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Road and Bridge design

1. INTRODUCTION

For rapid economic, industrial and cultural growth of any country, agood system of transportation is very essential. One of thetransportation systems that is economical for developing countries like

Ethiopia, is road. A well designed road network plays an importantroll in transporting people and other industrial products to anydirection with in short time. Roads, to satisfy their intended purpose,must be constructed to be safe, easy, and economical environmentallyfriendly and must full fill the needs of inhabitants. Being safe, thenumber of accidents that can occur will be minimized. Easinessdecreases operation cost, pollution and even time cost. Economicalroads assure their feasibility according to their plans and initiatefurther construction of roads; environmentally friendly roads avoidexcessive and unnecessary deforestation, agricultural land usage andpollution from high gradients. Schemes that do not satisfy the needs of

localities may not get the maximum utilization of the surplus manpower that is really to exit in the rural community and also itseconomical value may also decrease.

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2.GEOMETIC DESIGN

High way geometric design involves the design of geometric elementsof a high way fixation of standards with respect to various components.

Its dictated with in economic limitations to satisfy the requirement oftraffic in designing elements such as, cross section, horizontalalignment, vertical alignment, sight distance, lateral and verticalalignment, intersections, etcThe safety, efficiency and economic operation of a high way isgoverned to a large extent by the care with which the geometricdesign is worked out.The engineer has to consider the following points when selecting thedesign standards.

o Volume and composition of traffic in the design year should

be the basis of design.

o Faulty geometrics are costly to rectify at a later date.o The design should be consistent and the standards used for

the different elements should be compatible with oneanother.

o The design should embrace all aspects of design including

signs, markings, lightings, etco The road should be considered as an element of the total

environment and ats location and design should enhancerather than degrade the environment.

o The design should minimize the total transportation cost.

o Safety should be built in the design.

o The design should be enable all road users to use the facility

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3.HORIZONTAL ALIGNMENT

3.1.Details of the first horizontal curve

Available informationsDesign standard -Ds4Terrain classification flatFrom table 2-6 of ERA design manualSpeed = 85Km/hTransition curve is requiredMaximum supper elevation 8%

Rmin = VD2/(127(e+f)) =852/127(0.08+0.14) = 270m

3.1.1 Calculations of the horizontal curve setting out

information

1) Length of transtionLt= V3/(cR) = 853/(3.63*0.3*270) = 146.25m

C is radial acceleration and taken as 0.3m/s3

2) shift distance (s)S = Lt2/24R =3.3m

3) Tangent length

T= Lt/2 +(R+S) tan/2 =180.78m =43deg.

4) from the plan chianage of PI = 11+186.425) Station of TS = station of PI-T

= 11+186.42-180.78=11+005.64

o

CS

SC

=430

ch.TS =11+005.64 ch. PI = 11+186.42

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6) station of SC =ch.TS +Lt= 11+151.89m

offset calculation table for the first transtion curve of first

curve

deflection angle = 430

transtion curve length =146.25m

ch.TS =11+005.64m

ch.SC=11+151.89m

offset distance from the tangent to the curve is calculated as y=X3/(6RLt)where:

Y is offset distanceX is distance measured along the tangentLt is length of transition

R is the radius of the curve

Station X(m) Y(m)

TS 11+005.64 0.00 0.00

11+020 14.36 0.012

11+040 34.36 0.171

11+060 54.36 0.678

11+080 74.36 1.735

11+100 94.36 3.546

11+120 114.36 6.313

11+140 134.36 10.238

11+151.89 146.25 13.203

Offset calculation for the circular curve of the first curve

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Offset distancefrom long chord

-2

the angle subtended by the circular curve is -2but

=Lt/2R=1503131

-2= 1105658 length of the circular curve l=R(-2)/180

=56.3m

station of CS=Station of Sc+ l(curve length) = 11+151.89+56.3=11+208.19

station of ST = station of CS +Lt =11+208.19+146.25= 11+354.44

Half of the circular curve length = 56.3/2 =28.15

The chianage of the mid point of the curve= ch.SC+L/2 = 11+382.89

o Let l distance measured along the curve from N

o - angle subtended by lo X distance measured along long chord from E

o Y offset distance from long chord to cuve points

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From these formulae l = R/180 and X= 180l/R = Rsin

EN- offset distance = ON-OE = since ON =R=R-OE

R(1-cos(-2/2)But Y=(R2-X2) - OE

=(R2-X2) - R cos(-2)/2According to these formulas we have the following tables showingoffset distance from long chord.

