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Geometric Design

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Homework Ch 3 # 1,3,7,8,9

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Geometric Design Roadway is designed using v, R, e – Need to know the required Stopping Sight Distance for the roadway – D = 1.47vt +v 2 /(30(f+g) v is in mph (1.47, 30 are conversion factors) SSD – distance required to make a full stop Decision Sight Distance – adds in a maneuver time (3.5 – 4.5s) to the reaction time based on AASHTO guidelines – D = 1.47v(tr+tm) +v 2 /(30(f+g) Gives Decision Sight Distance

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Alignment 2 types of alignment – Vertical – Horizontal

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Vertical Curves 2 types of curves – Crest – Sag Different criteria for SSD – crest - 6” object over hill – sag headlights

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Vertical Curves Grade controls – Trucks and f control – Max grade Interstate = 9% – Max grade local street = 15% – min grade = 0.5% - drainage – Fig 3.15 – length of grade is important as well – Also look at length of curve and change in grade Don’t want cars grabbing air

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Vertical Curves Truck performance Note crawl speed Fig 3.16, 3.17, 3.18

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Vertical Curves

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Vertical Curve Design Vertical Curves are Parabolas A = G 2 - G 1 –algebraic diff in grades (in %) E = AL/8 – L is in stations (ft/100) – gives E in ft y = [(G2-G1)/2L]*x^2 + G1x+ Y0 – Gives y in ft – x is in stations, G1, G2 in %

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Vertical Curve Design Hi/Lo point occurs at dy/dx = 0 – x = -LG 1 /(G 2 -G 1 ) – Not in middle of L – All calcs in stations

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Vertical Curve Design SSD for Crest curves (dead cat distance) – 2 equations - L>=SSD, – L=(!G2-G1!*(SSD^2))/(200*(h1^0.5+h2^0.5)^2 – L<=SSD – L=2*SSD - (200*(h1^0.5+h2^0.5)^2/(!G2-G1!) – These provide minimum values – only one is true – eye height h 1 = 3.5 ft – object height h 2 = 0.5 ft – 200*(h1^0.5+h2^0.5)^2 = 1329 (not what is in the book)

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Given a design speed of 45 mph, approach grade of +3% and a departure grade of -4% meeting at VPI Station = 100+50, VPI Elevation = 347.85. – What is the PVC & PVT stations and elevations – Where is the high point located and what is elevation

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Vertical Curve Design L for Sag curves – L>SSD L=(!G2-G1!*(SSD^2))/(200*(h+SSD*tan ) – L

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Given a design speed of 45 mph, approach grade of -4% and a departure grade of -1% meeting at VPI Station = 100+50, VPI Elevation = 347.85. – What is the PVC & PVT stations and elevations – Where is the high point located and what is elevation

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Horizontal Alignment

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3 parts to alignment – tangents - straight sections – curves – transitions between curves and tangents

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Curves L=2 R /360 D - degree of curve D different for RR and highways

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Curves External Distance PI -> LE = R(sec(I/2)-1) Middle Ordinate L->LC M = R(1-cos(I/2) Tangent PC->PI T = Rtan(I/2) Curve Length PC->PT L = 100(I/D) Chord Length PC-> PT LC = 2Rsin(I/2) Degree of curve D = 5729.58/R – How many degrees to create an arc of 100’ for a given radius

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Horizontal Distance Need to maintain SSD around curve Look at M and L to determine safe speed – L = SSD – I changes to I=L*D/100 – M = R(1- cos (I/2)) – If M > available M then NO GOOD

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Example Given the following info for a horizontal curve, determine if a speed limit of 60 mph is safe. – I = 45 degrees, R =1500 ft, 2-12 foot lanes, 6 foot shoulder, e = 0.04, f=0.12, barn located 14 feet off centerline of inside lane.

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Design Vehicles Fig 2.4 Gives tightest turn radius for a low speed turn by a vehicle 10 specific types of vehicles used – determine speeds, curb radius, pavement width

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Delineation of Vehicle Paths Flow Control pavement markings, medians, islands Channelization provide improved flow through an intersection

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Channelization

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Lab Develop an Excel Spreadsheet which will solve horizontal and vertical curve problems

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Exercise 1: Basic Design Criteria 1) Route 17 is a minor arterial and it has a design ADT of 4000. Assume rolling terrain. Determine the following information:

Exercise 1: Basic Design Criteria 1) Route 17 is a minor arterial and it has a design ADT of 4000. Assume rolling terrain. Determine the following information:

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