Module 3 Geometric Design For Highways And Railways

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Module 3 Geometric Design for Highways and Railways Introduction Geometric design of highways and railways refers to their dimensional and spatial characteristics. Highways and railways vary in shape, elevation, size, etc. depending on their purpose and location. Geometry also defines the position of a highway with reference to the objects around it.

Topic Outcomes At the end of this module, the student will be able to: 1. Design and analyze a vertical curve 2. Design and analyze a horizontal curve 3. Consider the importance of balance between geometry and safety in designing highway curves This module will be divided into different topics to ensure that the outcomes will be attained: 1. Vertical alignment 2. Horizontal alignment 3. Super elevation

Topic 1: Vertical Alignment Vertical alignment means the elevation of different points along a roadway. Elevations are designed to provide an acceptable level of driver safety, driver comfort, and proper drainage. The main concern in designing the vertical alignment of roads is by establishing a transition of roadway elevations between to grades. This transition is achieved by means of a vertical curve. Vertical curves are classified into crest vertical curves and sag vertical curves. Figure 3.1.1 shows the two curves.

Figure 3.1.1. Crest and sag vertical curves (Image courtesy of F.L. Mannering, S. S. Washburn)

G1

= initial roadway grade in percent or m/m; referred to also as the initial tangent grade G2 = final roadway grade in percent or m/m A = absolute value of the difference in grades (initial minus final) L = length of curve in meters measured in a constant elevation horizontal plane PVC = point of the vertical curve (initial point of the curve) PVI = point of vertical intersection (intersection of initial and final grades) *Note: in a symmetrical vertical curve, PVI is located at L/2 PVT = point of vertical tangent (final point of the curve)

It is important to know that in connecting roadway grades, a parabolic curve is found to be suitable because it provides a constant change of slope. The general form of the parabolic equation is

𝑦 = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐

(Eq. 3.1.1)

where y = roadway elevation at distance x from the beginning of the vertical curve (PVC), in meters x = distance from the beginning of the curve, in meters c = elevation of PVC (since at PVC, x = 0)

In defining b, we will be taking the first derivative of y, that is

(Eq. 3.1.2)

At PVC, x = 0 so

(Eq. 3.1.3)

In defining a, the second derivative of y will now be taken, that is

(Eq. 3.1.4)

The second derivative of a parabolic function can also be defined as the rate of change of slope of the tangent line, so it can be written in terms of G1, G2, and L, that is

(Eq. 3.1.5) Combining Eq. 3.1.4 and Eq. 3.1.5

(Eq. 3.1.6)

Example 3.1 A 200-m symmetrical vertical curve must be designed such that its initial and final grades are -5% and +4.5% respectively. If PVC is located at STA 1 + 11.5 with an elevation of 50m, determine the stationing and elevation of PVI, PVT, and the lowest point of the curve. 1 station = 20 meters. Given: G1 = 2% G2 = -1.5% PVC @ STA 0 + 120 Elev PVC = 50m Required: Elevation and location of PVI Elevation and location of PVT Elevation and location of the lowest point Solution: Location of PVT: STA 11 + 11.5 Location of PVI: STA 6 + 11.5 (symmetrical)

Recall eq. 3.1.1,

𝑦 = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 Solving for a

Solving for b 𝑏 = 𝐺1 = βˆ’0.05 Substituting the values of a and b to the general formula 𝑦 = 0.0002375π‘₯ 2 βˆ’ 0.05π‘₯ + 50 @PVT, where x = 200, y = 49.50 Thus, the elevation of PVT is Elev PVT = 49.50 m

@PVI, where x = 100, y = 49.738 Thus, the elevation of PVI is Elev PVI = 47.375 m

Since the grades are of different signs, the location of the lowest point is defined by x where dy/dx = 0

π‘₯ = 105.263

Lowest point is located at STA 6 + 16.763 @ x = 105.263, y = 47.3684 Elev Lowest Point = 47.368 m

Final Answers: Location of PVT: STA 11 + 11.5, Elev PVT = 49.50 m Location of PVI: STA 6 + 11.5, Elev PVI = 47.375 m Lowest point is located at STA 6 + 16.763, Elev Lowest Point = 47.368 m

