1 / 20

Islamic University of Gaza Civil Engineering Department Surveying II ECIV 2332 By B elal A lmassri

Islamic University of Gaza Civil Engineering Department Surveying II ECIV 2332 By B elal A lmassri. Chapter 9 Route Surveying – Part 4 . Obstacles to curve location. Inaccessible point of intersection. Chaining obstacles. Vision obstacles. Transition curves. Super elevation .

madge
Télécharger la présentation

Islamic University of Gaza Civil Engineering Department Surveying II ECIV 2332 By B elal A lmassri

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Islamic University of GazaCivil Engineering DepartmentSurveying IIECIV 2332ByBelalAlmassri

  2. Chapter 9 Route Surveying – Part 4 Obstacles to curve location. Inaccessible point of intersection. Chaining obstacles. Vision obstacles. Transition curves. Super elevation. Extra Practice.

  3. Obstacles To Curve Location • In order to lay out a simple circular curve there are 3 parts should be known: • The Point of Intersection PI. • The Central Angle I. • The Radius R. • Under some certain conditions an inaccessibility happens due to the terrain features.

  4. 1. Inaccessible PI • In some cases it is impossible to occupy the Point of Intersection due to some obstacles i.e. Sea, river, lake or any others. • In these cases the following procedure is followed: • Mark two points A and B on the two tangents (forward and backward) so that the line AB will clear the obstruction.

  5. Measure the angles a and b using the theodolite at point A and B. Measure the line AB. Compute AV and BV as follows:

  6. 2. Chaining Obstacle • In some cases and if there is an obstacle on the curve line it is impossible to measure the chord distance on the curve and in that case it is preferable to measure the distance from the PC. • The distance between the PC and the new station by using the sine rule equals 2R.sin(d1+2d).

  7. 3. Vision Obstacle • Due to some obstructions along the sight or due to the large length of the curve there is a need for an intermediate setup, to solve these types of obstructions we follow the following steps: • Move the theodolite to the new point. • Direct it to PC then reverse it. • Measure d1+2d from that line then measure the distance of chord C.

  8. Underground . . .

  9. Transition Curves • The transition curves overcome gradually the centrifugal force. • The transition curves connect the straight line gradually with the curve. • The transition curves introduce gradually the super elevation (Raising the outer edge of the road at the curve). • The transition curves work as safety traffic signs which ask the drivers to slow down because of the coming curve.

  10. Types of Transition Curves • Cubic Parabola: Used for railways design. • Spiral Parabola: Used for highways design. Cubic Parabola Spiral Parabola

  11. Super Elevation • By lifting the outer edge of the road at the curve this will divide the weight component of the vehicle into two terms and will compensate the centrifugal force.

  12. Without considering the friction forces: • Max super elevation is for circular curve: Slope = e = tan θ = (V² / g.R ) • Super elevation for transitional curve: ei = e.(R/ri) With considering the friction forces: ei = V²/(314 ri) (where V in km/hr) Super elevation value: h = b . e (where b: is the road width)

  13. Example 9.3 Calculate the super elevation value for: R=275 m, V=60 km/hr, and b= 7.5m Solution: e = V²/(314 R) = 0.042 h = b . e = 0.315 meter

  14. Extra Practice ! • The central angle of a circular curve is 45 degrees with radius of 500 meters and chainage of point PI equals 2400 meters, Find: • Ch of PC and PT. • E and M. • C1, C, C2 • d1 , d, d2

  15. Solution !

  16. Extra Practice ! • For the following simple circular curve the coordinates of points PC and X are (240,120) and (300,150) respectively, Find the radius and the degree of curvature for it.

  17. Solution !

More Related