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Transportation Engineering Vertical curve design March 23, 2011

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Presentation on theme: "Transportation Engineering Vertical curve design March 23, 2011"— Presentation transcript:

1 Transportation Engineering Vertical curve design March 23, 2011
CE 3500 Transportation Engineering Vertical curve design March 23, 2011

2 ANNOUNCEMENTS

3 Group assignment due today.
Homework 4 posted later today. No class Wednesday.

4 REVIEW

5 The shape of a vertical curve is usually parabolic. Why?
PVI A = –8 G2= -5 G1= +3 L PVT PVC

6 For convenience, let x = 0 at PVC
Elevation y For convenience, let x = 0 at PVC PVC PVT

7 PVC PVT Case I: SSD is less than the curve length.

8 PVC PVT Case II: SSD is greater than the curve length.

9 Case I: SSD is less than the curve length.

10 Case II: SSD is greater than the curve length.

11 The first part of vertical alignment is identifying the constant slope sections.

12 From this, we identify where vertical curves are needed and calculate their PVI.

13 So, usually we are given the PVI and have to design everything else based off of that.

14 EXAMPLE

15 Design a vertical curve that can be safely traversed at 70 mi/hr, assuming 2.5 s reaction time and coefficient of braking friction 0.348 G1= -2 G2= +4 PVI: Elevation 7250 ft

16 Step 1: Find SSD G1= -2 G2= +4 PVI: Elevation 7250 ft

17 Step 2: Find curve length
SSD < L: SSD > L: G1= -2 G2= +4 PVI: Elevation 7250 ft

18 Step 2: Find curve length
SSD < L: SSD > L: G1= -2 G2= +4 PVI: Elevation 7250 ft SSD is 759 ft, so we need 1130 ft of curve

19 How do we report our answer
How do we report our answer? Give station and elevation for PVC, PVT, and every whole station in between. PVI is at ; so PVC comes (1130/2) = 565 ft earlier: PVT comes 565 ft after PVI: What is the elevation of PVC and PVT? What is the elevation in between?

20 Station x (ft) Elevation (ft) (PVC) 7261.3 340+00 55 7260.3 341+00 155 7258.8 ... 350+00 1055 7269.7 (PVT) 1130 7272.6

21 EXAMPLE

22 Coefficient of friction is 0
Coefficient of friction is 0.348; only one direction of travel from left to right. What is maximum safe speed? G2= -3 G1= +5 500 ft

23 Step 1: Find SSD from curve length:
SSD < L: SSD > L: G2= -3 G1= +5 500 ft

24 Step 2: Find speed from SSD:

25 IN-CLASS EVALUATIONS

26 Pace: Slow, OK, Fast Most unclear topic right now. What is most helpful to you? How can I improve my teaching? Any other comments. (No need to provide names. Place papers on back table on the way out.)


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