Offset to the left of N

Chord length from SC to CS = 2Rsin (-2)/2)= 2*270 *sin(-2)/2)= 56m

L/2=28m

Curve lengthFrom N

I subtended byli(in deg.)

=180li/R

Xi= Rsin Offset from the longchord to curve points

Yi=(R2-X2)-Rcos(-2)/2)

2.59 003259.6 2.59 1.45

7.59 103641 7.59 1.36

12.59 204023 12.59 1.17

17.59 304405 17.58 0.89

22.59 40

4746 22.57 0.5228.0 505641 27.96 0.00

Offset to the right of N

Curve lengthFrom N


=180li/R


Yi=(R2-X2)-Rcos(-2)/2)

2.41 003042 2.41 1.46

7.41 10

3424 7.41 1.3612.41 203805 12.41 1.18

17.41 304147 17.40 0.90

22.41 404529 22.39 0.54

28.0 505641 27.96 0.00

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Offset calculation for the exit transtion curve of firstcurve

Station X(m) Y(m)

TS 11+354.44 0.00 0.00

11+340 14.44 0.013

11+320 34.44 0.172

11+300 54.44 0.681

11+280 74.44 1.741

11+260 94.44 3.555

11+240 114.44 6.326

11+220 134.44 10.256

11+208.19 146.25 13.203

3.1.2. Designing of widening on the curve

Design standard Ds4

terrian classifcation flat design speed 85Km/h R=270 km

Since R is > 195m no widening is required.

3.1.3.Design of supperelevation

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From table 8-1 of ERA manual e= 0.08B=6.7m (width of the road)Lt=146.25mE=e*B = 0.536

Raising of pavement due to super elevation

0.08 m/m (6.7m) =0.536m

Assuming rotation of super elevation about the center, raising of outer edge involved

= 0.536/2

=0.268m EUsing 2.5% camber slope

0.025X(3.35) =0.08m 2.5%

Total raising at the center 3.35m

0.08m + 0.268m = 0.348m

supper elevation ratio =E/Lt =0.536/146.25=1/272

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Available informationsDesign standard -Ds4Terrian classification flatFrom table 2-6 of ERA design manualSpeed = 85Km/h

Transtion curve is requuredMaximum supper elevation 8%

Rmin = VD2/(127(e+f)) =852/127(0.08+0.14) = 270m

3.2.1. Calculations of the horizontal cutve setting outinformations

3) Length of transtionLt= V3/(cR) = 853/(3.63*0.3*270) = 146.25m

C is radial acceleration and taken as 0.3m/s3

4) shift distance (s)S = Lt2/24R =3.3m

3) tangent length

T= Lt/2 +(R+S)tan/2 =186.33m =45deg.

7) from the plan chianage of PI = 11+7048) Station of TS = station of PI-T

= 11+704-186.33=11+517.67

o

CS

SC

=450

ch.TS =11+517.67 ch. PI = 11+704

9) station of SC =ch.TS +Lt= 11+663.92

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offset calculation table for the first transtion curve of firstcurve

deflection angle = 450

transtion curve length =146.25m

ch.TS =11+517.67m ch.SC=11+663.92m

offset distance from the tangent to the curve is calculated as y=X3/(6RLt)where:

Y is offset distanceX is distance measured along the tangentLt is length of transtionR is the radius of the curve

Station X(m) Y(m)

