BE Civil Engineering (IOE, TU) Transportation Engineering II (IOE, CE 704) Question Paper 2078 Nepal

(a) Cumulative standard axles (msa)

The IRC:37 design traffic formula is

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

where $A$ is the initial number of commercial vehicles per day in the design lane direction at the start of the design period, $D$ is the lane distribution factor, $F$ is the vehicle damage factor.

Step 1 – Two-way to design-direction traffic. Traffic counted both directions = 1500 CVPD. Assuming a 50:50 directional split, traffic in one direction:

$$A = \frac{1500}{2} = 750 \text{ CVPD}.$$

Step 2 – Growth factor.

$$\frac{(1+r)^n - 1}{r} = \frac{(1.075)^{15} - 1}{0.075}.$$

$(1.075)^{15} = 2.95888$, so

$$\frac{2.95888 - 1}{0.075} = \frac{1.95888}{0.075} = 26.1184.$$

Step 3 – Cumulative standard axles.

$$N = 365 \times 26.1184 \times 750 \times 0.75 \times 3.5.$$

Work stepwise:

• $365 \times 26.1184 = 9533.2$
• $9533.2 \times 750 = 7\,149\,900$
• $7\,149\,900 \times 0.75 = 5\,362\,425$
• $5\,362\,425 \times 3.5 = 18\,768\,488$ standard axles.

$$\boxed{N \approx 18.77 \text{ msa}}$$

(b) Design CBR and reading the IRC:37 charts

Design CBR from multiple pits: The subgrade soil is tested (soaked CBR, 4-day soak) at representative locations. Because the weakest soil governs but isolated low outliers should not penalise the whole section, IRC recommends using a statistically representative value — the lowest value applicable to a homogeneous stretch, or where many tests exist, the value such that not more than a chosen percentile (commonly the value exceeded by ~90% of samples for high-volume roads) is exceeded. The 90th-percentile / representative CBR for the homogeneous section is adopted as the design subgrade CBR.

Using the charts: With (i) cumulative traffic $N = 18.77 \approx 20$ msa and (ii) subgrade CBR = 6%, the relevant IRC:37 pavement design catalogue/chart gives the total pavement thickness (sum of bituminous + granular base + granular sub-base) and a recommended pavement composition. The chart/table for CBR 6% and 20 msa specifies the thicknesses of:

• Bituminous Concrete (BC) wearing course,
• Dense Bituminous Macadam (DBM) binder course (the bituminous-layer thickness),
• Granular base (WMM) and granular sub-base (GSB).

The bituminous-layer (BC + DBM) thickness is read directly from the composition table for the design traffic; higher traffic and lower CBR both increase the required thicknesses.

(a) Radius of relative stiffness

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

Use consistent units (N, mm): $E = 3.0\times10^4\text{ MPa} = 3.0\times10^4\text{ N/mm}^2$, $h = 250\text{ mm}$, $k = 60\text{ MN/m}^3 = 0.06\text{ N/mm}^3$.

Numerator: $E h^3 = 3.0\times10^4 \times 250^3 = 3.0\times10^4 \times 1.5625\times10^7 = 4.6875\times10^{11}$.

Denominator: $12(1-0.15^2)k = 12 \times (1-0.0225) \times 0.06 = 12 \times 0.9775 \times 0.06 = 0.7038$.

$$l^4 = \frac{4.6875\times10^{11}}{0.7038} = 6.6603\times10^{11}\text{ mm}^4.$$ $$l = (6.6603\times10^{11})^{0.25}.$$

$\sqrt{6.6603\times10^{11}} = 8.1610\times10^{5}$; $\sqrt{8.1610\times10^{5}} = 903.4$.

$$\boxed{l \approx 903\text{ mm}}$$

(b) Edge-load stress

Equivalent radius ($a = 150 < 1.724h = 431$ mm, so use the small-area formula):

$$b = \sqrt{1.6a^2 + h^2} - 0.675h = \sqrt{1.6(150)^2 + 250^2} - 0.675(250).$$

$1.6 \times 22500 = 36000$; $+250^2 = 36000 + 62500 = 98500$; $\sqrt{98500} = 313.85$. $0.675 \times 250 = 168.75$.

$$b = 313.85 - 168.75 = 145.10\text{ mm}.$$

Log term: $l/b = 903.4/145.10 = 6.226$; $\log_{10}(6.226) = 0.7942$.

$$4 \times 0.7942 + 0.359 = 3.1768 + 0.359 = 3.5358.$$

Stress: $P = 50\text{ kN} = 50000\text{ N}$, $h^2 = 250^2 = 62500\text{ mm}^2$.

$$\sigma_e = \frac{0.572 \times 50000}{62500} \times 3.5358 = \frac{28600}{62500} \times 3.5358 = 0.4576 \times 3.5358.$$ $$\boxed{\sigma_e \approx 1.618\text{ MPa} \;(\approx 1.62\text{ N/mm}^2)}$$

(c) Most critical loading and implication

For a typical jointed plain concrete pavement, the corner load produces the highest tensile stress, but where corners are protected by dowels/load transfer the edge load governs flexural design. Among the three, edge loading generally gives the largest mid-slab bottom tensile stress for design checks. Design implication: provide tie bars at longitudinal joints and adequate edge thickening / dowelled transverse joints, and ensure the flexural strength (modulus of rupture) exceeds the critical stress with the required factor of safety.

