Vertical Curve By D.M Siddique
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10/9/2012 1 ADVANCE ENGINEERING SURVEYING (3+1) Dr. Mohsin Siddique Asst. Prof. Dept. of Civil Engineering FAST-NU Lecture 7: Vertical Curve 04/10/2012 Vertical curves • When two different gradients meet at a point along a road surface, they form a sharp point at the apex. • Unless this apex point is rounded off to form a smooth curve, no vehicle can move along that portion of the road. So, for the smooth and safe running of vehicles, the meeting point of the gradients is rounded off to form a smooth curve in a vertical plane. This curve is known as vertical curve. Generally, the parabolic curves are preferred as it is easy to work out the minimum sight distance in their case, and the minimum sight distance is an important factor to be considered while calculating the length of the vertical curve.
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Transcript of Vertical Curve By D.M Siddique
10/9/2012
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ADVANCE ENGINEERING SURVEYING (3+1)
Dr. Mohsin Siddique
Asst. Prof.
Dept. of Civil Engineering
FAST-NU
Lecture 7: Vertical Curve
04/10/2012
Vertical curves
• When two different gradientsmeet at a point along a roadsurface, they form a sharppoint at the apex.
• Unless this apex point isrounded off to form a smoothcurve, no vehicle can movealong that portion of the road.
So, for the smooth and safe running of vehicles, the meeting point ofthe gradients is rounded off to form a smooth curve in a vertical plane.This curve is known as vertical curve.
Generally, the parabolic curves are preferred as it is easy to work outthe minimum sight distance in their case, and the minimum sightdistance is an important factor to be considered while calculating thelength of the vertical curve.
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Vertical Curves
Sight distance
• Defined as “the distance required, for a given design speed to safely stop a vehicle thus avoiding a collision with an unexpected stationary object in the roadway ahead” by AASHTO (American Association of State Highway and Transportation Officials)
• Types
▫ Stopping Sight Distance
▫ Passing Sight Distance
▫ Decision Sight Distance
▫ Horizontal Sight Distance
Vertical curves
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Vertical curvesGradient
• The gradient is expressed in two ways
▫ (a). As percentage
▫ (b). As 1 in n,
� where n is the horizontal distance and 1 represents the vertical distance
• Again the gradient may be rise or fall.
• An up gradient is known as rise and is denoted by positive sign.
• A down gradient is known as fall and it indicated by negative sign.
Vertical curves
Rate of change of grade
• The characteristics of a parabolic curve is that the gradient changes from point to point but the rate of change of grade remains constant.
• Hence, for finding the length of the vertical curve, the rate of change of grade should be an important consideration as this factor remains constant throughout the length of vertical curve.
• Generally, the recommended rate of change of grade is 0.1% per 30m at the summit and 0.05% per 30m at sags.
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Vertical curves
Length of vertical curve
• The length of vertical curve is calculate by considering the sight distance. To provide minimum sight distance, a certain permissible rate of change of grade is determined and the length of the vertical curve is calculated as follows;
• Where, g1 and g2= percentage of grade and r= rate of change
• Example: find the length of curve connecting two grades +0.5% and -0.4% where rate of change of grade is 0.1% per 30m.
r
g-g
grade of change of rate
grade of difference algebraic
grade of change of rate
grade of Changecurve verticalof Length
21==
=
( )( )
m270
0.1
300.4--0.5
r
g-gcurve verticalof Length 21
=
×==
Vertical curves
• 1. Summit curve/crest curve
▫ Figure shows a summitcurve where an upgrade isfollowed by a down grade
▫ Figure shows a summitcurve where a down grade isfollowed by anotherdowngrade
• 2. Sag curve
▫ Figure shows a sag curvewhere a down grade isfollowed by and upgrade
▫ Figure shows a sage curvewhere an upgrade isfollowed by anotherupgrade.
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Vertical curveSetting out of vertical curve
• The vertical curve may be set out by the following two methods
• The tangent correction method
• The chord gradient method
• The tangent correction method is preferred in practical situations as it involves simple calculations and curve setting.
Vertical curveSetting out of vertical curve
• THE TANGENTCORRECTION METHOD
• In figure shown, the tangentcorrection or tangent offset isthe difference of elevationbetween point P and P1, P beingpoint on the curve and P1 on thegradient.
• Then
y=RL of P1-RL of P=tangentcorrection
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Vertical curveSetting out of vertical curve
• THE TANGENTCORRECTION METHOD
• Let x be the horizontal distanceof point P along the origin. X1 isthe sloping distance along thegradient of point P1.
• Here x is taken to beapproximately equal to x1.
• The equation of curve is
• Where c is constant=
• l is the half length of verticalcurve
2cxy =
l
gg
400
21 −
Vertical curveSetting out of vertical curve
• Tangent Correction at Any Point
• Where
• x1, x2,…=distance taken along the slope measured from tangent point
• g1 and g2 =percentage of grade
( )( )1
2
121
400xx
l
xggy =
×−= Q
( )
( )on so and
400
4002
2212
2
1211
l
xggy
l
xggy
×−=
×−=
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Vertical curvePoints to be remembered
• Point to be remembered while calculating data required for setting out vertical curve
• (a). The length of vertical curve is assumed equal to the length of two tangents
BT1+BT2=T1B1+B1T2=2l
l=length of half curve
• (b). The curve is assumed equally long on either side of the apex point.
