Geometric & Pavement Design Example Problems

Problem 3.15 Due to crashes at a railroad crossing, an overpass (with a surface 24 ft above the existing road) is to be constructed on an existing level highway. The existing highway has a design speed of 50 mph. The overpass structure is to be level, centered above the railroad, and 200 ft long. What length of the existing level highway must be reconstructed to provide an appropriate vertical alignment? Find K values for the crest and sag vertical curve. Ksag = 96 for 50 mph Kcrest = 84 for 50 mph Because the combination sag & crest curve goes between two flat grades: G1 for the sag = 0 G2 for the sag = G1 for the crest G2 for the crest = 0 Therefore, A for the crest = A for the sag Use L = LA to express length in terms of K. We know that A is the same for both because A = G2 – G1 and Lsag = 96A Lcrest = 84A Use a vertical relationship to get the second equation: AL/200 = Yf = AL/200 Therefore, ALsag/200 + ALcrest/200 = 24 ft Substitute and solve for A A(96A)/200 + A(84A)/200 = 24 A = 5.16% = G2 for the sag = G1 for the crest Find curve lengths Lsag = KsagA = (96)(5.16) = 495.36 ft. Lcrest = KcrestA = (84)(5.16) = 433.44 ft Find total length of highway that must be reconstructed Need one sag and one crest curve on each end = 2(495.36) + 2(433.44) = 1857.6 ft Add the 200 ft long flat section at the top = 1857.6 + 200 = 2057.6 ft.

Problem 3.27 A developer is having a single-lane raceway constructed with a 100 mph design speed. A curve on the raceway has a radius of 1000 ft, a central angle of 30 degrees, and PI stationing at 1125+10. If the design coefficient of side friction is 0.20, determine the superelevation required at design speed (do not ignore the normal component of the centripetal force). Also, compute the degree of curve, length of curve, and stationing of the PC and PT. The proper equation is equation 3.33 tan(α) + fs = (V2/gRv)(1 – fstan(α)) Although In the typical formula the fstan(α) term is ignored because it is small, this is the normal component of the centripetal acceleration and will not be ignored here. Solve for α tan(α) + 0.20 = ((100 x 1.47)2/(32.2)(1000))(1 – 0.20tan(α) tan(α) + 0.20 = 0.671(1 – 0.20tan(α) tan(α) + 0.20 = 0.671 – 0.1342tan(α) 1.1342 tan(α) = 0.471 tan(α) = 0.4153 α = 22.55° e = 41.53 If we ignored the normal component of centripetal acceleration, e = 47.11 As we get into big superelevations then it is no longer practical to ignore the normal component Degree of curvature (π/180)RΔ = 100Δ/D D = 5.73 Curve length L = 100Δ/D = 523.56 ft Stationing T = Rtan(Δ/2) = 267.95 ft PC = PI - T = 112510 – 267.95 = 1122+42.05 PT = PC + L PT = 112510 + 523.56 = 1130+33.56

Geometric Design Example A 2-lane (10 ft wide lanes) combined horizontal and crest vertical curve is currently posted for a 35 mph speed limit and is designed for 40 mph. Curve data are: Horizontal: L = 580 ft, Δ = 60°, e = 6%, Ms = 30 ft Vertical: L = 580 ft, G1 = 3%, G2 = -2.5% Is this curve adequately designed for 40 mph? Because this curve is often covered in packed snow and ice, the county wants to post a recommended speed limit for these conditions. What speed limit do you recommend? Determine if horizontal curve is adequate for 40 mph R = 180L/πΔ = 553.9 ft (this is the centerline radius) Rv = centerline radius – 5 ft = 548.9 ft Rv = V2/(g(fs + e)) Use limiting value of fs from Table 3.5 = 0.150 V = 60.9 ft/s = 41.44 mph OK Determine adequacy of vertical curve K = L/A = 580/(3 - -2.5) = 105.5 From Table 3.2, K = 44 for 40 mph What is the available SSD? Horizontal curve: SSD = (πRv/90)(cos-1((Rv – Ms)/Rv)) SSD = (19.16)cos-1(0.9453) SSD = 364.6 ft This SSD consists of braking distance + reaction time (2.5 x initial speed) Vertical curve (assume SSD < L): SSD2 = 2158L/A = 2158(580)/5.5 = 227,571 SSD = 477 ft Therefore, horizontal is more limiting and SSDavailable = 364.6 ft On packed snow and ice the braking distance will be much greater due to reduced friction S (braking distance ignoring aerodynamic resistance) = γb(V2)/(2g(ηbμ + frl +/- sinθ) Assumptions Mass factor = 1.04 Braking efficiency = 0.80 (can’t really assume ABS at this juncture) Coefficient of road adhesion from Table 2.4 = 0.10 Grade = assume 0, since part will be uphill and part will be downhill S = (1.04)(V2)/(2(32.2)((0.80)(0.10)+0.01(1+V/294) + 0)) S = 0.0161V2/(0.08 + 0.01 + V/29400) S = 0.0161V2/(0.09 + V/29400) Calculate stopping sight distance SSD = reaction time + braking distance SSD = 2.5(V) + 0.0161V2/(0.09 + V/29400) = 364.6 32.81 + 0.0124V = 0.225V + 0.000085V2 + 0.0161V2 0 = 0.01619V2 + 0.2126V – 32.81 V = 38.93 or -52.06 V = 38.93 ft/s = 26.48 mph Check SSD SSD = 2.5(38.93) + (1.04)(38.932)/(2(32.2)((0.80)(0.10)+0.01(1+38.93/294) + 0)) SSD = 97.3 + 268.0 = 365.3 ft Recommended speed limit could be posted at 25 mph or even 20 mph

