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AN EDUSAT LECTURE ON

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T1

A

B

C

T2

O

Fig. 1 . A CURVE

CURVES

Curves are regularbends provided in

the lines of

communication like

roads, railways and

canals etc. to bring

about gradual

change of direction.

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T1

A

B

C

T2

O

Fig. 2. A CURVE

CURVES

They enable the

vehicle to pass fromone path on to another

when the two paths

meet at an angle. Theyare also used in the

vertical plane at all

changes of grade to

avoid the abrupt

change of grade at the

apex.

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HORIZONTAL CURVESCurves provided in the horizontal plane to have

the gradual change in direction are known ashorizontal curves.

VERTICAL CURVES

Curves provided in the vertical plane to obtain

the gradual change in grade are called as

vertical curves.

Curves may be circular or parabolic. Curves

are generally arcs of parabolas.Curves are laid out on the ground along the

centre line of the work.

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NEED OF PROVIDING CURVES

Curves are needed on Highways, railwaysand canals for bringing about gradual change

of direction of motion. They are provided for

following reasons:-i) To bring about gradual change in

direction of motion.

ii) To bring about gradual change in grade

and for good visibility.

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NEED OF PROVIDING CURVES Contd

iii) To alert the driver so that he may not fall

asleep.

iv) To layout Canal alignment.

v) To control erosion of canal banks by the

thrust of flowing water in a canal.

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CLASSIFICATION OF CIRCULAR CURVES

Circular curves are classified as :

(i) Simple Curves.

(ii) Compound Curves.

(iii) Reverse Curves.

(iv) Deviation Curves.

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T1

A

B

C

T2

O

Fig. 3. A SIMPLE CURVE

i) Simple Curve:

A simple curve

Consists of a

single arc of

circle connecting

two straights. It

has radius of the

same magnitudethroughout.

8

RR

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ii) COMPOUND CURVE

A compound Curve consists of two or

more simple curves having different radii

bending in the same direction and lying on

the same side of the common tangent. Their

centres lie on the same side of the curve.

A

T1

M P N

C

O1

O2

Fig.4 Compound Curve

9

R2

R1

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iii) REVERSE OR SERPENTINE CURVE

A reverse or serpentine curve is

made up of two arcs having equal

or different radii bending in

opposite direction with a common

tangent at their junction .

Fig. 5. A Reverse or Serpentine Curve.

MTheir centres lie on

opposite sides of the curve.

Reverse curves are used

when the straights areparallel or intersect at a

very small angle.

N

O2

O1

A

T1

T2

p

B

10

R1

R2

R2

R1

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REVERSE OR SERPENTINE CURVE

Fig.6 A Reverse or Serpentine Curve.

They are commonly used

in railway sidings andsometimes on railway

tracks and roads meant

for low speeds. Theyshould be avoided as far

as possible on main lines

and highways where

speeds are necessarily

high.

A

T1

T2O2

O1

M N

B

P

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iv) DEVIATION CURVE

T1 T2

O2

O1 O3

Building

A deviation curve is

simply a combinationof two reverse curves.

it is used when it

becomes necessary to

deviate from a given

straight path in order

to avoid intervening

obstructions such asbend of river, a

building , etc.

Fig. 7 A Deviation Curve

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B

B

T1 T2

O

R

CA

E

F

I

/2

Fig. 8 SIMPLE CIRCULAR CURVE

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NAMES OF VARIOUS PARTS OF CURVE

(i) The two straight lines AB and BC which

are connected by the curve are called thetangentsor straightsto the curve.

(ii) The point of intersection of the two

straights (B) is called the intersection pointor the vertex.

(iii) When the curve deflectsto the right side of

the progress of survey ,it is termed as righthanded curve and when to the left , it is

termed as lef t handed curve.

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(iv) The lines AB and BC are tangents to the

curve. AB is called the f irst tangent or therear tangent . BC is called the second

tangent or the forward tangent.

(v)The points ( T1and T2) at which thecurve touches the tangents are called

the tangent points. The beginning of

the curve ( T1) is called the tangent

curve pointand the end of the curve

(T2) is called the curve tangent point.

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(vi) The angle between the lines AB and BC

(ABC) is called the angle of intersection(I).(vii) The angle by which the forward tangent

deflects from the rear tangent (BBC) iscalled the deflection angle () of the curve.(viii) The distance from the point of intersection

to the tangent point is called tangent length( BT1and BT2).

(ix) The line joining the two tangent points (T1

and T2) is known as the long chord.16

( ) Th T1FT2 i ll d th l th f

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(x) The arc T1FT2is called the length of curve.

(xi) The mid point(F) of the arc (T1FT2) is called

the summit or apex of the curve.

