UNIT 5: ADVANCED PAVEMENT DESIGN - EXAM-FOCUSED SHORT NOTES

Based on analysis of CE-603(C) - Advance Pavement Design past papers (Jun 2025 & May 2024).

I. FUNDAMENTALS OF PAVEMENT DESIGN & LOADING

1.1. Pavement Types & Structural Components

• Flexible Pavement: Multi-layer system (Surface course → Base → Sub-base → Subgrade). Function: Distributes loads through grain-to-grain transfer. Requirement: Sufficient thickness to keep subgrade stress within bearing capacity.

• Rigid Pavement: Cement concrete slab (single layer or with base). Function: Distributes loads through slab action (high flexural strength). Requirement: Adequate thickness to resist flexural stresses.

[!TIP] Exam Focus: Be prepared to compare structural/functional requirements side-by-side. Flexible relies on layer strength, rigid on slab modulus of rupture.

1.2. Traffic Loading & Characterization

Equivalent Single Wheel Load (ESWL) / Equivalent Axle Load (EAL/EASL)

• Concept: Convert multiple wheel loads (dual, tandem) into an equivalent single wheel load causing the same pavement damage (stress/strain) at a critical depth.

• Necessity: Simplifies design for complex traffic. Lateral Distribution Factor (LDF) accounts for load spreading between wheels/lanes.

• Determination Methods:

  1. Equal Vertical Stress Criterion: ESWL at depth z is the single wheel load whose vertical stress equals the combined stress from multiple wheels at that depth.

$$ \sigma_{z,eq} = \sigma_{z,multi} $$

2.  **Equal Contact Pressure Criterion:** Assumes equal contact pressure for all wheels. Less accurate for deeper depths.

• Design EASL: Not just a simple sum. It's the cumulative damage factor considering:

  • Traffic Survey: Count vehicles by axle configuration.

  • ESWL Conversion: For each axle type, compute ESWL at a standard depth (e.g., 45 cm for flexible, top of subgrade for rigid).

  • Damage Exponent (n): Typically 4 for flexible (4th power law), ~1.5-2 for rigid. Damage ∝ (axle load)^n.

  • Estimation:

$$ \text{Design EASL} = \sum_{i=1}^{N} (n_i \times \text{ESWL}_i^n) $$

where n_i is number of repetitions of axle type i.

[!TIP] Common Pitfall: Students confuse EASL (simple equivalent load) with Design EASL (cumulative damage factor using an exponent). The latter is used in IRC:37.

Lateral Distribution Factor (LDF)

• Concept: Fraction of wheel load transmitted to a given layer at a specific lateral position. Accounts for load spreading through layers.

• Significance: Reduces effective load on a particular layer compared to the applied wheel load. Critical for base/subbase design.

• Application: Effective load on a layer = (Wheel Load) × LDF.

• Typical Values (IRC): For flexible pavement base/subbase under a standard wheel load, LDF may range from 0.4 to 0.8 depending on layer position and thickness.

• Sketch: Show a wheel load P on surface. As we go deeper (to base), the load spreads over a larger area. The load on a vertical section at the base is less than P; this fraction is LDF.

1.3. Design Variables & Factors

• Load Variables: Wheel load (single/dual), tyre pressure, axle configuration (single, tandem, tridem), load spacing.

• Structural Variables: Thickness of each layer, material properties (E, Mr, k, modulus of rupture), layer interface condition.

• Environmental/Climatic: Temperature (daily/seasonal gradients for rigid), precipitation (drainage, frost), frost action (frost heave, thaw weakening).

• Subgrade Soil Properties: Strength (CBR, Mr), stiffness (E), swelling potential, drainage characteristics.

II. FLEXIBLE PAVEMENT DESIGN

2.1. Subgrade Strength Assessment

California Bearing Ratio (CBR)

• Test Procedure:

  1. Sample Prep: Soaked (4 days for CBR) or unsoaked. Compact in CBR mould (density ~ 95% of Proctor).

  2. Soaking: Immerse sample for 96 hours (for soaked CBR, representing worst-case).

  3. Penetration: Apply load via piston (50 mm dia) at 1.25 mm/min. Record load at 2.5 mm and 5.0 mm penetration.

  4. Calculation:

$$ \text{CBR (\%)} = \frac{\text{Load at penetration}}{\text{Standard load}} \times 100 $$

Standard load: 1305 kg (2.5 mm), 2055 kg (5.0 mm). Take lower value.

