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CE 404-SEC11
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Transcript of CE 404-SEC11
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8/17/2019 CE 404-SEC11
1/20
2/28/20
GE201: Dr.
N.
A.
Siddiqui
PRACTICE 1
Horizonta l a l ignment
2016CE 404. Highway Engineering
1
HIGHWAY ENGINEERING
(CE 404)
سم هللا لرحن لرحيم
By
Sight Distances
2016CE 404. Highway Engineering
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
2016CE 404. Highway Engineering
5
Passing Sight Distance (PSD)
Sight Distances
2016CE 404. Highway Engineering
6Passing Sight Distance (PSD)
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Sight Distances
2016CE 404. Highway Engineering
7
Passing Sight Distance (PSD)
Sight Distances
2016CE 404. Highway Engineering
8Passing Sight Distance (PSD)
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Table
2016CE 404. Highway Engineering
9
Table
2016CE 404. Highway Engineering
10
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Table
2016CE 404. Highway Engineering
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Table
2016CE 404. Highway Engineering
12
Stopping Sight Distance in HorizontalCurve Design
To find R min , V max
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Superelevation
2016CE 404. Highway Engineering
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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
2016CE 404. Highway Engineering
14
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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Siddiqui
Superelevation runoff and Tangent runout
2016CE 404. Highway Engineering
15
Normal crown
Normal crown
Full
Superelevation
Circular Arc Tangent
Runout
Runoff
PC
PT
1/3 Runoff
1/3 Runoff
Attainment of superelevation
2016CE 404. Highway Engineering
16
Theoretical
point of
normal
crown
Theoretical
point of full
superelevation
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Siddiqui
Superelevation
2016CE 404. Highway Engineering
17
o Calculate superelevation runoff and Tangent runout
To determine Superelevation runoff
Step‐1: use Maximum Relative Gradients
Attainment of superelevation
2016CE 404. Highway Engineering
18
Step‐2: The Adjustment Factor for Number of Lanes Rotated is calculated by
Or from this table
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Attainment of superelevation
2016CE 404. Highway Engineering
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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
2016CE 404. Highway Engineering
20
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
2016CE 404. Highway Engineering
21
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
2016CE 404. Highway Engineering
22
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
2016CE 404. Highway Engineering
23
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
2016CE 404. Highway Engineering
24
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
2016CE 404. Highway Engineering
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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
2016CE 404. Highway Engineering
26
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
2016CE 404. Highway Engineering
27
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
2016CE 404. Highway Engineering
28
Calculate coefficient of side frictionf= 0 .11 from table 3.5
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EXAMPLE 5
2016CE 404. Highway Engineering
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calculate length
calculate degree of curvature
EXAMPLE 6
2016CE 404. Highway Engineering
30
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
2016CE 404. Highway Engineering
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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
2016CE 404. Highway Engineering
32
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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GE201: Dr.
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Siddiqui
EXAMPLE 8
2016CE 404. Highway Engineering
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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
2016CE 404. Highway Engineering
34
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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Siddiqui
EXAMPLE 9
2016CE 404. Highway Engineering
35
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
2016CE 404. Highway Engineering
36
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
2016CE 404. Highway Engineering
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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
2016CE 404. Highway Engineering
38
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
2016CE 404. Highway Engineering
39