BE Civil Engineering (IOE, TU) Transportation Engineering II (IOE, CE 704) Question Paper 2077 Nepal

(a) Cumulative Standard Axles (CSA)

IRC:37 design traffic equation:

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

where $A$ = initial traffic (CVPD) at completion of construction, $r$ = growth rate (fraction), $n$ = design life (years), $D$ = lane distribution factor, $F$ = VDF.

Step 1 — Cumulative growth factor:

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

$(1.075)^{15} = 2.95888$

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

Step 2 — Annual number of vehicles in first year (×365):

$$365 \times 1450 = 529{,}250 \text{ CV/year}$$

Step 3 — Apply all factors:

$$N = 529{,}250 \times 26.1184 \times 0.75 \times 3.5$$ $$529{,}250 \times 26.1184 = 1.38231 \times 10^{7}$$ $$\times 0.75 = 1.03673 \times 10^{7}$$ $$\times 3.5 = 3.6286 \times 10^{7} \text{ standard axles}$$ $$\boxed{N \approx 36.29 \times 10^{6} \text{ standard axles} = 36.29\ \text{msa}}$$

Thus the pavement must be designed for ≈ 36.3 msa (use the 30–50 msa / CBR 6% design chart/catalogue of IRC:37).

(b) Concepts

Effective CBR: When a select/improved subgrade (borrow) layer of higher CBR is laid over a weaker natural subgrade, IRC:37 does not simply use the top-layer CBR. Instead it uses an effective CBR of the two-layer system (read from the IRC:37 two-layer chart as a function of the thickness and CBR of the upper layer over the lower-layer CBR). This represents the composite supporting capacity that the pavement structure actually "sees," preventing an over-optimistic design that ignores the weak natural soil beneath.

Compaction in 200–250 mm layers: Rollers can transmit effective compactive energy only to a limited depth. If a thick lift is placed, the bottom of the lift receives little compaction and remains loose, giving non-uniform density and later differential settlement. Placing soil in thin (200–250 mm loose) layers ensures the full lift thickness reaches the specified relative compaction (typically ≥ 97% of MDD for the upper subgrade).

Three structural layers:

A flexible pavement carries load by grain-to-grain transfer, with stress diffusing in a cone; hence layer strength decreases with depth.

(a) Radius of relative stiffness

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

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: $12(1-0.15^{2})k = 12 \times 0.9775 \times 0.080 = 0.9384$

$$l^{4} = \frac{4.6875\times10^{11}}{0.9384} = 4.9952\times10^{11}\ \text{mm}^{4}$$ $$l = (4.9952\times10^{11})^{0.25}$$

$\sqrt{4.9952\times10^{11}} = 7.0676\times10^{5}$; $\sqrt{7.0676\times10^{5}} = 840.7$

$$\boxed{l \approx 840.7\ \text{mm}}$$

(b) Westergaard load stresses

Check equivalent radius: $1.724h = 1.724\times250 = 431\ \text{mm}$. Since $a = 150\ \text{mm} < 431\ \text{mm}$, use Westergaard's special equivalent radius:

$$b = \sqrt{1.6a^{2}+h^{2}} - 0.675h$$ $$= \sqrt{1.6(150)^{2}+250^{2}} - 0.675(250) = \sqrt{36{,}000+62{,}500} - 168.75$$ $$= \sqrt{98{,}500} - 168.75 = 313.85 - 168.75 = 145.1\ \text{mm}$$

Interior stress (Westergaard):

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

$\dfrac{l}{b}=\dfrac{840.7}{145.1}=5.793$; $\log_{10}(5.793)=0.7629$

Bracket $=4(0.7629)+1.069 = 3.0516+1.069 = 4.1206$

$\dfrac{0.316\times50{,}000}{250^{2}} = \dfrac{15{,}800}{62{,}500} = 0.2528\ \text{N/mm}^2$

$$\sigma_i = 0.2528 \times 4.1206 = \boxed{1.042\ \text{MPa}}$$

Edge stress (Westergaard):

$$\sigma_e = \frac{0.572\,P}{h^{2}}\left[4\log_{10}\!\left(\frac{l}{b}\right)+0.359\right]$$

Bracket $=4(0.7629)+0.359 = 3.0516+0.359 = 3.4106$

$\dfrac{0.572\times50{,}000}{62{,}500} = \dfrac{28{,}600}{62{,}500} = 0.4576\ \text{N/mm}^2$

$$\sigma_e = 0.4576 \times 3.4106 = \boxed{1.561\ \text{MPa}}$$

Edge stress > interior stress, as expected.

