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1 Quiz Answers What can be done to improve the safety of a horizontal curve? Make it less sharp Widen lanes and shoulders on curve Add spiral transitions Increase superelevation

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2 Quiz Answers 5.Increase clear zone 6.Improve horizontal and vertical alignment 7.Assure adequate surface drainage 8.Increase skid resistance on downgrade curves

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3 Some of Your Answers Decrease posted speed Add rumble strips Bigger or better signs Guardrail Better lane markers Sight distance Decrease radius

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4 Superelevation and Spiral Curves CE 453 Lecture 18

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5 Objectives 1.Define superelevation runoff length and methods of attainment (for simple and spiral curves) 2.Calculate spiral curve length

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6 Other Issues Relating to Horizontal Curves 1.Need to coordinate with vertical and topography 2.Not always needed MAXIMUM CENTERLINE DEFLECTION NOT REQUIRING HORIZONTAL CURVE Design Speed, mphMaximum Deflection 255°30' 303°45' 352°45' 402°15' 451°15' 501°15' 551°00' 601°00' 650°45' 700°45' Source: Ohio DOT Design Manual, Figure 202-1E

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7 Attainment of Superelevation - General 1.Tangent to superelevation 2.Must be done gradually over a distance without appreciable reduction in speed or safety and with comfort 3.Change in pavement slope should be consistent over a distance 4.Methods (Exhibit 3-37 p. 186) a.Rotate pavement about centerline b.Rotate about inner edge of pavement c.Rotate about outside edge of pavement

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8 Superelevation Transition Section Tangent Runout Section + Superelevation Runoff Section

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9 Tangent Runout Section Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero For rotation about centerline

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10 Superelevation Runoff Section Length of roadway needed to accomplish a change in outside-lane cross slope from 0 to full superelevation or vice versa For undivided highways with cross- section rotated about centerline

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11 Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, th Ed.

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12 Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, th Ed.

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14 Source: CalTrans Design Manual online,

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15 Source: Iowa DOT Standard Road Plans Same as point E of GB

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16 Attainment Location - WHERE 1.Superelevation must be attained over a length that includes the tangent and the curve (why) 2.Typical: 66% on tangent and 33% on curve of length of runoff if no spiral 3.Iowa uses 70% and 30% if no spiral 4.Super runoff is all attained in Spiral if used (see lab manual (Iowa Spiral length = Runoff length)

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17 Minimum Length of Runoff for curve L r based on drainage and aesthetics rate of transition of edge line from NC to full superelevation traditionally taken at 0.5% ( 1 foot rise per 200 feet along the road) current recommendation varies from 0.35% at 80 mph to 0.80% for 15mph (with further adjustments for number of lanes)

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18 Minimum Length of Tangent Runout L t = e NC x L r e d where e NC = normal cross slope rate (%) e d = design superelevation rate L r = minimum length of superelevation runoff (ft) (Result is the edge slope is same as for Runoff segment)

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19 Length of Superelevation Runoff α = multilane adjustment factor Adjusts for total width r

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20 Relative Gradient (G) Maximum longitudinal slope Depends on design speed, higher speed = gentler slope. For example: For 15 mph, G = 0.78% For 80 mph, G = 0.35% See table, next page

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21 Maximum Relative Gradient (G) Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, th Ed.

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22 Multilane Adjustment Runout and runoff must be adjusted for multilane rotation. See Iowa DOT manual section 2A-2 and Standard Road Plan RP-2

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23 Length of Superelevation Runoff Example For a 4-lane divided highway with cross- section rotated about centerline, design superelevation rate = 4%. Design speed is 50 mph. What is the minimum length of superelevation runoff (ft) L r = 12eα G

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24 Lr = 12eα = (12) (0.04) (1.5) G 0.5 Lr = 144 feet

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25 Tangent runout length Example continued Lt = (e NC / e d ) x Lr as defined previously, if NC = 2% Tangent runout for the example is: L T = 2% / 4% * 144’ = 72 feet

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26 From previous example, speed = 50 mph, e = 4% From chart runoff = 144 feet, same as from calculation Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, th Ed.

