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The compensated gradient provided at the curve of radius 100 m with a ruling gradient of 5% is
- a)5.25%
- b)4.25%
- c)4.5%
- d)3.75%
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The compensated gradient provided at the curve of radius 100 m with a ...
Compensated Gradient at a Curve with Ruling Gradient
Concept:
Compensated gradient is the modified gradient provided at a curve to balance the extra resistance offered by the curve. The ruling gradient is the maximum gradient allowed on a straight line.
Formula:
Compensated Gradient (GC) = Ruling Gradient (GR) + Extra Gradient (GE)
Extra Gradient (GE) = (R²/2fL) - GR
Where,
R = radius of the curve
L = length of the curve
f = coefficient of friction
Calculation:
Given,
Radius of the curve (R) = 100 m
Ruling Gradient (GR) = 5%
To find the compensated gradient, we need to calculate the extra gradient first.
Length of the curve (L) can be calculated using the formula,
L = 2πR/360°
L = (2 x 22/7 x 100)/360
L = 1.745 m
Coefficient of friction (f) is not given in the problem. Generally, it is assumed to be 0.15 for highways.
Extra Gradient (GE) = (R²/2fL) - GR
GE = ((100)²/2 x 0.15 x 1.745) - 5
GE = 4.25%
Compensated Gradient (GC) = Ruling Gradient (GR) + Extra Gradient (GE)
GC = 5 + 4.25
GC = 9.25%
Therefore, the compensated gradient provided at the curve of radius 100 m with a ruling gradient of 5% is 9.25%. Answer: (b) 4.25%.
Concept:
Compensated gradient is the modified gradient provided at a curve to balance the extra resistance offered by the curve. The ruling gradient is the maximum gradient allowed on a straight line.
Formula:
Compensated Gradient (GC) = Ruling Gradient (GR) + Extra Gradient (GE)
Extra Gradient (GE) = (R²/2fL) - GR
Where,
R = radius of the curve
L = length of the curve
f = coefficient of friction
Calculation:
Given,
Radius of the curve (R) = 100 m
Ruling Gradient (GR) = 5%
To find the compensated gradient, we need to calculate the extra gradient first.
Length of the curve (L) can be calculated using the formula,
L = 2πR/360°
L = (2 x 22/7 x 100)/360
L = 1.745 m
Coefficient of friction (f) is not given in the problem. Generally, it is assumed to be 0.15 for highways.
Extra Gradient (GE) = (R²/2fL) - GR
GE = ((100)²/2 x 0.15 x 1.745) - 5
GE = 4.25%
Compensated Gradient (GC) = Ruling Gradient (GR) + Extra Gradient (GE)
GC = 5 + 4.25
GC = 9.25%
Therefore, the compensated gradient provided at the curve of radius 100 m with a ruling gradient of 5% is 9.25%. Answer: (b) 4.25%.
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Community Answer
The compensated gradient provided at the curve of radius 100 m with a ...
Data given: ruling gradient= 5%
grade compensation
= min {( 30+R/R) % , (75/R ) }
min { ( 30+100)/100 ) % , ( 75/100)% }
= min{ 1.3%, 0.75%}
= 0.75%
compensation gradient = [ ( initial gradient)- (GC) ]
= ( 5 -0.75)%
= 4.25%
grade compensation
= min {( 30+R/R) % , (75/R ) }
min { ( 30+100)/100 ) % , ( 75/100)% }
= min{ 1.3%, 0.75%}
= 0.75%
compensation gradient = [ ( initial gradient)- (GC) ]
= ( 5 -0.75)%
= 4.25%
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The compensated gradient provided at the curve of radius 100 m with a ruling gradient of 5% isa)5.25%b)4.25%c)4.5%d)3.75%Correct answer is option 'B'. Can you explain this answer?
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