Presentation is loading. Please wait.

Presentation is loading. Please wait.

ENGR-36_Lec-20_Cables.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical.

Similar presentations


Presentation on theme: "ENGR-36_Lec-20_Cables.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical."— Presentation transcript:

1 ENGR-36_Lec-20_Cables.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 36 Chp 7: Flex Cables

2 ENGR-36_Lec-20_Cables.pptx 2 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Recall Chp10 Introduction Examine in Detail Two Important Types Of Engineering Structures: 1.BEAMS - usually long, straight, prismatic members designed to support loads applied at various points along the member 2.CABLES - flexible members capable of withstanding only tension, designed to support concentrated or distributed loads

3 ENGR-36_Lec-20_Cables.pptx 3 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Load-Bearing Cables Cables are applied as structural elements in suspension bridges, transmission lines, aerial tramways, guy wires for high towers, etc. Göteborg, Sweden Curved Straight WHY the Difference?

4 ENGR-36_Lec-20_Cables.pptx 4 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Concentrated Loads on Cables To Determine the Cable SHAPE, Assume: a)Concentrated vertical loads on given vertical lines b)Weight of cable is negligible c)Cable is flexible, i.e., resistance to bending is small d)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 ENGR-36_Lec-20_Cables.pptx 5 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Concentrated Loads (2) Consider entire cable as a free-body Slopes of cable at A and B are NOT known FOUR unknowns (i.e., A x, A y, B x, B y ) are involved and the equations of equilibrium are NOT sufficient to determine the reactions.

6 ENGR-36_Lec-20_Cables.pptx 6 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 7 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Concentrated Loads (4) The 4 Eqns Yield A x & A y Can Now Work our Way Around the Cable to Find VERTICAL DISTANCE (y-CoOrd) For ANY OTHER point Example Consider Pt C 2 known UNknown known

8 ENGR-36_Lec-20_Cables.pptx 8 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Concentrated Loads 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: a)The elevation of points B and D b)The maximum slope and maximum tension in the cable.

9 ENGR-36_Lec-20_Cables.pptx 9 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Concentrated Loads Solution Plan Determine reaction force components at pt-A from solution of two equations formed from taking entire cable as a free-body and summing moments about E, and from taking cable portion ABC as a free-body and summing moments about C. 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. Known CoOrds

10 ENGR-36_Lec-20_Cables.pptx 10 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Concentrated Loads Determine two reaction force components at A from solution of two equations formed from taking the entire cable as a free-body and summing moments about E:

11 ENGR-36_Lec-20_Cables.pptx 11 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Concentrated Loads Next take Cable Section ABC as a Free-Body, and Sum the Moments about Point C Recall from ΣM E Solving 2-Eqns in 2-Unknowns for A x & A y

12 ENGR-36_Lec-20_Cables.pptx 12 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 13 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Maximum Tension Analysis By the ΣF x = 0 Solving for T Thus T is Maximized by Maximum θ

14 ENGR-36_Lec-20_Cables.pptx 14 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Find Maximum Segment Angle Use the y-data just calculated to find the cable segment of steepest slope

15 ENGR-36_Lec-20_Cables.pptx 15 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example Concentrated Loads Use y D 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 T max

16 ENGR-36_Lec-20_Cables.pptx 16 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Distributed Loads on Cables For a negligible- Weight Cable carrying a Distributed Load of Arbitrary Profile a)The cable hangs in shape of a CURVE b)INTERNAL force is a tension force directed along the TANGENT to the curve

17 ENGR-36_Lec-20_Cables.pptx 17 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Distributed Loads (2) Consider the Free-Body for a portion of cable extending from LOWEST point C to given point D. Forces are T 0 (F H in the Text Book) at Lo-Pt C, and the tangential force T at D From the Force Triangle

18 ENGR-36_Lec-20_Cables.pptx 18 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 19 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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: Summing moments about D The shape, y, is PARABOLIC:

20 ENGR-36_Lec-20_Cables.pptx 20 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics T0 T0 T0 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 x B & x A Thus L

21 ENGR-36_Lec-20_Cables.pptx 21 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics T0 T0 T0 T0 for Uniform Vertical Load Factoring Out 2T 0 /w Isolating T 0 If WE design the Suspension System, then we KNOW L (Span) w (Load) y A & y B (Dims) Example L= 95 m (312 ft) w = 640 N/m (44 lb/ft) y A = 19m y B = 37m Then T 0 = N (5955 lb) T max by

