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Transportation Engineering

CEE 320 Transportation I

1

Dr. Muhammet Vefa Akpınar, PhD, P.E

Fall 2011


Syllabus

Instructor:

Muhammet Vefa Akpınar, PhD., P.E.

Assistant Prof.

e-mail: [email protected]

http://www.muhfak.ktu.edu.tr/insaat/akademisyenler/

mvakpinar/akpinar.htm

Textbook: Nadir Yaylalı, İTÜ, Birsen Yayınevi, 2010

2

http://www.muhfak.ktu.edu.tr/insaat/akademisyenler/mvakpinar/akpinar.htm












Grading Component Percentage

Midterm Exam 25%

Project Homework 25%

Final Exam 50%

3


Lesson Topic

1 Introduction to Transportation Engineering

2 Road Vehicle Design and Performance

3 Geometric Design of Highways (Horizontal Curve)

4 Geometric Design of Highways (Vertical Curve)

5 Fundamental Curve Properties: Parabolic Formulation

and Offset

6 Application of Superelevation

7 Midterm Exam

8 Stopping Sight Distance

9 Crest Vertical Curves and Sag Vertical Curves

10 Passing Sight Distance

11 Cross Sections and Volume Computation

Cut and Fill

12 Mass Balance

13 Class Demonstrations

14 Final Exam and Term Project Submission

4

http://en.wikibooks.org/wiki/Fundamentals_of_Transportation/Vertical_Curves










http://en.wikibooks.org/wiki/Fundamentals_of_Transportation/Earthwork









http://ocw.korea.edu/ocw/college-of-engineering/transportation-engineering/lecture-notes-1/1.pdf



Transportation Engineering 5

Overview of Transportation Engineering


What is TRANSPORTATION?


Transportation

Transportation:

movement of people and goods from one location to another.

Primary need: economic.

A B


Users / Content

People Passenger Transportation

Goods Freight Transportation

Sh

are o

f to

tal p

assen

gers o

r

ton

s-k

m

Distance

Commuting Shopping Recreation

Business Tourism

Migration

Waste disposal Local distribution

Trade Energy & Raw Materials


Users / Content

Passengers Freight

Board, get off and transfer

without assistance

Must be loaded, unloaded

and transferred

Process information and

act on it without assistance

The information must be

processed through logistics

managers

Make choices between

means of transport often

irrationally

Logistics managers make

choices between means of

transport rationally

Source: Dr. Jean-Paul Rodrigue, Dept. of Economics & Geography, Hofstra University.


Transportation modes:

- Land transportation

- Highway - Rail

- Air transportation

- Domestic - International

- Water transportation

- Inland - Coastal - Ocean

- Pipelines

- Oil - Gas - Other


Vehicles / Services


Infrastructure


Control System


Evolution of Transportation

1500-1840 Average speed of wagon and sail

ships: 16 km/hr

1850-1930 Average speed of trains: 100 km/hr.

Average speed of steamships: 25 km/hr

1950 Average speed of airplanes: 480-640 km/hr

1970 Average speed of jet planes: 800-1120 km/hr

1990 Numeric transmission: instantaneous


What is a mode? Major transportation subsystems


Example Modes


Transportation System

Major transportation subsystems

Land transportation: highway, rail

Air transportation: domestic, international

Water transportation: inland, coastal, ocean

Pipelines: oil, gas, other


Land

18


Air


Water

20


Road transport Advantages of Road transport

(i) It is a relatively cheaper mode of transport as compared to other modes.

(ii) Perishable goods can be transported at a faster speed by road carriers over a short

distance.

(iii) It is a flexible mode of transport as loading and unloading is possible at any

destination. It provides door-to-door service.

(iv) It helps people to travel and carry goods from one place to another, in places

which are not connected by other means of transport like hilly areas.

Limitations of Road transport

(i) Due to limited carrying capacity road transport is not economical for long distance

transportation of goods.

(ii) Transportation of heavy goods or goods in bulk by road involves high cost.


Advantages of Rail transport

(i) It is a convenient mode of transport for travlling long distances.

(ii) It is relatively faster than road transport.

(iii) It is suitable for carrying heavy goods in large quantities over long

distances.

(iv) Its operation is less affected by adverse weathers conditions like rain,

floods, fog, etc.

Limitations of Railway transport

(i) It is relatively expensive for carrying goods and passengers over short

distances.

(ii) It is not available in remote parts of the country.

(iii) It provides service according to fixed time schedule and is not flexible for

loading or

unloading of goods at any place.

(iv) It involves heavy losses of life as well as goods in case of accident.


Evolution of Transportation

100

500

1000

1800 1900 2000 1850 1950

50

250

750

Stage Coach

Rail

Automobile

HST

Propeller Plane

Jet Plane

Liner Clipper Ship Containership

Road

Maritime

Rail

Air


What is TRANSPORTATION

ENGINEERING ?



One of the specialty areas of civil

engineering

• Development of facilities for the

movement of goods and people

• Planning, design, operation and

maintenance

Multidisciplinary study



Definition of Transportation Modes

A transportation system is an

infrastructure that serves to move people

and goods efficiently. The transportation

system consists of fixed facilities, flow

entities, and a control component.

Efficient = safe, rapid, comfortable,

convenient, economical, environmentally

compatible.


28


consisted of the fixed facilities, the flow entities and the control systems that permit people, and goods to overcome the friction of geographical space efficiently in order to participate in a timely manner in some desired activity


Highway Transportation

Engineering

Definition

The application of technology and scientific principles to the planning, functional design, operation, and management of roads, streets and highways, their networks, terminals, abutting lands, and relationships with other modes of transportation.

Areas of highway transportation engineering:

• Planning of streets and highways

• Geometric design of road facilities

• Traffic operations and control

• Traffic safety

• Maintenance of road facilities and controls


Traffic

Concepts


Selected Table Factoids

Traffic Typically Peaks twice per day.

0

1000

2000

3000

4000

5000

6000

7000

12:30

AM

1:30

AM

2:30

AM

3:30

AM

4:30

AM

5:30

AM

6:30

AM

7:30

AM

8:30

AM

9:30

AM

10:30

AM

11:30

AM

12:30

PM

1:30

PM

2:30

PM

3:30

PM

4:30

PM

5:30

PM

6:30

PM

7:30

PM

8:30

PM

9:30

PM

10:30

PM

11:30

PM

12:30

AM

Time of Day

Flo

w i

n v

eh

icle

s p

er h

ou

r Highway Capacity

Highly Congested


Outline

1. Basic Concepts

a. Flow Rate

b. Spacing

c. Headway

d. Speed

e. Density

2. Relationships

3. Example


Flow Rate (q)

The number of vehicles (n) passing some designated

roadway point in a given time interval (t)

Units are typically vehicles/hour

Flow rate is different than volume

t

nq


Density (k)

The number of vehicles (n)

occupying a given length (l) of

a lane or roadway at a

particular instant

Unit of density is vehicles per

mile (vpm).

u

q

l

nk


Other Concepts

Free-flow speed (uf)

Jam density (kj)

Capacity (qm)

vehmtspacing

vehicleskmvehDensity

//

hourkmspeed

vehmtspacingvehsHeadway

/

//

vehsheadwayhrvehrateFlow

/

1/


Speed vs. Density

j

fk

kuu 1

Density (veh/km)

Spee

d (

km

/hour)

kj

Jam Density

uf

Free Flow Speed


Flow vs. Density

j

fk

kkuq

2

Density (veh/km)

FLow

(veh/h

r)

kj

Jam Density

Optimal flow,

capacity, qm

km

Optimal density Uncongested Flow

Congested Flow


Speed vs. Flow

f

ju

uukq

2

Flow (veh/hr)

Spee

d (

mph)

uf

Free Flow Speed

Optimal flow,

capacity, qm

Uncongested Flow

Congested Flow

um


Traffic – Time of Day Patterns

0.00%

1.00%

2.00%

3.00%

4.00%

5.00%

6.00%

7.00%

8.00%

9.00%

1 3 5 7 9 11 13 15 17 19 21 23

Hour of Day

Percen

t o

f D

ail

y T

ra

ffic

Rural Cars

Business Day Trucks

Through Trucks

Urban Cars


From WSDOT 2003 Annual Traffic Report


Volume Patterns (contd.)


Traffic Stream Characteristics


Volume

Traffic volume is defined as the number of vehicles that pass a point on a highway, or a given lane or direction of a highway, during a specified time interval.

A. Daily volumes:

- Average Annual Daily Traffic: (AADT):

- Average Annual Weekday Traffic (AAWT):

- Average Daily Traffic (ADT):

- Average Weekday Traffic (AWT):

Note: The unit is vehicles per day (vpd).

Daily volumes are used to establish trends over time and for planning purposes. Daily volumes generally are not differentiated by direction or lane but are totals for an entire facility at the specified location.


Daily Volumes

- Average Annual Daily Traffic (AADT): is the average 24

hour traffic volume at a given location over a full 365-day

year – that is the total number of vehicles passing the

site in a year divided by 365

- Average Annual Weekday Traffic (AAWT): is the average

24-hour traffic volume occurring on weekdays over a full

year. AAWT is computed by dividing the total weekday

traffic volume for the year by 260. This volume is of

considerable interest where weekend traffic is light, so

that averaging higher weekday volumes over 365 days

would mask the impact of weekday traffic.


Daily Volumes (contd.)

