Maximum sag with supports at different levels

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 $x_1 + x_2 = L$

                                    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.