BE Civil Engineering (IOE, TU) Transportation Engineering I (IOE, CE 652) Question Paper 2077 Nepal

Stages of Highway Planning Surveys

Highway planning surveys are carried out in four broad stages, each refining the route choice:

1. Economic studies — collect data on population (size, distribution, trends), agricultural and industrial products, existing road/transport facilities, per-capita income and land use. These establish the need and the traffic-generating potential of an area.
2. Financial studies — sources of revenue (taxes, tolls, fuel levy), funds available, cost of construction and maintenance, returns expected. They check the affordability and viability.
3. Traffic (road-use) studies — traffic volume, origin-and-destination (O-D), traffic flow patterns, accident records, growth-factor estimation, projected design traffic. They size the facility.
4. Engineering studies — topographic surveys, soil and material surveys, drainage, geological investigation, and special problems (slides, high embankments). They fix the alignment and design.

Role in route selection: the surveys progressively narrow choices from a map study (desk study of contoured maps) → reconnaissance (broad feasible corridors) → preliminary survey (comparison of alternatives on technical/economic merit) → final location & detailed survey (the chosen alignment is pegged on ground).

Saturation System (Maximum Utility) Method

The optimum road system is the one giving the maximum utility per unit length. Utility is computed from served population and productivity, weighted by assigned utility units.

Given utility weights: 0.50 unit per 1000 t of product; 0.25 unit per 1000 persons.

Alternative I

• Product utility $= 0.50 \times \dfrac{120{,}000}{1000} = 0.50 \times 120 = 60$ units
• Population utility $= 0.25 \times \dfrac{80{,}000}{1000} = 0.25 \times 80 = 20$ units
• Total utility $= 60 + 20 = 80$ units
• Utility per km $= \dfrac{80}{50} = 1.60$ units/km

Alternative II

• Product utility $= 0.50 \times \dfrac{100{,}000}{1000} = 0.50 \times 100 = 50$ units
• Population utility $= 0.25 \times \dfrac{110{,}000}{1000} = 0.25 \times 110 = 27.5$ units
• Total utility $= 50 + 27.5 = 77.5$ units
• Utility per km $= \dfrac{77.5}{40} = 1.9375 \approx 1.94$ units/km

Recommendation

Although Alternative I has a higher total utility (80 vs 77.5 units), the saturation system ranks by utility per unit length. Since 1.94 > 1.60, Alternative II is recommended as it delivers greater service per kilometre of road built.

Definitions

Stopping Sight Distance (SSD): the minimum distance visible to a driver so that the vehicle can be brought to a stop before hitting a stationary object on the road. It equals the lag (reaction) distance plus the braking distance.

Overtaking Sight Distance (OSD): the minimum distance open to the view of a driver on a two-way road that enables a vehicle to safely overtake a slower vehicle ahead without endangering an oncoming vehicle.

Derivation of SSD on a level road

Let $v$ = speed (m/s), $t$ = reaction time (s), $f$ = coefficient of friction, $g$ = 9.81 m/s².

• Lag distance $d_1 = v\,t$ (distance during the perception–reaction time).
• Braking distance $d_2$: by the work–energy principle, kinetic energy is dissipated by friction:

$$\tfrac{1}{2}\,\frac{W}{g}\,v^2 = f\,W\,d_2 \;\Rightarrow\; d_2 = \frac{v^2}{2 g f}$$

• Therefore $\boxed{SSD = v t + \dfrac{v^2}{2 g f}}$ on level ground. On a gradient of $\pm n\%$ ($+$ up, $-$ down) the friction term becomes $\dfrac{v^2}{2 g (f \pm 0.01 n)}$.

(a) SSD at 65 km/h on a 3% downgrade

Convert speed: $v = \dfrac{65}{3.6} = 18.056\ \text{m/s}$.

Downgrade reduces effective resistance, so use $(f - 0.01 n) = 0.36 - 0.03 = 0.33$.

