7.5% · design period n = 15 yr · construction period 16 months · subgrade CBR = 5% · EAC = 2500, EBTB = 1200, EGSB = 125 MPa · minimum AC 50 mm

    Traffic at the end of the construction period

    y = 16 months = 1.333 yr A = P(1 + r)ˣ = 0.95 × 10³ × (1.075)1.333 = 950 × 1.1013 = 1046 EAL/day

    Cumulative design traffic

    Two-lane single carriageway → LDF = 0.75; no VDF is needed because A is already expressed in ESAL.

    growth factor = [(1.075)15 − 1]/0.075 = (2.9589 − 1)/0.075 = 26.118 Ns = 365 × 1046 × 26.118 × 0.75 = 7.48 × 10⁶ EAL

    Design subgrade resilient modulus

    MR = 10.3 × CBR = 10.3 × 5 = 51.5 MPa

    Full depth of asphalt concrete

    From the Asphalt Institute design chart (full-depth AC) for Ns = 7.48 × 106 EAL and MR = 51.5 MPa → full depth ≈ 350 mm (chart read-off).

    Replace the lower part of the AC by the real layers

    AC surface course = 50 mm → AC to replace = 350 − 50 = 300 mm bituminous-treated base = 300 × (2500/1200)1/3 = 300 × 1.2772 = 383 mm provide BTB = 125 mm → remaining = 258 mm granular sub-base = 258 × (1200/125)1/3 = 258 × 2.1253 = 549 mm
    Cross-section (top → bottom): 50 mm asphalt concrete / 125 mm bituminous-treated base / 550 mm granular sub-base over the subgrade (CBR 5%).

    Step-for-step the Asphalt Institute procedure of Manual.pdf §2.5, worked in Example 8 (book pp.152-154) and solved problem 3 (book pp.180-182): A = P(1+r)y, Ns = 365·A·[((1+r)n−1)/r]·LDF, MR = 10.3×CBR, chart, then (E1/E2)1/3 substitution.

  6. A road pavement is to be designed for a stretch of road with the following pavement layers: (i) Minimum thickness of asphalt concrete on the surface course 50 mm; (ii) Well graded crushed stone aggregate for base course, CBR value 90%; (iii) Fairly graded gravel for sub-base course, CBR value 20%; (iv) Compacted soil, CBR value 10%; (v) 90 percentile sub-grade CBR value = 4%. The road has a single lane carriageway and caters for a present ADT of 1200 commercial vehicles per day with an annual growth of 6%. The pavement is to be designed for a 10 year period. Design the pavement section using the IRC recommendation for the CBR method. The road is to be completed within 6 months from the initial traffic count. (2072 Chaitra, Q6)
    Solution
    Single-lane carriageway · present ADT = 1200 commercial vehicles/day · growth r = 6% · design life n = 10 yr · road completed 6 months (y = 0.5 yr) after the count · minimum asphalt-concrete surfacing 50 mm · base = well-graded crushed stone CBR 90% · sub-base = fairly graded gravel CBR 20% · compacted soil CBR 10% · 90-percentile subgrade CBR = 4%

    Design traffic volume

    In the IRC CBR-curve method the design curve is chosen from the number of commercial vehicles per day at the end of the design life:

    A = P(1 + r)(n + y) = 1200 (1.06)10.5 = 1200 × 1.8438 = ≈ 2213 commercial vehicles/day

    For a single-lane road the design is based on the total number of commercial vehicles in both directions (Manual p.122), i.e. 2213 cvpd — which lies in the band 1500-4500 → design curve F.

    IRC traffic classification (Design Chart 1)
    CurveABCDEFG
    cvpd > 3 t laden0-1515-4545-150150-450450-15001500-4500> 4500

    Total crust thickness

    Enter Design Chart 1 on curve F at the subgrade CBR = 4% → total thickness above the subgrade = 55 cm.

    Thickness of each layer

    Re-enter the same curve F at the CBR of every material; the chart gives the cover that must be provided above that material, and the layer thickness is the difference of two successive covers:

    MaterialCBRCover required above it (curve F)Layer thickness
    Sub-grade4%55 cm—
    Compacted soil10%33 cm55 − 33 = 22 cm
    Fairly graded gravel sub-base20%21 cm33 − 21 = 12 cm
    Well-graded crushed stone base90%9 cm21 − 9 = 12 cm
    Asphalt concrete surfacing——9 cm

    Check

    surfacing 9 cm > 5 cm minimum ✓ total = 9 + 12 + 12 + 22 = 55 cm ✓
    Section (top → bottom): 90 mm asphalt concrete / 120 mm crushed-stone base (CBR 90%) / 120 mm gravel sub-base (CBR 20%) / 220 mm compacted soil (CBR 10%) over the subgrade of CBR 4% — total crust 550 mm.

    Exactly the procedure and the same chart read-offs as Manual.pdf §2.4.1, Examples 1 and 2 (book pp.122-127): A = P(1+r)n+y, curve F for 1500-4500 cvpd, 55 cm over CBR 4%, 33 cm over CBR 10%, 21 cm over CBR 20% and 9 cm over CBR 90%.

  7. What do you understand by legal axle load and standard axle load? Describe the stresses induced in the rigid pavement slab. How is the spacing between contraction joints calculated in jointed plain concrete pavement? In the figure below, a pavement system with the resilient moduli, layer coefficient of surface course and drainage coefficients are shown. If the predicted ESAL = 15 x 10^6, R = 90%, So = 0.4, present serviceability index = 4.2 and terminal serviceability index = 2.7, select the thickness of D1, D2 and D3. [Figure: E1 = 450,000 psi with a1 = 0.44 (D1); E2 = 25,000 psi with m2 = 1.2 (D2); E3 = 15,000 psi with m3 = 1.2 (D3); subgrade MR = 5,000 psi.] (2071 Chaitra, Q5)
    Solution

    (a) Legal axle load and standard axle load

    • Legal axle load — the maximum axle load that the law allows a vehicle to carry on a given highway network (in Nepal/India commonly 10.2 t for a single axle with dual wheels). It is a regulatory limit, enforced at weigh stations.
    • Standard axle load — the reference axle of 80 kN (8.16 t), single axle with dual wheels, to which every other axle is converted through (L/80)4. It is a design yardstick, not a legal limit.

