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For a circular curve of radius 200m, the coefficient of lateral friction is 0.15 and the design speed is 40 kmph. The equilibrium super elevation would be
  • a)
    21.3
  • b)
    7
  • c)
    6.3
  • d)
    4.6
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
For a circular curve of radius 200m, the coefficient of lateral fricti...
The super elevation at which pressure on outer and inner wheels is same is called equilibrium super elevation.
Friction force = 0
 
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Most Upvoted Answer
For a circular curve of radius 200m, the coefficient of lateral fricti...
Given:
Radius of circular curve (R) = 200m
Coefficient of lateral friction (f) = 0.15
Design speed (V) = 40 kmph

To find:
Equilibrium super elevation (e)

Formula used:
e = (V²)/(127Rf+3g)

where,
g = 9.81 m/s² (acceleration due to gravity)

Calculation:
Converting design speed from kmph to m/s
V = (40 * 1000) / 3600
V = 11.11 m/s

Substituting the given values in the formula, we get
e = (11.11²)/(127*200*0.15+3*9.81)
e = 6.3%

Therefore, the equilibrium super elevation is 6.3%.
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For a circular curve of radius 200m, the coefficient of lateral friction is 0.15 and the design speed is 40 kmph. The equilibrium super elevation would bea)21.3b)7c)6.3d)4.6Correct answer is option 'C'. Can you explain this answer?
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