Highway planning miscellaneous
Direction: For a portion of national highway where a descending gradient of 1 in 25 meets with an ascending gradient of 1 in 20, a valley curve needs to be designed for a vehicle travelling at 90 kmph based on the following conditions.
1. head light sight distance equal to the stopping sight distance (SSD) of a level terrain considering length of valley curve > SSD.
2. comfort condition with allowable rate of change of centrifugal acceleration = 0.5 m/sec³. Assume total reaction time = 2.5 seconds; co-efficient of longitudinal friction of the pavement = 0.35; height of head light of the vehicle = 0.75 m; and beam angle = 1°.
- What is the length of valley curve (in m) based on the comfort condition?
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L = NS² (1.50 0.035S) L = 0.09(153.6636)² = 308.9641 m (1.50 0.035 153.6636) Lv = 2 
NV³ 
1/2 C = 2 0.09 × (0.278 × 90)³ 1/2 = 106.066 m 0.5 Correct Option: B
L = NS² (1.50 0.035S) L = 0.09(153.6636)² = 308.9641 m (1.50 0.035 153.6636) Lv = 2 NV³ 1/2 C = 2 0.09 × (0.278 × 90)³ 1/2 = 106.066 m 0.5
Direction: A horizontal circular curve with a centre line radius of 200 m is provided on a 2-lane, 2-way SH section. The width of the 2-lane road is 7.0 m. Design speed for this section is 80 km per hour. The brake reaction time is 2.4 s, and the coefficients of friction in longitudinal and lateral directions are 0.355 and 0.15, respectively.
- The safe stopping sight distance on the section is
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V = 80 × 5 = 22.2 m/s 18 SSD = vt + v² 2gƒ = 22.2 × 2.5 + (22.2)² ≈ 125 m 2 × 9.81 × 0.355 Correct Option: C
V = 80 × 5 = 22.2 m/s 18 SSD = vt + v² 2gƒ = 22.2 × 2.5 + (22.2)² ≈ 125 m 2 × 9.81 × 0.355
- The set-back distance from the centre line of the inner lane is
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Setback distance from centre line = R – R cos α/2
α = S ⇒ α = S = S × 180 degree R 2 2R 2R π 
∴ Setback distance = R - R.cos S × 180 2πR = 200 - 200 cos 125 × 180 = 9.6 m 2π × 200 Correct Option: C
Setback distance from centre line = R – R cos α/2
α = S ⇒ α = S = S × 180 degree R 2 2R 2R π ∴ Setback distance = R - R.cos S × 180 2πR = 200 - 200 cos 125 × 180 = 9.6 m 2π × 200
- The perception-reaction time for a vehicle travelling at 90 km/h, given the coefficient of longitudinal friction of 0.35 and the stopping sight distance of 170 m (assume g = 9.81 m/s²), is ____________ seconds.
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3.1 to 3.2
SSD = 0.278Vt + (0.278V)² 2gf HO = 0.278 × 90 × t + (0.278 × 90)² 2 × 9.81 × 0.35
∴ t = 3.151 sCorrect Option: C
3.1 to 3.2
SSD = 0.278Vt + (0.278V)² 2gf HO = 0.278 × 90 × t + (0.278 × 90)² 2 × 9.81 × 0.35
∴ t = 3.151 s
- In a Marshall sample, the bulk specific gravity of mix and aggregates are 2.324 and 2.546 respectively. The sample includes 5% of bitumen (by total weight of mix) of specific gravity 1.10. The theoretical maximum specific gravity of mix is 2.441. The void filled with bitumen (VFB) in the Marshall sample (in%) is __________
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62 to 68
Vv = Gt - Gm × 100 Gm = 2.441 - 2.324 × 100 = 5.03 % 2.324 Vb = Gm Wb = 2.324 × 5 = 10.564 Gb 1.1
VMB = Vv + Vb = 15.594%VFB = Vb × 100 VMb = 10.564 × 100 = 67.7% 15.564 Correct Option: B
62 to 68
Vv = Gt - Gm × 100 Gm = 2.441 - 2.324 × 100 = 5.03 % 2.324 Vb = Gm Wb = 2.324 × 5 = 10.564 Gb 1.1
VMB = Vv + Vb = 15.594%VFB = Vb × 100 VMb = 10.564 × 100 = 67.7% 15.564