(xii) The distance from the point of intersectionto the apex of the curve BF is called the

apex distance.

(xiii) The distance between the apex of the curveand the mid point of the long chord (EF) is

called versed sine of the curve.

(xiv) The angle subtended at the centre of the

curve by the arc T1FT2 is known as

central angle and is equal to the deflection

angle () .17

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ELEMENTS of a Simple Circular Curve

(i) Angle of intersection +Deflection angle = 1800.

or I + = 1800

(ii)

T1OT2 = 180

0- I =

i.e the central angle = deflection angle.

(iii)Tangent length = BT1 =BT2= OT1tan /2

= R tan /2

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ELEMENTS of a Simple Circular Curve

(iv) Length of long chord =2T1E

=2R sin/2(v) Length of curve = Length of arc T1FT2

= R X (in radians)

= R/1800(vi) Apex distance = BF = BOOF

= R sec. /2 - R

= R (1cos /2)=R versine/2

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A curve may be designated either bythe radius or by the angle subtended at the

centre by a chord of particular length.

In India, a curve is designated by the

angle (in degrees)subtended at the centre by a

chord of 30 metres (100 ft.) length. This angle

is called the degree of curve (D).

The degree of the curve indicates thesharpness of the curve.

DESIGNATION OF CURVE

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DESIGNATION OF CURVES

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DESIGNATION OF CURVES.

In English practice , a curve is defined

by the radius of the curve in terms of chains,such as a six chain curve means a curve having

radius equal to six full chains, chain being 30

metres unless otherwise specified.

In America,Canada,India and some

other countries a curve is designated by the

degree of the curve. For example a 40 curve

means a curve having angle of 4 degrees at the

centre subtended by a chord of 30m length

unless otherwise specified.

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RELATION between the Radius of curve and

Degree of Curve.

The relation between the radius

and the degree of the curve may

be determined as follows:-

Let R = the radius of the curve in metres.

D = the degree of the curve.MN = the chord, 30m long.

P = the mid-point of the chord.

In OMP,OM=R,MP= MN =15m

MOP=D/2

Then, sin D/2=MP/OM= 15/R

MN

O

D

D/2

R R

Fig.9 Degree of Curve

P

PTO

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RELATION between the Radius of curve and

Degree of Curve.Then,sin D/2=MP/OM= 15/R

Or R = 15

sin D/2

But when D is small, sin D/2 may be

assumed approximately equal to

D/2 in radians.Therefore:

R = 15 X 360

D= 1718.87

D

Or say , R = 1719

D

MN

O

D

D/2

R R

Fig. 10 Degree of Curve

P

This relation holds good up to 50

curves.For higher degree curves the

exact relation should be used.

(Exact)

(Approximate)

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METHODS OF CURVE RANGING

A curve may be set out

(1) By linear Methods, where chain and tapeare used or

(2) By Angular or instrumental methods,

where a theodolite with or without a chain isused.

Before starting setting out a curve by any

method, the exact positions of the tangents

points between which the curve lies ,must be

determined. Following procedure is adopted:-

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METHODS OF SETTING OUT A CURVE

Procedure :-

i) After fixing the directions of the straights,produce them to meet in point (B)

ii) Set up the Theodolite at the intersection

point (B) and measure the angle ofintersection (I) .Then find the deflection

angle ( ) by subtracting (I) from 1800 i.e

=1800I.

iii) Calculate the tangent length from the

following equation

Tangent length = R tan/2

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Instrumental Methods

(i) Tape and Theodolite Method (Rankine's

method) and

(ii) Two Theodolite Method

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RankinesMethod

In this method, both the linear and angularmeasurements are carried out simultaneouslyto stake points along which curve will be set

out. A tape is used for the linear measurements,

whereas a theodolite is used for the angularmeasurements.

This method is quite accurate and iscommonly used in practice

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RankinesMethod

In this method, curves are staked out by use of deflectionangles turned at the point of curvature from the tangent topoints along the curve.

The curve is set out by driving pegs at regular intervalequal to the length of the normal chord.

Usually, the sub-chords are provided at the beginning andend of the curve to adjust the actual length of the curve.

The method is based on the assumption that there is nodifference between length of the arcs and their

corresponding chords of normal length or less. The underlying principle of this method is that the

deflection angle to any point on the circular curve ismeasured by the one-half the angle subtended at thecentre of the circle by the arc from the P.C. to that point

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Let points a, b, c, d, e are to be identifiedin the field to layout a curve between T1and T2to change direction from thestraight alignment AV to VB.

To decide about the points, chords ab, bc,

cd, de are being considered havingnominal length of 30m. To adjust theactual length of the curve two sub-chordshave been provided one at the beginning,T1a and other, eT2at the end of the curve.

