PE Exam Review - Surveying Demonstration Problem Solutions

Transcription

1 PE Exam Review - Surveying Demonstration Problem Solutions I. Demonstration Problem Solutions Circular Curves Part A.... Circular Curves Part B Vertical Curves Part A Vertical Curves Part B Spiral Curves Grade Separation Earthwork... 49

2 Solved Problems Surveying Demo Problem Solutions Module 1 Part a - Circular Curves Situation A circular curve is required to make the transition between two tangent sections in a highway alignment. Use the data and sketch below to solve the following six requirements. Design Speed 60 mph Side Friction Factor 0.15 Maximum Superelevation Rate 0.08 Requirements ai. Based on the design speed, what is the minimum curve radius permitted? aii. What is the maximum degree of curvature? aiii. Using a degree of curvature of , what is the station of the P.C.? aiv. Using a degree of curvature of , what is the station of the P.T.? av. If a theodolite is set up at the P.C. of the curve with a backsight on the P.I. what is the required deflection angle to set a stake at the station on the curve? avi. What is the bearing of the forward tangent?

3 Solved Problems Surveying 3 Requirement 1ai 1. V (speed) 60 mph. e (Superelevation rate) f (Side friction factor) R min (Minimum curve radius) ( e f ) 15 V + (AASHTO green book: A Policy on Geometric Design of Highways and Streets, 1994, p 151) ( ) 1, ft. (ft.)

4 Solved Problems Surveying 4 Requirement 1aii 1. R min (Minimum curve radius) 1, (from 1AB). D max (Maximum degree of curvature) 5, R min 5, ,

5 Solved Problems Surveying 5 Requirement 1aiii 1. D max (Maximum degree of curvature) 5.5. (Deflection angle) Station of P.I. (Point of intersection) R min (Minimum curve radius) 5, D max 5, , ft. 5. T (Length of subtangent) R min tan 1, tan (1.375) 8.56 ft. 6. Station of P.C. (Point of curvature) Station of P.I. T

6 Solved Problems Surveying 6 Requirement 1aiv 1. D max (Maximum degree of curvature) 5.5. (Deflection angle) Station of P.C. (Point of curvature) (from 1AC) 4. L (Length of circular arc) Dmax ft. 5. Station of P.T. Station of P.C + L

7 Solved Problems Surveying 7 Requirement 1av 1. Station at given point Station of P.C (from 1AC) 3. (Deflection angle between tangents) arc (Segment of arc length) Station at given point Station of P.C L (Length of circular arc) 450 ft. (from 1AD) arc 6. Deflection Angle L

8 Solved Problems Surveying 8 Requirement 1avi Figure 1avi N Back Tangent N E α S? E Forward Tangent S 1. Angle of back tangent (Deflection angle between tangents) α (Bearing of forward tangent in Figure 1AF) 180 Angle of back tangent S E

9 Solved Problems Surveying 9 Module 1 Part b - Circular Curves Situation The present highway layout (Curve 1) is too close to a historic building. The proposed alignment (Curve ) will shift the forward tangent 50 west and parallel to the original tangent. Using the data and sketch provided, solve the following seven requirements. Design Speed is 55 miles per hour The P.C. of both curves is at the same station. Curve 1 has a degree curvature of The railroad track is 80 feet south of and parallel to the back tangent Requirements bi. What is the subtangent length for Curve 1? bii. What is the station of the P.C.? biii. What is the degree of curvature for Curve? biv. What is the station of the P.T. for Curve? bv. Using a degree of curvature of , what is the horizontal sight distance for Curve? bvi. Using a 55 mile per hour design speed, what is the maximum horizontal sight distance (rounded for design) for Curve? Bvii. Using a degree of curvature of , what is the station at the intersection of the railroad track and Curve?

10 Solved Problems Surveying 10 Requirement 1bi 1. D (Degree of curvature) , R (Curve radius) D 5, ,637.0 ft, 3. (Deflection angle between tangents) T (Length of subtangent) R tan 1,637.0 tan (16.3) ft.