    (b) Stresses induced in a rigid pavement slab

    • Wheel-load stresses at the three critical positions — interior, edge and corner (Westergaard)
    • Warping (temperature) stresses from the temperature differential across the slab
    • Frictional stresses from the subgrade restraining the uniform expansion/contraction of the slab
    Si = 0.316 P/h² [4 log₁₀(l/b) + 1.069] interior load Se = 0.572 P/h² [4 log₁₀(l/b) + 0.359] edge load Sc = 3 P/h² [1 − (a√2/l)0.6] corner load σti = (EαΔt/2)(Cx+μCy)/(1−μ²) interior warping σte = C · EαΔt/2 edge warping σtc = EαΔt √(a/l)/[3(1−μ)] corner warping σf = W · L · f/(2×10⁴) frictional

    (c) Spacing of contraction joints in JPCP

    The joint spacing is fixed so that the frictional (tensile) stress developed in the slab as it contracts does not exceed the allowable tensile stress of the concrete (σc ≈ 0.8 kg/cm² for plain concrete):

    σc = W · L · f/(2×10⁴) ⇒ L = 2×10⁴ · σc/(W · f) L = 2×10⁴ × 0.8 / (2400 × 1.5) = 4.44 m

    with W = unit weight of concrete = 2400 kg/m³ and f = coefficient of subgrade friction = 1.5 — which is why IRC limits the contraction-joint spacing of plain slabs to 4.5 m. (For a reinforced slab the steel carries the friction force and the spacing may be increased to 13-14 m.)

    (d) AASHTO-1993 layer design

    W18 = 15 × 106 · R = 90% → ZR = −1.282 · S0 = 0.4 · pi = 4.2, pt = 2.7 → ΔPSI = 1.5 · E1 = 450 000 psi with a1 = 0.44 · E2 = 25 000 psi, m2 = 1.2 · E3 = 15 000 psi, m3 = 1.2 · subgrade MR = 5000 psi

    Layer coefficients

    a₁ = 0.44 (given) a₂ = 0.249 log₁₀(25 000) − 0.977 = 0.249×4.39794 − 0.977 = 0.118 a₃ = 0.227 log₁₀(15 000) − 0.839 = 0.227×4.17609 − 0.839 = 0.109

    Required structural numbers

    Solve the AASHTO equation three times, each time with the resilient modulus of the material directly below the interface being protected:

    Interface protectedMR usedRequired SN
    Top of base25 000 psiSN1 = 3.42
    Top of sub-base15 000 psiSN2 = 4.18
    Top of subgrade5 000 psiSN3 = 6.08

    Thicknesses (rounded up to the next practical value)

    D₁ = SN₁/a₁ = 3.42/0.44 = 7.78 → D₁ = 8 in (≈ 200 mm) SN₁* = 0.44 × 8 = 3.52 D₂ = (SN₂ − SN₁*)/(a₂m₂) = (4.18 − 3.52)/(0.118×1.2) = 4.65 → D₂ = 5 in (≈ 125 mm) SN₂* = 0.118 × 1.2 × 5 = 0.708 D₃ = (SN₃ − SN₁* − SN₂*)/(a₃m₃) = (6.08 − 3.52 − 0.708)/(0.109×1.2) = 14.1 → D₃ = 15 in (≈ 375 mm)

    Check

    SNprovided = 0.44×8 + 0.118×1.2×5 + 0.109×1.2×15 = 3.52 + 0.708 + 1.962 = 6.19 ≥ 6.08 ✓ safe
    D1 = 8 in (200 mm) asphalt concrete, D2 = 5 in (125 mm) base, D3 = 15 in (375 mm) sub-base.

    Stress formulae, joint-spacing equation and the AASHTO layered procedure are those of Manual.pdf §2.6-2.8 (book pp.155-159, 168) and §2.4.4 (book pp.144-149). The three SN values were obtained by solving the AASHTO-1993 equation numerically (the paper supplies the corresponding nomograph).

  8. a) An existing two lane single carriageway gravel road has to be upgraded by a bituminous pavement to cater for the growing traffic demand. Present traffic in terms of ESA is 0.8 x 10^3 per day. The regional traffic growth rate is taken as 6.5% per annum. Data required for pavement design are as given below. [8] i) Design period = 10 years; ii) Construction period = 1 year; iii) 87.5 percentile CBR value of sub grade soil from 7 sample locations = 5%; iv) Elastic modulus of asphalt concrete for surface course E_ac = 2000 MPa; v) Elastic modulus of crushed stone base E_base = 350 MPa; vi) Elastic modulus of granular sub-base E_sub-base = 250 MPa. You are required to design the pavement by the Asphalt Institute Method. Draw the cross section of the final pavement layers considering the minimum thickness of asphalt concrete on the surface course equal to 50 mm. b) What are the factors affecting pavement design? Write down the steps of the IRC design guidelines for rigid pavement. [8] (2068 Baishakh, Q3)
    Solution

    (a) Asphalt Institute design [8]

    Existing two-lane single-carriageway gravel road · present traffic P = 0.8 × 103 ESA/day · growth r = 6.5% · design period n = 10 yr · construction period y = 1 yr · 87.5-percentile subgrade CBR = 5% · EAC = 2000, Ebase = 350, Esub-base = 250 MPa · minimum AC surfacing 50 mm

    Traffic at the end of the construction period

    A = P(1 + r)ˣ = 0.8×10³ × (1.065)¹ = 852 EAL/day

    Cumulative design traffic

    Two-lane single carriageway → LDF = 0.75; no VDF because A is already in ESAL.

    growth factor = [(1.065)10 − 1]/0.065 = (1.87714 − 1)/0.065 = 13.494 Ns = 365 × 852 × 13.494 × 0.75 = 3.15 × 10⁶ ≈ 3.1 × 10⁶ ESAL

    Design subgrade resilient modulus

    MR = 10.3 × CBR = 10.3 × 5 = 51.5 MPa

    Full depth of asphalt concrete

    From the Asphalt Institute design chart (full-depth AC) attached to the paper, for 3.1 × 106 EAL and MR = 51.5 MPa → full depth = 280 mm.

    Replace the lower asphalt by base and sub-base

    AC surface course = 50 mm → AC to replace = 280 − 50 = 230 mm crushed-stone base = 230 × (2000/350)1/3 = 230 × 1.7878 = 412 mm provide base = 125 mm → remaining = 287 mm granular sub-base = 287 × (350/250)1/3 = 287 × 1.1187 = 321 mm

    Cross-section of the final pavement

    LayerModulusThickness
    Asphalt concrete surface2000 MPa50 mm
    Crushed-stone base350 MPa125 mm
    Granular sub-base250 MPa321 mm
    SubgradeMR = 51.5 MPa (CBR 5%)—
    Total crust = 50 + 125 + 321 = 496 mm.

