Exercise 1: Basic Design Criteria 1) Route 17 is a minor arterial and it has a design ADT of 4000. Assume rolling terrain. Determine the following information:

Exercise 1: Basic Design Criteria

1) Route 17 is a minor arterial and it has a design ADT of 4000. Assume rolling terrain. Determine the following information:

Design Speed

Drawing # for typical section

Degree of curve or radius

Min. SSD

Min. R/W required

Foreslope and min. ditch depth

Max. percent grade

Range of K value

50 mph

D-62

7 1/2

400 ft

150 ft

6:1 , 2ft

5%

90-110

Exercise 1: Basic Design

Criteria2) You are realigning a portion of I-44, assuming

rolling terrain, determine the following:Design Speed

Drawing # for typical section


Min. SSD

Min. R/W required

Foreslope and min. ditch depth

Max. percent grade

Range of K value

What type of R/W access do you think I-44 requires? Fully Controlled

70 mph

D-61

3

625 ft

250 ft dual

6:1, 4 ft

4%

150-220

Exercise 1: Basic Design Criteria

3) You are designing a 4-lane principal arterial, find the following design criteria. This is an urban setting.

Design Speed

Median Width


Min. SSD

C&G

Protected left turn lanes?

Lane width, parking lane

Sidewalk

Max. % grade

Range of K value

Width of shoulder

Superelevation rate

40 mph

14

10

275 ft

None Required

Yes

12 ft, none

5

5%

60-70

10 or 12

0.04 or 4%

Exercise 2: Horizontal Alignment, Simple and

Spiral CurvesGiven Information to fix substandard curve:

• Road is principal arterial with ADT=1500• Back bearing is N 85^ 49’ 13” E• Ahead bearing is S 56^ 39’ 24” E = 37^ 31’ 23” (RT)

A) Design curve, find L & T

B) If the PC station starts at 15+63.34, what are the PI & PT stations for the curve you designed?

C) When does MoDOT require spiral curves?

PI = 1563.34 + 389.00 = 1952.34 PI sta = 19+52.34

PT = 1563.34 + 750 = 2313.34 PT sta = 23+13.34

1. When degree of curve > 2

2. V >50 mph

3. ADT > 400

L = 750 ft & T = 389.00 ft


Spiral Curves

D) Design a spiral curve based on MoDOT Design Criteria

1) What’s the transition length, Ls?

2) If the curve is 617.88 ft long, what’s the overall length of the SCS?

3) If the station at the SC point is 16+50.42, what are the stations at TS and ST points?

Ls=192

617.88 +192+192=1001.88 ft

TS= 1650.42 – 192 = 1458.42, so TS station = 14+58.42

CS = 1650.42 + 617.88 = 2268.30, so CS station = 22+68.30

ST = 2268.30 + 192 = 2460.30, so ST station = 24+60.30


Spiral Curves

Given information to design a temporary bypass:

• Minor arterial with ADT = 1500

• 20 ft requirement between the EOP of main line and EOP of temporary bypass

• d = 500 ft in length on each end of bypass

A) Find , R, and L for this reverse curve

B) If the first PC station on the bypass starts at 20+00, what are the stations for the PRC and PT of the first reverse curve you designed?

D = 10^ 3’ 29” (10.058^), R = 1431.47 ft, L = 251.29 ft

PRC = 2000.00 + 251.29 = 2251.29, so PRC station = 22+51.29

PT = 2251.29 + 251.29 = 2502.58, so PT station = 25+02.58

Exercise 4: Horizontal Alignment, Superelevation

Fix the substandard superelevation transitions on the following project without re-aligning the curve:

• Principal arterial with ADT = 22940,

• NC(%) = 2, rural flat terrain

• Existing Curve Data

PC station = 388+72.21 D (degree of curve) = 3

PT station = 394+42.77 L = 570.56 ft

PI station = 391+60.46 T = 288.25 ft

2. Find the SE rate and runoff length, specify the standard plan usedStandard Drawing 203.21 pg 3/5

Runoff length, L = 356 ft

Exercise 4: Horizontal Alignment, Superelevation

Fix the substandard superelevation transitions on the following project without re-aligning the curve:

• Principal arterial with ADT = 22940, rural setting, NC(%) = 2

• Existing Curve Data

PC station = 388+72.21 D (degree of curve) = 3

PT station = 394+42.77 L = 570.56 ft

PI station = 391+60.46 T = 288.25 ft

3. Calculate the superelevation transition points.

Before PC Station SE Transition Section

After PT Station

385+44.75 A-A 397+70.23

386+34.88 B-B 396+80.10

387+25.01 C-C 395+89.97

389+90.88 D-D 393+24.10

Exercise 5: Horizontal Alignment, Horizontal

Sight Distance

Check for horizontal sight distance:

