Traffic engineering miscellaneous


Traffic engineering miscellaneous

Traffic Engineering

  1. A linear relationship is observed between speed and density on a certain section of a highway. The free flow speed is observed to be 80 km per hour and the jam density is estimated as 100 vehicles per km length. Based on the above relationship, the maximum flow expected on this section and the speed at the maximum flow will respectively be









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    qmax =
    Free flow speed × Jam density
    4

    =
    80 × 100
    = 2000 vehicles / hour
    4

    Speed( At qmax ) =
    Free flow speed
    2

    =
    80
    = 40 kmph
    2

    Correct Option: D

    qmax =
    Free flow speed × Jam density
    4

    =
    80 × 100
    = 2000 vehicles / hour
    4

    Speed( At qmax ) =
    Free flow speed
    2

    =
    80
    = 40 kmph
    2


  1. It is proposed to widen and strengthen an existing 2-lane NH section as a divided highway. The existing traffic in one direction is 2500 commercial vehicles (CV) per day. The construction will take 1 year. The design CBR of soil subgrade is found to be 4 percent. Given: traffic growth rate for CV = 8 percent, vehicle damage factor = 3.5 (standard axles per CV), design life = 10 years and traffic distribution factor = 0.75. The cumulative standard axles (msa) computed are









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    Cumulative axle load, (CSA)

    Ns =
    365A[(1 + r)n - 1]
    × F
    r

    = 365 × 3750 × 1 +
    8
    10 - 1 × 3.5 ≈ 70 msa
    100
    8
    100

    Correct Option: D

    Cumulative axle load, (CSA)

    Ns =
    365A[(1 + r)n - 1]
    × F
    r

    = 365 × 3750 × 1 +
    8
    10 - 1 × 3.5 ≈ 70 msa
    100
    8
    100



  1. A roundabout is provided with an average entry width of 8.4 m, width of weaving section as 14 m and length of the weaving section between channelizing islands as 35 m. The crossing traffic and total traffic on the weaving section are 1000 and 2000 PCU per hour respectively. The nearest rounded capacity of the roundabout (in PCU per hour is)









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    Proportion of weaving traffic =
    1000
    = 0.5
    2000

    Capacity , QP = 280w1 +
    e
    (1 - P / 3)
    w
    1 +
    w
    L

    = 280 × 141 +
    8.4
    1 - 0.5
    143
    1 +
    14
    35

    = 3733.3 ≈ 3700 pcu

    Correct Option: B

    Proportion of weaving traffic =
    1000
    = 0.5
    2000

    Capacity , QP = 280w1 +
    e
    (1 - P / 3)
    w
    1 +
    w
    L

    = 280 × 141 +
    8.4
    1 - 0.5
    143
    1 +
    14
    35

    = 3733.3 ≈ 3700 pcu


  1. The capacities of "One-way 1.5 m wide sidewalk (persons per hour)" and "One-way 2-lane urban road (PCU per hour, with no frontage access, no standing vehicles and very little cross traffic) are respectively









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    1 way side walk – 1200 persons/hour 1 way 2 lane side walk – 2400 persons/hour

    Correct Option: A

    1 way side walk – 1200 persons/hour 1 way 2 lane side walk – 2400 persons/hour