    (b) Factors affecting pavement design and the IRC steps for rigid pavement [8]

    Factors: traffic and loading (axle load, wheel/axle configuration, repetitions/CSA, contact pressure, speed); subgrade strength (CBR / MR, moisture, drainage); material properties (moduli of the layers); environment (temperature, rainfall, frost); and the failure criteria adopted (fatigue cracking, rutting).

    IRC steps for rigid-pavement design:

    1. Collect the design data — wheel load P, contact radius a, modulus of subgrade reaction k (on the sub-base), E, μ, α, flexural strength (modulus of rupture) of the concrete and the temperature differential Δt.
    2. Work out the design traffic Ad = P(1+r)n+y and its traffic group.
    3. Fix the joint spacings — expansion joint from the permitted movement (Le = δ/2 ÷ αΔT) and contraction joint from the frictional stress (Lc = 2×104σc/Wf).
    4. Assume a trial slab thickness h and compute the radius of relative stiffness l = [Eh³/(12k(1−μ²))]0.25.
    5. Read/compute the edge and corner load stresses (IRC stress charts).
    6. Compute the warping stresses with Bradbury's coefficients.
    7. Check the critical combinations — (load + warping − friction) at the edge in summer, (load + warping + friction) at the edge in winter and (load + warping) at the corner — against the modulus of rupture; revise h until the factor of safety is just above 1.
    8. Apply the traffic-intensity adjustment to h, then design the dowel bars, tie bars and the reinforcement.

    This is Example 8 of Manual.pdf §2.5 (book pp.152-154) reproduced with the paper's own data; the design steps in (b) follow Manual.pdf §2.8, book pp.164-173.

Empirical Methods: CBR Methods

CBR Method Recommended by

California State Highway Department

IRC (37-1970)

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  1. Evaluate the soak CBR value of the subgrade
  2. Choose appropriate curve considering the design traffic
  3. Determine the pavement thickness for the CBR value of the subgrade
  4. To determine thickness of sub-base, base and surface course, use the curves repeatedly
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Road Note 31 /Catalogue Method

Pavement Layers:

Design steps:

  1. Estimating the amount of traffic and the cumulative number of equivalent standard axles that will use the road over the selected design life;

Normal Traffic: Traffic that would pass along the existing road or track even if no new pavement were provided

Diverted Traffic: Traffic that changes from another route (or mode of transport) to the project road because of the improved pavement, but still travels between the same origin and destination

Generated Traffic: Additional traffic which occurs in response to the provision or improvement of the road.

b. Determine the average daily one-directional traffic flow for each class of vehicle. c. Determine the mean equivalency factor of each class of vehicle and for each direction from the results of axle load survey and any other surveys that have recently been carried out to convert the daily cvpd to 80kN equivalent standard axle load. Corrections are suggested to consider distribution of heavy vehicles between lanes.

d. The products of the cumulative one-directional traffic flows for each class of vehicle over the design life of the road and the mean equivalence factor for that class added together is the cumulative ESAL for each direction. The larger of the two directional values should be used for design purposes.

[Determine the one-directional number of commercial vehicles per day for each class of vehicle for each year during the design life. The sum of number of commercial vehicles per day multiplied by 365 for each year of life is the cumulative traffic flow for the class in the design life. The sum of product of this cumulative traffic flow and corresponding equivalence factor for each class is the cumulative number of axles over the design life.] e) Data from the lane with the highest traffic loading, as determined, are then used to calculate the total traffic over the design life.

  1. Assessing the strength of the subgrade soil and foundation layer over which the road is to be built.
  1. Selecting the most economical combination of pavement materials and layer thicknesses that will provide satisfactory service over the design life of the pavement using structural catalogues. For subgrade classes S1 and S2, capping layers using table must be applied before using the subsequent pavement design catalogues. This means that they must be converted to an effective Subgrade Class S3 (also known as Foundation Class F1) or higher. Subgrades of classes S3 to S5 may also be converted to higher effective classes using the capping options.

Charts and catalogues

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Empirical Method

AASHTO Method

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Design steps:

  1. Evaluate design traffic (W18):
Lane distribution factor (AASHTO, 1993)
  1. Determine reliability factors (S0, ZR):
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  1. Evaluate resilient modulus (MR):
  1. Determine the overall structural number (SN)
  1. Determine structural layer coefficients (𝛼1, 𝛼2, 𝛼3)
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  1. Determine drainage coefficients (𝑚2, 𝑚3)
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Problem Design a pavement for a rural highway using AASHTO flexible design procedure. The expected design ESAL is 13x106. The pavement structure is to consist of asphalt concrete with an elastic modulus of 400,000 psi, a granular base with an elastic modulus of 26,000 psi and a granular sub base with an elastic modulus of 12,000 psi. The elastic modulus of the subgrade is 8,000 psi. The drainage conditions of base and subbase are fair. It is estimated that the pavement will be saturated about 15% of the time. Assume that a reliability level of about 90% is required and that the initial serviceability index is 4.2 and the final serviceability index is 2.5.

IRC Flexible Pavement Design Guidelines

IRC (37-2018)

Scope: Design of pavements for roads with design traffic >2 msa. Design Approach and Criteria

Pavement Type Selection:

Structural Model and Pavement Analysis

Critical Stress/Strain:

Reliability Criteria:

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Vehicle Damage Factor (VDF):

⁻ A multiplier for converting the number of commercial vehicles of different axle loads and axle configurations to the number of standard axle-load repetitions ⁻ Equivalent number of standard axles per commercial vehicle ELAF ⁻ The conversion of one repetition of a particular type of axle carrying a specific axle load into equivalent repetitions of 80 kN single axle can be done using a) b) c) d).

⁻ VDF for a typical vehicle can be obtained as the sum of the equivalency factor of all the axles in the vehicle.