• Lag distance $d_1 = v t = 18.056 \times 2.5 = 45.14\ \text{m}$
• Braking distance $d_2 = \dfrac{v^2}{2 g (f-0.01n)} = \dfrac{18.056^2}{2 \times 9.81 \times 0.33} = \dfrac{326.0}{6.4746} = 50.35\ \text{m}$
• $SSD = 45.14 + 50.35 = 95.49\ \text{m}$

SSD ≈ 95.5 m (round up to 96 m for design).

(b) OSD at 70 km/h

Convert speeds:

• Overtaking vehicle $v = \dfrac{70}{3.6} = 19.444\ \text{m/s}$
• Overtaken vehicle $v_b = \dfrac{42}{3.6} = 11.667\ \text{m/s}$

OSD $= d_1 + d_2 + d_3$.

$d_1$ (reaction phase): $d_1 = v_b\, t = 11.667 \times 2 = 23.33\ \text{m}$

$d_2$ (overtaking manoeuvre): $d_2 = v_b\,T + 2 s$, where $s = 0.7 v_b + 6 = 0.7(11.667) + 6 = 14.17\ \text{m}$ is the spacing, and the overtaking time is

$$T = \sqrt{\frac{4 s}{a}} = \sqrt{\frac{4 \times 14.17}{0.92}} = \sqrt{61.61} = 7.85\ \text{s}$$

$d_2 = 11.667 \times 7.85 + 2 \times 14.17 = 91.58 + 28.34 = 119.92\ \text{m}$

$d_3$ (opposing vehicle): assumed to travel at design speed $v$ during time $T$: $d_3 = v\,T = 19.444 \times 7.85 = 152.64\ \text{m}$

$$OSD = 23.33 + 119.92 + 152.64 = 295.9\ \text{m}$$

OSD ≈ 296 m.

(a) Superelevation

$v = \dfrac{80}{3.6} = 22.222\ \text{m/s}$, $R = 220$ m, $g = 9.81$ m/s².

(i) Superelevation for 75% of design speed, friction ignored

Design per IRC: provide $e$ for 75% of design speed neglecting $f$.

$$e = \frac{(0.75 v)^2}{g R} = \frac{V^2}{225\,g R}\Big|_{V\text{ in m/s? use km/h form}}$$

Using the standard IRC form with $V$ in km/h: $e = \dfrac{V^2}{225 R} = \dfrac{80^2}{225 \times 220} = \dfrac{6400}{49500} = 0.1293$.

This 0.129 (12.9%) exceeds the IRC limit of 0.07, so cap the superelevation at $e_{max}=0.07$.

(ii) Equilibrium check with $e=0.07$, find friction developed at full speed

With $e + f = \dfrac{v^2}{gR} = \dfrac{22.222^2}{9.81 \times 220} = \dfrac{493.8}{2158.2} = 0.2288$.

If $e = 0.07$, required $f = 0.2288 - 0.07 = 0.159$. This is slightly above the allowable $f = 0.15$, so the safe (restricted) speed with $e=0.07$, $f=0.15$ is

$$v_s = \sqrt{gR(e+f)} = \sqrt{9.81 \times 220 \times 0.22} = \sqrt{474.8} = 21.79\ \text{m/s} = 78.4\ \text{km/h}.$$

Adopt $e = 0.07$ (7%); restrict/sign the curve to ≈ 78 km/h, marginally below 80 km/h.

(b) Length of transition curve

(i) Rate of change of centrifugal acceleration

$C = \dfrac{80}{75 + V} = \dfrac{80}{75 + 80} = \dfrac{80}{155} = 0.516\ \text{m/s}^3$ (within IRC bounds 0.5–0.8).

$$L_s = \frac{v^3}{C R} = \frac{22.222^3}{0.516 \times 220} = \frac{10973.9}{113.5} = 96.7\ \text{m}$$

(ii) Rate of introduction of superelevation (rotation about inner edge)

Raise of outer edge $= e \times (\text{full width}) = 0.07 \times 7.0 = 0.49$ m (about inner edge, total width is raised).

$$L_s = N \times e \times W = 150 \times 0.49 = 73.5\ \text{m}$$

(iii) Empirical (IRC, plain/rolling) check

$L_s = \dfrac{2.7 V^2}{R} = \dfrac{2.7 \times 80^2}{220} = \dfrac{17280}{220} = 78.5\ \text{m}$.

