CE 404-SEC11

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PRACTICE 1

Horizonta l a l ignment

2016CE 404. Highway Engineering

1

HIGHWAY ENGINEERING

(CE 404)

سم هللا لرحن لرحيم

By

Sight Distances


2Stopping Sight Distance(SSD)

brake reaction distance = d1 = v x t (meters) . t=2.5 sec

If‘ V’ is the design speed in (m/sec) and ‘t’ is thetotal reaction time of the driver in seconds,

brake reaction distance

braking distance braking distance on grade

a deceleration rate of at least

3.4 m/s2

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Sight Distances


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Passing Sight Distance (PSD)

Sight Distances


6Passing Sight Distance (PSD)

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Sight Distances


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Passing Sight Distance (PSD)

Sight Distances


8Passing Sight Distance (PSD)

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Table


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Table


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Table


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Table


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Stopping Sight Distance in HorizontalCurve Design

To find R min , V max

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Superelevation


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oDesign of Superelevation

Step‐1:The Superelevationfor75 percent of design speed (v m/sec/kmph) is calculated

neglecting the friction.

Step‐2:If the calculated value of ‘e’ is less than 6% (rural road),4%(urban road) the value

so obtained is provided. If the value of ‘e’ as step‐1 exceeds emax then provides maximum

Superelevation equal to emax and proceed with step‐3 or 4.

Step‐3:Check the coefficient of friction of friction developed for the maximum value of e

= emax at the full value of design speed.

If the value of f thus calculated is less than 0.12 the Superelevation of emax is safe for the

design speed. If not, calculate the restricted speed as given in step ‐4.

Superelevation


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oDesign of SuperelevationStep‐4 The allowable speed (Va m/sec. or Va Kmph) at The curve is calculated by

considering the design coefficient of lateral friction and the maximum Superelevation.

If the allowed speed, as calculated above is higher than the design speed, then the design

is adequate and provides a Superelevation of ‘e’ equal to emax.

•If the allowable speed is less than the design speed, the speed is limited to the allowed

speed Va kmph

calculated

above

and

Appropriate

warning

sign

and

speed

limit

regulation

sign are installed to restrict and regulate the speed.

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Superelevation runoff and Tangent runout


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Normal crown

Normal crown

Full

Superelevation

Circular Arc Tangent

Runout

Runoff

PC

PT

1/3 Runoff

1/3 Runoff

Attainment of superelevation


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Theoretical

point of

normal

crown

Theoretical

point of full

superelevation

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Superelevation


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o Calculate superelevation runoff and Tangent runout

To determine Superelevation runoff

Step‐1: use Maximum Relative Gradients



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Step‐2: The Adjustment Factor for Number of Lanes Rotated is calculated by

Or from this table

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Step‐3: The Superelevation runoff is calculated by

Step‐4: The Tangent runout is calculated by

Step‐5: The Theoretical point of normal crown is calculated by

Step‐6: Theoretical point of full superelevation is calculated by

Step‐7: Draw Diagram

EXAMPLE 1


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You are asked to design a horizontal curve for a two-lane road . The road has 3.65m lanes . Due toexpensive excavation, it is determined that amaximum of 10.4 m can be cleared from the road'scenterline toward the inside lane to provide forstopping sight distance . Also, local guidelines dictatea maximum superelevation of 0 .08 m/m . What is the

highest possible design speed for this curve ?

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EXAMPLE 1


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necessary middle ordinate distance is thedistance from the center1ine minus 1/2 theinside lane

Assume the SSD = 105 m

From table 3.1 at SSD =105 m calculate speed V= 70 km/h

EXAMPLE 1


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Assume the SSD = 130 m

calculate middle ordinate distance

this is less than 8.57 m ok

From table 3.1 at SSD =130 m calculate speed V= 80 km/h

calculate middle ordinate distance

this is larger than 8.57m, so design speed is too high

so 70 km/h is the maximum design speed

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EXAMPLE 2


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A horizontal curve on a single-lane highway has its PC at station 24 + 10 and its PI atstation 31 + 40 . The curve has a superelevation of 0 .06 m/m and is designed for 110km/h . What is the station of the PT?

calculate radiusf= 0.11 from table 3.5R = 560m or we can use this equationcalculate tangent lengthT=PI station -. PC station = 730m

knowing tangent length and radius, solve for central angle

EXAMPLE 2


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A horizontal curve on a single-lane highway has its PC at station 24 + 10 and its PI atstation 31 + 40 . The curve has a superelevation of 0 .06 m/m and is designed for 110km/h . What is the station of the PT?

calculate length

calculate PT stationPT station =PC station +. L =24+36.3

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EXAMPLE 3


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surface is determined to have a coefficient of sidefriction of 0 .08, and the curve's superelevation is 0 .0 9 m/m . What is the stationing of the PC and PT and what is the safe vehicle speed ?

calculate PC station

PC station =PI station -. T = 840-155.5 = 684.5 m = 6+84.5

EXAMPLE 3


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calculate safe vehicle speed

calculate radius

calculate length

calculate PT stationPT station =PC station +. L =9+84.5

Since the road is 4 lanes with 3 m lanes, the distance from the

centerline to centerline of first lane is 3 m+1.5 m

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EXAMPLE 4


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A new interstate highway is being built with a design speed of 110 km/h . For oneof the horizontal curves, the radius (measured to the innermost vehicle path) istentatively planned as 275 m . What rate of superelevation is required for thiscurve?