- Average Daily Traffic (ADT): is an average 24-hour traffic

volume at a given location for some period of time less

than a year. While an AADT is for a full year, an ADT

may be measured for six months, a season, a month, a

week, or as little as two day. an ADT is a valid number

only for the period over which it was measure.

- Average Weekday Traffic (AWT): is an average 24-hour

traffic volume occurring on weekdays for some period of

time less than one year, such as for a month or a

season. The relationship between AAWT and AWT is

analogous to that between AADT and ADT


Daily Volumes


Sub hourly Volumes

veh/hn timeobservatio

nobservatio during vehiclesofnumber =qflow of rate


Sub hourly Volumes (contd.)

Example of volumes and rate of flow

Time Volume Rate of flow

interval (vehicles) (vehicles/h)

5:00-5:15 PM 950 950*4(15 minutes)=3800

5:15-5:30 PM 1150 1150*4=4600

5:30-5:45 PM 1250 1250*4=5000

5:45-6:00 PM 1000 1000*4=4000

For the hour

5:00-6:00 PM 4350 (veh/h)

A facility may have capacity adequate to serve the peak-hour demand, but short-term peaks of flow within the peak hour may exceed capacity, thereby creating a breakdown.


Relationships among Flow Rate,

Speed and Density

v = S * D

Where

v – rate of flow, veh/h or veh/h/ln

S – Space Mean Speed (mi/s)

D – Density, veh/mi or veh/mi/ln

Space mean speed and density are measures that refer to a specific section of a lane or highway, while flow rate is a point measure

This relationship is most often used to estimate density, which is difficult to measure directly, from measured values of flow rate and space mean speed

t


Vehicle Dynamics


Main Concepts

Resistance

Tractive effort

Vehicle acceleration

Braking

Stopping distance

grla RRRmaF


Resistance

Resistance is defined as the force impeding vehicle

motion

1. What is this force?

2. Aerodynamic resistance

3. Rolling resistance

4. Grade resistance

grla RRRmaF


Grade Resistance Rg

Composed of

• Gravitational force acting on the

vehicle gg WR sin

gg tansin

gg WR tan

Ggtan

WGRg

For small angles,

θg W

θg

Rg


Available Tractive Effort

The minimum of:

1. Force generated by the engine, Fe

2. Maximum value that is a function of

the vehicle’s weight distribution and

road-tire interaction, Fmax

max,mineffort tractiveAvailable FFe


Diagram

θg


Braking Distance

Theoretical

• ignoring air resistance

Practical

Perception

Total

grlb

b

fg

VVS

sin2

2

2

2

1

Gg

ag

VVd

2

2

2

2

1

pp tVd 1

ps ddd

a

VVd

2

2

2

2

1

For grade = 0


Stopping Sight Distance , SSD

Length of roadway that should be visible

ahead of you in order to ensure that you will

be able to stop if there is an object in your

path

Calculate the SSD for a vehicle traveling on

your roadway at the design speed, and then

make sure the actual sight distance that you

provide is at least as great as the stopping

sight distance


Stopping Sight Distance (2)

Assume

• Driver eye height of 3.5 feet

• Height of object between 2.0 and 3.5 feet

Reaction distance + braking distance

Design standard: tr=2.5, a=11.2

Ga

VtVSSD

sft

mph

srmphft

2.3230

47.1)/(

2

)(

)()()(2


Other Sight Distances

Decision sight distance

• Allow longer tr for information processing for different maneuver

conditions

Passing sight distance

• Ensure safe passing maneuver

• 4 distance components

At 70 mph

• SSD = 730 ft

• DSD = 1445 ft (maneuver E)

• PSD = 2480 ft


Stopping Sight Distance

(SSD)

Worst-case conditions

• Poor driver skills

• Low braking efficiency

• Wet pavement

Perception-reaction time = 2.5 seconds

Equation rtV

Gg

ag

VSSD 1

2

1

2



(SSD)

from ASSHTO A Policy on Geometric Design of Highways and Streets, 2004

Note: this table assumes level grade (G = 0)


SSD – Quick and Dirty

a

VVV

V

Ggag

VVd

222

22

1

2

2

2

1 075.12.11

075.12.11

1

2

47.1

02.322.112.322

047.1

2

1. Acceleration due to gravity, g = 32.2 ft/sec2

2. There are 1.47 ft/sec per mph

3. Assume G = 0 (flat grade)

ppp VttVd 47.147.1 1

V = V1 in mph

a = deceleration, 11.2 ft/s2 in US customary units

tp = Conservative perception / reaction time = 2.5 seconds

ps Vta

Vd 47.1075.1

2


Sight Distance


Lecture Outline

The ability to see ahead is critical for

traffic safety and efficiency.

Four cases are to be discussed:

1. Stopping sight distance,

2. Passing sight distance,

3. Sight distance in complex situations,

and

4. Provided sight distance.



Stopping Sight Distance =

Reaction Distance + Braking Distance

The reaction distance is calculated as:

where: dr = break reaction distance, m; tr = reaction time, s; V = initial speed, km/h.

The median reaction time is 0.7 s, and the 90th percentile is 1.5 s. In unexpected situations the 90th percentile tends to be one second longer. The Policy recommends the 2.5-second reaction time.

Vtd rr 278.0


Braking distance on a level roadway is

calculated as:

where:

db = braking distance, m; V = initial speed,

km/h; and a = 3.4 m/s2, deceleration

rate.

a

Vdb

2

039.0




4.3039.05.2278.0

2VVd


Stopping on Grades

A stopping distance on grades G is calculated as follows:

where G is the percent of graded divided by 100 with the minus sign for downgrades and the plus sign for upgrades.

)81.9

(254

278.02

Ga

VVtd


Stopping on Grades


Trucks Stopping

A stopping distance of trucks is longer than of smaller vehicles. A higher position of seats in trucks than in other vehicles recompenses the longer stopping distance.

Above-minimum design for trucks is recommended where sight distance is reduced by horizontal obstructions, particularly at downgrades.


Sight Distance in Complex

Conditions


Passing Sight Distance Sight distance is determined for a single vehicle passing a single vehicle

with the assumption that cover majority of situations observed in the real-

world conditions.

d t v m at d vt d d d1 1 1 2 2 3 4 20278 2 0278 30 90 2 3. ( / ); . ; /m;

where: t1 = time of initial maneuver, s; a = average acceleration, km/h/s; v = average

speed of passing vehicle, km/h; m = difference in speed between passing and passed

vehicles, km/h; t2 = time passing vehicle occupies the left lane, s.

Exhibit 3-4


Provided Sight Distance

Potential sight obstructions

• On vertical curves: road surface at some

point on a crest vertical curve, range of

head lights on a sag curve

• On horizontal curves: barriers, bridge-

approach fill slopes, trees, back slopes of

cut sections


Provided Sight Distance

Assumed heights:

Height of the drivers eye = 1080 mm

Height of the object for stopping distance =

600 mm (the lowest object that can create

hazardous conditions)

Height of the object for passing distance =

1330 mm (15th percentile height of passenger

car body)


Sight Distance


Consider a typical example of a driver approaching a STOP sign. The

driver first sees the sign (perception), then recognizes it as a STOP

sign (intellection), then decides to STOP (emotion), and finally puts

his or her foot on the brake (volition).

Why is perception-reaction time important in design? (i) Used to determine safe stopping distance

(ii) Used to determine minimum sight distance

(iii) Used to determine the length of the yellow phase at a signalized

intersection

AASHTO recommends a perception-reaction time of 2.5 seconds for

design

E:/program files/1-inactive documents/courses/EGN 1004 -- First Year Engineering/Lecture Material/Driver Reaction Time.ppt




Sight Distance:

Sight distance = length of highway visible to the driver

Stopping sight distance = the sight distance required to safely stop a vehicle

traveling at design speed

Passing sight distance = the sight distance required (two-lane highway) for a

vehicle to execute a normal passing maneuver as related to design conditions

and design speed

Decision sight distance = the sight distance required for a driver to detect an

unexpected or difficult-to-perceive information source or hazard, interpret the

information, recognize the hazard, select and appropriate maneuver

Entering sight distance = the sight distance along a roadway that an object of

specified height is continuously visible to a driver entering a roadway from a

driveway or cross street.


Stopping Distance

Distance ahead of the driver in which the driver can bring

the vehicle to a stop after seeing an object in the

vehicle’s path without hitting the object

perception

SSD = reaction + braking

distance distance

Perception reaction distance: distance traveled during

the perception reaction time process

Braking distance: distance to stop a vehicle once the

brakes are applied


Stopping Distance (contd.)

(metric)

Where

v – speed (km/h)

t - perception reaction time (typically 2.5 s)

f - longitudinal coefficient of friction

G - upgrade (+) or downgrade (-)

Table 1.2.5.3 – minimum stopping sight distance for vehicle and trucks on level terrain and wet pavement

Table 1.2.5.4 – minimum stopping sigh distance for trucks with conventional braking systems

Tables from TAC Geometric Design Guide for Canadian roads

)(2546.3

22

0

Gf

vvtvSSD


Stopping Distance (contd.)

or

Where

u – speed (mile/h)

t - perception reaction time (typically 2.5 s)

f - longitudinal coefficient of friction (f = a/g)

G - upgrade (+) or downgrade (-)

Table 3.4– minimum stopping sight distance for different

design speeds (grade considered is zero)

Table from class textbook

)(30*47.1

22

0

GftSSD


Passing Distance (contd.)