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27 Spiral Curve Transitions

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28 Spiral Curve Transitions Vehicles follow a transition path as they enter or leave a horizontal curve Combination of high speed and sharp curvature can result in lateral shifts in position and encroachment on adjoining lanes

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29 Spirals 1.Advantages a.Provides natural, easy to follow, path for drivers (less encroachment, promotes more uniform speeds), lateral force increases and decreases gradually b.Provides location for superelevation runoff (not part on tangent/curve) c.Provides transition in width when horizontal curve is widened d.Aesthetic

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30 Minimum Length of Spiral Possible Equations: Larger of (1)L = 3.15 V 3 RC Where: L = minimum length of spiral (ft) V = speed (mph) R = curve radius (ft) C = rate of increase in centripetal acceleration (ft/s 3 ) use 1-3 ft/s 3 for highway)

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31 Minimum Length of Spiral Or (2)L = (24p min R) 1/2 Where: L = minimum length of spiral (ft) R = curve radius (ft) p min = minimum lateral offset between the tangent and circular curve (0.66 feet)

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32 Maximum Length of Spiral Safety problems may occur when spiral curves are too long – drivers underestimate sharpness of approaching curve (driver expectancy)

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33 Maximum Length of Spiral L = (24p max R) 1/2 Where: L = maximum length of spiral (ft) R = curve radius (ft) p max = maximum lateral offset between the tangent and circular curve (3.3 feet)

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34 Length of Spiral oAASHTO also provides recommended spiral lengths based on driver behavior rather than a specific equation. See Table of text and the associated tangent runout lengths in Table oSuperelevation runoff length is set equal to the spiral curve length when spirals are used. oDesign Note: For construction purposes, round your designs to a reasonable values; e.g. Ls = 147 feet, round it to Ls = 150 feet.

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35 Source: Iowa DOT Design Manual

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36 Source: Iowa DOT Design Manual

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37 Source: Iowa DOT Design Manual

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38 Source: Iowa DOT Design Manual SPIRAL TERMINOLOGY

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39 Attainment of superelevation on spiral curves See sketches that follow: Normal Crown (DOT – pt A) 1.Tangent Runout (sometimes known as crown runoff): removal of adverse crown (DOT – A to B) B = TS 2.Point of reversal of crown (DOT – C) note A to B = B to C 3.Length of Runoff: length from adverse crown removed to full superelevated (DOT – B to D), D = SC 4.Fully superelevate remainder of curve and then reverse the process at the CS.

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40 Source: Iowa DOT Standard Road Plans RP-2 With Spirals Same as point E of GB

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41 With Spirals Tangent runout (A to B)

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42 With Spirals Removal of crown

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43 With Spirals Transition of superelevation Full superelevation

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45 Transition Example Given: station D = 4º (R = 1,432.4 ft) = º L = ft

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46 With no spiral … T = ft PC = PI – T =

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47 For: Design Speed = 50 mph superelevation = 0.04 normal crown = 0.02 Runoff length was found to be 144’ Tangent runout length = 0.02/ 0.04 * 144 = 72 ft.

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48 Where to start transition for superelevation? Using 2/3 of Lr on tangent, 1/3 on curve for superelevation runoff: Distance before PC = Lt + 2/3 Lr =72 +2/3 (144) = 168 Start removing crown at: PC station – 168’ = = Station =

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49 Location Example – with spiral Speed, e and NC as before and = º Station R = 1,432.4’ Lr was 144’, so set Ls = 150’

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50 Location Example – with spiral See Iowa DOT design manual for more equations: m#Chapter_2 Spiral angle Θs = Ls * D /200 = 3 degrees P = 0.65 (calculated) Ts = (R + p ) tan (delta /2) + k = ft

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51 TS station = PI – Ts = – = Runoff length = length of spiral Tangent runout length = Lt = (e NC / e d ) x Lr = 2% / 4% * 150’ = 75’ Therefore: Transition from Normal crown begins at ( ) – ( ) = Location Example – with spiral

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52 With spirals, the central angle for the circular curve is reduced by 2 * Θs Lc = ((delta – 2 * Θs) / D) * 100 Lc = ( *3)/4)*100 = ft Total length of curves = Lc +2 * Ls = Verify that this is exactly 1 spiral length longer than when spirals are not used (extra credit for who can tell me why, provide a one-page memo by Monday) Location Example – with spiral

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53 Also note that the tangent length with a spiral should be longer than the non-spiraled curve by approximately ½ of the spiral length used. (good check – but why???) Location Example – with spiral

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54 Notes – Iowa DOT Source: Iowa DOT Standard Road Plans Note: Draw a sketch and think about what the last para is saying

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