22 ENGR-36_Lec-20_Cables.pptx 22 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics T max T max for Uniform Vertical Load Find x max from In this case x max = 55.3 m (181 ft) And finally T max = N (9 943 lbs) Buy Cable rated to 20 kip for Safety factor of 2.0 >> 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 = e+04 >> xB = sqrt(2*TO*yB/w) xB = >> Tmax = sqrt(TO^2 + (w*xB)^2) Tmax = e+04 MATLAB Calcs

23 ENGR-36_Lec-20_Cables.pptx 23 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Load-Bearing Cables Tension in the Straight-Section is roughly Equal to the Parabolic-Tension at the Tower-Top. So the Support Tower does Not Bend Göteborg, Sweden Vertically Loaded End Loads NO Deck-Support Cables Deck-Support COLUMNS

24 ENGR-36_Lec-20_Cables.pptx 24 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 25 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics UNloaded Cable Catenary (2) Next, relate horizontal distance, x, to cable-length s But by Force Balance Triangle Thus Also From last slide recall

26 ENGR-36_Lec-20_Cables.pptx 26 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics UNloaded Cable Catenary (3) Factoring Out c in DeNom Integrate Both Sides using Dummy Variables of Integration: σ: 0x η: 0s Finally the Differential Eqn

27 ENGR-36_Lec-20_Cables.pptx 27 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics UNloaded Cable Catenary (4) Using σ: 0x η: 0s Now the R.H.S. AntiDerivative is the argSINH Noting that

28 ENGR-36_Lec-20_Cables.pptx 28 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 29 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics UNloaded Cable Catenary (6) Finally, Eliminate s in favor of x & y. From the Diagram So the Differential Eqn From the Force Triangle And From Before

30 ENGR-36_Lec-20_Cables.pptx 30 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics UNloaded Cable Catenary (7) Recall the Previous Integration That Relates x and s Integrating the ODE with Dummy Variables: Ω: cy σ: 0x Using s(x) above in the last ODE

31 ENGR-36_Lec-20_Cables.pptx 31 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics UNloaded Cable Catenary (8) Noting that cosh(0) = 1 Where –c = T 0 /w –T 0 = the 100% laterally directed force at the y min point –w = the lineal unit weight of the cable (lb/ft or N/m) Solving for y yields the Catenary Equation:

32 ENGR-36_Lec-20_Cables.pptx 32 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Catenary Comments With Hyperbolic-Trig ID: cosh 2 – sinh 2 = 1 Recall From the Differential Geometry Or: or

33 ENGR-36_Lec-20_Cables.pptx 33 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Loaded and Unloaded Cables Compared

34 ENGR-36_Lec-20_Cables.pptx 34 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics y = 0 at Cable Minimum Translate the CoOrd System Vertically from Previous: Recall Eqn for yc Sub y = y O +c Thus with Origin at cable Minimum

35 ENGR-36_Lec-20_Cables.pptx 35 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics y = 0 at Cable Minimum (2) Then Recall c = T 0 /w Thus Next, Change the Name of the Cables Lineal Specific Weight (N/m or lb/ft) L

36 ENGR-36_Lec-20_Cables.pptx 36 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics y = 0 at Cable Minimum (3) With µ replacing w In Summary can Use either Formulation based on Axes Origin: L

37 ENGR-36_Lec-20_Cables.pptx 37 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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

38 ENGR-36_Lec-20_Cables.pptx 38 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 39 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 ENGR-36_Lec-20_Cables.pptx 40 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics T(y) T(y) for Catenary Also ReCall Solve above for cosh Sub cosh into T(x) Expression

41 ENGR-36_Lec-20_Cables.pptx 41 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Catenary Summary y(x) T(x) T(y) S(x) Slope at any pt Angle θ at any pt

42 ENGR-36_Lec-20_Cables.pptx 42 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work Lets Work These Nice Problems

43 ENGR-36_Lec-20_Cables.pptx 43 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Registered Electrical & Mechanical Engineer Engineering 36 Appendix

44 ENGR-36_Lec-20_Cables.pptx 44 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work Lets Work These Nice Problems

45 ENGR-36_Lec-20_Cables.pptx 45 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

46 ENGR-36_Lec-20_Cables.pptx 46 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics


Download ppt "ENGR-36_Lec-20_Cables.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical."

Similar presentations


Ads by Google