TS 11+517.67 0.00 0.00

11+520 2.33 0.001

11+540 22.33 0.047

11+560 42.33 0.32

11+580 62.33 1.022

11+600 82.33 2.355

11+620 102.33 4.523

11+640 122.33 7.727

11+660 142.33 12.690

11+663.92 146.25 13.203

Offset calculation for the circular curve of the first curve

Offset distancefrom long chord

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-2

the angle subtended by the circular curve is -2but

=Lt/2R=1503131-2= 1305657 length of the circular curve l=R(-2)/180

=65.7m

station of CS=Station of Sc+ l(curve length) = 11+663.92+65.7=11+729.62

station of ST = station of CS +Lt =11+729.92+146.25= 11+875.87

Half of the circular curve length = 65.7/2 =32.85

The chianage of the mid point of the curve= ch.SC+L/2 = 11+696.77

o Let l distance measured along the curve from N

o - angle subtended by lo X distance measured along long chord from E

o Y offset distance from long chord to cuve points

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From these formulae l = R/180 and X= 180l/R = Rsin

EN- offset distance = ON-OE = since ON =R=R-OE

=R(1-cos(-2/2)

But Y=(R2-X2) - OE=(R2-X2) - R cos(-2)/2

According to these formulas we have the following tables showingoffset distance from long chord.

Offset to the left of N

Chord length from SC to CS = 2Rsin (-2)/2)= 2*270 *sin(-2)/2)= 65.6m

L/2=32.8m

Curve lengthFrom N


=180li/R


Yi=(R2-X2)-Rcos(-2)/2)

1.77 002233 1.77 2.00

6.77 102615 6.77 1.91

11.77 202956 11.77 1.74

16.77 303338 16.77 1.48

21.77 403719 21.76 1.12

26.77 504101 26.74 0.67

31.77 604443 31.71 0.13

32.8 605750 32.74 0.00

Offset to the right of N

Curve lengthFrom N


=180li/R


Yi=(R2-X2)-Rcos(-2)/2)

3.23 004109 0.00 1.98

8.23 104450 3.23 1.87

12.23 203641 8.23 1.72

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17.23 303929 12.23 1.45

22.23 404311 22.22 1.08

27.23 504653 27.20 0.624

32.80 605750 32.74 0.00

Offset calculation for the exit transtion curve of secondcurve

Station X(m) Y(m)

TS 11+875.87 0.00 0.00

11+860 15.87 0.017

11+840 35.87 0.195

11+820 55.87 0.736

11+800 75.87 1.843

11+780 95.87 3.719

11+760 115.87 6.567

11+740 135.87 10.587

11+729.62 146.25 13.2

3.2.2. Designing of widening on the curve

Design standard Ds4

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terrian classifcation flat design speed 85Km/h R=270 km

Since R is > 195m no widening is required.

3.2.3.Design of supperelevation

From table 8-1 of ERA manual e= 0.08B=6.7m (width of the road)Lt=146.25mE=e*B = 0.536

Raising of pavement due to super elevation

0.08 m/m (6.7m) =0.536mAssuming rotation of super elevation about the center, raising of outer edge involved

= 0.536/2=0.268m E

Using 2.5% camber slope

0.025X(3.35) =0.08m 2.5%

Total raising at the center 3.35m

0.08m + 0.268m = 0.348m

supper elevation ratio =E/Lt =0.536/146.25=1/272

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In road alignment, when two different or contrary gradient meet theyform curve in the vertical plane called vertical curve.

Vertical curve are classified as

Summit( crest) curveSag curves

The first vertical curve curve is a sag curve with -3.88% and -1.96%gradientThe second curve is a crest curve with -1.96% and -2.63% gradient.And the third curve is a sag curve with -2.63% and -1.24%.

