Transportation Learning HubTransportation Learning Hub · PE Civil Transportation · Area 9 of 10
Geotechnical and Pavement — PE Civil Transportation
The ground is part of the road. This area connects soil testing and properties to compaction, earthwork balance, traffic loading, and the design and rehabilitation of flexible and rigid pavements.
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Start with the soil: testing and evaluation
Pavement and earthwork questions begin with a material description. Classification organises a soil by particle size and plasticity; strength and stiffness tests then describe how that material behaves under load. Keep the purpose of each measure separate:
| Classification | What the soil is — grading and plasticity, used to group materials with similar engineering behaviour |
|---|---|
| CBR | California Bearing Ratio — an empirical penetration-strength comparison used as a subgrade support input |
| R-value | A stabilometer resistance value used as another empirical subgrade support input |
| Resilient modulus, \(M_r\) | Recoverable stiffness under repeated loading — the subgrade stiffness used in pavement design |
| Field tests | In-place checks of the material actually built, such as density and moisture condition after compaction |
CBR, R-value and \(M_r\) are related in purpose but are not interchangeable numbers. Use the measure the question gives, and convert only with a relationship the question or reference supplies. For fine-grained subgrade soils, a commonly used estimate is:
\[M_r \approx 1{,}500 \times CBR \quad \text{(psi, for } CBR \le 10\text{)}\]
Worked example Resilient modulus from CBR
Given: A fine-grained subgrade has \(CBR = 5\). Estimate its resilient modulus using \(M_r \approx 1{,}500 \times CBR\).
Solution:
- Check the stated range first: \(CBR = 5 \le 10\), so the estimate applies.
- Substitute: \(M_r \approx 1{,}500 \times 5 = 7{,}500\) psi.
Answer: \(M_r \approx 7{,}500\) psi.
Stabilization changes the material itself — for example by mixing in a binder or replacing unsuitable soil — while slope stability questions ask whether a mass of soil stays in place under gravity and water pressures. In both cases, identify the mechanism first: a question about a failing slope is not answered by a pavement-thickness calculation.
Soil properties: the behaviour behind the numbers
Four property families carry most of the conceptual questions in this area:
| Phase relationships | How solids, water and air share the total volume and mass — the bookkeeping behind density, moisture content, saturation and void space |
|---|---|
| Strength | Resistance to shearing — the property that governs bearing, slope and lateral-pressure behaviour |
| Permeability | How readily water flows through the soil — it drives drainage, seepage and pore-pressure questions |
| Compressibility | How much the soil volume changes under load over time — the property behind settlement of embankments and fills |
A reliable exam habit: name the property before reaching for a formula. Strength, flow and settlement are three different behaviours, and the specification lists them separately because the exam tests them separately.
Compaction, excavation and mass balance
Earthwork is tracked in three volume states, and confusing them is the classic mass-balance error. Bank volume is material in place before excavation; loose volume is the same material after digging, when it bulks up; compacted volume is what remains after placement and compaction in an embankment. A shrink factor given in a problem converts bank volume to compacted volume:
\[\text{Compacted fill} = \text{bank cut} \times \text{shrink factor}\]
| Bank cut | volume measured in place before excavation, CY |
|---|---|
| Shrink factor | compacted volume per unit of bank volume — a factor below 1.0 means the fill occupies less volume than the cut it came from |
| Compaction | densifying placed soil so the embankment reaches the required strength and settlement behaviour; verified by field density and moisture checks |
A mass balance compares what the cuts supply with what the fills demand in the same volume state. Convert first, compare second — never compare bank cut directly with compacted fill.
Worked example Mass balance with shrinkage
Given: A grading operation excavates 12,000 CY of bank material. The shrink factor is given as 0.90. How much compacted fill can it supply?
Solution:
- Apply the factor to the bank volume: \(12{,}000 \times 0.90 = 10{,}800\) CY.
- Interpretation: the cut supplies 10,800 CY of compacted fill — any fill demand above that needs borrow; any surplus goes to waste.
Answer: \(\approx 10{,}800\) CY of compacted fill.
Traffic, pavement design and rehabilitation
Pavement design characterises traffic first — the loads the structure must carry over its life — and then gives the structure enough capacity for them. The two pavement families solve the problem differently:
| Flexible | Asphalt layers spread wheel loads progressively downward, each layer weaker and cheaper than the one above; the subgrade must never be overstressed. Capacity is summarised by the structural number \(SN\) |
|---|---|
| Rigid | A concrete slab carries load by bending and spreads it over a wide area of subgrade; characterised by slab thickness and concrete strength rather than \(SN\) |
| Rehabilitation | Restoring a deteriorated pavement — the choice between surface treatment, overlay and reconstruction turns on whether the failure is functional (surface) or structural (in the layers or subgrade) |
For flexible design, each layer contributes \(a_i D_i\), with the drainage coefficient \(m_i\) applied to unbound base and subbase layers only — not to the asphalt surface.
\[SN = a_{1}D_{1} + a_{2}D_{2}m_{2} + a_{3}D_{3}m_{3}\]
Worked example Structural number of a three-layer flexible pavement
Given: Layer coefficients and thicknesses are given as: surface \(a_1 = 0.44\), \(D_1 = 4\) in; base \(a_2 = 0.14\), \(D_2 = 6\) in, \(m_2 = 1.0\); subbase \(a_3 = 0.11\), \(D_3 = 8\) in, \(m_3 = 0.9\). Find \(SN\).
Solution:
- Surface: \(0.44 \times 4 = 1.76\).
- Base: \(0.14 \times 6 \times 1.0 = 0.84\).
- Subbase: \(0.11 \times 8 \times 0.9 = 0.792\).
- Total: \(SN = 1.76 + 0.84 + 0.792 = 3.392\).
Answer: \(SN \approx 3.39\).