The amount of deflection angles that are

to be set from the tangent line at the P.C.are computed before setting out thepoints. The steps for computations are asfollows:

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let the tangential angles forpoints a, b, c, be d1, d,, d, dnand their deflection angles (fromthe tangent at P.C.) be Da, Db, .., Dn.

Now, for the first tangentialangle d1, from the property of acircle

Arc T1a = R x 2d1radians

Assuming the length of the arc issame as that of its chord, if C1 isthe length of the first chord i.e.,chord T1 a, then

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Thus, the deflection angle for any point on the

curve is the deflection angle upto previous

point plus the tangential angle at the previous

point.

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Procedure : RankineSMethod1. A theodolite is set up at the point of curvature T1, and get it temporary adjusted.

2. The vernier A is set to zero, and get the upper plate clamped. After opening the lowerplate main screw, sight the point of intersection, V. Then the lower plate main screwgets tightened and get the point V bisected exactly using the lower plate tangent screw.Now the line of sight is in the direction of the rear tangent T1V and the vernier A readszero.

3. Open the upper plate main screw, and set the vernier A to the deflection angle D a. Theline of sight is now directed along the chord T1a. Clamp the upper plate.

4. Hold the zero end of the tape of a steel tape at T1. Note a mark equal to the first chordlength C1on the tape and swing an arrow pointed at the mark around a' till it isbisected along the line of sight. The arrow point then indicates the position of the firstpeg a'. Fix the first peg at a'.

5. Unclamp the upper plate, and set the vernier A to the deflection angle Db. The line ofsight is now directed along T1b.

6. With the zero end of the tape at a, and an arrow at a mark on the tape equal to thenormal chord length C, swing the tape around b until the arrow is bisected along theline of sight. Fix the second peg at the point b at the arrow point.

It may be noted that the deflection angles are measured from the tangent point T1butthe chord lengths are measured from the preceding point. thus, deflection anglesobserved are cumulative in nature but chord lengths swung are individual in nature.

7. Repeat steps (5) and (6) till the last point is reached. The last point so located must

coincide with the tangent point T2already fixed from the point of intersection. 43

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Procedure :-

iv) Measure the tangent length (BT1)

backward along the rear tangent BA from

the intersection point B, thus locating theposition of T1.

vi) Similarly, locate the position of T2 by

measuring the same distance forward

along the forward tangent BCfrom B.

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Procedure (contd) :-

After locating the positions of the tangentpoints T1 and T2 ,their chainages may be

determined. The chainage of T1is obtained by

subtracting the tangent length from the knownchainage of the intersection point B. And the

chainage of T2 is found by adding the length

of curve to the chainage of T1.

Then the pegs are fixed at equal intervals

on the curve.The interval between pegs is

usually 30m or one chain length. ...............45

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This distance should actually be measured alongthe arc ,but in practice it is measured along

the chord ,as the difference between the chord

and the corresponding arc is small and hencenegligible. In order that this difference is

always small and negligible ,the length of the

chord should not be more than 1/20th of the

radius of the curve. The curve is then obtained

by joining all these pegs. ...............

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The distances along the centre line of thecurve are continuously measured from the

point of beginning of the line up to the end .i.e

the pegs along the centre line of the workshould be at equal interval from the beginning

of the line up to the end. There should be no

break in the regularity of their spacing in

passing from a tangent to a curve or from a

curve to the tangent. For this reason ,the first

peg on the curve is fixed .47

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at such a distance from the first tangent point(T1) that its chainage becomes the whole

number of chains i.e the whole number of peg

interval. The length of the first sub chord isthus less than the peg interval and it is called a

sub-chord. Similarly there will be a sub-chord

at the end of the curve. Thus a curve usually

consists of two sub-chords and a no. of ful l

chords.

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Example : A simple circular curve is to have a radius

of 573 m .the tangents intersect at chainage 1060 m

and the angle of intersection is 1200. Find,

(i) Tangent Distance.

(ii) Chainage at beginning and end of the curve.

(iii) Length of the long chord.

(iv) Degree of the curve.(v) Number of full and sub chords.

Solution: Please see fig.11

Given,The deflection angle,= 18001200 =600

Radius of curve = R = 573 mPTO

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Fig.11

1060 m

O

729.15 1329.15

T1 T2

600

1200

330.85

R=573m

=

L=600m

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(i) We know ,tangent length = R tan /2

= 573 x tan 300

= 573x 0.5774

= 330.85 m (Ans.)

(ii) Length of curve is given by: R1800

= x 573x600

1800

= 600 m (Ans.)

Chainage of first tangent point (T1)

= Chainage of intersection pointtangent length.= 1060330.85= 729.15 m (Ans.)

