CIVE 319: Transportation Engineering Winter 2014

ASSIGNMENT # 5

Assigned: April 2, 2014 Due: April 11, 2014

HOMEWORK IS DUE AT 4:00 P.M. AND NO LATE ASSIGNMENTS WILL BE ACCEPTED!

Solve the following problems: 1. Solve problem 3.23:

An existing equal-tangent sag vertical curve is design for 60mi/h. The initial grade is -3% and the elevation of the PVT is 754ft. The PVC of the curve is at station 13,416ft and the PVI is at 13,732ft. An overpass is being constructed directly above the PVI. The highway is for cars ONLY and AASHTO minimum and recommended structure clearance do not apply. Also the overpass design assumes the drivers eye height is set conservatively to 5ft. What is the lowest possible elevation of the bottom of the overpass structure to ensure sufficient stopping sight distance at 60 mi/h?

H1 =5, H2 =2 G1 = -3%

SPVC = 13,416 ft, SPVI = 13,732 ft and EPVT = 754 ft Ks = 136 for 60 mi/h L = 2 * (SPVI - SPVC) = 2(13,732 13,416) = 632 ft L = Ks * A, therefore, A = L / Ks = 632 / 136 = 4.65 SSD = 570 ft for SD =60mi/h

Now, get L =A*SSD2 / [800 * (Hc (H2 + H2)/2] From this, get Hc = 6.5 ft Then, we get also: Ym = AL/800 = 3.67 ft Minimum elevation: Emin = EPVI + Ym + Hc G2 = A + G1 = 1.65 EPVI = EPVT + (G2/100) * (L/2) = 748.8 ft Emin = EPVI + Ym + Hc = 758.9 ft

2. Solve problem 3.32: A sag curve and crest curve connect a -3.5% tangent section of highway (to the west) with a +2.5% tangent section of highway to the east. The +2.5% tangent is at a higher elevation than the -3.5% section. The two tangents are separated by 1150ft of horizontal distance. If the design speed of he curves is 50mi/h, what is the common grade between the sag and crest curves (G2 of sag and G1 of the crest from the west to east)? And what is the elevation difference between the PVCs and PVTc

Solution L = 1275 kc = 84 from table 3.2, speed = 50mi/h ks = 96 from table 3.3, speed = 50mi/h Then Ls = ks * As Lc = kc * Ac With As = |G1 G2| and And Ac = |G2 G3| Where G2 = ?, and G1 = -3% and G3 = +2%. (We do not know the sing of G2, but we know that this should be positive and > 2% to a crest curve to exist). Therefore, to guarantee that this becomes an addition (G1=-3 and G2 =+) and to eliminate the absolute number: As = G2 G1 And L = Ls + Lc Solving for G : L = ks * As + kc * Ac Solving for G2: G2 = [L + ks * G1 + kc * G2] / [ks + kc] = 6.4%

1150ft

Then, As = | -3.0 - 6.4 | = 9.4 Ac = | -6.4 2.0 | = 4.4 and Ls = 904 ft Lc = 372 ft Using the offset, calculate elevation difference This is obtained as: Yf = AL/200 [AsLs]/200 + [G1/100] Ls + [G3/100] Lc + [Ac Lc]/200 = 31 ft

3. Solve problem 3.46 A horizontal curve on a two-lane highway (12-ft lanes) has PC at station 123+80 and PT at station 129 + 60. The central angle is 35 degrees, the super-elevation is 0.08, and 20.6 ft is cleared (for sight distance) from the inside edge of the innermost lane. Determine a maximum safe speed (assuming current design standards) to the nearest 5 mi/h.

NOTE: Alternatively, you can work with the international system.

Solution

V = 60 H1 = 5 and H2 =2 (as before) G1 =-3% PVC =13,416 PVI = 13,732 EPVT = 754 ks= 136 for 60mi/h Curve length: L = 2(13,732 - 13,416) = 632 Then: A = L / Ks = 4.65 Then, using Eq 3.27 L = (A * SSD2) / {800 *[ Hc (H1 + H2)/2} With SSD = 570ft from table and for V = 60mi/h Solving for Hc Hc = 6.5ft Finally, And EPVI = 748 using the offset formula.