Governing formula

For Indian Railways practice the cant (superelevation) is

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

Using the dynamic gauge $G = 1.750\text{ m}$, $R = 1000\text{ m}$.

(a) Cant for maximum speed and for equilibrium speed

Cant for maximum speed (110 km/h):

$$e_{max} = \frac{1.750 \times 110^2}{127 \times 1000} = \frac{1.750 \times 12100}{127000} = \frac{21175}{127000} = 0.16673\text{ m} = 166.7\text{ mm}.$$

Cant for equilibrium speed (80 km/h):

$$e_{eq} = \frac{1.750 \times 80^2}{127 \times 1000} = \frac{1.750 \times 6400}{127000} = \frac{11200}{127000} = 0.08819\text{ m} = 88.2\text{ mm}.$$

Adopt actual provided cant $e_a = 88.2$ mm (equilibrium-speed cant).

(b) Cant deficiency and cant excess

Cant deficiency (fast train, 110 km/h): Cant deficiency = (cant needed for max speed) − (cant provided)

$$C_d = e_{max} - e_a = 166.7 - 88.2 = 78.5\text{ mm}.$$

Since $78.5\text{ mm} \le 100\text{ mm}$, the cant deficiency is within the permissible limit.

Cant excess (goods train, 50 km/h): Cant required for the slow speed:

$$e_{slow} = \frac{1.750 \times 50^2}{127 \times 1000} = \frac{1.750 \times 2500}{127000} = \frac{4375}{127000} = 0.03445\text{ m} = 34.4\text{ mm}.$$

Cant excess = (cant provided) − (cant needed for slow speed)

$$C_e = e_a - e_{slow} = 88.2 - 34.4 = 53.8\text{ mm}.$$

Since $53.8\text{ mm} \le 75\text{ mm}$, the cant excess is within the permissible limit.

Summary

$$\boxed{C_d = 78.5\text{ mm (OK)},\quad C_e = 53.8\text{ mm (OK)}}$$

ICAO corrections (sequential)

Step 1 – Elevation correction. Increase basic length by 7% per 300 m of elevation above MSL.

$$\text{Elevation factor} = \frac{7}{100} \times \frac{1300}{300} = 0.07 \times 4.3333 = 0.30333.$$

Corrected length after elevation:

$$L_1 = 2100 \,(1 + 0.30333) = 2100 \times 1.30333 = 2737.0\text{ m}.$$

Step 2 – Temperature correction. First find the Airport Reference Temperature (ART):

$$ART = T_a + \frac{T_m - T_a}{3} = 24 + \frac{32-24}{3} = 24 + 2.667 = 26.667^\circ\text{C}.$$

Standard atmospheric temperature at 1300 m elevation (lapse $0.0065\,^\circ\text{C/m}$ from $15^\circ\text{C}$ at MSL):

$$T_{std} = 15 - 0.0065 \times 1300 = 15 - 8.45 = 6.55^\circ\text{C}.$$

Temperature rise above standard:

$$\Delta T = ART - T_{std} = 26.667 - 6.55 = 20.117^\circ\text{C}.$$

Increase length by 1% per $1^\circ\text{C}$ rise:

$$L_2 = L_1 (1 + 0.01\,\Delta T) = 2737.0 \times (1 + 0.20117) = 2737.0 \times 1.20117 = 3287.6\text{ m}.$$

Check on combined elevation + temperature correction: total = $(3287.6 - 2100)/2100 = 56.6\%$. ICAO requires a specific site study if elevation+temperature correction exceeds 35%; here it is exceeded, so the result must be verified by site-specific data, but we proceed with the standard computation as required.

Step 3 – Gradient correction. Increase length after temperature correction by 20% per 1% effective gradient.

$$\text{Gradient factor} = 0.20 \times 0.35 = 0.07.$$ $$L_3 = L_2 (1 + 0.07) = 3287.6 \times 1.07 = 3517.7\text{ m}.$$

Final corrected runway length

$$\boxed{L \approx 3518\text{ m}}$$

(Elevation correction = +637.0 m; temperature = +550.6 m; gradient = +230.1 m; total = +1417.7 m over the basic 2100 m.)

