to accompany principles of highway engineering and traffic...

Solutions Manual to accompany

Principles of Highway Engineering and Traffic Analysis, 3e

By Fred L. Mannering, Walter P. Kilareski, and Scott S. Washburn

Chapter 3 Geometric Design of Highways

Metric Units

Copyright © 2004, by John Wiley & Sons, Inc. All rights reserved.

Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 3e, by Fred L. Mannering, Walter P. Kilareski, and Scott S. Washburn. Copyright © 2004, by John Wiley & Sons, Inc. All rights reserved.

Preface The solutions to the third edition of Principles of Highway Engineering and Traffic Analysis were prepared with the Mathcad1 software program. You will notice several notation conventions that you may not be familiar with if you are not a Mathcad user. Most of these notation conventions are self-explanatory or easily understood. The most common Mathcad specific notations in these solutions relate to the equals sign. You will notice the equals sign being used in three different contexts, and Mathcad uses three different notations to distinguish between each of these contexts. The differences between these equals sign notations are explained as follows.

• The ‘:=’ (colon-equals) is an assignment operator, that is, the value of the variable or expression on the left side of ‘:=’is set equal to the value of the expression on the right side. For example, in the statement, L := 1234, the variable ‘L’ is assigned (i.e., set equal to) the value of 1234. Another example is x := y + z. In this case, x is assigned the value of y + z.

• The ‘==’ (bold equals) is used when the Mathcad function solver was used to find the value of a variable in the equation. For example, in the equation

, the == is used to tell Mathcad that the value of the expression on the left side needs to equal the value of the expression on the right side. Thus, the Mathcad solver can be employed to find a value for the variable ‘t’ that satisfies this relationship. This particular example is from a problem where the function for arrivals at some time ‘t’ is set equal to the function for departures at some time ‘t’ to find the time to queue clearance.

• The ‘=’ (standard equals) is used for a simple numeric evaluation. For example, referring to the x := y + z assignment used previously, if the value of y was 10 [either by assignment (with :=), or the result of an equation solution (through the use of ==) and the value of z was 15, then the expression ‘x =’ would yield 25. Another example would be as follows: s := 1800/3600, with s = 0.5. That is, ‘s’ was assigned the value of 1800 divided by 3600 (using :=), which equals 0.5 (as given by using =).

Another symbol you will see frequently is ‘→’. In these solutions, it is used to perform an evaluation of an assignment expression in a single statement. For example, in the following

statement, , Q(t) is assigned the value of Arrivals(t) – Departures(t), and this evaluates to 2.2t – 0.10t2. Finally, to assist in quickly identifying the final answer, or answers, for what is being asked in the problem statement, yellow highlighting has been used (which will print as light gray). 1 www.mathcad.com


solve for Gcon

As = | Gcon |As = | Gcon - 0 |Ac = | 4.0 - Gcon |

substitute for Ac and As

Kc Ac2

⋅

200

4.0 Kc⋅ Ac⋅( )100

−Gcon Ltotal Kc Ac⋅( )− Ks As⋅( )−⎡⎣ ⎤⎦⋅

100+

Ks As2

⋅

200+ ∆elev

Lcon = 1070 - Kc*Ac - Ks*AsLc + Lcon + Ls = Ltotal

L = K*A(Eq. 3.10)

substitute for Lcon Lc, and Ls

Ac Lc⋅

200

4.0 Lc⋅

100−

⎛⎜⎝

⎞⎠

Gcon Lcon⋅

100+

As Ls⋅

200+ ∆elev

substitute in for final offsets using Equation 3.9

(Yfc - ∆Yc) + ∆Ycon + Yfs = 12

set change in elevation equal to sum of offsets and change in elevation of constant grade section

Ltotal 1070=Ltotal staPVTs staPVCc−:=

calculate total length of alignment

K c 4 Gcon+( )2⋅

200

4.0 K c⋅ 4 Gcon+( )⋅⎡⎣ ⎤⎦100

−G con L total K c 4 Gcon+( )⋅⎡⎣ ⎤⎦− K s Gcon⋅−⎡⎣ ⎤⎦⋅

100+

K s G con2

⋅

200+ ∆ elev


solve for Gcon

Ac G2c Gcon−As Gcon G1s−

substitute in for A s and Ac

As Ls⋅

200

G1s100

Ls⋅−Gcon100

Ltotal Ls− Lc−( )⋅+Ac Lc⋅

200+

Gcon100

Lc⋅− elevdiff

substitue values for final offsets using equation 3.9

Gcon G2s G1c

G2c 1−:=G1s 1−:=

Yfs ∆ys− ∆ycon+ Yfc+ ∆yc− elevdiff

set total elevation change equal to sum of offsets and changes in elevation from constant grade

(Table 3.2)Kc 26:=

(Table 3.3) Ks 30:=

elevdiff 15=elevdiff elevPVTc elevPVCs−:=

Ltotal 1200=Ltotal PVTc PVCs−:=

calculate total length and change in elevation

elevPVTc 40:=PVTc 1345:=

(given) elevPVCs 25:=PVCs 145:=

Determine the lowest grade possible for the constant-gradesection that will still complete this alignment.

