BE Civil Engineering (IOE, TU) Transportation Engineering I (IOE, CE 652) Question Paper 2076 Nepal

Historical development

Tresaguet (France, ~1764): First scientific road. Cross-section of about 30 cm thickness with a sub-base of large foundation stones laid on edge, an intermediate layer of smaller broken stones, and a wearing surface of gravel. Emphasised subgrade drainage and cross-slope.

Telford (England, ~1820): Provided a flat subgrade and built a stone foundation (large flat stones hand-set), increasing thickness towards the centre to give a cambered surface (about 1 in 90). Heavy, expensive but durable; drainage by side ditches.

Macadam (England, ~1827): Recognised that the subgrade itself carries the load if kept dry. Used a compacted subgrade raised above ground with cross-slope, then layers of broken angular stone of small uniform size (< 25 mm) compacted by traffic — the first economical, well-graded construction (the basis of water-bound macadam).

Four stages of road development (Highway Development Plans)

1. Nagpur Road Plan (first 20-year plan) — classified roads as National Highways, State Highways, Major District Roads, Other District Roads and Village Roads; based on star-and-grid pattern.
2. Bombay Road Plan (second 20-year plan) — aimed to double road density; introduced expressways.
3. Lucknow Road Plan (third 20-year plan) — target road length tied to population and area.
4. Road Development Plan / Vision documents — modern stage emphasising network optimisation, expressways and rural connectivity. (In the Nepalese context the analogous stages are reflected in successive National Transport / Road Master Plans.)

Saturation-system computation (utility per unit length)

The saturation (maximum-utility) system ranks roads by utility obtained per unit length:

$$\text{Utility per km} = \frac{\text{Total utility units served}}{\text{Length of alignment}}$$

Alignment X:

$$\frac{64}{22} = 2.91\ \text{units/km}$$

Alignment Y:

$$\frac{80}{30} = 2.67\ \text{units/km}$$

Recommendation

Alignment X gives the higher utility per unit length (2.91 > 2.67 units/km), so under the saturation system Alignment X should be constructed first as it provides maximum service per kilometre of road built.

Definition

Stopping sight distance is the minimum sight distance available to a driver travelling at design speed to safely stop the vehicle without colliding with a stationary object on the road. It equals the lag (reaction) distance plus the braking distance.

Factors affecting SSD: design speed, reaction time of driver, friction (tyre–pavement) condition, gradient of the road, condition of brakes/vehicle, and presence of rain/wet surface.

Derivation on a gradient

Let $v$ = speed in m/s, $t$ = total reaction time, $f$ = longitudinal friction coefficient, $g$ = grade as a fraction (+ for up, − for down).

Lag distance $= v\,t$.

During braking, energy dissipated by friction plus work against gravity:

$$\tfrac{1}{2}m v^2 = m g_0 (f \pm n) d_b \quad\Rightarrow\quad d_b = \frac{v^2}{2\,g_0\,(f \pm n)}$$

where $g_0 = 9.81\ \text{m/s}^2$ and $n$ = grade fraction (+ up, − down).

$$\boxed{SSD = v\,t + \frac{v^2}{2\,g_0\,(f \pm n)}}$$

Numerical (V = 65 km/h, t = 2.5 s, f = 0.36, +3%)

$$v = \frac{65}{3.6} = 18.06\ \text{m/s}$$

Lag distance:

$$v t = 18.06 \times 2.5 = 45.14\ \text{m}$$

Braking distance (upgrade, use $f+n = 0.36 + 0.03 = 0.39$):

$$d_b = \frac{18.06^2}{2 \times 9.81 \times 0.39} = \frac{326.2}{7.652} = 42.63\ \text{m}$$

SSD:

$$SSD = 45.14 + 42.63 = 87.77\ \text{m}$$

SSD ≈ 87.8 m (upgrade).

Comment on a 3% downgrade

On a downgrade, gravity assists motion so $f - n = 0.36 - 0.03 = 0.33$, giving a larger braking distance:

$$d_b = \frac{326.2}{2\times9.81\times0.33} = 50.38\ \text{m},\quad SSD = 45.14 + 50.38 = 95.5\ \text{m}.$$

Thus the downgrade requires a longer SSD (≈ 95.5 m); for design of two-way roads the worst (downgrade) value governs.

