Show complete solution A 50-m steel tape that weighs 0.04 kg/m is used to lay-out a distance of 275 m. The tape is supported at its end points and atthe 15-m and 25-m marks of every tape length. The tape is of standard length under a pull of 6 kgs. During the survey if a pull of5.5 kgs. is applied, determine the following:a. Correction due to sag for each unsupported segment of the tape.b.C.Total correction due to sag for one tape length.Correct distance to be laid-out in order to have the desired length of the line.

Answered: Image /qna-images/answer/e7fecc52-db14-4edf-b79d-1e18019294f7.jpg

A 50-m steel tape that weighs 0.04 kg/m is used to lay-out a distance of 275 m. The tape is supported at its end points and the 15-m and 25-m marks of every tape length. The tape is of standard length under a pull of 6 kgs. During the survey if a pull 5.5 kgs. is applied, determine the following: a. Correction due to sag for each unsupported segment of the tape. b. Total correction due to sag for one tape length. C. Correct distance to be laid-out in order to have the desired length of the line.

A 50-m steel tape that weighs 0.04 kg/m is used to lay-out a distance of 275 m. The tape is supported at its end points and the 15-m and 25-m marks of every tape length. The tape is of standard length under a pull of 6 kgs. During the survey if a pull 5.5 kgs. is applied, determine the following: a. Correction due to sag for each unsupported segment of the tape. b. Total correction due to sag for one tape length. C. Correct distance to be laid-out in order to have the desired length of the line.

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter10: Shear Strength Of Soil

Section: Chapter Questions

Problem 10.24CTP

See similar textbooks

Similar questions

can you pls help me understand how this is solve.. Topic: Tape corrections (fundamentals of surveying) •Pls answer with Complete Solution •3 decimal places all answer 1. A 30m steel tape standardized at 20degC and supported throughout under a tension of 5.45 kg was found to be 30.012m long. The tape had a cross-sectional area of 0.050 sq.cm and a weight of 0.03967 kg/m. This tape was held horizontal, supported at the ends only, with a constant tension of 9.09kg, to measure a line from A to B in three segments. The data listed in the following table were recorded. Apply corrections for tape length, temperature, pull, and sag to determine the correct length of the line. *image attached for reference* Thank you. I'll give your answer a LIKE for helping me.
elementary surveying A 30-m tape weighing 1.5kg is of standard length under a pull of 6kg, supported for full length. The tape was used in laying a line 2,893.4m long on smooth level ground under a steady pull of 17kg. Assuming E = 3.2x10⁶ kg/cm² and the unit weight of steel to be 7.9x10⁻³ kg/cm³. 1.Determine the correction for increase in tension. 2.Determine the correct length of the line measured.
A baseline was measured using 100m tape which is standardized at 15°C with standard pull of 10kg. It is required to have a line with length of 925.15m. At the time during measurement the temperature was 28°C and the pull exerted along the ground was 16kg. The weight of steel is 7.86 g/m3, weight of tape is 2.67kg, E = 2 x 106 kg/cm?, a =7 x 10-7/°C.
A 50m tape weighing 0.03kg/m is used to measure distance AB. The tape is supported at both end points and at its midspan and then applied tension is 6kg. Determine the total sag correction if the measured distance is 1,320.50m.
i need clear ans by hand and solve very very fast in 20 min and thank you | ᴅʏʙᴀʟᴀ ?✨؛
Calculate the cost required to construct a premix layer having bulk specific gravity of 2.239 and thickness of 46mm. The material cost is RM170/ton and cost to lay and compact is RM5/m2. The width and length are 7m and 2km respectively. O RM 315 126 ORM 245 126 ORM 299 125 RM 300 317
A surveyor used a 30 m tape in measuring an inclined distance. The measured length onthe slope was recorded to be 459.20 m long. The difference in elevation between theinitial point and the end point was found to be 1.25 m. The 30 m tape is of standardlength at a temperature of and a pull of 50 N. During measurement, the temperaturereading was and the tape was supported at both ends with an applied pull of 75N.The cross sectional area of the tape is and the modulus of elasticity is 200 GPA. The tape has a mass of 0.075kg/m. K = 0.0000116 Determine the correction for slope and horizontal distance.
The data in the table below is for a closed-polygon traverse. Course Azimuth Length (m) AB 75°14'47" 519.674 BC 455.265 CD DE EA Interior Angles A = 12°25'31" B 48°19'25" 433.426 C = 126°37'24" 425.558 D = 46°03'36" 109.434 E = 306° 34'19" Part A Compute the preliminary azimuths. O AZAB = 57°14'47". Azpc = 303°34′09", AzcD=250°11'30". AZDE=116°15'03". AZEA = 242° 49'19" O AZAB = 75°00'47", AZ BC = 303°00'09", AzcD = 250°00'30", AZDE = 116°00'03", AZEA = 242°00'19" O AZAB = 75°14'47". AzBc = 313° 34'09", AzcD = 25°11'30", AZDE = 116°10'00", AZEA = 224° 49'19" O AZAB = 75°14'47". AzBC = 303°34'09", AzcD = 250°11'30". AZDE = 116°15'03". AZEA = 242° 49'19" Part B Compute the unadjusted departures. Express your answers in meters to six significant figures and separated by commas. AB, BC, CD, DE, EA= Part C vec IVE ΑΣΦΠ ← A Compute the unadjusted latitudes. Express your answers in meters to six significant figures and separated by commas. AB, BC, CD, DE, EA= ΠΑΠΑΣΦ Π vec ^ A ? ? m, m, in mi,…
Please solve the follow up qestion.
A 50 m tape was standardized and found to be 0.0015 m longer at an observed temperature of 31.5°C under a pull of 10 kg. The same tape is used to measure a line found to be 523.61 m at an average temperature of 24.5 °C using a pull 12 kg. What is the correct length of the line? detailed solution pls thank you
Answer the do (0.0026 L'= L te=) 30 + (-0.0026) s)? (L+= 29.9974 and Explain types of errors in me nents, 1 (L'/L)ME TL = ( 29.9974/30) *500 H TL = 499.9566657 382) Arizontal distance of (500 m) must be lay out with a steel tape, length of the steel tape der standard conditions is (30 m). Field conditions indicate that standard conditions apply except the measured temperature is 50 °C and applied tension 120'N, Determine the correct T= (T-Ts) Lx T= (50-2032) A =T= 0.0lounder P= (P-P₂)L AE P-(120-50)3 The standard temperature 20 °C. 0.02*21*Coefficient of thermal expansion of steel tape = 11.6x10-6 °C 70.005) Standard tension = 50N ∙W C-W² Mass of the tape = 0.05 kg/m 2p² -(0.05 jan Total = 0. olo44 +0.005-0. Total (€) = mans. له میشه راه horizontal distance to be laid out, if Modulus of elasticity (E) = 210*106 N/m2 24 (120) Cross-sectional area of the tape = 0.02 cm² 0.022 022 1.05 +X5 +1·15=3₁3 X5=3-3-22 X5=11 1.1-2.4 -1.3 567 5-1.3= 55.4 fall 1₁4.5-2.28 = -0183
The flanged steel cantilever beam with riveted bracket is subjected to the couple and two forces shown, and their effect on the design of the attachment at A must be determined. Replace the two forces and couple by an equivalent couple M and resultant R at A. The couple is positive if counterclockwise, negative if clockwise. 1.90 KN 0.69 m 2.27 m- 66° A 1 L--x 490 N·m 1.34 KN 0.13 m 0.13 m