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ENGINEERING SURVEY (CC201)
TOPIC 4CURVE RANGING
OBJECTIVEOBJECTIVE
At the end of the unit you should be able to :-
• Explain the basic concept of curves.• To identify the terminologies of curves. • To differentiate between circular curves, transition curves and vertical curves.• Explain the methods of setting out circular curves.• Calculate setting out of circular curves.• Apply the method of setting out of a circular curves.
INTRODUCTIONINTRODUCTION
• In the geometric design of roads, railways and pipelines, the design and setting out of curves is an important aspect of an engineer’s work. • Curve ranging is a design of 2 straight ways with a curve. • In the design of roads or railways, the parts of straight line/ways connected with a curve whose its radius is constant or variables.
θ
Straig
ht w
ay
(left)
Straight way (right)
Curve
Radius
PURPOSESPURPOSES
i) To connected 2 straight ways with curve through a deflection angle.
ii) To allow the slowly movement on straight ways and curve in horizontal and vertical direction.
REASON OF CURVE RANGINGREASON OF CURVE RANGING
i) The physical conditions of earth surface is hilly land and swampy.
ii) To avoid from fixed items such as building mosques, cemeteries and so on.
TYPE OF CURVETYPE OF CURVE
1. HORIZONTAL CURVEi) Circular Curveii) Transition Curveiii) Combined Curve
2. VERTICAL CURVE
HORIZONTAL CURVECircular Curve
HORIZONTAL CURVECircular Curve
• Curve that have same radius.
• Type of circular curve
i) Simple curve (mudah)
- Same radius
i) Simple curve (mudah)
- Same radius
Radiu
sii) Compound curve (berbagai)
- 2 and more circular curve
- Different circular curve and radius.
ii) Compound curve (berbagai)
- 2 and more circular curve
- Different circular curve and radius.
iii) Reverse curve (songsang)
- Same radius
- 2 inverse curve
iii) Reverse curve (songsang)
- Same radius
- 2 inverse curve
HORIZONTAL CURVE Circular Curve
HORIZONTAL CURVE Circular Curve
Compound Curve
Reverse Curve
HORIZONTAL CURVE Transition Curve
HORIZONTAL CURVE Transition Curve
Transition Curve Transition Curve
HORIZONTAL CURVE Transition Curve
HORIZONTAL CURVE Transition Curve
• The radius is vary from one point to another.
• The purpose of this circular is to maintain comfortable and safety while the driver through on the straight line to the curve.
• A vehicle moving from the straight with no centrifugal force acting upon it, into a curve would suddenly receive the maximum amount of centrifugal force for that radius of curve.
• To prevent this sudden lateral shock on passengers in the vehicle, a transition curve is inserted between the straight circular curve
HORIZONTAL CURVE Combined Curve
HORIZONTAL CURVE Combined Curve
• Combined curve is a combination of circular curve and transition curve.
• Usually there are two transition curves, in beginning and ending, while circular curve is in the center.
Transition Curve Transition Curve
Circular Curve
CIRCULAR CURVE TERMINOLOGIESCIRCULAR CURVE TERMINOLOGIES
TERMINOLOGIES AND ITS FORMULA :
Tangent Line/Subtangent (T) = R Tan θ/2
Long Chord (BC – EC) = 2 R Sin θ/2
Mid-Ordinate (M) = R(1 - Cos θ/2 )
Curve’s Length (L) = Rπθ / 180º
External (E) = R (sec θ/2 – 1)
Curve Deflection Angle (α) = 1718.9 (Sub Chord / R)Sub ChordSub Chord
α = curve deflection angleα = curve deflection angle
TangentTangent
CALCULATION FOR SETTING OUT CIRCULAR CURVE
CALCULATION FOR SETTING OUT CIRCULAR CURVE
Type of setting out curve :
1. Setting Out With Curve Deflection Angle
2. Setting Out Offset From Tangent Line
3. Setting Out Offset From Long-Chords
4. Setting Out Offset From Chord Produced
Setting Out With Deflection AngleSetting Out With Deflection Angle
T1 T2
A B C
α2α1
Setting Out With Deflection AngleSetting Out With Deflection Angle
Example :
The centre-line of two straights is projected forward to meet at I, the tangent deflection angle being 30°. If the straight ways are to be connected by a circular curve of radius 200 m, tabulate all the setting-out data, assuming 20-m chords on a through chainage basis, the chainage of I being 2259.59 m.
