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

  2. 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

  3. UNloaded Cable → Catenary (2) • Next, relate horizontal distance, x, to cable-length s • But by Force Balance Triangle • Also From last slide recall • Thus

  4. 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

  5. UNloaded Cable → Catenary (4) • Using σ: 0→x η: 0→s • Now the R.H.S. AntiDerivative is the argSINH • Noting that

  6. UNloaded Cable → Catenary (5) • Thus the Solution to the Integral Eqn • Then • Solving for s in terms of x

  7. 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

  8. 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→y σ: 0→x

  9. 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)

  10. Catenary Tension, T(y) • With Hyperbolic-Trig ID: cosh2 – sinh2 = 1 • Thus: • Recall From the Differential Geometry

  11. 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

  12. 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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