7.4 Cables Flexible cables and chains are used to support and transmit loads from one member to another In suspension bridges and trolley wheels, they carry majority of the load In force analysis, weight of cables is neglected as it is small compared to the overall weight Consider three cases: cable subjected to concentrated loads, subjected to distributed load and subjected to its own weight

7.4 Cables Assume that the cable is perfectly flexible and inextensible Due to its flexibility, the cables offers no resistance to bending and therefore, the tensile force acting in the cable is always tangent to the points along its length Being inextensible, the length remains constant before and after loading, thus can be considered as a rigid body

7.4 Cables Cable Subjected to Concentrated Loads For a cable of negligible weight supporting several concentrated loads, the cable takes the form of several straight line segments, each subjected to constant tensile force

7.4 Cables Cable Subjected to Concentrated Loads Known: h, L1, L2, L3 and loads P1 and P2 Form 2 equations of equilibrium at each point A, B, C and D If the total length L is given, use Pythagorean Theorem to relate the three segmental lengths If not, specify one of the sags, yC and yD and from the answer, determine the other sag and hence, the total length L

7.4 Cables Example 7.13 Determine the tension in each segment of the cable.

7.4 Cables Solution FBD for the entire cable

7.4 Cables Solution

7.4 Cables Solution Consider leftmost section which cuts cable BC since sag yC = 12m

7.4 Cables Solution

7.4 Cables Solution Consider point A, C and E,

7.4 Cables Solution Point A

7.4 Cables Solution Point C

7.4 Cables Solution Point E

7.4 Cables Solution By comparison, maximum cable tension is in segment AB since this segment has the greatest slope For any left hand side segment, the horizontal component Tcosθ = Ax Since the slope angles that the cable segment make with the horizontal have been determined, the sags yB and yD can be determined using trigonometry

7.4 Cables Cable Subjected to a Distributed Load Consider weightless cable subjected to a loading function w = w(x) measured in the x direction

7.4 Cables Cable Subjected to a Distributed Load For FBD of the cable having length ∆

7.4 Cables Cable Subjected to a Distributed Load Since the tensile force in the cable changes continuously in both the magnitude and the direction along the cable’s length, this change is denoted on the FBD by ∆T Distributed load is represented by its resultant force w(x)(∆x) which acts a fractional distance k(∆x) from point O where o < k < 1

7.4 Cables Cable Subjected to a Distributed Load

7.4 Cables Cable Subjected to a Distributed Load Divide by ∆x and taking limit, Integrating,

7.4 Cables Cable Subjected to a Distributed Load Integrating, Eliminating T, Perform second integration,

7.4 Cables Example 7.14 The cable of a suspension bridge supports half of the uniform road surface between the two columns at A and B. If this distributed loading wo, determine the maximum force developed in the cable and the cable’s required length. The span length L and, sag h are known.

7.4 Cables Solution Note w(x) = wo Perform two integrations Boundary Conditions at x = 0

7.4 Cables Solution Therefore, Curve becomes This is the equation of a parabola Boundary Condition at x = L/2

7.4 Cables Solution For constant, Tension, T = FH/cosθ Maximum tension occur at point B for 0 ≤ θ ≤ π/2

7.4 Cables Solution Slope at point B Or Therefore Using triangular relationship

7.4 Cables Solution For a differential segment of cable length ds Determine total length by integrating Integrating yields

7.4 Cables Cable Subjected to its Own Weight When weight of the cable is considered, the loading function becomes a function of the arc length s rather than length x For loading function w = w(s) acting along the cable,

7.4 Cables Cable Subjected to its Own Weight FBD of a segment of the cable

7.4 Cables Cable Subjected to its Own Weight Apply equilibrium equations to the force system Replace dy/dx by ds/dx for direct integration

7.4 Cables Cable Subjected to its Own Weight Therefore Separating variables and integrating

7.4 Cables Example 7.15 Determine the deflection curve, the length, and the maximum tension in the uniform cable. The cable weights wo = 5N/m.

7.4 Cables Solution For symmetry, origin located at the center of the cable Deflection curve expressed as y = f(x) Integrating term in the denominator

7.4 Cables Solution Substitute So that Perform second integration or

7.4 Cables Solution Evaluate constants or dy/dx = 0 at s = 0, then C1 = 0 To obtain deflection curve, solve for s

7.4 Cables Solution Hence Boundary Condition y = 0 at x = 0 For deflection curve, This equations defines a catenary curve

7.4 Cables Solution Boundary Condition y = h at x = L/2 Since wo = 5N/m, h = 6m and L = 20m, By trial and error,

7.4 Cables Solution For deflection curve, x = 10m, for half length of the cable Hence Maximum tension occurs when θ is maximum at s = 12.1m

7.4 Cables Solution

Chapter Summery Internal Loadings If a coplanar force system acts on a member, then in general a resultant normal force N, shear force V and bending moment M will act at any cross-section along the member These resultants are determined using the method of sections The member is sectioned at the point where the internal loadings are to be determined

Chapter Summery Internal Loadings A FBD of one of the sectioned parts is drawn The normal force is determined from the summation of the forces normal to the cross-section The shear force is determined from the summation of the forces tangent to the cross section The bending moment is found from the summation of the moments about the centroid of the cross-sectional area

Chapter Summery Internal Loadings If the member is subjected to 3D loading, in general, a torisonal loading will also act on the cross-section It can be determined by the summation of the moments about an axis that is perpendicular to the cross section and passes through its centroid

Chapter Summery Shear and Moment Diagrams as Functions of x Section the member at an arbitrary point located distance x from one end to construct the shear and moment diagrams for a member Unknown shear and moment are indicated on the cross-section in the positive direction according to the established sign convention Applications of the equilibrium equations will make these loadings functions of x which then, can be plotted

Chapter Summery Shear and Moment Diagrams as Functions of x If the external loadings or concentrated forces and couple moments act on the member, then different expressions of V and M must be determined within regions between these regions Graphical Methods for Establishing Shear and Moment Diagrams Plot the shear and moment diagrams using differential relationships that exist between the distributed loading w and V and M

Chapter Summery Graphical Methods for Establishing Shear and Moment Diagrams Slope of the shear diagram = distributed loading at any point dV/dx = -w Slope of the moment diagram = shear at any point V = dM/dx Change in shear between an y two points = area under the shear diagram between points, ∆M = ∫Vdx

Chapter Summery Cables When a flexible and inextensible cable is subjected to a series of concentrated forces, the analysis of the cable can be performed by using the equations of equilibrium applied to the FBD of wither segments or points of application of the loading If the external distributed loads or weight of the cables are considered, forces and shape of the cable must be determined by analysis of the forces on a differential segment of the cable and integrating the result

Chapter Review

Chapter Review

Chapter Review

Chapter Review

Chapter Review