BE Civil Engineering (IOE, TU) Transportation Engineering II (IOE, CE 704) Question Paper 2080 Nepal

(a) Cumulative Standard Axles

Step 1 – Project the traffic to the year of completion of construction.

The count is 3 years old; project it forward at $r = 7.5\%$:

$$A = P(1+r)^x = 1500\,(1.075)^3$$ $$(1.075)^3 = 1.242297$$ $$A = 1500 \times 1.242297 = 1863.4 \approx 1863 \text{ CVPD}$$

Step 2 – Apply the IRC:37 cumulative ESAL formula.

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

where $A$ = CVPD at completion year, $n$ = design life = 15.

$$(1.075)^{15} = 2.958877 \Rightarrow (1.075)^{15}-1 = 1.958877$$ $$\frac{(1+r)^n-1}{r} = \frac{1.958877}{0.075} = 26.1184$$

Step 3 – Combine.

$$N = 365 \times 1863 \times 26.1184 \times 0.75 \times 3.5$$

Compute progressively:

• $365 \times 1863 = 679{,}995$
• $679{,}995 \times 26.1184 = 1.77603\times10^{7}$
• $\times 0.75 = 1.33202\times10^{7}$
• $\times 3.5 = 4.6621\times10^{7}$ standard axles

$$\boxed{N \approx 4.66\times10^{7} \text{ ESAL} = 46.6 \text{ msa}}$$

Design traffic is therefore taken as 46.6 msa (rounded up to 50 msa if selecting a standard catalogue category).

(b) Design CBR and reading the catalogue

Design subgrade CBR. When a number of CBR values are obtained along the alignment, the data are grouped. The recommended practice is to take the lowest value of the most frequent (modal) group, or for a more rigorous approach plot CBR against percentage of values equal-to-or-greater and adopt the value corresponding to the 90th-percentile of the data (i.e. the value exceeded by 90% of samples). The CBR is determined on samples remoulded at the field density and soaked for 4 days to simulate the worst (monsoon) moisture condition. Here the effective design CBR = 6%.

Reading thickness from the IRC:37 catalogue. For a given (msa, CBR) pair the IRC:37 plates / pavement-composition charts directly give:

1. Total pavement thickness above the subgrade for the design traffic of ~50 msa and CBR 6% (typically of the order of 660–700 mm).
2. The composition split into:
   • Bituminous surfacing (Bituminous Concrete wearing course + Dense Bituminous Macadam binder), thickness fixed by the msa band;
   • Granular base (Water Bound Macadam / Wet Mix Macadam);
   • Granular sub-base (GSB), with a specified minimum thickness governed by CBR.

Individual layer thicknesses are not freely chosen; the catalogue specifies them so that each interface stress and the subgrade strain remain within the allowable fatigue and rutting limits used to generate the charts. The designer simply adopts the tabulated layer thicknesses (rounding up to the nearest practical 5–10 mm) for the (50 msa, 6% CBR) entry.

Summary: Design traffic 46.6 msa → catalogue 50 msa, CBR 6% → total ≈ 670 mm comprising BC + DBM surfacing over WMM base over GSB sub-base.

(a) Radius of relative stiffness

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

Use consistent SI units (N, mm). $E = 3.0\times10^4 \text{ MPa} = 3.0\times10^4 \text{ N/mm}^2$, $h = 250$ 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: $12(1-0.15^2)(0.06) = 12 \times 0.9775 \times 0.06 = 0.7038$

$$l^4 = \frac{4.6875\times10^{11}}{0.7038} = 6.6603\times10^{11} \text{ mm}^4$$ $$l = (6.6603\times10^{11})^{0.25}$$ $$\sqrt{6.6603\times10^{11}} = 8.1611\times10^{5}; \quad \sqrt{8.1611\times10^{5}} = 903.4$$ $$\boxed{l \approx 903 \text{ mm}}$$

(b) Edge-load stress

Check b. $1.724h = 1.724 \times 250 = 431$ mm. Since $a = 150 \text{ mm} < 431$ mm, use the equivalent radius:

$$b = \sqrt{1.6a^2 + h^2} - 0.675h$$ $$1.6a^2 = 1.6 \times 150^2 = 1.6 \times 22500 = 36000$$ $$h^2 = 62500;\quad 1.6a^2 + h^2 = 98500$$ $$\sqrt{98500} = 313.85$$ $$b = 313.85 - 0.675\times250 = 313.85 - 168.75 = 145.1 \text{ mm}$$

Apply Westergaard edge equation. $P = 50{,}000$ N, $h = 250$ mm.

