BE Civil Engineering (IOE, TU) Transportation Engineering II (IOE, CE 704) Question Paper 2079 Nepal

(a) Group Index method and computation

The Group Index (GI) is an empirical number (range 0–20) used in the AASHTO/HRB classification to rate a subgrade soil for highway use. A lower GI means a better subgrade; a higher GI indicates poorer load-supporting capacity and therefore a thicker pavement is required. It is computed from the fines content, liquid limit and plasticity index.

$$GI = 0.2a + 0.005\,ac + 0.01\,bd$$

where, taking $P=$ % passing 0.075 mm:

• $a = (P-35)$, limited to $0\le a\le 40$
• $b = (P-15)$, limited to $0\le b\le 40$
• $c = (\mathrm{LL}-40)$, limited to $0\le c\le 20$
• $d = (\mathrm{PI}-10)$, limited to $0\le d\le 20$

Step 1 — evaluate the partial terms ($P=60$, $LL=45$, $PI=22$):

• $a = 60-35 = 25$ (within 0–40) → $a=25$
• $b = 60-15 = 45 \Rightarrow$ capped at $40$ → $b=40$
• $c = 45-40 = 5$ (within 0–20) → $c=5$
• $d = 22-10 = 12$ (within 0–20) → $d=12$

Step 2 — substitute:

$$GI = 0.2(25) + 0.005(25)(5) + 0.01(40)(12)$$ $$GI = 5.0 + 0.625 + 4.8 = 10.425$$

Rounded to the nearest whole number, GI ≈ 10.

(b) Quality, thickness and pavement layers

A GI of about 10 falls in the poor-to-fair range (good subgrades have GI 0–1; GI of 9–20 = poor). For medium traffic the GI design chart calls for a comparatively large total thickness (of the order of 50–60 cm) of pavement above this subgrade because of its high fines and plasticity.

Four structural layers of a flexible pavement (top to bottom):

Answer: GI ≈ 10.4 (poor-to-fair subgrade, large pavement thickness required).

Westergaard wheel-load stresses

(a) Radius of relative stiffness

$$l=\left[\dfrac{E h^{3}}{12\,(1-\mu^{2})\,k}\right]^{0.25}$$ $$l=\left[\dfrac{3\times10^{5}\times 22^{3}}{12(1-0.15^{2})(6.0)}\right]^{0.25}=\left[\dfrac{3\times10^{5}\times 10648}{12(0.9775)(6.0)}\right]^{0.25}$$ $$l=\left[\dfrac{3.1944\times10^{9}}{70.38}\right]^{0.25}=(4.5388\times10^{7})^{0.25}$$ $$\boxed{l \approx 82.08\ \text{cm}}$$

(b) Equivalent radius of resisting section

Since $a = 15\ \text{cm} < 1.724\,h = 1.724(22)=37.93\ \text{cm}$, use

$$b=\sqrt{1.6a^{2}+h^{2}}-0.675h=\sqrt{1.6(15)^2+22^2}-0.675(22)$$ $$b=\sqrt{360+484}-14.85=\sqrt{844}-14.85=29.05-14.85$$ $$\boxed{b \approx 14.20\ \text{cm}}$$

(c) Edge and corner stresses

Edge load stress (Westergaard, loaded circular area at edge):

$$\sigma_{e}=\dfrac{0.572\,P}{h^{2}}\Big[4\log_{10}\!\Big(\dfrac{l}{b}\Big)+0.359\Big]$$ $$\dfrac{l}{b}=\dfrac{82.08}{14.20}=5.780,\qquad \log_{10}(5.780)=0.7619$$ $$\sigma_{e}=\dfrac{0.572(5100)}{22^{2}}\big[4(0.7619)+0.359\big]=\dfrac{2917.2}{484}\,[3.0476+0.359]$$ $$\sigma_{e}=6.027\times 3.4066$$ $$\boxed{\sigma_{e}\approx 20.53\ \text{kg/cm}^2}$$

Corner load stress (Westergaard):

$$\sigma_{c}=\dfrac{3P}{h^{2}}\left[1-\left(\dfrac{a\sqrt{2}}{l}\right)^{0.6}\right]$$ $$\dfrac{a\sqrt2}{l}=\dfrac{15(1.4142)}{82.08}=\dfrac{21.213}{82.08}=0.2585,\quad (0.2585)^{0.6}=0.4396$$ $$\sigma_{c}=\dfrac{3(5100)}{484}\,[1-0.4396]=31.61\times0.5604$$ $$\boxed{\sigma_{c}\approx 17.71\ \text{kg/cm}^2}$$

(Using $0.6$ exponent gives $\sigma_c\approx 17.7\ \text{kg/cm}^2$.) The edge stress (≈ 20.5 kg/cm²) is the most critical for this load case and governs the flexural-strength check of the slab.

