GATE Civil Engineering Transportation Engineering Syllabus

Transportation Engineering syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Transportation Engineering in GATE Civil Engineering, not a summary of it.

1. 1

   Transportation Infrastructure

   3 topics

   • Geometric design of highways
     • Cross-sectional elements
     • Sight distances
     • Horizontal alignments
     • Vertical alignments
   • Geometric design of railway Track
     • Speed
     • Cant
   • Concept of airport runway length, calculations and corrections
     • Taxiway and exit taxiway design
2. 2

   Highway Pavements

   3 topics

   • Highway materials
     • Desirable properties and tests
   • Desirable properties of bituminous paving mixes
   • Design factors for flexible and rigid pavements
     • Design of flexible pavement using IRC codes
     • Design of rigid pavement using IRC codes
3. 3

   Traffic Engineering

   6 topics

   • Traffic studies on flow and speed
     • Peak hour factor
     • Accident study
     • Statistical analysis of traffic data
   • Microscopic and macroscopic parameters of traffic flow
     • Fundamental relationships
   • Traffic signs
   • Signal design by Webster’s method
   • Types of intersections
   • Highway capacity

Transportation Engineering flashcards for GATE Civil Engineering

19 of 50 cards from the Transportation Engineering deck — real questions with worked answers.

1. In highway geometric design, what is the difference between gradient compensation at curves and ruling gradient?

   Ruling gradient (design gradient) is the maximum gradient used in ordinary design (e.g., $1 \text{ in } 30$ in plains). Grade compensation at a horizontal curve reduces the gradient to offset the extra resistance from the curve, given by $\text{Grade compensation} = \frac{30 + R}{R}\,\%$, subject to a maximum of $\frac{75}{R}\,\%$, and is not applied on grades flatter than $4\%$.

2. Define camber (cross slope) of a road and give the IRC-recommended value for bituminous/concrete pavements in heavy rainfall areas.

   Camber is the transverse slope provided to the road surface to drain off rainwater to the sides. For bituminous and cement concrete pavements, IRC recommends a camber of about $1.7\%$ to $2\%$ (i.e., $1 \text{ in } 60$ to $1 \text{ in } 50$) in heavy rainfall regions.

3. What are the cross-sectional elements of a highway?

   The main cross-sectional elements are: carriageway (traffic lanes), shoulders, camber, kerbs, medians (central reserve), road margins (footpath, parking lane, cycle track, frontage road), right of way, side drains, and the building line / control line. A single traffic lane is taken as $3.75\,\text{m}$ wide for a single-lane road.

4. State the formula for Stopping Sight Distance (SSD) on a level road and define each term.

   $$\text{SSD} = v\,t + \frac{v^{2}}{2gf}$$ where $v$ = design speed in $\text{m/s}$, $t$ = reaction time (usually $2.5\,\text{s}$), $g$ = $9.81\,\text{m/s}^{2}$, and $f$ = design coefficient of longitudinal friction (about $0.35$ to $0.40$). In $\text{km/h}$: $\text{SSD} = 0.278\,Vt + \frac{V^{2}}{254f}$.

5. How is Stopping Sight Distance modified on a gradient?

   On a gradient the braking distance changes because gravity adds to or resists braking: $$\text{SSD} = vt + \frac{v^{2}}{2g(f \pm 0.01n)}$$ where $+$ is used for an ascending gradient and $-$ for a descending gradient, with $n$ being the gradient in percent.

6. What is Overtaking Sight Distance (OSD) and its formula for a two-way road?

   OSD is the minimum distance visible to a driver to safely overtake another vehicle. $$\text{OSD} = d_{1} + d_{2} + d_{3}$$ where $d_{1} = v_{b}t$ (reaction), $d_{2} = v_{b}T + 2s$ with $s = 0.7v_{b} + 6$ and $T = \sqrt{\frac{4s}{a}}$, and $d_{3} = vT$ (opposing vehicle). $v_b$ is the speed of the overtaken vehicle, $v$ the design speed.

7. Define superelevation and give its design equation for highways.

   Superelevation ($e$) is the transverse raising of the outer edge of a road at a horizontal curve to counteract centrifugal force. $$e + f = \frac{v^{2}}{gR} = \frac{V^{2}}{127R}$$ where $V$ is in $\text{km/h}$, $R$ in metres, and $f$ is the lateral friction coefficient ($\approx 0.15$). For design, $e$ is found for $75\%$ of design speed ignoring $f$.

8. What is the IRC maximum value of superelevation and how is it computed in design practice?

   Maximum superelevation in plain/rolling terrain is $e_{max} = 0.07$ ($7\%$); in hilly (non-snowbound) areas up to $0.10$. Design superelevation is taken as $e = \frac{V^{2}}{225R}$ (i.e., for $75\%$ of design speed, neglecting friction). If this exceeds $e_{max}$, $e_{max}$ is used and friction checked.