11 Solved Problems Surveying 11 Requirement 1bii 1. Station of P.I. (Point of intersection) T (Length of subtangent) ft. (from 1BA) 3. Station of P.C. (Point of curvature) Station of P.I. T

12 Solved Problems Surveying 1 Back Tangent Requirement 1biii Figure 1BC-1 c ft. 1. (Deflection angle between tangents) 3.6 Forward Tangent. c sin 50 ( ) (Figure 1BC-1) 50 sin (3.6) 9.80 ft. 3. T 1 (Length of subtangent of Curve 1) ft. (from 1BA) 4. T (Length of subtangent of Curve ) T 1 c ft. 5. R (Radius of Curve ) T tan tan( 16.3) 1, ft. 5, D (Degree of curvature for Curve ) R 5, ,

13 Solved Problems Surveying 13 Requirement 1biv 1. (Deflection angle between tangents) 3.6. D (Degree of curvature for Curve ) 4.34 (from 1BC) 3. L (Length of Curve ) D ft. 4. Station of P.C. (Point of curvature for Curve ) (From 1BB) 5. Station of P.T. (Point of tangency for Curve ) Station of P.C. + L

14 Solved Problems Surveying 14 Requirement 1bv 1. D (Degree of curvature for Curve ) 4.5 5, R (Radius of Curve ) D 5, , ft. 3. Sight distance for Curve (S ) Assume S < L M (distance from the centerline of roadway to obstruction) 5 ft. S (Sight distance) 8 R M (if S < L ) 8 1, ft. (Deflection angle between tangents) 3.6 D (Degree of curvature) 4.5 L (Length of Curve) 100 D ft < , so assumption and S formula correct. S ft.

15 Solved Problems Surveying 15 Requirement 1bvi Design Speed (mph) Assumed Speed (mph) Table 1BF-1 Computed (ft) Stopping Sight Distance Rounded for Design (ft) From A Policy on Geometric Design of Highways and Streets, Copyright 1994 by the American Association of State Highway and Transportation Officials, Washington, D.C. Used by permission. 1. Maximum stopping sight distance 550 ft. (Table 1BF-1)

16 Solved Problems Surveying 16 Requirement 1bvii Figure 1BG-1 Back Tangent P.C. 80 ft. A α Curve Railroad Track Center of Circle Radius 1. R (Radius of Curve ) 1, ft.. A (Length from center to railroad track) R 80 1, ,68.14 A 3. α (Angle at center of Curve ) arccos (Figure 1BG-1) R 1,68.14 arccos 1, D (Degree of curvature for Curve ) L (Length of Curve from P.C. to Railroad track) α D Station of P.C. (Point of curvature for Curve ) 7. Station where Curve ft (From 1BB)

17 Solved Problems Surveying 17 intersects Railroad track Station of P.C. + L

18 Solved Problems Surveying 18 Module Part a - Vertical Curves Situation A parabolic curve is used to make the vertical transition between the two tangent sections in a highway alignment for an equal tangent crest curve. Using the design data and the information in the sketch below, solve the following requirements. Design Speed 60 m.p.h. Elevation on top of 36 pipe ft. Requirements ai. Using the design speed and stopping sight distance (rounded for design), what is the minimum length of the vertical curve required to make the grade transition? aii. What is the elevation on the vertical curve at station ? aiii. What is the elevation of the high point on the curve? aiv. If six feet of cover is required over the pipe at station 34+00, what is the maximum length of vertical curve that can be used? av. What is the actual passing sight distance for the vertical curve?

19 Solved Problems Surveying 19 Requirement ai Table ai Stopping Sight Distance over a Crest Vertical Curve Design Speed (mph) Assumed Speed for Condition (mph) Stopping Sight Distance Rounded for Design (ft.) Rate of Vertical Curvature, K (length [ft] per percent of A) Computed Rounded for Design From A Policy on Geometric Design of Highways and Streets, Copyright 1994 by the American Association of State Highway and Transportation Officials, Washington, D.C. Used by permission. 1. K (Rate of vertical curvature) 190 (minimum value in Table AA-1). g 1 (Gradient 1) +3.5% 3. g (Gradient ) -.5% 4. A (Total change in grade of the curve) g 1 g 3.5% (.5%) 6.0% 5. L (Length of curve) K A ,140 ft.

