3_Vertical Alignment.pdf

8/14/2019 3_Vertical Alignment.pdf

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KAAF UNIVERSITY COLLEGE

Civil Engineering DepartmentCollege of Engineering

__________________________________

Highway Engineering 1CIV 368

Lecture 3_ Vertical Alignment

Kwasi Agyeman Boakye ( [email protected])

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Vertical Gradients

In the same way that horizontal curves are used to connect intersecting straights in thehorizontal plane, vertical curves are used to connect intersecting straights in the vertical plane.These straights are usually referred to as gradients and the combination of the gradients andvertical curve is known as vertical alignment.

Maximum Gradients

One of the most important considerations in designing a highway is the gradient. The cost ofoperation of vehicles and the capacity of a highway are profoundly affected by the gradesprovided. On mountainous and steep terrain, the grades are not only influenced by the ability ofvehicles to negotiate them, but also by the altitude of a road above the sea level sincerarefaction of air causes loss of engine power. Based on this maximum gradient of roads arerecommended based on road class, design, terrain condition or vehicle performance.

By most design standards, including Ghana, there are recommended standard gradients foreach given design speed. However if the recommended standard gradient is not suitable onemay go a few steps above it but should not exceed a certain critical length of road. Thus,application of gradient also depends on amount of stretch over which it is acceptable. In anexample in the GHA design guide, for a design speed of 100km/hr, standard gradient is 3%.Where one wants to go a step above, thus 4%, it should not be applied over a length more than700m.

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Vertical CurvesTypes

A crest curve which can also be referred toas a summit or hogging curve , is one for whichthe algebraic difference of the gradients ispositive.

A sag curve which can also be referred to asa valley or sagging curve , is one for which thealgebraic difference of the gradients is negative

The algebraic difference A is obtained as; A= (entry grad %) (exit grad%)

This gives 6 possible combinations of gradients

4

+n%

+m%-n%

A=(+m)-(+n)positive

+m%

A=(+m)-(-n)positive

-m%

-n% A=(-m)-(-n)negative

+n%-m%

A=(-m)-(+n)negative

+m%

+n%

A=(+m)-(+n)negative

-m%

-n% A=(-m)-(-n)positive


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Purpose of Vertical Curves

Passenger comfort and safety As a motorist traverses a vertical curve a radial force acts on the vehicle and tries to force itaway from the centre of curvature and this may give the driver a feeling of discomfort; inextreme cases, as at hump-backed bridges, vehicles travelling at high speed may leave thecarriageway. Also in sag design the underside of the vehicle could come into contact with theroad surface, particularly where the gradients are steep and opposed. Both scenarios result indiscomfort and danger to passengers travelling in the vehicle and hence their possibility must beminimised. This is first achieved by restricting the gradients and secondly by choosing a suitabletype and length of curve such that this reduced force is introduced as gradually and uniformly aspossible.

Adequate VisibilityIn order that vehicles travelling at the design speed can stop or overtake safely, it is essentialthat oncoming vehicles or any obstructions in the road can be seen clearly and in good time.

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Length of Vertical Curves

The minimum length of vertical curve for any given road is obtained from the equation

L = KA (metres) 1

A is the algebraic difference of the gradients, the absolute value being used.

K is a constant derived due to the curvature of the vertical curve for any particular road type anddesign speed. Since the cubic parabola is used and not circular curves in the design of verticalcurves, radius is not used but rather the K value. The K value ensures that the length of verticalcurve obtained from the equation above contains adequate visibility and provides sufficientcomfort. The unit of K is in metres.

It must be noted that the length obtained is the minimum required and it is perfectly acceptableto increase the value obtained. This may be necessary when trying to phase the verticalalignment with the horizontal.

Also it has been established that changes in gradients must be connected by vertical curves.

