Class_19_Vertical_Curvature_I

# Class_19_Vertical_Curvature_I - The Pennsylvania State...

This preview shows pages 1–14. Sign up to view the full content.

The Pennsylvania State University Department of Civil and Environmental Engineering CE 321: Highway Engineering Class 19 – Vertical Curvature (Part 1) Spring 2008

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Vertical Curves
Crest Curve Design

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Sag Curve Design
Types of Vertical Curves

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Vertical Curve Fundamentals y = ax 2 + bx + c assuming a constant rate of change of slope and equal tangent lengths. When x = 0, y = C = elevation on curve dy / dx = 2ax + b = slope (rise / run) at x = 0 => slope is dy/dx = b = G 1 (the initial slope)
Vertical Curve Fundamentals d 2 y/dx 2 = 2a = rate of change of slope 2a = (G 2 - G 1 ) / L a = (G 2 - G 1 ) / 2L therefore: y x = a x 2 + G 1 x + Elevation at PVC Where: L (sta.), x (sta.)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
High / Low Point on Curve When dy/dx = 0 then a = (G 2 - G 1 ) / 2L therefore, dy/dx = 2ax + b = 0 2ax + G 1 = 0 2ax = - G 1 2(G 2 - G 1 )(x)/2L = - G 1 then: x = - G 1 L / (G 2 - G 1 )
Vertical Curve Design Vertical Curves maximum grades depend on: Design Speed Type of Terrain Type of Highway Length of Grade

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Vertical Curve Design Grades also affect: fuel consumption speed safety or crashes (speed differential)
Vertical Curve Design Design Speed Flat / Rolling Mountainous 50 mi/hr 5% 7% 60 mi/hr 4% 6% 70 mi/hr 3% 5% (+ 2 % for secondary / local roads)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Example Problem 1 A 600 ft equal tangent sag vertical curve has the PVC at station 170+00 and elevation 1000 ft. The initial grade is -3.5 % and the final grade is 0.5 %. Determine the elevation and stationing of the PVI, PVT, and the lowest point on the curve.
Example 1 (con’t) Solution: Station PVI = 170+00 + 300 = 173+00 Station PVT = 170+00 +

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern