Tollway Barrier Guidelines Examples Hands-On Workshop III July 2013 1.

Tollway Barrier GuidelinesExamples

Hands-On Workshop IIIJuly 2013

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Each AOC should be analyzed separately.

a. Establish EOTW j. Warrant Analysis Level

b. Design Speed k. Lateral Offset of Barrier

c. Design ADT l. Upstream End of Guardrail

d. Length of Runout m. Upstream End of Barrier Wall

e. Shy Line Offset n. Downstream Terminal

f. Foreslope / Backslope o. Barrier Obstacle

g. Clear Zone p. Compare existing length to proposed

h. Clear Zone Adjustment q. Prepare warrant text and exhibits

i. Lateral Extent of the Area of Concern

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Level 2 Example- Location 9

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Each AOC shall be analyzed separately.

a. Establish EOTW

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a. Establish EOTW

b. Design Speed

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a. Establish EOTW

b. Design Speed

c. Design ADT

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

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d. Length of Runout

L2

Figure 5-39. Approach Barrier Layout Variables

LR

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10


a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset

f. Foreslope / Backslope

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone

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g. Clear Zone

Select highest value in range

Table 3-1Clear Zone Distances in Feet from EOTW

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone

h. Clear Zone Adjustment

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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L2


LA

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Level 2 Example- Location 9 Lateral Extent of the Area of ConcernLA =

LA = 23.9’ + 1.5’ = 25.4’

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b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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Flowchart for Analysis of Existing Obstacles (Figure 5.5a)

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c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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k. Lateral Offset of Barrier

L2


L2

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Level 2 Example- Location 9 LA = 23.9’ + 1.5’ = 25.4’Lateral Offset of BarrierL2 =

L2 = 1’ + 11’ + 1’ = 13.0’

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d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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Level 2 Example- Location 9 LA = 25.4’L2 = 13.0’Y=Y = L2 + 0.75’Y = 13.0’ + 0.75’Y = 13.75’

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l. Upstream End of Guardrail

L2


X= (LA – Y) / (LA / LR )LA

LR

LA - Y X

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Level 2 Example- Location 9 LA = 25.4’L2 = 13.0’Y = 13.75’LR = 360.0’

Length of Need – FormulaSolve for XX = (LA – Y) / (LA / LR)X = (25.4’ – 13.75’) / (25.4’ / 360.0’)X = 165.12’

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Level 2 Example- Location 9 ObstacleSFSouthface_Foundation_Station = SFSF = (3945+98.50)LAOC = Length of AOC = 3.0’

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L2


Point of Need

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Level 2 Example- Location 9 Point of Need PON = SF – X PON = (3945+98.50) – 165.12’PON = (3944+33.38)

Length of Need = LON = X + LAOC

LON = 165.12’ + 3.0’LON = 168.12’

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Level 2 Example- Location 9 Distance from back of guardrail to foundation=25.4’(LA)-3.0’(Fnd.)-13.0’(L2 )-1.76’(Depth of GR)= 7.64’

Therefore, LON ends at downstream end of obstacle.

Condition 2, TBG Figure 5.1232

Level 2 Example- Location 9 TerminalsTerminal = Type T2T2 = 12.5’ (does not count toward length of need)Terminal = Type T1 (Special) T1 = 50.0’ (37.5’ counts toward length of need)Barrier Limits DeterminationLength of Guardrail LGuardrail = LON – T1LGuardrail = 168.12’ – 37.5’

LGuardrail = 130.62’ LGuardrail, R = LGuardrail (Round up to the nearest 12.5’ increment)

LGuardrail, R = 137.5’

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e. Shy Line Offset


g. Clear Zone



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g. Clear Zone



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Level 2 Example- Location 9 T2End_Station = SF + LAOC + T2T2End_Station = (3945+98.50) + 3’ + 12.5’T2End_Station = (3946+14.00)GREnd_Station = Begin T2Station = End T2Station – 12.5’GREnd_Station = (3946+14.00) – 12.5’GREnd_Station = (3946+01.50)

GRBegin_Station = GREnd_Station – LGuardrail,R

GRBegin_Station = (3946+01.50) – 137.5’GRBegin_Station = (3944+64.00)T1Begin_Station = GRBegin_Station – 50’T1Station = (3944+64.00) – 50’T1Begin_Station = (3944+14.00)

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g. Clear Zone



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Barrier Limits Check

12.5’ < |PON- T1Begin_Station|< 25’12.5’ < |(3944+33.38) – (3944+14.00)|< 25’12.5’ < 19.38’ < 25’

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Level 2 Example – Location 9 – Site Plan

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Questions?

