FS 2.875 m 4.252 m 5.478 m 6.142 m BS ТР 1 ТР2 4.325 m 9.542 m 9.631 m 8.133 m вм 1 Elev. = 56.287 m ТР 3 Not to scale BM 2 Given the traverse illustration above, fill up and complete the table below. Make sure to show your solution in solving for HI, Elevation, Correction to be applied, and Corrected Elevation. BS HI FS Elevation Corrected Station (m) (m) (m) (m) Elevation (m) BM1 TP1 BM2 TP3 TP4 BM1 7.258 m 2.841 m

FS 2.875 m 4.252 m 5.478 m 6.142 m BS ТР 1 ТР2 4.325 m 9.542 m 9.631 m 8.133 m вм 1 Elev. = 56.287 m ТР 3 Not to scale BM 2 Given the traverse illustration above, fill up and complete the table below. Make sure to show your solution in solving for HI, Elevation, Correction to be applied, and Corrected Elevation. BS HI FS Elevation Corrected Station (m) (m) (m) (m) Elevation (m) BM1 TP1 BM2 TP3 TP4 BM1 7.258 m 2.841 m

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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Situation5. A line is recorded as 472.90 m. long. It is measured with a 0.65 kg. tape which is 30.005 m. long at 20°C under a 50 N pull supported at both ends. During measurement, the temperature is 5 °C and the tape is suspended under a 75 N pull. The line is measured on 3 % grade. Cross-sectional area of the tape is 3 mm^2. a. Compute the actual length of tape during measurement. b. Compute the total error to be corrected for the inclined distance. c. What is the true horizontal distance?
P (kN)=80 L (m)=0.65 A1 (mm2)=540 A2 (mm2)= 650 Eal (GPa)=74 Est (GPa)=205
Q#1: A profile levelling is carried out and consecutive readings were observed on staff rod along a chain line at common interval of 10 meter which are listed below in table. First reading is taken on bench mark of 105.615 meter reduce level. Station Points Chainage (m) Staff Readings (m) Remarks BS IS FS 1 00 3.005 BM 2 10 2.285 3 20 1.560 4 30 1.785 5 40 2.105 6 50 2.875 3.105 CP 1 X 53.35 3.465 7 60 3.955 8 70 3.120 9 80 3.015 10 90 2.580 11 100 1.955 CP 2 Determine the reduce levels of all points using both line of collimation and rise and fall method. (Present your answer in table form).
Can you check back parts b & c My answers are: using r > 3n method where r= reactions & n= nb of parts b) (16) > (3)(4)statically indeterminate by 4th degreec) (12)> (3)(3) statically indeterminate by 3rd degree let me know where is my mistakethank you
shouldnt R1=8.666 & R2=11.334??
0 a 2178 kN O b. 3726 kN 1| select one: WG o weiox92
F (kN)=36 L1 (m)= 37.0053 Eal (GPa)=74
An 80 m spiral is used to connect a 4o simple at both ends. 1. Compute the long tangent. a. 53.37 m b. 42.51 m c. 26.73 m d. 36.41 m 2. Compute the deflection angle at second quarter. a. 2 deg b. 0.67 deg c. 1.33 deg d. 0.33 deg 3. Compute the tangent distance if I = 50deg. a. 261.51 m b. 185.62 m c. 128.14 m d. 174.02 m
P (N)=570 L1 (mm)=75 L2 (mm)=180 Eal (GPa)=74 Est (GPa)=205
8.TAPE MEASUREMENTSa. A line was recorded as 1650.00 meters long. It is measured with a 0.65 kg standard tape30.00m long at 20°C under a 50N pull supported at both ends. During the measurementthe temperature was 5°C and the tape was suspended under a 75N pull. The line wasinclined on a 3% grade. What is the true inclined distance? E = 200GPa, A = 3.0 mm² andthe coefficient of linear expansion is 0.0000116 m/°C.b. A line was recorded as 2000.00 meters long. It is measured with a 0.65 kg standard tape30.00m long at 20°C under a 50N pull supported at both ends. During the measurementthe temperature was 5°C and the tape was suspended under a 75N pull. The line wasinclined on a 3% grade. What is the true inclined distance? E = 200GPa, A = 3.0 mm² andthe coefficient of linear expansion is 0.0000116 m/°C.
The distance between points is exactly 100 ft. This measurement is read correctly on steel surveyors tape at 70°F and with a 10-lb tension. If the tape measures another distance as 100ft at 100°F and with 20-lb tension, what is the actual distance? The tape is 3/8-inch wide and 1/32-inch thick. A. 100.0443ft B. 100.0332ft C. 100.0223ft D. 100.0112ft
Solve for: a.) sum = θ1 + θ2 + θ3 + θ4 b.) horizon error of closure: e = 360° - sum c.) correction = e/n