Adopt

The largest governs: $L_s = 96.7\ \text{m}$. Adopt $L_s = 100$ m (rounded to a convenient design value).

Set-up

Deviation angle (algebraic difference of grades):

$$N = g_1 - g_2 = (+0.035) - (-0.025) = 0.060$$

Design sight distances: SSD $S = 120$ m, OSD $S = 470$ m.

For a summit curve the standard length formulas are:

• When $L > S$: $\;L = \dfrac{N S^2}{(\sqrt{2 H} + \sqrt{2 h})^2}$
• When $L < S$: $\;L = 2S - \dfrac{(\sqrt{2 H} + \sqrt{2 h})^2}{N}$

where $H$ = driver eye height, $h$ = object/oncoming-vehicle height (m).

(a) For SSD = 120 m ($H=1.2$ m, $h=0.15$ m)

$(\sqrt{2H} + \sqrt{2h})^2 = (\sqrt{2.4} + \sqrt{0.30})^2 = (1.549 + 0.5477)^2 = (2.0967)^2 = 4.396$.

Assume $L > S$:

$$L = \frac{N S^2}{4.396} = \frac{0.060 \times 120^2}{4.396} = \frac{0.060 \times 14400}{4.396} = \frac{864}{4.396} = 196.5\ \text{m}$$

Check: $L = 196.5\ \text{m} > S = 120\ \text{m}$ ✓ (assumption valid).

Length for SSD ≈ 197 m (adopt 200 m).

(b) For OSD = 470 m ($H = h = 1.2$ m)

$(\sqrt{2H} + \sqrt{2h})^2 = (\sqrt{2.4} + \sqrt{2.4})^2 = (2\sqrt{2.4})^2 = 4 \times 2.4 = 9.6$.

Assume $L > S$:

$$L = \frac{N S^2}{9.6} = \frac{0.060 \times 470^2}{9.6} = \frac{0.060 \times 220900}{9.6} = \frac{13254}{9.6} = 1380.6\ \text{m}$$

Check: $L = 1380.6\ \text{m} > S = 470\ \text{m}$ ✓.

Length for OSD ≈ 1381 m.

Comment

Providing full OSD on a summit curve needs ~1381 m, which is impractically long and costly. In practice, summit curves are designed for SSD (≈197 m) and overtaking is prohibited (no-passing zones / signs) where OSD cannot be economically achieved. Thus the SSD-based length of about 200 m is adopted.

Objectives of Traffic Volume Studies

• Determine the magnitude, classification and composition of traffic flow.
• Establish design hourly volume and capacity needs for geometric design.
• Identify peak-hour and directional distribution for signal timing and lane allocation.
• Provide data for planning, economic justification, and accident analysis.
• Establish traffic trends (growth) for forecasting future demand.

Methods of Conducting Volume Studies

1. Manual count — observers with tally sheets/mechanical counters; good for classified counts and turning movements.
2. Automatic count — pneumatic tube detectors, inductive loops, axle/contact counters; for continuous long-period counts.
3. Moving observer (Floating car) method — a test vehicle records overtaking/opposing flows to derive volume and speed.
4. Video/photographic and modern sensor (radar, CCTV with AI) methods — for permanent data collection.

(a) Traffic Volume in PCU/hour

$$\text{Volume} = 600 + 450 + 360 + 160 + 60 = 1630\ \text{PCU/hr}$$

Total traffic volume = 1630 PCU/hr (both directions).

(b) Number of Lanes Required

This 1630 PCU/hr is taken as the design hourly volume (30th-highest hour).

Design-direction volume (60% split):

$$V_{design} = 0.60 \times 1630 = 978\ \text{PCU/hr}$$

Lanes required in the design direction:

$$n = \frac{V_{design}}{\text{lane service volume}} = \frac{978}{1800} = 0.543 \Rightarrow 1\ \text{lane}$$

Opposing direction (40%): $0.40 \times 1630 = 652$ PCU/hr → $652/1800 = 0.36$ → 1 lane.