20 Solved Problems Surveying 0 Requirement aii 1. L (Length of vertical curve) 1,00 ft Station at PVI (Point of vertical intersection) Station of PVC L (Point vertical curvature) Station of PVI Elev PVI (Elevation PVI) g 1 (Gradient 1) +3.5% L 6. Elev PVC (Elevation PVC) Elev PVI g ( 3.5 6) P Point at X p (Horizontal distance from PVC to P) g (Gradient ).5% 10. R g g1 (Rate of change of grade) L Y p (Elevation of P) ( ) R XP YPVC + g1 XP ( )

21 Solved Problems Surveying 1 Requirement aiii 1. P Highest point on curve.. L (Length of vertical curve) 1,00 ft g 1 (Gradient 1) +3.5% 4. g (Gradient ).5% 5. X p (Horizontal distance g1 L from PVC to P) g1 g (.5) 7.0 stations 6. Station of PVC (Point vertical curvature) (from AB) 7. Station of P Station of PVC + X p Elev PVC (Elevation PVC) (from AA) 9. R (Rate of change of grade) 0.5 (from AA) R X 10. Y p (Elevation of P) Y ( ) PVC + g1 X ( )

22 Solved Problems Surveying Requirement aiv 1. Elev TopPipe (Elevation at top pipe) ft.. D (Depth of cover) 6 ft. 3. Elev curve (Elevation of curve) Elev TopPipe + D ft. 4. L Length of new vertical curve. 5. P Point at top of new curve. 6. Station at P Station at PVI (Point vertical intersection) Elev PVI (Elevation of PVI) ft. 9. X P Horizontal distance from PVC to P. L (Station at PVI Station at P) L ( ) L (4.5 stations) 10. g 1 (Gradient 1) +3.5% 11. g (Gradient ).5% 1. R (Rate of change g g1 of grade) L L 6 L R X P 13. Y p (Elevation of P) Y + ( ) + PVC g1 XP New PVC is unknown, so express in terms of PVI, g 1, and L L R X P Elev + ( ) + PVI g1 g1 XP

23 Solved Problems Surveying 3 X P unknown, so express in terms of L L R 4.5 L g L g Elev 1 1 PVI + + L 4.5 L L 3.5 L Simplifying equation for Y P L (19.79 L) Solving for L with ( ) a c a 4 b b X ± L 14 stations or 5.8 stations 1,400 ft. (max length as requested)

24 Solved Problems Surveying 4 Requirement av 1. Sight distance for Curve (S) Assume S < L L (Curve length) A (Total change in grade of the curve) 1,00 ft. 6.0% (from AA) S Book) L (if S < L; AASHTO Green A , ft < 1,00, so formula is correct. S 786 ft. (rounded down)

25 Solved Problems Surveying 5 Module Part b - Vertical Curves Situation A parabolic vertical sag curve is used to make the transition between two equal tangent sections in a highway alignment. Using the design data and the information on the sketch below, solve the following requirements: Design Speed 50 m.p.h. Elevation on underside of bridge structure ft. Bridge located at station Requirements B. At what station on the vertical would a pair of catch basins or drainage inlet structures be required? C. What is the curve elevation at the lowest point on the vertical curve? D. If the river flowing under the highway floods to elevation 133.0, what is the minimum station to which the flooding would extend? E. If the river flowing under the highway floods to elevation 133.0, what is the maximum station to which the flooding would extend? F. What is the bridge clearance at station ? G. Using the design speed and stopping sight distance (rounded for design), what is the maximum length of the vertical curve required to make the grade transition?

26 Solved Problems Surveying 6 Requirement bi 1. Station of PVI (Point of vertical intersection) L (Length of curve) 8 stations. 3. Station of PVC L (Point of vertical curvature) Station of PVI g 1 (Gradient 1).% 5. g (Gradient ) +1.8% 6. P Point at lowest point of curve. 7. X P (Horizontal distance g1 L from PVC to P) g g stations 8. Station at P Station of PVC + X P

27 Solved Problems Surveying 7 Requirement bii 1. g 1 (Gradient 1).%. g (Gradient ) +1.8% 3. L (Length of curve) 8 stations. 4. R (Rate of change in grade) g g1 L 1.8 (.) Elev PVI (Elevation of Point of vertical intersection) ft. 6. Elev PVC (Elevation of Point L of vertical curvature) Elev PVI g (. 4) ft. 7. P Point at lowest point of curve. 8. X P (Horizontal distance from PVC to P) 4.4 stations 9. Y P (Vertical distance R X P from PVC to P) ( ) Elev + g X + PVC 1 P ( ) ft.