$$\frac{0.572P}{h^2} = \frac{0.572 \times 50000}{62500} = \frac{28600}{62500} = 0.45760 \text{ N/mm}^2$$ $$\frac{l}{b} = \frac{903}{145.1} = 6.2233;\quad \log_{10}(6.2233) = 0.79392$$ $$4\log_{10}\frac{l}{b} + 0.359 = 4(0.79392) + 0.359 = 3.17568 + 0.359 = 3.53468$$ $$\sigma_e = 0.45760 \times 3.53468 = 1.6175 \text{ N/mm}^2$$ $$\boxed{\sigma_e \approx 1.62 \text{ MPa}}$$

This edge stress would be checked against the allowable flexural strength of the concrete (modulus of rupture / factor of safety) to confirm the slab thickness is adequate.

(a) Benkelman Beam Deflection survey

Principle. A standard truck (rear axle 8160 kg = 80 kN, dual wheels, tyre pressure 5.6 kg/cm²) is positioned so that one set of dual rear wheels straddles the test point. The probe tip of the lever-arm beam is inserted between the dual tyres and rests on the pavement. As the truck creeps forward, the pavement rebounds (recovers elastically); the upward movement of the probe tip is magnified by the lever ratio and read on a dial gauge. The rebound deflection represents the elastic (recoverable) deflection of the pavement structure under the standard load.

Procedure (IRC:81). Mark test points at regular intervals along the wheel path; place probe; take initial dial reading $D_i$; drive truck forward 2.7 m and take intermediate reading $D_x$ (to check creep); drive a further distance until rebound is complete and take final reading $D_f$. The rebound deflection at the point is computed (with leg-ratio correction if intermediate reading shows significant movement).

Corrections.

• Temperature correction: deflection increases with pavement temperature; readings are corrected to a standard pavement temperature of 35 °C using the prescribed correction factor (about 0.01 mm per °C per 25 mm of bituminous layer).
• Seasonal / moisture correction: subgrade moisture varies seasonally; a seasonal correction factor (depending on subgrade type, climate and the season of testing) is applied to bring the deflection to the most critical (post-monsoon, wettest) condition.

(b) Characteristic deflection and overlay

For National Highways and major roads (high-volume), IRC:81 uses the 99-percentile value:

$$D_c = \bar{x} + 2\sigma$$ $$D_c = 1.05 + 2(0.30) = 1.05 + 0.60 = 1.65 \text{ mm}$$ $$\boxed{D_c = 1.65 \text{ mm}}$$

(For other/low-volume roads $D_c = \bar{x} + \sigma$ is used.)

Use for overlay design. The characteristic deflection (after temperature and seasonal correction) is entered into the IRC:81 design chart together with the design traffic (cumulative msa). The chart gives the required thickness of bituminous macadam overlay. If a different overlay material is used, the BM thickness is converted using the equivalency factors. Thus a higher $D_c$ (weaker, more-deflecting pavement) demands a thicker overlay for the same traffic.

Governing formula

For a railway curve the cant (superelevation) is

$$e = \frac{G\,V^2}{127\,R}$$

with $e$ in metres (or mm), $G$ in metres, $V$ in km/h, $R$ in metres. Use $G = 1.750$ m (dynamic gauge).

(a) Equilibrium cant for booked (goods) speed, V = 60 km/h

$$e_{eq} = \frac{1.750 \times 60^2}{127 \times 875} = \frac{1.750 \times 3600}{111125} = \frac{6300}{111125}$$ $$e_{eq} = 0.05669 \text{ m} = 56.7 \text{ mm}$$ $$\boxed{e_{eq} \approx 57 \text{ mm (for 60 km/h)}}$$

(b) Cant for maximum speed, V = 110 km/h

Theoretical (equilibrium) cant at max speed:

$$e_{th} = \frac{1.750 \times 110^2}{127 \times 875} = \frac{1.750 \times 12100}{111125} = \frac{21175}{111125} = 0.19055 \text{ m} = 190.6 \text{ mm}$$

Cant actually provided is limited by cant deficiency. The provided cant plus permissible deficiency must cover the requirement of the fastest train, but the provided cant itself need only equal:

$$e_{provided} = e_{th} - C_d = 190.6 - 75 = 115.6 \text{ mm}$$

Check maximum cant: $115.6 \text{ mm} < 165 \text{ mm}$ (allowable). ✔

Check against equilibrium for slow goods train (excess cant). Excess cant for the 60 km/h train if 115.6 mm is laid:

$$C_{excess} = 115.6 - 56.7 = 58.9 \text{ mm}$$

This is within the permissible cant excess for BG (75 mm). ✔

Cant to be provided:

$$\boxed{e \approx 116 \text{ mm}}$$

This value satisfies the cant-deficiency limit for the fastest (110 km/h) train, the maximum-cant limit, and the cant-excess limit for the slowest booked goods train.