One lane per direction is theoretically sufficient (2-lane road). However, since a single carriageway with one lane each way is the practical minimum and to provide for overtaking, lateral clearance and future growth, a standard 2-lane two-way carriageway (one lane each direction) is adopted; if growth is expected to push the design direction above 1800 PCU/hr, provide 2 lanes in the design direction.

Desirable Properties of Road Aggregates

• Strength — resistance to crushing under traffic load.
• Hardness — resistance to abrasion/wear from traffic.
• Toughness — resistance to impact/sudden shock.
• Durability (soundness) — resistance to weathering action.
• Shape — preferably cubical; low flakiness/elongation.
• Adhesion / hydrophobic nature — good bonding with bitumen, resistance to stripping.

Three Laboratory Tests

1. Aggregate Crushing Value (ACV) test — measures resistance to crushing under a gradually applied compressive load (load 40 t over 10 min). Lower value = stronger aggregate. Limit: ≤ 30% for surface/bituminous courses (≤ 45% for base).
2. Los Angeles Abrasion test — measures resistance to abrasion and impact using steel charge in a rotating drum (% wear). Limit: ≤ 30% for bituminous surface courses (≤ 40–50% for bases).
3. Aggregate Impact Value (AIV) test — measures toughness/resistance to sudden impact (15 blows of standard hammer). Limit: ≤ 30% for surface courses (≤ 35% for bituminous bound, ≤ 45% for water-bound).

(Other valid tests: Soundness test, Flakiness & Elongation index, Stripping value/Bitumen adhesion, Polished Stone Value, Water absorption.)

Penetration Test

• Measures the hardness/consistency of bitumen as the depth (in tenths of a millimetre, i.e. 1 dmm = 0.1 mm) to which a standard needle penetrates the sample under a load of 100 g for 5 s at 25°C.
• A higher penetration value = softer bitumen; a lower value = harder.
• It is the basis of the penetration grading system (e.g. 60/70, 80/100).

Softening Point (Ring & Ball) Test

• Measures the temperature at which bitumen attains a particular degree of softening — the temperature (°C) at which a standard steel ball sinks through a bitumen disc in a ring and touches the base plate of the apparatus while the bath is heated at 5°C/min.
• Indicates temperature susceptibility; higher softening point = less likely to soften/flow at high service temperatures (good for hot climates).

Grade and Suitability

• Penetration = 65 at 25°C → falls in the 60/70 penetration grade bitumen.
• Hot climate: 60/70 (harder, higher softening point) is suitable — it resists softening, bleeding and rutting under high pavement temperatures.
• Cold hill region: 60/70 is comparatively stiff/brittle; in very cold conditions a softer grade (e.g. 80/100) is preferred to avoid cracking. Hence 60/70 suits warmer terai/plains better than cold high-altitude roads.

Formula

Total extra widening $W_e = W_m + W_{ps}$ where:

• Mechanical widening $W_m = \dfrac{n\,l^2}{2R}$
• Psychological widening $W_{ps} = \dfrac{V}{9.5\sqrt{R}}$

with $n$ = number of lanes, $l$ = wheelbase (m), $R$ = radius (m), $V$ = speed (km/h).

Given

$n = 2$, $l = 6.1$ m, $R = 180$ m, $V = 65$ km/h.

Mechanical Widening

$$W_m = \frac{n l^2}{2R} = \frac{2 \times 6.1^2}{2 \times 180} = \frac{2 \times 37.21}{360} = \frac{74.42}{360} = 0.2067\ \text{m}$$

Psychological Widening

$$W_{ps} = \frac{V}{9.5\sqrt{R}} = \frac{65}{9.5\sqrt{180}} = \frac{65}{9.5 \times 13.416} = \frac{65}{127.45} = 0.510\ \text{m}$$

Total Extra Widening

$$W_e = 0.2067 + 0.510 = 0.717\ \text{m}$$

Total extra widening ≈ 0.72 m (mechanical ≈ 0.21 m, psychological ≈ 0.51 m). Provide about 0.7–0.75 m of widening, applied gradually over the transition curve.

