Surveying FE Review. Fall CIVL 4197 FE Surveying Review 1/9

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1 CIVL 4197 FE Surveying Review 1/9 Surveying FE Review Fall 017 Problem 18.01: Two sides of a triangular-shaped parcel are 80 ft. and 100 ft. with a 60 angle between them. The length of the third side of the parcel (ft.) is most nearly: A ft. B ft. C ft. D ft. Problem 18.01: Two sides of a triangular-shaped parcel are 80 ft. and 100 ft. with a 60 angle between them. The length of the third side of the parcel (ft.) is most nearly: A ft. B ft. C ft. D ft. Problem 18.01: Two sides of a triangular-shaped parcel are 80 ft. and 100 ft. with a 60 angle between them. The length of the third side of the parcel (ft.) is most nearly: A ft. B ft. C ft. D ft. Two triangle sides, b and c, and angle A, are known (see figure). Use the Law of Cosines to determine the length of side a. The formula for the Law of Cosines is given in the NCEES Handbook, Mathematics, page. Problem 18.01: Two sides of a triangular-shaped parcel are 80 ft. and 100 ft. with a 60 angle between them. The length of the third side of the parcel (ft.) is most nearly: A ft. B ft. C ft. D ft. Problem 18.0: Two sides of a triangular-shaped parcel are C. 0, 90 Using the Law of Cosines: a = b + c - bc cos A o a = cos ft.

2 CIVL 4197 FE Surveying Review /9 Problem 18.0: Two sides of a triangular-shaped parcel are Problem 18.0: Two sides of a triangular-shaped parcel are C. 0, 90 a = C. 0, 90 a = Three triangle sides, a, b, c, and angle A, are known (see figure). Since the three sides of the triangle are given, use the Law of Sines to determine the associated angles. The formula for the Law of Sines is given in the NCEES Handbook, Mathematics, page. Problem 18.0: Two sides of a triangular-shaped parcel are C. 0, 90 a = Problem 18.0: Two sides of a triangular-shaped parcel are C. 0, 90 a = a b c = = sina sinb sinc 100 = C = 70.9 sin C o = = = sin60 sinb sinc 80 sin B o = B = 49.1 Problem 18.0: In an equilateral triangular-shaped parcel, the height (altitude) of the triangle is 5 ft. less than its side length. The side length (ft.) of the triangle is most nearly equal to: A. 10 ft. B. 0 ft. C. 0 ft. D. 40 ft. Problem 18.0: In an equilateral triangular-shaped parcel, the height (altitude) of the triangle is 5 ft. less than its side length. The side length (ft.) of the triangle is most nearly equal to: A. 10 ft. B. 0 ft. C. 0 ft. D. 40 ft. An equilateral triangle has three sides of equal length. This implies that each internal angle is 60. The figure illustrates the relationship between the side and the altitude.

3 CIVL 4197 FE Surveying Review /9 Problem 18.0: In an equilateral triangular-shaped parcel, the height (altitude) of the triangle is 5 ft. less than its side length. The side length (ft.) of the triangle is most nearly equal to: A. 10 ft. B. 0 ft. C. 0 ft. D. 40 ft. CD a-5 sin A = sin B = = = sin 60 AC a o a - 5 = a 0.14 a = 5 a = 7.1 ft. Problem 18.04: Which of the following statements is NOT correct? A. A leveling staff is a crew assigned to a route surveying task under the supervision of a licensed professional surveyor. B. A leveling staff is a rod used in surveying. C. A level line is one where all points are normal to the direction of the force of gravity. D. A freely-suspended plumb-bob shows the direction of the gravitational force. Problem 18.04: Which of the following statements is NOT correct? A. A leveling staff is a crew assigned to a route surveying task under the supervision of a licensed professional surveyor. B. A leveling staff is a rod used in surveying. C. A level line is one where all points are normal to the direction of the force of gravity. D. A freely-suspended plumb-bob shows the direction of the gravitational force. Statements B, C and D are correct. Statement A is incorrect. Problem 18.05: A surveyor takes several leveling readings. The instrument is on a known point of elevation of 1.45 ft., and the height of the instrument (HI) is 5.15 ft. To determine the elevation of the underside of a beam, an inverted sight (IS) reading of.1 ft. is obtained. To determine the elevation of a point on a slope, a reading of 4. ft. is obtained. The elevations of the underside of the beam and the point on the slope (ft.) are respectively most nearly: A ft., ft. B ft., 14.8 ft. C ft., 1.9 ft. D ft., ft. Problem 18.05: A surveyor takes several leveling readings. The instrument is on a known point of elevation of 1.45 ft., and the height of the instrument (HI) is 5.15 ft. To determine the elevation of the underside of a beam, an inverted sight (IS) reading of.1 ft. is obtained. To determine the elevation of a point on a slope, a reading of 4. ft. is obtained. The elevations of the underside of the beam and the point on the slope (ft.) are respectively most nearly: Problem 18.05: Elevation at Sta. B = Elevation at Sta. A + HI + IS = = 11.7 ft. Elevation at Sta. C = Elevation at Sta. A + HI - FS = = 14.8 ft.

