important point for the method of curve settingprocedure for setting out the curve 1) locate ∆find...

Important point for the method of curve setting

Assumption:-1) The length of chord is assumed to be equial to that

of arc .2) The difference between the arc and chord is

negligible . Peg Interval :-1) Peg are normally fixed at equal interval on curve as

along the straight .2) The interval between the peg is normally 20m to

30m.

Mr.R.S.Sonawane

Mr. .R.S. Sonawane DCE,BE(Civill),ME(Geotechnical Engg) ,AMIEI

Chord Length:-1) The unit chord length for flat curve is 30m o 20m2) The unit chord length for sharp curve is 10m or less

than 10m.3) The chord length is limited to ( 1/20 th ) of the radius

of the the curve

Mr.R.S.Sonawane


Designation of curve by Radius & Degree of curve

A curve is Designated By

The Degree of curve Radius of curve

Mr.R.S.Sonawane


1)The Degree of Curve :-The angle a unit chord of 30m length subtended at the

centre by standard chord 30m is called degree of curve.

Where D is the degree of curve this common method to designated the degree of the curve & more advantageous setting out of the curve.

For Example If the unit chord ( 30m chord ) subtended on angle of 10

then it is called one degree Curve when the angle is 20

then it is called two degree curve.


2) Radius of curve :-A curve is also designated in term of its radius such

as,80m curve,100m,200 curve etc. or 2 chain curve ,4 chain curve, 5 chain curve ,etc.

Mr.R.S.Sonawane


The relation between degree of curve & its radius

Mr.R.S.Sonawane


Let “R” be the radius of the curve in meter ‘D’ be the degree of curve.

Relation between the degree of curve and its radius is given by

R = ½ length of standard chord sin D/2

for the chord 30m long = R = 1719/ D (10 = 1719 ) for the chord 20m long = R = 1145/ D (10 = 1145.45 )

Mr.R.S.Sonawane


Method of setting out of curve

Mr.R.S.Sonawane

Method

Linear method Instrumental method


linear Method 1) By ordinates or offset from Long chord.2) By successive bisection of arc.3) By offset from tangent.

i) Radial method.ii) Perpendicular method.

4) By offset from chord produced.

Mr.R.S.SonawaneMr. .R.S. Sonawane DCE,BE(Civill),ME(Geotechnical Engg) ,AMIEI

Instrumental Method or angular Method1) Rankine Method of tangential angle .2) Two theodolite Method.3) Tacheometric Method.


1) By ordinates or offset from Long chord.

Mr.R.S.Sonawane


1) Let AB and BC be the tangent to the curve at the two tangent point T1 & T2 resp.

2) Length of the chord ( L ) = T1T2.3) Offset at mid point of chord T1 T2 = DE = O0.4) The offset at distance “x” from D = PQ =Ox

Dp = x5) Radius of the curve ( R ) = OE = OT1 = OT2.

Mr.R.S.Sonawane


Derivation :-let QQ1 be parallel to the long chord “T1T2”And join OQ so as to intersect the long chord

“T1T2” at point “ P” as shown in fig.let DE =O0

In ∆ OT1DOT1 =R andT1D = L/2OD = OE – DE

= R- O0 ( OE = R and DE = Oo )Mr. .R.S. Sonawane DCE,BE(Civill),ME(Geotechnical Engg) ,AMIEI

By Pythagoras ( OT1)2 = ( T1D) 2 +( OD ) 2

R2 = ( L/2 ) 2 + ( R – Oo ) 2

Reaarranging above Eqn.( R – Oo ) 2 = R2 - ( L/2 ) 2

Square root ( R – Oo ) = √ R2 - ( L/2 ) 2

(Oo ) = R – √ R2 - ( L/2 ) 2 ……………………… IIf L & R Or L and O0 are known then remaining

terms can be found out equation

Mr.R.S.Sonawane

In ∆ OQQ1Use Pythagoras theorem.( OQ ) 2 = ( QQ1) 2 + ( OQ1) 2

But ( OQ1) = OD + DQ1 =OD + Ox = ( R – Oo) + Ox but( OD= R-Oo )

( OQ1) = x & OQ = RR2 = x 2 + [( R – Oo) + Ox ] 2

[ox + ( R – Oo) ] 2 = R2 - x 2

ox + ( R – Oo) =√ R2 - x 2

Ox = √ R2 - x 2 - ( R – Oo) -----------------------IIEquation II is the exact formula for Ox

Mr.R.S.Sonawane


Procedure of setting out of curve Split up the long chord into even number of equal

part. Set out the offset which are calculated from

Ox = √ R2 - x 2 - ( R – Oo)& obtain the required points on the curve ,note that

the curve is symmetrical along DE , hence the offset for the right half of the curve will be the same as that on the left half.