Where:

• d1 = distance traversed during perception-reaction time and during initial acceleration to the point where the passing vehicle just enters the left lane

• d2 = distance traveled during the time the passing vehicle is traveling in the left lane

• d3 = distance between the passing vehicle and the opposing vehicle at the end of the passing maneuver

• d4 = distance moved by the opposing vehicle during two thirds of the time the passing vehicle is in the left lane (usually taken to be 2/3 d2 )


Geometric Design


Km P = 0+170Kot P = 764,69 m

14

e = 0,15 m

14

6

14

4

10

1

04

4

15

7

16

9

00

4

21

2


Outline

1. Concepts

2. Horizontal Alignment a. Fundamentals

b. Superelevation

3. Vertical Alignment a. Fundamentals

b. Crest Vertical Curves

c. Sag Vertical Curves

d. Examples

4. Volume Table and Earthwork


What are the four basic elements of geometric design?

a. Horizontal alignment

b. Vertical alignment

c. Cross-section design

Horizontal and vertical alignment are controlled by two basic

design criteria:

a. Design speed

b. Sight distance


Design speed

Design speed is defined as the maximum safe speed that can

be maintained over a specified section of a highway when

conditions are so favorable that the design features of the

highway govern.

Sight distance

There are two types of sight distance used in designing

highways:

a. Stopping sight distance

b. Passing sight distance


Stopping sight distance

Stopping sight distance is the distance required to see an

object 15 cm high on the roadway. It is intended to allow

drivers to stop safely after sighting an object on the

roadway large enough to cause damage to the vehicle or

loss of control

Passing sight distance

Passing sight distance is the distance required to see an

oncoming vehicle of a certain minimum size. A passing

driver must have sight distance to observe an oncoming

vehicle at a distance sufficient to allow him or her to enter

the opposing lane, pass a moving vehicle, and return to the

travel lane safely.


Highway Components

Highway plan and profile


Concepts

Alignment is a 3D problem broken

down into two 2D problems

• Horizontal Alignment (plan view)

• Vertical Alignment (profile view)

Stationing

• Along horizontal alignment

• 12+00 = 1,200 mt.

Piilani Highway on Maui


Vertical Alignment

Objective:

• Determine elevation to ensure

• Proper drainage

• Acceptable level of safety

Primary challenge

Transition between two grades

• Vertical curves G1 G2

G1 G2

Crest Vertical Curve

Sag Vertical Curve


Vertical Curves

To provide transition between two

grades

Consider

• Drainage (rainfall)

• Driver safety (SSD)

• Driver comfort

Use parabolic curves

Crest vs Sag curves


Vertical Curves

Controlling factor: sight distance

Stopping sight distance should be provided as a minimum

Rate of change of grade should be kept within tolerable limits

Drainage of sag curves is important consideration, grades not less than 0.5% needed for drainage to outer edge of roadway


Specifies the elevation of points along a roadway

Provides a transition between two grades

Sag curves and crest curves

Equal-tangent curves - half the curve length positioned before

the PVI; half after


Vertical Curves

Vertical curves provide a gradual change between two adjacent road grades

Components of the equal tangent vertical curve


Vertical Curves

Given

– G1, G2: initial & final grades in percent

– L: curve length (horizontal distance)

Develop the actual shape of the vertical curve

PVI

point of vertical curvature

point of vertical intersection

point of vertical

tangency G2%

G1%


Vertical Curves

• Define curve so that PVI is at a horizontal distance of L/2 from PVC and PVT

• Provides constant rate of change of grade: L

GGr 12

G1%

G2%

A

L

Axx

GEE PVCP

200100

2

1


Vertical Curve Fundamentals

G1

G2

PVI

PVT

PVC

L

L/2

δ

cbxaxy 2

x

Choose Either: • G1, G2 tangent grades

•y is the roadway elevation x stations

from the beginning of the curve


Relationships

Choose Either: • G1, G2 in decimal form, L in mt

• G1, G2 in percent, L in stations

G1

G2

PVI

PVT

PVC

L

L/2

δ

x

1 and 0 :PVC At the Gbdx

dYx

cYx and 0 :PVC At the

L

GGa

L

GGa

dx

Yd

22 :Anywhere 1212

2

2


Vertical Alignment Relationships

1

2

200

800

200

GKx

A

LK

ALY

ALY

xL

AY

hl

f

m

L

GGa

adx

yd

Gdx

dyb

xatPVC

baxdx

dy

cbxaxy

2

2

:0,

2

12

2

2

1

2


Example

A 400 ft. equal tangent crest vertical curve has a PVC station of 100+00 at

800 m elevation. The initial grade is 2.0 percent and the final grade is -4.5

percent. Determine the elevation and stationing of PVI, PVT, and the high

point of the curve.

PVI

PVT

PVC: STA 100+00

EL 800 m.


PVI

PVT

PVC: STA 100+00

EL 59 ft.


PVI

PVT

PVC: STA 100+00

EL 59 ft.

Determine the elevation and stationing of PVT, and the

high point of the curve.

400 ft. vertical curve PVT is at STA 104+00

Equal tangents:

Elevation of the PVI is 59’ + 0.02(200) = 63 ft.

Elevation of the PVT is 63’ – 0.045(200) = 54 ft.


High point elevation requires figuring out the equation for a vertical curve

At x = 0, y = c => c=59 ft.

At x = 0, dY/dx = b = G1 = +2.0%

a = (G2 – G1)/2L = (-4.5 – 2)/(2(4)) = - 0.8125

y = -0.8125x2 + 2x + 59

High point is where dy/dx = 0

dy/dx = -1.625x + 2 = 0

x = 1.23 stations

Find elevation at x = 1.23 stations

y = -0.8125(1.23)2 + 2(1.23) + 59

y = 60.23 ft

PVI

PVT

PVC: STA 100+00

EL 59 ft.


Other Properties

G1

G2

PVI

PVT PVC

x

Ym

Yf

Y

21 GGA

•G1, G2 in percent

•L in mt

A is the absolute value in grade differences,

if grades are -3% and +4%, value is 7


Crest Vertical Curves

G1 G2

PVI

PVT PVC

h2 h1

L

SSD

2

21

2

22100 hh

SSDAL

A

hhSSDL

2

212002

For SSD < L For SSD > L

Line of Sight


Sag Vertical Curves

G1 G2

PVI

PVT PVC

h2=0 h1

L

Light Beam Distance (SSD)

tan200 1

2

Sh

SSDAL

A

SSDhSSDL

tan2002 1

For SSD < L For SSD > L

headlight beam (diverging from LOS by β degrees)


Sag Vertical Curves

Four criteria for establishing length of

sag curves

• Headlight sight distance

• Passenger comfort

• Drainage control

• General appearance


Sag VC - Design Criteria

Headlight sight distance

Rider comfort

Drainage control

Appearance


Example 1

A car is traveling at 30 mph in the country at night on a wet road through a

150 ft. long sag vertical curve. The entering grade is -2.4 percent and the

exiting grade is 4.0 percent. A tree has fallen across the road at approximately

the PVT. Assuming the driver cannot see the tree until it is lit by her

headlights, is it reasonable to expect the driver to be able to stop before hitting

the tree?


Example 2

Similar to Example 1 but for a crest curve.

A car is traveling at 30 mph in the country at night on a wet road through a

150 ft. long crest vertical curve. The entering grade is 3.0 percent and the

exiting grade is -3.4 percent. A tree has fallen across the road at

approximately the PVT. Is it reasonable to expect the driver to be able to stop

before hitting the tree?


Example 3

A roadway is being designed using a 45 mph design speed. One section of the

roadway must go up and over a small hill with an entering grade of 3.2

percent and an exiting grade of -2.0 percent. How long must the vertical

curve be?


Example Problem: Vertical

Curve

A vertical curve crosses a 4’ diameter pipe at

right angles. Pipe at sta 110+85 with

centerline elevation of 1091.60’. PVI at sta

110+00 elevation 1098.4’. Equal tangent

curve, 600’ long with initial and final grades of

+1.2% and -1.08%. Using offsets determine

the depth below the surface of the curve the

top of the pipe and determine the station of

the highest point of the curve.


Example

G1 = 2%

G2 = -4%

Design speed = 70 mph

Is this a crest or sag curve?

What is A?


Horizontal Alignment

Objective:

• Geometry of directional transition to ensure:

• Safety

• Comfort

Primary challenge

• Transition between two directions

• Horizontal curves

Fundamentals

• Circular curves

Δ


Horizontal Curves

Provide transition of a roadway between

two straight sections

Two key factors

• Superelevation е – number of vertical

feet of rise per 100 feet of horizontal

distance

• Coefficient of side friction fs - function

of design speed


Horizontal Curve Fundamentals

R

T

PC PT

PI

M

E

R

Δ

Δ/2 Δ/2

Δ/2 L

Based on circular curve

• R: radius of curve

• D: degree of curve

• : central angle

• T: length of tangent

• L: length of curve

• LC: long chord

• M: middle ordinate dist

• E: external dist

Point of Curvature

Point of Tangency


Horizontal Curve

Fundamentals

12cos

1RE

2cos1RM

R

T

PC PT

PI

M

E

R

Δ

Δ/2 Δ/2

Δ/2 L

2tanRT

DRL

100

180

RRD

000,18

180100


Minimum Curve

Radius

• Curve requiring the most

centripetal force for the

given speed

• Given emax, umax, Vdesign

Horizontal Curve

ue

VR

mph

ft15

min

2

)(

)(

R


Stopping Sight Distance &

Horizontal Curve Design

Adequate sight distance must be provided in

the design of horizontal curves

Cost of right of way or the cost of moving

earthen materials often restrict design options

When such obstructions exist, stopping sight

distance is checked and measured along the

horizontal curve from the center of the

traveled lane



Rv

Δs

Obstruction

Ms

v

sR

SSD180

DRSSD s

sv

100

180

SSD

v

vsR

SSDRM

90cos1

v

svv

R

MRRSSD 1cos

90


Horizontal Curve Sight Distance

Sight line is a chord

of the circular curve

Sight Distance is

curve length

measured along

centerline of inside

lane R

SDRM

65.28cos1

R

Ga

VVtSSD r

2.3230

47.12

Recall


Basic controlling expression

e = rate of superelevation

u = side friction factor (dep. on pavement, speed, …)

V = vehicle speed

R = radius of curve


)(

2

15

)(

ftR

Vue

mph



Overall design procedure

• Determine a reasonable maximum

superelevation rate.

