Question #191235

A line is recorded as 475.25 m long. It is measured with a 6.5 N tape which is 30.492 m long at 68^{0}F under a 10 lb. pull supported at end points. During measurement the temperature is 4.5^{0}C and the tape is suspended under a 71 N pull. The line is measured on 3% grade. What is the true horizontal distance? Modulus of elasticity of the tape is 1.93 X 10^{8 }KPa and cross-sectional area of tape is 0.0284 cm^{2}. Coefficient of expansion of the tape material is 0.0000116 m/^{0}C.

Expert's answer

Following corrections are required:

Temperature corrections, Pull correction, Sag Correction, Slope correction.

A)

1) Field temperature 4.5 degree C

2) Tape standard temperature=68 degree F=20 degree C

Temperature Correction per tape length =0.0000116 x (4 -20) 30.492

=- 0.005659 m (+ ve)

B) Pull correction per tape length = (P_{m }- P_{o})L / AE

= {(71 â€“ 6.5)(30.492}/{(0.00000284 )(1.93 X 10^11)}

= 0.003588 m (+ve)

Combined correction = -0.005659 + 0.003588 m.=-0.00207m

C) correction due to sag

= C_{s} = *l*_{1} (Mg)^{2} / 24 P^{2}

*l1*_{ }= 30.492 m; M =10 lb= 4.53 kilograms; P = 71N.

C_{s} = 30.492 x (4.53 x 9.81)^{2} / (24 x 71^2)= 0.4977m.

Thus Corrected length of the tape = *l*

= 30.492â€“ 0.00207-0.4977

= 29.99223m

True horizontal length of the line = (29.99223 / 30.492)x 475.25

= 467.46m.

Since gradient is 3%,

Since horizontal distance is measured on slope of 3% so the slope distance

=âˆš(467.46^2+(467.46*3/100)^2)

=âˆš(218519.37+196.66)

=âˆš218716.03

=467.67

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