CE 453 Sample Problem for Spiral and superelevation computations
The following problem will be started in class. Each student is expected to complete the computations that are not finished in class and be prepared to turn the solution in for grading on Wednesday, October 5. Several equations for spirals and superelevation are provided in the Iowa Design Manual. Prepare a table to summarize your results.
Given Pi = Station 32+00.00Delta = 36° 00' 00" / D = 6°e= 0.06'/' / NC = -.02'/'
Design Speed = 50 mph / Spiral (if used) = 150 ft.
2 lane roadway with one 12' lane in each direction Rotation about the centerline will be used.
A)Verify that Degree of curvature is satisfactory for design speed with the superelevation shown.
B)Determine:
Without a spiral curve : what are the stations for the following
Beginning of curve:
End of Curve:
Starting station for tangent runout using Iowa DOT Manual procedures
Starting location for superelevation runoff
Ending station for superelevation runoff
C) Part B with Spiral curve, using Length of spiral = 150 feet on each end of circular curve. Consider only the first spiral at this time.
Find the station where the tangent runout begins and the station where the superelevation runoff begins and ends.
Verify that:
a)For the spiral stationing the length of the circular plus spiral arcs is one spriral length loner that if no spiral is used.
b)for a spiraled curve, the tangent length is approximately'/2 of a spiral length longer than when a simple circular curve is considered.
c)the k length on the spiral is approximately %2 of a spiral length
d)the 150 foot spiral falls within the range of minimum and maximum length spiral, based on the "throw" distance (p) being between 0.33 and 3.3, as per Iowa DOT approach.
Given Pi = Station 32+00.00Delta = 36° 00' 00" / D = 6°e= 0.06'/' / NC = -.02'/'
Design Speed = 50 mph / Spiral (if used) = 150 ft.
2 lane roadway with one 12' lane in each direction Rotation about the centerline will be used.
Given Pi = Station 32+00.00Delta = 36° 00' 00" / D = 6°e= 0.06'/' / NC = -.02'/'
Design Speed = 50 mph / Spiral (if used) = 150 ft.
2 lane roadway with one 12' lane in each direction Rotation about the centerline will be used.
Given Pi = Station 32+00.00Delta = 36° 00' 00" / D = 6°e= 0.06'/' / NC = -.02'/'
Design Speed = 50 mph / Spiral (if used) = 150 ft.
2 lane roadway with one 12' lane in each direction Rotation about the centerline will be used.
A: Verify that design conditions are acceptable
Rmin = V2 /[15*(e+f) ] where e = 0.06, f for 50 mph = 0.14 (AASHTO and Table 3.3 of text)
R min = 502/[15*(0.06+0.14)] = 833 ft.
R ( design)= 5729.58 / D = 5729.58 / 6 = 954.93
Therefore Design radius exceeds R min - OK
B. Find key stations for curve points and superelevation:
Find BC: BC = PI - T where T = Rtan(delta/2) = 310.28
= 32+00.00 – 3+10.28 = 28+89.72 (BC)
EC: EC = BC + L where L = delta/D*100 = 600.00 ft
= 28 +89.72 + 6+00.00 =34+89.72 (EC)
Find starting station for tangent runout:
For Iowa procedure: (See related slide 10 of lecture for the 2/3 : 1/3 rule sketch)
Tangent runout begins at BC– 70% of superelev.runoff – tangent runout
Superelev. Runoff rate, G, for 50 mph design speed = 0.50% ( or 1’ per 200 ft.)
(from AASHTO standards)
Using runoff length equation LR = 12( e%/G%)*alpha
Where alpha = 1 because we only have a 2 lane road here.