Problem 3.37 Two straight sections of freeway cross at a right angle. At the point of crossing, the east-west highway is at elevation 150 ft and is a constant +5.0% grade (upgrade in the east direction), the north-south highway is at elevation 125 ft and is a constant -3.0% grade (downgrade in the north direction). Design a 90-degree ramp that connects the northbound direction of travel to the eastbound direction of travel. Design the ramp for the highest design speed (to the nearest 5 mph) without exceeding a minimum D of 8.0. (Assume the PC of the horizontal curve is at station 15+00, and the vertical curve PVIs are at the PC and PT of the horizontal curve). Maximum allowed superelevation by design standards is 12%. Determine the following: Stationing and elevations of the horizontal curve PC and PT Horizontal curve radius and length Stationing and elevations of the vertical curve PVCs and PVTs Vertical curve lengths Length of constant grade in between the 2 vertical curves (if any)

Elevation difference = 25 ft North Downhill (-3.0%) Uphill (+5.0%) EW freeway Elevation difference = 25 ft NS freeway

Pavement Design Example The pavement on Stevens way is slowly being switched to PCC a few thousand yards at a time. Assuming all the heavy vehicle traffic on Stevens Way is due to bus traffic, estimate a rigid pavement design using the following data: Design life = 40 years Articulated bus average axle weights: Front = 15,000 lb Middle = 18,500 lb Rear = 11,500 lb Regular (non-articulated) bus average axle weights: Front = 12,200 lb Rear = 20,000 lb Typical school day = 80 regular, 40 articulated in one direction Bus traffic growth rate = 2% per year Use undowel joints and crushed stone base material Convert each type of bus to an ESAL (use 4th power law as approximation) Articulated = 0.4823 + 1.1158 + 0.1667 = 1.7647 Regular = 0.211 + 1.52 = 1.7352 Calculate total ESALs per day 1.7647(40) + 1.7352(80) = 209 ESALs/day Calculate total lifetime ESAL loading ESALs per day during the school year is probably much higher than not during Assuming ESAL loading is about half during the summer and during weekends Assume summer = 3 months = 100 days Assume school year = 265 days (of which about 80 are weekends/holidays) ESALs = (185)(209) + (80 + 100)(209/2) = 57,475/year Grow the ESALs over 40 years from present year Total ESALs = (57,475)((1 + 0.02)40 – 1)/0.02 = 57,475(60.40) = 3,471,604 ESALs Use WSDOT Table to determine PCC design Need to select a reliability (might want to use a high one since it is a highly visible road) If reliability = 75% then design is 9” PCC over 4” or 6” of base course If reliability = 85% then design is 9.5” PCC over 4” or 6” of base course If reliability = 95% then design is 10.5” PCC over 4” or 6” of base course

Pavement Design Example What would a flexible pavement design be for the same loading? Assume: Two layers (HMA and a granular base course) Subgrade MR = 10,000 psi Grow the ESALs over 40 years from present year Total ESALs = (57,475)((1 + 0.02)40 – 1)/0.02 = 57,475(60.40) = 3,471,604 ESALs Use WSDOT Table to determine flexible design Need to select a reliability (might want to use a high one since it is a highly visible road) If reliability = 75% then design is 8” HMA over 4” of base course If reliability = 85% then design is 8.5” HMA over 4” of base course If reliability = 95% then design is 10.0” HMA over 4” of base course Keep in mind the WSDOT assumptions in the table Note that pavement depths of HMA and PCC are quite similar Mention that ESALs for rigid and flexible pavement are not the same but it is okay for the broad assumptions of this example