9. What is the formula for the minimum radius of a horizontal curve?

   $$R_{min} = \frac{V^{2}}{127(e_{max} + f)}$$ where $V$ is the design speed in $\text{km/h}$, $e_{max}$ the maximum allowable superelevation, and $f$ the maximum lateral friction coefficient ($\approx 0.15$).

10. Why are transition curves provided on highways and what are their functions?

    A transition curve gradually changes radius from infinity (straight) to the circular curve radius. Functions: (1) gradual introduction of centrifugal force / superelevation, (2) enable smooth steering, (3) allow gradual application of superelevation and extra widening, (4) improve aesthetics and comfort. The ideal transition is a spiral (clothoid).

11. State the formula for the length of a transition curve by the rate of change of centrifugal acceleration method.

    $$L_{s} = \frac{0.0215\,V^{3}}{C\,R}$$ where $V$ is in $\text{km/h}$, $R$ in metres, and $C = \frac{80}{75 + V}$ (with $0.5 \leq C \leq 0.8$) is the allowable rate of change of centrifugal acceleration in $\text{m/s}^3$.

12. Give the formula for extra widening of pavement on horizontal curves.

    $$W_{e} = W_{m} + W_{ps} = \frac{n l^{2}}{2R} + \frac{V}{9.5\sqrt{R}}$$ where $W_m$ is mechanical widening, $W_{ps}$ psychological widening, $n$ = number of lanes, $l$ = wheelbase ($\approx 6\,\text{m}$), $R$ = radius (m), $V$ = speed ($\text{km/h}$).

13. Classify vertical curves and state which equation governs a summit curve length when SSD < L.

    Vertical curves are summit (convex) and valley (sag/concave) curves, designed as parabolas. For a summit curve when length $L >$ SSD ($S$): $$L = \frac{N S^{2}}{2(\sqrt{H} + \sqrt{h})^{2}}$$ where $N$ = algebraic deviation of grades, $H = 1.2\,\text{m}$ (driver eye height), $h = 0.15\,\text{m}$ (object height).

14. For a summit curve, give the length formula when SSD > L.

    When $S > L$: $$L = 2S - \frac{2(\sqrt{H} + \sqrt{h})^{2}}{N}$$ With $H = 1.2\,\text{m}$ and $h = 0.15\,\text{m}$, this becomes $L = 2S - \frac{4.4}{N}$, and the case $S < L$ gives $L = \frac{N S^{2}}{4.4}$.

15. What criteria govern valley (sag) curve length design?

    Valley curves are designed for two criteria, and the larger length is adopted: (1) comfort criterion (limiting rate of change of centrifugal acceleration), and (2) headlight sight distance criterion (driver must see the road illuminated by headlights at night). Headlight height is taken as $0.75\,\text{m}$ with a $1^{\circ}$ upward beam.

16. Give the comfort (allowable rate) formula for valley curve length.

    By the comfort criterion: $$L = 2\left(\frac{N v^{3}}{C}\right)^{1/2} = 0.38\,(N V^{3})^{1/2}$$ where $N$ is the deviation angle, $V$ in $\text{km/h}$, and $C$ the allowable rate of change of centrifugal acceleration ($\approx 0.6\,\text{m/s}^3$).

17. In railway track geometry, define gauge and give the Indian Broad Gauge value.

    Gauge is the clear minimum distance between the inner (running) faces of the two rails of a track. Indian Broad Gauge (BG) is $1676\,\text{mm}$ ($1.676\,\text{m}$). Metre Gauge (MG) is $1000\,\text{mm}$ and Narrow Gauge is $762\,\text{mm}$ or $610\,\text{mm}$.

18. Define cant (superelevation) in railways and give its formula.

    Cant is the amount by which the outer rail is raised above the inner rail on a curve to counteract centrifugal force. $$e = \frac{G v^{2}}{g R} = \frac{G V^{2}}{127 R}$$ where $G$ = dynamic gauge (m), $V$ = speed ($\text{km/h}$), $R$ = radius (m). For BG, $G \approx 1.75\,\text{m}$.

19. What is equilibrium cant and cant deficiency in railways?

    Equilibrium cant is the superelevation at which the resultant of weight and centrifugal force is perpendicular to the track for the equilibrium (weighted average) speed. Cant deficiency ($C_d$) is the difference between the cant required for the maximum (booked) speed and the equilibrium cant actually provided; IRC/IRPWM limits $C_d$ to $76\,\text{mm}$ (BG) and $51\,\text{mm}$ (MG).

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