Help asap b.) RE =perimeter of all coursesPerimeter = 95 + 135 + 47 + 68Perimeter = 345 m1.295RE =345266Technology Drven by (nnProblem 21• A close traverse has the following data:CourseBearingDistance1-2N 9.27° E58.7 m2-3S 88.43° E27.3 m3-4N 86.78° E35.2 m4-5S 5.2° E35.0 m5-110-11 / 27• What is the bearing of line 5-1?Technology Driven by JnnSolutionCourseBearingDistanceLatDep1-2N 9.27° E58.7 m+ 57.933+ 9.4562-3S 88.43° E27.3 m0.748+ 27.2903-4N 86.78° E35.2 m1.977+ 35.1444-5S 5.2° E35.0 m34.8563.172+5-1-----The respective sums of Departure and Latitude must be zero.Latz-1 = - (57.933 - 0.748 + 1.977 - 34.856) =- 24.307Deps1 = - (9.456 + 27.290 + 35.144 + 3.172) = - 75.062Distance = v24.3072 +75.0622 = 78.900 mBearing = S arctan pe w= S arctan(25.062) w24.307)Dep(75.062YLatBearing = S 72.057° WTechnolagy Driven by (nn.Compass Rule• the correction to be applied in the latitude or departureof any course is the total correction in latitude ordeparture multiply to the ratio length of the course is to

Close traverse data: Course Bearing Distance 1-2 N 9.27o E 58.7 m 2-3 S 88.43o E 27.3 m 3-4…

A close traverse has the following data: Course Bearing Distance 1-2 N 9.27° E 58.7 m 2-3 S 88.43° E 27.3 m 3-4 N 86.78° E 35.2 m 4-5 S 5.2° E 35.0 m 5-1 10-11 / 27 What is the bearing of line 5-1?

A close traverse has the following data: Course Bearing Distance 1-2 N 9.27° E 58.7 m 2-3 S 88.43° E 27.3 m 3-4 N 86.78° E 35.2 m 4-5 S 5.2° E 35.0 m 5-1 10-11 / 27 What is the bearing of line 5-1?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A closed traversed has the following data. LINE DISTANCE BEARING AB 185.33 N 43 E BC 258.28 S 67 E CD ? S 38 W DE ? S 81 W EA 150.87 N 27 W The summation of the latitude and departure of the traverse are -26 and 55, respectively. Using COMPASS RULE, determine the follwowing:(Round off into four decimal places)WHAT IS THE LENGTH OF THE LINE CD?
Answers a) 0.436 b) 17.44 kPa c) 31.98 kPa d) 3.57 kN/m3
Compute for the given closed traverse the following:a. Bearing of each lineb. Latitude and Departure of each line,c. Linear error of closured. Precision e. Adjust the traverse using the transit rule. If the coordinates of point B are Northing =3000.00 m. and EASTING = 1500.00 m., determine the coordinates of all other points. LINE INTERIOR ANGLE LENGTH (m.) BEARING A-B A = 125 deg 54 min 162.60 N 28o 17’ E B-C B = 135 deg 00 min 253.98 C-D C = 114 deg 27 min 256.32 D-E D = 121 deg 52 min 305.94 E-F E = 88 deg 59 min 335.40 F-A F =133 deg 48 min 182.04
A closed field traverse has the following lengthsAB = 636.45; BC = 654.49; CD = 382.85; and DA = 512.77.The interior angles are as follows (measured as angles to the right):A = 81º 23'; B = 72º 33'; C = 89º 40'; and D =116º 24'.The bearing of AB is S 33º 19' W and the side BC is in theSE quadrant.a) Which is the bearing of line CD? b) Which is the bearing of line BC?c) Which nearly gives the value of LEC?Choose the correct ansS 65deg 18min ES 74deg 08min Es 48deg 04min WN 15deg 32E
A closed field traverse has the following lengthsAB = 636.45; BC = 654.49; CD = 382.85; and DA = 512.77.The interior angles are as follows (measured as angles to the right):A = 81º 23'; B = 72º 33'; C = 89º 40'; and D =116º 24'.The bearing of AB is S 33º 19' W and the side BC is in theSE quadrant.a) Which is the bearing of line CD? b) Which is the bearing of line BC?c) Which nearly gives the value of LEC?
PROBLEM 1 Given the following interior angles (angles-to-the-right) of a five-sided closedpolygon traverse and if the azimuth of side AB is fixed at 74°31'17", determine the adjusted azimuth of line EA.A =105°13'14"; B = 92°36'06"; C = 67°15'22"; D = 217°24'30"; E = 57°30"38". (Note: line BC bears NW.) NEEDED TO ANS Compute the linear misclosure for the traverse of Problem 1 if the lengths of the sides (in feet) are as follows: AB = 2157.34; BC =1722.58;CD =1318.15; DE =1536.06; and EA =1785.58. (Note: Assume units of feet for all distances.) Round off the answer to three decimal places
Problem #1 For the given data of a close traverse, determine the missing bearings of courses DE and EF then solve the AREA using DPD. Sketch the traverse. LINES DISTANCE (m) BEARING AB 60 S 49⁰16’ E BC 59 S 67⁰48’ E CD 95 N 32⁰28’ W DE 110 ? EF 128 ? FA 71 S 18⁰56’ E
A closed traverse has the following data: Course Length Bearing Interior Angle to the Right AB 125.98 N42°15′E A = 112°07′ BC 75.25 N35°20′W B = 102°25′ CD 80.34 S74°44′W C = 110°04′ DE 112.68 S19°56′W D = 125°12′ EA 78.35 S69°52′E E = 90°12′ a.) Adjust the traverse using transit rule. b.) Compute the area (m2) of the traverse using DPD method.
PROBLEM 1 Given the following interior angles (angles-to-the-right) of a five-sided closedpolygon traverse and if the azimuth of side AB is fixed at 74°31'17", determine the adjusted azimuth of line EA.A =105°13'14"; B = 92°36'06"; C = 67°15'22"; D = 217°24'30"; E = 57°30"38". (Note: line BC bears NW.) THE ANSWER HERE IS Azimuth EA=149.300278 NEEDED TO ANSWER it is connected to problem 1: Compute the relative precision for the traverse of Problem 1 if the lengths of the sides (in feet) are as follows: AB = 2157.34; BC =1722.58;CD =1318.15; DE =1536.06; and EA =1785.58. (Note: Assume units of feet for all distances.) Round off the denominator to two decimal places.
PROBLEM 1 Given the following interior angles (angles-to-the-right) of a five-sided closedpolygon traverse and if the azimuth of side AB is fixed at 74°31'17", determine the adjusted azimuth of line EA.A =105°13'14"; B = 92°36'06"; C = 67°15'22"; D = 217°24'30"; E = 57°30"38". (Note: line BC bears NW.) NEEDED ANS: Compute the relative precision for the traverse of Problem 1 if the lengths of the sides (in feet) are as follows: AB = 2157.34; BC =1722.58;CD =1318.15; DE =1536.06; and EA =1785.58. (Note: Assume units of feet for all distances.) Round off the denominator to two decimal places.
Answers a) 0.864 b) 18.76 kN/m3 c) 5.91 meters
Answer are nearly around 2237N ,0 N and and 1887N