The vertical curves are deemed unnecessary where the changes in gradients are below 0.5%. 6


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Length of Vertical CurvesCrest Curves

The length of vertical curves for crest curves are based on stopping sight distance (SSD) ,overtaking sight distance (OSD) and the drivers comfort criteria . Whilst OSDconsiderations give very long vertical curves, the driver comfort criteria gives short verticallengths, especially for high speeds. Thus, the length of vertical crest curve is given by ;

Where Lcrest = The minimum length of vertical crest curveS = Sight distance required (m)

A = Algebraic difference in gradientsH = Drivers eye height ( Ghana, 1.2m)

h = Object height (Ghana, 0.1m)Putting equation 2 into equation 1 ;

Kcrest = S 2 .3 398

Where Kcrest= The K value for a crest curve

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Length of Vertical CurvesCrest Curves

The Length of vertical crest curve due to driver comfort criteria is given by;Lcrest/com = AV 2 = AV 2 .4

1300C 360Where L crest/com = The minimum length of vertical curve for comfort (m)

A = Algebraic difference in gradients

V = Design SpeedC = Vertical acceleration (0.028g m/sec 2, where g is 9.8 m/sec 2)

Putting equation 4 into equation 1;Kcrest/com = V 2 .5

360Where K

crest/com = The K value for comfort on a crest curve

V = Design speed of vehicle

ExampleFor a given road of design speed 80Km/hr having a crest curve with sight distance 110m,determine the suitable K value and hence the minimum length of curve for an A value of 2.5.

Ans: Kcrest = 17.8m, Kcrest/com = 30.2m, Suitable K = 30m, Length of Curve = 30x2.5=75m 8


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Length of Vertical CurvesSag Curves

It is assumed that adequate sight distance is available on sag curves during the day. Sag curveconsiderations are therefore based on headlight considerations and driver comfort. Thus theminimum sag length is given by;

Lsag = AS 2 = AS 2 ..6 200(h + Stan ) 120 + 3.49S

Where L sag = The minimum length of vertical sag curve

S = Sight distance required (m) A = Algebraic difference in gradientsh = Headlight height (0.6m)= Angle of upward divergence of light beam (1 )

Putting equation 6 into equation 1 ;Ksag = S 2 .7

120+3.49SWhere K sag = The K value for a sag curve

Note ! . The length required by the driver comfort condition are as for the crest curve. It isrecommended that sag curves are designed using the driver comfort criteria since the headlightcondition gives sight distances far in excess of the effective ranges of the headlamp beams.

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Length of Vertical CurvesSag Curves

Example An arterial road to be reconstructed has a design speed of 100km/hr. Determine the minimumsag length for a curve in a valley portion of the alignment where the sight distance is 160m andthe absolute change in vertical gradient is 3. Assume Headlight height of (0.6m).

Ans: Ksag = 37.7m, Ksag/com=Kcrest/com= 27.8m, Recommended K=28m, minimum length =28x3=84m.

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Vertical Curve ElevationsFormulae assumptions

In practice, a parabolic curve is used in thedesign of vertical curves. This is because it isable to achieve a uniform rate of change ofgradient and therefore a uniform introductionof vertical radial force. Parabolic equationis given by ;

X = cy 2

dx =2cydyd2x =2c=constant=rate of change ofdy 2 gradient

Main Assumption about parabola -Vertical curves are designed such that the two tangent lengths are equal, PQ=QR. Unequaltangent may occur but not often used.- Chord PWR=arc PSR=PQ+PR-Length along the tangents=horizontal length, PQ = PQ -QU=QW, thus there is no difference in dimension in either vertical plane or perpendicular planeto the entry tangent length

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P

Q

S=mid-point of curveW=mid-point of chord

Q'

+m % S

W

-n%

R

U


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Vertical Curve ElevationsEquations of the vertical curve

Level of Q above P= (m) (Lv) = mLv

100 2 200

Level of R above P

= (mLv) - (nLv) = (m-n)Lv200 200 200

PW=WR thereforeLevel of W above P= (m-n)Lv

400

From properties of parabola QS=SW=QW/2 Since QW=Level of Q above P Level of W above PQS = 1 [ (mLv) - (m-n)Lv) = (m+n)Lv = ALv

2 200 400 800 800 12

mLv

200

xx

P

Q


Q'

nLv200

+m % S

W

-n%

Y

Lv2

Datum

in either direction

R

U


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Vertical Curve ElevationsEquations of the vertical curve

Since parabolic equation, x=cy 2

At point QS=c(PQ) 2 ALv = c ( Lv ) 2

800 2c = ( ALv ) ( 4 )

800 Lv 2c = A

200LvHence substituting it into the parabolic

Equation;

x = Ay 2

200Lv

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mLv

200

xx

P

Q


Q'

nLv200

+m % S

W

-n%

Y

Lv2

Datum

in either direction

R

U


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Vertical CurvesExample

SolutionV= 100Km/hr

A= (2)-(-1.5) = 3.5 (positive, crest curve)K=?