(a) Runway length corrections

Basic length $L_b = 2100$ m.

Step 1 – Elevation correction (ICAO). Increase basic length by 7% per 300 m rise above MSL.

$$\text{Elev. factor} = \frac{0.07 \times 1300}{300} = 0.07 \times 4.3333 = 0.30333$$ $$L_1 = L_b(1 + 0.30333) = 2100 \times 1.30333 = 2737.0 \text{ m}$$

Step 2 – Temperature correction. First find the standard atmospheric temperature at this elevation:

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

Rise of reference temperature over standard:

$$\Delta T = 28 - 6.55 = 21.45\,^\circ C$$

Correction = 1% per 1 °C rise, applied to the elevation-corrected length $L_1$:

$$L_2 = L_1(1 + 0.01\Delta T) = 2737.0 \times (1 + 0.2145) = 2737.0 \times 1.2145 = 3324.1 \text{ m}$$

Check the 35% combined elevation+temperature limit.

$$\text{Combined increase} = \frac{3324.1 - 2100}{2100} = \frac{1224.1}{2100} = 0.583 = 58.3\%$$

This exceeds 35%, so per ICAO the design should be verified by a specific (aeroplane performance) study rather than the standard percentage rule. We continue with the standard method for the exam answer, noting this flag.

Step 3 – Gradient correction. Increase by 10% per 1% of effective gradient, applied to $L_2$:

$$L_3 = L_2(1 + 0.10 \times 0.5) = L_2(1 + 0.05) = 3324.1 \times 1.05 = 3490.3 \text{ m}$$ $$\boxed{\text{Corrected runway length} \approx 3490 \text{ m}}$$

Summary table

(b) Airport reference temperature; order of corrections

Airport reference temperature is the monthly mean of the daily maximum temperatures of the hottest month of the year (the hottest month being the month with the highest monthly mean of daily maxima), averaged over a period of years. It represents the high-temperature condition the runway must perform under.

Order: The elevation correction reduces air density and lift, lengthening the takeoff run; the temperature correction further reduces density. Because both effects compound on the already-lengthened run, ICAO specifies that the temperature correction be applied to the elevation-corrected length (a multiplicative sequence), not to the basic length, so the cumulative reduction in engine/aerodynamic performance is correctly captured.

(a) Air voids

$$V_v = \left(1 - \frac{G_m}{G_t}\right)\times100 = \left(1 - \frac{2.380}{2.495}\right)\times100$$ $$\frac{2.380}{2.495} = 0.95391 \Rightarrow V_v = (1-0.95391)\times100 = 4.61\%$$ $$\boxed{V_v = 4.61\%}$$

(b) Volume of bitumen

Volume of bitumen per 100 units of mix volume:

$$V_b = \frac{G_m \times W_b}{G_b} = \frac{2.380 \times 5.2}{1.02}$$ $$= \frac{12.376}{1.02} = 12.13\%$$ $$\boxed{V_b = 12.13\%}$$

(Here $W_b$ = 5.2% bitumen by weight; the formula gives volume fraction of bitumen in the compacted mix.)

(c) VMA and VFB

VMA = volume of voids in the mineral aggregate = $V_v + V_b$:

$$VMA = 4.61 + 12.13 = 16.74\%$$ $$\boxed{VMA = 16.74\%}$$

VFB = voids in mineral aggregate filled with bitumen:

$$VFB = \frac{V_b}{VMA}\times100 = \frac{12.13}{16.74}\times100 = 72.5\%$$ $$\boxed{VFB = 72.5\%}$$

These values (air voids ~4.6%, VFB ~72%) fall within the typical Marshall design ranges for BC (air voids 3–5%, VFB 65–75%).

(a) Types of joints in CC pavement

1. Expansion joints – full-depth transverse gaps (with a compressible filler and dowel bars) that allow the slab to expand in hot weather without buckling; relieve compressive stress.
2. Contraction joints – transverse joints (a weakened plane, often a sawn groove ~1/4 depth, with dowel bars for load transfer) that control/relieve tensile stresses from contraction and shrinkage by inducing cracks at chosen locations.
3. Warping (hinge) joints – longitudinal or transverse joints with tie bars that relieve stresses due to temperature warping while keeping the faces in contact (no movement across joint).
4. Construction joints – formed where paving stops at the end of a day's work or due to interruption; provide a clean, load-transferring junction between successive pours.

