Sagging Cable (Small Sag)

T_{h} = \frac{w L ^{2}}{8 d} T_{ma x} = T_{h} 1 + (4 d / L)^{2}

A flexible cable under uniform horizontal load (or self-weight with small sag) is a parabola. Horizontal tension Th = w L² / (8d) where d is midspan sag. Support tension is larger because of the slope.

$w$ (kN/m): Load per horizontal length
$L$ (m): Span
$d$ (m): Sag
$T_{h}$ (kN): Horizontal tension

Step-by-step solved example

w=2 kN/m, L=40 m, d=2 m. Th?

Sagging cable: T ≈ wL²/(8d) for small sag.

1. Formula
Th = 2×1600/(8×2)=3200/16=200 kN.
$T_{h} = 200 kN$

Answer: 200 kN

10 practice questions

0/10 correct

1.Th = wL²/(8d) is the same form as

2.Larger sag d means

3.A true catenary is for

4.Tmax > Th because

5.If d→0, Th

6.w=1 kN/m, L=20 m, d=0.5 m. Th=

7.Horizontal equilibrium of half the cable gives

8.Vertical reaction at each level support is

9.Point load on a cable produces

10.Units: w in kN/m, L,d in m → Th in