5.2 Traffic Engineering, Flow Fundamentals, and Pavement Design

5.2 Traffic Engineering, Flow Fundamentals, and Pavement Design

Traffic stream characteristics and pavement structural design form the analytical core of transportation engineering in the CELE syllabus. Traffic engineering evaluates the macroscopic interactions of vehicles traversing roadway facilities, establishing highway capacity and Level of Service (LOS). Pavement engineering designs structural layers capable of withstanding millions of repetitive heavy vehicular wheel loads over multi-decade design lives without excessive fatigue cracking or rutting.

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Traffic Stream Variables & Fundamental Relations

Macroscopic traffic flow theory describes vehicular movement using three continuous stream variables:

1. Flow Rate ($q$): The number of vehicles passing a designated cross-section of a roadway per unit time, conventionally expressed in vehicles per hour (veh/h) or passenger cars per hour (pc/h): $q = \frac{N}{T} = \frac{3600}{\bar{h}_t}$ Where $\bar{h}_t$ is the average time headway (seconds between successive front vehicle bumpers).
2. Density ($k$): The number of vehicles occupying a unit length of roadway lane at a given instant, conventionally expressed in vehicles per kilometer (veh/km): $k = \frac{N}{L} = \frac{1000}{\bar{s}}$ Where $\bar{s}$ is the average space headway or vehicle spacing (meters between front bumpers).
3. Speed ($u$): The rate of vehicle movement, expressed in kilometers per hour (km/h).

The Fundamental Equation of Traffic Flow

Through dimensional analysis and continuity of mass:

$q = u \cdot k$ $\left[ \frac{\text{vehicles}}{\text{hour}} \right] = \left[ \frac{\text{kilometers}}{\text{hour}} \right] \times \left[ \frac{\text{vehicles}}{\text{kilometer}} \right]$

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Time-Mean Speed ($u_t$) vs Space-Mean Speed ($u_s$)

A major conceptual topic in traffic engineering is the distinction between point-based and space-based speed measurements:

1. Time-Mean Speed ($u_t$)

Time-mean speed is the arithmetic average of instantaneous vehicle spot speeds recorded by an observer or detector (e.g., radar gun or inductive loop) at a single fixed roadway cross-section over time interval $T$:

$u_t = \frac{1}{n} \sum_{i=1}^{n} v_i$

2. Space-Mean Speed ($u_s$)

Space-mean speed is the average speed of all vehicles occupying a designated roadway length $L$ at a specific instant. Operationally, it is calculated as the total vehicle-distance divided by total vehicle-travel time, which is mathematically identical to the harmonic mean of spot speeds:

$u_s = \frac{n L}{\sum_{i=1}^{n} t_i} = \frac{n}{\sum_{i=1}^{n} \frac{1}{v_i}}$

Wardrop's Relationship

J.G. Wardrop established the fundamental mathematical identity linking time-mean and space-mean speed:

$u_t = u_s + \frac{\sigma_s^2}{u_s}$

Where $\sigma_s^2$ is the variance of vehicle speeds about the space-mean speed:

$\sigma_s^2 = \frac{1}{n} \sum_{i=1}^{n} (v_i - u_s)^2$

Because the variance $\sigma_s^2 \ge 0$, it is a mathematical axiom that $u_t \ge u_s$ always. The two speed values are equal if and only if all vehicles travel at the identical speed ($\sigma_s^2 = 0$). A fixed observer sees fast vehicles more often per hour than slow ones, while a snapshot of a road section captures slow vehicles longer. This is why the time-mean speed weights fast vehicles more heavily than the space-mean speed does.

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Greenshields Macroscopic Stream Model

Introduced by Bruce Greenshields in 1935, the Greenshields model hypothesizes a linear relationship between traffic speed $u$ and traffic density $k$:

$u = u_f \left( 1 - \frac{k}{k_j} \right)$

Where:

• $u_f$ = free-flow speed (km/h) as density approaches zero ($k \to 0$)
• $k_j$ = jam density (veh/km) at complete standstill when speed drops to zero ($u = 0$)

Derivation of Flow-Density Parabola

Substituting Greenshields speed into the fundamental flow equation $q = u \cdot k$:

$q = k \left[ u_f \left( 1 - \frac{k}{k_j} \right) \right] = u_f k - \frac{u_f}{k_j} k^2$

This is the equation of a downward-opening parabola passing through the origin $(0,0)$ and jam density $(k_j, 0)$. To find the density that maximizes flow, differentiate $q$ with respect to $k$ and set the derivative to zero:

$\frac{dq}{dk} = u_f - \frac{2 u_f}{k_j} k = 0 \implies k_c = \frac{k_j}{2}$

Where $k_c$ is the critical density. Substituting $k_c$ into the speed-density equation yields the critical speed ($u_c$):

