BE Civil Engineering (IOE, TU) Transportation Engineering II (IOE, CE 704) Question Paper 2076 Nepal

(a) Cumulative Standard Axles (msa)

The cumulative number of standard axles is given by IRC:37 as:

$$N = \frac{365 \times A \times \big[(1+r)^n - 1\big]}{r} \times D \times F$$

where $A$ = initial number of commercial vehicles per day at completion, $r$ = growth rate (fraction), $n$ = design life, $D$ = lane distribution factor, $F$ = VDF.

Step 1 — Cumulative growth factor.

$$(1+r)^n = (1.075)^{15}$$

$\ln(1.075) = 0.072321$, so $15 \times 0.072321 = 1.084813$, and $e^{1.084813} = 2.9589$.

$$\frac{(1.075)^{15} - 1}{0.075} = \frac{2.9589 - 1}{0.075} = \frac{1.9589}{0.075} = 26.119$$

Step 2 — Cumulative vehicles over design life (one direction, all commercial).

$$365 \times A \times 26.119 = 365 \times 1200 \times 26.119 = 1.1440 \times 10^{7}\ \text{vehicles}$$

Step 3 — Apply VDF and lane distribution factor.

$$N = 1.1440 \times 10^{7} \times 3.5 \times 0.75 = 3.003 \times 10^{7}\ \text{standard axles}$$

Design traffic $N \approx \mathbf{30.0\ msa}$.

(b) Pavement composition and role of factors

With $N \approx 30$ msa and subgrade CBR = 6 %, the IRC:37 design catalogue / Pavement Analysis & Design software is entered to read off a balanced section. A representative composition for 30 msa, CBR 6 % is:

Total crust ≈ 580–590 mm. The design is verified so that the horizontal tensile strain at the bottom of the bituminous layer (fatigue) and the vertical compressive strain at the top of the subgrade (rutting) are within the allowable values for 30 msa.

• VDF converts the mixed commercial traffic into equivalent passes of the standard 80 kN single axle (fourth-power law); a higher VDF (heavier/overloaded axles) sharply increases msa and hence the required thickness.
• Lane distribution factor $D$ apportions the directional commercial traffic to the most heavily loaded lane; for a two-lane single carriageway 0.75 of traffic in each direction is assumed to use the design lane, governing the critical strains.

(a) Radius of relative stiffness

$$l = \left[\frac{E\,h^{3}}{12\,(1-\mu^{2})\,k}\right]^{0.25}$$

With $E = 30000$ N/mm², $h = 250$ mm, $\mu = 0.15$, $k = 0.08$ N/mm³:

$$E h^{3} = 30000 \times 250^{3} = 30000 \times 1.5625\times10^{7} = 4.6875\times10^{11}$$ $$12(1-\mu^2)k = 12 \times (1-0.0225) \times 0.08 = 12 \times 0.9775 \times 0.08 = 0.9384$$ $$l = \left[\frac{4.6875\times10^{11}}{0.9384}\right]^{0.25} = \left[4.9952\times10^{11}\right]^{0.25}$$ $$= (4.9952\times10^{11})^{0.25} = 839.6\ \text{mm}$$

$l \approx \mathbf{839.6\ mm}$ (≈ 0.84 m).

(b) Interior load stress (Westergaard)

$$\sigma_i = \frac{0.316\,P}{h^{2}}\left[4\log_{10}\!\left(\frac{l}{b}\right) + 1.069\right]$$

Since $a = 150$ mm and $h = 250$ mm, check $a/h = 0.6 < 1.724$, so the equivalent radius $b$ is:

$$b = \sqrt{1.6a^{2} + h^{2}} - 0.675h = \sqrt{1.6(150)^2 + 250^2} - 0.675(250)$$ $$= \sqrt{1.6\times22500 + 62500} - 168.75 = \sqrt{36000 + 62500} - 168.75$$ $$= \sqrt{98500} - 168.75 = 313.85 - 168.75 = 145.10\ \text{mm}$$

Now $\log_{10}(l/b) = \log_{10}(839.6/145.10) = \log_{10}(5.786) = 0.7625$.

$$\sigma_i = \frac{0.316 \times 51000}{250^{2}}\big[4(0.7625) + 1.069\big]$$ $$= \frac{16116}{62500}\big[3.050 + 1.069\big] = 0.25786 \times 4.119 = 1.062\ \text{N/mm}^2$$

Maximum interior load stress $\sigma_i \approx \mathbf{1.06\ MPa}$.

