True length of suspended distance AB (66)

Assumptions. The tape is standardised fully supported and used suspended (catenary) in 3 full tapes of 30 m plus a last part of 24.095 m. The ground slopes uniformly at 2.5 m per 100 m. $g = 9.81$ m/s$^2$.

Data

• $A = 0.028$ cm$^2 = 2.8$ mm$^2$; density $= 7750$ kg/m$^3$; $w = 7750\times 2.8\times 10^{-6}\times 9.81 = 0.21288$ N/m ($W = 6.386$ N per 30 m)
• $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 2.8\times 2.1\times 10^{5} = 588{,}000$ N
• Standard: 30.015 m at $20^\circ$C, $P_0 = 70$ N; field: $T = 32^\circ$C, $P = 100$ N; $\alpha = 1.15\times 10^{-5}$/$^\circ$C
• $L = 114.095$ m

1. Tape length

$$C_a = \frac{114.095\times 0.015}{30} = +0.0570\ \text{m}$$

2. Temperature

$$C_t = 1.15\times 10^{-5}\times(32 - 20)\times 114.095 = +0.0157\ \text{m}$$

3. Pull

$$C_p = \frac{(100 - 70)\times 114.095}{588{,}000} = +0.0058\ \text{m}$$

4. Sag

Full tape: $\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = \dfrac{0.21288^{2}\times 30^{3}}{24\times 100^{2}} = 0.005098$ m; three tapes $= 0.015294$ m. Last part: $\dfrac{0.21288^{2}\times 24.095^{3}}{24\times 100^{2}} = 0.002641$ m.

$$C_s = -(0.015294 + 0.002641) = -0.0179\ \text{m}$$

5. Slope

For each tape length $l$, the height difference is $h = 0.025\,l$, so summing $h^{2}/2l$ over all lengths gives:

$$C_h = -\sum\frac{h^{2}}{2l} = -\frac{0.025^{2}\times 114.095}{2} = -0.0357\ \text{m}$$

6. True length

$$L_{true} = 114.095 + 0.0570 + 0.0157 + 0.0058 - 0.0179 - 0.0357 = 114.1200\ \text{m}$$

Answer: true horizontal length of AB $= 114.120$ m.