Mr.R.S.Sonawane


2) By successive bisection of arcs of chord

Mr.R.S.Sonawane


Joint the tangent T1,T2 and Bisect the long chord at O. Erect the perpendicular ON & make it equal to the

versed sine of the curve Thus,

NO = R ( 1 – Cos Θ/2 )= R - √ R 2 - ( L/2)

join the ‘T1N’ and ‘T2N’ and bisect at O1 & O2 respectively at O1 and O2 set out perpendicular offset N1O1= N2 O2 = R ( 1 – Cos Θ/4) to get the point O1 & O2 on the curve. By the successive bisection of these chords more point

may be determine.Mr.R.S.Sonawane

3)By offset from tangent

Mr.R.S.Sonawane

The curve can be set out by offset from the tangent .if the deflection angle and radius of curvature both are small .

The offset from the tangent can be two types1) Radial offsets2) Perpendicular offsets.


1) Radial offsetsLet , ox = Radial offset PN at any

distance x along he tangent.T1 p = xFrom ∆ T1 P O ,Use pythagorous(PO)2 = T1O2 + T1 P2

Or (PN+NO) 2 = T1O2 + T1 P2

(Ox+R) 2 = R2 + x2

(Ox+R) =√ R2 + x2

Ox = √ R2 + x2 - R

Mr.R.S.Sonawane


2) Perpendicular offsets.Let, DN = Ox =offset

perpendicular to the tangentT1D =x= Measured along the

tangent,Drawn NN1 parallel to the tangent.

∆N1NO ,use pythagorousN1O 2 = NO 2 + NN1 2

( R – Ox ) 2 = R 2 + x 2( R – Ox ) = √ R 2 + x 2

Ox = √ R 2 + x 2 – R

Mr.R.S.Sonawane


Procedure to set curve1. The distance x1,x2,x3 etc are measured from the first

tangent point T1 along the tangent.2. The perpendicular offset calculate ,are erected with the

help of an optical square at the corresponding point.3. When the distance ‘x’ increase the offset becomes too

large to set out accurately.4. In such case ,the central point position of the curve may

be set out from a third tangent drawn through apex of the curve.

5. This method is useful only for small curve.

Mr.R.S.Sonawane


4) By offset from chord produced ( By deflection distance )

When the curve is long ,this method is useful ( generally Highway)

When the theodolite is not availableLet , T1 L1 = T1L =initial sub-chord

= C1, L,M,N . Point on the curve.LM = C2 , MN = C3 etcT1x = rear tangent<L1T1L = δ = deflection angle of

first chord.L1L = O1 = first offset.M2M = O2 = Second offsetN2N = L1L = O1 = T1 L δ ………I

Mr.R.S.Sonawane


Now,Since T1x is the tangent to the circle at T1

angle T1 O L = 2 < L1T1L= 2 δ

T1L = R2 .δδ = T1 L / 2R ………………..II

Substituting the value of the δ in Eqn in ---IWe get Arc L1L = O1 = T1 L .(T1 L / 2R )

= T1 L2 / 2RTaking arc T1 L =chord T1 L ( very nearly )

Mr.R.S.Sonawane


O1 = C1/2R ………………..IIIO2 = C2/2R (C1+ C2) ………………..VIO3 = C3/2R (C2+ C3) ………………..VThe last on the offset given On = Cn/2R (Cn+ Cn+1) ………………..III

Mr.R.S.Sonawane


Procedure for setting out the curve1) Locate ∆ find the out change point T1 ∆ T2.2) Calculate the length of first sub-chord (c ) so that first

peg is the fall station.3) Measure the length T1 ∆ as with the help of chain is at T1

now T1 ∆ 1 = C = length of the first sub-chord.4) with T1 as centre and T1 ∆ 1 as the radius ,swing the

chain such that the arc L1L = calculate offset O1.

5) Now fig.the point L on the curve.6) With zero of the chain at ∆,spread the chain along T1 L

and pull .it straight towards M2 the distance ∆M 2 = C

Mr.R.S.Sonawane


Length normal chord.7) With zero of the chain at station L and ∆M 2 as radious

swinging the chain to point M .Such that M2M = O2 = length of second offset .fix point on the curve.

8) Spread the chain along MN & repeat the above step till the point of tangency ( T2 ) is reached.

Mr.R.S.Sonawane


important point for the method of curve settingprocedure for setting out the curve 1) locate ∆find...

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