(c) Assumptions of Westergaard's analysis

1. The slab acts as a homogeneous, isotropic, elastic thin plate (plate theory applies).
2. The subgrade behaves as a dense liquid (Winkler foundation): reaction at any point is proportional only to the deflection at that point, $p = k\delta$.
3. The slab has uniform thickness and full contact with the subgrade.
4. The wheel load is applied over a circular (or semi-circular at edge) area of uniform pressure.
5. Load positions analysed are interior, edge, and corner — each treated separately.

(a) Equilibrium superelevation for $V_{max}$

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

Using dynamic gauge $G = 1.750\ \text{m}$ (standard Indian Railways practice for BG cant):

$$e = \frac{1.750 \times 110^{2}}{127 \times 875} = \frac{1.750\times12{,}100}{111{,}125} = \frac{21{,}175}{111{,}125} = 0.1906\ \text{m}$$ $$\boxed{e_{Vmax} \approx 190.6\ \text{mm}}$$

This exceeds the BG maximum permissible cant of 165 mm, so the curve will be canted for the equilibrium speed, not $V_{max}$, and the remainder taken as cant deficiency.

(b) Cant for equilibrium speed and checks

Equilibrium (weighted average) speed — using the simplified relation that the provided cant equals deficiency-limited cant. We design cant $C_a$ so that at $V_{max}$ the deficiency stays within 75 mm.

Required theoretical cant at $V_{max}$ = 190.6 mm. Maximum cant deficiency allowed = 75 mm, so provide:

$$C_a = 190.6 - 75 = 115.6\ \text{mm} \Rightarrow \text{provide } C_a = 115\ \text{mm (within 165 mm limit).}$$

Check cant deficiency at $V_{max}$:

$$C_d^{actual} = e_{Vmax} - C_a = 190.6 - 115 = 75.6\ \text{mm} \approx 75\ \text{mm (just at limit, OK).}$$

Check cant excess for goods train (cant excess = provided cant − equilibrium cant for goods speed). Equilibrium cant for goods:

$$e_{goods} = \frac{1.750\times50^{2}}{127\times875} = \frac{1.750\times2500}{111{,}125} = \frac{4375}{111{,}125} = 0.03937\ \text{m} = 39.4\ \text{mm}$$ $$C_{excess} = C_a - e_{goods} = 115 - 39.4 = 75.6\ \text{mm}$$

Permissible cant excess for BG = 75 mm. The computed 75.6 mm is marginally over; in practice round provided cant down to 110 mm giving deficiency 80.6 mm (slightly high) — a trade-off. A practical balanced choice is $C_a = 115\ \text{mm}$ with the section booked so the slowest train speed is raised, OR accept the marginal excess. Design is essentially safe but at the limits; provided cant ≈ 115 mm.

(c) Length of transition curve

Use the larger of:

(i) Rate of change of cant / cant deficiency (BG, $V$ in km/h):

$$L = 0.008\,C_a\,V_{max} = 0.008 \times 115 \times 110 = 101.2\ \text{m}$$

and on cant deficiency: $L = 0.008\,C_d\,V_{max} = 0.008\times75\times110 = 66.0\ \text{m}$

(ii) Cant gradient (max 1 in 720 for BG):

$$L = 720 \times C_a = 720 \times 0.115\ \text{m} = 82.8\ \text{m}$$

Governing value = largest = 101.2 m.

$$\boxed{L_{transition} \approx 102\ \text{m (round up)}}$$

(a) Runway length corrections

Step 1 — Elevation correction (7% per 300 m):

$$\Delta_{elev} = 1900 \times 0.07 \times \frac{1350}{300} = 1900 \times 0.07 \times 4.5 = 598.5\ \text{m}$$