• Decide upon a maximum side-friction

factor.

• Calculate the minimum radius.

• Iterate and test several different radii until

you are satisfied with your design.

• Make sure that the stopping sight

distance is provided. Adjust your design if


Side Friction

Design based on point where centrifugal

force creates feeling of discomfort for

driver Speed umax udesign

20 0.50 0.17

30 0.35 0.16

40 0.32 0.15

50 0.30 0.14

60 0.29 0.12

70 0.28 0.10


Example 4

A horizontal curve is designed with a 1500 ft. radius. The tangent length is

400 ft. and the PT station is 20+00. What are the PI and PT stations?


Superelevation


Superelevation

Tilting the roadway to help offset centripetal forces developed as the vehicle goes around a curve

General Practice • Highways, no ice/snow

emax = 0.10

• Highways, snow/ice

emax = 0.06

• Traffic congestion or roadside development, limit speeds

emax = 0.04 ~ 0.06

e

1


Centripetal or Centrifugal?

As a vehicle moves in a circular path

• Centripetal acceleration acts on the vehicle in the direction of the center of the curve

The acceleration is sustained by

• Component of the vehicle’s weight related to the roadway superelevation

• Side friction developed between the vehicle’s tires and the pavement surface

• Or a combination of the two


Centrifugal Force

Imaginary force that drivers believe is

pushing them outward while

maneuvering a curve

In fact, the force they feel is the vehicle

being accelerated inward towards the

center of the curve


Centripetal Acceleration

Is counter-balanced by two factors: • Superelevation

• Side Friction Factor

Research has been conducted (dated) that has established limiting values for superelevation rate (e max) and side friction demand (f max)

Applying the limiting values results in the minimum curve radius for various design speeds


Superelevation

Limits of the rate superelevation are related

to

• Climate

• Ice and snow can slow vehicles. Should not create a

situation where these vehicles slide into the center of the

curve when traveling slowly or standing still.

• Constructability (cost)

• Adjacent land use

• Frequency of slow moving vehicles


Superelevation

Too much super

• When traveling slowly, must steer up the slope or

against the horizontal curve to maintain proper

path

• Undesirable to have such situations when slow

traveling traffic can occur often (urban areas with

congestion)

• Considerations for SUV traffic, high center of

gravity, can cause roll-overs on such designs


Side Friction Factor

The vehicle’s need for side friction to

maintain path on curve

Upper limit of side friction is the point at

which a tire would begin to skid, point of

impending skid

We design for safety, so f values

substantially less than this


Side Friction Factor

How do we choose maximum side friction factors for use in design?

We measure the level of centripetal or lateral acceleration that causes drivers to react instinctively to choose a lower speed.

We set this as the maximum side friction factor.


Maximum Rates of

Superelevation

Controlled by four factors:

• Climate conditions (snow/ice regions)

• Terrain conditions (flat, rolling, mountainous)

• Type of area (rural, urban, suburban)

• Frequency of very slow-moving vehicles

Conclusion: no universal e max can be set

However, for similar areas, a consistent

maximum superelevation should be selected


Recommended Practice

12 percent superelevation should not be exceeded

4 or 6 percent superelevation is applicable for urban design with little constraints

Superelevation may be omitted on low-speed urban streets where severe constraints exist


Minimum Radius

Controls design speed

Can be determined from the max

superelevation and the max side friction

factor

Can be calculated from equation 3.34 or

determined from Table 3.5


Example – Minimum Radius

70 mph design speed; e = 8%; fs = 0.10

Determine the minimum radius of curve

(measured to the traveled path).


Superelevation

cpfp FFW

cossincossin22

vv

sgR

WV

gR

WVWfW

α

Fc

W 1 ft

e

≈

Rv


Superelevation

cossincossin22

vv

sgR

WV

gR

WVWfW

tan1tan2

s

v

s fgR

Vf

efgR

Vfe s

v

s 12

efg

VR

s

v

2


Selection of e and fs

Practical limits on superelevation (e)

• Climate

• Constructability

• Adjacent land use

Side friction factor (fs) variations

• Vehicle speed

• Pavement texture

• Tire condition


Minimum Radius Tables


Design Side Friction Factors

fro

m t

he

20

05

WS

DO

T D

esig

n M

an

ua

l, M

22

-01

For Open Highways and Ramps


Design Superelevation Rates - AASHTO

from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004


Design Superelevation Rates -

WSDOT

from the 2005 WSDOT Design Manual, M 22-01

emax = 8%


Supplemental Stuff

Cross section

Superelevation Transition

• Runoff

• Tangent runout

Spiral curves

Extra width for curves



from the 2001 Caltrans Highway Design Manual


Superelevation

Runoff/Runout

fro

m A

AS

HT

O’s

A P

oli

cy o

n G

eom

etri

c D

esig

n o

f H

igh

wa

ys a

nd

Str

eets

20

01


Superelevation Runoff -

WSDOT

from the 2005 WSDOT Design Manual, M 22-01

New Graph


Spiral Curves

No Spiral

Spiral



No Spiral


Spiral Curves

WSDOT no longer uses spiral curves

Involve complex geometry

Require more surveying

Are somewhat empirical

If used, superelevation transition should

occur entirely within spiral


Desirable Spiral Lengths



With Transition Curves

Transition Curves

Gradually changing the curvature from

tangents to circular curves

Without Transition Curves


Transition Curves

Gradually changing the curvature from

tangents to circular curves

• Use a spiral curve

L: min length of spiral (ft)

V: speed (mph)

R: curve radius (ft)

C: rate of increase of centrifugal accel

(ft/sec3), 1~3 RC

VL

315.3


Transitional Curves Gradually changing the cross-section of the roadway from

normal to superelevated

Keep water drainage in mind while

considering all of the available

cross-section options


Vertical Alignment

Grade

• measure of inclination or slope, rise over

the run

• Cars: negotiate 4-5% grades without

significant speed reduction

• Trucks: significant speed changes

• 5% increase on short descending grades

• 7% decrease on short ascending grades


Grade Considerations

Maximum grade – depends on terrain type,

road functional class, and design speed

Terrain 60mph 70mph

Level 3% 3%

Rolling 4% 4%

Mountainous 6% 5%

Rural Arterials


Example

horizontal curve with 2000’ radius; 400’

tangent length; PI is at station 103+00

Determine the stationing of the PT


Example continued

Determine the central angle, ∆. Next determine

the Length of Curve, L.

ftL

RL

RT

58.789)62.22(2000180

1416.3

180

62.22

2tan2000400

2tan


Example continued

Knowing tangent length is 400’ and PI is at 103+00:

stationing PC=103+00 minus 4+00=99+00

Horizontal curve stationing is measured along the alignment of the road:

stationing of PT = stationing of PC+L

=99+00 plus 7+89.58 = 106+89.58


In-Class Problems

Calculate the maximum degree of curve

and minimum radius of a simple circular

curve with an external angle of 100º.

Design speed of 50mph; fmax 0.14; max

e =0.10.


Sight Distance Example

Horizontal curve with 2000’ radius;

12’lanes; 60mph design speed.

Determine the distance that must be

cleared from the inside edge of the

inside lane to provide sufficient stopping

sight distance.


Example Problem

70mph design speed; equal tangent

vertical curve needed to connect +1.0%

with -2.0%.

Determine min length of curve to meet

SSD requirements.