(b) Maximum contraction-joint spacing

Contraction-joint spacing is governed by the frictional resistance the slab can develop without the tensile stress exceeding the allowable value:

$$S_c = \frac{2 \times f_t \times 10^4}{W \times f}$$

where $S_c$ in m, $f_t$ = allowable tensile stress (kg/cm²), $W$ = unit weight (kg/m³? ) — to avoid unit confusion work in SI from first principles.

First-principles SI derivation. The slab tends to contract; subgrade friction resists, building tensile force. Maximum tensile stress occurs at mid-length:

$$\sigma_t = \frac{W \cdot \gamma \cdot f \cdot (S_c/2)}{ \text{... }}$$

Using the standard slab-restraint expression, the longitudinal tensile stress = $\gamma \cdot f \cdot \dfrac{S_c}{2}$ (force per unit area = weight per unit plan area × friction × half-length, divided by thickness — thickness cancels because both weight and section scale with it):

$$\sigma_t = \gamma \cdot f \cdot \frac{S_c}{2}$$

Set $\sigma_t = 0.8 \text{ MPa} = 800 \text{ kN/m}^2$, $\gamma = 24 \text{ kN/m}^3$, $f = 1.5$:

$$800 = 24 \times 1.5 \times \frac{S_c}{2} = 18\,S_c$$ $$S_c = \frac{800}{18} = 44.4 \text{ m}$$ $$\boxed{S_c \approx 44 \text{ m (friction-governed maximum)}}$$

In practice the spacing is far smaller (4.5 m for plain slabs) for crack control; the friction calculation only gives the theoretical upper bound below which the slab will not crack from restraint. (The thermal data confirm a temperature drop of 30 °C; the contraction strain $\alpha\Delta T = 10\times10^{-6}\times30 = 3\times10^{-4}$ would, if fully restrained, give $E\varepsilon$ far above $f_t$, which is exactly why joints are required.)

(a) Common flexible-pavement distresses

1. Alligator (fatigue) cracking – interconnected map-like cracks in the wheel path. Cause: repeated traffic loading exceeding the fatigue life of the bituminous layer, often with inadequate structural thickness or weak/wet subgrade.
2. Rutting – longitudinal depressions along the wheel paths. Cause: permanent (plastic) deformation in one or more layers due to heavy/channelised loading, unstable mix (excess bitumen, rounded aggregate) or insufficient compaction.
3. Ravelling – progressive loss of aggregate from the surface. Cause: insufficient binder, poor adhesion (stripping), aged/oxidised bitumen, or inadequate compaction.
4. Potholes – bowl-shaped cavities. Cause: progression of cracking and water ingress weakening the layers, then disintegration under traffic.

(Other acceptable: bleeding, shoving, edge cracking, depressions.)

(b) Maintenance categories

• Routine maintenance: frequent, small-scale, day-to-day works to keep the road serviceable. Example: crack sealing, pothole patching, drain/shoulder clearing.
• Periodic maintenance: planned works at intervals of a few years to restore surface characteristics. Example: renewal surfacing / fog or slurry seal, thin bituminous overlay.
• Special / rehabilitation works: major works to restore structural capacity or correct serious deterioration. Example: strengthening (structural) overlay designed by BBD/msa, reconstruction of failed sections.

(a) Rail creep

Definition. Rail creep is the longitudinal (forward) movement of rails with respect to the sleepers in the direction of dominant traffic.

Causes.

• Wave action / wave motion of rails under moving wheels (the depression in front of and behind the wheel pushes the rail forward).
• Starting, accelerating and braking (drag) of trains.
• Thermal expansion and contraction with poor anchorage.
• Predominant direction of traffic on a line, uneven/gradient sections, ballast resistance variation.

Prevention measures.

• Anchors / anti-creepers (rail anchors) clamped to the rail foot bearing against sleepers.
• Adequate, well-compacted ballast giving longitudinal resistance.
• Proper fishplate/fastening tightness and use of elastic fastenings.
• Provision of expansion gaps and pull-back of rails when creep is excessive.

(b) Crossing of 1 in 8.5

Crossing number $N = 8.5$ (the crossing is described as 1 in 8.5).

Crossing angle (centre-line / right-angle method). $\cot\alpha = N$:

$$\tan\alpha = \frac{1}{N} = \frac{1}{8.5} = 0.117647$$ $$\alpha = \tan^{-1}(0.117647) = 6.7090^\circ \approx 6^\circ 43'$$ $$\boxed{\alpha \approx 6.71^\circ \;(6^\circ 43'),\;\; N = 8.5}$$

Theoretical lead. Using the common approximation for the lead of a turnout (BG, gauge $G = 1.676$ m) with crossing number $N$:

$$L = 2N\,G$$

(approximate centre-line lead).

$$L = 2 \times 8.5 \times 1.676 = 28.49 \text{ m}$$ $$\boxed{\text{Lead } L \approx 28.5 \text{ m}}$$

(If the switch lead and curve lead are computed separately a slightly different value results; the $2NG$ form is the standard quick estimate of the overall lead.)