Definitions

• Spot speed: the instantaneous speed of a vehicle at a specified point/section of road.
• Running speed: distance divided by the time the vehicle is actually in motion (stopped time excluded).
• Journey (overall) speed: total distance divided by total time of the journey, including all stopped delays.

Individual Speeds (base $L = 30$ m)

$v = L/t$, then convert m/s → km/h ($\times 3.6$):

Time Mean Speed (arithmetic mean of spot speeds)

$$\bar v_t = \frac{15+12+10+7.5+6}{5} = \frac{50.5}{5} = 10.1\ \text{m/s} = 36.36\ \text{km/h}$$

Space Mean Speed (harmonic mean)

$$\bar v_s = \frac{n}{\sum (1/v_i)} = \frac{5}{\tfrac{1}{15}+\tfrac{1}{12}+\tfrac{1}{10}+\tfrac{1}{7.5}+\tfrac{1}{6}}$$

$\sum(1/v_i) = 0.06667 + 0.08333 + 0.10000 + 0.13333 + 0.16667 = 0.55000$.

$$\bar v_s = \frac{5}{0.55} = 9.091\ \text{m/s} = 32.73\ \text{km/h}$$

Result

Time mean speed = 10.1 m/s (36.36 km/h); Space mean speed = 9.09 m/s (32.73 km/h). The time mean speed is larger than the space mean speed (always $\bar v_t \ge \bar v_s$).

Camber — Definition and Functions

Camber (cross-slope) is the transverse slope provided to the road surface to drain rainwater quickly to the sides.

Functions:

• Rapidly drains surface water, preventing infiltration that weakens the pavement and subgrade.
• Reduces hydroplaning and improves skid resistance/safety.
• Keeps the carriageway dry, prolonging pavement life.

Fall from Crown to Edge (parabolic camber)

For a parabolic camber the surface equation about the crown is $y = \dfrac{2 x^2}{n W}$ where $n$ is the camber denominator. Simpler: the total cross-fall over half-width equals the camber rate × half-width.

Half width $= 7.0 / 2 = 3.5$ m. Camber rate $= 2.5\% = 0.025$.

$$\text{Fall from crown to edge} = 0.025 \times 3.5 = 0.0875\ \text{m} = 87.5\ \text{mm}$$

Drop from crown to each edge ≈ 87.5 mm (8.75 cm).

Cross-section (described)

        crown (highest)
          /\
         /  \        2.5% each side
        /    \
  edge /      \ edge   (each edge is 87.5 mm below crown)
  |<--3.5 m-->|<--3.5 m-->|
         7.0 m total

The surface is a smooth parabola, steepest near the edges and flat at the crown.

Typical Recommended Camber Values

Requirements of an Ideal Highway Alignment

An ideal alignment should be:

1. Short — as straight/direct as possible to reduce travel distance and cost.
2. Easy — easy to construct and maintain; easy gradients and curves for vehicle operation.
3. Safe — safe for traffic at design speed; safe against slope instability and embankment failure.
4. Economical — minimum total cost of construction, maintenance and vehicle operation.

(Mnemonic: an alignment should be Short, Easy, Safe and Economical.)

Factors Controlling Alignment in Hilly Terrain

1. Stability of slopes (geology): the alignment must avoid landslide-prone, unstable and steep slopes; preferably along stable formations and well-knit rock to prevent slips and slope failure. This is the governing factor in hills.
2. Drainage / hydrology: minimize cross-drainage works; avoid water-logged reaches; the alignment should keep hill-side and valley-side drainage easy and reduce the number of bridges/culverts.
3. Geometric standards:
   • Ruling/limiting gradient must not be exceeded; provide hairpin bends where height has to be gained, with compensated grades on curves (grade compensation).
   • Adequate radius, sight distance and extra widening on sharp curves.
4. Resisting length / rise and fall: keep the actual length close to the aerial distance; control needless rise and fall to reduce fuel and time costs.
5. Other site controls: obligatory points (passes, bridge sites, towns), monsoon/snow conditions, and availability of construction materials.

In hills the priorities reorder: stability and drainage take precedence over directness, and gradients/curve geometry are tightened to suit the terrain.