Problem 3.17

Gcon G 1s−( )2 K s⋅

200

G1s100

Gcon G 1s−⋅ K s⋅−Gcon100

L total Gcon G 1s− K s⋅− G2c Gcon− K c⋅−( )⋅+G2c G con−( )2 K c⋅

200+

Gcon100

G2c Gcon− K c⋅( )⋅− elev diff


(Eq. 3.10)

Lcon 915 Ls− Lc−:= Lcon 497=

Using final offset equation 3.9, calculate the total elevation difference

Yfc ∆Yc−( ) ∆Ycon+ Yfs ∆Ys−( )+ = elevation difference

YfcAc Lc⋅

200:= Yfc 8.32= ∆Yc G1c Lc⋅:= ∆Yc 6.24=

YfsAs Ls⋅

200:= Yfs 7.35= ∆Ys G2s Ls⋅:= ∆Ys 4.2=

∆Ycon Gcon Lcon⋅:= ∆Ycon 24.85=

∆Y Yfc ∆Yc− ∆Ycon+ Yfs+ ∆Ys−:=

Ac Lc⋅

200G1c Lc⋅( )−

⎡⎢⎣

⎤⎥⎦

Gcon Lcon⋅( )+As Ls⋅

200+ G2s Ls⋅( )−

∆Y 30.08= m

Problem 3.18 Determine the elevation difference.

G1c 3%:= Gcon 5− %:= G2s 2%:=

Kc 26:= (Table 3.2)

Ks 30:= (Table 3.3)

Ac 3.0 5.0−−:= Ac 8=

As 5.0− 2.0−:= As 7=

calculate lengths of crest and sag curve, subtract from total length to find length of constant grade

Lc Kc Ac⋅:= Lc 208= (Eq. 3.10)

Ls Ks As⋅:= Ls 210=


solve for G

Ac G G2c−As G G1s−

substitute for A s and Ac

Ltotal Ks As⋅ Kc Ac⋅+L KA

substitute for L s and Lc

Ltotal Ls Lc+

(Table 3.3)Ks 30:=

(Table 3.2)Kc 26:=

Ltotal 390:=

(given)G2s G1cG2c 2.0:=G1s 3.0−:=

Determine the common grade between the sag and crest curves anddetermine the elevation difference between the PVCs and PVTc.

Problem 3.19


Problem 3.22 Determine how many feet below the railway the curvePVI should be located.

First find clearance based on SSD

K 23:= (Table 3.3)

G1 2−:= G2 2:= (given)

A G1 G2−:= A 4=

L K A⋅:= L 92= (Eq. 3.10)

SSD 105:= (Table 3.1)

Since SSD >L, use Eq. 3.30 to get Hc

Hc 2.09= m

2.09 m is less than the AASHTO desirable clearance height of 5 m, so 5 m will be provided

Hc 5:=

now find necessary elevation of the PVI

elevPVI Hc− Ym−

YmA L⋅800

:= Ym 0.46= (Eq. 3.8)

elevPVI Hc− Ym−:= elevPVI 5.46−= m


Ms809.17= m (Eq. 3.42)

this is larger than 8.2 m, so design speed is too high

try 70 km/h V 70:=

fs .145:= (Table 3.5)

calculate radius to vehicle travel path

RvV 0.2778⋅( )2

g e fs+( )⋅:= Rv 171.373= (Eq. 3.34)

SSD70 105:= (Table 3.1)

calculate necessary middle ordinate for 70 km/h

Ms70Rv 1 cos

90 SSD70⋅

π Rv⋅

⎛⎜⎝

⎞

⎠deg⋅

⎡⎢⎣

⎤⎥⎦

−⎡⎢⎣

⎤⎥⎦

⋅:= Ms707.98= m (Eq. 3.42)

this is less than 8.2 m, so 70 km/h is the maximum design speed

Problem 3.23 Determine the highest possible design speed for the curve.

necessary middle ordinate distance is the distance from the centerline minus 1/2 the inside lane

Ms 103.62

−:= Ms 8.2= (given)

first try 80 km/h

e .08:= g 9.807:= V 80:= (given)

fs .14:= (Table 3.5)

calculate radius to vehicle travel path

RvV 0.2778⋅( )2

g e fs+( )⋅:= Rv 228.921= (Eq. 3.34)

