Problem 01 - Symmetrical Parabolic Curve

From the grade diagram:
$\dfrac{S_1}{0.042} = \dfrac{260}{0.042 + 0.03}$

$S_1 = 151.67 \, \text{ m}$
 

 

$d = S_1 - 130 = 151.67 - 130$

$d = 21.67 \, \text{ m}$
 

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 ]