Engineering 25
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Engineering 25. Catenary Tutorial Part-1. Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected] UNloaded Cable → Catenary. Consider a cable uniformly loaded by the cable itself, e.g., a cable hanging under its own weight.

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Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected]

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Bruce mayer pe registered electrical mechanical engineer bmayer chabotcollege

Engineering 25

Catenary Tutorial Part-1

Bruce Mayer, PE

Registered Electrical & Mechanical [email protected]


Unloaded cable catenary

UNloaded Cable → Catenary

  • Consider a cable uniformly loaded by the cable itself, e.g., a cable hanging under its own weight.

  • With loading on the cable from lowest point C to a point D given by W = ws, the Force Triangle on segment CD reveals the internal tension force magnitude, T

  • Where


Unloaded cable catenary 2

UNloaded Cable → Catenary (2)

  • Next, relate horizontal distance, x, to cable-length s

  • But by Force Balance Triangle

  • Also From last slide recall

  • Thus


Unloaded cable catenary 3

UNloaded Cable → Catenary (3)

  • Factoring Out c

  • Finally the Integral Eqn

  • Integrate Both Sides using Dummy Variables of Integration:

    • σ: 0→x η: 0→s


Unloaded cable catenary 4

UNloaded Cable → Catenary (4)

  • Using σ: 0→x η: 0→s

  • Now the R.H.S. AntiDerivative is the argSINH

  • Noting that


Unloaded cable catenary 5

UNloaded Cable → Catenary (5)

  • Thus the Solution to the Integral Eqn

  • Then

  • Solving for s in terms of x


Unloaded cable catenary 6

UNloaded Cable → Catenary (6)

  • Finally, Eliminate s in favor of x & y. From the Diagram

  • From the Force Triangle

  • And From Before

  • So the Differential Eqn


Unloaded cable catenary 7

UNloaded Cable → Catenary (7)

  • Recall the Previous Integration That Relates x and s

  • Using s(x) above in the last ODE

  • Integrating with Dummy Variables:

    • Ω: c→yσ: 0→x


Unloaded cable catenary 8

UNloaded Cable → Catenary (8)

  • Noting that cosh(0) = 1

  • Solving for y yields theCatenary Equation in x&y:

  • Where

    • c = T0/w

    • T0 = the 100% laterally directed force at the ymin point

    • w = the lineal unit weight of the cable (lb/ft or N/m)


Catenary tension t y

Catenary Tension, T(y)

  • With Hyperbolic-Trig ID: cosh2 – sinh2 = 1

  • Thus:

  • Recall From the Differential Geometry


Bruce mayer pe registered electrical mechanical engineer bmayer chabotcollege

Catenary Cabling Contraption

  • Shape is defined by the Catenary Equation

  • Note that the ORIGIN for y is the Distance “c” below the HORIZONTAL Tangent Point

y = c


The problem

An 8m length of chain has a lineal unit mass of 3.72 kg/m. The chain is attached to the Beam at pt-A, and passes over a small, low frictionpulley at pt-B.

Determine the value(s) of distance a for which the chain is in equilibrium (does not move)

The Problem


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