• Limitations:

  • Empirical, not stress-based.

  • Does not account for repeated loading (resilient modulus is better).

  • Soaking condition may not represent field for arid regions.

  • Poor correlation with pavement performance in some soils.

  • Not suitable for stabilized/improved soils.

Other Parameters:

• Modulus of Elasticity (E): Stress/Strain in elastic range. Used in elastic layer theory.

• Poisson's Ratio (ν): Lateral strain/Axial strain. Typically 0.4-0.5 for saturated clays, 0.3-0.4 for granular.

• Resilient Modulus (Mr): Most fundamental for modern mechanistic design. Ratio of recoverable axial stress to axial strain under repeated load.

$$ M_r = \frac{\sigma_d}{\epsilon_r} $$

where σ_d is deviator stress, ε_r is recoverable strain.

[!TIP] Exam Trend: CBR test procedure and limitations of CBR are frequently asked. Know the 96-hour soaking and the standard load values.

2.2. Design Methodologies

IRC Method (Flexible Pavements - IRC:37)

• Step-by-Step:

  1. Determine Design Traffic in terms of Design EASL (from traffic survey, using 4th power law).

  2. Assess Subgrade CBR (soaked for general design).

  3. Select ** pavement type** (flexible, semi-rigid, rigid) based on traffic & CBR.

  4. Use IRC:37 design charts/graphs (Fig. 2 to 5). Plot Design EASL vs. Subgrade CBR to get Total Pavement Thickness (in cm).

  5. Distribute total thickness among layers (Bituminous Macadam, Base, Sub-base) based on material specifications and CBR values of each layer. Ensure each layer's CBR > required.

  6. Check for drainage requirements (permeability, cross-slope).

AASHTO Method (1993/1998 Guide)

• Core Concept: Structural Number (SN).

$$ SN = a_1 D_1 + a_2 D_2 + a_3 D_3 + ... $$

where a_i = layer coefficient (strength), D_i = thickness (inches).

• Design Equation:

$$ \log_{10} W_{18} = Z_1 S_0 + Z_2 (1 - \frac{29.4}{SN}) + ... $$

Complex, includes reliability, standard deviation, initial/final serviceability.

• Discussion: More mechanistic-empirical than IRC. Uses layer coefficients derived from resilient modulus or CBR. Less common in Indian exams but good to know for comparison.

Elastic Layer Theory (Boussinesq)

• Single Layer: Stress, strain, deflection under circular load (wheel) calculated using Boussinesq equations for homogeneous, isotropic, elastic half-space.

• Multi-layer System: More realistic. Assumes each layer is homogeneous, isotropic, elastic, infinite in horizontal extent, and perfectly bonded.

• Key Outputs: Vertical stress σ_z, horizontal strain ε_x (critical at top of subgrade), vertical deflection δ.

• Application in Design: Often used to compute design strain (e.g., horizontal tensile strain at bottom of bituminous layer, vertical compressive strain on top of subgrade) which is then related to allowable repetitions (using fatigue/ rutting equations).

[!TIP] Numerical Focus: Past papers ask for flexible pavement thickness using single-layer elastic theory. Given: wheel load P, tyre pressure q, permissible deflection Δ_all, modulus E. Use formula for deflection under circular loaded area:

$$ \Delta = \frac{3P(1 - \nu^2)}{2\pi E^2} \left[ \left(1 - \frac{a^2}{c^2}\right) \sin^{-1}\frac{a}{c} + \frac{a}{c} \sqrt{1 - \frac{a^2}{c^2}} \right] $$

where a = radius of loaded area, c = radius of rigid base (or depth of interest). Often simplified for design depth.

2.3. Pavement Evaluation & Overlay Design

Benkelman Beam Method

• Principle: Measures surface rebound deflection under a standard dual-wheel load (8160 kg, tyre pressure 7 kg/cm²). Deflection is inversely related to pavement strength.

