Statics

83/3848–12Not a printed equation in Statics pp. 95–101. Th·d = wL²/8 follows from ΣM = 0 on p. 95 for a parabolic cable under uniform horizontal load.

Handbook formula

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 with uniform load per horizontal length is a parabola. Horizontal tension satisfies Th·d = wL²/8, the same moment balance as a simple-beam Mmax. That equation is not printed in Statics pp. 95–101. Support tension is larger than Th because the tangent is inclined. The figure exaggerates the sag; the numbers stay L = 40 m and d = 2 m.

$w$ (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