(c) Assumptions and critical positions

Assumptions (any two): (i) the slab behaves as an elastic, homogeneous, isotropic thin plate; (ii) the subgrade reacts as a dense liquid (Winkler foundation) — reaction proportional to deflection ($p = k\delta$); (iii) full contact between slab and subgrade; (iv) the wheel load distributes over a circular area.

Three critical load positions: Interior, Edge, and Corner loading.

(a) Characteristic deflection

$$D_c = \bar{x} + 1.28\,\sigma = 1.05 + 1.28 \times 0.30 = 1.05 + 0.384 = 1.434\ \text{mm}$$

$D_c = \mathbf{1.434\ mm}$ (uncorrected).

(b) Apply corrections

Temperature correction (test at 42 °C, standard 35 °C): excess = $42-35 = 7\,°C$. Correction $= -0.020 \times 7 = -0.140$ mm (deflection at higher temperature is reduced to standard 35 °C... applied to bring to standard):

$$D_{c,T} = 1.434 - 0.140 = 1.294\ \text{mm}$$

Seasonal (moisture) correction (factor 1.10, dry-season test corrected up to worst wet condition):

$$D_{c,\text{corr}} = 1.294 \times 1.10 = 1.423\ \text{mm}$$

Corrected characteristic deflection $D_c \approx \mathbf{1.42\ mm}$.

(c) Overlay decision

The corrected characteristic deflection (1.42 mm) exceeds the allowable deflection (1.25 mm) for 1.0 msa, so the existing pavement is structurally deficient and an overlay IS required.

The required overlay thickness is read from the IRC:81 design chart, which relates the corrected characteristic deflection and the design traffic (msa) to the thickness of a bituminous (or granular-equivalent) overlay. Entering the chart with $D_c = 1.42$ mm and 1.0 msa gives a bituminous-macadam overlay thickness; this granular-equivalent thickness is then converted to the actual bituminous layer using equivalency factors. The overlay must reduce the in-service deflection below the allowable value.

(a) Equilibrium cant for maximum speed

The cant (superelevation) for full equilibrium at speed $V$ (km/h) is:

$$e = \frac{G\,V^{2}}{127\,R}\quad (e\ \text{in m, }G\ \text{in m, }R\ \text{in m})$$

Using $G = 1.750$ m (dynamic), $V = 110$ km/h, $R = 875$ m:

$$e_{max} = \frac{1.750 \times 110^{2}}{127 \times 875} = \frac{1.750 \times 12100}{111125} = \frac{21175}{111125} = 0.19055\ \text{m}$$

Equilibrium cant for 110 km/h $= \mathbf{0.1906\ m \approx 191\ mm}$.

(This exceeds the BG maximum permissible cant of 165 mm, so the provided cant is capped — see part b.)

(b) Cant for equilibrium (average) speed and cant deficiency

The cant actually provided is normally that for the weighted average / a chosen design speed. Provide cant for, say, the booked-average; here we provide the maximum permissible cant $e_a = 165$ mm (BG limit).

Cant deficiency for the fastest train:

$$C_d = e_{theoretical,max} - e_{provided} = 191 - 165 = 26\ \text{mm}$$

Since $C_d = 26$ mm $<$ permissible 75 mm, the cant deficiency is within limits and the 110 km/h speed is permitted with the capped cant of 165 mm.

