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Engineering 25. Catenary Tutorial Part-1. Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. 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 BMayer@ChabotCollege

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Engineering 25

Catenary Tutorial Part-1

Bruce Mayer, PE

Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

### 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)

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

• But by Force Balance Triangle

• Also From last slide recall

• Thus

### 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)

• Using σ: 0→x η: 0→s

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

• Noting that

### UNloaded Cable → Catenary (5)

• Thus the Solution to the Integral Eqn

• Then

• Solving for s in terms of x

### 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)

• 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)

• 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)

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

• Thus:

• Recall From the Differential Geometry

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

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)