Question #155309

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

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*10^{11 }N/m^{2}

Coefficient of Linear Expansion, "\\alpha" = 0.0000116 m/^{o}C

C/S Area of the tape, A= 0.0284 cm^{2 }=2.84*10^{-6} m^{2}

Standard Length, L_{o}=30.492 m

Standard Temperature, T_{o}=68^{o}F=20^{o}C

Standard Pull, P_{o}= 10 lb = 44.5 N

**During measurement**

Weight of the Tape, W=6.5 N.

Temperature, T_{m}=4.5^{o}C

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\n 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"

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