Question

A civil engineer used a 30-m tape in measuring an inclined distance. The measured length on the slope was recorded to be 459.20 m long. The difference in elevationbetween the initial and the end point was found to be 1.25 m. The 30-m tape is of standard length at a temperature of 10ºC and a pull of 5 kg. During measurement thetemperature reading was 15ºC and the tape was supported at both ends with an applied pull of 7.5 kg. The cross-sectional area of the tape is 0.065 cm2 and E = 2.0 x106 kg/cm2. The tape weighs 0.075 kg/m and has a linear expansion is 0.0000116 / ºC. Apply corrections for temperature, pull, and sag to determine the correcthorizontal distance considering a gentle slope.

Answer 1

→ Solution. Given data → ML= 459.20 m . L = 30m. 459.20 TI: 1.25 The difference between two points tm= 15°c. to=10°c A = 0.08 sema Po = 5kg Im = 7.5kg.. E = 2.00x106 Kglem? w = 0.075 kg/m. d= 0. OOOO 6/'c . 0 Temperaturze Correction → at a(tm-to) XL = 0.0000116 X (15-10)x 30 =+1.74 x 10-3 m. CInerenze) ♡ Pull correction. + (Pm-POL (7.5-5:)X30 Cp = AE 0.06 X 7.0X106 = +6.25*10:4 m 24 x(7-5) [n = 1). 22 (1005 Correction → mo 'Llu)? 30x2.25)? es - 24 m2 Pm² W = $,x 574% boy 6061/78 0.0757 30 - 2.251g cs = ·1125 m Total correction = Cz+ Cp + Cs 0.74X10-2) + (625x104)+(--1125) 1102 m. Actual length of Tape 30- 11102 L'.29.89 m. True length of line = 4 x ML X459.20 = 457.52M lingwer). 29.89 30

Slope Correafion (Co)- en hoe 459.20 .3.48103 m. Total correction stope CA+Cp - Cs-ch > (174X103)+(6-25210-4)-(-1125)-(3-4810*3 Actual rength of the tape True longth of line LXML- → 457-43M (shus) : 1135 30.1135 e 29.8865. 29.8865 30 X 459,20
in first page total correction is wrong due to not adding slope correction .2nd page slope correction is adding then finding the correct length. Thank you