BE Civil Engineering (IOE, TU) Transportation Engineering I (IOE, CE 652) Question Paper 2079 Nepal

a) Importance of highway planning

Highway planning is essential to:

• Develop a road network that gives maximum utility at minimum cost.
• Phase construction according to priority and available funds.
• Fix geometric and structural design standards for anticipated future traffic.
• Plan financing and the future development of an area in an integrated manner.

Road development plans:

• Nagpur Road Plan (1943–63): First 20-year plan in the subcontinent. Used the star-and-grid pattern and target road density of $16\ \text{km}$ per $100\ \text{km}^2$. Classified roads into National Highways, State Highways, Major District Roads, Other District Roads and Village Roads.
• 20-Year Road Development Plan (Lucknow Plan): Targeted a density of $32\ \text{km}$ per $100\ \text{km}^2$ and introduced the Expressway category.

In Nepal, the road network is similarly planned hierarchically (Strategic Road Network — National Highways and Feeder roads — and Local Road Network) to connect district headquarters and economic centres.

b) Computations

Total road length $L = 180 + 320 + 540 = 1040\ \text{km}$.

Road density by area:

$$D_A = \frac{L}{\text{Area}} \times 100 = \frac{1040}{1500}\times 100 = 69.33\ \text{km per }100\ \text{km}^2$$

Road density by population: Population $= 4.5\ \text{lakh} = 450{,}000$.

$$D_P = \frac{L}{\text{Population (in lakh)}} = \frac{1040}{4.5} = 231.1\ \text{km per lakh persons}$$

Density per 1000 persons: $\dfrac{1040}{450} = 2.31\ \text{km per 1000 persons}$.

Utility / saturation (vehicle utility per unit length):

$$\text{Vehicles per km of road} = \frac{12{,}000}{1040} = 11.5\ \text{veh/km}$$

Comment: Density by area ($69.33\ \text{km}/100\ \text{km}^2$) exceeds the Nagpur target ($16$) and even the Lucknow target ($32$), so geometric coverage is more than adequate for a developed region. With only $\approx 11.5$ vehicles per km the traffic loading is light, indicating spare capacity. The network is adequate in extent; emphasis should shift to quality/upgradation rather than new construction.

a)

• SSD is the minimum distance visible to a driver, equal to the distance required to stop a vehicle travelling at design speed safely without collision with any obstruction. It equals the lag (reaction) distance + braking distance.
• OSD is the minimum distance open to the vision of a driver of a vehicle intending to overtake a slower vehicle safely against traffic in the opposite direction.

Assumptions: constant design speed, constant friction, level/known gradient, single obstruction on the same lane (SSD); for OSD, the overtaken vehicle moves at uniform reduced speed, overtaking is done with uniform acceleration, and an opposing vehicle approaches at design speed.

b) SSD on descending grade

Convert speed: $v = \dfrac{80}{3.6} = 22.22\ \text{m/s}$.

Gradient $n = 3\% = 0.03$ (descending, so use $f - 0.03$).

$$SSD = vt + \frac{v^2}{2g(f - n)}$$ $$= 22.22 \times 2.5 + \frac{(22.22)^2}{2 \times 9.81 \times (0.35 - 0.03)}$$

Lag distance $= 55.56\ \text{m}$.

Braking distance $= \dfrac{493.8}{2 \times 9.81 \times 0.32} = \dfrac{493.8}{6.278} = 78.65\ \text{m}$.

$$SSD = 55.56 + 78.65 = \mathbf{134.2\ m}$$

c) OSD (two-way road)

$$OSD = d_1 + d_2 + d_3$$

Speeds: $V = 22.22\ \text{m/s}$, $V_b = \dfrac{64}{3.6} = 17.78\ \text{m/s}$.

$d_1 = V_b \cdot t = 17.78 \times 2 = 35.56\ \text{m}$.

Time for overtaking manoeuvre: $T = \sqrt{\dfrac{4s}{a}}$, where $s = 0.7 V_b + 6 = 0.7(17.78) + 6 = 18.45\ \text{m}$ (spacing).

$$T = \sqrt{\frac{4 \times 18.45}{0.92}} = \sqrt{80.2} = 8.96\ \text{s}$$

$d_2 = V_b T + 2s = 17.78 \times 8.96 + 2 \times 18.45 = 159.3 + 36.9 = 196.2\ \text{m}$.

$d_3 = V \cdot T = 22.22 \times 8.96 = 199.1\ \text{m}$.

$$OSD = 35.56 + 196.2 + 199.1 = \mathbf{430.9\ m}$$

a) Superelevation

Design speed $V = 70\ \text{km/h}$, $v = \dfrac{70}{3.6} = 19.44\ \text{m/s}$.

IRC method — superelevation for 75% of design speed, neglecting friction:

$$e = \frac{(0.75 V)^2}{127\,R} = \frac{V^2}{127 \times 1.78\,R}$$

Using the standard form $e = \dfrac{(0.75V)^2}{127R}$ with $V$ in km/h:

$$e = \frac{(0.75 \times 70)^2}{127 \times 250} = \frac{(52.5)^2}{31750} = \frac{2756.25}{31750} = 0.0868$$

This exceeds $e_{max} = 0.07$, so adopt $e = 0.07$.