Why summit curves are designed for sight distance

On a summit curve the road surface itself obstructs the driver's line of sight; centrifugal force acts upward, relieving pressure on springs, so passenger discomfort is not critical. The governing criterion is therefore adequate sight distance (SSD, and OSD where overtaking is allowed).

On a valley (sag) curve, in daylight there is no sight obstruction, but at night the headlight beam limits the visible distance; also centrifugal force acts downward, adding to gravity and causing discomfort/impact. Hence valley curves are designed for headlight sight distance and rider comfort.

Data

Deviation (algebraic difference) of grades:

$$N = +0.03 - (-0.02) = 0.05$$

SSD, $S = 120\ \text{m}$, $H = 1.2\ \text{m}$ (eye), $h = 0.15\ \text{m}$ (object).

Denominator constant:

$$\left(\sqrt{2H}+\sqrt{2h}\right)^2 = \left(\sqrt{2.4}+\sqrt{0.30}\right)^2 = (1.549+0.548)^2 = 4.40$$

Case 1: $L > S$

$$L = \frac{N S^2}{\left(\sqrt{2H}+\sqrt{2h}\right)^2} = \frac{0.05 \times 120^2}{4.40} = \frac{720}{4.40} = 163.6\ \text{m}$$

Check: $L = 163.6\ \text{m} > S = 120\ \text{m}$ — valid.

Case 2: $L < S$

$$L = 2S - \frac{\left(\sqrt{2H}+\sqrt{2h}\right)^2}{N} = 2(120) - \frac{4.40}{0.05} = 240 - 88 = 152.0\ \text{m}$$

Check: $L = 152.0\ \text{m} > S = 120\ \text{m}$ — invalid (contradicts assumption $L<S$).

Adopted length

Only Case 1 is consistent, so the design length of the summit curve = 163.6 m (≈ 164 m).

(a) Minimum radius

$$R_{min} = \frac{V^2}{127\,(e+f)} = \frac{80^2}{127\,(0.07+0.15)} = \frac{6400}{127 \times 0.22} = \frac{6400}{27.94} = 229.1\ \text{m}$$

$R_{min} \approx 229\ \text{m}$ (adopt $R = 229\ \text{m}$).

(b) Transition length by rate of change of radial acceleration

Allowable rate:

$$C = \frac{80}{75+V} = \frac{80}{75+80} = \frac{80}{155} = 0.516\ \text{m/s}^3 \quad (0.5 \le C \le 0.8\ \text{OK})$$ $$L_s = \frac{0.0215\,V^3}{C\,R} = \frac{0.0215 \times 80^3}{0.516 \times 229} = \frac{0.0215 \times 512000}{118.2} = \frac{11008}{118.2} = 93.1\ \text{m}$$

$L_s \approx 93.1\ \text{m}$.

(c) Transition length for rate of introduction of superelevation

For rotation about the centre line, the outer edge is raised by $e\,(W+W_e)/2$ relative to the centre. Total width $= W + W_e = 7.0 + 0.9 = 7.9\ \text{m}$. Rise of outer edge above centre line:

$$\Delta = \frac{e\,(W+W_e)}{2} = \frac{0.07 \times 7.9}{2} = 0.2765\ \text{m}$$

At a longitudinal slope of 1 in 150:

$$L_s = 150 \times \Delta = 150 \times 0.2765 = 41.5\ \text{m}$$

$L_s \approx 41.5\ \text{m}$.

Governing transition length

The largest of the computed values governs:

$$L_s = \max(93.1,\ 41.5) = 93.1\ \text{m}$$

Adopt transition length $L_s \approx 95\ \text{m}$ (rounded up for design).

Definitions

• Traffic volume (flow), $q$: the number of vehicles passing a section of road per unit time (veh/h).
• Traffic density (concentration), $k$: the number of vehicles present per unit length of road at an instant (veh/km).
• Space-mean speed, $v$: the average speed of vehicles obtained by averaging over the vehicles occupying a length of road (harmonic mean of spot speeds).

Fundamental relation: $\;q = k \times v.$

(a) Free-flow speed and jam density

Given $v = 70\left(1 - \dfrac{k}{200}\right)$.

