TGPSC HMWSSB Manager Transportation Engineering previous year questions with answers
Every Transportation Engineering question asked in the TGPSC HMWSSB Manager (Engineering) papers on this site, grouped by topic. Answers come from TSPSC's final key; each has a short explanation.
Practise Transportation Engineering onlineHighway Engineering
In functional classification of highways, which one of the following highway type have highest mobility and less accessibility
- National Highways
- State Highways
- Major District roads
- Street and Village Roads
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Answer: (1) National Highways
In the functional hierarchy, arterial roads such as National Highways are meant for long-distance through traffic, so they give the highest mobility with restricted access. Mobility falls and accessibility rises down the hierarchy, with village roads and local streets giving the most access.
For the calculation of stopping distance, the longitudinal friction co-efficient values of ______ have been recommended by Indian Roads Congress.
- 0.35 to 0.40
- 0.15 to 0.20
- 0.40 to 0.45
- 0.45 to 0.50
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Answer: (1) 0.35 to 0.40
IRC recommends a longitudinal friction coefficient ranging from 0.40 at low speeds down to 0.35 at high speeds for calculating braking distance in SSD. Values of 0.15 to 0.20 are of the order used for lateral friction on curves, not braking.
Calculate the value of lag distance in SSD for a highway with a design speed of 65 kmph?
- 45.18 m
- 36.14 m
- 32.50 m
- 451.80 m
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Answer: (1) 45.18 m
Lag distance = 0.278 V t, with the IRC reaction time t = 2.5 s. So lag distance = 0.278 × 65 × 2.5 = 45.18 m.
The psychological widening of pavement is calculated using which one of the following formula (take V = speed of vehicle in kmph; R = radius of the curve in m)
- V²/(9.5√R)
- V/(9.5√R)
- V/(9.5R)
- V³/(9.5R²)
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Answer: (2) V/(9.5√R)
IRC gives psychological (extra) widening on horizontal curves as W_ps = V/(9.5√R) m. It is added to the mechanical widening nl²/(2R) to get the total widening.
Find out the rate of change of centrifugal acceleration for a design speed of 75 kmph, using IRC recommended formula.
- 0.63 m/sec³
- 0.53 m/sec³
- 0.48 m/sec³
- 0.73 m/sec³
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Answer: (2) 0.53 m/sec³
IRC formula: C = 80/(75 + V) = 80/(75 + 75) = 0.53 m/sec³. This lies within the IRC limits of 0.5 to 0.8 m/sec³, so it is adopted.
In a flexible pavement, the different materials with the CBR values are available as follows: 80%, 60%, 15% and 4%. Indicate the order (top to bottom) in which the materials are to be placed for making a good pavement
- 4%, 15%, 60%, 80%
- 4%, 60%, 80%, 15%
- 80%, 60%, 15%, 4%.
- 4%, 80%, 15%, 60%
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Answer: (3) 80%, 60%, 15%, 4%.
In a flexible pavement the wheel stresses are highest at the surface and spread out with depth, so the strongest material (highest CBR) goes on top and weaker layers below. Hence 80% (base), 60%, 15% (sub-base) and 4% (subgrade) from top to bottom.
Calculate the Equivalent Axle load factor (EALF) for single axle load of 10 tons using fourth power formula.
- 2.26
- 22.60
- 1.23
- 11.23
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Answer: (1) 2.26
By the fourth power rule, EALF = (axle load/standard axle load)⁴. With the standard single axle of 8.16 t (80 kN), EALF = (10/8.16)⁴ = 1.2255⁴ ≈ 2.26.
The hardness and toughness properties of a road aggregate will be obtained from
- Aggregate Impact Test
- Aggregate crushing test
- Los Angeles Abrasion Test
- Aggregate Shape test
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Answer: (3) Los Angeles Abrasion Test
In the Los Angeles abrasion test the aggregate is tumbled with steel balls in a rotating drum, so it is subjected to abrasion (testing hardness) and impact (testing toughness) at the same time. The impact test measures toughness only and the crushing test measures strength.