Project Example 1


Tablo 1. Örnek yol geometrik standartları

YıllıkO r t a l a m a . G ü n . T r a f i k

Y . O . G . T . T a ş ı t / G ü n

Proje Saatlik T rafiği P . S . T . ( T a ş ı t / S a a t )

100 80 80 70 70 60 80 70 70 60 60 40 70 60 60 50 50 30

400 250 250 200 200 150 250 200 200 150 150 60 200 150 150 90 90 30

160 130 130 120 120 100 130 120 120 100 100 60 120 100 100 70 70 30

4 4 6 6 7 7 5 5 7 7 8 8 6 6 8 8 9 9

kapalı kurb Kk ( - ) 107-56 44-26 44-26 29-20 29-20 17-15 44-26 29-20 29-20 17-15 17-15 6-6 29-20 17-15 17-15 10-9 10-9 5-5

açık kurb Ka ( - ) 51-35 30-23 30-23 22-19 22-19 16-15 20-23 22-19 22-19 16-15 16-15 8-8 22-19 16-15 16-15 12-11 12-11 7-7

8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

155 110 110 90 90 70 110 90 90 70 70 40 90 70 70 55 55 25

670 550 550 480 480 420 550 480 480 420 420 270 480 420 420 340 340 190

3,50 3,50 3,50 3,50 3,50 3,50 3,50 3,50 3,25 3,25 3,25 3,25 3,00 3,00 3,00 3,00 3,00 3,00

2,50 2,50 2,00 2,00 2,00 2,00 1,50 1,50 1,50 1,50 1,50 1,50 1,00 1,00 1,00 1,00 1,00 1,00

12,00 12,00 11,00 11,00 11,00 11,00 10,00 10,00 9,50 9,50 9,50 9,50 8,00 8,00 8,00 8,00 8,00 8,00

Kısa köprüler

(0 - 45 m) Wk (m)

Uzun köprüler

( 45 m) Wu

(m)

5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00 5,00

Toplam genişlik K G ( m )

Eksen uzaklığıL e

( m )

*** Dağlık arazideki karışık kesitlerde banket genişlikleri dolgularda 50 cm. fazla , yarma tarafında 50 cm. eksik uygulanır.

Normal 60.00 m. Projenin gerektirdiği kadar Normal 40.00 m. Projenin gerektirdiği kadar Normal 15.00 m. Projenin gerektirdiği kadar

KS 23.00 E 37.00 KS KS 20.00 E 20.00 KS KS 7.50 E 7.5 KS

S1KENT DIŞI

İKİ ŞERİTLİ YOLLARKARAYOLU GEOM ETRİK STANDARTLARI

* Projelendirilen yolun yapımın bitiminden itibaren 20 yıl sonra ulaşılacağı hesaplanan trafik değeri

250

Düz Dalgalı Dağlık

450

D

8000 4500 2500

D

Dalgalı Dağlık

İK İNCİ SINIF

D

800

ÜÇÜNCÜ SINIF

3000

1100 550 300

BİRİNCİ SINIF

Dağlık

D

4000

400

12000

D

Düz

1200

D D D

11000 5500

Düz

** Kar ve Buzlanma olmayan kesimlerde % 10"a kadar artırılabilir.

Platform genişliği PG (m )

Köprü genişliği K ö p r ü p r o j e y ü k ü

H: 20 - S: 16

Alt geçit (m inim um h:5) h (m )

Kam ulaştırm a g e n i ş l i ğ i

Gabari

Em niyetli duruş uzaklığı Ld (m )

Em niyetli geçiş uzaklığı Lg (m )

Şerit genişliği L (m )

Banket genişliği*** b (m )

Minim um klotoit param etresi A(-)

Maksim um boyuna eğim m (%)

Düşeyk u r b

katsayısı L=GK

Maksim um dever** n (%)

PROJE ELEM ANLARI

Proje Hızı Vp (km /saat)

M inim um kurb yarıçapı R(m )

D

6500

650

DalgalıTopografik Model TM(Dz,Dl,Dğ)

Trafik*

Hizm et Sev iyesi HS(A,B,C,D,E,F)