Railway superelevation

(a) Definitions

• Superelevation (cant): the amount by which the outer rail is raised above the inner rail on a curve so that the resultant of the train weight and centrifugal force acts perpendicular to the plane of the rails, balancing the lateral thrust at the design speed.
• Cant deficiency: the additional cant (over the actual provided cant) that would be needed to fully balance the centrifugal force when a train runs faster than the equilibrium speed. It represents the unbalanced lateral acceleration felt by passengers and is limited (75 mm for BG) for comfort and safety.

(b) Equilibrium cant

$$e=\dfrac{G\,V^{2}}{127\,R}\quad (V\text{ in km/h},\ e,G\text{ in m})$$ $$e=\dfrac{1.750\times 90^{2}}{127\times 875}=\dfrac{1.750\times 8100}{111125}=\dfrac{14175}{111125}=0.1276\ \text{m}$$ $$\boxed{e \approx 127.6\ \text{mm}}$$

(c) Maximum permissible speed (with cant deficiency)

The maximum speed corresponds to actual cant $e$ plus deficiency $C_d=0.075\ \text{m}$:

$$e+C_d=\dfrac{G\,V_{max}^{2}}{127R}\ \Rightarrow\ V_{max}=\sqrt{\dfrac{(e+C_d)\,127\,R}{G}}$$ $$V_{max}=\sqrt{\dfrac{(0.1276+0.075)(127)(875)}{1.750}}=\sqrt{\dfrac{(0.2026)(111125)}{1.750}}$$ $$V_{max}=\sqrt{\dfrac{22513.9}{1.750}}=\sqrt{12865.1}$$ $$\boxed{V_{max} \approx 113.4\ \text{km/h}}$$

(d) Martin's safe-speed check

$$V=4.4\sqrt{R-70}=4.4\sqrt{875-70}=4.4\sqrt{805}=4.4(28.37)$$ $$\boxed{V_{Martin} \approx 124.8\ \text{km/h}}$$

Comment: Martin's safe speed (≈ 124.8 km/h) is greater than the cant-deficiency-limited maximum speed (≈ 113.4 km/h). Hence the governing maximum permissible speed for this curve is ≈ 113 km/h, controlled by the cant-deficiency (passenger-comfort) criterion rather than by curve safety.

IRC:37 design traffic

(a) ESAL concept and fourth-power law

Different axle loads damage a pavement by different amounts. The Equivalent Standard Axle Load (ESAL) converts every axle to an equivalent number of passes of a standard 80 kN (8.16 t) single axle that would cause the same damage. By the fourth-power law, the damaging effect varies roughly with the fourth power of the axle load:

$$\text{Damage factor} = \left(\dfrac{\text{axle load}}{\text{standard axle load}}\right)^{4}$$

The vehicle damage factor (VDF, $F$) is the average number of standard axles per commercial vehicle obtained from axle-load surveys.

(b) Cumulative standard axles

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

Step 1 — growth factor:

$$(1+0.075)^{15}=1.075^{15}=2.9589\ \Rightarrow\ (1+r)^n-1=1.9589$$ $$\dfrac{(1+r)^n-1}{r}=\dfrac{1.9589}{0.075}=26.119$$

Step 2 — cumulative repetitions of CVPD over life:

$$365\times1500\times26.119=1.4300\times10^{7}\ \text{commercial-vehicle passes}$$

Step 3 — apply $D$ and $F$:

$$N=1.4300\times10^{7}\times0.75\times2.5=2.681\times10^{7}\ \text{standard axles}$$ $$\boxed{N \approx 26.8\ \text{msa}}$$

(c) Use in design

The pair (design traffic $N \approx 26.8$ msa, subgrade CBR) is entered into the IRC:37 design catalogue/charts. For the given CBR (e.g. 5–8%) and ~27 msa, the chart fixes the total pavement thickness and the thickness of each layer — bituminous surfacing (BC + DBM), granular base (WMM) and granular sub-base (GSB) — such that the cumulative fatigue (bottom of bituminous layer) and rutting (top of subgrade) strains stay within allowable limits for the design life.

Benkelman beam evaluation

(a) Principle

The Benkelman beam is a simple lever (probe + pivot + dial gauge). The probe tip is placed between the dual rear wheels of a loaded standard truck (8.16 t rear axle, specified tyre pressure). As the truck slowly moves away, the loaded pavement surface rebounds (recovers) and the upward movement of the probe is magnified by the lever and read on the dial gauge. The rebound deflection is a measure of the elastic (recoverable) response of the whole pavement–subgrade system; a large rebound deflection means a structurally weak pavement with inadequate remaining life.