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  1. Selecting a trial composition.
  2. Bituminous mix design and the mix resilient modulus.
  3. Selecting layer thickness
  4. Structural analysis of the selected pavement structure (IITPAVE)
  5. Computing the allowable strains/stresses
  6. Doing the iterations
  7. Check for cumulative fatigue damage (For CTB)
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DoR (2021): Flexible Pavement

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ELAF

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DoR (2021): Flexible Pavement (>2msa)

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Excel sheet for main volume design (>2msa):

Practice Tasks from the Lecture

3.5 Rigid pavement design (Westergaard theory)2081207620722071

Past Exam Questions on this Topic

  1. A CC pavement slab of thickness 20 cm is constructed over a granular subbase having a modulus of subgrade reaction of 15 kg/cm3. The maximum temperature difference between the top and bottom of the slab during a summer day and night is found to be 18 degC. The spacing between the transverse contraction joints is 4.5 m and that between the longitudinal joints is 3.5 m. The design wheel load is 5100 kg; radius of contact area is 15 cm; E value of CC is 3 x 10^5 kg/cm2; Poisson's ratio is 0.15; coefficient of thermal expansion is 10 x 10^-6 per degC and the friction coefficient is 1.5. Find the worst combination of stresses at the edge. (Bradbury's warping stress coefficient chart and the IRC edge load stress chart are attached with the paper.) [8] (2081 Baishakh, Q6)
    Solution
    h = 20 cm · k = 15 kg/cm³ · Δt = 18°C (top–bottom, summer day–night) · transverse contraction joints Lx = 4.5 m · longitudinal joints Ly = 3.5 m · P = 5100 kg · a = 15 cm · E = 3×105 kg/cm² · μ = 0.15 · α = 10×10−6/°C · friction coefficient f = 1.5 (W = 2400 kg/m³)

    Radius of relative stiffness

    l = [E h³ / (12 k (1 − μ²))]0.25 = [3×10⁵ × 20³ / (12 × 15 × (1 − 0.15²))]0.25 = [2.4×10⁹ / 175.95]0.25 = [1.364×10⁷]0.25 = 60.77 cm

    Equivalent radius of the resisting section

    a = 15 cm < 1.724 h = 34.5 cm → use the second form b = √(1.6a² + h²) − 0.675h = √(1.6×225 + 400) − 0.675×20 = √760 − 13.5 = 14.07 cm

    Warping (temperature) stress at the edge

    Lx/l = 450/60.77 = 7.41 → Cx = 1.05 (Bradbury chart) Ly/l = 350/60.77 = 5.76 → Cy = 0.87 the larger coefficient governs → C = 1.05 σte = C · E · α · Δt / 2 = 1.05 × 3×10⁵ × 10×10⁻⁶ × 18 / 2 = 28.3 kg/cm²

    Load stress at the edge

    From the IRC Edge Load Stress Chart supplied with the paper (drawn for exactly P = 5100 kg, a = 15 cm, E = 3×105 kg/cm², μ = 0.15), at h = 20 cm and k = 15 kg/cm³ → σle ≈ 24 kg/cm².

    check by formula — the chart plots the IRC (Teller & Sutherland) equation: σle = 0.529 (P/h²)(1 + 0.54μ)[4 log₁₀(l/b) + log₁₀b − 0.4048] = 0.529 × (5100/400) × 1.081 × [4×0.6355 + 1.1483 − 0.4048] = 23.95 kg/cm² (Westergaard's original edge formula 0.572(P/h²)[4 log₁₀(l/b)+0.359] gives 21.16)

    Frictional stress

    slab length between contraction joints, L = 4.5 m σf = W · L · f / (2×10⁴) = 2400 × 4.5 × 1.5 / 20 000 = 0.81 kg/cm²

    Critical combinations at the edge

    SeasonCombinationValue (kg/cm²)
    Summer, mid-dayload + warping − friction (slab expanding, friction is compressive)23.95 + 28.3 − 0.81 = 51.4
    Winter, mid-dayload + warping + friction (slab contracting, friction is tensile)23.95 + 28.3 + 0.81 = 53.1
    Worst combination of stresses at the edge = ≈ 53 kg/cm² (winter mid-day); the summer value is ≈ 51 kg/cm². (With Westergaard's unmodified edge stress the pair becomes 48.7 and 50.3 kg/cm².)

    Since ordinary pavement-quality concrete has a modulus of rupture of only about 40-45 kg/cm², a 20 cm slab on k = 15 kg/cm³ is not safe for this wheel load — the thickness would have to be increased.

    Radius of relative stiffness, b, the IRC edge formula, Bradbury's chart, the frictional-stress equation and the summer/winter critical combinations are all taken from Manual.pdf §2.6-2.7, book pp.155-159.

  2. Write down the assumptions and analysis of Westergaard's Theory. How are the warping stress and friction stress developed in the rigid pavement? [8] (2076 Ashwin, Q6)
    Solution

    (a) Assumptions of Westergaard's theory

    • The cement-concrete slab is a homogeneous, isotropic, thin elastic plate of uniform thickness.
    • The subgrade behaves as a dense liquid (Winkler foundation): the reaction at any point is vertical and proportional to the deflection at that point — p = k·Δ, where k is the modulus of subgrade reaction.
    • The slab remains in full contact with the subgrade throughout.
    • The wheel load is distributed over a circular area of radius a; the load is static.
    • The slab is of infinite extent in plan for interior loading; the three critical load positions are interior, edge and corner.

    Analysis. The stiffness of the slab relative to the subgrade is expressed by the radius of relative stiffness, and the pressure is assumed to be resisted over an equivalent radius b:

    l = [E h³ / (12 k (1 − μ²))]0.25 b = a for a ≥ 1.724 h b = √(1.6a² + h²) − 0.675h for a < 1.724 h
    PositionCritical load stress
    InteriorSi = 0.316 P/h² · [4 log10(l/b) + 1.069]
    EdgeSe = 0.572 P/h² · [4 log10(l/b) + 0.359]
    CornerSc = 3 P/h² · [1 − (a√2/l)0.6]

    IRC uses the versions later modified by Teller & Sutherland (edge) and Kelley (corner).

    (b) How the warping stress is developed

    During the day the top of the slab is heated by the sun while the bottom stays cool, so a temperature differential Δt exists across the depth. The warmer face tries to expand more than the cooler face and the slab tends to warp (curl) — downwards at mid-day and upwards at night. This warping is restrained by the self-weight of the slab and by the subgrade friction, and the restraint produces warping stresses:

    interior : σ = (EαΔt/2) · (Cx + μCy)/(1 − μ²) edge : σ = C · EαΔt / 2 corner : σ = EαΔt √(a/l) / [3(1 − μ)]

    where C, Cx, Cy are Bradbury's coefficients, read from the chart against L/l for the slab length and width. At mid-day the slab warps downwards, so the warping tension at the bottom adds to the load tension — this is the critical combination at the edge.