4 CIVL 4197 FE Surveying Review 4/9 Problem 18.05: Elevation at Sta. B = Elevation at Sta. A + HI + IS = = 11.7 ft. Elevation at Sta. C = Elevation at Sta. A + HI - FS = = 14.8 ft. Problem 18.06: A closed traverse has six segments and four interior angles measuring 90 each. The sum of the remaining interior angles is most nearly: A. 0 o B. 90 o C. 180 o D. 60 o A ft., ft. B ft., 14.8 ft. C ft., 1.9 ft. D ft., ft. Problem 18.06: A closed traverse has six segments and four interior angles measuring 90 each. The sum of the remaining interior angles is most nearly: A. 0 o B. 90 o C. 180 o D. 60 o For a polygon of n sides, the internal angle is expressed as: Sum of the interior angles 180 n Sum of two unknown angles = 70 o 60 o = 60 o Problem 18.07: The table below shows differential leveling data using a transit level. The starting station is of known elevation. Find the elevation of station D. Station BS (m) FS (m) Elevation (m) Notes A Benchmark B C D 6.78 A m B m C m D m Problem 18.07: The table below shows differential leveling data using a transit level. The starting station is of known elevation. Find the elevation of station D. Station BS (m) FS (m) Elevation (m) Notes A Benchmark B C D 6.78 Problem 18.07: The table below shows differential leveling data using a transit level. The starting station is of known elevation. Find the elevation of station D. Station BS (m) HI (m) FS (m) Elevation (m) A B C D A m B m C m D m BS FS ELED ELEA BS FS A m B m C m D m

5 CIVL 4197 FE Surveying Review 5/9 Problem 18.08: A pile is made of recycled aggregate material. The pile has a diameter of 100 ft. at the base, and 5 ft. at the top. It is 5 ft. high. The volume of the pile (yd ) is most nearly: A. 105,000 yd B. 5,787 yd C.,860 yd D.,576 yd Problem 18.08: A pile is made of recycled aggregate material. The pile has a diameter of 100 ft. at the base, and 5 ft. at the top. It is 5 ft. high. The volume of the pile (yd ) is most nearly: A. 105,000 yd B. 5,787 yd C.,860 yd D.,576 yd Generally, if the properties of two end sections are given from which the end areas can be readily computed, the volume equals the average cross-sectional area multiplied by the height. Problem 18.08: A pile is made of recycled aggregate material. The pile has a diameter of 100 ft. at the base, and 5 ft. at the top. It is 5 ft. high. The volume of the pile (yd ) is most nearly: A. 105,000 yd B. 5,787 yd C.,860 yd D.,576 yd In the NCEES Handbook, Civil Engineering, page 176, is the Average End Area method and is expressed as: L A A V 1 Problem 18.08: A pile is made of recycled aggregate material. The pile has a diameter of 100 ft. at the base, and 5 ft. at the top. It is 5 ft. high. The volume of the pile (yd ) is most nearly: A. 105,000 yd B. 5,787 yd C.,860 yd D.,576 yd d L A A V 1 A1 7,854 ft 4 4 d 5 A 491ft ft. 7,854ft 491ft 104,1ft,86 yd Problem 18.09: What is the southern azimuth of a line with a bearing of S1 4' 56"E? A. 1 4'56" B. 77 5'04" C '04" D. 47 5'04" Problem 18.09: What is the southern azimuth of a line with a bearing of S1 4' 56"E? A. 1 4'56" B. 77 5'04" C '04" D. 47 5'04" Azimuths are generally expressed from the north. However, there are exceptions: some navigation systems use south as the reference plane. The rotation of the azimuth is always clockwise. In this case, the bearing needs to be subtracted from 60 to determine the azimuth from the south.

6 CIVL 4197 FE Surveying Review 6/9 Problem 18.09: What is the southern azimuth of a line with a bearing of S1 4' 56"E? A. 1 4'56" B. 77 5'04" C '04" D. 47 5'04" Problem 18.10: Line AB bears N 1 4' 56'' E, and line AC bears S 1 4' 56'' E. The deflection angle between the lines is: A. Straight East or 90 B. 5 09' 5" (Left) C ' 04" (Left) D ' 08" (Right) The azimuth from the south is: ' 56" = 47 5' 04" Problem 18.10: Line AB bears N 1 4' 56'' E, and line AC bears S 1 4' 56'' E. The deflection angle between the lines is: A. Straight East or 90 B. 5 09' 5" (Left) C ' 04" (Left) D ' 08" (Right) A deflection angle is the difference in angle from the prolongation of the back line to the forward line along a traverse. The difference in angle between AB' and AB is the sum of the angles 1 4'56" and 1 4'56" which is 5 09' 5". Problem 18.10: Line AB bears N ' E, and line AC bears S 1 4' 56'' E. The deflection angle between the lines is: A. Straight East or 90 B. 5 09' 5" (Left) C ' 04" (Left) D ' 08" (Right) A deflection angle is the difference in angle from the prolongation of the back line to the forward line along a traverse. Since line AC is located at the left hand side of the prolongation line, it is deflected to the left. Problem 18.11: The table below shows length and is the correction to the departure of CD, using the transit AB ' BC ' CD ' DA ' A ft B ft C ft D ft Problem 18.11: The table below shows length and is the correction to the departure of CD, using the transit AB ' BC ' CD ' DA ' Departures L sin 850sin80.5 1,50sin16.5 1,000sin 0.5 1,850sin