Corrected length after elevation:

$$L_1 = 1900 + 598.5 = 2498.5\ \text{m}$$

Step 2 — Temperature correction (1% per °C of (ART − standard atmosphere temp at elevation)): Standard atmospheric temperature at 1350 m = $15 - 0.0065\times1350 = 15 - 8.775 = 6.225\ \text{°C}$. Temperature rise $= ART - 6.225 = 28 - 6.225 = 21.775\ \text{°C}$.

$$\Delta_{temp} = L_1 \times 0.01 \times 21.775 = 2498.5 \times 0.21775 = 544.05\ \text{m}$$

Length after temperature:

$$L_2 = 2498.5 + 544.05 = 3042.6\ \text{m}$$

Screening check (elevation + temperature ≤ 35%):

$$\frac{598.5 + 544.05}{1900}\times100 = \frac{1142.55}{1900}\times100 = 60.1\%$$

This exceeds 35%, so a specific site study (wind-tunnel / manufacturer data) would normally be required; for this exam we proceed with the ICAO arithmetic correction.

Step 3 — Gradient correction (20% per 1% effective gradient, applied on $L_2$):

$$\Delta_{grad} = L_2 \times 0.20 \times 0.85 = 3042.6 \times 0.17 = 517.2\ \text{m}$$

Final corrected runway length:

$$L_3 = 3042.6 + 517.2 = 3559.8\ \text{m}$$ $$\boxed{\text{Corrected runway length} \approx 3560\ \text{m}}$$

(b) Clearway and Stopway

Clearway: A defined rectangular area beyond the runway end, on the ground or water under airport control, free of obstructions above a defined plane (slope ≤ 1.25%), over which an aircraft may make a portion of its initial climb to a specified height. It has no pavement strength requirement.

Stopway: A defined rectangular area beyond the runway end, paved/prepared and capable of supporting the aircraft, on which an aircraft can be stopped during an aborted (rejected) take-off.

Relation to declared distances:

• TODA (Take-Off Distance Available) = runway length (TORA) + clearway.
• ASDA (Accelerate-Stop Distance Available) = runway length (TORA) + stopway.
• LDA (Landing Distance Available) = runway length available for landing (clearway/stopway not added).

(a) Statistical analysis of deflections

Data ($n = 10$): 0.95, 1.10, 1.25, 0.88, 1.40, 1.05, 1.20, 1.32, 1.00, 1.15

Mean:

$$\sum x = 0.95+1.10+1.25+0.88+1.40+1.05+1.20+1.32+1.00+1.15 = 11.30$$ $$\bar{x} = \frac{11.30}{10} = 1.130\ \text{mm}$$

Deviations and squares:

Standard deviation (sample, $n-1$):

$$\sigma = \sqrt{\frac{\sum (x-\bar{x})^2}{n-1}} = \sqrt{\frac{0.24780}{9}} = \sqrt{0.027533} = 0.1659\ \text{mm}$$

Characteristic deflection (major road, $\bar{x}+2\sigma$):

$$D_c = 1.130 + 2(0.1659) = 1.130 + 0.3318 = 1.462\ \text{mm}$$ $$\boxed{\bar{x}=1.130\ \text{mm},\quad \sigma = 0.166\ \text{mm},\quad D_c \approx 1.46\ \text{mm}}$$

This $D_c$ is then entered into the IRC:81 overlay-thickness chart against 30 msa to read the required bituminous overlay thickness (the chart gives roughly a 100+ mm BM/BC overlay for this deflection level and traffic).

(b) BBD Test Procedure & Limitations

Procedure (IRC:81):

1. A standard loaded truck with a rear axle load of 8.17 tonnes (80 kN) on dual tyres inflated to 5.6 kg/cm² is used.
2. The probe (toe) of the Benkelman beam is inserted between the dual rear tyres at the test point; the initial dial reading is taken.
3. The truck is driven forward slowly (≈ 2.7 m); intermediate and final dial readings are taken as the wheel moves away and the pavement rebounds.
4. Rebound deflection is computed (×2 for the beam lever ratio) and corrected for temperature (to 35 °C standard) and seasonal/moisture variation.
5. Pavement temperature and subgrade moisture are recorded for corrections.

Limitations (any two):

• Slow, labour-intensive, and disrupts traffic — gives point measurements only (now largely superseded by FWD).
• Measures only the total surface rebound; it cannot separate the contribution of individual pavement layers, so it does not directly indicate which layer is deficient.
• Sensitive to temperature and seasonal corrections, which introduce uncertainty if not carefully measured.