    (c) How the frictional stress is developed

    A uniform rise or fall of the mean slab temperature makes the whole slab expand or contract. The subgrade resists this movement by friction at the interface, so the slab is put into compression when it expands (summer) and into tension when it contracts (winter). Taking the friction force on half the slab length:

    σf = W · L · f / (2 × 10⁴) kg/cm² W = 2400 kg/m³, L = slab length between joints (m), f ≈ 1.5

    Because σf grows with L, it is this stress that fixes the spacing of the contraction joints (L = 2×104σc/Wf ≈ 4.5 m for plain slabs).

    Manual.pdf §2.6 and §2.7 — Westergaard's theory, temperature differential and subgrade friction, book pp.155-159.

  3. Differentiate between flexible pavement design and rigid pavement design. Describe Westergaard's concept for temperature stresses (2072 Chaitra, Q5)
    Solution

    (a) Flexible-pavement design vs rigid-pavement design

    Flexible pavement designRigid pavement design
    Load transfer assumedGrain-to-grain contact; the load spreads through the layers in a truncated coneSlab (beam/plate) action; the concrete slab bridges over the subgrade
    Design criterionLimit the deformation — vertical compressive strain on the subgrade (rutting) and tensile strain at the bottom of the bituminous layer (fatigue)Limit the flexural tensile stress in the slab to the modulus of rupture of the concrete
    Subgrade parameterCBR or resilient modulus MRModulus of subgrade reaction k
    MethodEmpirical/mechanistic-empirical — CBR curves, Road Note 31 catalogue, AASHTO SN, IRC-37 / DoR catalogues, Asphalt Institute chartsAnalytical — Westergaard/IRC stress equations and charts, checked for the critical load + warping (± friction) combinations
    OutputThickness of several layers (surface, base, sub-base)One slab thickness plus the joint, dowel-bar, tie-bar and reinforcement design
    TemperatureEnters only through the modulus of the bituminous layerEnters directly as warping and frictional stresses

    (b) Westergaard's concept of temperature stresses

    Westergaard showed that temperature produces two distinct stresses in a concrete pavement:

    1. Warping (curling) stress — caused by the differential temperature Δt between the top and the bottom of the slab. The hotter face wants to expand more, so the slab tries to curl; its own weight and the subgrade prevent this and bending stresses appear. With Bradbury's coefficients Cx, Cy (read against Lx/l and Ly/l):

    interior : σti = (EαΔt/2)(Cx + μCy)/(1 − μ²) edge : σte = C · EαΔt / 2 corner : σtc = EαΔt √(a/l) / [3(1 − μ)]

    During the day the slab warps down (tension at the bottom, adds to the load stress at the edge); at night it warps up (tension at the top at the corner).

    2. Frictional stress — caused by the uniform change of the mean slab temperature. The subgrade resists the free expansion/contraction of the slab:

    σf = W · L · f / (2×10⁴)

    compressive in summer, tensile in winter, and it is the quantity that fixes the contraction-joint spacing.

    Critical combinations (used for design):

    • Summer mid-day, edge: load + warping − friction
    • Winter mid-day, edge: load + warping + friction
    • Mid-night, corner: load + warping (there is no frictional stress at the corner)

    Manual.pdf §2.7 — Temperature differential and subgrade friction on rigid pavement, book pp.155-159.

  4. What do you understand by legal axle load and standard axle load? Describe the stresses induced in the rigid pavement slab. How is the spacing between contraction joints calculated in jointed plain concrete pavement? In the figure below, a pavement system with the resilient moduli, layer coefficient of surface course and drainage coefficients are shown. If the predicted ESAL = 15 x 10^6, R = 90%, So = 0.4, present serviceability index = 4.2 and terminal serviceability index = 2.7, select the thickness of D1, D2 and D3. [Figure: E1 = 450,000 psi with a1 = 0.44 (D1); E2 = 25,000 psi with m2 = 1.2 (D2); E3 = 15,000 psi with m3 = 1.2 (D3); subgrade MR = 5,000 psi.] (2071 Chaitra, Q5)
    Solution

    (a) Legal axle load and standard axle load

    • Legal axle load — the maximum axle load that the law allows a vehicle to carry on a given highway network (in Nepal/India commonly 10.2 t for a single axle with dual wheels). It is a regulatory limit, enforced at weigh stations.
    • Standard axle load — the reference axle of 80 kN (8.16 t), single axle with dual wheels, to which every other axle is converted through (L/80)4. It is a design yardstick, not a legal limit.

    (b) Stresses induced in a rigid pavement slab

    • Wheel-load stresses at the three critical positions — interior, edge and corner (Westergaard)
    • Warping (temperature) stresses from the temperature differential across the slab
    • Frictional stresses from the subgrade restraining the uniform expansion/contraction of the slab
    Si = 0.316 P/h² [4 log₁₀(l/b) + 1.069] interior load Se = 0.572 P/h² [4 log₁₀(l/b) + 0.359] edge load Sc = 3 P/h² [1 − (a√2/l)0.6] corner load σti = (EαΔt/2)(Cx+μCy)/(1−μ²) interior warping σte = C · EαΔt/2 edge warping σtc = EαΔt √(a/l)/[3(1−μ)] corner warping σf = W · L · f/(2×10⁴) frictional

    (c) Spacing of contraction joints in JPCP

    The joint spacing is fixed so that the frictional (tensile) stress developed in the slab as it contracts does not exceed the allowable tensile stress of the concrete (σc ≈ 0.8 kg/cm² for plain concrete):

    σc = W · L · f/(2×10⁴) ⇒ L = 2×10⁴ · σc/(W · f) L = 2×10⁴ × 0.8 / (2400 × 1.5) = 4.44 m

    with W = unit weight of concrete = 2400 kg/m³ and f = coefficient of subgrade friction = 1.5 — which is why IRC limits the contraction-joint spacing of plain slabs to 4.5 m. (For a reinforced slab the steel carries the friction force and the spacing may be increased to 13-14 m.)