7 CIVL 4197 FE Surveying Review 7/9 Problem 18.11: The table below shows length and is the correction to the departure of CD, using the transit AB ' BC ' CD ' DA ' DepCD Correction DepCD 1.004ft Departures ft Problem 18.11: The table below shows length and is the correction to the departure of CD, using the transit AB ' BC ' CD ' DA ' A ft B ft C ft D ft 0.19ft Problem 18.11a: The table below shows length and is the correction to the departure of CD, using the compass AB ' BC ' CD ' DA ' A ft B ft C ft D ft Problem 18.11a: The table below shows length and is the correction to the departure of CD, using the compass AB ' BC ' CD ' DA ' Departures L sin 850sin80.5 1,50sin16.5 1,000sin 0.5 1,850sin Problem 18.11a: The table below shows length and is the correction to the departure of CD, using the compass AB ' BC ' CD ' DA ' Correction Dep CD LCD 1.004ft perimeter 1, ft 850 1,50 1,000 1,850 Problem 18.11a: The table below shows length and is the correction to the departure of CD, using the compass AB ' BC ' CD ' DA ' A ft B ft C ft D ft 0.0ft

8 CIVL 4197 FE Surveying Review 8/9 Problem 18.1: A,500-m long trapezoidal open channel is constructed at a specified slope of 1.5%. If the base elevation at the channel inlet is m, the base elevation at the channel outlet should be most nearly: A m B m C. 5.8 m D. 7.5 m Problem 18.1: A,500-m long trapezoidal open channel is constructed at a specified slope of 1.5%. If the base elevation at the channel inlet is m, the base elevation at the channel outlet should be most nearly: A m B m C. 5.8 m D. 7.5 m The problem concerns open channel flow; therefore, the slope should be directed downwards. 1.5 EL EL1,500 m EL 90.57m,500m 5.857m 100 Problem 18.1: Two cross-sections of a proposed roadway are located at Station and One crosssection needs 00 ft of cut and the other needs 15 ft of fill. The net excavation (yd ) required between the two sections is most nearly: A. 0,000 yd B. 1,800 yd C. 1,100 yd D. 57 yd Problem 18.1: Two cross-sections of a proposed roadway are located at Station and One crosssection needs 00 ft of cut and the other needs 15 ft of fill. The net excavation (yd ) required between the two sections is most nearly: A. 0,000 yd B. 1,800 yd C. 1,100 yd D. 57 yd The volume is calculated using the average end area method from the Earthwork Formulas section of the NCEES Handbook, Civil Engineering, page 176: L A A V 1 Problem 18.1: Two cross-sections of a proposed roadway are located at Station and One crosssection needs 00 ft of cut and the other needs 15 ft of fill. The net excavation (yd ) required between the two sections is most nearly: A. 0,000 yd B. 1,800 yd C. 1,100 yd D. 57 yd Problem 18.14: Excavated soil from a barrow pit is stockpiled in a conical shape. The stockpile has a diameter of 0 m at its base and a height of 15 m. Its volume in m is most nearly: A. 5,000 m B.,500 m C. 1,600 m D. 1,000 m V ,98 ft 57 yd

9 CIVL 4197 FE Surveying Review 9/9 Problem 18.14: Excavated soil from a barrow pit is stockpiled in a conical shape. The stockpile has a diameter of 0 m at its base and a height of 15 m. Its volume in m is most nearly: A. 5,000 m B.,500 m C. 1,600 m D. 1,000 m The shape of the stockpile in this problem is conical. As given in the NCEES Handbook, Civil Engineering, page 176, the volume of a cone V is: V base areaheight Problem 18.14: Excavated soil from a barrow pit is stockpiled in a conical shape. The stockpile has a diameter of 0 m at its base and a height of 15 m. Its volume in m is most nearly: A. 5,000 m B.,500 m C. 1,600 m D. 1,000 m base areaheight V 0m 15m m 15 m 1,571m Problem 18.15: A 00 m long runway measures 0 mm on an aerial photo. The scale of the photo is most nearly: Problem 18.15: A 00 m long runway measures 0 mm on an aerial photo. The scale of the photo is most nearly: A. 1:1,500 B. 1:15,000 C. 1:15 D. 1:666 A. 1:1,500 B. 1:15,000 C. 1:15 D. 1:666 Photo scale: Distance at ground 1: Distance in photo 00m 1,000 1: 0 mm mm m 1:15,000 Surveying FE Review Fall 017 Questions?

NCEES FS Practice Exam

NCEES FS Practice Exam Terrametra Resources Lynn Patten 1. One corner of a 60-ft. 120-ft. lot, otherwise rectangular, is a curve with a radius of 20 ft. and a central angle of 90. The area (ft. 2 ) of