$u_c = u_f \left( 1 - \frac{k_j / 2}{k_j} \right) = \frac{u_f}{2}$

Maximum Roadway Capacity ($q_{\max}$)

The maximum flow rate (roadway capacity) is the product of critical speed and critical density:

$q_{\max} = u_c \cdot k_c = \left( \frac{u_f}{2} \right) \left( \frac{k_j}{2} \right) = \frac{u_f k_j}{4}$

Speed-Flow Relationship

Expressing density in terms of speed $k = k_j (1 - u / u_f)$ and substituting into $q = u \cdot k$ yields the speed-flow parabolic equation:

$q = k_j \left( u - \frac{u^2}{u_f} \right)$

For any flow rate below capacity ($q < q_{\max}$), there exist two operational regimes:

1. Uncongested / Free-Flow Regime: High speed ($u > u_c$) and low density ($k < k_c$). Traffic is stable, and drivers maintain desired travel speeds.
2. Congested / Forced-Flow Regime: Low speed ($u < u_c$) and high density ($k > k_c$). Flow is constrained by downstream bottlenecks, producing stop-and-go waves.

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Highway Capacity & Level of Service (LOS)

The Highway Capacity Manual (HCM) defines Level of Service (LOS) as a qualitative measure describing operational driving conditions within a traffic stream, graded from LOS A (best) to LOS F (worst):

Each facility type has its own LOS measure. For basic freeway segments, the HCM uses density in passenger cars per mile per lane (pc/mi/ln). The thresholds below follow the 6th edition:

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Pavement Engineering: Flexible vs Rigid Pavements

Highway pavements are broadly classified into two categories based on how they distribute structural loads into the subgrade:

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Flexible Pavement Design: AASHTO Structural Number (SN)

Equivalent Single Axle Load (ESAL)

To evaluate mixed traffic composed of passenger cars, buses, and multi-axle heavy haul trucks, traffic loads are converted into an equivalent standard reference axle. The standard reference is the 18-kip (18,000 lbs / 80 kN / 8.2 metric tonnes) Equivalent Single Axle Load (ESAL).

Under the AASHO Road Test fourth-power damage law, the Load Equivalency Factor (LEF) for an axle load $L_x$ is approximately:

$\text{LEF} \approx \left( \frac{L_x}{18\text{ kips}} \right)^4 = \left( \frac{L_x}{80\text{ kN}} \right)^4$

AASHTO Structural Number Equation

The Structural Number (SN) is an empirical index representing the overall structural capacity required of a flexible pavement to support cumulative design ESALs:

$\text{SN} = a_1 D_1 + a_2 D_2 m_2 + a_3 D_3 m_3$

Where:

• $a_1, a_2, a_3$ = structural layer coefficients for the surface course, base course, and subbase course
  • Typical values: $a_1 \approx 0.40\text{ to }0.44$ (dense-graded asphalt concrete), $a_2 \approx 0.12\text{ to }0.14$ (crushed stone base), $a_3 \approx 0.10\text{ to }0.12$ (granular subbase)
• $D_1, D_2, D_3$ = thickness of respective layers in inches
• $m_2, m_3$ = drainage modification coefficients (dimensionless, typically $0.80\text{ to }1.15$ based on rainfall and drainage quality)

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Rigid Pavement Joints & Load Transfer Detailing

Portland cement concrete slabs expand and contract with temperature and moisture cycles. Without engineered joints, unrestrained shrinkage induces irregular, destructive thermal curling and cracking.

1. Transverse Contraction Joints

• Function: Relieve tensile stresses induced by drying shrinkage and thermal contraction.
• Spacing: Commonly about $4.5\text{ to }5.0\text{ m}$ for plain jointed slabs. A frequently used rule of thumb limits spacing to roughly 21 to 24 times the slab thickness.
• Joint Sawing: Cut to a depth of $D/4$ to $D/3$ (where $D$ is slab thickness) within 6 to 12 hours of concrete placement.

2. Dowel Bars

• Mechanics: Smooth, round solid steel bars installed across transverse joints at slab mid-depth.
• Load Transfer: Provide vertical shear transfer across adjacent slabs, preventing joint faulting and pumping, while their lubricated surface permits free longitudinal expansion and contraction.
• Sizing: Typically $28\text{ to }36\text{ mm}$ diameter, $450\text{ mm}$ length, spaced at $300\text{ mm}$ on center.

