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Example #4: Designing Superelevation Design the superelevation transitions for a curve with the following parameters: 4-lane divided urban principal arterial.

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Presentation on theme: "Example #4: Designing Superelevation Design the superelevation transitions for a curve with the following parameters: 4-lane divided urban principal arterial."— Presentation transcript:

1 Example #4: Designing Superelevation Design the superelevation transitions for a curve with the following parameters: 4-lane divided urban principal arterial PC station = Design Speed = 50 mphDegree of Curvature = 3Original normal crown = 2% Calculate the tangent runout & runoff lengths E max = 4% table should be used because it is an urban facility Calculate the radius R R = /D = /3 = ft Runoff Length: Using R = 1800, e max from table = 3.3% L = 119 ft Tangent Runout Length: x = NC(%) x L runoff x = 2.0%x 119 e(%)3.3% x = ft Developing the SE transitions: SE should be developed two thirds before the curve and one third after. Calculate beginning of SE sta /3*L sta – 2/3* Superelevation begins at: 8+72 Calculate beginning of tangent runout sta 8+72 – Tangent runout begins at sta

2 Example #6: Locating Vertical Curve Elevations A crest vertical curve with a length of 400 ft connects grades +1.0% and %. The VPI station and elevation ft. What are the elevations and stations of VPC & the VPT? VPI Elevation = VPC elev +G 1 *L/2 VPI Station = VPC station + L/2 VPC = *400/2 VPC elevation = ft VPC station = 3500–400/2 VPC station = VPI Elevation = VPT elev - G 2 *L/2 VPT Station = VPC station + L VPT = ( )*400/2 VPT elevation = ft VPT station = /2 VPT station = 37+00

3 Example #7: Designing a Crest Vertical Curve Design a length of vertical curve for a crest curve based on SSD and the following information: Design Speed = 40 mphg1 = 1.25%g2 = -2.75% Solution: Determine the change in grade elevation A = |g2-g1| = | | A = 4.0% AASHTO Design Controls for SSD (Exhibit 3-76) - Min. SSD = 305 ft Assuming S > L L = 2S -2158/A L = 2* /4 L = L = 70.5 ft Assumption is correct S > L, We have found the minimum length of curve! Remember MoDOT recommends a length of curve of at least 300 ft when possible. What if we had started with this assumption ? S < L L = AS 2 /2158 L = 4*305 2 /2158 L = ft S < L, Assumption is wrong, would have had to use the other formula to obtain the correct answer

4 Example #8: Designing a Sag Vertical Curve Design a length of vertical curve for a sag curve based on 1.Comfort Ride 2.Headlight Sight Distance Design Speed = 40 mphSSD = 305 g1 = -3.2%g2 = +2.4% Solution: Determine the change in grade elevation A = |g2-g1| = |2.40-(-3.2)| A = 5.6 % Comfort Ride Equation L=AV 2 /46.5 L = 5.6*40 2 /46.5 L = ft Remember, MoDOT rounds length of vertical curve to the nearest 10 ft Headlight SD: Assume S > L: L = 2S – [( S)/A] L = 2*305-[( *305)/5.6] L = 610 – L = ft S < L Assumption is wrong, let’s try again Headlight SD: Assume S < L: L=AS 2 /( S) L = 5.6*305 2 /( *305) L = / L = ft S < L Assumption is correct


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