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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Engineering 36. Chp 7: Flex Cables. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Recall Chp10 Introduction. Examine in Detail Two Important Types Of Engineering Structures:

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Engineering 36 Chp7:Flex Cables Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. Recall Chp10 Introduction • Examine in Detail Two Important Types Of Engineering Structures: • BEAMS - usually long, straight, prismatic members designed to support loads applied at various points along the member • CABLES - flexible members capable of withstanding only tension, designed to support concentrated or distributed loads

  3. Göteborg, Sweden http://www.nmt.edu/~armiller/bridgefu.htm Load-Bearing Cables Straight Curved WHY the Difference? • Cables are applied as structural elements in suspension bridges, transmission lines, aerial tramways, guy wires for high towers, etc.

  4. Concentrated Loads on Cables • To Determine the Cable SHAPE, Assume: • Concentrated vertical loads on given vertical lines • Weight of cable is negligible • Cable is flexible, i.e., resistance to bending is small • Portions of cable between successive loads may be treated as TWO FORCE MEMBERS • Internal Forces at any point reduce to TENSION Directed ALONG the Cable Axis

  5. Concentrated Loads (2) • Consider entire cable as a free-body • Slopes of cable at A and B are NOT known • FOUR unknowns (i.e., Ax, Ay, Bx, By) are involved and the equations of equilibrium are NOT sufficient to determine the reactions.

  6. Concentrated Loads (3) • To Obtain an additional equation • Consider equilibrium of cable-section AD • Assume that CoOrdinates of SOME point D, (x,y), on the cable have been Determined (e.g., by measurement) • Then the addedEqn: • With pt-D info, the FOUR Equilibrium Eqns

  7. Concentrated Loads (4) • The 4 Eqns Yield Ax & Ay • Can Now Work our Way Around the Cable to Find VERTICAL DISTANCE (y-CoOrd) For ANY OTHER point known • Example  Consider Pt C2 known UNknown known known known known

  8. The cable AE supports three vertical loads from the points indicated. Point C is 5 ft below the left support For the Given Loading & Geometry, Determine: The elevation of points B and D The maximum slope and maximum tension in the cable. Example  Concentrated Loads

  9. Solution Plan Determine reaction force components at pt-A from solution of two equations formed from taking entire cable as a Example  Concentrated Loads • free-body and summing moments about E, and from taking cable portionABC as a free-body and summing moments about C. KnownCoOrds • Calculate elevation of B by considering AB as a free-body and summing moments B. Similarly, calculate elevation of D using ABCD as a free-body. • Evaluate maximum slope and maximum tension which occur in DE.

  10. Determine two reaction force components at A from solution of two equations formed from taking the Example  Concentrated Loads entire cable as a free-body and summing moments about E:

  11. Next take Cable Section ABC as a Free-Body, and Sum the Moments about Point C Example  Concentrated Loads • Recall from ΣME • Solving 2-Eqns in 2-Unknowns for Ax & Ay

  12. Example  Concentrated Loads • Determine elevation of Bby considering AB as a free-body and summing moments about B. • Similarly, Calc elevation at D using ABCD as a free-body

  13. Maximum Tension Analysis • By the ΣFx = 0 • Solving for T • Thus T is Maximized by Maximum θ

  14. Find Maximum Segment Angle • Use the y-data just calculated to find the cable segment of steepest slope

  15. Evaluate maximum slope and maximum tension which occur in the segment with the STEEPEST Slope (large θ); DE in this case Example  Concentrated Loads • Use yD to Determine Geometry of tanθ • Employ the Just-Determined θto Find Tmax

  16. Distributed Loads on Cables • For a negligible-Weight Cable carrying a Distributed Load ofArbitrary Profile • The cable hangs in shape of a CURVE • INTERNAL force is a tension force directed along the TANGENT to the curve

  17. Distributed Loads (2) • Consider the Free-Body for a portion of cable extending from LOWEST point C to given point D. Forces are T0 (FH in the Text Book) at Lo-Pt C, and the tangential force T at D • From the Force Triangle

  18. Distributed Loads (3) • Some Observations based on • Horizontal component of T is uniform over the Cable Length • Vertical component of T is equal to the magnitude of W • Tension is minimum at lowest point (min θ), and maximum at A and B (max θ)