SSD80 130:= (Table 3.1)

calculate necessary middle ordinate for 80 km/h

Ms80Rv 1 cos

90 SSD80⋅

π Rv⋅

⎛⎜⎝

⎞

⎠deg⋅

⎡⎢⎣

⎤⎥⎦

−⎡⎢⎣

⎤⎥⎦

⋅:=


sta PT = 4+205.85staPT 4205.848=staPT staPC L+:=

(Eq. 3.39)L 425.8=Lπ

180R⋅ ∆( )⋅:=

calculate length

(Eq. 3.36)∆ 41:=∆ 40.577deg=∆ 2 atanTR

⎛⎜⎝

⎞⎠

⋅:=T R tan∆

2⎛⎜⎝

⎞⎠

⋅

knowing tangent length and radius, solve for central angle

T 220=T staPI staPC−:=

R 595.1=

R Rv:=since road is single-lane,

(Eq. 3.34)Rv 595.1=RvV 0.2778⋅( )2

g e fs+( )⋅:=

calculate radius

(Table 3.5)fs 0.10:=

g 9.807:=V 110:=e 0.06:=

(given) staPI 4000:=staPC 3780:=

Determine the station of the PT.Problem 3.24


Problem 3.27 Determine the superelevation required at the design speed. Also,compute the degree of curve, length of curve, and stationing of the PC annd PT.

V 160:= R 305:= ∆ 30:= (given)

staPI 34300:= fs 0.20:= g 9.807:=

Since the racetrack is single-lane, Rv R:= Rv 305=

sta PT = 34+377.97staPT 34377.973=staPT staPC L+:=

sta PC = 34+218.28staPC 34218.275=staPC staPI T−:=

(Eq. 3.36)T 81.725=T R tan∆

2deg⋅⎛⎜

⎝⎞⎠

⎛⎜⎝

⎞⎠

⋅:=

calculate tangent length

mL 159.7=L30.5∆⋅

D:=

(Eq. 3.39)

Lπ

180R⋅ ∆⋅R

18000π D⋅

use this and Equation 3.39 to solve for length of curve

(Eq. 3.35)degreesD 5.73=D180 30.5⋅

π R⋅:=

solve for degree of curve

e 0.407=


Problem 3.28 Determine the radius and stationing of the PC and PT.

staPI 7640:= g 9.807:= V 105:= (given)

e 0.08:= ∆ 35:=

(Table 3.5)fs 0.11:=

calculate radius

RvV 0.2778⋅( )2

g fs e+( )⋅:= Rv 456.62= m (Eq. 3.34)

calculate length and tangent length of curve

Lπ

180R⋅ ∆⋅:= L 280.032= m (Eq. 3.39)

T R tan∆

2⎛⎜⎝

⎞⎠

deg⋅⎡⎢⎣

⎤⎥⎦

⋅:= T 144.539= (Eq. 3.36)

staPC staPI T−:= staPC 7495.461= sta PC = 7+495.5

staPT staPC L+:= staPT 7775.493= sta PT = 7+775.5

(Eq. 3.35)degreesD 3.12=D180 30.5⋅

π R⋅:=

(Eq. 3.39)mL 390.95=Lπ

180R⋅ ∆⋅:=

(Table 3.5)mR 560:=for a 110 km/h design speed with e restricted to 0.06,

2 3-m lanes∆ 40:=(given)

Give the radius, degree of curvature, and length of curve that youwould recommend.

Problem 3.29


since Ms and Rv are known, we can solve Equation 3.42 to find SSD

Alternative Solution

from Table 3.1, SSD for 60 km/h is 85 m, design speed is 60 km/h

SSD 84.6=

(Eq. 3.43)SSDπ Rv⋅

90 deg⋅acos

Rv Ms−

Rv

⎛⎜⎝

⎞

⎠

⎛⎜⎝

⎞

⎠⋅:=

since Ms and Rv are known, we can use Equation 3.43 to find SSD

Rv 130.5=Rv R:=

(Eq. 3.39)R 130.5=RL 180⋅

π ∆⋅:=L

π

180R⋅ ∆⋅

using L and ∆, solve for R

(given)Ms 6.8:=L 205:=∆ 90:=

Since the ramp is a single 3.6-m lane, center of roadway is center of traveled path

Determine the design speed used.

Problem 3.31

Cornering Check

V 60 0.2778⋅:= V 16.7=

for 60 km/h, fs 0.15:= (Table 3.5)

g 9.807:=

eV2

g Rv⋅fs−:= e 0.067=

So this combination of speed, radius, and superelevation is OK


Rv 293.105=

First, try 80 km/h

SSD 130:= (Table 3.1)

Ms Rv 1 cos90 SSD⋅

π Rv⋅deg⋅⎛

⎜⎝

⎞

⎠−⎛

⎜⎝

⎞

⎠⋅:= Ms 7.18= (Eq. 3.42)

7.18 m is less than 7.8 m so 80 km/h is acceptable, but can speed be higher?

try 90 km/h

(Table 3.1)SSD 160:=


π Rv⋅deg⋅⎛

⎜⎝

⎞

⎠−⎛

⎜⎝

⎞

⎠⋅:= Ms 10.85= m (Eq. 3.42)

this value is greater than 7.8 m, therefore 80 km/h is the design speed

Check values vs. Table 3.5 - Minimum radius for e = 0.08 is 230, R exceeds this value.

