CELE Transportation & Highway Engineering — Pavement Design (Flexible and Rigid)Revision Notes

Exam Tips

• Board questions often ask you to identify which pavement type is appropriate for a given scenario — think: heavy loads + weak subgrade + long design life → rigid pavement.
• Remember the mnemonic: Flexible = Flexible layers share the load; Rigid = Rigid slab bends to carry the load.
• Know the typical thickness ranges: AC surface 50–100 mm; PCC slab 150–300 mm.
• Distress patterns are testable: rutting and fatigue cracking → flexible; corner cracking and faulting → rigid.

Key Points

• Flexible pavement consists of an asphalt concrete (AC) surface layer over a granular base and subbase, all resting on the compacted subgrade.
• Load is distributed through the pavement layers by grain-to-grain contact; the pavement deflects (flexes) under load — hence the name.
• Primary failure modes of flexible pavement: rutting (permanent deformation), fatigue cracking (alligator/crocodile cracking), and longitudinal cracking.
• Rigid pavement is a Portland Cement Concrete (PCC) slab, typically 150–300 mm thick, placed on a prepared subbase.
• The stiff PCC slab distributes wheel loads over a wide area through bending (beam/plate action), dramatically reducing pressure on the subgrade.
• Primary failure modes of rigid pavement: corner cracking, transverse cracking, and joint faulting.
• Joints (transverse, longitudinal, expansion, contraction) are essential in rigid pavements to control thermally induced cracking.
• Dowel bars transfer load across transverse joints; tie bars hold longitudinal joints together.
• Rigid pavement is preferred where subgrades are weak but traffic is heavy and long design lives are needed (e.g., airport aprons, NLEX, SLEX).
• Flexible pavement is more easily staged (overlaid) but requires more frequent maintenance.

Definitions

Section Title

1. Pavement Types: Flexible vs. Rigid

Common Mistakes

• Confusing dowel bars (smooth, allow movement, transverse joints) with tie bars (deformed, no movement, longitudinal joints).
• Stating that flexible pavement is 'weak' — it is designed to carry the same design load as rigid pavement, just through a different structural mechanism.
• Forgetting that rigid pavement can also crack — joints are designed to control where cracks occur, not eliminate cracking entirely.
• Assuming rigid pavement never deflects — it does deflect slightly on the subgrade, which is why k (modulus of subgrade reaction) is still needed.

Formulas

Exam Tips

• In the formula k = p/δ, make sure p is in kPa (= kN/m²) and δ is in meters → result is in kN/m³. Converting: 56,000 kN/m³ = 56 MN/m³.
• A quick conversion rule: 1 pci (lb/in³) ≈ 271.3 kN/m³ — useful when US references are cited.
• Board problems typically give p and δ directly and ask for k. Less common but possible: given k and δ, find p — just rearrange.
• Know that subbase improves k (increases the design k value used in thickness charts) — this is a conceptual question.

Key Points

• The subgrade is the in-situ or compacted soil that supports the entire pavement structure; its strength governs required pavement thickness.
• CBR (California Bearing Ratio) is the standard subgrade strength index for flexible pavement design.
• CBR is expressed as a percentage: CBR = (test load / standard load) × 100%, where the standard is a high-quality crushed stone base.
• Typical CBR values: expansive clay 2–3%, sandy clay 5–8%, well-graded gravel 20–80%.
• Higher CBR → stronger subgrade → thinner pavement required.
• The modulus of subgrade reaction k is used in rigid pavement design (Westergaard theory).
• k is determined from a plate load test using a 762 mm (30-inch) diameter plate.
• k represents the pressure required to produce a unit deflection of the subgrade: k = p/δ.
• Typical k values: soft clay 13–27 MN/m³; medium soil 27–82 MN/m³; dense gravel 82–271 MN/m³.
• A subbase layer beneath a PCC slab increases the effective k value used in design.

Definitions

Section Title

2. Subgrade Strength Parameters

Common Mistakes

• Using CBR for rigid pavement design or k for flexible pavement design — they are not interchangeable without conversion.
• Forgetting to convert δ from mm to m when computing k — the most common unit error in board exams.
• Using the CBR at 2.5 mm penetration only — always compare 2.5 mm and 5.0 mm values; the higher governs (if the 5 mm value is higher, retest).
• Confusing k units: kN/m³ is the SI unit; some references use MN/m³ or psi/in — always check and convert.

Formulas

Exam Tips

• The formula A = P/p is simple but unit consistency is where marks are lost. Always state units at each step.
• A common board variation: given contact area and inflation pressure, find wheel load: P = A × p.
• Another variation: given wheel load and contact area, find inflation pressure: p = P/A.
• If p is given in psi and P in lbs, convert first: 1 psi = 6.895 kPa; 1 kN = 224.8 lbs.

Key Points

• The interface between a pneumatic tire and the pavement is called the tire contact area (also called the tire footprint).
• For design purposes, the contact area is assumed to be circular or rectangular with uniform pressure equal to the tire inflation pressure.
• The relationship: Contact area = Wheel load / Tire inflation pressure.
• Higher tire inflation pressure → smaller contact area → higher stress concentration on the surface layer.
• Typical truck tire inflation pressures: 0.55–0.80 MPa (80–115 psi).
• Typical equivalent circular contact radius a = √(A/π), used in mechanistic-empirical pavement analysis.
• Dual tires (tandem/tridem axles) distribute load differently; each tire has its own contact area.
• AASHTO considers the tire contact area implicitly in its equivalency factors.

