Answer to Question #306381 in Civil and Environmental Engineering for ale

Question

A 50m tape of standard length has a weight of 0.05 kg/m, with a cross sectional area of 0.04 sq.cm. It has a modulus of elasticity of 2.10 x 106 kg/cm2 . The tape is of standard length under a pull of 5.5 kg when supported throughout its length and temperature of 20°C. This tape was used to measure a distance between A and B and was recorded to be 458.65 meters long. At the time of measurement, the pull applied was 8kg with the tape supported only at its end points and the temperature observed was 18°C. Assuming coefficient of linear expansion of the tape is 0.0000116/°C.

(a) Compute the correction due to the applied pull of 8 kg.

(b) Compute the correction due to weight of tape. 

(c) Compute the true length of the measured line AB due to the combined effects of tension, sag and temperature. 

Answer 1

l = Length of tape = 50 m

L = Length of measurement = 458.650 m

Po = Pull at the time of measurement = 8 kg

Ps = Pull at the time of standardization = 5. 5 kg

w = Weight of tape = 0.05 kg/m

A = Cross — sectional area of tape = 0.04 cm²

E = Modulus of Elasticity = 2. 1 x 10⁶ Kg/cm² 

Temperature (°C) = 18°C

linear expansion coefficient = 0.0000116/°C

(a.) Correction due to the applied pull of 8 kg

$Cp=\frac{(Po-Ps)L}{AE}$

$Cp=\frac{(8-5.5)458.65}{0.04(2. 1 x 10^6)}$

$Cp=\frac{1146.625}{84,000}$

$Cp= 0.01365$

(b.) Correction due to weight of tape

$Cs= \frac{W^2L}{24Po}$

$Cs= \frac{0.05^2(50)}{24(8)}$

$Cs= \frac{0.125}{192}$

$Cs= 0.000651$

(c.) True length of the measured line AB due to the combined effects of tension, sag and temperature. 

L= 50+0.0000116(20-18)(50)+$\frac{(8-5.5)(50)}{0.04(2. 1 x 10^6)}$

L= 50.00116+0.00119

L= 50.00235m