(a) Characteristic deflection

$$D_c = \bar{x} + 2\sigma = 1.10 + 2(0.25) = 1.10 + 0.50 = 1.60\text{ mm}.$$ $$\boxed{D_c = 1.60\text{ mm}}$$

(b) Temperature correction to 35°C

Temperature difference from standard: $\Delta T = 40 - 35 = 5^\circ\text{C}$. Bituminous thickness in units of 25 mm: $100/25 = 4$.

Correction magnitude:

$$\Delta D = 0.0065 \times \Delta T \times \frac{t}{25} = 0.0065 \times 5 \times 4 = 0.130\text{ mm}.$$

Since measured at $40^\circ\text{C}$ (higher than standard, deflection is larger), subtract to reduce to $35^\circ\text{C}$:

$$D_{c,35} = 1.60 - 0.130 = 1.47\text{ mm}.$$ $$\boxed{D_{c,35} \approx 1.47\text{ mm}}$$

(c) Overlay decision and other methods

The corrected characteristic deflection $D_{c,35} = 1.47\text{ mm}$ exceeds the allowable design deflection of $1.00\text{ mm}$. The existing pavement is structurally deficient for the projected traffic, so a strengthening (bituminous) overlay is required. The overlay thickness is read from the IRC:81 overlay design chart using the characteristic deflection (1.47 mm) and projected cumulative traffic (msa); a thicker overlay is needed because the deflection is well above the allowable value.

Two other evaluation methods:

1. Falling Weight Deflectometer (FWD) — dynamic deflection bowl for back-calculation of layer moduli (structural).
2. Roughness measurement by Bump Integrator / IRI, or pavement condition / distress survey (PCI) for functional evaluation; skid-resistance testing for surface friction.

(a) Bulk specific gravity

Using SSD (saturated surface-dry) method:

$$G_{mb} = \frac{W_{air}}{W_{SSD} - W_{water}} = \frac{1230}{1233 - 705} = \frac{1230}{528} = 2.3295.$$ $$\boxed{G_{mb} \approx 2.330}$$

(b) Air voids

$$V_a = \left(1 - \frac{G_{mb}}{G_{mm}}\right) \times 100 = \left(1 - \frac{2.3295}{2.480}\right) \times 100.$$ $$\frac{2.3295}{2.480} = 0.93932.$$ $$V_a = (1 - 0.93932) \times 100 = 0.06068 \times 100 = 6.07\%.$$ $$\boxed{V_a \approx 6.1\%}$$

This is slightly above the typical desirable range of 3–5% for a wearing course, indicating the mix may need a small increase in binder content or improved compaction.

Bradbury edge warping stress

The edge warping stress is

$$\sigma_{te} = \frac{C\,E\,\alpha\,t}{2}.$$

Substitute $C = 0.72$, $E = 3.0\times10^4\text{ MPa}$, $\alpha = 10\times10^{-6}\,/^\circ\text{C}$, $t = 16^\circ\text{C}$:

Step 1: $E\,\alpha\,t = 3.0\times10^4 \times 10\times10^{-6} \times 16$.

• $3.0\times10^4 \times 10\times10^{-6} = 0.30$.
• $0.30 \times 16 = 4.80\text{ MPa}.$

Step 2: $\sigma_{te} = \dfrac{0.72 \times 4.80}{2} = \dfrac{3.456}{2} = 1.728\text{ MPa}.$

$$\boxed{\sigma_{te} \approx 1.73\text{ MPa}}$$

Note: The edge formula does not contain $\mu$ (Poisson's ratio appears only in the interior warping-stress expression $\sigma_{ti} = \frac{E\alpha t}{2}\left(\frac{C_x + \mu C_y}{1-\mu^2}\right)$). The slab thickness $h$ is needed for selecting $C$ from the ratio $L/l$ but is not in the final edge-stress formula. This warping stress must be combined with the wheel-load stress to check the slab in the most critical season/time of day.

(a) Permanent-way cross-section and functions

        Rail   Rail
         |      |        <-- rails
   ======================  <-- sleepers (cross-ties)
   \\::::::::::::::::::://   <-- ballast (crib + shoulder)
    \\__________________//
   ----- formation (subgrade) -----

• Rail: provides a smooth, continuous, hard running surface for wheels; guides the train and transmits wheel loads to the sleepers.
• Sleeper: holds the two rails at correct gauge, distributes the rail load over a wider area of ballast, and provides resistance against longitudinal and lateral movement.
• Ballast: transmits and distributes load from sleepers to the formation, provides drainage, holds sleepers in position, and gives elasticity to the track.
• Formation (subgrade): the prepared earthwork (embankment/cutting) that ultimately bears all the track loads and provides a stable base.

(b) Turnout and components

A turnout is the complete arrangement of points and crossing with lead rails by which a train is diverted from one track to another.