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Notes: 1. There is a crash wall in

front of the bridge piers.2. Regrading of the bridge

cone is not a feasible alternative.


a. Establish EOTW

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a. Establish EOTW

b. Design Speed

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a. Establish EOTW

b. Design Speed

c. Design ADT

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

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d. Length of Runout

L2


LR

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d. Length of Runout

50


a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone

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g. Clear Zone

Select highest value in range

Table 3-1Clear Zone Distances in Feet from EOTW

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone


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Adjusted Clear Zone= 32’ x 1.2= 38.4’

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a. Establish EOTW

b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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L2


LA

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Lateral Extent of the Area of ConcernLA =LA = 32’ x 1.2LA = 38.4’

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b. Design Speed

c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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Flowchart for Analysis of Existing Obstacles (Figure 5.5a)

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c. Design ADT

d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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k. Lateral Offset of Barrier

L2


L2

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LA = 38.4’L2 =L2 = 12.0’ (11’ + 1’)

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d. Length of Runout

e. Shy Line Offset


g. Clear Zone



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LA = 38.4’Lateral Offset of BarrierL2 = 12.0’ (11’ + 1’)Y = Y = L2 + 0.75’Y = 12.0’ + 0.75’Y = 12.75’

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< LR

LA

L2

Y

X

Figure 5-48. Example of Barrier Design for Fixed Object on Horizontal Curve

Graphical Determination of “X”

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Level 2 Example- Location 17LA = 38.4’ L2 = 12.0’ (11’ + 1’) Y=Y = L2 + 0.75’Y = 12.0’ + 0.75’ = 12.75’ ObstaclesBPBridge_Pier_Station = BP BP = (132+35.72)ECEmbankment_Cone_Station = EC EC = (132+18.58) LAOC = BP-EC LAOC = 17.14’ (assumes that EC ends at BP)Length of Need – Graphical SolutionTangent Runout Path from EC = 474.85’Tangent Runout Path = 474.85 > LR = 300.0’Therefore, use LR to determine upstream end of the runout pathX = Calculated Graphically = 166.65’

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Point of Need = PON

PON = EC – XPON = (132+18.58) – 166.65’PON = (130+51.93)Length of Need = LON LON = X + LAOC

LON = 166.65’ + 17.14’LON = 183.79’

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e. Shy Line Offset


g. Clear Zone



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g. Clear Zone



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Level 2 Example- Location 17Terminals

Terminal = Type T6B T6B = 43.15’ (counts toward length of need)Terminal = Type T1 (Special) = T1 = 50.0’ (37.5’ counts toward length of need)Barrier Limits DeterminationLength of Guardrail

LGuardrail = LON – T6B – T1LGuardrail = 183.79’ – 43.15’ – 37.5’

LGuardrail = 103.14’LGuardrail, R = LGuardrail (Round up to the neareast 12.5’ increment)

LGuardrail, R = 112.5’

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Level 2 Example- Location 17T6BEnd_Station = BPT6BEnd_Station = (132+35.72) GREnd_Station = End T6BStation – 43.15’GREnd_Station = (132+35.72) – 43.15’GREnd_Station = (131+92.57) GRBegin_Station = GREnd_Station – LGuardrail,R

GRBegin_Station = (131+92.57) – 112.5’GRBegin_Station = (130+80.07) T1Begin_Station = GRBegin_Station – 50’T1Begin_Station = (130+80.07) – 50’T1Begin_Station = (130+30.07)

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Barrier Limits Check:

12.5’ < |PON – T1Begin_Station |< 25’

12.5’ < |(130+51.93) – (130+30.07)| < 25’

12.5’ < 21.86’ < 25’

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g. Clear Zone



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Level 2 Example- Location 17 – Site Plan

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Questions?

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Example-Speed Profile

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Site Plan

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Acceleration Table

Table 10-3Minimum Acceleration Lengths for Entrance Terminals with Flat Grades of Two Percent or Less

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Example-Speed Profile-Site Plan

30MPH

40MPH

50MPH

60MPH

70MPH

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Level 3 Example – Overhead Sign Truss

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Level 3 Example – Plan

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Flowchart for Analysis of Proposed Obstacles (Figure 5.5b)

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New Overhead Sign Truss (Span and Cantilever) Article 5.5.9 The Designer shall perform a Level 3 Analysis to

determine the most cost-effective design for all new overhead sign installations. The alternatives at minimum to evaluate shall include: (1) base condition – shortest span or arm with shielding; (2) span or arm length which places foundation outside defined clear zone with no shielding or if clear zone is undefined places foundation such that the criteria in Table 5.6.1 is met; (3) longer span or arm which places foundation well outside the clear zone with no shielding.