In the rigid pavement fatigue analysis, the ratio of flexural stress due to load and the flexural strength due to concrete is less than 0.45 indicates
- the allowable number of repetitions of the axle loads is infinity
- the allowable number of repetitions of the axle loads is zero
- the allowable number of repetitions of the axle loads is 4500
- the allowable number of repetitions of the axle loads is 45000
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Answer: (1) the allowable number of repetitions of the axle loads is infinity
In IRC:58 fatigue analysis, when the stress ratio (flexural stress/modulus of rupture) is below 0.45 the concrete can withstand an unlimited number of load repetitions. Such axle loads therefore consume no fatigue life.
Traffic Engineering
Webster's equation for computing saturation flow rate in signal design (S = saturation flow rate PCU/hour; W = carriage way width in meters.)
- S = 225 W PCU/hour
- S = 525 W PCU/hour
- S = 550 W PCU/hour
- S = 250 W PCU/hour
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Answer: (2) S = 525 W PCU/hour
For an approach width of about 5.5 m to 18 m, Webster's method takes the saturation flow as S = 525 W PCU per hour of green, where W is the approach width in metres.
In total reaction time of the driver, the time required for the sensations received by the eyes/ears to be transmitted to the brain through the nervous system and spinal cord is called ______.
- Intellection time
- Emotion time
- Volition time
- Perception time
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Answer: (4) Perception time
In the PIEV theory, perception time is the time taken for the sensations received by the eyes or ears to be carried to the brain through the nervous system and spinal cord. Intellection is the time to understand the situation, emotion the time spent on fear, anger, etc., and volition the time to decide and act.
Estimate the theoretical capacity of a traffic lane with traffic flow stream speed of 50 kmph (take average centre to centre spacing of vehicles as 10 m)
- 500 vehicles /hour/lane
- 5000 vehicles /hour/lane
- 3000 vehicles /hour/lane
- 300 vehicles /hour/lane
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Answer: (2) 5000 vehicles /hour/lane
Theoretical capacity C = 1000 V/S, with V in kmph and spacing S in m. C = 1000 × 50/10 = 5000 vehicles/hour/lane.
Railways & Airports
The gauge widths (in meter) for broad, standard and narrow gauges respectively are
- 1.767, 1.650 and 0.760
- 1.676, 1.500 and 0.676
- 1.676, 1.435 and 0.762
- 1.876, 1.656 and 0.800
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Answer: (3) 1.676, 1.435 and 0.762
Broad gauge is 1676 mm, standard gauge is 1435 mm and narrow gauge (2 ft 6 in) is 762 mm; metre gauge is 1000 mm.
For airports serving big aircrafts, ICAO recommends the cross wind component should not exceed ______.
- 25 kmph
- 35 kmph
- 15 kmph
- 23 kmph
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Answer: (2) 35 kmph
ICAO limits the crosswind component to about 37 km/h (20 knots) for large aircraft that need long runways (reference field length of 1500 m or more), which Indian textbooks quote as 35 kmph. Smaller aircraft are allowed lower crosswind limits.
The monthly mean of average daily temperature for the hottest month of the year is 30°C and the monthly mean of the maximum daily temperature for the same month of the year is 45°C, Find out the airport reference temperature
- 25°C
- 35°C
- 40°C
- 15°C
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Answer: (2) 35°C
Airport reference temperature = Ta + (Tm − Ta)/3, where Ta is the mean of average daily temperature and Tm the mean of maximum daily temperature of the hottest month. ART = 30 + (45 − 30)/3 = 35°C.
The minimum value of Composite Sleeper Index (CSI) prescribed on Indian railways for track sleeper is
- 783
- 1352
- 1455
- 873
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Answer: (1) 783
Indian Railways prescribe a minimum composite sleeper index of 783 for timber track sleepers. The higher values 1352 and 1455 are the minimums for crossing sleepers and bridge sleepers respectively.
The formula used for calculation of superelevation on railways is ______ (take G = gauge in meters, V = Speed in kmph; R = radius of curve in meters)
- GV²/(127R) m
- GV²/(127R) cm
- GV³/(127R) m
- GV²/(225R) m
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Answer: (1) GV²/(127R) m
For equilibrium on a curve e/G = v²/(gR); with v = V/3.6 m/s and g = 9.81 m/s² this gives e = GV²/(127R), since 3.6² × 9.81 ≈ 127. With G and R in metres the superelevation e also comes out in metres, so option (2) has the wrong unit.