(b) Characteristic deflection

Step 1 — mean:

$$\bar{x}=\dfrac{1.20+1.35+1.10+1.50+1.25+1.40+1.30+1.15+1.45+1.55}{10}=\dfrac{13.25}{10}=1.325\ \text{mm}$$

Step 2 — standard deviation (sample, $n-1$):

Deviations squared (mm²): $0.0156,\ 0.0006,\ 0.0506,\ 0.0306,\ 0.0056,\ 0.0056,\ 0.0006,\ 0.0306,\ 0.0156,\ 0.0506$

$$\sum(x-\bar x)^2 = 0.2065\ \text{mm}^2$$ $$\sigma=\sqrt{\dfrac{0.2065}{10-1}}=\sqrt{0.022944}=0.1515\ \text{mm}$$

Step 3 — characteristic deflection:

$$D_c=\bar{x}+2\sigma=1.325+2(0.1515)=1.325+0.303$$ $$\boxed{D_c \approx 1.63\ \text{mm}}$$

(c) Overlay requirement

Since the corrected characteristic deflection $D_c \approx 1.63\ \text{mm}$ exceeds the allowable deflection of $1.00\ \text{mm}$, the existing pavement is structurally deficient and a bituminous overlay is required.

Overlay thickness procedure (IRC:81): the corrected characteristic deflection $D_c$ and the design traffic (msa) are entered in the IRC:81 deflection–traffic overlay design chart, which directly reads off the required bituminous-macadam-equivalent overlay thickness. Greater $D_c$ (relative to the allowable value) and higher traffic both demand a thicker overlay; the thickness is then converted to the chosen surfacing layers (e.g. DBM + BC).

Runway length corrections (ICAO)

Given: basic length $L_0 = 1800\ \text{m}$, elevation $=600\ \text{m}$, ART $=27.0^{\circ}\text{C}$, effective gradient $=0.5\%$.

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

$$\text{increase}=0.07\times\dfrac{600}{300}=0.14\;(14\%)$$ $$L_1=1800\,(1+0.14)=2052.0\ \text{m}$$

Step 2 — Temperature correction (1% per °C of rise above standard)

Standard atmospheric temperature at 600 m:

$$T_{std}=15-0.0065\times600=15-3.9=11.1^{\circ}\text{C}$$

Temperature rise $=27.0-11.1=15.9^{\circ}\text{C}$

$$\text{increase}=0.01\times15.9=0.159\;(15.9\%)$$ $$L_2=2052.0\,(1+0.159)=2378.3\ \text{m}$$

Step 3 — Check combined elevation + temperature limit (≤ 35%)

$$\dfrac{L_2-L_0}{L_0}=\dfrac{2378.3-1800}{1800}=0.3213=32.13\% \ \le\ 35\%\ \;\text{✓ OK}$$

Step 4 — Gradient correction (20% per 1% of effective gradient)

$$\text{increase}=0.20\times0.5=0.10\;(10\%)$$ $$L_3=2378.3\,(1+0.10)=2616.1\ \text{m}$$ $$\boxed{\text{Corrected runway length} \approx 2616\ \text{m}\ (\text{say } 2620\ \text{m})}$$

Note: the elevation+temperature correction (32.13%) is within the 35% advisory limit, so the design is acceptable without a special site investigation.

(a) Economic span

For a long bridge of fixed total length, as the span increases the number of piers decreases (substructure cost falls) but each superstructure span becomes more expensive. The economic span is the one giving minimum total cost per unit length.

Let $L$ = span (m). Per unit length of bridge:

• Substructure cost contributes $\dfrac{P}{L}$ (one pier of cost $P$ spread over each span length $L$).
• Superstructure cost per metre of deck $= c\,L$ (given $c=$ Rs 8000 per m of deck per m of span).

Total cost per unit length:

$$C(L)=\dfrac{P}{L}+cL$$

Minimise: $\dfrac{dC}{dL}=-\dfrac{P}{L^{2}}+c=0$

$$\Rightarrow \boxed{L_{eco}=\sqrt{\dfrac{P}{c}}}$$

This is the classic result: the economic span occurs when the cost of the superstructure of one span equals the cost of one substructure (pier).

Computation:

$$L_{eco}=\sqrt{\dfrac{32{,}00{,}000}{8000}}=\sqrt{400}=20\ \text{m}$$ $$\boxed{L_{eco}=20\ \text{m}}$$

(b) Main components of a bridge

        ___parapet/railing___
       |  deck slab + wearing coat  |   <-- SUPERSTRUCTURE
   [bearing]====girders/beams====[bearing]
   ============================================
   ||  pier      ||           || abutment ||    <-- SUBSTRUCTURE
   ||  + cap     ||           || + wing wall||
   [pier foundation]        [abutment foundation]

Superstructure (carries and transmits traffic load to bearings): deck slab + wearing coat, girders/beams (or trusses/arch), parapet & railing, expansion joints, bearings.

