Determine the correct length of a line reduced to the mean sea level when the recorded length with a tape hanging in catenary at a tension of 85 N and at a temperature of 22°C is 30.071 m. The difference between the ends is 0.42 m and the site is 2000 m above M.S.L. The tape had been previously standardised in catenary at a tension of 50 N and at a temperature of 27°C, and the distance between zeros was 30.035 m. Weight of tape Cross-sectional area Coefficient of expansion Young's modulus Radius of earth = 7N = 3.9 mm² = 1.15 x 10-5 per °C = 2×105 N/mm² = 6370 km
Determine the correct length of a line reduced to the mean sea level when the recorded length with a tape hanging in catenary at a tension of 85 N and at a temperature of 22°C is 30.071 m. The difference between the ends is 0.42 m and the site is 2000 m above M.S.L. The tape had been previously standardised in catenary at a tension of 50 N and at a temperature of 27°C, and the distance between zeros was 30.035 m. Weight of tape Cross-sectional area Coefficient of expansion Young's modulus Radius of earth = 7N = 3.9 mm² = 1.15 x 10-5 per °C = 2×105 N/mm² = 6370 km
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Determine the correct length of a line reduced to the mean sea level when the
recorded length with a tape hanging in catenary at a tension of 85 N and at a temperature of 22°C is 30.071 m.
The difference between the ends is 0.42 m and the site is 2000 m above M.S.L.
The tape had been previously standardised in catenary at a tension of 50 N and at a temperature of 27°C, and
the distance between zeros was 30.035 m.
Weight of tape
Cross-sectional area
Coefficient of expansion
Young's modulus
Radius of earth
= 7N
= 3.9 mm²
= 1.15 x 10-5 per °C
= 2×105 N/mm²
= 6370 km
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