(a) Tunnel ventilation

Necessity. Ventilation is required to (i) supply fresh air (oxygen) to occupants and workers; (ii) dilute and remove exhaust gases (CO, NOx, smoke) and dust from vehicles or construction; (iii) control temperature and humidity; (iv) clear smoke in case of fire to permit safe evacuation; and (v) during construction, remove blasting fumes and supply air to the working face.

Common mechanical ventilation systems.

1. Longitudinal ventilation – air is driven along the axis of the tunnel from one portal to the other (or by jet fans mounted on the soffit / by a Saccardo nozzle). Air moves parallel to traffic. Suited to shorter tunnels and unidirectional traffic.

   Portal IN →[ jet fan ]→ ... →[ jet fan ]→ Portal OUT

2. Semi-transverse ventilation – fresh air is supplied uniformly along the length through a supply duct (above false ceiling or below roadway) and exhausted through the portals (or vice-versa). Gives uniform fresh-air distribution.

3. Fully transverse ventilation – separate supply and exhaust ducts run the full length; fresh air is injected and vitiated air extracted uniformly along the tunnel. Air flows transversely (across the section), independent of length. Used for long tunnels.

(b) Tunnel vs deep open cut; shield method

Advantages of tunnelling over deep open cut:

• Shorter, more direct alignment with easier grades; avoids huge excavation/spoil disposal of a deep cut.
• No disturbance/acquisition of large surface land; surface activities and ecology above remain undisturbed.
• Protection from snow, landslides and weather; lower long-term maintenance of slopes.
• Often more economical and stable than a very deep cut in steep terrain.

Shield method. The shield (tunnelling shield / TBM shield) is a rigid steel cylindrical frame pushed forward through soft, water-bearing or unstable ground; it supports the freshly excavated face and surrounding soil while the permanent lining (segmental rings) is erected within its tail, preventing collapse. It is used for soft-ground tunnels (e.g. metros under cities, sub-aqueous tunnels).

(a) Lacey regime waterway and normal scour depth

(i) Regime / Lacey waterway (effective linear waterway):

$$P = 4.8\,\sqrt{Q} = 4.8\,\sqrt{1800}$$ $$\sqrt{1800} = 42.426$$ $$P = 4.8 \times 42.426 = 203.6 \text{ m}$$ $$\boxed{P \approx 203.6 \text{ m}}$$

(ii) Normal scour depth (regime depth):

$$R = 0.473\left(\frac{Q}{f}\right)^{1/3}$$ $$\frac{Q}{f} = \frac{1800}{1.1} = 1636.4$$ $$(1636.4)^{1/3}:\quad 11.78^3 = 1635 \Rightarrow (1636.4)^{1/3} \approx 11.78$$ $$R = 0.473 \times 11.78 = 5.572 \text{ m}$$ $$\boxed{R \approx 5.57 \text{ m}}$$

(b) Design (maximum) scour depth and founding level

The normal scour depth $R$ is the mean regime depth. Local scour at obstructions is deeper. For piers, the maximum scour depth is taken as:

$$D_{max} = 2.0\,R$$

(IRC/standard practice: factor 2.0 for piers; 1.27R for noses of guide banks, 1.5R near abutments/upstream of guide bunds, etc.)

$$D_{max} = 2.0 \times 5.572 = 11.14 \text{ m below HFL}$$ $$\boxed{D_{max} \approx 11.14 \text{ m below HFL}}$$

Founding level / grip length. Foundations must be carried below the maximum scour level by a grip (anchorage) of at least one-third of the maximum scour depth below HFL (i.e. minimum 1.33 × scour from HFL is a common rule). Provide grip = $\tfrac{1}{3}D_{max}$:

$$\text{Grip} = \frac{11.14}{3} = 3.71 \text{ m}$$ $$\text{Depth of founding below HFL} = D_{max} + \text{grip} = 11.14 + 3.71 = 14.85 \text{ m}$$ $$\boxed{\text{Found the pier} \approx 14.85 \text{ m below HFL (say 15 m)}}$$

Reason for the factor on scour. The regime depth $R$ is an average; flow concentration, contraction of the waterway by piers, and obstruction-induced vortices scour locally to roughly twice the mean. Founding below twice the regime scour plus a grip length ensures the foundation is not undermined during the design flood, providing the required safety against scour for piers.