Vertical Alignment and Gradients Questions and Answers

Q: At sharp horizontal curves on highways radius 'R' (in meters), the percentage reduction in gradient, provided to compensate the loss of traction force due to curvature is:

The maximum grade compensation is 75/R per cent.On a horizontal curve, extra tractive effort is needed to overcome curve resistance, so IRC requires the gradient to be eased. The grade compensation is (30 + R)/R per cent, subject to a maximum of 75/R per cent, where R is the radius in metres. Compensation is not needed for gradients flatter than 4 per cent.50/R, 100/R and 125/R are not the IRC limit.

Q: As per the recommendations of Indian Road Congress, the ruling gradient in plain is:

The IRC ruling gradient in plain terrain is 1 in 30.IRC recommends, for plain and rolling terrain, a ruling gradient of 3.3 per cent (1 in 30), a limiting gradient of 5 per cent (1 in 20) and an exceptional gradient of 6.7 per cent (1 in 15). The ruling gradient is the one adopted for design wherever possible.1 in 20 and 1 in 15 are the limiting and exceptional values for plains, and 1 in 45 is flatter than any IRC design category.

Q: An ascending gradient 1 in 100 meets a descending gradient of 1 in 120. The deviation angle for designing of summit curve is:

The deviation angle is 11/600.The deviation angle is the algebraic difference of the gradients. An ascending gradient is positive and a descending one negative:\[ \begin{aligned} N &= \frac{1}{100} - \left(-\frac{1}{120}\right) \\ &= \frac{6 + 5}{600} = \frac{11}{600} \end{aligned} \]Subtracting the magnitudes instead gives \(\frac{1}{600}\), the trap for forgetting the sign. \(\frac{1}{12000}\) is the product of the two gradients, which has no meaning here.

At sharp horizontal curves on highways radius 'R' (in meters), the percentage reduction in gradient, provided to compensate the loss of traction force due to curvature is:

A.50/R
B.75/R
C.100/R
D.125/R

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

The maximum grade compensation is 75/R per cent.

On a horizontal curve, extra tractive effort is needed to overcome curve resistance, so IRC requires the gradient to be eased. The grade compensation is (30 + R)/R per cent, subject to a maximum of 75/R per cent, where R is the radius in metres. Compensation is not needed for gradients flatter than 4 per cent.

50/R, 100/R and 125/R are not the IRC limit.

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As per the recommendations of Indian Road Congress, the ruling gradient in plain is:

A.1 in 15
B.1 in 20
C.1 in 30
D.1 in 45

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

The IRC ruling gradient in plain terrain is 1 in 30.

IRC recommends, for plain and rolling terrain, a ruling gradient of 3.3 per cent (1 in 30), a limiting gradient of 5 per cent (1 in 20) and an exceptional gradient of 6.7 per cent (1 in 15). The ruling gradient is the one adopted for design wherever possible.

1 in 20 and 1 in 15 are the limiting and exceptional values for plains, and 1 in 45 is flatter than any IRC design category.

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An ascending gradient 1 in 100 meets a descending gradient of 1 in 120. The deviation angle for designing of summit curve is:

A.1/600
B.11/600
C.1/12000
D.- 1/12000

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

The deviation angle is 11/600.

The deviation angle is the algebraic difference of the gradients. An ascending gradient is positive and a descending one negative:

\[ \begin{aligned} N &= \frac{1}{100} - \left(-\frac{1}{120}\right) \\ &= \frac{6 + 5}{600} = \frac{11}{600} \end{aligned} \]

Subtracting the magnitudes instead gives \(\frac{1}{600}\), the trap for forgetting the sign. \(\frac{1}{12000}\) is the product of the two gradients, which has no meaning here.

Discuss in Forum

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Vertical Alignment and Gradients Questions and Answers

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