For the slowest goods train (50 km/h), equilibrium cant $= \dfrac{1.750\times50^2}{127\times875} = \dfrac{4375}{111125} = 0.0394$ m $= 39.4$ mm, which is well below the provided 165 mm, giving cant excess $= 165 - 39.4 = 125.6$ mm — this should be checked against the permissible cant excess (BG 75 mm). Here it exceeds, so the provided cant would in practice be reduced; the controlling design balances both. Comment: design cant is governed by both the fast-train deficiency and slow-train excess limits.

(c) Equilibrium speed and negative superelevation

Equilibrium speed relation (Martin's / Indian Railways approximate maximum speed on transitioned curve):

$$V = 0.27\sqrt{R \times (e+C_d)} \quad\text{(}e,C_d\text{ in mm)}$$ $$V = 0.27\sqrt{875 \times (165+75)} = 0.27\sqrt{875 \times 240} = 0.27\sqrt{210000} = 0.27 \times 458.3 = 123.7\ \text{km/h}$$

So the curve can safely carry up to ≈ 124 km/h, comfortably above the 110 km/h sanctioned speed.

Negative superelevation: on the main line of a curve where a branch turns out in the opposite direction, the outer rail of the branch (which is the inner rail of the main curve) must be raised; this gives the main line a cant of the wrong sign — a negative superelevation — which limits the main-line speed.

(a) Runway length corrections

Step 1 — Elevation correction (ICAO): increase basic length by 7 % per 300 m rise above MSL.

$$\text{Elevation factor} = \frac{1400}{300}\times 0.07 = 4.6667 \times 0.07 = 0.32667\ (32.67\%)$$ $$L_1 = 1800 \times (1 + 0.32667) = 1800 \times 1.32667 = 2388.0\ \text{m}$$

Step 2 — Temperature correction: the standard atmospheric temperature at 1,400 m elevation is

$$T_{std} = 15 - 0.0065 \times 1400 = 15 - 9.1 = 5.9\ °C$$

Rise of ART above standard $= 32 - 5.9 = 26.1\ °C$. Correction = 1 % per °C:

$$\text{Temperature factor} = 26.1 \times 0.01 = 0.261\ (26.1\%)$$ $$L_2 = L_1 \times (1 + 0.261) = 2388.0 \times 1.261 = 3011.2\ \text{m}$$

Check combined elevation + temperature correction: $32.67\% + 26.1\% = 58.8\% > 35\%$.

Since the combined correction exceeds 35 %, ICAO requires that the basic length be checked by a specific (aircraft-manufacturer) study; the 7 %/1 % rule alone is no longer reliable. For this exam computation we proceed with the corrected value but flag the warning.

Step 3 — Gradient correction: increase the elevation-and-temperature-corrected length by 20 % per 1 % effective gradient.

$$\text{Gradient factor} = 0.5 \times 0.20 = 0.10\ (10\%)$$ $$L_3 = L_2 \times (1 + 0.10) = 3011.2 \times 1.10 = 3312.3\ \text{m}$$

Corrected runway length $\approx \mathbf{3312\ m}$ (with the caveat that, because the elevation + temperature correction = 58.8 % > 35 %, a manufacturer's performance check is mandatory).

(b) Definitions

Airport Reference Temperature (ART): the monthly mean of the daily maximum temperatures for the hottest month of the year (the hottest month being the month having the highest monthly mean temperature), defined as

$$\text{ART} = T_a + \frac{T_m - T_a}{3}$$

where $T_a$ = monthly mean of daily mean temperature of the hottest month and $T_m$ = monthly mean of daily maximum temperature of the same month.

Effective gradient: the maximum difference in elevation between the highest and lowest points of the runway centre line divided by the total runway length, expressed as a percentage.