T1 T2
A B C
α2α1
2259.59m θ = 30º
I
Setting Out By Offsets With Deflection Angle
Setting Out By Offsets With Deflection Angle
Solution :
1. Tangent length = R (tan θ/2) = 200 tan 15°
= 53.59 m
2. Chainage of T1 = Chainage of I – Tangent Length= 2255.59 - 53.59
= 2202 m
3. Length of circular arc (L)= Rπθ/180° = 200(π)(30°)/180= 104.72 m
4. Chainage of T2 = Chainage of T1 + Length of circular arc = 2202 m + 104.72 m = 2306.72 m
Setting Out With Curve Deflection AngleSetting Out With Curve Deflection Angle
Solution :
5. Table of tabulation setting out dataChord number
Chordlength (c )
Chainage (m) Curve Deflection angleo , „
Setting-out angleo , „
Remarks
T1 0 2202. 00 0 00 00 0 00 00 T1
A 18 2220.00 2 34 42 2 34 42 peg 1
B 20 2240.00 2 51 53 5 26 35 peg 2
C 20 2260.00 2 51 53 8 17 58 peg 3
D 20 2280.00 2 51 53 11 09 51 peg 4
E 20 2300.00 2 51 53 14 01 44 peg 5
T2 6.72 2306.72 0 57 45 14 59 29 T2
Check: Last data of setting out angle = θ/2 = 14° 59' 29“15° Check: Last data of setting out angle = θ/2 = 14° 59' 29“15°
1718.9 (c/R) / 60
+=
+=
Setting Out Offset From Tangent LineSetting Out Offset From Tangent Line
Setting Out Offset From Tangent LineSetting Out Offset From Tangent Line
Example :
The tabulation of data required to setting out curve. Radius of circular curve is 600m length whose 2 straight ways connected with its tangent deflection angle is 182400. The chainage of I being 2140.00m .
Solution :
1. Tangent length = R (tan θ/2) = 600 tan 1824/2
= 97.20m
Example :
The tabulation of data required to setting out curve. Radius of circular curve is 600m length whose 2 straight ways connected with its tangent deflection angle is 182400. The chainage of I being 2140.00m .
Solution :
1. Tangent length = R (tan θ/2) = 600 tan 1824/2
= 97.20m
Syarat yang dikenakan iaitu panjang bagi garis rentas pendek hendaklah lebih kecil daripada R/20, tetapi sekarang R/20 = 30m, di mana sela ini adalah mencukupi untuk tujuan ‘peg spacing’. Walau bagaimana pun untuk lebih mudah, garis rentas sepanjang 20m akan digunakan.
Syarat yang dikenakan iaitu panjang bagi garis rentas pendek hendaklah lebih kecil daripada R/20, tetapi sekarang R/20 = 30m, di mana sela ini adalah mencukupi untuk tujuan ‘peg spacing’. Walau bagaimana pun untuk lebih mudah, garis rentas sepanjang 20m akan digunakan.
Setting Out Offset From Tangent LineSetting Out Offset From Tangent Line
Point Y X = R - √ (R2 – Y2)
1 20 0.333
2 40 1.335
3 60 3.008
4 80 5.357
5 97.20 7.926
2. Tabulation of data 2. Tabulation of data
Setting Out Offset From Tangent Line (Subtangent)
Setting Out Offset From Tangent Line (Subtangent)
Y2Y2
Y1Y1 X1X1
X2X2
T1T1T2T2
Setting Out procedure :
1. Started from T1 and end at I.
2. From T1, Measure Y1 = 20m with measuring tape. From point of 20m, offset X1 = 0.333m. Mark as point A.
3. From T1, measure Y2 = 40. From point 40m, offset X2 = 1.335. Mark as point B.
4. Tabulation of data is only fixing half curve (T1 – I). Its will be use for setting out T2 – I.
Setting Out procedure :
1. Started from T1 and end at I.
2. From T1, Measure Y1 = 20m with measuring tape. From point of 20m, offset X1 = 0.333m. Mark as point A.
3. From T1, measure Y2 = 40. From point 40m, offset X2 = 1.335. Mark as point B.
4. Tabulation of data is only fixing half curve (T1 – I). Its will be use for setting out T2 – I.
AA BB
II
Setting Out Offset From Long ChordSetting Out Offset From Long Chord
Example :
The tabulation of data required to setting out curve. Radius of circular curve is 600m length whose 2 straight ways connected with its tangent deflection angle is 182400. The chainage of I being 2140.00m .
Solution :
1. Long Chord Length = 2R (Sin θ/2)= 2(600) (Sin
182400/2)= 191.857m
2. Therefore ; Long Chord(LC) / 2 = 191.857/2= 95.929 m
Example :
The tabulation of data required to setting out curve. Radius of circular curve is 600m length whose 2 straight ways connected with its tangent deflection angle is 182400. The chainage of I being 2140.00m .
