Conductor tension under varying conditions

Why tension must be computed for varying conditions

The tension and sag of a strung conductor change whenever the temperature or the loading (wind, ice) changes. The design must therefore check every condition the line will meet:

1. Strength: in the toughest condition (minimum temperature, maximum wind, ice), the tension must not exceed $\text{UTS}/\text{FS}$. Otherwise the conductor or the towers can fail.
2. Ground clearance: in the easiest condition (maximum temperature, no wind), the sag is largest. This sag fixes the tower height and the minimum ground and crossing clearances.
3. Stringing: at erection, the conductor must be pulled to a particular tension or sag, at the temperature on that day, so that conditions 1 and 2 are met later. Erection crews use stringing charts of tension and sag against temperature.
4. Tower loads and vibration: the tension at every condition sets the angle and dead-end tower loads and the aeolian vibration risk. Everyday tension is usually limited to about 18–25% of UTS.

Derivation of the change-of-state equation

Let condition 1 have tension $T_1$, effective weight per unit length $W_1$ and temperature $\theta_1$. Let condition 2 have $T_2$, $W_2$ and $\theta_2$. The span is $L$, the area $A$, the modulus $E$ and the coefficient of linear expansion $\alpha$.

Step 1 – conductor length. For a parabola, the length of conductor in a span is

$$l = L + \frac{8D^2}{3L}, \quad D = \frac{WL^2}{8T} \;\Rightarrow\; l = L + \frac{W^2L^3}{24T^2}$$

So $l_1 = L + \dfrac{W_1^2L^3}{24T_1^2}$ and $l_2 = L + \dfrac{W_2^2L^3}{24T_2^2}$.

Step 2 – physical change in length. Going from condition 1 to condition 2, the length changes in two ways:

• by thermal expansion: $\alpha(\theta_2-\theta_1)\,l_1$
• by elastic stretch: $\dfrac{(T_2-T_1)\,l_1}{AE}$

Since $l_1 \approx L$:

$$l_2 - l_1 = \alpha(\theta_2-\theta_1)L + \frac{(T_2-T_1)L}{AE}$$

Step 3 – equate the two expressions for the length change.

$$\frac{W_2^2L^3}{24T_2^2} - \frac{W_1^2L^3}{24T_1^2} = \alpha(\theta_2-\theta_1)L + \frac{(T_2-T_1)L}{AE}$$

Step 4 – multiply by AE/L and rearrange.

$$\begin{aligned} \frac{W_2^2L^2AE}{24T_2^2} &= T_2 - T_1 + \alpha(\theta_2-\theta_1)AE + \frac{W_1^2L^2AE}{24T_1^2}\\ T_2^2\left[T_2 + K_2\right] &= K_1 \end{aligned}$$

where

$$K_2 = -T_1 + \alpha(\theta_2-\theta_1)AE + \frac{W_1^2L^2AE}{24T_1^2}, \qquad K_1 = \frac{W_2^2L^2AE}{24}$$

This is the stringing (change-of-state) equation. It is a cubic in $T_2$ with one positive real root. Normally condition 1 is the toughest condition, with $T_1 = \text{UTS}/\text{FS}$ known. The equation is then solved for $T_2$ in the stringing or easiest condition, and the sag follows from $D_2 = W_2L^2/(8T_2)$.

The effective weights are $W = \sqrt{(w_c + w_{ice})^2 + w_{wind}^2}$, where $w_{ice} = \rho_{ice}\,\pi t(d+t)$ and $w_{wind} = p(d+2t)$ for ice thickness $t$.