28 Solved Problems Surveying 8 Requirement biii 1. P Point where elevation is 133 ft.. g 1 (Gradient 1).% 3. g (Gradient ) +1.8% 4. R (Rate of change in grade) 0.5 (from BB) 5. Elev PVC (Elevation of Point of vertical curvature) ft (from BB) 6. X P Horizontal distance PVC to P 7. Y P (Vertical distance PVC to P) ( ) R XP ElevPVC + g1 XP X P (-. X ) + P 8. Simplifying equation for X P X P (8.8 X) X P Solving for X P with 9. Station of PVC (Point of vertical curvature) X ( b 4 a c) b ± a stations & stations (from BA) 10. Minimum station of flooding Station of PVC + Minimum X P

29 Solved Problems Surveying 9 Requirement biv 1. Station of PVC (Point of vertical curvature) (from BA). X P (Maximum) stations (from BC) 3. Maximum station of flooding Station of PVC + Maximum X P

30 Solved Problems Surveying 30 Requirement bv 1. P Point on curve at Station Station of PVC (Point of vertical curvature) (from BA) 3. Elev PVC ft (from BB) 4. g 1 (Gradient 1).% 5. R (Rate of change in grade) 0.5 (from BB) 6. X P Horizontal distance from PVC to P Station at P Station at PVC stations 7. Y P (Vertical distance PVC to P) ( ) R XP ElevPVC + g1 XP ( ) ft 8. Elev Bridge (Elevation bridge structure) ft. 9. Bridge clearance Elev Bridge Y P ft.

31 Solved Problems Surveying 31 Requirement bvi Table bvi Stopping Sight Distance on a Sag Vertical Curve Design Speed (mph) Assumed Speed for Condition (mph) Stopping Sight Distance Rounded for Design (ft.) Rate of Vertical Curvature, K (length [ft] per percent of A) Computed Rounded for Design From A Policy on Geometric Design of Highways and Streets, Copyright 1994 by the American Association of State Highway and Transportation Officials, Washington, D.C. Used by permission. 1. K (Rate of vertical curvature) 110 (maximum value in Table BF-1). g 1 (Gradient 1).% 3. g (Gradient ) +1.8% 4. A (Total change in grade of curve) g g 1 1.8% (.%) 4.0% 5. L (Length of curve) K A ft.

32 Solved Problems Surveying 3 Module 3 - Spiral Curves Situation Two tangent sections in a horizontal highway alignment study intersect at an angle of The change in direction will be accomplished using a circular curve with a transition spiral at each end. Using the data and sketch below, solve the following requirements: Design Speed 60 m.p.h. Station of the PI Degree of curvature 6 00 Requirements a. Based on the design speed, what is the minimum length of the spiral required? b. Using a 400 foot long spiral curve, what is the value of the spiral angle (θ s )? c. What is the total tangent distance (from PI to TS)? d. What is the station of the SC? e. What is the length of the circular arc between the SC and CS? f. What is the station of the ST? g. What is the value of the total external distance (E S )? h. A theodolite is set up at the TS with a backsight on the PI. What is the deflection angle to set a stake at station ?

33 Solved Problems Surveying 33 Requirement 3a 1. D (Degree of curvature) 6. V (Design speed) 60 mph 3. C (Rate of increase of centripetal acceleration) (for comfort and safety) 4. R C (Curve radius) 5, D ft. 5. L (Minimum length of spiral) V R C (AASHTO Green book) C ft (rounded down)

34 Solved Problems Surveying 34 Requirement 3b 1. L S (Length from TS to SC) 400 ft.. D C (Degree of curvature) 6 L D 3. θ S (Spiral angle) S C