Calculate coefficient of side frictionf= 0 .11 from table 3.5

Calculate superelevation e

EXAMPLE 5


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Calculate coefficient of side frictionf= 0 .11 from table 3.5

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EXAMPLE 5


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calculate length

calculate degree of curvature

EXAMPLE 6


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A horizontal curve on a single-lane freeway ramp is 122 m long, and the design speed of

the ramp is 70 km/h . If the superelevation is 10% and the station of the PC is 17 + 35,

what is the station of the PI and how much distance must be cleared from the center of the

lane to provide adequate stopping sight distance ?

station of the PC = 17 + 35 superelevation e = 10% ramp long L = 122m

Design speed = 70km/h

since the ramp is single-lane, R=R v

calculate cleared distance M v

since the stopping sight distance SSD=105m fromtable 3.1 is less than ramp long

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EXAMPLE 7


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Design speed = 60 km/h =16.7 m/s

calculate the required SSD.calculate cleared distance M v

A horizontal curve with a radius of 245 m connects the tangents of a two-lane highway that has aposted speed limit of 60 km/h. If the highway curve is not superelevated, , determine thehorizontal sightline offset that a large billboard can be placed from the centerline of the insidelane of the curve, without reducing the required SSD. Perception-reaction time is 2.5 sec, and f=0.35.

Radius = 245 m Perception-reaction time = 2.5 sec

f= 0.35.

EXAMPLE 8


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A horizontal curve is to be designed for a two-lane road in mountainous terrain. Thefollowing data are known: Deflection angle: 40 degrees, tangent length= 133.12 m,station of PI: 2700+10.65, fs = 0.12, e = 0.08.Determine:(a) design speed(b) station of the PC(c) station of the PT(d) deflection angle and chord length to the first 100 ft station

(a) From the given horizontal curve data, the radius can be calculated,

calculate the design speed

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EXAMPLE 8


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A horizontal curve is to be designed for a two-lane road in mountainous terrain. Thefollowing data are known: Deflection angle: 40 degrees, tangent length= 133.12 m,station of PI: 27+10.65, fs = 0.12, e = 0.08.Determine:(a) design speed(b) station of the PC(c) station of the PT

(b) station of the PC c) station of the PT

PC station =PI station - T. = 25+77.5 PT station =PC station + L = 25+77.5

EXAMPLE 9


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Building is located 5.8 m from the centerline of the inside lane of a curved section of highway with a 122 m radius. The road is level; e = 0.10. Determine the appropriatespeed limit (to the nearest 10 km/h) considering the following conditions: stopping sightdistance and curve radius.

For stopping sightFor curve radius

M v =5.8 m Radius = 122 m e = 0.10

to the nearest 10 km/h

to the nearest 10 km/h

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EXAMPLE 9


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The stopping sight distance for a speed of 100 km/h and a down grade of 4.5% is mostnearly:(A) 194.5 m(B) 191.4 m(C) 199 m(D) 182.3

Answer: (C)

EXAMPLE 10


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lane 7.3 m single carriageway road has a horizontal curve of radius of 600 m. If theminimum sight stopping distance required is 160 m, calculate in metres the requireddistance to be kept clear of obstructions if the length of the curve is:(a) 200 m;(b) 100 m.

(a) The length of the curve 200 m > 160 m. So the required sight distance S lies wholly within the length of the curve.

(b) The length of the curve 100 m < 160 m. So the required sight distance S lies

outside the length of the curve. Applying equation (4.9), the required offset

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EXAMPLE 11


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Calculate the safe stopping for design speed of 50 kmph for (a)two way traffic on atwo lane road (b) two-way traffic on a single lane road.? Assume coefficient offriction as 0.37 and reaction time of driver as 2.5 sec

Stopping sight distance with single lane= 2*64.1 =122.8m

SSD = 0.278vt + (v²/254f)= 0.278*50*2.5 + (50²/254*0.37)=61.4m

Stopping sight distance when there are two lanes= stopping distance= 61.4m

EXAMPLE 12


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For second car, SD2= 0.278vt + (V²/254f)= 0.278*60*2.5 + (60²/(254*0.35))

=82.2m

Calculate the minimum sight distance required to avoid a head-oncollision of two cars approaching from the opposite direction at 90 and60 kmph. Assume a reaction time of 2.5 sec, coefficient of friction of 0.7and brake efficiency 50%?

Sight distance to avoid head-on collision of twoapproaching cars = SD1+SD2= 153.2+82.2=235.4m

f= 0.5*0.7 = 0.35

Stopping sight distance for first car, SD1=0.278vt + (V²/254f)= 0.278*90*2.5 +(90²/(254*0.35))=153.6m

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Thank You


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Documents

CE 404-SEC11