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

Distance Measured on the ground, D_m=475.25 m

Grade on which line is measured = 3%

Modulus of Elasticity of tape, E= 200 GPa =2*10¹¹ N/m²

Coefficient of Linear Expansion, $\alpha$ = 0.0000116 m/^oC

C/S Area of the tape, A= 0.0284 cm² =2.84*10^-6 m²

Standard Length, L_o=30.492 m

Standard Temperature, T_o=68^oF=20^oC

Standard Pull, P_o= 10 lb = 44.5 N

During measurement

Weight of the Tape, W=6.5 N.

Temperature, T_m=4.5^oC

Pull, P_m=71 N

Correction due to temperature is given by; $C_{temp}=L_o \alpha (T_m-T_o)$

$C_{temp}=30.492\times0.0000116(4.5-20)= -0.0054824616m$

Correction due to Pull is given by; $C_{pull}=\frac{(P_m-P_o)L_o}{A.E}$

$C_{pull}=\frac{(71-44.5)\times 30.492}{2.84\times10^{-6} \times2\times10^{11}}=0.00142260211m$

Correction due to Sag is given by$C_{sag}=-\frac{W^2L_0}{24P_m^2}$

$C_{sag}=-\frac{6.5^2 \times 30.492}{24 \times 71^2}=-0.01064840805 m$

Net Correction applied on the tape,$C_{net}=C_{temp}+C_{pull}+C_{sag}$

$C_{net}=-0.0054824616m+0.00142260211+0.01064840805= -0.014708268 m$

Hence, the Actual Length of the tape is given by:$L=L_o+C_{net}=30.492 -0.014708268=30.47729173 m$

Total Error to be corrected for inclined distance:$E_{slope}=D_t\times(1-cos\theta)$

But $D_t=\frac{30.492}{30.47729173}\times475.25=475.47835454303$

Grade =3% =$\frac{3}{100}\implies \theta=tan^{-1}(0.03)$

$E_{slope}=475.47835454303\times(1-cos(tan^{-1}0.03))=+0.21382094123 m$

True Horizontal Distance: $D=D_t+C{slope}$

$D_t+E{slope}=475.47835454303+0.21382094123=475.6921755m$