• Switch (points): a pair of tongue (movable) rails and stock rails that guide the wheels from the main track toward the turnout.
• Crossing / frog: the assembly (a V-shaped point rail with two wing rails) provided where one rail crosses another to allow wheel flanges to pass.
• Crossing number $N$: defines the flatness of the crossing; a $1\text{ in }N$ crossing spreads 1 unit laterally for $N$ units along the track. A larger $N$ means a flatter crossing and permits higher speed.

Relation (right-angle / RDSO method):

$$\tan\alpha = \frac{1}{N} \quad\Rightarrow\quad \alpha = \tan^{-1}\!\left(\frac{1}{N}\right),$$

where $\alpha$ is the crossing angle. (By the centre-line method, $2\tan(\alpha/2) = 1/N$.)

(a) Objectives and methods of tunnel ventilation

Objectives:

• Supply fresh air (oxygen) to workers and, in service, to passengers.
• Dilute and remove dust, smoke, blasting fumes and exhaust gases (CO, $\text{CO}_2$, $\text{NO}_x$).
• Control temperature and humidity for comfort and safety; remove heat.
• Provide smoke control / escape conditions in case of fire.

Two methods of mechanical ventilation:

1. Longitudinal ventilation (air pushed along the tunnel axis, e.g. by jet fans).
2. Transverse / semi-transverse ventilation (separate supply and exhaust ducts deliver and extract air across the cross-section). (Blowing-in and exhaust systems are the basic forced types.)

(b) New Austrian Tunnelling Method (NATM)

NATM is an observational, sequential excavation method in which the surrounding rock mass is mobilised as a primary load-bearing ring rather than merely being a load on the lining. The excavation is supported quickly with a flexible primary support — shotcrete, rock bolts, steel ribs/lattice girders and wire mesh — and the deformation/convergence of the opening is continuously monitored (instrumentation). Support is adjusted based on observed behaviour, allowing controlled stress redistribution; a secondary (final) lining is added once movements stabilise.

Two suitable situations:

1. In weak, jointed or squeezing rock / mixed ground where the rock can still contribute to support if controlled deformation is allowed.
2. For tunnels of varying or non-circular cross-section (e.g. road/metro tunnels, caverns, station chambers) where rigid mechanised TBM methods are uneconomic or impractical.

(a) Classification and super/sub-structure

Classification (any two bases):

1. By material: timber, masonry, RCC, prestressed concrete, steel, composite bridges.
2. By structural form: slab, beam/girder, arch, truss, cantilever, suspension, cable-stayed bridges. (Other bases: by function — road, rail, pedestrian, aqueduct; by span length; by inter-span continuity — simply supported, continuous; by alignment — square/skew.)

Superstructure vs substructure:

• Superstructure: the portion above the bearings — deck slab, girders/trusses, parapets — that directly carries the moving (live) and dead loads and transmits them to the supports.
• Substructure: the portion below the bearings — abutments, piers, and their foundations — that receives the load from the superstructure and transfers it safely to the founding strata, also resisting water/earth pressure and scour.

(b) Afflux, economical span, and Lacey waterway

• Afflux: the rise (heading-up) of the upstream water level above the normal flood level caused by the contraction of the natural waterway by the bridge piers and abutments.
• Economical span: the span for which the total cost of the bridge is minimum — approximately the span at which the cost of one span of the superstructure equals the cost of one pier/substructure (the substructure cost equals the superstructure cost per span).

Lacey's regime (linear) waterway:

$$P = 4.8\sqrt{Q} = 4.8 \times \sqrt{500} = 4.8 \times 22.3607 = 107.3\text{ m}.$$ $$\boxed{\text{Linear waterway} \approx 107.3\text{ m}}$$

Fourth-power equivalency law

The load equivalency factor (LEF) for an axle of load $L$ relative to the standard 80 kN axle is

$$\text{LEF} = \left(\frac{L}{80}\right)^4.$$

Total equivalent standard axles (ESAL) per day = $\sum n_i \,\text{LEF}_i$.

Step 1 – LEF for each group.

• 100 kN: $(100/80)^4 = (1.25)^4 = 2.44141.$
• 80 kN: $(80/80)^4 = 1.0000.$
• 60 kN: $(60/80)^4 = (0.75)^4 = 0.31641.$

Step 2 – ESAL contribution of each group.

Details:

• $200 \times 2.44141 = 488.28$
• $300 \times 1.0 = 300.00$
• $500 \times 0.31641 = 158.20$
• Sum $= 488.28 + 300.00 + 158.20 = 946.48$ standard axles/day.

Step 3 – Vehicle Damage Factor. VDF = (equivalent standard axles per day) / (number of commercial vehicles per day)

$$VDF = \frac{946.48}{500} = 1.893.$$ $$\boxed{\text{Total } \approx 946.5 \text{ ESAL/day}, \quad VDF \approx 1.89 \text{ standard axles per commercial vehicle}}$$