Simple Circular Curve and its uses in the fieldpptx

Simple Circular Curve

1. Simple Circular Curve 1

2. Obstruction Tangent 1 Tangent 2 R R Δ : Deflection Angle Simple Circular Curve ( Simple Horizontal Curve) or (Simple Curve ) A curve with single arc and constant radius connecting two tangents. This type of curve is mostly used in roads and highways. R = Radius of simple curve (it is in meters or feet) If degree of curve (D) is given then : R = meters  

3. Sketch & Elements of a Simple Circular Curves PC PT PI R R I Back tangent (T 1 ) Forward tangent (T 2 ) PI : Point of Intersection of two tangents.. 3. Points: PC : Point of curve or point of commencement of curve; start point of the curve. PT : Point of tangency; end point of the curve. Tangents: Back Tangents or Original Straight Line (T 1 ) Forward Tangents or Deflected line (T2) T T LC O Δ : Deflection Angle 2. Deflection Angle ( Δ ): It is angle of change in direction. It is an external angle from line T 1 to T 2 as shown in sketch. Δ + I = 180 Where I = Angle of Intersection

4. Sketch & Elements of a Simple Circular Curves PC PT PI R R I Back tangent (T 1 ) Forward tangent (T 2 ) T T 5. Tangent length (T): The distance from point PC to PI (or PI to PT ) is called the tangent length , T. Distance from PC to PI and PI and PT are equal. T = R tan ( )   LC O Δ : Deflection Angle 4 . Length of Curve (L): It the curved length from point PC to PT. It the length (arc length) of simple circular curve. L = R is in meters or feet Δ = Deflection angle in degrees.  

5. PC PT PI R R Δ I M E Δ Back tangent Forward tangent Δ /2 T T LC Sketch & Elements of a Simple Circular Curves 6. Long Chord (LC): The line connecting the PC and PT is the long chord LC. LC = 2 R Sin ( )   7 . Mid Ordinate (M): The middle ordinate M is the perpendicular distance from the midpoint of the long chord to the curve’s midpoint . M= R (1 - Cos ( ) )  

6. PC PT PI R R I Back tangent (T 1 ) Forward tangent (T 2 ) T T LC O Δ : Deflection Angle 8 . Chainages): If chainage of pint PI (point of intersection) is given, then Chainage of PC = Chainage of PI – T (tangent length) Chainage of PT = Chainage of PC + L (length of curve)

7. Problem No: 1 Compute all the setting out Parameters (length of curve, tangent length, long chord and mid ordinate) of a simple circular curve and Chainages at start and end point of curve with the help of following data. Radius of simple curve = 500 meters Angle of intersection= 60 ᵒ Chainage of point of intersection (PI)= 1560 meters. 7

8. Problem No: 1 Compute all the setting out Parameters (length of curve, tangent length, long chord and mid ordinate) of a simple circular curve and Chainages at start and end point of curve with the help of following data. Radius of simple curve = 500 meters Angle of intersection= 60 ᵒ Chainage of point of intersection (PI)= 1560 meters. 8

9. PC PT PI R R I T 1 T 2 T T LC O Δ : Deflection Angle Solution of Problem 1 : Radius of simple curve = R= 500 meters Angle of intersection= I = 60 ᵒ Chainage of point of intersection (PI)= 1560 meters. Deflection Angle = Δ Δ + I = 180ᵒ Δ = 180 - I = 180 - 60 = 120ᵒ Settiong Out Parameters: 1. Length of Curve (L): L = L = = 1046.67 meters.  

10. 10 2. . Tangent length (T): T = R tan ( ) = 500 x tan (120/2) = 866.02 meters   3 . Long Chord (LC): LC = 2 R Sin ( ) LC = 2x500x Sin ( ) = 866.02 meters   4 . Mid Ordinate (M): M= R (1 - Cos ( ) ) = 500 (1 - Cos ( ) ) = 250 meters.   5 . Chainages: Chainage of PC = Chainage of PI – T (tangent length) = 1560-866.02= 693.98 meters Chainage of PT = Chainage of PC + L (length of curve) = 693.98 + 1046.67 = 1740.65 meters.

11. Problem No: 2 Draw the sketch and calculate all the setting out Parameters and Chainages at start and end point of a simple horizontal curve of a railway track with the help of following data. Degree of simple curve = 5 ᵒ curve Angle of intersection= 100 ᵒ Chainage of point of intersection (PI)= 1610 meters. 11

12. Problem No: 3 In a simple circular curve, an angle of deflection is 28 ° 2 4’, the station of the PI is 6+340 meters, and terrain conditions require the minimum radius permitted by the specifications is 286 meters. Calculate the PC and PT stationing and the external and middle ordinate distances for this curve.