# Problem 01 - Symmetrical Parabolic Curve

**Problem**

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 gradelines are to be connected by a 260 meter vertical parabolic curve.

- At what station is the cross-drainage pipes be situated?
- Sta 11 + 493.42 km
- Sta 11 + 509.67 km
- Sta 11 + 515.14 km
- Sta 10 + 600.00 km

- 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?
- 25.76 m
- 22.15 m
- 27.32 m
- 21.83 m

**Solution**

$\dfrac{S_1}{0.042} = \dfrac{260}{0.042 + 0.03}$

$S_1 = 151.67 \, \text{ m}$

$d = S_1 - 130 = 151.67 - 130$

$d = 21.67 \, \text{ m}$

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:

$\text{Sta } A = \text{Sta } PI + d$

$\text{Sta } A = 11488 + 21.67 = 11509.67$

$\text{Sta } A = 11 + 509.67 \, \text{ km}$ Answer for Part 1: [ B ]

Vertical distance between PC and PI:

$a = 130(0.042)$

$a = 5.46 \, \text{ m}$

Vertical distance between PC and the lowest point A:

$A_1 = \frac{1}{2}S_1(0.042) = \frac{1}{2}(151.67)(0.042)$

$A_1 = 3.18 \, \text{ m}$

Elevation of the lowest point A:

$\text{Elev } A = \text{Elev } PI + a - A_1$

$\text{Elev } A = 20.80 + 5.46 - 3.18$

$\text{Elev } A = 23.08 \, \text{ m}$

$\text{Elev of invert } = \text{Elev } A - 0.30 - 0.95$

$\text{Elev of invert } = 23.08 - 0.30 - 0.95$

$\text{Elev of invert } = 21.83 \, \text{ m}$ Answer for Part 2: [ D ]