---
title: "Conductor tension under varying conditions"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:tension-varying-conditions-equation"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tension-varying-conditions-equation
site: BCT Engineering
---

# Conductor tension under varying conditions

Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection

- Asked in: 2072 Magh, 2+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tension-varying-conditions-equation
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

Why is it necessary to compute the tension in conductor for transmission line under varying condition? Derive the equation correlating the tension in line in two different conditions.

## Short answer

It is necessary to verify that maximum tension under worst loads does not exceed strength and that sag at highest temperature meets clearance. The change-of-state equation T2^2[T2 + K2] = K1 is derived by equating parabolic length change to thermal expansion and elastic stretch, with K2 = -T1 + α(θ2-θ1)AE + W1^2L^2AE/(24T1^2), K1 = W2^2L^2AE/24.

## Answer

#### 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 $t$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In which condition must the tension not exceed UTS/FS?
   A. The easiest condition: maximum temperature, no wind
   B. The toughest condition: minimum temperature with maximum wind and ice
   C. At erection only
   D. The everyday condition at average temperature
   Answer: B. Tension is highest when it is cold and loaded, so that case governs strength. (Key point: Strength check)

2. Which condition fixes the tower height and clearances?
   A. Minimum temperature with ice, where tension is highest
   B. Maximum wind condition
   C. Maximum temperature without wind, where the sag is largest
   D. Stringing at low temperature
   Answer: C. Largest sag occurs when the conductor is hot and lightly loaded. (Key point: Ground clearance)

3. What is the conductor length in a parabolic span in terms of W, L and T?
   A. l = L + W L^2 / (8T)
   B. l = L + 8 D / (3L)
   C. l = L + W^2 L / (24 T)
   D. l = L + W^2 L^3 / (24 T^2)
   Answer: D. Substituting D = WL^2/(8T) in l = L + 8D^2/(3L) gives this. (Key point: Length of conductor)

4. What two effects change the length between condition 1 and 2?
   A. Thermal expansion alpha(theta2-theta1)L and elastic stretch (T2-T1)L/(AE)
   B. Wind pressure and ice weight only
   C. Thermal expansion and sag only
   D. Elastic stretch and tower deflection
   Answer: A. Equating this physical change with the difference of l2 and l1 gives the change-of-state equation. (Key point: Length change)

5. What is the form of the resulting equation?
   A. A linear equation in T2
   B. A cubic in T2 with one positive real root
   C. A quadratic in T2 with two valid roots
   D. A cubic in T1 solved for sag directly
   Answer: B. T2^2(T2 + K2) = K1, solved for T2 and then D2 = W2L^2/(8T2). (Key point: Form of the equation)

## Related in Transmission and Distribution System Design

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