4.1.GRADIENT CALCULATIONS

TO CALCULATE THE FIRST GRADIENT

elevation of the first point =2454.186elevation of the second point =2441horizontal distance b/n the two points = 340m

h = -13.186slope (gradient) = elevation difference/H.Dslope (gradient) g1 = -3.88%

TO CALCULATE THE SECOND GRADIENT

elevation of the first point =2441elevation of the second point =2430horizontal distance b/n the two points = 560m

h = -11slope (gradient) = elevation difference/H.Dslope (gradient) g2 = -1.96%

TO CALCULATE THE THIRD GRADIENT

elevation of the first point =2430elevation of the second point =2420

horizontal distance b/n the two points = 380mh = -10

slope (gradient) = elevation difference/H.Dslope (gradient) g3 = -2.63%

TO CALCULATE THE FOURTH GRADIENT

elevation of the first point =2420

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elevation of the second point =2417.276horizontal distance b/n the two points =220m

h = -10slope (gradient) = elevation difference/H.Dslope (gradient) g4 = -1.24%

4.2. CALCULATION OF VERTICAL CURVE LENGTH

parameters

velocity (85Km/h)

gradiento For the first sag curve the change in gradient is -2.63 - (-1.96)

=0.67reading from ERA table which relates change in gradient withspeed the curve length of 50m is found.

o For the second crest curve the change in gradient is -1.96 - (-3.88)

=1.92 reading from ERA table which relates change in gradient with

speed the curve length of 100m is found.o For the third sag curve the change in gradient is -1.24 - (-2.63)

=1.39 reading from ERA table which relates change in gradient withspeed the curve length of 60m is found.

4.3.Calculation of vertical curve setting outinformations

The First sag curve

L=100mg1=-3.88%r = (g2-g1)/100Lr = 0.000192g2=-1.96%station of BVC = station of PVI - L/2

= 11+340 - 50 =11+290Elevation of BVC = EPVI - (g1*L/2)/100

= 2441-(-3.88*50)/100= 2442.94

STATION

LENGTH(X) m

g1*X

r/2(X2)

EBVC

Y (Elevationon the curve)

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11+290 0 0 0.00000000 2442.94 2442.940

11+300 10 -0.388 0.00960000 2442.94 2442.562

11+320 30 -1.164 0.08600000 2442.94 2441.862

11+340 50 -1.94 0.24000000 2442.94 2441.240

11+360 70 -2.716 0.47000000 2442.94 2440.694

11+380 90 -3.492 0.77700000 2442.94 2440.225

11+390 100 -3.88 0.96000000 2442.94 2440.020

The second crest curve

L=50mg1=-1.96%r = (g2-g1)/100Lr = - 0.000134g2=-2.63%

station of BVC = station of PVI - L/2= 11+900 - 25 =11+875Elevation of BVC = EPVI - (g1*L/2)/100

= 2430-(-1.96*25)/100= 2430.49

STATION

LENGTH(X) m

g1*X

r/2(X2)

EBVC


11+875 0.000 0.000 0.000 2430.49 2430.490

11+880 5 -0.098 -0.001675 2430.49 2430.390

11+900 25 -0.49 -0.041875 2430.49 2429.95811+920 45 -0.882 -0.135675 2430.49 2429.472

11+925 50 -0.98 -0.1675 2430.49 2429.343

Third sag curve

L=60mg1=-2.63%r = (g2-g1)/100Lr = 0.000231667

g2=-1.24%station of BVC = station of PVI - L/2

= 12+280 - 30 =11+250Elevation of BVC = EPVI - (g1*L/2)/100

= 2420-(-2.63*30)= 2420.789

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STATION

LENGTH(X) m

g1*X

r/2(X2)

EBVC


12+250 0.000 0.000 0.000 2420.789 2420.789

12+260 10 -0.263 0.012 2420.789 2420.538

12+280 30 -0.789 0.104 2420.789 2420.104

12+300 50 -1.315 0.290 2420.789 2419.764

12+310 60 -1.578 0.417 2420.789 2419.628

5.Calculation of crossectional area and volume

The calculations are calculated below in a tabular fashion.

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