• 2-lane minor arterial with ADT = 1890, rural setting on a rolling terrain, assume lane width = 12’

• Existing Curve DataPC station = 0+00 = 29^ 30’ 6” (RT) D =

4^46’29”

PT station = 6+17.88 L = 570.56 ft

PI station = 3+15.65 T = 288.25 ft

1. Find the SSD for a sight of obstruction of 25 ft

2. Determine the distance of obstruction M based on Figure 4-04.2 & 4-04.3

SSD = 489.49 ft

Round to 490 ft according to Fig. 4-01.1

M = 23.55 ft

Exercise 6: Vertical Alignment

Design a crest curve that will meet MoDOT standards based on the given information:• Road is principal arterial with ADT=19,800 on a rolling terrain• Divided highway• Plus grade = 1.03%• Minus grade = 2.99%• PI station = 683+50.00 with elevation = 676.99 ft

1) Based on the length of curve designed, calculate SSD, PSD and K value if appropriate

2) What is the station at the VPC and VPT?

3) Calculate the high point of the curve

VPC station = 677 + 45 VPC elevation = 670.76 ft

VPT station = 689 + 55 VPT elevation = 658.90 ft

Xm = 310.02 ft

HP station = 680+55.02; HP elevation = 675.55 ft

For “my” L 1210 ft, K = 301 and SSD = 650 ft

Exercise 7: At-Grade Intersections (Handout)

Given: Route 123Principal ArterialDesign ADT = 18000, Rolling TerrainDesign Speed = 60 mphT = 12%4-Lane divided Width = 12 ftShoulder width = 8 ft

Route AMinor ArterialDesign ADT = 2500Rolling TerrainDesign Speed = 50 mphT = 6%2-lane undividedLane width = 12 ftShoulder width = 8 ft

Product Volume

Std. Dwg. to be used

Driveway Type

Min. Radius

Min. Surface Width

“d” along main road

“d” along minor road

SSD along main road

SSD along minor road

Find: P.V.=18000x2500 = 45,000,000203.65

Type V

90 ft

24 ft

260 ft

220 ft

1050-1300 ft

400-475 ft

Comprehensive Lab: Horizontal, Vertical

Alignment and Sight Distance

Given: Rural Collector, 2-12’ lanes (undivided)Rolling TerrainDesign ADT = 770, Construction ADT = 530Design Speed = 40 mphDHV = 12% T = 7%NC = -2%

Project Description: Bridge replacement and re-alignment. Project starts at station 226+30, and ends at station 244+69.35. Total length of project = 1839.35 ft.Beginning tie-in bearing is S 48°40’57” E and ending tie-in bearing is S 41°01’12” E

1st Curve 2nd Curve

Incoming Grade -0.3% +5.07%

Outgoing Grade +5.07% +1.06%

VPI station & elevation 236+30, 721.14 240+22.74, 741.04



1. Find L, R, and T for each curveCurve 1 Curve 2

R = 2721.67 ft

D = 2.105°

T = 301.22 ft

= 12.6.31° RT = 4.9684° LT

R = 6919.22 ft

D = 0.8281º

T = 300.19 ft

2. Find PC, PI, and PT stationing for each curveCurve 1 Data

PC Station 227+79.84

PI Station 230+81.06

PT Station 233+79.84

Curve 2 Data

237+99.00

240+99.19

243+99.00



3. Find superelevation transition pointsCurve 1

Curve 2:

According to Standard Plan 203.20 page 25 no superelevation is required.



4. Find VPI and VPT stationing and elevations for e/vertical curve

Vertical Curve 1 Vertical Curve 2

A = 5.37%

VPC sta = 234+60.00

VPC elevation = 721.65 ft

L = 340 ft for K = 63.3

VPT sta = 238+00.00

VPT elevation = 729.76 ft

A = 4.01%

L = 320 ft SSD = 325.71 and V = 40 mph

VPC sta = 238+62.74

VPC elevation = 732.93 ft

VPT sta = 241+82.74

VPT elevation = 742.74 ft



5. Calculate the low and/or high point for each vertical curve

Vertical Curve 1 Vertical Curve 2

Xm = 18.99 ft from VPC

E18.99= 721.62 ft

VPT station is low point = 241+82.74

Elevation = 742.74 ft

Documents

Exercise 1: Basic Design Criteria 1) Route 17 is a minor arterial and it has a design ADT of 4000. Assume rolling terrain. Determine the following information: