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ENGINEERING SURVEY (CC201) TOPIC 4 CURVE RANGING

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Page 1: Curve

ENGINEERING SURVEY (CC201)

TOPIC 4CURVE RANGING

Page 2: Curve

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.

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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

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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.

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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.

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TYPE OF CURVETYPE OF CURVE

1. HORIZONTAL CURVEi) Circular Curveii) Transition Curveiii) Combined Curve

2. VERTICAL CURVE

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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

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HORIZONTAL CURVE Circular Curve

HORIZONTAL CURVE Circular Curve

Compound Curve

Reverse Curve

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HORIZONTAL CURVE Transition Curve

HORIZONTAL CURVE Transition Curve

Transition Curve Transition Curve

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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

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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

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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

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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

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Setting Out With Deflection AngleSetting Out With Deflection Angle

T1 T2

A B C

α2α1

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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

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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

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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

+=

+=

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Setting Out Offset From Tangent LineSetting Out Offset From Tangent Line

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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.

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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

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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

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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

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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

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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

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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

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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.

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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

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SETTING OUT COMBINED CURVESETTING OUT COMBINED CURVE

Transition CurveTransition Curve

Circular Curve

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SETTING OUT COMBINED CURVESETTING OUT COMBINED CURVE

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SETTING OUT COMBINED CURVESETTING OUT COMBINED CURVE