: LR= 12( 6%/0.5%) = 144 feet
Tangent run out found by proportion : Runout = ( Normal crown rate / Super rate) * LR
LT = Tangent run out length = (0.02 / 0.06) * 144
LT = 48 ft
* STATION TO BEGIN TANGENT Runout
BC – 70% LR - LT =
28+89.72 – 0.70* ( 1+44) - 0+48) =27+ 40.92 Start Tangent run out
* Station to begin superelevation run off
BC – 70% LR = 28+89.72 – 0.70* ( 1+44) =27+ 88.92 Start se runoff
Attain Full superelevation at BC + 30% (LR) = 29+ 32.92 Achieve full e
Similar calculations for runoff and runout stationing would be applied at the EC.
C. Find key stations if 150 foot spiral is used but other design elements remain the same.
1) Spiral stationing is needed - ΘS , TS , p, k will be needed
from: LS =( ΘS / D) *200 , ΘS = 40 30’
from other Spiral curve equations found in lecture notes and Iowa Design manual
p = 0.98 ft.
k = 74.98 ft.
TS = 385.58 ft.
Therefore spiral begins at station TS = PI - TS
TS = 32+00.00 – 3+85.58 = 28+14.22 (TS station)
Superelevation runoff length: By policy for spirals, the superelevation run off length will be equal to spiral length = 150 feet. (Note that Table 16.12 of text indicates a desirable length of spiral for V = 50 mph is 147 feet. Our designer selected a 150 foot length).
Now, LT = (NC / e ) * LR = (0.02’/’ / 0.06’/’ ) * 150’ = 50 ft.
(compare the 50 ft to the tabled value in Table 16.13
where LT = 49 ft. when LS = 147 ft.
* STATION TO BEGIN TANGENT Runout
TS - LT =
28+14.22 0+50.00 =27+ 64.22 start Tangent run out
* Station to begin superelevation run off = TS = 28+14.22 (TS station)
* Attain Full superelevation at TS + (LR) = 29+ 64.22 Achieve full e
Similar computations for stationing of the runout and runoff would be needed for the departing spiral.
PART D - Verification - Original problem sheet does not identify this section as Part D, but to help with organization, Part D is identified here.
a) verify that total spiraled curve is one spiral length longer than simple curve:
When using spiral, circular curve portion is: L =( (delta - 2ΘS)/ D )*100
= ((36O– 2 * 4.5O)/6O )* 100 =450.00 ft
Total curved length = 150.00 + 450.00 + 150.00 = 750.00’
When only a circular curve was used L = delta / D * 100 = 600.00’
Additional curved length = 750 – 600 = 150’ which is one spiral length
b) Verify that tangent length for spiral is approximately 1/2 a spiral length longer than T for circular curve.
T in circular curve was = 310.28 TS was 385.58’
Difference = 75.30’ which is approximately LS/2 = 75’
c) Verify that k is approximately 1/2 of LS
k = 74.98 or approximately 75 ‘ = LS /2
d) verify that p is within range of 0.66 to 3.33: p = 0.98 which is in desired range.
Table 1 Tabular summary of key stations and spiral components
Sect. / Key point / Location or value checkA. / Minimum R = 833 ft / Available R = 955 - OK
B.
BC Station (tangent = 310.28’) / 28+89.72
EC Station ( L = 600.00’) / 34+89.72
Superelevation runoff length / 144.00’
Tangent runout length / 48.00’
Begin tangent runout / 27+40.92
Begin superelevation runoff / 27+88.92
Attain full superelevation / 29+32.92
C. / With 150 foot spiral - Key spiral comps
Theta for spiral / 4O 30’
P = / 0.98’
k= / 74.98
TS = / 385.58
TS (tangent to spiral location = start of superelevation runoff / 28+14.22
Tangent runout begins / 27+64.22
Full superelevation attained at / 29+64.22
D. / Verify length elements
a) Simple Curved length = 600’; spiraled curve = 750’ / Diff = 150’ = LS check
b) T = 310.28 TS = 385.58 Diff = 75.30’ / Diff ~ LS/2 check
K = 74.98’ / k ~ LS/2 check
p = 0.98 / P is in range of
0.66 to 3.33 check