Download Vertical Curves Symmetrical parabolic Curves and more Lecture notes Civil Engineering in PDF only on Docsity!
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
VERTICAL ALIGNMENT
In road construction, to avoid sudden change in direction when two tangents with slopes
greater than or less than zero intersects, it is customary to introduce a vertical curve at
every such point where the angle is large enough to warrant it. The vertical curve that
may be used to connect tangents to gradually change slope from positive to negative
(summit curve) or from negative to positive (sag curve) are compound curves and
parabolic curves. Generally parabolic curves are being used because parabola effects
the transition rather better theoretically than the circle, but its selection for the purpose is
due principally to its greater simplicity of application.
Types of Parabolic Curves
- Symmetrical Parabolic Curves
- Unsymmetrical Parabolic Curves
Elements of a parabolic curve
L = length of curve, the horizontal projection of the parabolic curve from PC to PT.
H = vertical tangent offset from the vertex, V, to a point below it (summit curve) or
above it (sag curve) on the curve.
G
1
= grade (slope) of the back tangent in %
g 1
= grade (slope) of the back tangent in decimal
G 2 = grade (slope) of the forward tangent
g 2 = grade of the forward tangent in decimal
A = change in grade from PC to PT in %
Summit curves
Sag curves
PC
PC
V
V
PT
PT
L
L
H
H
PC
PC
V
V
PT
PT
L
L
H
H
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
Symmetrical Parabolic Curves
Symmetrical parabolic curve are curves with their equation as the equation of a parabola
either opening upward or downward. They are called symmetrical parabolic curve
because the length of curve to the left and to the right of the vertex, V, are equal and
equal to one-half of the total length of curve, L.
The figure shown below is a profile. The vertical axis represents the elevation at any
stations and horizontal axis represent the stations. For example, sta. PC is known, sta. V
equals sta. PC plus L/2 and sta. PT equals sta. PC plus L. Also, if elev. PC is known,
Elev. V equals elev. PC plus Y.
Relationships of the elements of a symmetrical parabolic curve
PC
V
PT
H
L
L/
L/
Elevations
Stations
Elev. V
Elev. PC
Sta. PC
Sta. V Sta. PT
Y
PC
V
PT
H
L
L/
L/
g 1
(L/2)
2
(L/2)
(g 1
)(L/2)
H
B
C
Figure X
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
Squared Property of Parabola
Y’s are the vertical tangent offsets from the tangents and X’s are the horizontal distances
from the point of tangency of the curve and tangent, where the vertical tangent offset is
reckoned, to the location of the point.
Y 1 and Y 2 are tangent offsets reckoned from the back tangents, Y 3 and Y 4 are tangent
offsets reckoned from the forward tangent.
The back tangent and the parabolic curve are tangent at point PC, thus the horizontal
distance of Y 1
and Y 2
shall be reckoned from point PC to the point location. Similarly, the
forward tangent and the parabolic curve are tangent at point PT, thus the horizontal
distance of Y 3 and Y 4 shall be reckoned from point PT to the point location. The vertical
tangent offset, H, is either reckoned from the back tangent or forward tangent, and its
horizontal distance either from PC or PT is L/2.
The square property of parabola states that in the same tangent, the ratio of the vertical
tangent offset and the square of its equivalent horizontal distance of a point is equal to
the ratio of the vertical tangent offset and the square of its equivalent horizontal distance
of another point.
Applying squared property of parabola in Figure Y, above.
Considering the back tangent.
1
1
2
2
2
2
2
1
2
2
Considering the forward tangent.
3
3
2
4
4
2
2
PC
V
PT
H
L
L/2 L/
g 1
(L/2)
2
(L/2)
(g 1
)(L/2)
C
X
3
Y
3
Y
4
X
4
Figure Y
M
N
Curve Elevation
Grade Elevation
Vertical tangent
offset
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
Therefore, for symmetrical parabolic curve.
1
1
2
2
2
2
3
3
2
4
4
2
2
Meaning the squared property of parabola can be applied to any tangent for a symmetrical
parabolic curve. You can get a tangent offset from the back tangent and can be
proportioned to any tangent offset from the forward tangent.
Using squared property of parabola to relate the elements of the curve. Considering the
vertical tangent offset through V and PT.
2
1
2
2
2
1
2
1
2
1
2
Point M is on the tangent, the elevation of points on the tangents are called grade
elevations. Point N is on the curve, the elevation of points on the curve are called curve
elevations. Y 2
is a vertical tangent offset. The relationship of grade elevation, curve
elevation, and vertical tangent offset is
Curve Elevation = Grade Elevation minus vertical Tangent offset (summit curves)
Location of the highest or lowest point on the curve.
