1. Geometric & Pavement Design Example Problems

2. 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) = ft.
Lcrest = KcrestA = (84)(5.16) = 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) = ft
Add the 200 ft long flat section at the top = = ft.

4. 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 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) = ((100 x 1.47)2/(32.2)(1000))(1 – 0.20tan(α)
tan(α) = 0.671(1 – 0.20tan(α)
tan(α) = – tan(α)
tan(α) = 0.471
tan(α) =
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 = ft
Stationing
T = Rtan(Δ/2) = ft
PC = PI - T = – =
PT = PC + L
PT = =

6. 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 = ft (this is the centerline radius)
Rv = centerline radius – 5 ft = ft
Rv = V2/(g(fs + e))
Use limiting value of fs from Table 3.5 = 0.150
V = 60.9 ft/s = mph OK
Determine adequacy of vertical curve
K = L/A = 580/( ) = 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 = 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 = 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 = V2/( V/29400)
S = V2/( V/29400)
Calculate stopping sight distance
SSD = reaction time + braking distance
SSD = 2.5(V) V2/( V/29400) = 364.6
V = 0.225V V V2
0 = V V – 32.81
V = or
V = ft/s = mph
Check SSD
SSD = 2.5(38.93) + (1.04)(38.932)/(2(32.2)((0.80)(0.10)+0.01( /294) + 0))
SSD = = ft
Recommended speed limit could be posted at 25 mph or even 20 mph

7. 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 (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)

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

11. 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 = =
Regular = =
Calculate total ESALs per day
1.7647(40) (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) + ( )(209/2) = 57,475/year
Grow the ESALs over 40 years from present year
Total ESALs = (57,475)(( )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

12. 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)(( )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