Combiend curves of surveying

combiend curves ppt.fig draw in power pointr

1. DEPRATMENT OF CIVIL ENGINEERING SHREE S’AD VIDHYA MANDAL INSTITUTE OF TECHNLOGY

2. Faculty :- Prof. RUCHI GUPTA SUBJECT :- SURVEYING GROUP NO : - M9 ENROLLMENT NO. NAME 130450106043 Rashidbhai Badare Munir 130450106044 Shah Ishani Milankumar 130450106045 Shah Richaben Umeshbhai 130450106046 Shrimali Pritesh Bharatbhai 130450106047 Taira Naznin Iqbal

3. CURVES

4. COMBIEND CURVES Combined curves are a combination of simple circuler curve and transition curves. These types of curves are preferred in railways and highways. When transition curves are introduced at a both ends of a circular curve, the resulting curve is known as a combined curve or composite curve.

5. B T1 F E ∆ ∆ ∆s ∆s ∆-2∆s O D1 F1 R R D T1 ′ T2 ′ T2 A C Transition curve Transition curve SHIFT Circular curve

6. NOTATIONS AB = Back tangent BC = Forward tangent 2T 1 = Point of curve T 2 = Point of tangency ∆ = Deflection angle ∆s = Spiral angle T 1 D or T 2 F =Length of the transition curve

7. DEF = Length of circular curve T 1 ʹ D 1 or T 2 ʹF 1 = Shift of the curve T 1 DEFT 2 = Length of the combined curve ∆ -2∆ s = Central angle

8. COMPUTATIONS FOR A COMBINED CURVE For setting out a combined curve,the following data are required. The data given are: 1) Deflection angle (∆) 2) Radius of curve (R) 3) Length of transition curve(L) 4) Chainage of point of intersection B.

9. 1) Spiral angle (∆s) ∆s = L 2r radians = L X 180 Degrees 2r π 2) Shift (s) S = L 2 24R

10. 3) Tangent length T T = BT 1 =(R+S) TAN ∆ + L 2 2 4) Length of circular curve Lc = π R (∆ - 2∆ S ) 180 5) Chainages of various points Chainage of T 1 = Chainage of B –BT 1 Chainage of D =Chainage of T 1 + L Chainage of T 2 = Chainage of F + L

11. 6) Length of the normal chords Generally,10 m for the transition curve and 20 m for the circuler curve are used. 7)Length of sub chords = It is calculated after knowing the chainages of salient points . 8) Deflection Angles = For transition curve α = 1800 2 minutes π RL

12. L is measured from the tangent point T1. The deflection angle at the junction of transition curve and circular curve for l=L is α n = 1800L π R b) For circular curve δ =1718.9 C minutes R c = length of the chord δ =tangential angles from tangent at D.

13. The total deflection angles are= ∆ 1 = δ 1 ∆ 2 = ∆ 1 + δ 2 ∆ 3 = ∆ 2 + δ 3 ∆ n = ∆n-1 + δ n ∆ n should be equal to 1 (∆ -2 ∆ s) 2

14. TYPES OF VERTICAL CURVES There are two types of vertical curves. 1) Summit curve or convex curve 2) valley curve or concave curve 1) summit curve or convex curve Summit curve is provided in following situations. An a upgrade (+g1) followed by a down grade(-g2).

15. A T1 T2 F E=- g1=+ g2=-

16. An upgrade (+g1) followed by another upgrade (+g2) g1 >g2. A down grade (-g1) followed by another down grade(-g2). g2> g1. A plane surface followed by down grade(-g1).

17. 2) Valley curve or concave curve Valley curve is provided in following situations. A down grade (-g1) followed by a upgrade.(+g2). A down grade (-g1) followed by another down grade(-g2). g1 > g2. An upgrade(+g1) followed by another upgrade (+g2). g2 > g1. A plane surface followed by upgrade (+g1).

18. T1 T2 A F g1=- g2=+ E=+

19. Length of a vertical curve The length of vertical curve can be obtained by dividing the algebric difference of the two grades by the rate of change of grade Length of curve(L)= total change of grade rate of change of grade = g2 – g1 r Where, g1,g2 = grades in % r=rate of change of grade (%)

20. 1) in respect to calculation for setting out of combined curve if ∆ =34⁰30′, radius of circular curve is 400m,length of transition curve 60m. calculate shifts, tangent length and spiral angle. solution L =60 m ∆= 34⁰30′ R=400m

21. 1) shift (s) S = L 2 = 60 2 = 0.375m 24R 24x400 2) Tangent length T = (R+S) tan ∆ + L 2 2 = (400+0.375)tan 34⁰30′ + 60 2 2 = 124.32+30 =154.32 m

22. 3) Spiral angle (∆s) ∆s = L 2R = 60 2x400 = 0.075 radian =0.075 x 180 π = 4⁰17′50″

23. 2) Calculate length of parabolic curve joining curve two uniform grades of +1.2% per 100m. solution g1=+1.2% g2= -0.8% r =0.15% (rate of change of grade) Length of curve = L = (g1- g2 ) x 100 r = (-0.8-.1.2) x 100 0.15 = 1333.33m

24. 3) A parabolic vertical curve is to be set out connecting two uniform grades of +1.25 and 0.8%. The Rate of change of grade is 0.05% per 20 m chain. The R.L of point of intersection is 120 m ,calcuate R.L of beginning and end of Curve. g1 =1.2% g2 =-0.8% r= 0.05% (for 20 m)

25. Length of curve = L = g2 –g1 x 20 r = -0.8-1.2 x20 0.05 =800m R.L of beginning point of curve =120- 1.2 x 800 100 2 =120 – 4.80 = 115.20 m

26. R.L. of end point of curve = 120 – 0.8 x 800 100 2 = 120 – 3.2 = 116.8m

27. An upgrade of 3% followed by down grade 5% are connected by a vertical parabolic curve of 500m length. If R.L. of point of intersection is 400m, calculate the R.L. of beginning point, end point and the mid point of curve. Solution g1= +3% g2= -5% Length of curve =L= 500m R.L. of beginning point of curve = 400- 3 x 500 100 2 = 392.5m

28. R.L. of end point of curve = 400 - 5 x 500 100 2 = 387.5m R.L. of mid point of long chord = 392.5 + 387.5 2 =390m R.L. of mid point of curve = R.L. of point of intersection+ R.L. of mid point of long chord 2 400+390 2 =395m