To determine K;Kcrest = S 2 = 160 2 = 64.3m

398 398Kcrest/com = V 2 = 100 2 = 27.8m

360 360Recommended K= 64

i)Minimum Curve Length, L= 64x3.5=224m

ii) Position of height of crest pointLevel of P = 100 mLv = 100 (+2x224)

200 200= 97.76m

Question 1 A rural 2 lane dual carriageway has a designspeed of 100km/h. A vertical curve with sightdistance 160m is to be designed to connect2 tangents of gradients +2% and -1.5%

which intersect at a level of 100.00m.Determine

i) The minimum curve lengthii)The position and height of the crest point

for the minimum curve length.

P

R

I = 100

-1.5%

Lv/2

Crest Point, Hmax

+2%


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Distance of crest point from P; (same asturning point distance)y= mLv = 2x224 = 128m

A 3.5Hmax = m 2L = 2 2x 224 = 1.28m

200A 200x3.5

Height of crest point = P + Hmax= 97.76 + 1.28= 99.04m


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SolutionV= 120Km/hr

A= (-3)-(2.5) = -5.5 (negative , sag curve)K=?

To determine K;Ksag = S 2 = 210 2 = 51.71m

120+3.49S 120+(3.49x210)Ksag/com = 120 2 = 120 2 = 40m

360 360

Recommended K= 40m

Minimum Curve Length, L= 40x5.5=220m

Level at beginning of vertical curveLevel of P = 96.20 mLv = 96.20 (-3x220)

200 200= 99.5m

Question 2 A rural motorway, which has a design speedof 120km/h, is to be designed to have aminimum curve linking two gradients -3%and +2.5%. Determine the height of the

curve at 30m intervals if the verticaltangents intersect at a level of 96.20m andat a chainage of 13+85.00 (1385m). Thesight distance for the curve is 210m.


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Chainage at beginning of vertical curve;Ch P = I Lv/2 = 1385 220/2=1275m

Chainage at end of vertical curve;Ch R = Chp + Lv = 1275 +220 =1495m

To determine the height of curve at 30m intervals the ff equation is used;

H = [ my - Ay 2 ] 100 200Lv

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Chainage y(m)my/100 (m)

(1)Ay2/200L (m)

(2)H (m)(1)-(2) Reduced Level

1275 0 0 0 0 99.5

1280 5 -0.15 0.00 -0.15 99.35

1310 35 -1.05 -0.15 -0.90 98.60

1340 65 -1.95 -0.53 -1.42 98.08

1370 95 -2.85 -1.13 -1.72 97.78

1400 125 -3.75 -1.95 -1.80 97.70

1430 155 -4.65 -3.00 -1.65 97.85

1460 185 -5.55 -4.28 -1.27 98.23

1490 215 -6.45 -5.78 -0.67 98.83

1495 220 -6.6 -6.05 -0.55 98.95


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The distance of the sag point from the initial point of vertical curvature;y= mLv = -3x220 = 120m

A -5.5Sag Chainage = 1275+120=1395mHeight of sag point relative to the initial point of vertical curvature;Hmax = m 2L = (-3) 2x 220 = -1.80m

200A 200x-5.5Reduced Level of Sag point = 99.5 1.80 = 97.7m

Check for level at end of curve.

= 96.20 + (2.5x220) =98.95m200

There could be a small difference in the table calculated level (98.95m) at the end of the curveand that from the equation(98.95) in some scenarios. This is due to the several assumptionsand rounding offs done in the table.


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.

Sag, Ch=1395m, Lev=97.7m

P, Ch=1275m,Lev=99.5m

R, Ch=1495m, Lev=98.95m


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Trial Question A single carriageway , which has a designspeed of 60km/h, is to be designed to havea minimum curve of length 50m linking twogradients +1% and -0.4%.

i) Determine the height of the curve at 5mintervals if the vertical tangentsintersect at a level of 205m and at achainage of 11+03.70 (1103.71m).

ii) Determine height of crest point andchainage and compare with highest

point on curve.

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