9.50 9.50 7.00

8.50 8.50 7.00

%2

b b2L

PG

%2

b b2L

PG

%2

b b2L

PG


B

772

A

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756

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760

762

764

766

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770

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784

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780

782

784

786

788

790

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796

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K

0 20 40 60 80 100

TEPE


B

A

B

772

A

K5

75

0

75

2

75

4

75

6

75

8

76

0

76

2

76

4

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6

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8

77

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K

0 20 40 60 80 100

TEPE

Sıfır Poligonu

%5

%5

%5 %5%4

%4 %4

%4

%4

%4%4


183

B

A

B

772

A

K5

75

0

75

2

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4

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6

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8

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0 20 40 60 80 100

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B

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B

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0 20 40 60 80 100

TEPE

Geçkiler

A1

A2

A3

A4

S


O

B

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B

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752

754

756

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760

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764

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770

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K

0 20 40 60 80 100

TEPE

Yatay Kurp

R=2

00m

TFTO

S

51,7

5°

Bs

? = 51,75°R = 200mT = 97,55mD = 180,55mBs = 22,2m

R=200m


186

O

A

772

A

K5

77

2

77

4

77

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2

79

0

79

2

79

4

79

6

79

8

80

0

80

0

TEPE

R=200m

TFTO

SB

s

= 51,75°R = 200mT = 97,55mD = 180,55mBs = 22,2m

90°

Kurp BaşlanğıcıKurp Bitişi

Kurp Merkezi

51,7

5°

90°

26° 26°

51,75°

R=200m


B

A

B

772

A

K5

750

752

754

756

758

760

762

764

766

768

770

772

774

776

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754

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K

0 20 40 60 80 100

TEPE

Yol Güzergahı

O


188

B

A

B

772

A

K5

75

0

75

2

75

4

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6

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76

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734

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716

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K

0 20 40 60 80 100

TEPE

Yol Güzergahı

O

ALİYMAN

ALİYMAN

KURP

Yol Ekseni

Platform Genişliği

Yol Platformu


B

A

0+000

A

T

0+037

R=200m

S

O

0

S0

+127

T

0+217

F

H3

0+300

H4

0+400

B

0+476

K5

27

54

H1

0+

100

H2

0+

200

75

0

75

2

75

4

75

6

75

8

76

0

76

2

76

4

76

6

76

8

77

0

77

2

77

4

77

6

74

8

74

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754

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78

0

78

2

78

4

78

67

88

79

0

79

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6

79

8

80

0

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0

744

744

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742

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734

732

730

728726

724

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716

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712

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716

718

R=200m

K

0 20 40 60 80 100

TEPE

En Kesit Çizgileri

= 51,75°

R = 200m

T = 97,55mD = 180,55m

Bs = 22,2m


190

K5

K

0 20 40 60 80 100

TEPE

En Kesit Çizgileri

= 51,75°

R = 200m

T = 97,55mD = 180,55m

Bs = 22,2m

B

A

0+000

1

0+020

A

T

0+037

R=200m

S

O

0

S0

+127

T

0+217

F

8

0+240

9

0+260

10

0+280

H3

0+300

11

0+320

12

0+340

13

0+360

14

0+380

H4

0+400

15

0+420

16

0+440

17

0+460

B

0+476

K5

27

54

2

0+060

3

0+

080

H1

0+

100

H2

0+

200

40

+120

50

+140

60

+160

7

0+

180

75

0

75

2

75

4

75

6

75

8

76

0

76

2

76

4

76

6

76

8

77

0

77

2

77

4

77

6

74

8

74

6

778776774772770768

766764

758756

754

752

750748

746

744

742

740

738

780782

784

786788790792794796

798

77

8

78

0

78

2

78

4

78

67

88

79

0

79

2

79

4

79

6

79

8

80

0

80

0744

744

744

742

740738736

734

732

730

728726

724

722

720

718

716

714

712

712714

716

718R

=200m

26° 26°

51,75°


KIRMIZI KOTLAR

SİYAH KOTLAR

EN KESİT NO

ARA MESAFELER

METRELER

HEKTOMETRELER

KİLOMETRELER

EĞİM VE EĞİM

DEĞİŞME NOKTALARI

YATAY KURBA

ELEMANLARI

0 1 2 3 4

760,00

758,00

756,00

754,00

752,00

749,00

759,00

757,00

755,00

753,00

751,00

765,00

763,00

761,00

764,00

762,00

769,00

767,00

768,00

766,00

772,00

770,00

771,00

1/1000

1/100

0

+0

00

0+

020

1

772

A

00

2000

771

77

0+

037

771

37

1700

T0

0+

060

2

770

40

2300

0+

080

3

769

65

2000

0+

100

H1

769

02

2000

0+

120

4

768

34

2000

S

768

00

0+

12727

0+

140

5

767

00

1273

727

0+

160

6

764

63

2000

0+

180

7

762

56

2000

0+

200

761

48

2000

H2

0+

21754

762

35

TF17

54

0+

240

8

763

33

2246

0+

260

9

762

80

2000

0+

280

10

761

73

2000

0+

300

759

65

2000

H3

0+

320

2000

11

757

56

H4

0+

340

2000

12

754

67

0+

360

2000

13

752

67

0+

380

2000

14

750

77

0+

420

2000

15

749

32

0+

440

2000

16

749

55

0+

460

2000

17

750

00

0+

400

749

78

2000

0+

476

B

750

00

1600

750,00

Km P = 0+170Kot P = 764,69 m

14

e = 0,15 m

772

00

771

15

770

42

769

57

768

59

760

13

759

10

758

07

757

04

756

01

754

98

753

95

751

88

750

85

749

82

752

92

749

00

L= 37 m L= 258,46 mR=200 m T=97,55 m D=180,55 m BS=22,2 m? =51,75°

767

76

766

88

766

56

765

99

765

07

764

13

763

17

762

31

761

21

g1= % 4 g2=% 5 L1=170 m L2=306 m30 10

30 26

12

6

10

6

09

6

09

5

06

2

00

0

14

6

14

4

10

1

04

4

15

7

16

9

00

4

21

2

26

7

26

3

15

8

05

2

13

4

23

1

31

8

31

4

25

6

13

0

01

8

10

0


G1 G2

PVI

PVT

PVC

L

L/2

δ

x

1 0 and dY

x b Gdx

PVC: 0 and x Y c

2

2 1 2 1

2Herhangi bir nokta : 2

2

G G G Gd Ya a

dx L L

2y ax bx c

y=((g2 - g1)/2L) x x² + g1 x x

En Kesit Km x yKırmızı

Kot

T1 0+09640

3,60

0 767,915

H1 0+100

23,60

-0,155 767,760

4 0+120

30,87

-1,030 766,885

S 0+12727

43,60

-1,353 766,562

5 0+140

0

-1,925 765,990

6 0+160

75,00

-2,843 765,072

C 0+17140

83,60

-3,375 764,540

7 0+180

103,60

-3,781 764,134

0+200

121,14

-4,741 763,174H2

0+21754

143,60

-5,600 762,315T

0+240

150,00

-6,701 761,2148

0+24640

63,60

-7,050 760,865T2

F

DÜŞEY KURBA GEÇİŞİ PARABOL DENKLEMİ


g

2 = -0.051

g1 = -0,043

T2

T1

P

L=150 mt=75 m

BOY KESİTTEN KIRMIZI KOTLAR

KO

TLAR

MESAFELER

A = 772,00 m

B = 749,00 m

P = 764,69 m

A = 0+000 Km

B = 0+476 Km

P = 0+171 Km

g1 = (kotA-kotP)/(kmA-kmP) = (772-764,69)/(0-171,40) = -0,043

g2 = (kotP-kotB)/(kmP-kmB) = (764,69-749)/(171,40-476) = -0,051

G = g1-g2 = -0,043-(-0,051) = 0,008

kot C = kotP - e = 764,69 - 0,15 = 764,54 m

kot T1= kotP + g1xt = 764,69+0,043x75 = 767,915 m

kot T2= kotP - g2xt = 764,69-0,051x75 = 760,865 m

C = 764,54 m

T1 = 767,915 m

T2 = 760,865 m

C = 0+171 Km

T1= 0+096 Km

T2= 0+246 Km

40

e = (L x G)/8 = (150x0,008)/8 = 0,15 m

40

40

40

C

t=75 m


R e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4

( m ) % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit

7000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0

6000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0

3000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 16 25

2500 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 15 23 TE 16 25

2000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 14 22 TE 15 23 2.2 18 27

1500 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 13 20 TE 14 22 2.3 16 26 2.6 21 32

1400 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 13 20 2.1 15 23 2.4 18 28 2.7 22 33

1300 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 13 20 2.2 16 24 2.5 19 29 2.8 23 34

1200 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 12 18 TE 13 20 2.3 17 25 2.6 20 30 2.9 24 36

1000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 12 18 2.2 14 22 2.5 18 27 2.8 21 32 3.2 25 39

900 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 11 17 TE 12 18 2.4 16 24 2.7 19 29 3.0 23 34 3.4 26 42

800 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 11 17 2.1 13 19 2.5 18 25 2.8 20 30 3.2 25 37 3.5 29 43

700 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 11 17 2.3 14 21 2.7 18 27 3.0 22 32 3.4 26 39 3.7 30 45

600 ÇE 0 0 ÇE 0 0 TE 10 15 2.1 12 17 2.5 15 23 2.9 19 28 3.2 23 35 3.6 28 41 3.9 32 48

500 ÇE 0 0 ÇE 0 0 TE 10 15 2.3 13 19 2.7 16 24 3.1 20 30 3.5 26 38 3.8 29 44 4.0 33 49

400 ÇE 0 0 ÇE 0 0 2.1 11 18 2.5 14 21 3.0 18 27 3.4 22 33 3.7 27 40 4.0 31 46

300 ÇE 0 0 TE 10 14 2.4 12 19 2.8 16 23 3.3 20 30 3.8 25 37 4.0 29 43

250 ÇE 0 0 TE 10 14 2.6 13 20 3.0 17 25 3.6 22 32 3.9 26 38

200 ÇE 0 0 2.3 11 17 2.8 14 22 3.3 18 27 3.8 23 34

175 ÇE 0 0 2.4 12 17 2.9 15 22 3.5 19 29 3.9 23 35

150 TE 9 14 2.5 12 18 3.1 16 24 3.7 20 31 4.0 24 36

140 TE 9 14 2.6 12 19 3.2 16 25 3.8 21 32

130 TE 9 14 2.6 12 19 3.3 17 26 3.8 21 32

120 TE 9 14 2.7 13 19 3.4 17 26 3.9 22 32

110 TE 9 14 2.8 13 20 3.6 18 27 4.0 22 33

100 2,1 9 14 2.9 14 21 3.6 19 28 4.0 22 33

90 2,2 10 16 3.0 14 22 3.7 19 29

80 2,4 11 16 3.2 15 23 3.8 20 29

70 2,5 11 17 3.3 16 24 3.9 20 30

60 2,6 12 18 3.5 17 25 4.0 21 31

50 2,8 13 19 3.7 18 27

40 3.0 14 20 3.9 19 28

30 3.3 15 22

20 3.8 17 28

Rmin= 16

Rmin= 35

Rmin= 80

Tasarım Hızına ve Yarıçapa Bağlı Olarak Uygulanacak Dever Oranları

Vt=100 km/sa

Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m )

Vt=60 km/sa Vt=70 km/sa Vt=80 km/sa Vt=90 km/saVt=20 km/sa Vt=30 km/sa Vt=40 km/sa Vt=50 km/sa

Rmin= 100

Rmin= 150

Rmin= 215

Rmin= 280

R = Kurp yarıçapı ( m )

Vt = Tasarım hızı ( km/sa )

Rmin= 375

Rmin= 490

emax = %4

Şehir Geçişlerinde uygulanacak dever oranı ( emax = %4)

Lr = %0 'dan tasarım deverine ulaşmak için gerekli mesafe ( m )

ÇE = Çatı eğimi ( % )

TE = Çatı eğiminin tek yönlü dever durumu ( % )

e = Dever oranı


R e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4 e 2 4

( m ) % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit % şerit şerit

7000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0

6000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0

3000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 16 25 TE 18 26 2.3 22 33 2.5 26 39

2500 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 15 23 TE 16 26 2,3 20 30 2.7 26 38 3.0 31 46

2000 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 14 22 2.1 18 24 2.5 20 31 2,8 25 37 3.3 31 47 3.7 38 67

1500 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 13 20 2.2 16 24 2.7 21 31 3.1 25 38 3,6 32 47 4.2 40 60 4.7 48 73

1400 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 13 20 2.4 17 25 2.8 21 32 3.3 27 41 3,8 33 50 4.4 42 63 5.0 51 77

1300 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 12 18 2.1 14 21 2.5 18 27 3.0 23 34 3.5 29 43 4.0 35 53 4.7 46 67 5.3 55 82

1200 ÇE 0 0 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 12 18 2.2 14 22 2.7 19 29 3.2 25 37 3.7 30 45 4.2 37 55 5.0 47 71 5.5 58 86

1000 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 11 17 2.1 13 19 2.6 17 26 3.1 22 33 3.6 28 41 4.2 34 52 4.8 42 63 5.5 53 80 6.0 62 93

900 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 11 17 2.3 14 21 2.8 18 27 3.4 24 37 3.8 30 45 4.5 37 55 5.1 45 67 5.8 55 82

800 ÇE 0 0 ÇE 0 0 ÇE 0 0 TE 11 17 2.5 15 23 3.1 20 30 3.6 26 39 4.2 32 48 4.9 40 60 5.4 47 71 6.0 57 85

700 ÇE 0 0 ÇE 0 0 TE 10 15 2.1 12 17 2.9 17 25 3.4 22 33 4.0 29 43 4.5 35 53 5.2 43 64 5.8 51 76

600 ÇE 0 0 ÇE 0 0 TE 10 15 2.4 13 20 3.1 19 28 3.8 26 37 4.3 31 46 5.0 38 57 5.6 46 69 6.0 53 79

500 ÇE 0 0 ÇE 0 0 2.1 11 16 2.6 16 23 3.5 21 32 4.2 27 41 4.6 35 52 5.4 41 62 5.9 46 72

400 ÇE 0 0 TE 10 14 2.5 13 19 3.3 18 27 4.0 24 36 4.7 31 46 5.3 38 57 5.9 45 68

300 ÇE 0 0 TE 10 14 3.1 16 24 3.9 22 32 4.6 28 41 5.4 35 53 5.9 42 64

250 ÇE 0 0 2.3 11 17 3.5 18 27 4.2 23 35 5.0 30 45 5.8 38 57 6.0 43 66

200 ÇE 0 0 2.8 13 20 3.9 20 30 4.7 26 39 5.5 33 50 6.0 39 58

175 TE 9 14 3.0 14 22 4.1 21 32 5.0 28 42 5.8 35 52

150 TE 9 14 3.3 16 24 4.4 23 34 5.3 29 44 6.0 38 54

140 TE 9 14 3.5 17 25 4.5 23 35 5.4 30 45 6.0 38 54

130 2.1 9 14 3.6 17 26 4.6 24 35 5.6 31 47

120 2.2 10 15 3.8 18 27 4.9 25 37 5.7 32 47

110 2.4 11 16 3.9 19 28 5.0 26 38 5.8 32 48

100 2.5 11 17 4.1 20 30 5.2 27 40 6.0 33 50

90 2.7 12 18 4.2 20 30 5.4 28 42 6.0 33 50

80 3.0 14 20 4.5 22 32 5.6 29 43

70 3.2 14 22 4.7 23 34 5.8 30 45

60 3.3 16 24 5.0 24 35 6.0 31 46

50 3.8 17 26 5.4 26 39

40 4.2 19 30 5.8 28 42

30 4.7 21 32 6.0 29 43

20 5.6 25 37

Vt=130 km/sa

Lr ( m )

