Highway planning miscellaneous Difficult Questions and Answers | Page - 2

Highway planning miscellaneous

The extra widening required for a two-lane national highway at a horizontal curve of 300 m radius, considering a wheel base of 8 m and a design speed of 100 kmph is

1. 0.42 m
1. 0.62 m
1. 0.82 m
1. 0.92 m

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Extra widening = nL² + V
2R 9.5√R

= 2 × 8² + 100
2 × 300 9.5√300

= 0.82 m

Correct Option: C

Extra widening = nL² + V
2R 9.5√R

= 2 × 8² + 100
2 × 300 9.5√300

= 0.82 m

Using IRC: 37-1984 “Guidelines for the Design of Flexible Pavements” and the following data, choose the total thickness of the pavement.
No. of commercial vehicles when construction is completed = 2723 veh/day
Annual growth rate of the traffic = 5.0%
Design life of the pavement = 10 years
Vehicle damage factor = 2.4
CBR value of the subgrade soil = 5%
Data for 5% CBR value

1. 620 mm
1. 640 mm
1. 670 mm
1. 700 mm

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Axle load = 365 × A × [(1 + r)ⁿ - 1] × F
Y

= 365 × 2723 × [(1 + 0.05)¹⁰ - 1] × F
0.05

= 30 × 10⁶axle load = 30 msa
For 30 msa, thickness = 670 mm

Correct Option: C

Axle load = 365 × A × [(1 + r)ⁿ - 1] × F
Y

= 365 × 2723 × [(1 + 0.05)¹⁰ - 1] × F
0.05

= 30 × 10⁶axle load = 30 msa
For 30 msa, thickness = 670 mm

A vehicle moving at 60 kmph on an ascending gradient of a highway has to come to stop position to avoid collision with a stationary object.The ratio of lag to brake distance is 6: 5. Considering total reaction time of the driver as 2.5 seconds and the coefficient of longitudinal friction as 0.36, the value of ascending gradient (%) is

1. 3.3
1. 4.8
1. 5.3
1. 6.8

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Lag distance = 60 × 5 × 2.5 = 41.67 m
18

Lag distance = 6
Brake distance 5

∴ Brake distance(s) = 41.67 × 5 = 34.725 m
6

S = v²
2gƒ + g sin θ

∴ 2ƒ + sinθ = v²
Sg

∴ sinθ = v² - 2ƒ
Sg

= 16.672 - 2 × 0.36 = 34.75
34.75 × 9.81

∴ θ = 4.8%

Correct Option: B

Lag distance = 60 × 5 × 2.5 = 41.67 m
18

Lag distance = 6
Brake distance 5

∴ Brake distance(s) = 41.67 × 5 = 34.725 m
6

S = v²
2gƒ + g sin θ

∴ 2ƒ + sinθ = v²
Sg

∴ sinθ = v² - 2ƒ
Sg

= 16.672 - 2 × 0.36 = 34.75
34.75 × 9.81

∴ θ = 4.8%

The co-efficient of friction in the longitudinal direction of a highway is estimated as 0.396. The breaking distance for a new car moving at a speed of 65 km/hr is

1. 87 m
1. 45 m
1. 42 m
1. 40 m

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Braking distance,
S = V²
2gƒ

= 18.055²
2 × 9.81 × 0.396

≈ 42 m

Correct Option: C

Braking distance,
S = V²
2gƒ

= 18.055²
2 × 9.81 × 0.396

≈ 42 m

For a 25 cm thick cement concrete pavement, analysis of stresses gives the following values Wheel load stress due to corner loading 30 kg/cm² Wheel load stress due to edge loading 32 kg/cm²
Warping stress at corner region during summer 9 kg/cm²
Warping stress at corner region during winter 7 kg/cm²
Warping stress at edge region durng summer 8 kg/cm²
Warping stress at edge region during winter 6 kg/cm²
Frictional stress during winter 5 kg/cm²
Frictional stress during winter 4 kg/cm²
The most critical stress value for this pavement is

1. 40 kg/cm²
1. 42 kg/cm²
1. 44 kg/cm²
1. 45 kg/cm²

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During Summer At edge
S_R = Wheel load (edge) + warping stress (edge) – friction resistance (–ve)
= 32 + 8 – 5 = 35 km/m²
At Corner
S_R = Wheel load + warping stress
= 30 + 9 = 39 kg/cm²
During winter
At edge
S_R = Wheel load + warping stress + frictional resistance (+ve)
= 32 + 6 + 4 = 42 kg/m²
At corner
S_R = Wheel load + warping stress
= 30 + 7 = 37 kg/m²
The critical value is 42 kg/m²

Correct Option: B

During Summer At edge
S_R = Wheel load (edge) + warping stress (edge) – friction resistance (–ve)
= 32 + 8 – 5 = 35 km/m²
At Corner
S_R = Wheel load + warping stress
= 30 + 9 = 39 kg/cm²
During winter
At edge
S_R = Wheel load + warping stress + frictional resistance (+ve)
= 32 + 6 + 4 = 42 kg/m²
At corner
S_R = Wheel load + warping stress
= 30 + 7 = 37 kg/m²
The critical value is 42 kg/m²