DEPRATMENTOF CIVIL ENGINEERING

SHREE S’AD VIDHYA MANDAL

INSTITUTE OF TECHNLOGY

Faculty:- Prof. RUCHI GUPTASUBJECT:- SURVEYINGGROUP 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

CURVES

COMBIEND CURVESCombined 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.

B

T1

FE

∆

∆

∆s ∆s ∆-2∆s

O

D1 F1R R

DT1′ T2′

T2

AC

Transition

curveTransition curve

SHIFT

Circular curve

NOTATIONSAB = Back tangentBC = Forward tangent2T1 = Point of curveT2 = Point of tangency∆ = Deflection angle∆s = Spiral angleT1D or T2F =Length of the transition curve

DEF = Length of circular curveT1ʹ D1 or T2ʹF1 = Shift of the curve

T1DEFT2 = Length of the combined curve

∆ -2∆s = Central angle

COMPUTATIONS FOR A COMBINED CURVEFor 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.

1) Spiral angle (∆s) ∆s = L 2r radians = L X 180 Degrees 2r π

2) Shift (s) S = L2

24R

3) Tangent length TT = BT1 =(R+S) TAN ∆ + L 2 24) Length of circular curve Lc = πR (∆ - 2∆S) 1805) Chainages of various pointsChainage of T1 = Chainage of B –BT1Chainage of D =Chainage of T1+ LChainage of T2 = Chainage of F + L

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= a) For transition curve α =18002 minutes πRL

L is measured from the tangent point T1.The deflection angle at the junction of

transition curve and circular curve for l=L is αn =1800L πRb) For circular curve δ =1718.9 C minutes R c = length of the chord δ =tangential angles from tangent at D.

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

TYPES OF VERTICAL CURVESThere 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).

A

T1T2F

E=-g1=+ g2=-

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).

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).

T1 T2

A

F

g1=- g2=+E=+

Length of a vertical curveThe 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 rWhere, g1,g2 = grades in % r=rate of change of grade (%)

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

1) shift (s) S = L2 = 602 = 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

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

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

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)

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- 1.2x800 100

2 =120 – 4.80 = 115.20 m

R.L. of end point of curve = 120 – 0.8 x 800

100 2

= 120 – 3.2 = 116.8m

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.

Solutiong1= +3%g2= -5%Length of curve =L= 500mR.L. of beginning point of curve = 400- 3

x500 100

2 = 392.5m

R.L. of end point of curve = 400 - 5 x 500 100 2 = 387.5mR.L. of mid point of long chord = 392.5 +

387.5 2 =390mR.L. of mid point of curve =R.L. of point of intersection+ R.L. of mid point

of long chord2

400+390 2

=395m

http://www.google.co.in/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&docid=IBmw4nN3EPjydM&tbnid=BlVdBCOhf7sGSM:&ved=0CAcQjRw&url=http%3A%2F%2Fwww.searchenginepeople.com%2Fblog%2Foptimizing-post-conversion-maximimizing-thank-you-page.html&ei=Pm0iVKfcJpCNuAT4pYHYCw&bvm=bv.76180860,d.c2E&psig=AFQjCNFgknwtZsCt89AzVowBvt2hrQQAsw&ust=1411628662365960