35 Solved Problems Surveying 35 Requirement 3c 1. θ S (Spiral angle) 1 (from 3A). L (Length of spiral) 400 ft. (from 3B) 3. p (Offset distance from initial tangent to Point Curvature) 4. k (Abscissa of shifted PC referred to Tangent to spiral point [TS]) L (From Hickerson, Route location and design, 5 th Edition, McGraw-Hill.) ft L (From Hickerson) R C (Curve radius) ft. (from 3A) 6. (Angle between tangents) D C (Degree of curvature) 6 8. T S (Total tangent distance) ( R C p) tan + k o 40 0'00" ft + ( 6.97) tan

36 Solved Problems Surveying 36 Requirement 3d 1. Station of PI (Point of intersection) T S (Total tangent distance) ft. (from 3C) 3. Station of TS (Tangent to spiral point) Station of PI T S L S (Length of spiral) 400 ft. (from 3B) Station of SC (Spiral to curve) Station of TS + L S

37 Solved Problems Surveying 37 Requirement 3e 1. (Angle between tangents) θ S (Spiral angle) 1 (from 3A) 3. D C (Degree of curvature) 6 4. C (Central angle of circular arc) ( θ S ) L C (Length of circular arc) C 100 D C ft.

38 Solved Problems Surveying 38 Requirement 3f 1. Station of SC (Spiral to curve) (from 3D). L C (Length of circular arc) +7. (from 3E) 3. L S (Length of spiral) (from 3B) 4. Station of CS (Curve to spiral) Station of SC + L C Station of ST (Spiral to tangent) Station of CS + L S

39 Solved Problems Surveying 39 Requirement 3g 1. R C (Curve radius) ft. (from 3A). p (Offset distance from initial tangent to Point Curvature) 6.97 ft. (from 3C) 3. (Angle between tangents) E S (Total external distance) ( R C p) sec 1 + p + ( ) ( sec[ 0.167] 1) ft.

40 Solved Problems Surveying 40 Requirement 3h 1. Station of Point P Station of TS (Tangent to spiral point) (from 3D) 3. L S (Length of spiral) (from 3B) 4. θ S (Spiral angle) 1 (from 3A) 5. L (Length along spiral curve) Station of P Station of TS L 6. θ (Deflection angle) θ S 3 L S

41 Solved Problems Surveying 41 Module 4 - Grade Separation Situation As part of a design process for a new by-pass highway, a grade separation structure (bridge) is required to carry traffic on an existing road (Smith Lane) over the by-pass. Based on the design data and sketch below, solve the following requirements: By-pass Grade 4.5% from West to East Smith Lane Grade Crest Vertical Curve PVI Station PVI Elevation Curve Length 600 ft. g 1 +3.% g 3.% Intersection Equation: Bypass Station ; Smith Lane 6+10 Minimum Bridge Clearance 16 ft. By-pass Transverse or Cross Slope 1/4 per foot Smith Lane Cross Slope 3/8 per foot

42 Solved Problems Surveying 4 Requirements a. What is the centerline station on Smith Lane opposite Point X? b. What is the centerline elevation on the vertical curve on Smith Lane at Station ? c. What is the elevation at X on Smith Lane bridge? d. Using the minimum bridge clearance permitted, what is the maximum elevation at X on the by-pass? e. What is the maximum elevation on the centerline of the by-pass at station ?

43 Solved Problems Surveying 43 Requirement 4a Figure 4a 1. θ (Angle between roads) A (Angle DXE) W B (Width by-pass lane) 30 ft. 4. W S (Width Smith Lane) 16 ft. 5. Station at C CD W B cos( A) 30 cos(.5) 3.47 ft 7. DE W S tan(a) 16 tan(.5) 6.63 ft. 8. Center of Smith opposite X Station at C CD DE

44 Solved Problems Surveying 44 Requirement 4b 1. Elev PVI (Elevation of Point of vertical intersection) ft.. Station of PVI g 1 (Grade 1) +3.% 4. g (Grade ) 3.% 5. L (Curve length) 600 ft. g g1 6. R (Rate of change of grade) L P Point at Station Elev PVC (Elevation of Point of L vertical curvature) Elev PVI g ft. L 9. Station of PVC Station of PVI 10. X P (Horizontal distance from PVC to P) Station of P Station of PVC R X P 11. Y P (Elevation at P) ElevPVC + (g 1 R) ( ) ft.

