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# Highway Design C E 427

GPA 3.96

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This 22 page Class Notes was uploaded by Charley Wintheiser DVM on Monday October 5, 2015. The Class Notes belongs to C E 427 at California State University - Long Beach taught by Shadi Saadeh in Fall. Since its upload, it has received 21 views. For similar materials see /class/218763/c-e-427-california-state-university-long-beach in Civil Engineering at California State University - Long Beach.

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Date Created: 10/05/15

HORIZONTAL ALIGN MENT Horizontal Alignment Geometric Elements of T S I Horizontal Curves rans39t39on or p39ra Curves Superelevation Design Sight Distance Simple Curve Circular Curve Tangent PC PT Point of Curvature POint 0f Tangency Curve with Spiral Transition Circular Curve S ira Tangent p 30 CS TS Curve to Spiral ST Spiral to Curve Spiral to Tangent Tangent to Spiral Design Elements of Horizontal Curves Pl gt A T L A L E n engn PC in PT X CIRCULAR CURVE EQUATIONS F s 9 95 A D 72 77 m duh C ZR sin 7 c 7 21rRA l L P 360 P mo39 Ft tune Fur any tangent diatuncc x l A D ngilRlixljlJ 1 T R WA For any arc length L 2 V D 39 Rsln 9 IV R A y I R sac 7 1 y V D Centramnute 39 y f far too Arc M RI We Rtl 030 2 NOTES in This vannhle uscd nnly for cum de nition in traditional us mm 2 Um lm mesa variables can bc cxprcsscd in minor lcet or muturs VARIABLES PC Point of curvature Beginning of curve Pl Point of intersection PT 2 Point of lungcncy End ol39curvc D Degree of curvature I Pl Point of intersection R Radius of curve 2 A Central tingle E External distance 3 erght and DIXOn L Length of curve PC to PT 2 M Middle ordinate 2 I r Lenglh of nrc PC to P 12 C Chord length 2 ti Central angle for arc length l T Tangent length PC to P1 amp PT to Pl 3 1 2 De ection angle at PC between tangent and chord for P a De ection angle at P1 between tangent and line from Pl 0 P x Tangent distance ti39om PC to P 2 y Tangent offsct P 2 Basic Formulas Basic Formula that governs vehicle operation on a curve 00le f V2 a f V2 1 00lef 15R 15R Where e superelevation 00le f side friction factor V vehicle speed mph R radius of curve ft Basic Formulas Minimum radius 2 R V m 1500lemax fmax Where e superelevation percent f side friction factor V vehicle speed mph R radius of curve ft Minimum Radius with Limiting Values of e and f US Custamary I DEsign Calcu39atad Ru1n39ad Speed Maximum 01ax1391 1m Tats Radius Fiuliiu 11mph 49 iiu f 2130 r F3 it iii 10 40 030 150 15 15 40 03 030 41 42 20 40 021quot 0 31 000 80 25 40 0225 029quot 1513 154 30 40 020 034 22500 250 35 40 0 13 022 312 3T1 40 40 0123 0 3920 533 45 40 015 01 139105 33911 50 40 011 010 Eli59 020 55 40 013 01 1115150 1190 130 40 012 0 10 quot15000 1500 10 60 030 04 15 15 00 032 030 305 39 20 60 02 0233 000 01 25 00 023 029 143 144 30 00 020 020 2303 231 35 50 013 02 403 7340 410 50 015 02 430 105 39 45 60 015 021 EARS 513 3930 00 014 020 0333 83933 55 80 013 019 10014 1000 I 00 60 012 015 133 1330 1 05 60 01 01 1V 3 1050 1 7 0 60 010 0 i ii 2040 3 5 60 009 015 2500 30 60 0 00 0 1i 3050 Superelevation Design Desirable superelevation edZV2fm forRgtRmin gR ax Where V design speed in fts or ms g gravity 981 ms2 or 322 ftsZ R radius in ft or m Various methods are available for determining the desirable superelevation but the equation above offers a simple way to do it The other methods are presented in the next few overheads Methods for Estimating Desirable Superelevation Method 1 Superelevation and side friction are directly proportional to the inverse of the radius straight relationship between 1RO and 1R 1Rmin Method 2 Side friction is such that a vehicle traveling at the design speed has all the acceleration sustained by side friction on curves up to those requiring fmax Superelevation is introduced only after the maximum side friction is used Method 3 Superelevation is such that a vehicle traveling at the design speed has all the lateral acceleration sustained by superelevation on curves up to those required by emax No side friction is provided on flat curves May result in negative side friction Method 4 Same approach as Method 3 but use average running speed ratherthan design speed Uses speeds lower than design speed Eliminate problems with negative side friction Method 5 Superelevation and side friction are in a curvilinear relationship with the inverse of the radius of the curve with values between those of methods 1 and 3 Represents a practical distribution for superelevation over the range of curvature This is the method used for computing values shown in exhibits 325 to 329 Side Friction Factor Five Methods Reciprocal of Radius 1R Side Friction Factor Selection of faesign and ealesign Method 5 fmax for the design speed design e for the design speed e 0 f quot max Reciprocal of Radius 1R Side Friction Factor Selection of fdesign and edesign Rf V2gfmax Rmin VZgfmax emax R0 V2ge max 1R Reciprocal of Radius R0 f 0 e emax Side Friction Factor Selection of fdesign and edesign fdesign 01R2 fmax for the design speed a fmamein139RminRO39Rmin B fmamein3RO39R min emax for the design speed Reciprocal of Radius 1R Superelevation Design for High Speed Rural and Urban Highways in ow V4 35mm R r m mpr ini um 1i n a Ehihil 27 Minimum Radii rm Design Supmimiiun Rams Design 5mm and 3 Continual Example Design Speed 100 kmh fmax 0128 emax 006 Question What should be the design friction factor and design superelevation for a curve with a radius of 600 m 1 Compute Rf R0 and Rmin Rf VZgfmax 27782 981 x 0128 615 m R0 V2gemax 27782 981 x 006 1311 m Rmin V2gfmax emax 27782 9810128006 R 418m min Side Friction Factor Selection of fdesign and edesign example 11311 1615 1418 2 Compute o and 3 o 0128 x 418 x 1 418 1311 418 2845 m 3 0128 x 4183 1311 418 10502 m2 3 Compute fdesign and edesign First estimate the righthand side of equation for designing superelevation e f V2gR 27782 981 X 600 0131 Then fdesign 2845 600 10502 6002 0076 edesign 0131 0076 0055 lt emax 006 Selection of fdesign and edesign exa m ple f 5 O E fmax 0128 C U 0 g e0 g fdesig emax 0 06 A 0076 V 11311 1615 1418 R 1600

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