• Procedure (Existing Pavement Evaluation):

  1. Place Benkelman Beam (BB) in front of rear wheel.

  2. Apply test load, measure initial dial reading.

  3. Release load, measure final dial reading (rebound).

  4. Deflection (D) = Initial - Final (corrected for beam length, etc.).

  5. Take multiple readings (at least 17 per km) at regular intervals.

  6. Compute Mean Deflection (D_m) and Standard Deviation (S).

  7. Characteristic Deflection (D_c) = D_m + 1.64 S (for 95% confidence).

• Overlay Design using BBD Data:

  1. Determine Characteristic Deflection (D_c) of existing pavement.

  2. Determine Design Deflection (D_d) for the proposed overlay + existing system, based on future traffic (Design EASL) and subgrade CBR (using IRC:37 charts or equations).

  3. Overlay Thickness (T_o): Found from the relationship between deflection and thickness. Often uses the Burmister's two-layer theory or empirical correlations.

$$ T_o = C \times (D_c - D_d) $$

where C is a constant (empirical, from experience or calibration).

4.  Alternatively, compute the **effective thickness** of existing pavement in terms of a standard material (e.g., granular base equivalent). Then, find total required thickness for new traffic, subtract existing effective thickness to get overlay thickness.

[!TIP] High-Frequency Question: "Steps of overlay design using BBD data." Memorize the 4-5 key steps: D_c calculation → D_d determination → Effective thickness concept → T_o calculation.

III. RIGID (CEMENT CONCRETE) PAVEMENT DESIGN

3.1. Stress Analysis in Rigid Pavements

Westergaard's Theory

• Fundamental Concepts: Considers concrete slab as a thin elastic plate resting on a Winkler foundation (springs with modulus k). Loads cause bending.

• Assumptions:

  1. Slab is homogeneous, isotropic, elastic.

  2. Foundation is elastic (modulus of subgrade reaction k), no tension, infinite depth.

  3. Slab is thin (length/width >> thickness).

  4. No shear transfer between slab and foundation.

• Key Parameter: Radius of Relative Stiffness (l)

$$ l = \left[ \frac{E h^3}{12 k (1 - \nu^2)} \right]^{1/4} $$

*   `E` = Modulus of elasticity of concrete (kg/cm²)

*   `h` = Slab thickness (cm)

*   `k` = Modulus of subgrade reaction (kg/cm³)

*   `ν` = Poisson's ratio of concrete (~0.15)

*   **Significance:** `l` represents the **relative stiffness** of slab vs. subgrade. Larger `l` means slab is stiffer relative to foundation.

Modulus of Subgrade Reaction (k) vs. Radius of Relative Stiffness (l)

• k: Foundation property. Pressure per unit deflection (kg/cm³). Measured by plate load test on subgrade. Indicates strength of subgrade.

• l: Slab-foundation system property. Depends on slab properties (E, h) and foundation property (k). Indicates overall system stiffness. Used directly in Westergaard stress equations.

3.2. Critical Stress Combinations & Temperature Effects

Critical Stress Combinations:

Occur when maximum load stress coincides with maximum warping stress (due to temperature/moisture gradient).

• Interior Region: Critical during daytime (top in compression due to temperature gradient, bottom in tension from load). Load + negative gradient.

• Edge Region: Critical during nighttime (top in tension from gradient, edge in tension from load). Load + positive gradient.

• Corner Region: Critical under corner loading (load + warping). Often governs thickness.

Thermal Stresses:

• Daily (Gradient) Variation: Top and bottom of slab at different temperatures → curvature → warping stresses.

  • Warping Stress (σ_t):

$$ \sigma_t = \frac{E \alpha \Delta T}{2} \left[ 1 - \frac{1}{\cosh(\beta x)} \right] $$

for interior, simplified. Where α = thermal coefficient, ΔT = temperature difference between top & bottom, β = coefficient related to l.

• Seasonal Variation: Uniform temperature change → uniform expansion/contraction → frictional stresses if restrained.

$$ \sigma_f = \frac{E \alpha \Delta T}{1 - \nu} $$

for fully restrained slab. Usually less critical than warping.

