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**Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu**

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

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**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

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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.

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**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

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**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.

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**Maximum Tension Analysis**

By the ΣFx = 0 Solving for T Thus T is Maximized by Maximum θ

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**Find Maximum Segment Angle**

Use the y-data just calculated to find the cable segment of steepest slope

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**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

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**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

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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

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**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

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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:

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**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

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**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

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**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

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**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

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**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

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**UNloaded Cable → Catenary (2)**

Next, relate horizontal distance, x, to cable-length s But by Force Balance Triangle Also From last slide recall Thus

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**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

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**UNloaded Cable → Catenary (4)**

Using σ: 0→x η: 0→s Now the R.H.S. AntiDerivative is the argSINH Noting that

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**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

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**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

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**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

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**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)

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**Catenary Comments With Hyperbolic-Trig ID: cosh2 – sinh2 = 1 Or:**

Recall From the Differential Geometry or

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**Loaded and Unloaded Cables Compared**

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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

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**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)

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**y = 0 at Cable Minimum (3) With µ replacing w**

In Summary can Use either Formulation based on Axes Origin: L

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**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

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**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

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**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

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**T(y) for Catenary Also ReCall Solve above for cosh**

Sub cosh into T(x) Expression

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

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**Let’s Work These Nice Problems**

WhiteBoard Work Let’s Work These Nice Problems H13e P7-119, => ENGR36_H13_Tutorial_Catenary_Cables_1207.pptx

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**Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu**

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

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**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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