Correct length of 1350 m line

Reading of the data. The tape is stated as "30 m" but the standard length is printed "19.985 m". This is taken as 29.985 m (a 30 m tape 0.015 m short). The tape is fully supported at standardisation and used in catenary in 30 m spans, so 1350 m $= 45$ tapes.

Data

• $W = 11.775$ N for 30 m, $w = 0.3925$ N/m; density of steel $0.081$ N/cm$^3$
• $A = \dfrac{W}{0.081\times 3000} = \dfrac{11.775}{243} = 0.04846$ cm$^2 = 4.846$ mm$^2$
• $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 4.846\times 2.1\times 10^{5} = 1.0176\times 10^{6}$ N
• Standard: 29.985 m at $T_0 = 20^\circ$C and $P_0 = 100$ N; field: $T = 17^\circ$C, $P = 120$ N, $L = 1350$ m
• $\alpha = 11\times 10^{-6}$/$^\circ$C

1. Tape length

$$C_a = 1350\times\frac{29.985 - 30}{30} = -0.6750\ \text{m}$$

2. Temperature

$$C_t = 11\times 10^{-6}\times(17 - 20)\times 1350 = -0.0446\ \text{m}$$

3. Pull

$$C_p = \frac{(120 - 100)\times 1350}{1.0176\times 10^{6}} = +0.0265\ \text{m}$$

4. Sag

Per tape: $\dfrac{11.775^{2}\times 30}{24\times 120^{2}} = 0.012036$ m; $C_s = -45\times 0.012036 = -0.5416$ m.

5. Correct length

$$L = 1350 - 0.6750 - 0.0446 + 0.0265 - 0.5416 = 1348.7654\ \text{m}$$

Answer: correct length of the line $= 1348.765$ m (with the standard length read as 29.985 m).