A line was determined to be 2538.45 m when measured with a 45m tape supported throughout its length under a pull of 3.4 kg at a mean temperature of 36⁰C. Tape used is of standard length at 20.2⁰C under pull of 4.1 kg. Cross sectional area of tape is 0.03 cm^2. Coefficient of thermal expansion is 0.0000116/⁰C. Modulus of elasticity of tape is 2.3x10^6kg/cm2 ▪ Determine the error of tape due to change in temperature. ▪ Determine due the error of tension. ▪ Determine the corrected length of the line.
A line was determined to be 2538.45 m when measured with a 45m tape supported throughout its length under a pull of 3.4 kg at a mean temperature of 36⁰C. Tape used is of standard length at 20.2⁰C under pull of 4.1 kg. Cross sectional area of tape is 0.03 cm^2. Coefficient of thermal expansion is 0.0000116/⁰C. Modulus of elasticity of tape is 2.3x10^6kg/cm2 ▪ Determine the error of tape due to change in temperature. ▪ Determine due the error of tension. ▪ Determine the corrected length of the line.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
A line was determined to be 2538.45 m when measured with a 45m tape supported throughout its length under a pull of 3.4 kg at a mean
temperature of 36⁰C. Tape used is of standard length at 20.2⁰C under pull of 4.1 kg. Cross sectional area of tape is 0.03 cm^2. Coefficient of
thermal expansion is 0.0000116/⁰C. Modulus of elasticity of tape is 2.3x10^6kg/cm2
▪ Determine the error of tape due to change in
temperature.
▪ Determine due the error of tension.
▪ Determine the corrected length of the line.
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where did the highlighted part came from?
Transcribed Image Text:length of the
L + 1₁ + 1₂
= 25 38-45 +0.46525
= 2538.45 +0.46525- 0.02572
= 2538 88953
Correct
=
m
T
0.000456x 56.41
Solution
by Bartleby ExpertFollow-up Question
where did 56.11 or 56.41 in 2nd line on corrected length of the line come from after the value of - 0.000456?
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