[!TIP] Numerical Alert: Warping stress calculation is a sure-shot question. Know the simplified formulas for interior, edge, corner from Westergaard. Given: h, l, α, ΔT (daily gradient), E, ν. Compute β = (1/l) * sqrt(2) or similar. Then use:

• Interior:

$$ \sigma_t = \frac{E \alpha \Delta T}{2} \left( 1 - \frac{1}{\cosh(\beta x)} \right) \approx \frac{E \alpha \Delta T}{2} \text{ for small } x/l $$

• Edge:

$$ \sigma_t = \frac{E \alpha \Delta T}{2} \left( 1 - \frac{2}{\pi} \cos^{-1} e^{-\beta x} \right) \text{ at edge (x=0)} \approx 0.3 \frac{E \alpha \Delta T}{2} $$

• Corner:

$$ \sigma_t = \frac{E \alpha \Delta T}{2} \left( 1 - \frac{2}{\pi} \cos^{-1} e^{-\beta l} \right) \approx 0.7 \frac{E \alpha \Delta T}{2} \text{ (max)} $$

3.3. IRC Recommendations for CC Pavement Thickness (IRC:58)

• Design Basis: Westergaard's stress analysis for critical corner loading (most severe).

• Procedure:

  1. Determine Design Traffic (EASL for rigid, using exponent ~1.5-2).

  2. Determine Subgrade Strength (CBR or k). Convert CBR to k if needed: k ≈ 0.5 * CBR (approx, for design).

  3. Select Concrete Properties: E, α, modulus of rupture f_r (typically 0.7 * sqrt(f_ck) for flexure).

  4. Assume trial thickness h. Compute l.

  5. Calculate corner stress σ_c due to corner load (load = 6300 kg for design single wheel, contact radius 15 cm).

  6. Calculate warping stress σ_w at corner (using ΔT daily gradient, typically 0.5 °C/cm).

  7. Total Stress σ_total = σ_c + σ_w.

  8. Check if σ_total ≤ f_r / FoS (FoS ~ 1.1-1.2). If not, increase h and repeat.

  9. Also check edge stress for edge loading (load = 4200 kg/wheel, contact radius 15 cm).

• IRC:58 Charts: Provides design curves relating required thickness to k and EASL.

3.4. Pavement Joints

Types of Joints:

• Transverse Joints:

  • Expansion Joints: Provide space for slab expansion. Filled with pre-moulded joint filler (e.g., bitumen-treated fibre board). Dowels not provided. Spacing: 50-90 m.

  • Contraction (Control) Joints: Induce controlled cracking. Formed by saw cut (depth 1/4-1/3 h). May have dowel bars for load transfer.

  • Construction Joints: Where concreting stops. May be header (vertical face) or keyed. Dowels provided if intended as contraction joint.

• Longitudinal Joints: Separate lanes. Tie bars provided to hold faces together, prevent separation.

• Isolation Joints: Around structures (manholes, bridges) to allow independent movement.

Tie Bars vs. Dowel Bars

$$ A_s = \frac{\mu W L}{f_s} $$

where μ=coeff. friction, W=slab weight, L=spacing, f_s=allowable steel stress. | Based on shear stress from wheel load.

$$ \tau = \frac{P}{n \times \pi d^2 / 4} \leq \text{allowable} $$

where P=load on wheel, n=number of dowels, d=diameter. |

| Diameter/Length | Smaller (e.g., 12-16 mm dia), longer (0.6-0.8L). | Larger (e.g., 20-25 mm dia), shorter (0.2-0.3L). | | Installation | Placed before concreting, tied to reinforcement. | Placed after concrete placement, inserted into pre-drilled holes or using assemblies. | | Difficulties | Misalignment, congestion, corrosion if not coated. | Misalignment, spalling during insertion, improper alignment causing stress concentrations. |

Joint Fillers vs. Sealing Compounds

IV. PAVEMENT MATERIALS & TESTING

4.1. Road Aggregates

Key Tests (as per IS codes):

• Crushing Strength: Aggregate crushing value (ACV), 10% fines value. Lower = stronger.

