---
title: "Maximum sag with supports at different levels"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:max-sag-derivation-unequal-levels"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/max-sag-derivation-unequal-levels
site: BCT Engineering
---

# Maximum sag with supports at different levels

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

- Asked in: 2074 Magh, 6 marks
- Video: [Lecture#30: Sag Calculation When Supports are at Equal Level and Unequal Levels | Dr. J. A. Laghari](https://www.youtube.com/watch?v=U6FkyPhCpXg) — Dr. J. A. LAGHARI
- 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/max-sag-derivation-unequal-levels
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

Derive the expression for maximum sag when the supports are at different level from the reference line.

## Short answer

With supports at heights differing by h, the conductor follows parabola y = w x²/(2T). The distances to the lowest point are x₁ = L/2 - Th/(wL) and x₂ = L/2 + Th/(wL). The maximum sags from lower and higher supports are S₁ = w x₁²/(2T) and S₂ = w x₂²/(2T); for equal levels S=wL²/(8T).

## Answer

When the supports are at different levels, the conductor still hangs as a parabola (for spans that are not too long). The lowest point O, the point of maximum sag, lies nearer the lower support.

Let:
- $L$ = span, $h$ = difference in levels (B higher than A)
- $w$ = conductor weight per metre, $T$ = tension at the lowest point
- $x_1$, $x_2$ = horizontal distances of O from A and B, so $x_1 + x_2 = L$

```text
                                    B
                                 .' |
     A                        .'    | S2
     |`.                   .'       |
  S1 |  `.______O______.'..........|
     |<--- x1 --->|<----- x2 ------>|
```

#### Equation of the conductor
Consider a portion OP of length about $x$, with O at the origin. It carries weight $wx$ acting at $x/2$. Taking moments about P:

$$
T\,y = w x \cdot \frac{x}{2} \quad\Rightarrow\quad y = \frac{w x^2}{2T}
$$

#### Maximum sags from each support
$$
S_1 = \frac{w x_1^2}{2T}, \qquad S_2 = \frac{w x_2^2}{2T}
$$

#### Finding $x_1$ and $x_2$
$$
\begin{aligned} S_2 - S_1 &= h = \frac{w}{2T}(x_2 - x_1)(x_2 + x_1) = \frac{wL}{2T}(x_2 - x_1) \\ x_2 - x_1 &= \frac{2Th}{wL} \\ x_1 &= \frac{L}{2} - \frac{Th}{wL}, \qquad x_2 = \frac{L}{2} + \frac{Th}{wL} \end{aligned}
$$

#### Final expressions
$$
S_1 = \frac{w}{2T}\left(\frac{L}{2} - \frac{Th}{wL}\right)^2, \qquad S_2 = \frac{w}{2T}\left(\frac{L}{2} + \frac{Th}{wL}\right)^2
$$

- For equal levels ($h = 0$): $S = \dfrac{wL^2}{8T}$.
- Under wind and ice, $w$ is replaced by the resultant load $w_r = \sqrt{(w + w_i)^2 + w_w^2}$, and the sag is inclined.
- If $x_1$ comes out negative, the lowest point is outside the span (virtual sag), and the lower support has uplift.

## Check yourself

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

1. Taking moments about P for the portion OP gives which conductor equation?
   A. y = w x^2 / (2T)
   B. y = w x / T
   C. y = T x^2 / (2w)
   D. y = w x^2 / (8T)
   Answer: A. T y = (w x)(x/2) gives the parabola y = w x^2/(2T). (Key point: Conductor shape)

2. Which result gives the distance x1 of the lowest point from the lower support A?
   A. x1 = L/2 + Th/(wL)
   B. x1 = L/2 - Th/(wL)
   C. x1 = L/2
   D. x1 = Th/(wL)
   Answer: B. From S2 - S1 = h we get x2 - x1 = 2Th/(wL), so x1 = L/2 - Th/(wL); the lowest point is nearer the lower support. (Key point: Locating lowest point)

3. What do S1 and S2 reduce to when h = 0?
   A. wL^2/(2T)
   B. wL^2/(8T)
   C. wL/T
   D. 0
   Answer: B. With h = 0, x1 = x2 = L/2, so S = w(L/2)^2/(2T) = wL^2/(8T). (Key point: Equal level check)

## Related in Transmission and Distribution System Design

- [Location of maximum sag, supports at different levels](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Amax-sag-location-derivation) — Transmission and Distribution System Design, chapter 4, asked 2×
- [Position of maximum sag, unequal supports](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Amax-sag-position-unequal-levels) — Transmission and Distribution System Design, chapter 4, asked 4×
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