Highway planning miscellaneous


Highway planning miscellaneous

  1. The data given below pertain to the design of a flexible pavement.
    Initial trafic = 1213 cvpd
    Traffic growth rate = 8 percent per annum
    Design life = 12 years
    Vehicle damage factor = 2.5
    Distribution factor = 1.0
    The design traffic in terms of million standard axles (msa) to be catered would be









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

    =
    365 × 1213 [ (1 + 0.08)12 - 1]
    × 2.5
    0.08

    = 21.005 × 106 ≈ 21 msa

    Correct Option: C

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

    =
    365 × 1213 [ (1 + 0.08)12 - 1]
    × 2.5
    0.08

    = 21.005 × 106 ≈ 21 msa


  1. The design value of lateral friction coefficient on highway is









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    As per IRC, Coeff of lateral friction = 0.15
    Coeff of longitudinal friction = 0.35 to 0.4

    Correct Option: D

    As per IRC, Coeff of lateral friction = 0.15
    Coeff of longitudinal friction = 0.35 to 0.4



  1. Design rate of super elevation for horizontal highway curve of radius 450 m for a mixed traffic condition, having a speed of 125 km/hour is









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    For mixed traffic,

    e =
    (0.75V)2
    ; V in m / s
    gR

    Design speed = 125 kmph
    = 125 ×
    5
    = 34.7 m /s
    18

    e =
    (0.7V)2
    gR

    =
    (0.75 × 34.7)2
    = 0.154
    9.81 × 450

    Correct Option: D

    For mixed traffic,

    e =
    (0.75V)2
    ; V in m / s
    gR

    Design speed = 125 kmph
    = 125 ×
    5
    = 34.7 m /s
    18

    e =
    (0.7V)2
    gR

    =
    (0.75 × 34.7)2
    = 0.154
    9.81 × 450


  1. The ruling minimum radius of horizontal curve of a national highway in pljain terrain for a design speed of 100 km/hour with e = 0.07 and f = 0.15 is close to









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    Design speed = 100 kmph

    = 100 ×
    5
    m/s = 28 m /s
    18

    e + f =
    v2
    gR

    0.07 + 0.15 =
    282
    9.8 × R

    ∴ R ≈ 360 m

    Correct Option: B

    Design speed = 100 kmph

    = 100 ×
    5
    m/s = 28 m /s
    18

    e + f =
    v2
    gR

    0.07 + 0.15 =
    282
    9.8 × R

    ∴ R ≈ 360 m



  1. Width of carriage way for a single lane is recommended to be









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    Cland Carriage wayWidth
    Single wad 3.75 m
    Multilane parement 3.5 m/lane
    Two lane (without raised Kerb) 7m
    Two lane (with raised kerb) 7.5 m
    Intermediate carriage way 5.5 m

    Correct Option: C

    Cland Carriage wayWidth
    Single wad 3.75 m
    Multilane parement 3.5 m/lane
    Two lane (without raised Kerb) 7m
    Two lane (with raised kerb) 7.5 m
    Intermediate carriage way 5.5 m