Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

These are the Lecture Slides of Engineering Mechanics Statics which includes Free Body Diagrams, Magnitude and Direction of Forces, Coordinate System, Newton's Third Law, Structural Supports, Sliding and Free Vectors, Center of Gravity, Center of Gravity, Plane of Symmetry etc.Key important points are: Flex Cables, Structural Elements, Equations of Equilibrium, Segment of Steepest Slope, Concentrated Loads, Geometry of Tangent, Parabolic Cable, Internal Tension Force, Dummy Variables of Integra

Typology: Slides

2012/2013

1 / 41

Download Flex Cables - Engineering Mechanics Statics - Lecture Slides and more Slides Mechanical Engineering in PDF only on Docsity! Engineering 36 Chp 7: Flex Cables Docsity.com 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 Docsity.com 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. Docsity.com 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: 0DM 33221103 xPxPxPLBdBM yxA With pt-D info, the FOUR Equilibrium Eqns xxx BAF 01̀ 12102 PPPBAF yyy xAyAxxPM yxD 1104 Docsity.com 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 2 yields02 yMC yxyx TTFF , yields 0,0 Example Consider Pt C2 constantcos xx ATT known known known UNknown known known known Docsity.com 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: 333 OR 06606020 041512306406020 :0 yx yx yx E AA AA AA M Docsity.com Example Concentrated Loads • Next take Cable Section ABC as a Free- Body, and Sum the Moments about Point C Recall from ΣME 126OR 0610305 :0 yx yx C AA AA M 333 yx AA Solving 2-Eqns in 2-Unknowns for Ax & Ay kips 5 kips 18 y x A A Docsity.com 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 020518:0 BB yM ft 56.5By 0121562554518 :0 D D y M ft83.5Dy Docsity.com Example Concentrated Loads • Evaluate maximum slope and maximum tension which occur in the segment with the STEEPEST Slope (large θ); DE in this case Use yD to Determine Geometry of tanθ 15 17.14 tan 4.43 Employ the Just-Determined θ to Find Tmax cos kips 18 max T kips 8.24max T Docsity.com 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 x y Docsity.com 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 0 22 0 0 tan sincos T W WTT WTTT Docsity.com 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 L AB xxL 0 2 2T wx y wyTx 02 wyTx wyTx AA BB 0 0 2 2 wyTwyTL AB 00 22 Docsity.com 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 = 26 489 N (5955 lb) • Tmax by AB yywTL 20 2 2 0 2 AB yy wL T 2 max 22 0max xwTT Docsity.com Tmax for Uniform Vertical Load • Find xmax from • In this case xmax = 55.3 m (181 ft) • And finally Tmax = 44 228 N (9 943 lbs) – Buy Cable rated to 20 kip for Safety factor of 2.0 • MATLAB Calcs wyTx BB 02 >> 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 Docsity.com UNloaded Cable → Catenary (2) Next, relate horizontal distance, x, to cable-length s cosdsdx But by Force Balance Triangle Thus T T0cos Also From last slide recall wcTscwT 0 22 and ds sc c scw wc ds T T dsdsdx 2222 0cos Docsity.com UNloaded Cable → Catenary (3) Factoring Out c in DeNom Integrate Both Sides using Dummy Variables of Integration: • σ: 0→x η: 0→s ds csccc c ds sc c dx 222222 Finally the Differential Eqn ds cs dx 221 1 Docsity.com UNloaded Cable → Catenary (4) Using σ: 0→x η: 0→s Now the R.H.S. AntiDerivative is the argSINH Noting that sx d c d 0 220 1 1 s sxx c cd c d 0 0 22 0 0 sinharg 1 1 00sinh0sinharg 1 Docsity.com UNloaded Cable → Catenary (7) Recall the Previous Integration That Relates x and s Integrating the ODE with Dummy Variables: • Ω: c→y σ: 0→x x xy c y c c cd c d 0 0 coshsinh c x c s sinh Using s(x) above in the last ODE dx c x dx c s dx wc ws dx T W dxdy o sinhtan Docsity.com UNloaded Cable → Catenary (8) Noting that cosh(0) = 1 • 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) Solving for y yields the Catenary Equation: cxcy cosh c c x c c ccy x y c coshcosh 0 Docsity.com Catenary Comments With Hyperbolic-Trig ID: cosh2 – sinh2 = 1 Recall From the Differential Geometry Or: cxcy cosh 222222 222222 sinhcosh sinhcosh ccxcxcsy cxccxcsy 222222 sycandscy yTwyywscwscT 222, wTc 0 or 0 0 0 0 coshcosh T wx T T wx w T wwyxT Docsity.com y = 0 at Cable Minimum (2) • Then Oy Recall c = T0/w Thus Next, Change the Name of the Cable’s Lineal Specific Weight (N/m or lb/ft) 1 c x cyO cosh 1 0 0 wT x w T yO cosh w L Docsity.com y = 0 at Cable Minimum (3) • With µ replacing w Oy 1 0 0 T xT yO cosh In Summary can Use either Formulation based on Axes Origin: 0 0 cosh T wx w T y 1 0 0 T xT yO cosh wT0 L Docsity.com 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 Oy L W = µs W = µs Docsity.com T(y) for Catenary • Also ReCall • Solve above for cosh • Sub cosh into T(x) Expression 1cosh 0 0 T xT y 11cosh 000 T µy T µ y T x 0 0 0 0 0 1 cosh Tµy T µy TT T µx TT Docsity.com Catenary Summary • y(x) • T(x) T(y) • S(x) • Slope at any pt • Angle θ at any pt 1cosh 0 0 T xT y 0 0 0 cosh TµyT T µx TT 00 0 sinhsinh T µx T µx µ T S AB 00 sinh T µs T µx dx dy 00 sinhtan T µs T µx Docsity.com