TE = Çatı eğiminin tek yönlü dever durumu ( % )

e = Dever oranı

Vt = Tasarım hızı ( km/sa )

R = Kurp yarıçapı ( m )

emax = %6

ÇE = Çatı eğimi ( % )

Lr = %0 'dan tasarım deverine ulaşmak için gerekli mesafe ( m )

Rmin= 550

Vt=110 km/sa

Lr ( m )

Vt=120 km/sa

Lr ( m )

Rmin= 335

Rmin= 435

Rmin= 755

Rmin= 950

Rmin= 90

Rmin= 135

Rmin= 195

Rmin= 250

Vt=20 km/sa Vt=30 km/sa Vt=40 km/sa Vt=50 km/sa Vt=60 km/sa Vt=70 km/sa Vt=80 km/sa Vt=90 km/sa Vt=100 km/sa

Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m ) Lr ( m )

Rmin= 15

Rmin= 30

Rmin= 65

Tasarım Hızına ve Yarıçapa Bağlı Olarak Uygulanacak Dever Oranları


1

0+020A

T

0+037

S

0

S

0+127

T

0+217

F

8

0+240

9

0+260

10

0+280

27

54

2

0+060

30+080

H10+100

H2

0+200

4

0+120

5

0+1406

0+160

7

0+180

M1

ÜA1

ÜE1

0+066

ÜE2

0+188

ÜA

2

0+246

54

50

M2

0+26450

0+28250

K2

K1

0+008

0+000

0- 010

0- 028

R=200m

O

R=200m

K

Lg

Geçiş

Eğrisi

-2-2

-20

-2 +2

-2-2

-20

-2 +2

ENKESİT NO KM b B+b %d

Dış kenar

h=B+b *d 2

İç kenar

h=B+b *d 2

K1 -0+028 0,000+0,000 11,000 -2,00 -0,11 -0,11

M1 -0+010 0,00

A 0+000

1 0+020

To 0+037

ÜA1 0+008

2 0+060

ÜE1 0+066

3 0+080

H1 0+100

4 0+120

S27

0+127

5 0+140

6 0+160

7 0+180

H2 0+200

TF 0+217

ÜE2 0+18854

54

8 0+240

ÜA2 0+24654

9 0+260

M2 0+26454

K2 0+28254

10 0+280

0,000+0,000

0,000+0,000

0,000+0,000

0,085+0,085

0,205+0,205

0,368+0,368

0,410+0,410

0,410+0,410

0,410+0,410

0,410+0,410

0,410+0,410

0,410+0,410

0,410+0,410

0,410+0,410

0,410+0,410

0,000+0,000

0,000+0,000

0,000+0,000

0,000+0,000

0,000+0,000

0,329+0,329

0,205+0,205

0,046+0,046

11,000

11,000

11,000

11,170

11,410

11,772

11,820

11,658

11,410

11,092

11,000

11,000

11,000

11,000

11,000

11,820

11,820

11,820

11,820

11,820

11,820

11,820

11,820

2,00

3,24

5,00

7,38

8,00

8,00

8,00

8,00

8,00

8,00

8,00

8,00

8,00

5,00

6,81

2,28

2,00

1,11

-2,00

0,00

0,50

0,00 -0,11

-0,11

+0,11 -0,11

+0,18 -0,18

+0,28 -0,28

+0,43 -0,43

0,06

+0,47 -0,47

+0,47 -0,47

+0,47 -0,47

+0,47 -0,47

+0,47 -0,47

+0,47 -0,47

+0,47 -0,47

+0,47 -0,47

-0,11

-0,11 -0,11

+0,11 -0,11

-2,00 -0,11 -0,11

+0,40 -0,40

-0,11+0,03

0,00

+0,28 -0,28

+0,13 -0,13

+0,47 -0,47


1/500

1/50

ÖLÇEK

K M A0+

000

ÜA

0+

008

10+

020

ÜE

0+

066

TO

0+

037

2

0+

060

18 18 29 2958

1/500

1/50

ÖLÇEK

K A0+

000

M ÜA

0+

008

10+

020

TO

0+

037

20+

060

-0,1

1-0

,11

-0,1

1 0

,00

-0,1

1

+0,0

6

-0,1

1

+0,1

1

-0,1

8

+0,1

8

-0,2

8

+0,2

8

-0,4

3

+0,4

3

ÜE

0+

066

-0,4

7

+0,4

7

K A0+

000

M ÜA

0+

008

10+

020

TO

0+

037

20+

060

ÜE

0+

066

1/500

1/50

ÖLÇEK

-0,1

8+

0,1

8

+0,2

8-0

,28

-0,4

3+

0,4

3

-0,4

7+

0,4

7

-0,1

1+

0,1

1

-0,1

1+

0,0

6


Cross-sections


8

0+240

FY= 39,4221 m²FD= 0 m²

761,08-5,55

761,34

5,55760,58

-7,05

760,84

7,05

761,21

0,00

763,33

0,00

764,00-5,50

762,64

5,50761,8512,00

762,28

8,49

765,00-12,60

760,39-11,28

150+420

FY= 0 m²FD= 40,8431 m²

751,88

0,00751,77-5,50

751,77

5,50

749,320,00

750,24-5,50

748,36

5,50

747,00

13,40

747,1712,40

751,00

-10,20

750,54

-7,35


Volume Table YARMA DOLGU YARMA DOLGU YARMA DOLGU

Σ 6879,38 5832,67 419,71 6459,66 5412,95

1046,71

1046,71

6879,38

5832,67

17,06

21,61

4,09

1,11

5,72

2

0+000

0+020

0+037

0+060

HACİMLER (m³)

A

1

To

14,33

0+160

5

6

7

H2

3

H1

4

S

0+217,54

0+240

TF

8

9

0+080

0+100

0+120

0+127,27

0+140

0+260

20,00

17,00

23,00

20,00

20,00

20,00

7,27

0+180

0+20017,54

22,46

20,00

12,73

20,00

20,00

20,00

49,85

27,75

22,05

5,40

0,34

22,25

26,64

21,89

28,95

0,58

0,16

0,00

0,10

7,69

39,42

143,31

315,58

432,25

478,46

28,61

3,74

0,23

0,99

10,45

30,26

318,18

64,45

6,77

1,82

532,82

437,79

394,72

277,52

95,06

836,93

996,96

40,86

20,54

114,50

39,08

11,68

3,26

0,02

0,00

0,00

2,27

14,29

124,70

605,19

40,86

20,54

114,50

39,08

504,72

46,23

14,29

64,45

6,77

1,82

11,68

3,26

0,02

2,27

46,23

0,00

0,00

102,45

295,05

317,75

439,38

521,14

275,25

303,89

434,54

394,71

60,26

598,43

502,91

102,45

397,49

715,25

1154,63

1675,77

2110,31

2505,01

2808,33

3805,29

2780,27

3084,15

3023,90

2425,47

1922,56

1971,40

EN KESİT NO EN KESİT KMARA

MESAFEGEÇİT NOKTASI

ALANLAR (m²)

20,0010

0,00

0,00

1,82

12

13

14

H3

11

0+280

0+300

0+320

20,00

20,00

20,00

20,00

20,000+340

0+360

0+380

20,00

4) ΣDF + ΣKKK=ΣD =

0,00 53,92

0,04 18,53

0,00 35,1920,00

20,00

0+460

2) ΣYF + ΣDF =

3) ΣYF + ΣKKK=ΣY =

0,00 50,73H4 0+400

15 0+42020,00

30,14 0,00

1) ΣY + ΣD =

KONTROLLER:

B 0+47616,00

16 0+440

17

0,42 18,38

0,00 40,84

30,38 0,00

48,83

836,93

996,96

DOLGU

FAZLASI

(m³)

CEBRİK TOPLAM

0,00

72,25

15,72 0,12

KENDİ

KESİTİNDE

KULLANILA

YARMA

FAZLASI

(m³)

7,49 3,94

9,76 0,50

602,80

130,10

0,00

308,180,62

4408,08

607,63 0,00 0,00 607,63 5015,71

0,00 0,00 602,80

6,70 6,70 123,40

308,79 0,62

3048,89

703,71 0,00 703,71

4830,94

1014,64 0,00 1014,64

0,00 1078,35 0,00 1078,35

0,00 2034,24

4127,23

5139,11

7,62 337,25 7,62 329,63 887,76

816,86 0,00

34,22

1217,38816,86

921,97

125,74 1,00 1,00 124,74 1046,71

38,04 38,04

20,00

20,00

13,64

10,00

10,00

18,50

20,00

21,50

12,36

21,23

20,00

20,00

14,43

11,94

20,00

17,64

20,00

18,35

20,00

13,34

16,67

20,00

9,65

8,00

TATBİK

MESAFESİ (m)

20,00

20,00

21,06

20,00 5,049

3,87

16,13 3,87

16,13

28,51

4,221

17,54

2,26

15,281

2,26

15,28

9,254

20,00 18,493

13,33

6,67 13,33

6,67

17,964

3,545

20,00

3,30

16,701

3,30

16,70

5,049

18,493


Project

Example 2


GENEL DOĞRULAR TABLOSU

DOĞRU UZUNLUKLAR KESİŞME AÇISI

1. DOĞRU ( AS ) 312,03 m.

45 º

2. DOĞRU ( SB ) 182,11 m.

KESİN DOĞRULAR VE KURBALAR TABLOSU

ARA UZAKLIK TOPLAM UZAKLIK

1. DOĞRU A-T0 275 m 275 m

KURBA ToTf 8011 m 35511 m

2. DOĞRU Tf-B 13534 m 49045 m

t = R x tg ( Δ / 2 ) = 90 x tg ( 45 / 2 ) = 3728m

D= ( 2 π R Δ ) / 360 = ( 2 x π 90 x 45 ) / 360 = 7069 m

BS ( b) = R x { 1 / ( Cos Δ / 2 ) - 1 } = 90 x { 1 / ( Cos 45 / 2 ) – 1

} = 742 m


g1 = (629.6-636.20)x100/ (220.00-0) = -3.00% Km. P= 0 + 22000m.

g2 = (615.22-629.6)x100/(490.45-220.00) = -5.317% Kot P = 629.60m.

