Simple circular curve and its setting out By D.M Siddique

9/14/2012

1

ADVANCE ENGINEERING SURVEYING (3+1)

Lecture 2: Simple circular curve and its setting out

Dr. Mohsin Siddique

Asst. Prof.

Dept. of Civil Engineering

FAST-NU

13/09/2012

9/14/2012

2

Properties of Simple Circular Curve

• If the angle of intersection is given as I, then

I−=180φ

• If radius is not given, then

DR /1719= DR /1719=

Where D is degree of curve

• Tangent length BT1 or BT= ( )2/tan φR

• Length of curve=length of arc T1ET2

180

oR

Rφπ

φ =

• Again length of curve

given is curve of D degree if

30

D

φ=

9/14/2012

3

Properties of simple circular curve

• Length of long chord

( ) ( )2sin22sin1212 φφ ROTDT ==

• Apex distance=BE=OB-BE

( ) ( ) )12(sec2sec −=− φφ RRR

• Versed sine of curve

)2cos1(2cos φφ −=−=

−=

RRRDE

ODOEDE

9/14/2012

4


• Full Chord (Peg Interval): pegs are fixed at regular interval along the curve. Each interval is said to equal the length of full chord or unit chord.

• Length of unit chord should not be more than 1/20th the radius be more than 1/20th the radius of the curve.

• In railways, the length of unit chord are generally taken between 20 and 30m.

• In roads length of unit chord should be 10m or less

• Initial subchord

• Final subchord

9/14/2012

5


• Initial subchord

• Final subchord

• Chainage of first tangent= Chainage of intersection point -tangent lengthtangent length

• Chainage of second tangent point= chainage of first tangtent point + curve length

9/14/2012

6

Circular curve

Numerical 1

• Two straights intersects at chainage 2056.44m and the angle of intersection is 120 degree. If the radius of simple curve to be introduced is 600m, find the following,

▫ 1. Tangent Length

▫ 2. Length of curve

▫ 3. Chainage of tangent points

▫ 4. length of long chord▫ 4. length of long chord

• SOLUTION

• Deflection angle

• 1. Tangent Distance

( )

( ) m

RBTBT

41.3462/60tan600

2/tan21

==

== φ

oI 60120180180 =−=−=φ

9/14/2012

7

Circular curve

Numerical 1

• Length of curve

• Chainage of T1(point of commencement)

• =Chainage of B-BT1

( )m

Ro

32.628180

60600

180

πφπ=

• =Chainage of B-BT1

• =2056.44-346.41=1710.03m

• Chainage of T2(point of tangency)

• =Chainage of T1+legth of curve

• 1710.03+628.32=2338.35m


( ) mR 6002/sin2 =φ

9/14/2012

8

Horizontal curve setting

• The following general methods are employed for setting out curves by

• Chain and tape

▫ A. taking offsets or ordinates from the long chord

▫ B. Taking offsets from chord produced

▫ C. Successively bisecting the arcs▫ C. Successively bisecting the arcs

▫ D. Taking offset from the tangents

• Instruments

▫ A. Deflection angle method or Rankine method

� One Theodolite method

� Two Theodolite method

9/14/2012

9


A. Offsets or ordinates from long chord

• Let AB and BC be two tangentsmeeting at a point B, with adeflection angle φ. Thefollowing data are calculatedfrom for the setting out thecurve.

• Calculate tangent lengths by

• Tangent points are marked

• Length of curve is calculatedfrom

• Chainage of T1 and T2 arefound out.

180

oRφπ

=

( )2/tan φR

9/14/2012

10



• Length of long chord iscalculated by

• Long chord is divided into twoequal halves, the right half andleft half. Here curve is

( )2/sin2 φR

F

OxOo

left half. Here curve issymmetrical on both halves.

• The mid ordinate, Oo, iscalculated by

( )( )

oO-ROD and

2/cos1

==

−==

ROF

RDEOo

φ

x

Ox

Eq. 1

9/14/2012

11



• From Triangle OT1D

2

1

22

1 DTODOT +=

F

OxOo

( )2

22

2

+−=

LORR

o

( )2

2

−=−L

ROR

x

Ox( )

2

2

2

2

2

−−=

−=−

LRROo

LROR

o

Eq. 2

Thus the mid ordinate Oo can be calculated from Eq. 1 and Eq.2

9/14/2012

12



• Considering the left half of thelong chord, the ordinates O1, O2,O3….are calculated at distanceX1, X2, X3,…. Taken from Dtowards the tangent point T1.

• The formula for calculatingthe tangent point is deduced

P1

OxOo

P2

the tangent point is deducedas follows

• Let P be point at a distance xfrom D. Then PP1(Ox) is therequired ordinate. A line P1P2 isdrawn parallel to T1T2. fromtriangle OP1P2

x

Ox

R-Oo

P

( )[ ] 222

2

21

2

2

2

1

xOORR

PPOPOP

xo++−=

+= ( )

( )o

xo

ORxROx

xROOR

−−−=

−=+−

22

22

Note : Same procedure can be adopted for right half.

9/14/2012

13

Numerical 2

• Two tangents AB and BC intersects at a point B at chainage150.5m.Calculate all the necessary data for setting out a circular curve of radius 100m and deflection angle 30 degree by the method of offsets from the long chord

• SOLUTION:

• Tangent length• Tangent length

• Chainage of T1

=Chainage of B-Tangent length

=150.5-26.79=123.71m

• Curve length

( ) ( ) mR 79.262/30tan1002/tan === φ

( )m

Ro

36.52180

30100

180

πφπ=

9/14/2012

14

( ) ( ) mR 76.512/30sin100*22/sin2 === φ

Numerical 2

• Chainage of T2

=Chainage of T1+length of curve

=123.71+52.36=176.07m


( ) ( ) mR 76.512/30sin100*22/sin2 === φ

• Lets divide the curve into two equal halves

each half=length of long chord/2=25.88m

Lets take 5m interval and calculate the vertical ordinate at 0, 5, 10, 15, 20, 25 and 25.88 m from the center of curve towards T1.

9/14/2012

15

Numerical 2

• Mid ordinate=

• Ordinate at x=5m=





• Ordinate at x=25.88m=

( ) mLRRO 41.32/22

0 =−−=

( ) mORxROom

28.322

5 =−−−=

( ) mORxROom

91.222

10 =−−−=

( ) mORxROom

3.022

25 =−−−=( ) mORxRO

om38.122

20 =−−−=

( ) mORxROom

28.222

15 =−−−=

( ) mORxROom

022

88.25 =−−−=

9/14/2012

16

Assignment Problem

• Resolve the numerical if the radius of curve is 150m and deflection angle is 45 degrees.

• Plot the curve on a A4 sheet according to • Plot the curve on a A4 sheet according to scale.

9/14/2012

17

THANK YOU

• Questions….

• Comments…

• Suggestions…

• Feel free to contact

[email protected]

Simple circular curve and its setting out By D.M Siddique

Documents

simple circular

horizontal

long chord

deflection

unit chord

left half

equal halves

taking offsets