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Level 3 Example – Overhead Sign TrussAnalyze 4 Alternatives for Sign Foundation

(Assumes Clear Zone is defined)1. Base Condition: Foundation Shielded by

Guardrail2. Foundation Beyond Clear Zone, not shielded3. Foundation on Backslope, not shielded4. Foundation on Backslope, not shielded (max.

length truss)

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Level 3 Example – Plan

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Level 3 Example – Cross Section 91

Sign Structure Foundation

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Level 3 Example – Feature Sketch

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Level 3 Example – Guardrail Calculation

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Level 3 Example – Estimated Costs

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Cost Tab

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Highway Tab

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Segments Tab

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Features Tab

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View Pulldown

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View/ Crash Cost

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View/ Crash Cost/ KABCO

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Options/ Units

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Option/ Edit Feature

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Option/ Edit Feature

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Edit Feature

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Features/ Category

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Features / Category/ User Defined

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Feature/ Types

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RSAP Output

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RSAP Output

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RSAP Output

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RSAP Output

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RSAP Output

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RSAP Output

2 vs. 33 vs. 44 vs. 1

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3 is better4 is better4 is better

AOC’s in Close

Proximity

14+25

Culvert - AOC #1

14+00 17+22

Cantilever Sign- AOC #2

15+70

16+12.517+05

Upstream Terminal End

18+30

18+42.5

Downstream Terminal End

Upstream Terminal End Downstream

Terminal End

92.5’Gap

Total Length of Need = 405’

EOTW

AOC’s in Close Proximity

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PON PON

16+00 18+27

14+05

Example- AOC Table FormatAOC# UPSTREAM

US TERMPON DS End of

GuardrailDOWNSTREAM

DSTERM

GAP(FT)

OVERLAP(FT)

1 14+00 14+25 16+00 16+12.50

92.50 ---

2 17+05 17+22 18+30 18+42.50

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Summary CalculationLON = AOC #2 DS End of GR – AOC #1 PON = (18+30)-(14+25) = 405’TerminalsDS Terminal= TBT Type T2=T2US Terminal= TBT Type T1=T1

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Barrier Limits DeterminationLGuardrail = LON-T1 = 405’-37.5’ = 367.5’LG, rounded= 367.5/12.5’ = 29.4, round to 30 sections

30 x 12.5’ = 375’T2end = DS End of GR+ 12.5’=18.30 + 12.5’= 18+42.5

Grend = AOC #2 DS End of GR = 18+30Grbegin = Grend - LG, rounded = 18+30 - 375’ = 14+55T1end = Grbegin = 14+55T1begin = T1end - 50’ = 14+55-50’ = 14+05

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Barrier Limits Check12.5’ < AOC #1 PON – T1begin < 25’AOC #1 PON – T1begin

14+25-14.05= 20’

Provide a Summary Site Plan

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14+25

Culvert - AOC #1

14+00 17+72


15+70

16+12.517+55


18+80

18+92.5




142.5’Gap

EOTW


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PON PON

16+00 18+77

Length of Need

Length of Need

14+25

Culvert - AOC #1

14+12.5 16+22


15+70

16+12.5

16+05Upstream Terminal End

17+30

17+42.5


Upstream Terminal End Downstream

Terminal End

7.5’ Overlap

Total Length of Need

EOTW


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PON PON

16+00 17+27

14+17.5

Example- AOC Summary Table FormatLEGENDPOINT OF NEED STATIONDOWNSTREAM END OF GUARDRAIL STATION

AOC SUMMARY TABLEAOC 1 15+00 26+00AOC 2 15+00 22+50AOC 3 16+25 27+00AOC 4 19+00 23+00 SUMMARY 15+00 27+00TOTAL LON 15+00 to 27+00 = 1,200 FT

AOC- Color Coded System

Questions?

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Tollway Barrier Guidelines Examples Hands-On Workshop III July 2013 1.

Documents

Transcript of Tollway Barrier Guidelines Examples Hands-On Workshop III July 2013 1.