Correct horizontal distance 630 m, tape 8.9 N

Reading. The modulus is used as given, $E = 12\times 10^{11}$ N/m$^2$. Standardised fully supported (flat), so no standard sag is added back. Field supports are at the 0 and 30 m ends only, so each tape length is a 30 m catenary span.

Data

• $W = 8.9$ N for 30 m; $A = 2.58$ mm$^2 = 2.58\times 10^{-6}$ m$^2$; $AE = 2.58\times 10^{-6}\times 12\times 10^{11} = 3.096\times 10^{6}$ N
• Standard: 30.005 m at $20^\circ$C, $P_0 = 52$ N
• Field: $T = 31.2^\circ$C, $P = 110$ N, $L = 630$ m (21 tapes of 30 m)
• $\alpha = 12\times 10^{-6}$/$^\circ$C

1. Tape length

$$C_a = \frac{630\times 0.005}{30} = +0.1050\ \text{m}$$

2. Temperature

$$C_t = 12\times 10^{-6}\times(31.2 - 20)\times 630 = +0.0847\ \text{m}$$

3. Pull

$$C_p = \frac{(110 - 52)\times 630}{3.096\times 10^{6}} = +0.0118\ \text{m}$$

4. Sag

Per tape: $\dfrac{W^{2}\,l}{24\,P^{2}} = \dfrac{8.9^{2}\times 30}{24\times 110^{2}} = 0.008183$ m.

$$C_s = -21\times 0.008183 = -0.1718\ \text{m}$$

5. Correct horizontal distance

$$D = 630 + 0.1050 + 0.0847 + 0.0118 - 0.1718 = 630.0296\ \text{m}$$

Answer: correct horizontal distance $= 630.030$ m.