• Abrasion Resistance: Los Angeles abrasion value. Lower = more resistant to wear.

• Impact Strength: Aggregate impact value (AIV).

• Soundness: Sodium/magnesium sulfate soundness test. Loss % indicates weathering resistance.

• Shape & Texture: Flakiness index, elongation index, angularity number.

• Water Absorption: Indicates porosity & durability.

• Specific Gravity & Density: For mix design.

4.2. Subgrade Soil & CBR

Detailed CBR Test Procedure (IS: 2720 (Part 16)):

1. Sample Preparation: Remoulded or undisturbed. Compact in CBR mould (diameter 150 mm, height 125 mm) to 95% of Proctor density (for soaked CBR, use soaked CBR - field worst-case).

2. Soaking: Immerse sample in water for 96 hours (4 days). Water level 25 mm above top of sample.

3. Surfacing: Place porous plate and filter paper on sample.

4. Penetration: Apply load via 50 mm diameter piston at 1.25 mm/min. Record load at 2.5 mm and 5.0 mm penetration.

5. Calculation:

$$ \text{CBR (\%)} = \frac{\text{Load at penetration (kg)}}{\text{Standard load (kg)}} \times 100 $$

Standard load: 1305 kg (2.5 mm), 2055 kg (5.0 mm). Take lower value.

6. Expansion: Measure swell after soaking (should be < 5% for good subgrade).

Correlation of CBR:

• With ** soaked CBR ≈ 0.2 × soaked CBR** (for granular soils).

• With Modulus of Elasticity (E): E (kg/cm²) ≈ 10 × CBR (very approximate, for granular).

• With Resilient Modulus (Mr): Mr ≈ 10-15 × CBR (in psi/100? Be cautious with units). Better to use direct Mr tests.

V. DESIGN STANDARDS & COMPARATIVE ANALYSIS

5.1. Indian Roads Congress (IRC) Guidelines

• Flexible: IRC:37 - "Guidelines for the Design of Flexible Pavements." Uses Design EASL (4th power law) and CBR-based thickness charts. Considers traffic, subgrade, climate, materials.

• Rigid: IRC:58 - "Guidelines for the Design of Plain Jointed Cement Concrete Pavements." Uses Westergaard's corner stress for thickness design. Provides charts for k vs. EASL. Details on joints (IRC:15), dowels/tie bars.

5.2. AASHTO Guide

• AASHTO 1993/1998/2021: Mechanistic-Empirical (MEPDG is newer, but 1993 is classic).

• Structural Number (SN):

$$ SN = a_1 D_1 + a_2 D_2 + a_3 D_3 + ... $$

a_i = layer coefficients (0.44 for granular base, 0.14 for sub-base, 0.40-0.50 for HMA).

• Design Equation: Solves for SN given traffic (W18), reliability, subgrade support (MR), serviceability loss.

• Comparison with IRC: AASHTO is more flexible (allows different materials via a_i), uses reliability and serviceability index. IRC is more prescriptive, CBR-based. AASHTO 2021 is fully mechanistic-empirical (MEPDG) using software.

5.3. Mitigation of Stress-Related Issues in Rigid Pavements

• Reinforcement Techniques:

  • Steel Reinforcement (CRC): Continuous, no joints. Controls cracking. High initial cost.

  • Mesh/Bar Reinforcement in JRCP: For longer joint spacings (> 10 m). Controls crack width.

• Joint Design Strategies:

  • Optimal Joint Spacing: Balance between warping stress (longer slabs) and curling/faulting (shorter slabs). Typically 3.6-4.5 m for panels, 12-15 m for full-depth saw-cut joints.

  • Dowel Bars: Proper alignment (parallel to traffic, centered) to ensure effective load transfer, prevent faulting.

  • Tie Bars: Prevent lane separation.

• Material Selection:

  • Concrete Mix: Lower E and α (using aggregates with low CTE), higher flexural strength (f_r), good durability.

  • Subgrade/Base: High k (stiff foundation) reduces l and stresses. Good drainage to prevent pumping.