(a) Air voids, $V_a$

$$V_a = \left(1 - \frac{G_{mb}}{G_{mm}}\right)\times100 = \left(1 - \frac{2.380}{2.495}\right)\times100$$ $$= (1 - 0.95391)\times100 = 4.61\%$$ $$\boxed{V_a \approx 4.6\%}$$

(b) Volume of bitumen (per unit volume of compacted mix), $V_b$

Taking 1 cm³ of compacted mix, its mass $= G_{mb}\times1 = 2.380\ \text{g}$. Mass of bitumen $= 0.05 \times 2.380 = 0.119\ \text{g}$. Volume of bitumen $= \dfrac{0.119}{G_b} = \dfrac{0.119}{1.02} = 0.1167\ \text{cm}^3$. As a percentage of mix volume (this is $V_b$, the effective binder volume by total volume):

$$V_b = 0.1167 \times 100 = 11.67\% \approx \boxed{11.7\%}$$

(c) Voids in Mineral Aggregate, VMA

$$VMA = 100 - \frac{G_{mb}\,P_s}{G_{sb}}$$

where $P_s$ = % aggregate by total mass $= 100 - 5.0 = 95.0$.

$$VMA = 100 - \frac{2.380 \times 95.0}{2.650} = 100 - \frac{226.1}{2.650} = 100 - 85.32 = 14.68\%$$ $$\boxed{VMA \approx 14.7\%}$$

Comment: $V_a = 4.6\%$ lies within the desirable 3–5% range for a surface course, so the air-void criterion is satisfied. VMA of 14.7% is acceptable (typically VMA ≥ 14% for 20 mm nominal max aggregate). The design binder content is reasonable.

(a) Joints in CC pavements

(b) Dowel bars vs Tie bars

In short: dowels transfer load and allow movement; tie bars hold adjacent slabs together and resist separation.

(a) Crossing number / frequency

The crossing number $N$ is the ratio of the length of the crossing measured along the gauge face to the spread (the lateral divergence) at that length — i.e. it defines the sharpness of the crossing. A '1 in N' crossing means the two gauge faces diverge 1 unit laterally for every $N$ units measured along the crossing.

Right-angle (RA) method: spread is measured at right angles to the main gauge face.

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

Centre-line (CL) method: spread measured along the bisector of the crossing angle.

$$\cot\frac{\alpha}{2} = N\ \text{(i.e. } 2N = \cot(\alpha/2)\text{)}\quad\Rightarrow\quad \alpha = 2\tan^{-1}\!\left(\frac{1}{N}\right)...$$

More precisely (CL): $\tan(\alpha/2) = \dfrac{1}{2N}$.

(b) Numerical

1 in 12 (Right-angle method):

$$\alpha = \tan^{-1}\!\left(\frac{1}{12}\right) = \tan^{-1}(0.08333) = 4.764^{\circ}$$ $$\boxed{\alpha \approx 4^{\circ}45'\,49'' \approx 4.76^{\circ}}$$

1 in 8.5 (Right-angle method):

$$\alpha = \tan^{-1}\!\left(\frac{1}{8.5}\right) = \tan^{-1}(0.11765) = 6.709^{\circ}$$ $$\boxed{\alpha \approx 6^{\circ}42'\,33'' \approx 6.71^{\circ}}$$

The flatter (1 in 12) crossing gives a smaller angle and permits higher turnout speeds; the sharper 1 in 8.5 crossing is used where space is limited.

(a) Choice of tunnel shape

Factors governing cross-section shape:

1. Nature of ground / rock pressure — magnitude and direction of overburden and lateral pressure.
2. Purpose / functional clearance — railway, road, water-conveyance, or pedestrian dictates the required internal clearance envelope.
3. Method of construction & lining — drill-and-blast, TBM, cut-and-cover.
4. Hydraulic considerations (for water tunnels) — circular gives best flow efficiency.
5. Structural efficiency — ability to resist external pressure with minimum lining.

Best-suited shapes:

• Soft ground / high all-round (hydrostatic) pressure → Circular section (resists radial pressure most efficiently, no stress concentration).
• Hard, self-supporting rock → Horse-shoe / D-shape (flat invert gives convenient working/road floor, arch roof carries rock load).

