Answer to Question #191235 in Civil and Environmental Engineering for Hannah

Question

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 68⁰F under a 10 lb. pull supported at end points. During measurement the temperature is 4.5⁰C 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 10⁸ KPa and cross-sectional area of tape is 0.0284 cm². Coefficient of expansion of the tape material is 0.0000116 m/⁰C.

Answer 1

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 – 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 = l₁ (Mg)² / 24 P²

  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 

                    = 30.492– 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

=√(467.46^2+(467.46*3/100)^2)

=√(218519.37+196.66)

=√218716.03

=467.67