Surveying Essay - 550 Words

Surveying (Essay Sample) Content: SurveyingStudentà ¢Ã¢â€š ¬s NameInstitutionSurveyingQuestion 1Steel tapeCross-Sectional area of 0.04cm2Weight 1kgLength 30.01MRanges from 0 à ¢Ã¢â€š ¬ 30M markAt a temperature of 200 C and a tension of 6kgUsed to measure a grade of 5% under a plied tension of 12kgTemperature of tape is 50 C, and a measured slope distance of 380M.Given that K= 1.15*10-5 /0 per m and E=2.1*106 kg/cm2Required to calculate the corrected horizontal distanceSolutionsCorrection of the incorrect length of the table (Corrected length)= Measured length or length to be measuredÂCorr*Measured length * Nominal length as measured by the mark (Mastin and Kavanagh, 2013).Corrected length= 380 MÂ(30M-30.01M)380M/30.01M = 379.8734MCorrection due to temperature differencesLength of line= Measured length+ Coefficient of thermal expansion (Tape temperature at the time of measurement-Standardized tape temperature)Length of the tape= 380M+1.15*10-5 /0 per m (5-20)Correction due to slope=379.999827 5 MCorrection due to the tension (Mastin and Kavanagh, 2013).Elongation of the tape due to pull= (Tension applied during measurement-standard tension )* Measured length/(Cross-sectional area*modulus of elasticity)Elongation of the tape due to pull= (12kg*9.81N/kg-6kg*9.81N/kg)*380M/(2.1*106 kg/cm2 * 0.04cm2)= (117.72N-58.86N)*380/84000=0.26627143MCorrected distance= correct distance + Elongation of the tape due to pullCorrected distance= 380-0.26627143M= 379.733728571MCorrection due to slopeFor gentle slopes where slope Ë 20%Correction height= h2/2s where h is the vertical height and s is the measured distanceSlope = 5% = 0.05= tan ÃŽHence, ÃŽ= 2.86240Height= 380Msin 2.8624 = 18.976MCorrection factor = 18.9762/2*380=0.4738mCorrected height = measured height à ¢Ã¢â€š ¬ Correction factor= 380-0.4738= 379.5262MQuestion 2Steel tapeCross Sectional area of 0.008in2 = 0.00005556ft2Weight 10lbsLength 100.1 ftLength of the steel rule 100ftAt temperature of 660 F and a tension of 10l b.Used to measure a grade of 2% under a plied tension of 96lbs.Temperature of tape is 960 F, and a measured slope distance of 300ft.Weight of 0.03lb/ftGiven that K= 1.15*10-5 /0 per m= 1.184754*10-4/Fper ft and E=2.1*106 kg/cm2 = 4.30113*109 lbs/ft2Required to calculate the corrected horizontal distanceSolutionsCorrection of the incorrect length of the table (Corrected length)= Measured length or length to be measuredÂCorr*Measured length * Nominal length as measured by the mark.Corrected length= 300ftÂ(100ft-100.01ft)300Mft100.01ft = 299.97ftCorrection due to temperature differencesLength of line= Measured length+ Coeefficient of thermal expansion(Tape temperature at the time of measurement-Standardized tape temperature)Length of the tape= 30...