---
title: "Sag and tension in easiest condition"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:sag-tension-easiest-condition"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/sag-tension-easiest-condition
site: BCT Engineering
---

# Sag and tension in easiest condition

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

- Asked in: 2067 Chaitra (old course), 8 marks
- Video: [Sag||calculation of sag||Sag at equal & unequal levels||Effect of ice & wind loadings](https://www.youtube.com/watch?v=haTxrrrVutg) — Electrical HUB
- 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/sag-tension-easiest-condition
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

Evaluate the maximum sag and tension in easiest condition for a transmission line of span 300 m with following data: Conductor: cross-section: 226 mm²; Diameter: 20 mm; Weight: 0.844 kg/m; UTS: 8000 kg.

| Parameter | Toughest condition | Easiest condition |
|---|---|---|
| Temperature | 0°C | 60°C |
| Wind pressure | 100 kg/m² | 0 |
| Tension | UTS/2 | ? |
| Ice thickness | 0 | 0 |

## Short answer

Using the change-of-state equation from the toughest condition (tension 4000 kg, resultant weight 2.171 kg/m) to the easiest (60 °C, no wind), the tension is solved as 1728 kg. Maximum vertical sag = 0.844×300²/(8×1728) = 5.49 m.

## Answer

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

| $T_2$ (kg) | LHS |
|---|---|
| 1700 | $4.527\times10^9$ |
| 1750 | $4.950\times10^9$ |
| 1728.3 | $4.763\times10^9$ ✓ |

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

#### 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.**

## Check yourself

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

1. How is the tension in the easiest condition found?
   A. From T = wL^2/(8D)
   B. From the change-of-state (stringing) equation starting at the toughest condition
   C. From T = UTS/4
   D. From the weight alone
   Answer: B. The stringing equation T2^2(T2 + K2) = K1 relates the two states. (Key point: Method)

2. What resultant loading W1 is used at the toughest condition?
   A. 0.844 + 2.0 = 2.844 kg/m
   B. 2.0 kg/m
   C. sqrt(0.844^2 + 2.0^2) = 2.171 kg/m
   D. 0.844 kg/m
   Answer: C. Weight and wind load are at right angles, so they combine vectorially. (Key point: Toughest-state loading)

3. Which loading is used for K1 and for the sag at 60°C?
   A. W2 = 0.844 kg/m, no wind
   B. 2.171 kg/m
   C. 2.0 kg/m
   D. 0.844 + 2.0 kg/m
   Answer: A. At the easiest condition there is no wind, so only the weight acts. (Key point: Easiest-state loading)

4. Why does the easiest condition give the largest vertical sag?
   A. Because tension is higher
   B. Because there is more wind
   C. Because the toughest sag is mostly horizontal; its vertical part is only about 2.37 m
   D. Because the conductor shortens
   Answer: C. The toughest sag of 6.11 m is inclined at 67°, with vertical component 2.37 m, less than 5.49 m. (Key point: Largest vertical sag)

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