• Free-flow speed $v_f$ occurs at $k = 0$: $\;v_f = 70\ \text{km/h}.$
• Jam density $k_j$ occurs at $v = 0$: $\;0 = 70(1 - k_j/200) \Rightarrow k_j = 200\ \text{veh/km}.$

(b) Flow in terms of density

$$q = k\,v = 70k\left(1 - \frac{k}{200}\right) = 70k - \frac{70k^2}{200} = 70k - 0.35k^2$$

(c) Maximum flow (capacity)

Maximum flow occurs where $\dfrac{dq}{dk}=0$:

$$\frac{dq}{dk} = 70 - 0.70k = 0 \Rightarrow k_o = 100\ \text{veh/km}$$

This is $k_j/2$, as expected for the Greenshields model. The corresponding speed:

$$v_o = 70\left(1 - \frac{100}{200}\right) = 35\ \text{km/h}\quad(= v_f/2)$$

Maximum (capacity) flow:

$$q_{max} = k_o\,v_o = 100 \times 35 = 3500\ \text{veh/h}$$

Capacity $q_{max} = 3500\ \text{veh/h}$, occurring at $k = 100\ \text{veh/km}$ and $v = 35\ \text{km/h}$.

Need for extra widening

Extra widening is provided on horizontal curves because:

1. A vehicle's rear wheels track inside the front wheels on a curve (off-tracking), so the vehicle occupies a wider path — mechanical widening.
2. Drivers tend to keep a greater clearance from the edge and from other vehicles at curves and steer erratically — psychological widening.

Mechanical widening

$$W_m = \frac{n\,l^2}{2R} = \frac{2 \times 6^2}{2 \times 80} = \frac{72}{160} = 0.45\ \text{m}$$

Psychological widening

$$W_{ps} = \frac{V}{2.64\sqrt{R}} = \frac{70}{2.64\sqrt{80}} = \frac{70}{2.64 \times 8.944} = \frac{70}{23.61} = 2.96\ \text{m}$$

Total extra widening

$$W_e = W_m + W_{ps} = 0.45 + 2.96 = 3.41\ \text{m}$$

Total extra widening $W_e \approx 3.41\ \text{m}$ (mechanical 0.45 m + psychological 2.96 m).

Components of OSD

For a two-way road the OSD has three parts:

  d1            d2                       d3
|----|---------------------------|---------------------|
 A1   <- overtaking manoeuvre ->        opposing vehicle

• $d_1$ — distance travelled by the overtaking vehicle during the reaction time at the speed of the slow vehicle.
• $d_2$ — distance travelled during the actual overtaking manoeuvre (slow-vehicle speed × manoeuvre time + 2× spacing).
• $d_3$ — distance travelled by an opposing vehicle (at design speed) during the manoeuvre time.

$$OSD = d_1 + d_2 + d_3$$

Data

$V = 70\ \text{km/h} \Rightarrow v = 19.44\ \text{m/s}$. Speed of overtaken vehicle $V_b = 70 - 16 = 54\ \text{km/h} \Rightarrow v_b = 15.0\ \text{m/s}$. $t = 2\ \text{s}$, $a = 0.92\ \text{m/s}^2$.

$d_1$ (reaction)

$$d_1 = v_b\,t = 15.0 \times 2 = 30.0\ \text{m}$$

Spacing and manoeuvre time

Minimum spacing: $s = 0.2 v_b + 6 = 0.2(15.0) + 6 = 9.0\ \text{m}$. Manoeuvre time:

$$T = \sqrt{\frac{4s}{a}} = \sqrt{\frac{4 \times 9.0}{0.92}} = \sqrt{39.13} = 6.26\ \text{s}$$

$d_2$ (overtaking manoeuvre)

$$d_2 = v_b\,T + 2s = 15.0 \times 6.26 + 2(9.0) = 93.9 + 18.0 = 111.9\ \text{m}$$

$d_3$ (opposing vehicle)

$$d_3 = v\,T = 19.44 \times 6.26 = 121.7\ \text{m}$$

Overtaking sight distance

$$OSD = 30.0 + 111.9 + 121.7 = 263.6\ \text{m}$$

OSD ≈ 263.6 m (≈ 264 m).

Desirable properties of aggregates and corresponding tests

Los Angeles abrasion test

A specified mass of graded aggregate is placed in a hollow steel drum together with a charge of standard steel balls (abrasive charge). The drum is rotated for a specified number of revolutions (typically 500–1000). The combined action of impact and grinding between the aggregate and the steel balls causes wear. After the test the sample is sieved on the 1.7 mm sieve; the material passing the sieve represents the wear.

Significance: It measures the resistance of aggregate to abrasion/wear under traffic. A low LA abrasion value indicates a hard, durable aggregate suitable for wearing courses.