VI. NUMERICAL PROBLEM AREAS (HIGH FREQUENCY)

6.1. ESWL Calculation (Equal Vertical Stress Criterion)

• Given: Dual wheel load 2P, spacing S, contact radius a. Find ESWL at depth z.

• Steps:

  1. Compute vertical stress σ_z at depth z under each wheel using Boussinesq for circular load:

$$ \sigma_z = \frac{3P}{2\pi z^2} \left[ \frac{1}{1 + (a/z)^2} \right]^{5/2} \text{ (approx)} $$

or use influence charts.

2.  Use **principle of superposition** to get total `σ_z` from both wheels at the point midway between them (critical point).

3.  Find single wheel load `P_eq` whose stress at depth `z` equals this total `σ_z`.

4.  **ESWL = P_eq**.

• Formula for ESWL (if using equal stress criterion & assuming point loads at depth): For dual wheels spaced S, ESWL at depth z can be approximated by:

$$ \text{ESWL} = P \left[ 1 - \frac{1}{2} \left( \frac{S}{z} \right)^2 \right] \text{ for } S/z < 1.5 $$

(Check exact formula from textbooks like Khanna & Justo).

6.2. Flexible Pavement Thickness (Single Layer Elastic Theory)

• Given: Wheel load P, tyre pressure q, radius of loaded area a = \sqrt{P/(\pi q)}, permissible vertical compressive strain ε_z on subgrade OR permissible deflection Δ at surface, modulus E of pavement layer (or subgrade).

• Using Deflection (Boussinesq for circular load on half-space):

$$ \Delta = \frac{3P(1 - \nu^2)}{2\pi E^2} \left[ \left(1 - \frac{a^2}{c^2}\right) \sin^{-1}\frac{a}{c} + \frac{a}{c} \sqrt{1 - \frac{a^2}{c^2}} \right] $$

where c is depth of interest (for surface deflection, c → ∞, simplifies to Δ = (1-ν²)P/(π a E)? Clarify: For surface deflection under rigid circular plate of radius a on elastic half-space:

$$ \delta = \frac{(1-\nu^2) P}{a E} $$

. This is common.

• Example from Paper (Jun 2025): P=4200 kg, q=6 kg/cm², E=150 kg/cm², Δ_all=0.25 cm. Find a = sqrt(4200/(π*6)) ≈ 16.7 cm. Then Δ = (1-0.5²)*4200/(16.7*150) ≈ 0.24 cm. So total thickness ≈ depth where strain/deflection becomes acceptable? The question likely expects using the formula for deflection at the top of a finite layer. But given values n^{1/2}=1.77 etc., it might be using a different approach (maybe for multi-layer?). Standard approach: For single layer, thickness D is such that deflection at surface = Δ_all. Using Δ = (1-ν²)P/(a E) gives E as subgrade modulus? If E is pavement modulus, then thickness is not directly in formula. Likely interpretation: The given E is subgrade modulus, and we need to find pavement thickness D such that vertical strain at top of subgrade is within limit. Use ε_z = σ_z / E_subgrade. Compute σ_z at depth D from Boussinesq, set ε_z = permissible. Solve for D. This is iterative.

6.3. Rigid Pavement Stress Calculation (Westergaard)

• Given: h, E, ν, k, α, ΔT (daily gradient), load P and radius a (usually 15 cm for design wheel).

• Steps:

  1. Compute l = [E h³ / (12 k (1-ν²))]^{1/4}.

  2. Compute β = (1/l) * sqrt(2) or β = (1/l) * sqrt( (k h)/(6E) )? Standard Westergaard: β = (1/l) * sqrt( (k h)/(6E) )? Actually, β = (1/l) * sqrt(2) for interior, β = (1/l) * sqrt(1)? Use standard formulas:

     • Interior Stress (σ_i):

$$ \sigma_i = \frac{3P}{h^2} \left[ 1 - (a/l)^{0.5} \right] \text{ (approx)} $$

or exact:

$$ \sigma_i = \frac{3P}{h^2} \left[ 1 - (1.121 \frac{a}{l})^{0.5} \right] $$

    *   **Edge Stress (σ_e):** 

$$ \sigma_e = \frac{3P}{h^2} \left[ 1 - (1.121 \frac{a}{l})^{0.5} \right] \times \text{edge coefficient} $$

Edge coefficient ≈ 0.8 for a/l < 1.0.