Stopping Sight Distance (SSD) In designing a highway curve, it is important to consider not only its geometry but also its safety. Stopping Sight Distance can be defined as the minimum distance a vehicle can travel starting from the driver seeing an obstruction, reacting to it, hitting the brakes, and the vehicle coming to a full stop. It is a factor of the driver’s reaction time, vehicle’s speed, roadway grade, and road and tire friction. Eq. 3.1.7 shows the formula used to calculate SSD

(Eq. 3.1.7) where, SSD v t f G g

= stopping sight distance, in meters = vehicle’s speed, in meters per second = driver’s reaction time, in seconds = coefficient of friction = roadway grade = gravitational constant, 9.81 m/s2

Take note that the first term corresponds to the reaction distance and the second term corresponds to the braking distance.

Example 3.2 A vehicle is travelling at 35 kilometers per hour. Its driver is about to hit a 2-meter high wall 30 meters away if he did not react accordingly. Assuming the coefficient of friction between the road and tires is 0.35 and the driver steps on the brakes 2 seconds after seeing the obstruction, will he hit the wall? The road is perfectly horizontal. Given: v = 35 kph = 9.72 m/s t = 2 seconds f = 0.35 G=0 Required: If SSD > 30, will the vehicle hit the wall?

Solution: 𝑣2 𝑆𝑆𝐷 = 𝑣𝑑 + 2𝑔(𝑓 Β± 𝐺) 𝑆𝑆𝐷 = (9.72)(2) +

(9.72)2 2(9.81)(0.35)

𝑆𝑆𝐷 =33.20 Since SSD = 33.20 m > 30 m, the driver will hit the wall.

SSD and Crest Vertical Curve

Figure 3.1.2. SSD and crest vertical curve (Image taken from ascelibrary.com)

The equations used in designing a crest vertical curve are as follows: Assuming SSD < L

(Eq. 3.1.8) Assuming SSD > L

(Eq. 3.1.9) where, Lm

= minimum length of crest curve, in meters

S H1 H2 A

= stopping sight distance, in meters = driver’s eye level above roadway surface, in meters = height of obstruction above roadway surface, in meters = absolute value of the difference in grades, in percentage

Example 3.3 Determine the length of the vertical curve with a stopping sight distance of 230 meters. Its initial and final grades are +1.75% and -2.05% respectively. The driver’s eye level above the roadway surface is 150 centimeters and the height of obstruction is 100 centimeters. Given: S = 230 m G1 = +1.75% G2 = -2.05% H1 = 150 cm = 1.5 m H2 = 100 cm = 1.0 m Required: L Solution: Solve for A 𝐴 = |βˆ’2.05 βˆ’ 1.75| = 3.80

Next, we assume that S < L

𝐿=

𝐴𝑆 2 200(√𝐻1 + √𝐻2 )

𝐿=

2

(3.80)(2302 ) 200(√1.5 + √1.0)

2

𝐿 =203.07 Since 230 > 203.07, our assumption was wrong. We now move on to the next equation where S > L

200(√𝐻1 + √𝐻2 ) πΏπ‘š = 2𝑆 βˆ’ 𝐴

2

200(√1.5 + √1.0) πΏπ‘š = 2(230) βˆ’ 3.80

2

πΏπ‘š =199.5 m

The statement S > L holds true in this case, so the length of the vertical curve is 199.5m.

SSD and Sag Vertical Curve

Figure 3.1.3. SSD and sag vertical curve (Image taken from thetruthaboutcars.com)

The equations used in designing a sag vertical curve are as follows:

Assuming SSD < L

(Eq. 3.1.10) Assuming SSD > L

(Eq. 3.1.11)

where, Lm S H

= minimum length of sag curve, in meters = stopping sight distance, in meters = height of headlight above roadway, in meters

Ξ± A

= inclined angle of headlight beam, in degrees = absolute value of the difference in grades, in percentage

