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

The theory of cable deformation and tension, specifically in the case study of the George Washington Bridge. It explains how to determine the configuration of a cable from its deformation and how to calculate the tension in the cable. The document also provides equations for determining the shape of a cable under uniform load and the maximum tension in the cable.

Typology: Study Guides, Projects, Research

2021/2022

1 / 5

Download Theory of Cable Deformation and Tension and more Study Guides, Projects, Research Structural Design and Architecture in PDF only on Docsity! Draft Chapter 6 Case Study II: GEORGE WASHINGTON BRIDGE 6.1 Theory 1 Whereas the forces in a cable can be determined from statics alone, its configuration must be derived from its deformation. Let us consider a cable with distributed load p(x) per unit horizontal projection of the cable length (thus neglecting the weight of the cable). An infinitesimal portion of that cable can be assumed to be a straight line, Fig. 6.1 and in the absence of any horizontal load we have H =constant. Summation of the vertical forces yields (+ ?) ΣFy = 0 ⇒ −V + wdx + (V + dV ) = 0 (6.1-a) dV + wdx = 0 (6.1-b) where V is the vertical component of the cable tension at x (Note that if the cable was subjected to its own weight then we would have wds instead of wdx). Because the cable must be tangent to T , we have tan θ = V H (6.2) Substituting into Eq. 6.1-b yields d(H tan θ) + wdx = 0 ⇒ − d dx (H tan θ) = w (6.3) 2 But H is constant (no horizontal load is applied), thus, this last equation can be rewritten as − H d dx (tan θ) = w (6.4) 3 Written in terms of the vertical displacement v, tan θ = dv dx which when substituted in Eq. 6.4 yields the governing equation for cables − Hv′′ = w (6.5) 4 For a cable subjected to a uniform load w, we can determine its shape by double integration of Eq. 6.5 − Hv′ = wx + C1 (6.6-a) Draft6.1 Theory 125 −Hv = wx2 2 + C1x + C2 (6.6-b) and the constants of integrations C1 and C2 can be obtained from the boundary conditions: v = 0 at x = 0 and at x = L ⇒ C2 = 0 and C1 = −wL 2 . Thus v = w 2H x(L − x) (6.7) This equation gives the shape v(x) in terms of the horizontal force H, 5 Since the maximum sag h occurs at midspan (x = L 2 ) we can solve for the horizontal force H = wL2 8h (6.8) we note the analogy with the maximum moment in a simply supported uniformly loaded beam M = Hh = wL2 8 . Furthermore, this relation clearly shows that the horizontal force is inversely proportional to the sag h, as h ↘ H ↗. Finally, we can rewrite this equation as r def = h L (6.9-a) wL H = 8r (6.9-b) 6 Eliminating H from Eq. 6.7 and 6.8 we obtain v = 4h ( −x2 L2 + x L ) (6.10) Thus the cable assumes a parabolic shape (as the moment diagram of the applied load). 7 Whereas the horizontal force H is constant throughout the cable, the tension T is not. The maximum tension occurs at the support where the vertical component is equal to V = wL 2 and the horizontal one to H, thus Tmax = √ V 2 + H2 = √ ( wL 2 )2 + H2 = H √ 1 + ( wL/2 H )2 (6.11) Combining this with Eq. 6.8 we obtain1. Tmax = H √ 1 + 16r2 ≈ H(1 + 8r2) (6.12) 8 Had we assumed a uniform load w per length of cable (rather than horizontal projection), the equation would have been one of a catenary2. v = H w cosh [ w H ( L 2 − x )] + h (6.13) The cable between transmission towers is a good example of a catenary. 1Recalling that (a+b)n = an +nan−1b+ n(n−1) 2! an−2b2 + · or (1+b)n = 1+nb+ n(n−1)b2 2! + n(n−1)(n−2)b3 3! + · · ·; Thus for b2 << 1, √ 1 + b = (1 + b) 1 2 ≈ 1 + b 2 2Derivation of this equation is beyond the scope of this course. Victor Saouma Structural Concepts and Systems for Architects