Vertical Alignment of Highways - IES Previous Year Questions

1. A summit curve is formed at the intersection of a 3% up gradient and 5% down gradient. To provide a stopping distance of 128 m, the length of the summit curve needed will be _____ .

[IES 1995, IES 2005]

  1. 298 m
  2. 322 m
  3. 340 m
  4. 271 m
Answer: Option A
Explanation:
N1 = 0.03, N2 = - 0.05
Deviation angle, N = | N1 - N2 | = | 0.03 - (-0.05) | = 0.08
For Stopping Sight Distance,
Height of driver’s eye, H = 1.2 m
Height of obstacle, h = 0.15 m
Assuming L > SSD, L =
NS2 / (√  2H  + √  2h  )2

⇒ L =
0.08 x 1282 / (√  2 x 1.2  + √  2 x 0.15  )2
= 297.89 m > SSD
Hence, Length of summit curve = 297.89 m

2. Which of the following are the criteria associated with the design of sag vertical curve ?

  1. Provision of minimum stopping sight distance.
  2. Adequate drainage.
  3. Comfortable operation. 
  4. Pleasant appearance.

Select the correct answer using the codes given below:

[IES 1996]

  1. 2 and 3
  2. 2, 3 and 4
  3. 1 and 3
  4. 1, 2 and 4
Answer: Option B
Explanation:
The design of the valley (sag) curve depends on
  • Rider's comfort
  • Headlight Sight Distance
  • Drainage at lowest point
  • Aesthetic appearance

3.  Which of the criteria given below are used for the design of valley vertical curves on roads ?

  1. Rider comfort
  2. Headlight sight distance
  3. Drainage

Select the correct answer using the codes given below:

[IES 1998]

  1. 1 and 3
  2. 2 and 3
  3. 1 and 2
  4. 1, 2 and 3
Answer: Option D
Explanation:
The design of the valley curve depends on
  • Rider's comfort
  • Headlight Sight Distance
  • Drainage at lowest point
  • Aesthetic appearance

4.  If a descending gradient of 1 in 25 meets an ascending gradient of 1 in 40, then the length of valley curve required for a headlight sight distance of 100 m will be _____ .

[IES 1999]

  1. 130 m
  2. 310 m
  3. 630 m
  4. 30 m
Answer: Option A
Explanation:
N1 = - 1/25, N2 = 1/40
Deviation angle, N = | N1 - N2 | = | (-1/25) - (1/40) | = 0.065
For Headlight Sight Distance,
Height of headlight, h1 = 0.75 m
Angle of Headlight, α = 1°, tan 1° = 0.0175
Assuming L > HSD, L =
NS2 / 2h1 + 2S tan α

⇒ L =
0.065 x 1002 / (2 x 0.75) + (2 x 100 x 0.0175)
= 130 m > HSD
Hence, Length of valley curve = 130 m

5.  An ascending gradient of 1 in 100 meets a descending gradient of 1 in 50. The length of the summit curve required to provide an overtaking sight distance of 500 m will be _____ .

[IES 1999]

  1. 781 m
  2. 470 m
  3. 170 m
  4. 938 m
Answer: Option A
Explanation:
N1 = 1/100, N2 = - 1/50
Deviation angle, N = | N1 - N2 | = | (1/100) - (-1/50) | = 0.03
For Overtaking Sight Distance,
Height of driver’s eye, H = 1.2 m
Height of obstacle, h = 1.2 m
Assuming L > OSD, L =
NS2 / (√  2H  + √  2h  )2

⇒ L =
0.03 x 5002 / (√  2 x 1.2  + √  2 x 1.2  )2
= 781.25 m > OSD
Hence, Length of summit curve = 781.25 m

6.  The camber provided on a sloping road is 1 in 48. Which one of the following is the ruling gradient ?

[IES 2003]

  1. 1 in 20
  2. 1 in 24
  3. 1 in 30
  4. 1 in 15
Answer: Option C
Explanation:
As per IRC, the ruling gradient for plain and rolling (sloping) terrain is 1 in 30.

7.  Which one of the following are the accepted criteria for design of valley curve for highways ?

  1. Headlight sight distance
  2. Passing and non-passing sight distance
  3. Aesthetic consideration
  4. Motorist comfort
  5. Drainage control

Select the correct answer using the codes given below:

[IES 2003]

  1. 1, 3, 4 and 5
  2. 2, 3, 4 and 5
  3. 1 and 5
  4. 1, 2, 3 and 4
Answer: Option A
Explanation:
The design of the valley curve depends on
  • Rider's comfort
  • Headlight Sight Distance
  • Drainage at lowest point
  • Aesthetic appearance

8.  A rising gradient of 1 in 50 meets a falling gradient of 1 in 30. Which one of the following is the length of the vertical curve if the stopping sight distance is 120 m?

[IES 2006]

  1. 158 m
  2. 140 m
  3. 120 m
  4. 174 m
Answer: Option D
Explanation:
N1 = 1/50, N2 = - 1/30
Deviation angle, N = | N1 - N2 | = | (1/50) - (- 1/30) | = 0.053
For Stopping Sight Distance,
Height of driver’s eye, H = 1.2 m
Height of obstacle, h = 0.15 m
Assuming L > SSD, L =
NS2 / (√  2H  + √  2h  )2

⇒ L =
0.053 x 1202 / (√  2 x 1.2  + √  2 x 0.15  )2
= 174.5 m > SSD
Hence, Length of vertical curve = 174.5 m

9.  A 3% downgrade curve is followed by a 1% upgrade curve and rate of change of grade adopted is 0.1% per 20 m length. The length of the respective vertical curve is _____ .

[IES 2011]

  1. 200 m
  2. 100 m
  3. 400 m
  4. 800 m
Answer: Option D
Explanation:
N1 = - 0.03, N2 = 0.01
Thus, total deviation angle N = | N1 - N2 | = |-0.03 - 0.01 | = 0.04 = 4%
Rate of change of gradient = 0.1% per 20 m
For 0.1% change in gradient, length = 20 m.
Hence, for 4% change in gradient, length =
20 / 0.1
x 4 = 800 m

10.  While aligning a hill road with a ruling gradient of 6%, a horizontal curve of 75 m radius is encountered. The compensated gradient at the curve will be _____ .

[IES 2014]

  1. 2%
  2. 3%
  3. 5%
  4. 1%
Answer: Option C
Explanation:
Grade Compensation =
30 + R / R
≤
75 / R

30 + R / R
=
30 + 75 / 75
= 1.4%
75 / R
=
75 / 75
= 1.0%
Thus, grade compensation = 1.0%
Compensated gradient = Ruling gradient - grade compensation
⇒ Compensated gradient = 6 - 1 = 5%
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