  19. Parabolic Cable • Consider a cable supporting a uniform, horizontally distributed load, e.g., support cables for a suspension bridge. • With loading on cable from lowestpoint C to a point D given by internal tension force, T, and Vertical Load W = wx, find: • Summing moments about D • The shape, y, is PARABOLIC:

  20. T0for Uniform Vertical Load • Consider the uniformly Loaded Cable • In this case: w(x) = w • w is a constant • L is the Suspension Span: • From Last Slide • Or • Then xB & xA • Thus L

  21. T0for Uniform Vertical Load • Factoring Out 2T0/w • Isolating T0 • If WE design the Suspension System, then we KNOW • L (Span) • w (Load) • yA & yB (Dims) • Example • L= 95 m (312 ft) • w = 640 N/m (44 lb/ft) • yA = 19m • yB = 37m • Then T0= 26 489 N (5955 lb) • Tmaxby

  22. Tmax for Uniform Vertical Load • Find xmax from • In this casexmax = 55.3 m (181 ft) • And finallyTmax = 44 228 N(9 943 lbs) • Buy Cable rated to 20 kip for Safety factor of 2.0 • MATLABCalcs >> L = 95 L = 95 >> w = 640 w = 640 >> yA = 19 yA = 19 >> yB = 37 yB = 37 >> TO =L^2*w/(2*(sqrt(yB)+sqrt(yA))^2) TO = 2.6489e+04 >> xB = sqrt(2*TO*yB/w) xB = 55.3420 >> Tmax = sqrt(TO^2 + (w*xB)^2) Tmax = 4.4228e+04

  23. Göteborg, Sweden http://www.nmt.edu/~armiller/bridgefu.htm Load-Bearing Cables End Loads VerticallyLoaded NO Deck-Support Cables Deck-Support COLUMNS • Tension in the Straight-Section is roughly Equal to the Parabolic-Tension at the Tower-Top. • So the Support Tower does Not Bend

  24. 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 reveals the internal tension force magnitude at Pt-D: • Where

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

  26. UNloaded Cable → Catenary (3) • Factoring Out c in DeNom • Finally the Differential Eqn • Integrate Both Sides using Dummy Variables of Integration: • σ: 0→x η: 0→s

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

  28. UNloaded Cable → Catenary (5) • Thus the Solution to the Integral Eqn • Then • Solving for s in terms of x by taking the sinh of both sides

  29. 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

  30. UNloaded Cable → Catenary (7) • Recall the Previous Integration That Relates x and s • Using s(x) above in the last ODE • Integrating the ODE with Dummy Variables: • Ω:c→yσ: 0→x

  31. UNloaded Cable → Catenary (8) • Noting that cosh(0) = 1 • Solving for y yields theCatenary Equation: • 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)

  32. Catenary Comments • With Hyperbolic-Trig ID: cosh2 – sinh2 = 1 • Or: • Recall From the Differential Geometry • or

  33. Loaded and Unloaded Cables Compared

  34. Translate the CoOrd System Vertically from Previous: y = 0 at Cable Minimum • Recall Eqn for y−c • Thus with Origin at cable Minimum • Sub y = yO+c

  35. Then y = 0 at Cable Minimum (2) L • Recall c = T0/w • Thus • Next, Change the Name of the Cable’s Lineal Specific Weight (N/m or lb/ft)

  36. With µ replacing w y = 0 at Cable Minimum (3) • In Summary can Use either Formulation based on Axes Origin: L

  37. Cable Length, S, for Catenary • Using this Axes Set • With Cable macro-segment and differential-segment at upper-right • The Force Triangle for the Macro-Segment L W = µs W = µs

  38. Cable Length, S , for Catenary • By Force Triangle • But by Differential Segment notice • By Transitive Property • Now ReCall • Then dy/dx • Subbing into the tanθ expression

  39. Cable Length, S , for Catenary • Solving the last Eqn for s • From the CoOrds • So Finally • Now find T(y) • Recall T = f(x) L

  40. T(y) for Catenary • Also ReCall • Solve above for cosh • Sub cosh into T(x) Expression

  41. Catenary Summary • y(x) • T(x)T(y) • S(x) • Slope at any pt • Angle θ at any pt

  42. WhiteBoard Work Let’s WorkThese NiceProblems

  43. Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  44. WhiteBoard Work Let’s WorkThese NiceProblems

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