(a) Tests on bitumen

1. Penetration test — measures the consistency/hardness of bitumen as the depth (in 0.1 mm) to which a standard needle penetrates under 100 g for 5 s at 25 °C. Lower penetration = harder grade.
2. Softening point (Ring & Ball) test — measures the temperature at which bitumen attains a particular softness; an index of susceptibility to temperature and resistance to flow at high service temperature.
3. Ductility test — measures the distance (cm) a standard briquette stretches before breaking at 27 °C; indicates the binding/elongation (adhesive) property.
4. Viscosity / Flash & Fire point test — viscosity measures resistance to flow (workability at mixing/laying temperature); flash and fire point measure the safe heating temperature limits.

(Other acceptable: specific gravity, solubility, loss-on-heating/thin-film oven test.)

(b) Marshall parameters

The two key parameters read are Marshall Stability (maximum load in kN the specimen carries) and Flow value (deformation in mm = 0.25 mm units at failure).

Typical desirable range for heavy traffic: Marshall Stability $\ge$ 9.0 kN and Flow value 2–4 mm (8–16 units).

(a) Dowel bars vs tie bars

(b) Tie bar area

The tie steel must resist the frictional resistance of the slab to be tied (one lane), per metre of joint.

Area of steel per metre length of joint:

$$A_s = \frac{b \cdot f \cdot h \cdot \gamma_c}{S_s}$$

where $b$ = width of slab being tied = 3.5 m, $f = 1.5$, $h = 0.25$ m, $\gamma_c = 24000$ N/m³ (24 kN/m³), $S_s = 125\times10^{6}$ N/m².

Weight of slab per metre length (one lane):

$$W = b\,h\,\gamma_c \times 1\,\text{m} = 3.5 \times 0.25 \times 24000 = 21000\ \text{N/m}$$

Frictional force to be resisted:

$$F = f \times W = 1.5 \times 21000 = 31500\ \text{N/m}$$

Area of tie steel:

$$A_s = \frac{F}{S_s} = \frac{31500}{125\times10^{6}} = 2.52\times10^{-4}\ \text{m}^2 = 252\ \text{mm}^2\ \text{per metre}$$

Required tie-bar area $= \mathbf{252\ mm^2}$ per metre length of joint. (e.g. 12 mm bars, area 113 mm² each → spacing $= 113/252 \times 1000 \approx 448$ mm c/c.)

(a) Flexible pavement distresses and causes

(Other acceptable: corrugation, depression, longitudinal/transverse cracking from shrinkage.)

(b) Pavement Management System (PMS)

A PMS is a systematic, data-driven framework for monitoring the condition of a pavement network and for planning, programming and prioritizing maintenance, rehabilitation and reconstruction activities so as to keep the network serviceable at minimum life-cycle cost.

Main objectives (any two): (i) to provide cost-effective allocation of limited maintenance funds across the network; (ii) to predict future pavement condition and schedule timely interventions; (iii) to maintain serviceability/safety at acceptable levels; (iv) to support objective decision-making with condition and performance data.

(a) Creep of rails

Creep is the longitudinal (forward) movement of rails with respect to the sleepers in the direction of dominant traffic movement. It is measured along the track and is usually greater in the direction of the heavier traffic.

Causes (any three):

1. Wave motion / wave theory — the wheel creates a wave (depression ahead, crest behind), and the rail tends to move forward in the direction of travel.
2. Percussion / starting and stopping (braking) — acceleration pushes the rail back, braking pushes it forward; the net effect with predominant one-way traffic causes creep.
3. Temperature variation — expansion and contraction of rails, combined with restraint, gives a ratcheting forward movement.
4. (Also: ineffective/loose fastenings, unbalanced gradients, badly maintained joints, ineffective ballast resistance.)

Prevention/reduction (any two):

1. Provide anchors / anti-creepers clamped to the rail bearing against sleepers.
2. Use well-maintained, deep and well-compacted ballast to increase longitudinal resistance.
3. Use adequate and tight fastenings (elastic fastenings) and steel/heavier sleepers.

(b) Functions

Sleepers (any two): (i) hold the two rails to correct gauge and alignment; (ii) transfer and distribute the load from rails to the ballast over a wider area; (iii) provide elasticity/cushioning and resist longitudinal and lateral movement.

