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

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

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

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

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Chapter 9 Route Surveying – Part 3

Linear Methods for setting out simple circular curve.

Method of offsets on the long chord.

Method of offsets on tangents.

Method of radial offsets.

Method of deflection angles.

Notes and Examples.

- These methods use the chain surveying tools only.
- These methods are used for the short curves which doesn’t require high degree of accuracy.
- These methods are used for the clear situations on the road intersections.

There are three types of the linear methods to set out a simple circular curve.

- By offsets from the long chord.
- By offsets from the tangents.
- By radial offsets.

- In the first method, the value of x = Lc/2 = 7.654 < 8m so we had used x = 7.654m and y at this point equals zero.
- In the second method, the value of x = 8m < T(8.28), we had used x = 8m but we still know that the circular curve close at PT when T = 8.28m.

- In the second and the third methods the x value will not exceed the value of T BUT in the first method the value of x will not exceed the value of Lc/2.
- The third method is used when the centre of the circular curve is accessible while the first two methods can be used when there is an obstacle.

Working method:

- Fix the theodolite device to be at point PC and directed at point PI.
- Measure the deflection angles d and the chords C.
- Connect the ends of the chords to draw the curve.
Deflection Angles: the angles between the tangent and the ends of the chords from point PC.

- Calculations Steps:
This method is a geometric based method :

- Calculate the values of T and L.
- T = R tan (Δ/2)
- L = R (Δ in radians)
- Calculate the chainage of point PC and point PI.
- Ch of PC = Ch of PI – T
- Ch of PT = Ch of PC + L

- Calculate the partial chords C, C1,C2.
- Choose C ≤ R/20 (then round it)
- Chainage of the first station: Chainage of PC (rounded) + C
- C1 = Chainage of first station – Chainage of PC (original)
- C2 = L – ( C1 + n C ); n = number of intermediate chords
- Calculate the deflection angles d, d1, d2.
- d = (28.648 C)/R
- d1 = (28.648 C1)/R
- d2 = (28.648 C2)/R
- Calculate the cumulative deflection angles.

- The total d’s will equal Δ/2.
- The chainage of the all stations on the curve should be even number divisible of 5 or 10.
- When locating the last point before PT we measure the distance to the PT if it was equal C2 then it is correct or the difference is 5- 10 cm BUT if it is more than that we should repeat !

Two tangents intersect at PI with chainage of 2140.00 m and deflection angle of 10 ͦ 35 ` 2``

Da or D˳= 4 ͦ , Using the deflection angles method Arrange all the information needed.

Solution !