    (d) AASHTO-1993 layer design

    W18 = 15 × 106 · R = 90% → ZR = −1.282 · S0 = 0.4 · pi = 4.2, pt = 2.7 → ΔPSI = 1.5 · E1 = 450 000 psi with a1 = 0.44 · E2 = 25 000 psi, m2 = 1.2 · E3 = 15 000 psi, m3 = 1.2 · subgrade MR = 5000 psi

    Layer coefficients

    a₁ = 0.44 (given) a₂ = 0.249 log₁₀(25 000) − 0.977 = 0.249×4.39794 − 0.977 = 0.118 a₃ = 0.227 log₁₀(15 000) − 0.839 = 0.227×4.17609 − 0.839 = 0.109

    Required structural numbers

    Solve the AASHTO equation three times, each time with the resilient modulus of the material directly below the interface being protected:

    Interface protectedMR usedRequired SN
    Top of base25 000 psiSN1 = 3.42
    Top of sub-base15 000 psiSN2 = 4.18
    Top of subgrade5 000 psiSN3 = 6.08

    Thicknesses (rounded up to the next practical value)

    D₁ = SN₁/a₁ = 3.42/0.44 = 7.78 → D₁ = 8 in (≈ 200 mm) SN₁* = 0.44 × 8 = 3.52 D₂ = (SN₂ − SN₁*)/(a₂m₂) = (4.18 − 3.52)/(0.118×1.2) = 4.65 → D₂ = 5 in (≈ 125 mm) SN₂* = 0.118 × 1.2 × 5 = 0.708 D₃ = (SN₃ − SN₁* − SN₂*)/(a₃m₃) = (6.08 − 3.52 − 0.708)/(0.109×1.2) = 14.1 → D₃ = 15 in (≈ 375 mm)

    Check

    SNprovided = 0.44×8 + 0.118×1.2×5 + 0.109×1.2×15 = 3.52 + 0.708 + 1.962 = 6.19 ≥ 6.08 ✓ safe
    D1 = 8 in (200 mm) asphalt concrete, D2 = 5 in (125 mm) base, D3 = 15 in (375 mm) sub-base.

    Stress formulae, joint-spacing equation and the AASHTO layered procedure are those of Manual.pdf §2.6-2.8 (book pp.155-159, 168) and §2.4.4 (book pp.144-149). The three SN values were obtained by solving the AASHTO-1993 equation numerically (the paper supplies the corresponding nomograph).

Design of Rigid Pavements

Types of Rigid Pavements

  1. Jointed Plain Concrete Pavement (JPCP)
  2. Jointed Reinforced Concrete Pavement (JRCP)
  3. Continuously Reinforced Concrete Pavement (CRCP)
  4. Prestressed Concrete Pavement (PCP)

Jointed Plain Concrete Pavement (JPCP)

Jointed Reinforced Concrete Pavement (JRCP)

Continuously Reinforced Concrete Pavement (CRCP)

Prestressed Concrete Pavement (PCP)

Stresses in Rigid Pavement

Wheel Load Stresses

Westergaard’s Theory

  1. Concrete slab acts as homogeneous, isotropic, elastic solid in equilibrium.
  2. Thickness of slab is uniform.
  3. Rigid pavement is considered to be a thin elastic plate resting on soil sub-grade which is assumed as a dense liquid. The subgrade reaction is vertical only and is proportional to the deflection of slab.
  1. The load at the interior and at the corner of the slab is distributed over a circular area of contact
  2. The load at the edge of the slab is distributed uniformly over semi-circular area of contact.

Radius of Relative Stiffness (l )

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Westergaard - Loading Diagram

Modulus of Subgrade Reaction(k)

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Equivalent Radius of Resisting Section (b)

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Equations for Wheel Load Stress

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Temperature Stresses

Warping Stress

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Frictional Stress

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Practice Tasks from the Lecture

3.6 Design of rigid pavement by latest IRC and AASHTO method2081207820752074

Past Exam Questions on this Topic

  1. Calculate the length and spacing of dowel bars of dia 25 mm at an expansion joint of a concrete pavement of thickness 25 cm. Design wheel load 5000 kg. Load capacity of the dowel system is 40 percent of the design wheel load. Joint width is 2.0 cm and the permissible stresses in shear, bending and bearing in the dowel bars are 1000, 1400 and 100 kg/cm2 respectively. Modulus of elasticity of concrete 2.45 x 10^5 kg/cm2; modulus of subgrade reaction = 8 kg/cm3; Poisson ratio 0.15. [8] (2081 Bhadra, Q6)
    Solution
    Dowel bar diameter d = 25 mm = 2.5 cm · slab thickness h = 25 cm · design wheel load P = 5000 kg · the dowel system must carry 40% of P · expansion-joint width δ = 2.0 cm · permissible stresses Fs = 1000 (shear), Ff = 1400 (bending), Fb = 100 kg/cm² (bearing) · E = 2.45×105 kg/cm² · k = 8 kg/cm³ · μ = 0.15

    Length of the dowel bar

    Bradbury's design equates the load capacity of the bar in bending and in bearing, which gives the balanced embedment length Ld (solved by trial):

    Ld = 5d √[(Ff/Fb) · (Ld + 1.5δ)/(Ld + 8.8δ)] = 5 × 2.5 √[14 × (Ld + 3.0)/(Ld + 17.6)] = 12.5 √[14 (Ld + 3)/(Ld + 17.6)]
    Trial Ld (cm)RHS (cm)
    3038.6
    3840.2
    4040.4
    40.540.5 ✓
    Ld = 40.5 cm total length = Ld + δ = 40.5 + 2.0 = 42.5 cm → provide 43 cm long, 25 mm φ bars actual embedded length Ld = 43 − 2 = 41 cm (20.5 cm each side)

    Load-transfer capacity of one dowel bar (Ld = 41 cm)

    shear : Ps = 0.785 d² Fs = 0.785 × 2.5² × 1000 = 4906 kg bending : Pf = 2d³Ff/(Ld+8.8δ) = 43 750 / 58.6 = 747 kg bearing : Pb = FbLd²d/[12.5(Ld+1.5δ)] = 420 250 / 550 = 764 kg the smallest governs → capacity of one bar = 747 kg (bending)

    Capacity factor required from the dowel group

    load to be transferred = 40% of 5000 = 2000 kg required capacity factor = 2000 / 747 = 2.68

    Effective length of the dowel group

    l = [E h³ / (12 k (1 − μ²))]0.25 = [2.45×10⁵ × 25³ / (12 × 8 × 0.9775)]0.25 = [4.079×10⁷]0.25 = 79.9 cm load is transferred by the bars within 1.8 l on either side: 1.8 l = 1.8 × 79.9 = 143.9 cm

    Spacing of the dowel bars

    The bar under the load counts 1.0 and every other bar counts (1.8l − x)/1.8l, x being its distance from the load:

    Trial spacingCapacity factor available≥ 2.68 ?
    40 cm1 + 0.722 + 0.444 + 0.166 = 2.33No
    35 cm1 + 0.757 + 0.513 + 0.270 + 0.027 = 2.57No
    33 cm1 + 0.771 + 0.541 + 0.312 + 0.082 = 2.71Yes (just)
    30 cm1 + 0.791 + 0.583 + 0.374 + 0.166 = 2.91Yes ✓
    Provide 25 mm φ mild-steel dowel bars, 43 cm long (say 45 cm), at 30 cm centre-to-centre across the expansion joint — one half of every bar painted/greased and fitted with an expansion cap so that it is free to slide.

