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Licensed Electrical & Mechanical Engineer

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1 Licensed Electrical & Mechanical Engineer
Engineering 36 Chp 7: Flex Cables Bruce Mayer, PE Licensed Electrical & Mechanical Engineer

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

9 Example  Concentrated Loads
free-body and summing moments about E, and from taking cable portion ABC as a free-body and summing moments about C. Known CoOrds 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. Solution Plan Determine reaction force components at pt-A from solution of two equations formed from taking entire cable as a Since Tx = const, and T is on LoA of cable, steepest cable section yields highest T by vector addition: T = Tx + Ty

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

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

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

16 Distributed Loads on Cables
For a negligible-Weight Cable carrying a Distributed Load of Arbitrary 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) T = T0/cosɵ => minmum at ɵ = 0

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 lowest point C to a point D given by internal tension force, T, and Vertical Load W = wx, find: w in lb/ft or N/m Summing moments about D The shape, y, is PARABOLIC:

20 T0 for 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 T0 for 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 = N (5955 lb) Tmax by

22 Tmax for Uniform Vertical Load
Find xmax from In this case xmax = 55.3 m (181 ft) And finally Tmax = N (9 943 lbs) Buy Cable rated to 20 kip for Safety factor of 2.0 MATLAB Calcs >> 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 = >> Tmax = sqrt(TO^2 + (w*xB)^2) Tmax = 4.4228e+04

23 NO Deck-Support Cables
Load-Bearing Cables Göteborg, Sweden End Loads Vertically Loaded NO Deck-Support Cables Deck-Support COLUMNS The STRAIGHT part is a 2-Force member 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: s = cable-wt per unit-length 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→y σ: 0→x When y=c, then x=0 ** at an arbitrary pt when y = y, then x = x

31 UNloaded Cable → Catenary (8)
Noting that cosh(0) = 1 Solving for y yields the Catenary 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 y = 0 at Cable Minimum Translate the CoOrd System Vertically from Previous: Recall Eqn for y−c Thus with Origin at cable Minimum Sub y = yO+c

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

36 y = 0 at Cable Minimum (3) With µ replacing w
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 W = µs L The Force Triangle for the Macro- Segment 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 Let’s Work These Nice Problems
WhiteBoard Work Let’s Work These Nice Problems H13e P7-119, => ENGR36_H13_Tutorial_Catenary_Cables_1207.pptx

43 Registered Electrical & Mechanical Engineer
Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer

44 Let’s Work These Nice Problems
WhiteBoard Work Let’s Work These Nice Problems ST P10.2.[12,28] => ENGR-36_Lab-24_Fa07_Lec-Notes.ppt



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