L = 150 m.

G = | g1 – g2 | = | -0.03 + 0.0532 | = 0.0232

e = ( L x G ) / 8 = ( 150 x 0.0241) / 8 = 0.435 m.

t = L / 2 = 150 / 2 = 75 m.

Kırmızı kot B = 629.60 – 0.435 = 629,165 m.

Km. T1 = 220.00 - 75 = 0+14500m.

Kot T1 = 629.6 + ( 75 x 0.03 ) = 631.85 m.

Km. T2 = 220.00 + 75 = 0+29500m

Kot T2 = 629.60 - ( 75 x 0.0532 ) = 625.62 m.

y = { (g2 - g1 ) / 2L }X2 + g1X = { ( -0.0532 + 0.03 ) / 300 } X2 - 0.03 X

y = -0.000077 X2 - 0.03 X


NOKTA NO KM X Y KOTLAR

T1 0+145 0 0 631.85

9 0+156 11 -0.34 631.51

10 0+176 31 -1.00 630.85

H2 0+200 55 -1.88 629.97

11 0+216 71 -2.52 629.33

12 0+241 96 -3.59 628.26

13 0+257 112 -4.33 627.52

To 0+275 130 -5.20 626.65

14 0+289.14 144.14 -5.92 625.93

T2 0+295 150 -6.23 625.62


Far ışığı = S = 100 m alınmıştır.

•S küçük L için(S<L) :

L= (G * S) / (1.22* 0.035 ) = (0.0232 * 100) / 0.0427 = 54.3325 m

•S büyül L için(S>L) :

L= 2 * S – (1.22 * 0.035 * S) / G = 2 * 100 – (1.22 * 0.035 * 100) /

0.0232 = 15.9483 m

Konfor Kriteri :

Lmin = G * V² / 3.95 = 0.0232 * 40.31² / 3.95 = 9.5437 m

Esneklik Kriteri :

Lmin = 3048 * G = 3048 * 0.0232 = 70,7136 m

Drenaj Kriteri :

4360 * G = 4360 * 0.0232 = 101.152 m

Şartname :

Lmin = 120 m ise L=150 m alınmıştır.

Rdüşey = 6 * Ryatay = 6 * 90 = 540 m

L = Rdüşey * G Rdüşey =150 / 0.0232 = 6465.52 m L

UYGUNDUR

T = L / 2 = 150 / 2 = 75 m e= L * G / 8 = 150 * 0.0232 / 8 = 0 .435

m


Dever Hesabi :

d = 0.00443 x 60²/90 = 0.17 d = % 8 alınmıştır.

Hız Sınırlaması

0.08 = 0.00443 x v² / 90 => v sınır = 40.31 km / saat

Kurbada Yol Genişletmesi

b = (n l²/ 2R ) + ( 0.05 V / √R ) = ( 2 x 12²/ 2x 90) + ( 0.05 x 40 /

√90 )

= 1.81 m.

Dever Rampa Boyunun Hesabı

h 1 = B x d 0 / 2 h 1 = 12 x 0.02 / 2 = 0.12 m.

h 2 = (B + b ) du /2 h 2 = ( 12 + 1.81) x 0.08 /2 = 0.55 m.

L = V³/ ( 46.7 x R x þ ) = ( 40 )³/ ( 46.7 x 90 x 0.4 ) = 38.07 m.

K = ( 2 x L x h 1 ) / (h 2 - h 1 ) = ( 2 x 38.07 x 0.12)/ ( 0.55 – 0.12)

= 21.25 m.

Savrulma Hızı:

Vsav=11.3√(R*(µe+tga)/(1- µe*tga))=11.3*√(90(0.2+0.08)/(1-

0.2*0.08)) = 57.18 km/saat

Devrilme Hızı :

Vdev=11.3√(R*(h*tga+e/2)/(h-

tga*e/2))=11.3√(90(1.45*0.08+1.95/2)/(1.45-0.08*1.95/2))


ENKESİT NO KM X Y=(X³/ 6 R L)

ÜA 0+256 0.00 0

13 0+257 1.00 0.000048

T0 0+27500 19.00 0.33

14 0+28914 33.14 1.77

ÜE 0+294 38.00 2.67

e = (L²/6R)=(38.07²/6x90) = 2.67m

ΔR = e/4= 4.01/4 = 1.0025 m

m = ΔR/2 = 1.0025/2 = 0.501 m


ENKESİT

NO

KM

b

B+b

d %

DIŞKENAR

h=(B+b)/2*d

İÇKENAR

h=(B+b)/2*d

K1 0+23475 0.00 12.00 - 2 -0.12 -0.12

12 0+24100 0.00 12.00 -0.83 -0.05 -0.05

M1 0+24538 0.00 12.00 0 0 -0.12

ÜA1 0+25600 0.00 12.00 2 0.12 -0.12

13 0+25700 0.03+0.03 12.06 2.16 0.13 -0.13

To 0+27500 0.45+0.45 12.90 5 0.32 -0.32

14 0+28914 0.8+0.8 13.60 7.23 0.49 -0.49

ÜE1 0+29400 0.905+0.905 13.81 8 0.55 -0.55

H3 0+30956 0.905+0.905 13.81 8 0.55 -0.55

S 0+31584 0.905+0.905 13.81 8 0.55 -0.55

15 0+33155 0.905+0.905 13.81 8 0.55 -0.55

ÜE2 0+33611 0.905+0.905 13.81 8 0.55 -0.55

Tf 0+35511 0.452+0.452 12.91 5 0.32 -0.32

ÜA2 0+37411 0.00 12.00 2 0.12 -0.12

16 0+37511 0.00 12.00 1.89 0.11 -0.11

M2 0+39311 0.00 12.00 0 0.11 -0.11

H4 0+400 0.00 12.00 -0.74 -0.04 -0.04

K2 0+41211 0.00 12.00 -2 -0.12 -0.12


Highway

Components

Cross-section


HACİM (m³) KENDİ

KESİTİ

KULLA.

YARMA

FAZLA (m³)

DOLGU

FAZLASI (m³)

CEBRİK TOPLAM

YARMA DOLGU YARMA DOLGU

177.020 177.020 177.020

125.917 37.554 37.554 88.363 265.383

60.774 50.684 50.684 10.091 275.474

81.647 171.357 81.647 89.710 185.764

40.702 35.412 35.412 5.290 191.053

198.552 40.968 40.968 157.584 348.637

257.025 110.221 110.221 146.805 495.442

315.690 26.800 26.800 288.890 784.332

223.516 23.294 23.294 200.222 984.554

159.517 72.558 72.558 86.959 1071.513

73.371 105.315 73.371 31.944 1039.569

91.190 170.412 91.190 79.222 960.347

86.040 215.660 86.040 129.620 830.727

56.949 300.223 56.949 243.274 587.454

74.784 208.075 74.784 133.291 454.163

691.679 691.679 237.516

24.909 443.420 24.909 418.511 656.027

7.925 541.738 7.925 533.813 1189.840

12.856 644.445 12.856 631.589 1821.428

61.806 631.126 61.806 569.320 2390.748

41.744 702.030 41.744 660.286 3051.034

77.556 311.389 77.556 233.833 3284.867

134.951 80.533 80.533 54.418 3230.449

412.262 13.059 13.059 399.203 2831.246

354.131 7.821 7.821 346.310 2484.936

288.704 4.784 4.784 283.920 2201.016

293.040 17.960 17.960 275.080 1925.936

125.898 96.514 96.514 29.384 1896.552

11.317 55.194 11.317 43.877 1940.429

3869.793 5810.222 1320.255 2549.538 4489.968

3) ∑YF+∑KKK = ∑Y => 2549,538 +1320,255 = 3869,793 m3

4) ∑DF+∑KKK = ∑D => 4489,968 +1320,255 = 5810,222 m3


Tablo İş makinalarıyla Toprak Dağıtımı


Tarih Ad Soyad İmza Karadeniz Teknik Üniversitesi

Mühendislik Fakültesi

İnşaat Mühendisliği

Çizen 09 / 01 / 2009

1. Kontrol

2. Kontrol

ÖLÇEK

1 /100

1/ 1000

BOYKESİT Öğrenci No:

1796


Primary References

Mannering, F.L.; Kilareski, W.P. and Washburn, S.S.

(2005). Principles of Highway Engineering and

Traffic Analysis, Third Edition. Chapter 3

American Association of State Highway and

Transportation Officials (AASHTO). (2001). A

Policy on Geometric Design of Highways and

Streets, Fourth Edition. Washington, D.C.


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