(b) Tunnel ventilation

Necessity: To (i) dilute and remove vehicle exhaust gases (CO, NOx, smoke) and dust, (ii) supply fresh oxygen, (iii) control temperature and humidity, and (iv) remove smoke and heat during a fire for safe evacuation.

• Longitudinal system: Air is moved along the tunnel axis (by jet/booster fans or piston effect), entering at one portal and leaving at the other.
• Semi-transverse system: Fresh air is supplied uniformly along the tunnel through a duct while vitiated air escapes at the portals (or vice-versa) — only one duct.
• Fully-transverse system: Separate supply and exhaust ducts run the length of the tunnel, giving uniform fresh-air supply and exhaust extraction at every point (best for long, heavily trafficked tunnels).

(a) Lacey regime width and normal scour depth

Regime (wetted) perimeter / linear waterway:

$$P_w = 4.75\sqrt{Q} = 4.75\sqrt{1600} = 4.75 \times 40 = 190\ \text{m}$$ $$\boxed{\text{Linear waterway} \approx 190\ \text{m}}$$

Normal (regime) scour depth, Lacey:

$$R = 0.473\left(\frac{Q}{f}\right)^{1/3}$$ $$\frac{Q}{f} = \frac{1600}{1.10} = 1454.5$$ $$\left(1454.5\right)^{1/3} = 11.33\quad(\text{since }11.33^3 \approx 1454.4)$$ $$R = 0.473 \times 11.33 = 5.36\ \text{m}$$ $$\boxed{\text{Normal scour depth } R \approx 5.36\ \text{m}}$$

(b) Pier scour and foundation depth

Maximum scour depth at pier (factor 2.0 for pier nose):

$$D_{scour} = 2.0 \times R = 2.0 \times 5.36 = 10.72\ \text{m below HFL}$$ $$\boxed{D_{scour,pier} \approx 10.7\ \text{m}}$$

Depth of foundation: As per IRC practice, the foundation should be taken at least an additional grip length (≈ 1.33 × max scour for piers, but minimum embedment of about a 1/3 of max scour below the scour line). Taking foundation level = $1.33 \times R_{max-pier}$ below HFL as a conservative grip:

$$D_{found} \approx 1.33 \times 10.72 = 14.3\ \text{m below HFL}$$

(Equivalently, provide foundation a minimum grip of $\tfrac{1}{3}\times$ max scour, i.e. ≈ 3.6 m, below the maximum scour line at 10.7 m, giving ≈ 14.3 m.)

$$\boxed{\text{Found. depth} \approx 14.3\ \text{m below HFL}}$$

(Answer any three — all four given for completeness.)

1. Alligator (fatigue) cracking

Description: Interconnected cracks forming a pattern resembling an alligator's hide, in wheel paths. Cause: Repeated traffic-load-induced tensile fatigue at the underside of the bituminous layer, aggravated by weak/saturated subgrade and inadequate pavement thickness. Remedy: If localised — full-depth patch; if extensive — structural strengthening with a bituminous overlay after improving drainage/subgrade.

2. Rutting

Description: Longitudinal surface depressions along the wheel paths. Cause: Permanent (plastic) deformation accumulating in one or more layers due to heavy/channelised loading, high temperature, low-stability mix (excess binder / rounded aggregate), or inadequate compaction. Remedy: Mill the rutted surface and re-lay a stable, well-compacted bituminous mix; improve mix design (lower binder, angular aggregate).

3. Pumping (rigid pavements)

Description: Ejection of water and fine subgrade material through joints/cracks/edges under passing wheels, leaving voids beneath the slab. Cause: Presence of free water + fine-grained (erodible) subgrade + heavy repeated loads; voids cause loss of support and corner cracking. Remedy: Provide a non-erodible granular/treated subbase (DLC), ensure good drainage, and seal joints; restore support by sub-sealing (grouting).

4. Pothole formation

Description: Bowl-shaped holes in the bituminous surface. Cause: Progressive loss of surface material; begins where water enters cracks/ravelled spots, weakens the layer, and traffic dislodges the material. Remedy: Clean and dry the hole, apply tack coat, fill with premix and compact (proper pothole patching); seal surrounding cracks to stop water ingress.