Solution :
1. Long Chord Length = 2R (Sin θ/2)= 2(600) (Sin
182400/2)= 191.857m
2. Therefore ; Long Chord(LC) / 2 = 191.857/2= 95.929 m
Setting Out Offset From Long ChordSetting Out Offset From Long Chord
3. Tabulation of data3. Tabulation of data
Point Y X = [R2 – Y2] - [R2 – (LC/2)2]
0 0 7.718
1 20.00 7.552
2 40.00 6.383
3 60.00 4.711
4 80.00 2.361
5 95.929 0
Setting Out Offset From Long ChordSetting Out Offset From Long Chord
Setting Out procedure :
1. Started from center of Long Chord (Y0). At Y0, Offset X0 = 7.718m. Mark as an A.
2. From centre, Measure Y1 = 20m with measuring tape. From point of 20m, offset X1 = 7.552m. Mark as point B.
3. Tabulation of data is only fixing half curve (Center LC – T1). Its will be use for setting out from Center LC to T2.
Setting Out procedure :
1. Started from center of Long Chord (Y0). At Y0, Offset X0 = 7.718m. Mark as an A.
2. From centre, Measure Y1 = 20m with measuring tape. From point of 20m, offset X1 = 7.552m. Mark as point B.
3. Tabulation of data is only fixing half curve (Center LC – T1). Its will be use for setting out from Center LC to T2.
X0 = 7.718X0 =
7.718
Y2Y2
Y1= 20 Y1= 20
X1= 7.552
X1= 7.552
Y2Y2
Y1= 20 Y1= 20
X1= 7.552
X1= 7.552
AABB BB
T1T1 T2T2
Setting Out Offset From Chord Produced(Sub Chord)
Setting Out Offset From Chord Produced(Sub Chord)
Example :
The tabulation of data required to setting out curve. Radius of circular curve is 720m length whose 2 straight ways connected with its tangent deflection angle is 121314. The chainage of I being 855.94m . Subchord = 20m
Solution :
1. Tangent length = R (tan θ/2) = 720 tan 121314/2
= 77.076m2. Chainage of T1 = Chainage I – Tangent Length
= 855.94 – 77.076= 778.864m
3. Curve Length = Rπθ/180= 720 121314/180=153.568m
4. Chainage of T2 = 778.864 + 153.568= 932.432m
Example :
The tabulation of data required to setting out curve. Radius of circular curve is 720m length whose 2 straight ways connected with its tangent deflection angle is 121314. The chainage of I being 855.94m . Subchord = 20m
Solution :
1. Tangent length = R (tan θ/2) = 720 tan 121314/2
= 77.076m2. Chainage of T1 = Chainage I – Tangent Length
= 855.94 – 77.076= 778.864m
3. Curve Length = Rπθ/180= 720 121314/180=153.568m
4. Chainage of T2 = 778.864 + 153.568= 932.432m
Setting Out Offset From Chord Produced(Sub Chord)
Setting Out Offset From Chord Produced(Sub Chord)
Point Chainage Sub chord
Offset
T1 778.864 - -
A 780 1.136 (a) 1st Offset = a2 / 2R = 0.001 m
B 800 20 (c) 2nd Offset = c( c + a)/ 2R = 0.294 m
C 820 20 (c) Other Offset = c2 / R = 0.556 m
D 840 20 (c) Other Offset = c2 / R = 0.556 m
E 860 20 (c) Other Offset = c2 / R = 0.556 m
F 880 20 (c) Other Offset = c2 / R = 0.556 m
G 900 20 (c) Other Offset = c2 / R = 0.556 m
H 920 20 (c) Other Offset = c2 / R = 0.556 m
T2 932.432 12.432 (b) Last Offset = b (b + c)/2R = 0.280 m
5.5.
Setting Out Offset From Chord Produced (Sub Chord)
Setting Out Offset From Chord Produced (Sub Chord)
1st Offset = 0.0011st Offset = 0.001
2nd= 0.2942nd= 0.294 Other Offset= 0.556Other Offset= 0.556
Last Offset = 0.280Last Offset = 0.280
1.136m1.136m
20m20m20m20m
12.432m12.432m
T1T1
T2T2
II
SETTING OUT COMBINED CURVESETTING OUT COMBINED CURVE
Transition CurveTransition Curve
Circular Curve
SETTING OUT COMBINED CURVESETTING OUT COMBINED CURVE
SETTING OUT COMBINED CURVESETTING OUT COMBINED CURVE