---
title: "Conductor tension in stringing condition"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:stringing-tension-300m-span"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/stringing-tension-300m-span
site: BCT Engineering
---

# Conductor tension in stringing condition

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

- Asked in: 2069 Bhadra (old course), 10 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/stringing-tension-300m-span
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

Taking the factor of safety under toughest condition to be 2.5, compute the tension in the conductor to be given in the conductor for stringing condition for a span of 300 m. Assume the ultimate tensile strength of a conductor is 15910 kg and factor of safety of 2.5. Use the following data: The conductor specifications are: Area = 462.60 mm², Linear expansion coefficient α = 17.73×10⁻⁶/°C, Modulus of elasticity (E) = 0.789×10⁶ kg/cm², Weight (Wc) = 1726 kg/km, Radius = 1.40 cm.

| Condition | Temperature | Wind pressure | Ice loading (ice density 950 kg/m³) |
|---|---|---|---|
| Toughest condition | −5°C | 100 kg/m² | 1 cm |
| Stringing condition | 25°C | - | - |

[Given stringing equation: T2²[T2 + K2] − K1 = 0, where K2 = −T1 + α(θ2 − θ1)AE + W1²L²AE/(24T1²) and K1 = W2²L²AE/24.]

## Short answer

Using the stringing equation, the stringing tension at 25°C without wind/ice is found from the toughest condition (UTS/2.5=6364 kg, wind+ice). The required tension is 2211 kg, giving a sag of about 8.78 m.

## Answer

The stringing tension is found by solving the given stringing equation from the toughest condition (−5°C, wind and ice) to the stringing condition (25°C, still air).

#### Toughest condition (condition 1)
Allowable tension:

$$
T_1 = \frac{\text{UTS}}{FS} = \frac{15910}{2.5} = 6364\ \text{kg}
$$

Conductor diameter $d = 2\times1.40 = 2.8$ cm; ice thickness $t = 1$ cm.

$$
\begin{aligned}
w_c &= 1726\ \text{kg/km} = 1.726\ \text{kg/m}\\
w_i &= \rho\,\pi t(d+t) = 950\times\pi\times0.01\times(0.028+0.01) = 1.134\ \text{kg/m}\\
w_w &= p(d+2t) = 100\times(0.028+0.02) = 4.80\ \text{kg/m}\\
W_1 &= \sqrt{(1.726+1.134)^2 + 4.80^2} = 5.588\ \text{kg/m}
\end{aligned}
$$

#### Stringing condition (condition 2)
$W_2 = w_c = 1.726$ kg/m, $\theta_2 - \theta_1 = 25-(-5) = 30^\circ$C.

$$
AE = 4.626\ \text{cm}^2\times0.789\times10^6\ \text{kg/cm}^2 = 3.650\times10^6\ \text{kg}
$$

#### Constants
$$
\begin{aligned}
K_2 &= -T_1 + \alpha(\theta_2-\theta_1)AE + \frac{W_1^2L^2AE}{24T_1^2}\\
&= -6364 + 17.73\times10^{-6}\times30\times3.650\times10^6 + \frac{5.588^2\times300^2\times3.650\times10^6}{24\times6364^2}\\
&= -6364 + 1941.4 + 10551.0 = 6128.3\ \text{kg}\\
K_1 &= \frac{W_2^2L^2AE}{24} = \frac{1.726^2\times300^2\times3.650\times10^6}{24} = 4.0775\times10^{10}\ \text{kg}^3
\end{aligned}
$$

#### Solving $T_2^2(T_2 + 6128.3) = 4.0775\times10^{10}$
By trial (or Newton's method):

| $T_2$ (kg) | $T_2^2(T_2+K_2)$ |
|---|---|
| 2000 | $3.25\times10^{10}$ (too low) |
| 2200 | $4.03\times10^{10}$ (slightly low) |
| 2211.2 | $4.0775\times10^{10}$ (matches) |

$$
T_2 = 2211\ \text{kg}
$$

Check: the stringing sag is $D = \dfrac{1.726\times300^2}{8\times2211.2} = 8.78$ m. The sag in the toughest condition is $\dfrac{5.588\times300^2}{8\times6364} = 9.88$ m.

**Answer: the conductor should be strung with a tension of about 2211 kg (about 13.9% of UTS) at 25°C, giving a stringing sag of about 8.78 m on the 300 m span.**

## Check yourself

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

1. How is T1 found?
   A. UTS/FS = 15910/2.5 = 6364 kg
   B. 15910 x 2.5
   C. 15910/2
   D. 15910 kg
   Answer: A. Toughest condition tension. (Key point: T1)

2. How is W1 found?
   A. wc + wi + ww
   B. sqrt((wc+wi)^2 + ww^2) = 5.588 kg/m
   C. sqrt(wc^2+ww^2)
   D. wc only
   Answer: B. Ice and conductor add vertically; wind is horizontal. (Key point: Effective weight W1)

3. What is W2?
   A. 5.588 kg/m
   B. 4.80 kg/m
   C. Bare conductor weight 1.726 kg/m
   D. 1.134 kg/m
   Answer: C. No wind or ice at stringing. (Key point: Stringing weight)

4. What is T2?
   A. About 6364 kg
   B. About 6128 kg
   C. About 4078 kg
   D. About 2211 kg
   Answer: D. Solves T2^2(T2+6128.3)=4.0775e10; sag 8.78 m. (Key point: Result)

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