    Dowel-bar equations (balanced Ld, the three capacities, the 40% rule, the 1.8 l effective group and the spacing check) are exactly those worked in Manual.pdf §2.8, book pp.168 and 172.

  2. Explain the stresses in rigid pavement. Write down the design procedure of dowel bars at expansion joints. [8] (2078 Bhadra, Q10)
    Solution

    (a) Stresses in a rigid pavement

    1. Wheel-load stresses (Westergaard) at the three critical positions:

    PositionStressRemark
    InteriorSi = 0.316P/h²[4 log10(l/b) + 1.069]tension at the bottom
    EdgeSe = 0.572P/h²[4 log10(l/b) + 0.359]largest load stress; tension at the bottom
    CornerSc = 3P/h²[1 − (a√2/l)0.6]tension at the top
    l = [Eh³/(12k(1−μ²))]0.25 b = √(1.6a²+h²) − 0.675h for a < 1.724h

    2. Warping (temperature) stresses due to the differential Δt across the slab — interior (Cx+μCy)EαΔt/[2(1−μ²)], edge C·EαΔt/2, corner EαΔt√(a/l)/[3(1−μ)].

    3. Frictional stresses due to the subgrade restraining the uniform expansion/contraction: σf = WLf/(2×104) — compressive in summer, tensile in winter.

    Critical combinations: summer mid-day at the edge = load + warping − friction; winter mid-day at the edge = load + warping + friction; mid-night at the corner = load + warping. All are checked against the modulus of rupture of the concrete.

    (b) Design procedure for dowel bars at an expansion joint

    1. Fix the dowel diameter d (IRC: 25 mm for h = 20-25 cm) and the joint width δ. Dowels are used only when h ≥ 15 cm.
    2. Find the balanced embedment length by equating the bending and bearing capacities:
      Ld = 5d√[(Ff/Fb)(Ld+1.5δ)/(Ld+8.8δ)] — solved by trial. Total bar length = Ld + δ, rounded up.
    3. Compute the load capacity of a single bar in the three modes and take the least:
      shear Ps = 0.785d²Fs; bending Pf = 2d³Ff/(Ld+8.8δ); bearing Pb = FbLd²d/[12.5(Ld+1.5δ)].
    4. The dowel system is designed to transfer 40% of the design wheel load; hence the required capacity factor = 0.40P / (capacity of one bar).
    5. Compute l and take the effective load-transfer zone as 1.8 l on either side of the load.
    6. Assume a spacing s; the available capacity factor = 1 + Σ(1.8l − n·s)/1.8l for all bars inside the zone. Reduce s until the available factor exceeds the required factor.
    7. Detail the bars: one half greased/painted with an expansion cap, placed at mid-depth of the slab, parallel to the centre line and to the surface.

    Manual.pdf §2.6-2.8 — stresses (book pp.155-159) and dowel-bar design (book pp.168, 172).

  3. Estimate the thickness of a plain cement concrete pavement for a 7m wide highway following the design procedure recommended by the Indian Roads Congress (IRC), using IRC stress charts for edge and corner regions wherever applicable. Use the following data: Design wheel load = 5100 kg; Traffic growth rate = 7.5%; Present traffic intensity = 1050 cvpd; Design life = 20 years; Construction period = 3 years; Radius of contact area = 15 cm; Modulus of elasticity of concrete = 3.0 x 10^5 kg/cm2; Poisson's ratio of concrete = 0.15; Modulus of rupture of concrete = 40 kg/cm2; Thermal expansion of concrete = 10 x 10^-6 per degC; Modulus of subgrade reaction = 6 kg/cm3; Maximum temperature in summer = 50 degC; Maximum temperature in winter = 15 degC. The temperature differential in the slab for thickness of 15, 20 and 25 cm is 17.3, 19.0 and 20.3 degC respectively. (2075 Chaitra, Q6)
    Solution
    7 m wide plain cement-concrete pavement · design wheel load P = 5100 kg · contact radius a = 15 cm · E = 3.0×105 kg/cm² · μ = 0.15 · modulus of rupture = 40 kg/cm² · α = 10×10−6/°C · k = 6 kg/cm³ · summer 50°C, winter 15°C (ΔT = 35°C) · present traffic 1050 cvpd, r = 7.5%, design life 20 yr, construction 3 yr · Δt = 17.3, 19.0, 20.3°C for h = 15, 20, 25 cm

    Spacing of joints

    EXPANSION JOINT (gap δ = 2.5 cm, each slab may move δ/2 = 1.25 cm) Le = (δ/2)/(α · ΔT) = 1.25/(10×10⁻⁶ × 35) = 3571 cm = 35.7 m ≤ 40 m ✓ CONTRACTION JOINT (plain slab, σc = 0.8 kg/cm², W = 2400 kg/m³, f = 1.5) Lc = 2×10⁴ σc/(W · f) = 2×10⁴ × 0.8/(2400 × 1.5) = 4.44 m ≤ 4.5 m ✓

    Provide contraction joints at 4.44 m and an expansion joint at every 8th contraction joint (8 × 4.44 = 35.5 m ≈ 35.7 m).