Ballast (any two): (i) transfers and distributes load from sleepers to the formation; (ii) holds sleepers in position and provides lateral/longitudinal stability; (iii) provides good drainage and elasticity to the track.

(a) Advantages and disadvantages of tunnels

Advantages (any two):

1. Provide the shortest and most direct route through a hill/mountain, reducing distance, grade and operating cost.
2. Unaffected by surface weather (snow, landslides, storms) and free of surface land-acquisition/property disturbance.
3. Can be cheaper than a very deep/long open cut in hard rock for deep crossings; protect the route environmentally.

Disadvantages (any two):

1. High initial construction cost and long, technically difficult construction.
2. Require artificial ventilation, lighting and drainage, increasing maintenance/operating cost.
3. Difficult and hazardous to construct in poor/weak ground; problems of safety, smoke and gases.

(b) Tunnel ventilation

Common methods of (artificial) ventilation:

1. Natural ventilation (relies on air pressure/temperature difference and piston action of traffic — adequate only for short tunnels).
2. Longitudinal system (air is blown along the length, often using jet fans / Saccardo nozzle).
3. Transverse system (separate supply and exhaust ducts deliver fresh air and remove vitiated air uniformly along the tunnel).
4. Semi-transverse system (only supply OR only exhaust duct provided).

Brief explanation — Transverse system: parallel supply and exhaust air ducts run the full length above/below the roadway. Fresh air is fed in at regular intervals and polluted air is drawn off at intervals, giving uniform air quality independent of tunnel length and traffic; it is the most effective but most expensive system, used for long tunnels.

Why essential: Vehicles/locomotives discharge CO, CO₂, oxides of nitrogen, smoke and heat in a confined space; without ventilation these reach toxic/asphyxiating levels and reduce visibility. Lighting is essential because tunnels are dark; proper lighting (with transition zones at portals to allow the eye to adapt) ensures driver visibility, comfort and safety.

(a) Lacey's regime waterway

Lacey's regime wetted perimeter (≈ minimum stable linear waterway):

$$P = 4.75\sqrt{Q}$$ $$P = 4.75\sqrt{600} = 4.75 \times 24.495 = 116.35\ \text{m}$$

Minimum required linear waterway $P \approx \mathbf{116.4\ m}$.

(For information, Lacey's regime velocity and depth: $V=\left(\dfrac{Qf^2}{140}\right)^{1/6}=\left(\dfrac{600\times1}{140}\right)^{1/6}=(4.286)^{1/6}=1.273$ m/s; hydraulic radius $R = 2.46\dfrac{V^2}{f}=2.46\times\dfrac{1.273^2}{1}=3.99$ m.)

(b) Afflux (Molesworth's formula)

Molesworth's formula for afflux:

$$x = \left(\frac{V^{2}}{17.88} + 0.01524\right)\left[\left(\frac{A}{a}\right)^{2} - 1\right]$$

where $V$ = velocity through the obstructed (bridge) opening (m/s), $A$ = unobstructed natural waterway area-ratio term (here represented by the wider downstream width), $a$ = obstructed waterway (m).

Using widths as the area ratio (uniform depth assumed), $A = 60$ m, $a = 45$ m, $V = 2.4$ m/s:

$$\frac{V^{2}}{17.88} = \frac{2.4^{2}}{17.88} = \frac{5.76}{17.88} = 0.32215$$ $$\left(\frac{A}{a}\right)^{2} - 1 = \left(\frac{60}{45}\right)^{2} - 1 = (1.3333)^{2} - 1 = 1.77778 - 1 = 0.77778$$ $$x = (0.32215 + 0.01524) \times 0.77778 = 0.33739 \times 0.77778 = 0.2624\ \text{m}$$

Afflux $x \approx \mathbf{0.26\ m}$. This rise in upstream water level must be added when fixing the bridge's high-flood level and vertical clearance.