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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
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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=
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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
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Properties of simple circular curve
• 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
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Properties of simple circular curve
• 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
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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
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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
• Length of long chord
( ) mR 6002/sin2 =φ
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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
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Horizontal curve setting
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
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Horizontal curve setting
A. Offsets or ordinates from long chord
• 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
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Horizontal curve setting
A. Offsets or ordinates from long chord
• 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
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Horizontal curve setting
A. Offsets or ordinates from long chord
• 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.
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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=
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( ) ( ) mR 76.512/30sin100*22/sin2 === φ
Numerical 2
• Chainage of T2
=Chainage of T1+length of curve
=123.71+52.36=176.07m
• Length of long chord
( ) ( ) 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.
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Numerical 2
• Mid ordinate=
• Ordinate at x=5m=
• Ordinate at x=10m=
• Ordinate at x=15m=
• Ordinate at x=20m=
• Ordinate at x=25m=
• 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 =−−−=
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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.
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THANK YOU
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