• T1B1=B1T2=l so, BT1=BT2=l
Vertical curvePoints to be remembered
• (c). The length of vertical curve is given by the formula:
• (d). Chainage of T1=chainageof B-BT1
• (e). Chainage of T2=Chainageof B+BT2
• (f).
• (g).
• (h).
• (i).
( )
r
ggL 21 −
=
100
g Bof RLT of RL 1
1 l±=
100
g Bof RLT of RL 2
2 l±=
( ) 2/T of RLT of RL Bof RL 212 +=
( ) 2/ Bof RL Bof RL Bof RL 21 +=
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Vertical curvePoints to be remembered
• (j). The tangent correction at distance x
• (k). The tangent correction is deduced from the RL of a point on thegrade to get the corresponding point on the curve.
• (l). A setting out table is prepared.
• (m). IF rise and falling gradients are equal, the verticalcurve becomes symmetrical, so tangent corrections are calculatedfor one side of the point of intersection. The tangent corrections forthe other side will be exactly same.
221
400x
l
ggyx
−=
Numerical 12
• Calculate the RL of the various station pegs on a vertical curve connecting two grades of +0.6% and -0.6%. The chainage and the RL of intersection point are 550 and 325.5m respectively. The rate of change of grade is 0.1% per 30 m.
• Solution:
• (a). Length of curve
• Length of curve on eider
side of apex is 180m.
( ) ( )
mL
r
ggL
360
301.0
6.06.021
=
+=
−=
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Numerical 12
• (b). Chainage of T1=chainage of B-BT1• =550-180=370m
• (c). Chainage of T2=Chainage of B+BT2 =550+180=730m
• (d).
• (e).
• (f).
• (g).
ml 42.324100
8010.6-50.253
100
g Bof RLT of RL 1
1 =×
=±=
ml 42.324100
1800.6 5.253
100
g Bof RLT of RL 2
2 =×
−=−=
( ) ( ) m42.3242/42.24342.2432/T of RLT of RL Bof RL 212 =+=+=
( ) ( ) m96.3242/42.2435.2532/ Bof RL Bof RL Bof RL 21 =+=+=
Numerical 12
• (h). Tangent correction at the center
=325.5-324.96=0.54m
(i). Tangent correction are found out at 30 m interval from the relation
Where, l being the half curve length
221
400x
l
ggy
−=
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Numerical 12
• Tangent correction at Point 1:
• Tangent correction at Point 2:
• Tangent correction at Point 3:
• Tangent correction at Point 4:
• Tangent correction at Point 5:
• Tangent correction at Point 6:
( )
( )mx
l
ggy 015.030
180400
6.06.0
400
22
121
1 =−−
=−
=
( )
( )mx
l
ggy 060.060
180400
6.06.0
400
22
221
2 =−−
=−
=
( )
( )mx
l
ggy 135.090
180400
6.06.0
400
22
321
3 =−−
=−
=
( )
( )mx
l
ggy 240.0120
180400
6.06.0
400
22
421
4 =−−
=−
=
( )
( )mx
l
ggy 375.0150
180400
6.06.0
400
22
521
5 =−−
=−
=
( )
( )mx
l
ggy 540.0180
180400
6.06.0
400
22
621
6 =−−
=−
=
Checked
Numerical 12
• RL on Grades
• RL of Point 1=RL of T1+rise in 30m
• =324.42+0.6(30)/100=324.6m
• RL of Point 2=RL of P1+rise in 30m
• =324.6+0.6(30)/100=324.78m
• RL of Point 3=RL of P2+rise in 30m
• =324.78+0.6(30)/100=324.96m
• RL of Point 4=RL of P3+rise in 30m
• =324.96+0.6(30)/100=325.14m
• RL of Point 5=RL of P4+rise in 30m
• =325.14+0.6(30)/100=325.32m
• RL of Point 6=RL of P5+rise in 30m
• =325.32+0.6(30)/100=325.5m
• RL on curves
• RL of Point 1=RL of Point 1 on tangent –tangent correction at point 1
• RL of Point 1=324.6-0.015=324.585m
• RL of Point 2=324.78-0.06=324.72m
• RL of Point 3=324.96-0.135=324.585m
• RL of Point 4=325.14-0.24=324.9m
• RL of Point 5=325.32-0.375=324.945m
• RL of Point 6=325.5-0.54=324.96m
(RL of B). Checked.
(RL of B1). Checked.
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Numerical 12
Point Chainage(m)
Grade RL (m)
Tangent correction (m)
Curve RL (m)
Remarks
T1 370 324.42 0 234.42 Starting of curve
1 400 324.60 0.015 324.585
2 430 324.78 0.060 324.72
3 460 324.96 0.135 324.825
4 490 325.14 0.240 324.900
5 520 325.32 0.375 324.945
6 550 325.50 0.540 324.960 Vertex
5’ 580 325.32 .375 324.945
4’ 610 325.14 0.240 324.900
3’ 640 324.96 0.135 324.825
2’ 670 324.78 0.060 324.720
1’ 700 324.60 0.015 324.585
T2 730 324.42 0 324.42 Finishing point
Numerical 13
• Calculate the RL of the various station pegs on a vertical curve connecting two grades of +0.5% and -0.3%. The chainage and the RL of intersection point are 550 and 325.5m respectively. The rate of change of grade is 0.1% per 30 m.
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