Surveying (Essay Sample)

Surveying
Student’s Name
Institution
Surveying
Question 1
Steel tape
Cross-Sectional area of 0.04cm2
Weight 1kg
Length 30.01M
Ranges from 0 – 30M mark
At a temperature of 200 C and a tension of 6kg
Used to measure a grade of 5% under a plied tension of 12kg
Temperature 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/cm2
Required to calculate the corrected horizontal distance
Solutions
Correction of the incorrect length of the table (Corrected length)= Measured length or length to be measured±Corr*Measured length * Nominal length as measured by the mark (Mastin and Kavanagh, 2013).
Corrected length= 380 M±(30M-30.01M)380M/30.01M = 379.8734M
Correction due to temperature differences
Length 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=379.9998275 M
Correction 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=0.26627143M
Corrected distance= correct distance + Elongation of the tape due to pull
Corrected distance= 380-0.26627143M= 379.733728571M
Correction due to slope
For gentle slopes where slope Ë‚ 20%
Correction height= h2/2s where h is the vertical height and s is the measured distance
Slope = 5% = 0.05= tan α
Hence, α= 2.86240
Height= 380Msin 2.8624 = 18.976M
Correction factor = 18.9762/2*380=0.4738m
Corrected height = measured height – Correction factor= 380-0.4738= 379.5262M
Question 2
Steel tape
Cross Sectional area of 0.008in2 = 0.00005556ft2
Weight 10lbs
Length 100.1 ft
Length of the steel rule 100ft
At temperature of 660 F and a tension of 10lb.
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/ft
Given 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/ft2
Required to calculate the corrected horizontal distance
Solutions
Correction of the incorrect length of the table (Corrected length)= Measured length or length to be measured±Corr*Measured length * Nominal length as measured by the mark.
Corrected length= 300ft±(100ft-100.01ft)300Mft100.01ft = 299.97ft
Correction due to temperature differences
Length of line= Measured length+ Coeefficient of thermal expansion(Tape temperature at the time of measurement-Standardized tape temperature)
Length of the tape= 30...