S 1 = location of the lowest or highest point from PC
S 2 = location of the highest or lowest point from PT
Since the lowest or highest point is a single point on the curve, the length of curve, L,
shall be equal to the sum of S 1
and S 2
1
1
1
2
1
PC
V
PT
H
L
L/ L/
Y
LP X
S 1
S 2
M
N
Y
Curve Elevation
Grade Elevation
Vertical tangent offset
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
Solution:
Hint: To choose the tangent where the vertical tangent offset be reckoned, choose the
tangent of known grade. For point HP and X, their vertical tangent offsets are reckoned
from the back tangent, Y Z
and Y X
respectively.
Using squared property of parabola to solve the length of curve, L.
2
𝑋
2
2
𝑋
2
𝑋
2
Where:
Solve A and Y X
in terms of L,
Since the grade of the back tangent is given, use S 1 to solve for A,
1
1
2
1
2
Using the relationship of the curve elevation, grade elevation, and vertical tangent
offset
Curve Elevation = Grade Elevation minus vertical Tangent offset (summit curves)
Tangent offset = Grade Elevation minus curve elevation
𝑋
′
𝑋
1
𝑋
Substitute:
𝑋
2
1
2
2
1
2
2
2
2
2
There are two solutions of the problem: L = 91.789 m and L = 80.104 m.
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
Using: L = 80 m
Checking the figure: point X and point HP (highest point) locations are correct.
1
2
1
2
2
2
Solve elevation and station of HP (highest point)
2
Curve Elevation = Grade Elevation minus vertical Tangent offset
𝑍
Where:
2
1
𝑍
2
2
2
𝑍
2
2
2
Thus
SAMPLE PROBLEM NO. 2
A 6% descending grade meets an ascending grade at the vertex, V, at station 60 + 150
and at elevation 200 m. the tangents are connected by a symmetrical parabolic curve
passing through a point whose station is at station 60 + 120 having a curve elevation of
202.10 m. The lowest point on the curve is located 40 m from PC. The sag curve is also
intersected by 10 m wide overpass bridge perpendicularly such that the centerline of the
overpass bridge is at station 60 + 160 and bottom side elevation at elevation 206. 769 m.
Determine the length of the curve, the grade of the other tangent, the elevation of the
lowest point, and the maximum vertical clearance below the overpass bridge.
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
𝟐
Elevation of LP (lowest point)
𝑍
= [𝐸𝑙𝑒𝑣. 𝑉 +
1
1
] + 𝑌
𝑍
Using squared property of parabola (points LP and V)
𝑍
1
2
2
𝑍
1
2
2
Thus
𝑍
= [𝐸𝑙𝑒𝑣. 𝑉 + (
1
1
] + 𝑌
𝑍
Determine maximum vertical clearance below the overpass bridge
𝐿
𝑅
Using squared property of parabola (points M and V)
𝑀
2
2
𝑀
2
2
′
𝑀
= [𝐸𝑙𝑒𝑣. 𝑉 − ( 5 )𝑔
1
] + 𝑌
𝑀
Using squared property of parabola (points N and V)
𝑁
2
2
𝑁
2
2
′
𝑁
= [𝐸𝑙𝑒𝑣. 𝑉 − ( 15 )𝑔
1
] + 𝑌
𝑁
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
Thus
𝐿
𝐿
And
𝑅
𝑅
Since C R
is less than C L
, the maximum vertical clearance below the overpass bridge is
4.5 meters. Meaning the height of any motor vehicle that can pass through the overpass
is 4.50 m or less than 4.50 m.
SAMPLE PROBLEM NO. 3
An ascending grade meets a descending grade at the vertex, V, at station 10 + 100 and
at elevation 243.5 m. A symmetrical parabolic curve connects these tangents passing
through a point whose station is at station 10 + 120 having a curve elevation at elevation
242.205 m. The elevations of PC and PT respectively are at elevations 240.0 m and 241.
m. Determine the length of curve, the grades of the tangents, and the station and elevation
of the highest point on the curve.
Given:
Required: L, G 1 , G 2 , sta. and elev. Of HP (highest point)
Solution:
Solve the grades in terms of L
1
PC
V
PT
H
L
L/
L/
2
(L/2)
Elev. V = 243.5 m
Elev. PC = 240.0 m
Elev. PT = 241.5 m
Elev. V = 243.5 m
X
X’
HP
S 1 S 2
Sta. 10 + 100
Sta. 10 + 120
20 m
3.50 m
2.0 m
g 1
(L/2)
Elev. HP
Y X
Z
Y Z
CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVES
There are two solutions of the problem: L = 275 m and L = 100 m.
Using L = 100 m
Then
𝟏
2
𝟐
1
2
And
1
1
1
2
1
2
2
2
1
2
Check
1
2
Determine the station and elevation of HP (highest point)
𝟏
𝑍
= [𝐸𝑙𝑒𝑣. 𝑉 + (𝑆
1
1
] − 𝑌
𝑍
But
𝑍
1
2
2
𝑍
1
2
2
Then
𝐸𝑙𝑒𝑣. 𝐻𝑃 = [𝐸𝑙𝑒𝑣. 𝑉 + (𝑆
1
1
)] − 𝑌
𝑍
