Horizontal Alignment of Highways - IES Previous Year Questions

21. An ideal horizontal transition curve is a _____ .

[IES 2012]

  1. Parabola
  2. hyperbola
  3. Clothoid spiral
  4. Circle
Answer: Option C
Explanation:
IRC suggests the use of spiral as the horizontal transition curve for highways because,
  • The spiral curve satisfies the requirements of an ideal transition.
  • The setting out of the spiral curve is simpler and easier when compared with other transition curves.

22.  The transition property of a lemniscate curve is disrupted when its deflection angle is around _____ .

[IES 2012]

  1. 45°
  2. 60°
  3. 90°
  4. 30°
Answer:
Explanation:

23.  If R is the radius of the curve and L is the length of the long chord, the shift of the curve is _____ (all in meter units).

[IES 2012]

  1. L2 / R
  2. L2 / 2R
  3. L2 / 24R
  4. L2 / 6R
Answer: Option C
Explanation:
The shift of the curve, S =
L2 / 24R

Where,
L - Length of the long chord.
R - Radius of the curve.

24.  For a bituminous carriageway surface having 2% camber and design speed of 100 kmph, radius beyond which super elevation is not essential is nearly _____ .

[IES 2013]

  1. 1500 m
  2. 1800 m
  3. 2200 m
  4. 1100 m
Answer: Option C
Explanation:
Design speed, v = 100 kmph = 27.78 m/s
The camber provided will act as super elevation.
Super elevation, e =
(0.75v)2 / gR

⇒ 0.02 =
(0.75 x 27.78)2 / 9.81 x R

⇒ R = 2212.5 m
Here, the closest option is 2200 m.

25.  The rate of super elevation for a horizontal curve of radius 500 m in a national highway for a design speed of 100 kmph is _____ .

[IES 2013]

  1. 0.063
  2. 0.07
  3. 0.70
  4. 0.04
Answer: Option C
Explanation:
Design speed, v = 100 kmph = 27.78 m/s
Super elevation, e =
(0.75v)2 / gR

⇒ e =
(0.75 x 27.78)2 / 9.81 x 500
= 0.088
But, maximum super elevation for plain terrain, emax = 0.07
Since e > emax, e = 0.07.
Now, e + f =
v2 / gR

⇒ 0.07 + f =
27.782 / 9.81 x 500

⇒ f = 0.087 < 0.15
Hence, super elevation to be provided = 0.07

26.  The maximum super elevation to be provided on a road curve is 1 in 15. If the rate of change of super elevation is specified as 1 in 120 and the road width is 10 m, then the minimum length of the transition curve on each end will be _____ .

[IES 2015]

  1. 100 m
  2. 80 m
  3. 180 m
  4. 120 m
Answer: Option B
Explanation:
Super elevation, e = 1/15
Rate of change of super elevation, N = 120
Width of the road, W = 10 m
Thus, minimum length of transition length assuming that the pavement is rotated with respect to inner edge,
L = eNW =
1 / 15
x 120 x 10
⇒ L = 80 m

27.  The radius of a horizontal circular curve is 480 m and design speed therein 70 kmph. What will be the equilibrium super elevation for the pressures on the inner and outer wheels to be equal ?

[IES 2015]

  1. 6%
  2. 7%
  3. 8%
  4. 5%
Answer: Option C
Explanation:
Design speed, v = 70 kmph = 19.44 m/s
Equilibrium super elevation, eequil =
v2 / gR
=
19.442 / 9.81 x 480

⇒ eequil = 0.08 (or) 8%

28.  The following purposes served by a transition curve in a highway alignment include:

  1. Gradual introduction of the centrifugal force on moving vehicles from zero on the straight alignment to a constant final value on the circular curve. 
  2. Enabling the gradual introduction of super elevation on the roadway. 

Select the correct answer using the codes given below:

[IES 2017]

  1. 2 only
  2. Both 1 and 2
  3. Neither 1 nor 2
  4. 1 only
Answer: Option B
Explanation:
The purpose of providing transition curves are as follows:
  • Gradual introduction of centrifugal force.
  • Gradual introduction of super elevation.
  • Provision of comfort and safety to the passengers.

29.  The rate of equilibrium super elevation on a road is _____ .

  1. Directly proportional to the square of vehicle velocity.
  2. Inversely proportional to the radius of the horizontal curve.
  3. Directly proportional to the square of the radius of the horizontal curve.

Which of the above statements are correct ?

[IES 2018]

  1. 1 and 3 only
  2. 2 and 3 only
  3. 1, 2 and 3
  4. 1 and 2 only
Answer: Option D
Explanation:
The equilibrium super elevation, eequil =
v2 / gR

Where,
v - Design speed of vehicle.
g - Acceleration due to gravity.
R - Radius of the horizontal curve.

30.  What is the extra widening required (as nearest magnitude) for a pavement of 7 m width on a horizontal curve of radius 200 m, if the longest wheel of vehicle expected on the road is 6.5 m and the design speed is 65 kmph ?

[IES 2019]

  1. 0.5 m
  2. 0.7 m
  3. 0.9 m
  4. 0.3 m
Answer: Option B
Explanation:
Mechanical widening, Wem =
nL2 / 2R
=
2 x 6.52 / 2 x 200

⇒ Wem = 0.21 m
Pyschological widening, Wep =
v / 9.5 √  R 
=
65 / 9.5 x √  200 

⇒ Wep= 0.48 m
Thus, total extra widening required, We = Wem + Wep = 0.21 + 0.48
⇒ We = 0.69 m ≈ 0.70 m
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