45 Solved Problems Surveying 45 Requirement 4c 1. P Point X on Smith Lane Bridge. Y P (Elevation at P) ft. (from 4B) 3. θ (Cross slope) 3/8 per foot. 4. W (Pavement width) 3 ft. 5. Elev X (Elevation of X) W Y P θ ft.

46 Solved Problems Surveying 46 Requirement 4d 1. P Point X on Smith Lane Bridge. Elev X (Elevation of X) ft. (from 4C) 3. C Min (Minimum Clearance) 16 ft. 4. D Slab (Depth of slab) 1 inches. 5. D Beams (Depth of beams) 36 inches. D D 6. D (Depth of bridge structure) + Slab Beams ft. 7. Elev Max (Maximum elevation of X) Elev X D C Min ft

47 Solved Problems Surveying 47 Requirement 4e Figure 4E-1 1. P Point at Station A (Intersection angle) W S (Width Smith Lane) 16 ft. 4. W B (Width by-pass lane) 30 ft. 5. gbp (Grade by-pass lane) 4.5% 6. Station at C Station at IP (Intersection point) Elev EP (Elevation at edge pavement) ft. (from 4D) 9. θ (Cross slope) 1/4 per foot. 10. CF W S cos( A) 16 cos(.5) 17.3 ft 11. FG W B tan(a) 30 tan(.5) 1.43 ft. 1. Station at G Station at IP CF FG θ WB 13. Elev G (Elevation of G) Elev EP

48 Solved Problems Surveying ft. 14. X P (Horizontal distance G to P) Station at G Station at P Y P (Elevation at P) Elev G + g BP X P ( ) ft.

49 Solved Problems Surveying 49 Module 5 - Earthwork Situation The centerline of a proposed highway has been established in the field. Cross sections showing pre-construction conditions have been taken at every half station (50 feet), and other selected stations where there is any significant change in ground conditions. A template showing the final design of the highway has been plotted on each of the cross sections. Based on the design information plotted on each of the cross sections and using a planimeter, the areas of cut and fill have been determined. Shown below is a listing of these areas for a 400-foot section on the highway. Using the given design data, solve the following requirements. Earthwork Data Station Cut Area (ft ) Fill Area (ft ) , , , , , , Requirements a. If the shrinkage of the soil is 1%, how much of a surplus exists, if any, when using the average end area method to calculate the amount of material needed to construct the highway subgrade. b. What is the station of the balance point on the mass diagram? (Assume the mass diagram ordinate at station ). c. For this requirement, assume a freehaul of 1,000 feet, the center of mass of the excavated material is at station +85, and the center of mass of the embankment is at station Based on these assumptions and the portion of a mass diagram shown with the requirement, what is the overhaul in cubic yard stations?

50 Solved Problems Surveying 50 Requirement 5a Table 5a Station Cut Area Cut Volume Fill Area Fill Volume (ft ) (yds 3 ) (ft ) (yds 3 ) , ,375, , , , ,60, ,139, , Total 9,783 8,783 A i L Cut Volume i ( A A ) i + i- 1 B i Cut Area at Station i Length between stations 50 ft. L (divide by 7 to convert from ft. to yds 3 ) 7 Fill Area at Station i B + B 1 ( ) i i- Fill Volume i L 11 (add 1% for shrinkage) 1. Surplus Total Fill Volume Total Cut Volume (Table 5A-1) 9,783 8,783 1,000 yds 3

51 Solved Problems Surveying 51 Station Cut Volume (yds 3 ) Requirement 5b Table 5B-1 Fill Volume (yds 3 ) Mass Diagram Ordinate (yds 3 ) , ,650-3, ,380-5, , ,6 1+50, , , , , j 1. Mass Diagram Ordinate j ( Volume Fill ) i 0 Cut i Volume i yds 3. Balance Point Station where Mass diagram ordinate is zero

52 Solved Problems Surveying 5 Requirement 5c Figure 5C-1 1. CM Embankment (Center of mass for embankment) CM Excavation (Center of mass for excavation) L Freehaul (Length of freehaul) 10 stations 4. L Overhaul (Length of overhaul) CM Embankment CM Excavation L Freehaul HL Ordinate (Horizontal-line ordinate in Figure 5C-1),60 yds 3 6. V Overhaul (Volume of overhaul) HL Ordinate L Overhaul, ,984 cubic yard stations