Sag and tension in easiest condition

The tension in the easiest condition is found from the change-of-state (stringing) equation, starting from the toughest condition. The maximum (vertical) sag then follows.

Assumed (not given): $E = 0.789\times10^6$ kg/cm² and $\alpha = 17.73\times10^{-6}$ /°C, the usual values for ACSR.

Toughest condition (1): 0°C, wind 100 kg/m²

$$\begin{aligned} T_1 &= \text{UTS}/2 = 8000/2 = 4000\ \text{kg}\\ w_w &= p\,d = 100\times0.020 = 2.0\ \text{kg/m}\\ W_1 &= \sqrt{0.844^2 + 2.0^2} = 2.171\ \text{kg/m} \end{aligned}$$

Easiest condition (2): 60°C, no wind

$W_2 = 0.844$ kg/m, $\theta_2 - \theta_1 = 60^\circ$C, and

$$AE = 2.26\ \text{cm}^2\times0.789\times10^6 = 1.7831\times10^6\ \text{kg}$$

Stringing equation $T_2^2(T_2 + K_2) = K_1$

$$\begin{aligned} K_2 &= -T_1 + \alpha\Delta\theta\,AE + \frac{W_1^2L^2AE}{24T_1^2}\\ &= -4000 + 17.73\times10^{-6}\times60\times1.7831\times10^6 + \frac{2.171^2\times300^2\times1.7831\times10^6}{24\times4000^2}\\ &= -4000 + 1896.9 + 1969.4 = -133.7\ \text{kg}\\ K_1 &= \frac{W_2^2L^2AE}{24} = \frac{0.844^2\times300^2\times1.7831\times10^6}{24} = 4.7632\times10^9 \end{aligned}$$

Solve $T_2^2(T_2 - 133.7) = 4.7632\times10^9$:

$$T_2 = 1728\ \text{kg}$$

Maximum sag (easiest condition)

$$D_{max} = \frac{W_2L^2}{8T_2} = \frac{0.844\times300^2}{8\times1728.3} = 5.49\ \text{m}$$

For comparison, the sag in the toughest condition is $2.171\times300^2/(8\times4000) = 6.11$ m. That sag is inclined, at $\tan^{-1}(2.0/0.844) = 67^\circ$, so its vertical component is only $6.11\times0.844/2.171 = 2.37$ m. The easiest condition therefore gives the largest vertical sag.

Answer: in the easiest condition the tension is about 1728 kg (21.6% of UTS) and the maximum vertical sag is about 5.49 m.