Substructure (transfers load to the ground): abutments with wing walls (at ends), piers with pier caps (intermediate), and the foundations (open/pile/well) of piers and abutments. Approach embankments and river-training works are appurtenant structures.

Tunnel ventilation

(a) Need and methods

Need: A tunnel is an enclosed space; vehicle exhaust accumulates carbon monoxide (CO), oxides of nitrogen, smoke and heat, and oxygen is depleted. Ventilation is required to (i) dilute and remove toxic CO/exhaust gases to safe limits, (ii) clear smoke (especially in a fire emergency for safe evacuation), and (iii) remove heat and supply fresh oxygen for comfort and visibility.

Two methods of mechanical ventilation:

1. Longitudinal ventilation (air pushed along the tunnel axis by jet/booster fans or a portal blower).
2. Transverse ventilation (separate supply and exhaust ducts deliver/extract air uniformly along the length); a combination gives semi-transverse ventilation.

(b) Air quantity

Step 1 — vehicles inside the tunnel at any instant. Time each vehicle spends in the tunnel $=\dfrac{L}{v}=\dfrac{1.5\ \text{km}}{40\ \text{km/h}}=0.0375\ \text{h}$. Number inside $=$ flow $\times$ travel time:

$$N=1200\ \tfrac{\text{veh}}{\text{h}}\times0.0375\ \text{h}=45\ \text{vehicles}$$ $$\boxed{N=45\ \text{vehicles in the tunnel}}$$

Step 2 — total CO emission inside.

$$Q_{CO}=N\times0.0025=45\times0.0025=0.1125\ \text{m}^3/\text{min}$$

Step 3 — fresh-air requirement (dilute CO to permissible fraction $0.0025$):

$$Q_{air}=\dfrac{Q_{CO}}{\text{permissible fraction}}=\dfrac{0.1125}{0.0025}=45\ \text{m}^3/\text{min}$$

Converting to per second:

$$Q_{air}=\dfrac{45}{60}=0.75\ \text{m}^3/\text{s}$$ $$\boxed{Q_{air}=0.75\ \text{m}^3/\text{s}}$$

Sleepers and ballast

(a) Sleeper density

Sleeper density is the number of sleepers provided per rail length. It is expressed as $(M+x)$, where $M$ is the rail length in metres (numerically) and $x$ is a number (typically 3 to 7) that depends on traffic, axle load, type of ballast and the strength of the formation. A higher $x$ gives more sleepers per rail length (a stronger track).

(b) Number of sleepers

Rail length $M = 13$, $x = 7$:

$$\text{Sleepers per rail length}=M+x=13+7=13+7=20$$ $$\boxed{20\ \text{sleepers per 13 m rail}}$$

Number of rail lengths per km $=\dfrac{1000}{13}=76.92$.

$$\text{Sleepers per km}=\dfrac{1000}{13}\times20=1538.5\approx 1539$$ $$\boxed{\approx 1539\ \text{sleepers per km}}$$

(c) Functions

Sleepers: hold the two rails to correct gauge and alignment; receive load from the rails and distribute it over a wider area to the ballast; provide a firm, level seat for the rails (with bearing plates/fastenings); give longitudinal and lateral stability to the track and act as elastic support absorbing vibration.

Ballast: transmits and distributes the sleeper load to the formation over a still larger area; holds the sleepers in position resisting lateral, longitudinal and vertical movement; provides drainage of the track; gives elasticity/resilience and helps maintain level and line during packing/tamping.

Material tests for bituminous pavements

(a) Aggregate tests

(Other acceptable tests: water absorption, soundness, flakiness/elongation index, stripping/adhesion.)

(b) Bitumen (binder) tests

(Other acceptable tests: viscosity, flash & fire point, specific gravity.)

(a) Joints in cement-concrete pavement

(b) Common distresses

Flexible pavements:

• Cracking — alligator/fatigue cracking (interconnected cracks from repeated traffic load over a weak base/subgrade); also longitudinal/transverse cracks.
• Rutting — longitudinal depressions in the wheel paths from permanent deformation of bituminous/granular layers and subgrade under repeated loading. (Also: raveling, potholes, bleeding, shoving.)

Rigid pavements:

• Cracking — corner/transverse/longitudinal cracking from load + curling stresses or loss of subgrade support.
• Pumping & faulting — ejection of fines-laden water at joints under load, leading to loss of support and a step (faulting) between adjacent slabs. (Also: scaling, spalling at joints, blow-up.)