Definitions

Section Title

3. Tire Contact Pressure and Contact Area

Common Mistakes

• Unit mismatch is the #1 error: if P is in N and p is in MPa (= N/mm²), then A is in mm². If P is in kN and p is in kPa, then A is in m² — always be consistent.
• Forgetting to convert kN to N when p is in MPa: 40 kN = 40,000 N, not 40.
• Assuming the contact area equals the tire footprint seen visually — it is a calculated uniform-pressure equivalent area.
• Neglecting to find equivalent radius a when the problem asks for stress analysis — do not use the raw area in stress formulas.

Formulas

Exam Tips

• The most common board question form: 'Find the LEF of a W-kN axle.' Apply LEF = (W/80)⁴ directly.
• Verify your answer by noting: W = 80 kN → LEF = 1.0 (by definition). If your answer gives LEF ≠ 1.0 for W = 80, recheck.
• For a quick check: (1.25)⁴ = 2.44; (1.50)⁴ = 5.06; (2.00)⁴ = 16.0 — memorize these benchmark values.
• If the problem gives load in tonnes: 1 metric tonne-force = 9.81 kN. So 10 tonnes ≈ 98.1 kN → LEF = (98.1/80)⁴ = (1.226)⁴ ≈ 2.26.

Key Points

• Highway pavements carry mixed traffic (cars, jeepneys, buses, trucks) with vastly different axle loads.
• For design, all axle loads are converted to an equivalent number of passes of a standard axle load.
• The AASHTO standard single axle load is 80 kN (18,000 lb or 18-kip), also called the Equivalent Single Axle Load (ESAL).
• The Load Equivalency Factor (LEF) quantifies the damage ratio: how many standard 80-kN axle passes one pass of a given axle equals.
• The approximate fourth-power law applies for single axles: LEF = (W/80)⁴, where W is in kN.
• AASHTO 1993 provides refined LEF tables by axle type (single, tandem, tridem) and pavement structure (structural number SN or slab thickness D).
• Design ESAL = Σ (daily traffic in each axle category × LEF × growth factor × design lane factor × design period in days).
• The fourth-power law dramatically amplifies the effect of overloaded axles — doubling the axle load increases damage by 2⁴ = 16 times.
• Overloading is the primary cause of premature pavement failure in Philippine roads.
• The Philippines enforces axle load limits under RA 8794 and DPWH guidelines; the standard single axle limit is 80 kN.

Definitions

Section Title

4. Traffic Loading — ESAL and Load Equivalency Factor

Common Mistakes

• Using the fourth-power law for tandem or tridem axles without correction — the law strictly applies to single axles; AASHTO tables give tandem/tridem LEFs.
• Forgetting to raise to the 4th power: students often compute (W/80)² instead of (W/80)⁴ — always double-check the exponent.
• Not converting axle load to kN before applying the formula — if given in tonnes, convert: 1 tonne-force ≈ 9.81 kN.
• Confusing wheel load with axle load: the standard is an 80-kN axle (two wheels, 40 kN per wheel) — do not use 40 kN as W in the LEF formula when the standard is stated as 80 kN axle.

Formulas

Exam Tips

• In a 5-choice board exam, eliminate answers with wrong units first — this alone can narrow to 2 choices.
• For multi-step problems, write the formula, substitute with units, then compute — partial credit in essay problems, and error-checking in MCQ.
• Memorize the three core formulas as a set: k = p/δ | A = P/p | LEF = (W/80)⁴ — one formula for each major topic.
• If the problem mixes tandem axles with single axles, treat them separately using AASHTO Table values for tandem LEF (approximate tandem LEF ≈ 0.773 × (Wtandem/80)³·²⁸ from AASHTO — but most board exams use the simple 4th-power approximation for all axle types unless otherwise stated).

Key Points

• Board exam problems on pavement design typically test one formula per problem — identify which formula is needed from the given data.
• Data clue: given p (pressure) and δ (deflection) → use k = p/δ (rigid pavement, subgrade modulus).
• Data clue: given P (wheel load) and p (tire pressure) → use A = P/p (contact area).
• Data clue: given axle load W → use LEF = (W/80)⁴ (traffic damage equivalency).
• Always state and check units at every step to avoid the most common board-exam errors.
• Conceptual questions (flexible vs. rigid, CBR vs. k, dowel vs. tie bar) are equally likely as numerical problems.
• Read the problem carefully: 'wheel load' vs. 'axle load' changes the standard in the LEF formula.
• Practice back-solving: given LEF, find W — requires the 4th root: W = 80 × LEF^(1/4).

Section Title

5. Integrated Problem-Solving Approach

Common Mistakes

• Solving for k but using p in kN/m² and δ in mm → gives k in kN/m²/mm = MN/m³ directly, but students often misidentify units as kN/m³ — always convert δ to meters for SI consistency.
• In combined problems, applying the LEF formula to wheel load instead of axle load when the standard is specified as an axle load.