    *   **Corner Stress (σ_c):** 

$$ \sigma_c = \frac{3P}{h^2} \left[ 1 - (1.121 \frac{a}{l})^{0.5} \right] \times \text{corner coefficient} $$

Corner coefficient ≈ 1.0 for a/l < 1.0.

3.  Compute **warping stress** at each region:

    *   **Interior:** 

$$ \sigma_{t,i} = \frac{E \alpha \Delta T}{2} \left[ 1 - \frac{1}{\cosh(\beta x)} \right] \approx \frac{E \alpha \Delta T}{2} \text{ at } x=0 \text{ (mid-slab)} $$

    *   **Edge:** 

$$ \sigma_{t,e} = \frac{E \alpha \Delta T}{2} \left[ 1 - \frac{2}{\pi} \cos^{-1} e^{-\beta x} \right] \approx 0.3 \frac{E \alpha \Delta T}{2} \text{ at edge (x=0)} $$

    *   **Corner:** 

$$ \sigma_{t,c} = \frac{E \alpha \Delta T}{2} \left[ 1 - \frac{2}{\pi} \cos^{-1} e^{-\beta l} \right] \approx 0.7 \frac{E \alpha \Delta T}{2} \text{ at corner} $$

4.  **Total Stress:** `σ_total = σ_load + σ_warping`. Compare with allowable flexural strength `f_r`.

6.4. Tie Bar Design for Longitudinal Joints

• Purpose: Resist frictional force trying to separate lanes due to traffic/thermal effects.

• Force per unit length: f = μ * W where μ = coefficient of friction (0.8-1.5), W = weight of concrete per unit area = h * γ (γ = unit weight, ~2400 kg/m³).

• Tie Bar Area per meter:

$$ A_s = \frac{f \times 1}{f_s} = \frac{\mu h \gamma}{f_s} $$

where f_s = allowable tensile stress in steel (e.g., 1800 kg/cm²).

• Diameter (d): Choose standard bar (e.g., 12 mm, 16 mm). Check bond: L_b = A_s * f_s / (π d * τ_b) where τ_b = allowable bond stress (e.g., 24.6 kg/cm²).

• Spacing (L):

$$ L = \frac{A_s}{A_{s,bar}} = \frac{\text{Area per meter}}{\pi d^2/4} $$

• Length: Must be sufficient to develop bond on both sides of joint, typically L ≥ 0.5 * slab thickness but from bond calculation.

[!TIP] Example from Paper (May 2024): Given h=0.203 m, γ=25 kN/m³ (≈ 2500 kg/m³), μ=1.5, f_s=200 MPa=2039 kg/cm², τ_b=2.4 MPa=244.6 kg/cm². Compute stepwise.

6.5. Warping Stress Determination

• As in 6.3. Key is computing β and applying correct coefficient for region. Remember:

  • Interior: ~0.5 * (E α ΔT)

  • Edge: ~0.3 * (E α ΔT)

  • Corner: ~0.7 * (E α ΔT) (MAXIMUM)

6.6. Design EASL Estimation

• From Traffic Count: For each vehicle type i with axle load L_i (in kg) and repetitions n_i.

• Convert to EASL: For each axle,

$$ \text{EASL}_i = n_i \times \left( \frac{L_i}{L_{std}} \right)^n $$

where L_std = standard axle load (usually 80 kN or 8200 kg for flexible, 6300 kg for rigid?), n = exponent (4 for flexible, 1.5-2 for rigid).

• Sum:

$$ \text{Design EASL} = \sum \text{EASL}_i $$

over all vehicle types and axles.

• Growth Factor: If design life N years, annual growth rate r, then cumulative EASL = A * [(1+r)^N - 1]/r where A = initial annual EASL.

Final Exam Strategy: Focus on derivations (Westergaard, ESWL), step-by-step procedures (IRC design, BBD overlay), and numerical problems (warping stress, tie bar, ESWL). Always state assumptions and units clearly. For theory, use bullet points with bold key terms.