Vertical Curves - Solved Examples

Q1.A descending grade of 6% and an ascending grade of 2% intersect at Sta 12 + 200 km whose elevation is at 14.375 m. The two grades are to be connected by a parabolic curve, 160 m long. Find the elevation of the first quarter point on the curve? Ans:

From the grade diagram:

=

S1=120 m Horizontal distance from the lowest point to point Q SQ = S1−40 =120−40 SQ = 80 m Grade at point Q by ratio and proportion of triangles: =

gQ =

=

gQ = 0.04 Elevation of PC: Elev PC=Elev V+80*(0.06) Elev PC=14.375+80*(0.06) Elev PC=19.175 m

Difference in elevation between PC and Q:

DEPC−Q = shaded area in the grade diagram DEPC−Q = DEPC−Q =

*(gQ+0.06)*(40) (0.04+0.06)*(40)

DEPC−Q = 2 m Elevation of the first quarter point Q: Elev Q=Elev PC−DEPC−Q Elev Q=19.175−2 Elev Q=17.175 m Q2.A grade of -4.2% grade intersects a grade of +3.0% at Station 11 + 488.00 of elevations 20.80 meters. These two center grade lines are to be connected by a 260 meter vertical parabolic curve. a). At what station is the cross-drainage pipes be situated? b). If the overall outside dimensions of the reinforced concrete pipe to be installed is 95 cm, and the top of the culvert is 30 cm below the subgrade, what will be the invert elevation at the center? Ans:

From the grade diagram:

=

S1=151.67 m d = S1−130 = 151.67- 130 d = 21.67 m The cross-drainage pipe should be at the lowest point of the curve. Stationing of the lowest point indicated as point A in the figure: Sta A=Sta PI+d Sta A=11488+21.67=11509.67 Sta A=11+509.67 km Vertical distance between PC and PI: a = 130 * (0.042) a = 5.46 m Vertical distance between PC and the lowest point A:

A1=

S1 * (0.042) =

(151.67) * (0.042)

A1=3.18 m Elevation of the lowest point A: Elev A=Elev PI+a − A1 Elev A=20.80 + 5.46 − 3.18 Elev A=23.08 m

Elev of invert =Elev A−0.30−0.95 Elev of invert =23.08−0.30−0.95 Elev of invert =21.83 m Q3. A vertical curve joins a -1.2% grade to a +0.8% grade. The P.I. of the vertical curve is at station 75+00 and elevation 50.90 m above sea level. The centerline of the roadway must clear a pipe located at station 75+40 by 0.80 m. The elevation of the top of the pipe is 51.10 m above sea level. What is the minimum length of the vertical curve that can be used? Hint:

yꞌ = the offset between the critical point and the tangent passing through

the PC. z = the horizontal distance from the P.I. to the critical point. C = 0.8m (Given) = Critical clearance. Using the below equations yꞌ =

, r is given as

=

---------------------(1)

x=

+z

From (1) yꞌ =

------------------------------(2)

From (2) AL2 + (4Az - 8 yꞌ)L + 4Az2 = 0--------------------------(3) For simplicity of equation put w =

and equating (3) we get two roots

the smaller is discarded as it gives vertical curve i.e. tangent between PI and Critical point. After simplification larger root gives an expression for maximum or minimum vertical curve length. L = 4w – 2z + 4

-----------------------(4)

Ans:

Determine z: z = (75 + 40) - (75 + 00) = 0.40 sta. Determine yꞌ : Elevation of tangent = 50.90 - (-1.2)*(0.4) = 50.42 m Elevation of roadway = 51.10 - 0.80 = 51.90 m yꞌ = 51.90 - 50.42 = 1.48 m Determine w : A = g1 – g2 = (0.8) – (-1.2) = 2.0

w=

=

= 0.74

Determine L : L = 4w – 2z + 4 = 4*(0.74) – 2*(0.40) + 4 = 4.17 sta. = 417 m