Example 3.4 A vertical curve is to be designed with a stopping sight distance of 310 m. Its initial and final grades are -1.2% and +2.1% respectively. The average height of the headlights of the vehicles that will pass through this road is 60 centimeters and Ξ± is set at 1Β°. Determine the length of the curve. Given: SSD = 310 m G1

= -3.2%

G2

= +2.1%

H

= 60 cm = 0.6 m

Ξ±

= 1Β°

Required: L Solution: Solve for A 𝐴 = |2.1 βˆ’ (βˆ’3.2)| = 5.3 Next, we assume that S < L 𝐴𝑆 2 πΏπ‘š = 200(𝐻 + π‘†π‘‘π‘Žπ‘›π›Ό)

πΏπ‘š =

(5.3)(310)2 200(0.6 + 310π‘‘π‘Žπ‘›(1Β°))

𝐿 = 423.66 The statement S < L holds true in this case, thus L = 423.66 m.

Topic 2: Horizontal Alignment In its simplest description, horizontal alignment of a roadway is all about road corners/cornering. Horizontal curves provide a directional transition between two straight sections of a roadway. The main concern in designing a horizontal curve is considering the cornering capabilities of the vehicle. In connecting two straight roads, the most common is using a simple circular curve. Other options may be reverse curves, compound curves, and spiral curves. Reverse curves are just two simple circular curves that turn in opposite directions. Compound curves are circular curves in succession. Spiral curves have continuously changing radius.

Simple Circular Horizontal Curve

Figure 3.2.1 Elements of a simple circular curve (Image courtesy of F.L. Mannering & S.S. Washburn)

In designing a simple circular curve, the following factors are considered: R Ξ” T E M

= radius of the curve, usually measured to the centerline of the road, in meters = central angle of the curve, in degrees = tangent length, in meters = external distance, in meters = middle ordinate, in meters

PC PI PT L

= point of curve (beginning point of the horizontal curve) = point of tangent intersection = point of tangent (ending point of the horizontal curve) = length of curve, in meters

Another consideration is the degree of curve, D, which is the central angle subtended by one station (20m).

The following equations are used in designing a simple circular curve:

Length of Tangent, T

(Eq. 3.2.1) External Distance, E

(Eq. 3.2.2) Middle Ordinate, M

(Eq. 3.2.3) Length of Curve, L

(Eq. 3.2.4) Degree of curve, D w/ 1 station = 20 meters

(Eq. 3.2.5)

Example 3.5 A horizontal curve is to be designed with PI located at STA 23 + 00. The radius of the curve is 500 meters and the length of its tangent is 100 m. Determine the location of PT, external distance, middle ordinate, and degree of curve. 1 station = 20 meters

Given: R = 500 meters T = 100 meters PI @ STA 23 + 00 Required: Location of the PT, external distance, middle ordinate, degree of curve Solution: Use the formula for length of tangent to determine Ξ” 𝑇 = π‘…π‘‘π‘Žπ‘›

π›₯ 2

100 = 500π‘‘π‘Žπ‘›

π›₯ 2

π›₯ = 22.62Β° We can now solve for the length of the curve, L 𝐿=

πœ‹π‘…π›₯ 180

𝐿 = 197.40 π‘š Location of PT is @ STA 32 + 17.40 Solving for external distance, E 𝐸 = 𝑅𝑠𝑒𝑐

π›₯ βˆ’π‘… 2

E = 9.90 m Solving for middle ordinate, M 𝑀 = 𝑅 βˆ’ π‘…π‘π‘œπ‘  M = 9.71 m

Solving for the degree of the curve, D 𝐷=

20π›₯ 𝐿

D = 0.90Β°

π›₯ 2

Stopping Sight Distance and Horizontal Curves Stopping sight distance is also considered in designing horizontal curves to provide safety.