Calculation

$$\text{Abrasion value} = \frac{\text{Mass passing 1.7 mm}}{\text{Original mass}} \times 100$$

Mass passing $= 5000 - 3850 = 1150\ \text{g}$.

$$\text{Abrasion value} = \frac{1150}{5000} \times 100 = 23\%$$

Comment

The abrasion value 23% < 30% (permitted maximum for surface/wearing course), so the aggregate is satisfactory for use in the surface course.

Role of bitumen

Bitumen is a black, viscous, thermoplastic hydrocarbon used as a binder in flexible pavements. It coats the aggregate, binds the mineral particles together, provides waterproofing, and imparts flexibility so that the surfacing can withstand traffic loads and temperature variations without cracking.

Three standard tests on bitumen

1. Penetration test — measures the depth (in 1/10 mm) to which a standard needle penetrates the bitumen in 5 s at 25 °C under a 100 g load. Indicates the consistency/hardness of bitumen; lower penetration = harder grade.
2. Softening point (Ring and Ball) test — the temperature at which the bitumen attains a specified softness (ball falls 25 mm). Indicates temperature susceptibility; a higher softening point is better for hot climates.
3. Ductility test — the distance (in cm) a standard briquette of bitumen stretches before breaking at 27 °C. Indicates the ability to deform without cracking (flexibility/adhesion). Minimum is usually 50 cm. (Other tests: viscosity, flash & fire point, specific gravity, loss on heating.)

Penetration grade vs viscosity grade

Passenger Car Unit (PCU)

A PCU is a factor that expresses the relative interference/space requirement of different vehicle types in terms of an equivalent number of passenger cars. It converts a mixed (heterogeneous) traffic stream into an equivalent homogeneous flow so that capacity, level of service and design volumes can be assessed on a common basis. The PCU value of a vehicle depends on its size, speed, manoeuvrability and the traffic/roadway conditions.

Computation

Convert the 15-minute counts to PCU:

Flow in 15 minutes $= 492\ \text{PCU}$.

Convert to hourly flow (multiply by 4):

$$q = 492 \times 4 = 1968\ \text{PCU/h}$$

Flow = 1968 PCU/h.

(Attempt any two; each note carries 5 marks.)

(a) Requirements of an ideal highway alignment

An ideal alignment should be:

• Short — a straight alignment between terminals gives the least length and cost (subject to other constraints).
• Easy — easy to construct, maintain and for vehicles to operate; gentle gradients and curves.
• Safe — safe geometric elements (adequate sight distance, safe curves and gradients) and a stable alignment from the embankment/cut-slope point of view.
• Economical — least overall cost considering construction, maintenance and vehicle operation.

Factors controlling alignment: obligatory points (through which the road must/must not pass), traffic demand, geometric design requirements, economics, nature of terrain/gradient, drainage and water-table, geology and soil, political/administrative considerations, and environmental factors.

(b) Elements of a highway cross-section

 |<----------- Right of Way ----------->|
 |   side    shoulder  carriageway  shoulder   side  |
 |   drain  |        |____________|        |  drain  |
 |   \__/   |_______/   camber     \_______|   \__/  |
              (slopes from crown to edges)

• Carriageway: the paved width used by vehicles; width depends on number and width of lanes.
• Shoulder: the strips on either side of the carriageway for emergency stopping and lateral support; usually surfaced more roughly than the carriageway.
• Camber (cross-slope): the transverse slope (e.g. 2–2.5%) given to the surface to drain rain water to the sides.
• Right of way: the total land width acquired for the road including future widening, utilities and drains.
• Kerb: the boundary between the carriageway and the shoulder/footpath (low, mountable or barrier type).
• Side drains: longitudinal drains/ditches alongside the road to collect and carry away surface and subsurface water.

(c) Engineering surveys for highway location

1. Map study: topographic maps are studied to identify possible alignment corridors, obligatory points, ridge/valley lines and approximate gradients.
2. Reconnaissance survey: a field inspection of the corridors identified from the map study; rough data on terrain, soil, drainage, obstacles and existing roads are collected to drop unfeasible alternatives.
3. Preliminary survey: detailed instrument survey of the selected alternatives — topographic, soil, drainage and traffic data are collected, alternative alignments are compared technically and economically, and the best is selected.
4. Final location and detailed survey: the chosen centre line is pegged on the ground (centre-line/location survey) and detailed levelling, cross-sections, drainage and soil data are collected for preparation of working drawings and estimates.