    Trial slab thickness

    For each trial h compute l = [Eh³/(12k(1−μ²))]0.25, b = √(1.6a²+h²) − 0.675h, C from Bradbury's chart at Lc/l, the edge warping stress C·EαΔt/2 and the IRC edge load stress 0.529(P/h²)(1+0.54μ)[4log10(l/b)+log10b−0.4048]:

    h (cm)Δt (°C)l (cm)b (cm)Lc/lCEdge load σleEdge warping σteTotal (kg/cm²)≤ 40 ?
    1517.361.614.067.311.04442.8927.1069.99Fail
    2019.076.414.075.890.89826.8525.5952.44Fail
    2520.390.314.514.980.71518.3621.7640.12≈ 40, just adequate
    at h = 25 cm: residual strength for load = 40 − 21.76 = 18.24 kg/cm² edge load stress = 18.36 kg/cm² factor of safety = 18.24/18.36 = 0.99 ≈ 1.0 → adopt h = 25 cm summer check (friction is compressive): σf = 2400 × 4.44 × 1.5/(2×10⁴) = 0.80 kg/cm² 18.36 + 21.76 − 0.80 = 39.3 kg/cm² < 40 ✓

    A strictly > 1 check would take h = 26 cm (FS = 1.12).

    Check the corner region (h = 25 cm, no frictional stress at the corner)

    corner load stress (Kelley) = 3P/h² [1 − (a√2/l)0.6] = 3×5100/625 × [1 − (21.21/90.3)0.6] = 14.22 kg/cm² corner warping stress = EαΔt √(a/l)/[3(1−μ)] = 3×10⁵ × 10×10⁻⁶ × 20.3 × √(15/90.3)/(3 × 0.85) = 9.73 kg/cm² total at corner = 14.22 + 9.73 = 23.95 kg/cm² < 40 ✓ safe

    Adjustment for traffic intensity

    Ad = P(1 + r)(n + y) = 1050 (1.075)20+3 = 1050 × 5.277 = 5541 cvpd 5541 > 4500 → traffic group G → thickness adjustment = +2 cm design thickness = 25 + 2 = 27 cm
    Provide a 270 mm thick plain cement-concrete slab, contraction joints at 4.44 m, expansion joints at 35.5 m, with 25 mm φ dowel bars at the expansion joints and 10 mm φ tie bars at 30 cm c/c across the longitudinal joint (the 7 m carriageway > 4.5 m requires one).

    Step-for-step the IRC design worked in Manual.pdf §2.8, book pp.169-173 (joint spacing → trial thickness with warping + IRC load stress → corner check → traffic adjustment Ad = P(1+r)n+y and the group-G +2 cm correction → dowel and tie bars). Load stresses computed from the IRC (Teller & Sutherland / Kelley) equations that the IRC stress charts plot, book p.156.

  4. What are the factors affecting rigid pavement design? Write down the steps of IRC design guidelines for rigid pavement. (2074 Ashwin, Q5)
    Solution

    (a) Factors affecting rigid-pavement design

    • Wheel load and its repetitions — design wheel load (5100 kg for IRC charts), contact radius, and the number of commercial vehicles projected to the end of the design life Ad = P(1+r)n+y, which fixes the traffic group and the thickness adjustment.
    • Subgrade / sub-base support — modulus of subgrade reaction k (from a plate-bearing test, corrected for the sub-base); k enters the radius of relative stiffness l = [Eh³/(12k(1−μ²))]0.25.
    • Concrete properties — flexural strength (modulus of rupture, 40-45 kg/cm²), E, μ and the coefficient of thermal expansion α.
    • Temperature differential Δt across the slab (region- and thickness-dependent) which produces the warping stresses; and the seasonal temperature range which, with the coefficient of subgrade friction f, produces the frictional stresses.
    • Slab geometry / joint spacing — Lx, Ly control Bradbury's coefficients and hence the warping stress; the carriageway width decides whether a longitudinal joint is needed.
    • Drainage and shoulder type, and whether a tied concrete shoulder is provided (reduces the edge stress).

    (b) Steps of the IRC design guideline for rigid pavement

    1. Collect design data — P, a, k, E, μ, α, flexural strength, Δt, present traffic, growth rate, design life and construction period.
    2. Compute the design traffic Ad = P(1 + r)(n+y) and identify the traffic group (A … G).
    3. Fix the joint spacings — expansion joint Le = (δ/2)/(αΔT) (≤ 140 m rough / 90 m smooth sub-base) and contraction joint Lc = 2×104σc/(Wf) (≤ 4.5 m plain slab).
    4. Assume a trial thickness h and compute l and b.
    5. Warping stresses — read Bradbury's C from L/l and compute σte = C·EαΔt/2 (edge) and σtc = EαΔt√(a/l)/[3(1−μ)] (corner).
    6. Residual strength = flexural strength − warping stress.
    7. Load stresses — read the edge and corner load stress from the IRC stress charts for h and k (or compute from the Teller & Sutherland / Kelley equations).
    8. Check factor of safety = residual strength / load stress at the edge; if it is much greater or less than 1, revise h and repeat from step 4. Also verify (load + warping) at the corner ≤ flexural strength.
    9. Adjust for traffic intensity — add the thickness increment prescribed for the traffic group (e.g. +2 cm for group G).
    10. Design the joint furniture — dowel bars at the expansion joints (Bradbury's method, 40% load transfer, effective group 1.8 l), tie bars at the longitudinal joint (As = bfhW/Ss, length = dSs/2τ) and, for JRCP, the temperature reinforcement As = LWf/(2Ss).

    Manual.pdf §2.8 — Details of the IRC method of design of rigid pavements for highways, book pp.164-173.

DoR (2021) and IRC 58-2015 Rigid Pavement Design

Design of Slab Thickness (DoR, 2021, IRC 58-2015)

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Typical Cross Sections of Concrete Pavements

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Subgrade

Effective CBR

Sub-Base

drainage layers:

Characteristic of Concrete

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Axle Load Characteristics

Design Traffic

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Design Traffic (BUC)

Design Traffic (TDC)

Temperature Differential

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Calculation of Flexural Stresses (BUC)

IV (chart) & Appendix V(Equations). Sample charts:

Calculation of Flexural Stresses (TDC)

Calculation Fatigue Damage (CFD) Analysis

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safe from large scale cracking if: CFD (BUC) + CFD (TDC) < 1

Design of Slab Thickness (DoR, 2021, IRC 58-2015)

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AASHTO 1993 Rigid Pavement Design

Design Equation
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Practice Tasks from the Lecture

3.7 Pavement design (Surface dressing method) for low cost road, IRC method

IRC SP72-2015 and DoR, 2021

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Subgrade Strength Classification:

Thickness

Design (Design

Chart):

Thickness

Design (Design

Chart):

Design of Surface Dressing (DoR, 2021)

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DOR (2021): Flexible Pavement (Upto 2msa)

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Practice Tasks from the Lecture