Problem 3.32 Determine a maximum safe speed to the nearest 5 mi/h.

∆ 34:= e 0.08:=

(given) PT 3940:= PC 3765:=

L PT PC−:= L 175=

since this is a two-lane road with 3.6-m lanes, Ms 63.62

+:= Ms 7.8=

Lπ

180R⋅ ∆⋅ R

L 180⋅

π ∆⋅:= R 294.905= (Eq. 3.39)

Rv R3.62

−:=


Problem 3.33 Determine the distance that must be cleared from the inside edgeof the inside lane to provide adequate SSD.

V is 100 km/h

SSD 185:= (Table 3.1)

R 395.3:=

since the road is two 3.6-m lanes, Rv R3.62

−:= Rv 393.5=


π Rv⋅deg⋅⎛

⎜⎝

⎞

⎠−⎛

⎜⎝

⎞

⎠⋅:= Ms 10.82= m (Eq. 3.42)

this is more than the required distance, therefore the design speed is 100 km/h

(Eq. 3.42)mMs 10.769=Ms Rv 1 cos90 SSD⋅

π Rv⋅deg⋅⎛

⎜⎝

⎞

⎠−⎛

⎜⎝

⎞

⎠⋅:=


(Table 3.5)Rv 560:=

try 110 km/h

this is less than the required distance, try again

(Eq. 3.42)mMs 9.798=Ms Rv 1 cos90 SSD⋅

π Rv⋅deg⋅⎛

⎜⎝

⎞

⎠−⎛

⎜⎝

⎞

⎠⋅:=


(Table 3.5)Rv 435:=

try 100 km/h

mMs 10.6=Ms 16 3.6−3.62

−:=since the road is four-lane with 3.6-m lanes,

(given)e 0.06:=

Determine the design speed used to design the curve.

Problem 3.34


(Eq. 3.39)mL 488.72=Lπ R⋅ ∆⋅

180:=

R 756.8=R Rv13.62

+:=since the road is two-lane with 3.6-m lanes,

(Eq. 3.34)Rv2 755.439=Rv2V 0.2778⋅( )2

9.807 fs e+( )⋅:=

fs 0.09:=V 120:=

(Table 3.5)orRv1 755:=

(Table 3.3)safe design speed is 120 km/h (K = 63 for 120 km/h)

(Eq. 3.10)Ks 62=KsLsAs

:=

As 2=As G2 G1−:=

e 0.06:=∆ 37:=Ls 124:=

(given) G2 3:=G1 1:=

Determine the length of the horizontal curve.

Problem 3.35


(Table 3.5)Rv 665:=

PC 423=PC PVC 90−:=

PVC 513=PVC PVT L−:=

(Eq. 3.10)L 252=L K A⋅:=

PVT 765:=e 0.08:=∆ 38:=

(given)K 63:=A 4=A G2 G1−:=

G2 1.5:=G1 2.5−:=

Determine the station of the PT.Problem 3.36


calculate the elevations of the ramp connections using T and the grades

elevEW 45 TG2100⋅+:= elevEW 55.922=

elevNS 38 TG1100⋅−:= elevNS 44.553=

Ks 30:= (Table 3.3)

GelevEW elevNS−

L100⋅:= G 3.313=

calculate the lengths of the two sag curves using Equation 3.10

A1 G1 G−:= A1 6.313= A2 G G2−:= A2 1.687=

L1 Ks A1⋅:= L1 189.399= L2 Ks A2⋅:= L2 50.601=

Problem 3.37 Design the ramp and give the stationing and elevations of the PC, PT,PVCs, and PVTs.

G1 3−:= G2 5:= (given)

D 8.0:= ∆ 90:=

using D, solve for R

R 30.5180π D⋅⋅:= R 218.44= (Eq. 3.35)

From Table 3.5, maximum design speed is 80 km/h

T R tan∆

2deg⋅⎛⎜

⎝⎞⎠

⋅:= T 218.44= (Eq. 3.36)

(Eq. 3.39)L

30.5∆⋅D

:= L 343.125=

calculate the length of the connecting grade

Lcon LL1 L2+( )

2−:= Lcon 223.13= m

finally, calculate the station and elevation of all PVCs, PVTs, PCs, and PTs

to accompany principles of highway engineering and traffic...

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