Check friction mobilised at full design speed with $e = 0.07$:

$$e + f = \frac{V^2}{127R} = \frac{70^2}{127 \times 250} = \frac{4900}{31750} = 0.1543$$ $$f = 0.1543 - 0.07 = 0.0843$$

Since $f = 0.084 < 0.15$ (permissible), the design is safe with $e = 0.07$.

b) Transition curve length

(i) Rate of change of centrifugal acceleration:

$$C = \frac{80}{75 + V} = \frac{80}{75 + 70} = \frac{80}{145} = 0.552\ \text{m/s}^3$$ $$L_s = \frac{v^3}{C\,R} = \frac{(19.44)^3}{0.552 \times 250} = \frac{7349.7}{138.0} = 53.3\ \text{m}$$

(ii) Rate of introduction of superelevation (rotation about centre line): Raise of outer edge $E = e \times \dfrac{W}{2} = 0.07 \times \dfrac{7.0}{2} = 0.245\ \text{m}$.

With rate $1\text{ in }150$:

$$L_s = 150 \times E = 150 \times 0.245 = 36.75\ \text{m}$$

Adopt the larger value:

$$L_s = \mathbf{53.3\ m \;(\approx 55\ m)}$$

a) Length of summit curve for SSD

Deviation angle $N = g_1 - g_2 = 0.03 - (-0.02) = 0.05$.

$SSD = S = 90\ \text{m}$, $H = 1.2\ \text{m}$, $h = 0.15\ \text{m}$.

Case 1: Assume $L > S$:

$$L = \frac{N S^2}{\left(\sqrt{2H} + \sqrt{2h}\right)^2}$$

Compute denominator: $\sqrt{2 \times 1.2} = \sqrt{2.4} = 1.549$; $\sqrt{2 \times 0.15} = \sqrt{0.3} = 0.5477$. Sum $= 2.097$; squared $= 4.397$.

$$L = \frac{0.05 \times 90^2}{4.397} = \frac{0.05 \times 8100}{4.397} = \frac{405}{4.397} = 92.1\ \text{m}$$

Check: $L = 92.1 > S = 90$. Assumption valid.

Case 2 (for completeness, assume $L < S$):

$$L = 2S - \frac{\left(\sqrt{2H} + \sqrt{2h}\right)^2}{N} = 2(90) - \frac{4.397}{0.05} = 180 - 87.94 = 92.1\ \text{m}$$

This gives $L = 92.1 > S$, contradicting the assumption $L < S$, so Case 2 is rejected.

Adopt $L = \mathbf{92.1\ m}$ (say $95\ \text{m}$).

b) Why valley curves are governed by headlight distance and comfort

On a valley curve, during daytime the sight distance is unobstructed (the curve falls away and then rises — visibility is not restricted by the road surface), so daytime SSD is not critical. The governing criteria are:

1. Headlight sight distance (night): At night the visible length of road ahead is limited by the upward divergence of the vehicle headlight beam. The curve must be long enough that the headlight beam (height $0.75\ \text{m}$, $1^\circ$ upward spread) illuminates at least the SSD ahead.
2. Comfort: On a sag curve the centrifugal force acts downward, adding to gravity, causing discomfort/jerk. The allowable rate of change of centrifugal acceleration limits the curve length.

The larger of the two governs the design length.

a) PCU

A Passenger Car Unit (PCU) is a factor used to express the relative interference effect (in terms of road space and speed reduction) of different vehicle types in a traffic stream, expressed in equivalent passenger cars. It allows a mixed-traffic stream to be reduced to a single homogeneous unit so that capacity, volume and level of service can be consistently analysed.

b) Total flow in PCU/h

$$Q = 600 + 450 + 360 + 192 = \mathbf{1602\ PCU/h}$$

c) Peak rate of flow

The peak hour factor relates the hourly volume to the peak rate of flow within the hour:

$$PHF = \frac{\text{Hourly volume}}{\text{Peak rate of flow}}$$ $$\text{Peak rate of flow} = \frac{Q}{PHF} = \frac{1602}{0.85} = \mathbf{1884.7\ PCU/h}$$

Comment: $PHF = 0.85$ indicates moderately non-uniform flow within the hour (a PHF closer to $1.0$ would mean uniform flow). The design/analysis should use the higher peak rate of $\approx 1885\ \text{PCU/h}$ rather than the average $1602\ \text{PCU/h}$ to avoid under-design during the busiest 15-minute interval.

Road aggregates form the major structural component of a pavement and must satisfy several properties:

(Any four of the above, briefly explained, are acceptable.)