Figure 3.2.2 SSD and simple circular horizontal curve (Image courtesy of F.L. Mannering & S.S. Washburn)

where R Rv

= radius measured to the centerline of the road, in meters = radius to the vehicle’s path, usually measured to the center of the innermost lane, in meters Ξ” = central angle of the curve, in degrees Ξ”s = angle subtended by an arc equal in length to the SSD L = length of curve, in meters Ms = middle ordinate necessary to provide adequate SSD, in meters SSD = stopping sight distance, in meters PC = point of curve (beginning point of the horizontal curve) PT = point of tangent (ending point of the horizontal curve) The equations used in computing the relationship of horizontal curves and SSD are:

(Eq. 3.2.6)

rearranging the term,

(Eq. 3.2.7)

Using the equation of the middle ordinate (Eq. 3.2.3), we can come up with this formula for Ms:

(Eq. 3.2.8) Example 3.6 A 2-lane road curve has a radius of 450 meters. If the required stopping sight distance is 210 meters, what is the allowable distance of an obstruction from the edge of the road? Each lane has a width of 3 meters. Given: R = 450 m SSD = 210 meters lane width = 3 meters Required: Allowable distance of an obstruction from the edge of the road Solution: First, solve for Rv 𝑅𝑣 = 450 βˆ’ 1.5 Rv = 448.5 m Solving for Ms

90(210) 𝑀𝑠 = 448.5 (1 βˆ’ π‘π‘œπ‘  [ ]) πœ‹(448.5) Ms = 12.23 m dallow = 12.23 - 1.5 dallow = 10.73 meters

Topic 3: Superelevation Banking Curve

Figure 3.2.1 Interaction of forces in a banking vehicle (Image courtesy of F.L. Mannering & S.S. Washburn)

where Rv Ξ± e W Wn Wp Ff Fc Fcp Fcn

= radius of curvature of the vehicle’s path, in meters = angle of inclination, in degrees = vertical rise in meters per 100 m of horizontal distance, also called superelevation, usually expressed in percentage = weight of the vehicle, in Newtons = weight of the vehicle normal to the road surface, in Newtons = weight of the vehicle parallel to the road surface, in Newtons = side frictional force, in Newtons = centripetal force, in Newtons = centripetal force parallel to the road surface, in Newtons = centripetal force normal to the road surface, in Newtons

Using the concepts of mechanics, one can come up with the following equation:

(Eq. 3.3.1) where, fs g

= coefficient of side friction = gravitational constant 9.81 m/s2

Example 3.7 A roadway is designed to have a maximum speed of 75 kilometers per hour. One of its horizontal curves is designed to be simple circular having a radius of 195 meters. What should be the angle the roadway makes with the horizontal so that the vehicles will not overturn? The coefficient of side friction is set at 0.15. Given: v = 75 kph = 20.83 m/s Rv = 1050 meters fs = 0.15 Required: The angle the roadway makes with the horizontal Solution: Solve for the value of superelevation, e

20.832 195 = 9.81(0.15 + 𝑒) 𝒆 = 𝟎. πŸŽπŸ•πŸ”πŸ– or 7.68%

The relationship of superelevation and the angle the roadway makes with the horizontal, ΞΈ, is shown by the figure below

π‘‘π‘Žπ‘›πœƒ = 𝑒/100 π‘‘π‘Žπ‘›πœƒ = 7.68/100 ΞΈ = 4.39Β°

End of Module Exercises Answer the following exercises and submit them in our classroom. 1. To create a transition between two roadways, a simple circular curve will not be enough. Designing a compound curve is now being considered. See figure below:

You are provided the following information: R1 = 630 m I1 = 50Β° R2 = 410 m I2 = 35Β° If PI is located at STA 520 + 12, what is the location of the PC? location of the PT? Hints: Loc. PC = Loc PI - x - T1 Loc. PT = Loc PC + Lc1 + Lc2

2. A reversed curve is shown below. PRC is located at STA 427 + 09. The radius of the first curve is 750 meters and subtends an angle of 56Β°. The second curve has a radius of 520 meters with a central angle of 37Β°. Determine the locations of PC and PT.

3. You have been instructed to design a crest vertical curve that will connect a highway segment with +3.2% grade to another segment with -1.1% grade. The stopping sight distance of the highway is 345 meters if the elevation at the beginning of the curve is 75 meters, what will the elevation be at L/2? What is the elevation at the highest point of the curve?

Bibliography Mannering Fred, Washburn Scott,Kilaresky Walter. 2004. Principles of Highway Engineering & TrafficAnalysis. Muze Inc.

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