3. Longitudinal Joints & Tie Bars

• Mechanics: Deformed (corrugated) steel rebars installed across longitudinal construction joints between adjacent lanes.
• Function: Tie bars prevent lane separation and lateral slab migration while allowing joint hinge rotation. Unlike dowels, tie bars are bonded and do not permit sliding.

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Step-by-Step Worked Problem Examples

Worked Example 1: Greenshields Model Parameter Analysis

Problem: A multi-lane expressway has an estimated free-flow speed $u_f = 96\text{ km/h}$ and a jam density $k_j = 110\text{ veh/km/lane}$. Assuming Greenshields linear stream behavior:

1. Determine the maximum lane capacity ($q_{\max}$), critical speed ($u_c$), and critical density ($k_c$).
2. Calculate the speed and density when the flow rate is operating at $2,100\text{ veh/h/lane}$ in the uncongested regime.
3. Calculate the flow rate when density is observed at $k = 75\text{ veh/km/lane}$.

Solution:

1. Capacity and Critical Parameters: $u_c = \frac{u_f}{2} = \frac{96}{2} = 48.0\text{ km/h}$ $k_c = \frac{k_j}{2} = \frac{110}{2} = 55.0\text{ veh/km/lane}$ $q_{\max} = u_c \cdot k_c = 48.0 \times 55.0 = 2,640\text{ veh/h/lane}$
2. Operating Parameters at $q = 2,100\text{ veh/h}$: Using the parabolic speed-flow relationship: $q = k_j \left( u - \frac{u^2}{u_f} \right) \implies 2100 = 110 \left( u - \frac{u^2}{96} \right)$ $19.0909 = u - \frac{u^2}{96} \implies \frac{u^2}{96} - u + 19.0909 = 0$ $u^2 - 96 u + 1832.73 = 0$ Solving via quadratic formula: $u = \frac{96 \pm \sqrt{(-96)^2 - 4(1)(1832.73)}}{2} = \frac{96 \pm \sqrt{9216 - 7330.91}}{2} = \frac{96 \pm \sqrt{1885.09}}{2} = \frac{96 \pm 43.42}{2}$
   • In the uncongested regime ($u > u_c = 48\text{ km/h}$): $u = \frac{96 + 43.42}{2} = 69.71\text{ km/h}$
   • Corresponding density: $k = \frac{q}{u} = \frac{2100}{69.71} = 30.12\text{ veh/km/lane}$
3. Flow at Density $k = 75\text{ veh/km}$: Speed at $k = 75\text{ veh/km}$: $u = u_f \left( 1 - \frac{k}{k_j} \right) = 96 \left( 1 - \frac{75}{110} \right) = 96 \left( \frac{35}{110} \right) = 30.55\text{ km/h}$ $q = u \cdot k = 30.55 \times 75 = 2,291\text{ veh/h/lane}$ (Note: Since $k = 75 > k_c = 55$, this represents congested forced flow).

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Worked Example 2: Flexible Pavement Layer Thickness Design

Problem: A national highway requires an AASHTO Structural Number $\text{SN} = 4.30$ to accommodate cumulative 20-year design traffic loads. The pavement structure consists of:

• Surface Course: Dense-graded asphalt concrete, $a_1 = 0.42$, thickness $D_1 = 4.0\text{ inches}$.
• Base Course: Crushed aggregate base, $a_2 = 0.14$, drainage coefficient $m_2 = 1.0$, thickness $D_2 = 8.0\text{ inches}$.
• Subbase Course: Granular subbase, $a_3 = 0.11$, drainage coefficient $m_3 = 0.85$.

Calculate the required thickness of the granular subbase layer ($D_3$) rounded to the nearest half-inch.

Solution:

1. AASHTO Structural Number Equation: $\text{SN} = a_1 D_1 + a_2 D_2 m_2 + a_3 D_3 m_3$
2. Evaluate Contributions of Surface and Base Layers: $\text{SN}_1 = a_1 D_1 = 0.42 \times 4.0 = 1.68$ $\text{SN}_2 = a_2 D_2 m_2 = 0.14 \times 8.0 \times 1.0 = 1.12$ $\text{SN}_{1+2} = 1.68 + 1.12 = 2.80$
3. Compute Required Subbase Contribution: $\Delta \text{SN}_3 = \text{SN}_{\text{total}} - \text{SN}_{1+2} = 4.30 - 2.80 = 1.50$
4. Solve for Subbase Thickness $D_3$: $a_3 D_3 m_3 = 1.50 \implies D_3 = \frac{1.50}{a_3 m_3} = \frac{1.50}{0.11 \times 0.85} = \frac{1.50}{0.0935} = 16.04\text{ inches}$ Rounding to the nearest standard construction half-inch yields $D_3 = 16.0\text{ inches}$ (or $400\text{ mm}$).

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