Example explanation (Crushing): A lower Aggregate Crushing Value means a stronger aggregate; for surface courses the ACV should generally not exceed $30\%$.

a) Penetration test & viscosity grading

• Penetration test: Measures the consistency/hardness of bitumen. A standard needle loaded with $100\ \text{g}$ is allowed to penetrate the bitumen sample for $5\ \text{s}$ at $25^\circ\text{C}$. The depth of penetration in tenths of a millimetre (0.1 mm units) is the penetration value. A higher value means softer bitumen.
• Viscosity grading (VG): Grades bitumen by its absolute viscosity (measured at $60^\circ\text{C}$ in poise) and kinematic viscosity at $135^\circ\text{C}$. It better reflects in-service temperatures than the older penetration grading.

b) Grade 60/70

Grade $60/70$ means the bitumen has a penetration value between 60 and 70 (i.e., $6.0$–$7.0\ \text{mm}$ of needle penetration) at the standard test conditions. It is a medium-hard grade.

Use in Nepal: Grade $60/70$ is the most widely used paving bitumen for bituminous surfacing of highways in the plains and mid-hills with moderate-to-warm climates. (Softer grades like $80/100$ are preferred in colder high-altitude regions.)

Bitumen vs Tar:

Total extra widening $W_e = W_m + W_{ps}$, the mechanical plus psychological widening.

Mechanical widening (for $n$ lanes):

$$W_m = \frac{n\,l^2}{2R} = \frac{2 \times (6.5)^2}{2 \times 220} = \frac{2 \times 42.25}{440} = \frac{84.5}{440} = 0.192\ \text{m}$$

Psychological widening:

$$W_{ps} = \frac{V}{9.5\sqrt{R}} = \frac{75}{9.5\sqrt{220}} = \frac{75}{9.5 \times 14.83} = \frac{75}{140.9} = 0.532\ \text{m}$$

Total extra widening:

$$W_e = 0.192 + 0.532 = \mathbf{0.724\ m \;(\approx 0.73\ m)}$$

Greenshields model: $v = v_f\left(1 - \dfrac{k}{k_j}\right)$, and $q = k v$.

Maximum flow occurs at:

$$k_m = \frac{k_j}{2} = \frac{200}{2} = 100\ \text{veh/km}$$ $$v_m = \frac{v_f}{2} = \frac{60}{2} = 30\ \text{km/h}$$

(a) At capacity: speed $= 30\ \text{km/h}$, density $= 100\ \text{veh/km}$.

(b) Maximum flow (capacity):

$$q_{max} = k_m \times v_m = 100 \times 30 = 3000\ \text{veh/h}$$

Equivalently $q_{max} = \dfrac{v_f\,k_j}{4} = \dfrac{60 \times 200}{4} = \mathbf{3000\ veh/h}$.

Stages of highway location surveys

1. Map study (desk study): Using topographic maps, possible alternative routes are studied to identify approximate alignments, avoiding obstacles (rivers, ridges, steep valleys) and noting controlling points.
2. Reconnaissance survey: Field inspection of the broad belt identified in the map study using simple instruments (Abney level, barometer). Collects data on terrain, soil, drainage, number of crossings, etc. Eliminates impractical routes.
3. Preliminary survey: Detailed instrument survey (total station/level) of the selected alternative(s) to obtain topographic, soil, traffic and drainage data, and to compare alternatives on a techno-economic basis to select the best alignment.
4. Final / detailed location survey: The selected alignment is pegged on the ground; detailed levelling, cross-sections, and data for drainage and structural design are collected for preparation of working drawings.

Requirements of an ideal alignment

• Short — as straight/direct as practicable.
• Easy — easy to construct and to traverse (easy gradients and curves).
• Safe — safe for traffic operation and stable against landslides/erosion (especially in hills).
• Economical — minimum overall cost (construction + maintenance + vehicle operating cost).

Controlling factors: obligatory points (to pass through or to avoid), traffic demand, geometric design standards, terrain/topography, soil and geology, drainage, and economics/environmental considerations.

a) Types of gradient

• Ruling (design) gradient: The maximum gradient adopted in the general design of the highway; the alignment is designed to keep gradients within this value as far as possible (e.g., $1\text{ in }30$, $3.3\%$ in plains).
• Limiting gradient: A steeper gradient used over short lengths where adoption of the ruling gradient would greatly increase cost (broken/rolling terrain), used in restricted stretches.
• Exceptional gradient: Still steeper, used only in very short stretches under unavoidable/exceptional situations, never to exceed about $100\ \text{m}$ at a stretch.
• Minimum gradient: A small gradient (e.g., $1\text{ in }500$ on the carriageway) provided to ensure efficient drainage of surface water along side drains; not needed for traffic but for drainage.

b) Stopping check on level road

$v = \dfrac{50}{3.6} = 13.89\ \text{m/s}$.

$$SSD = vt + \frac{v^2}{2gf} = 13.89 \times 2.5 + \frac{(13.89)^2}{2 \times 9.81 \times 0.36}$$

Lag distance $= 34.72\ \text{m}$.

Braking distance $= \dfrac{192.9}{7.063} = 27.32\ \text{m}$.

$$SSD = 34.72 + 27.32 = 62.0\ \text{m}$$

Comparison: Required $SSD = 62.0\ \text{m}$ but available sight distance $= 60\ \text{m}$.

Since $62.0\ \text{m} > 60\ \text{m}$, the available sight distance is NOT adequate (deficient by $\approx 2\ \text{m}$). The vehicle cannot safely stop; either the speed must be reduced, the sight distance increased, or a higher friction (better surface) provided.