PTO

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(iii) The length of long chordis given by:

L = 2R sin /2

= 2 x 573 x sin 300

= 573 m ( Ans.)

(iv) Degree of Curve

We know the relation , R= 1719D

or D = 1719

R=30

Therefore , degree of curve is =30(Ans.)PTO

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(v) Number of Full and sub chords:

Assuming peg interval =30mChainage of T1= 729.15 m = 729.15

30

= 24 full chain lengths + 9.15 m

Chainage of Ist peg on the curve should be 25 full chain lengths.

The length of Ist sub chord= (25+00)(24 + 9.15)

= 20.85 mChainage of T2= 1329.15 Chain lengths.

30

= 44 full chain lengths + 9.15 m.

Chainage of last peg on the curve =44 full chains.

Therefore length of last sub chord = (44+9.15)(44+00)= 9.15m

PTO

Chain lengths.

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No. Of full chords = chainage of last pegchainage of Ist peg= 4425 = 19

So, there will be 19 full chords and two sub chords.

Check:Length of full chords = 19x30 =570.00m

Ist sub chord = 20.85m last sub chord = 9.15m

Total length of all chords = 600.00m

PTO

(Same as length of curve)

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LINEAR METHODS of setting out Curves

The following are the methods of setting outsimple circular curves by the use of chain

and tape :-

(i) By offsets from the tangents.(ii) By successive bisection of arcs.

(iii) By offsets from chords produced.

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1. By offsets from the tangents. When the

deflection angle and the radius of thecurve both are small, the curves are set out

by offsets from the tangents.

Offsets are set out either(i) radially or

(ii) perpendicular to the tangents

according as the centre of the curve is

accessible or inaccessible

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B

B

T1T2

O

R

CA

Fig. 12 By Radial Offsets


Oxx P

P1900

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B

By Radial Offsets


Offsets is given by :

Ox= R2+x2R .. (Exact relation.)

When the radius is large ,the offsets may be

calculated by the approximate formulawhich is as under

Ox= x2 (Approximate )

2R

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B

O

(ii) By offsets perpendicular to the Tangents


Oxx

P

P1

P2

B

A

B

T2T1

Fig. 13.

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LINEAR METHODS f tti t C

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1. (ii) By offsets perpendicular to the Tangents

Ox= R R2x2 (Exact)

Ox= x2 (Approximate )

2R

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By offsets from the tangents: Procedure

(i) Locate the tangent points T1and T2.

(ii) Measure equal distances , say 15 or 30 m

along the tangent fro T1.

(iii) Set out the offsets calculated by any of

the above methods at each distance ,thus

obtaining the required points on the

curve.

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By offsets from the tangents: Procedure.

(iv) Continue the process until the apex of

the curve is reached.

(v) Set out the other half of the curve from

second tangent.

(vi) This method is suitable for setting out

sharp curves where the ground outside

the curve is favourable for chaining.

62

Example. Calculate the offsets at 20m intervals along

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p gthe tangents to locate a curve having a radius of

400m ,the deflection angle being 600.

Solution . Given:

Radius of the curve ,R = 400m

Deflection angle,= 600

Therefore tangent length = R. tan/2= 400 x tan 600

= 230.96 m

Radial offsets. (Exact method)

Ox= R2+ x2 - R (Exact)

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Radial offsets. (Exact method)

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( )

Ox= R2+ x2 - R (Exact)

O20 = 4002+202 - 400 = 400.50 - 400 = 0.50 m

O40 = 4002+402 - 400 = 402.00 - 400 = 2.00 m

O60 = 4002+602 - 400 = 404.47 - 400 = 4.47 m

O80 = 4002+802 - 400 = 407.92 - 400 = 7.92 m

O100 = 4002+1002- 400 = 412.31 - 400 = 12.31 m

And so on.

64

B) Perpendicular offsets (Exact method)

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Ox = R R2x2 (Exact)

O20= 400 - 4002- 202 = 400 -399.50 = 0.50 m

O40= 400 - 4002- 402 = 400 -398.00 = 2.00 m

O60= 400 - 4002- 602 = 400 -395.47 = 4.53 m

O80= 400 - 4002- 802 = 400 -391.92 =8.08 m

O100= 400 - 4002-1002 = 400 -387.30 =12.70 m

And so on..65

B By t e approx mate Formu a(Both radial and perpendicular offsets)

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(Both radial and perpendicular offsets)Ox =

2R

Therefore O20 = 202 = 0.50 m2x400

x2

O40 = 402 = 2.00 m

2x400

O60 = 602 = 4.50 m2x400

O80 = 802 = 8.00 m

2x 400O100 = 1002 = 12.50 m

2 x 400

and so on.66

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