# Equivalent Axial Stiffness of Horizontal Stays
## 1 Introduction
When the chord length, \(a\), is much bigger than the sag of the cable, \(d\), \(L_{e}\) can be acceptably estimated as  
\[L_{e}\cong a\left(1+8\left(\frac{d}{a}\right)^{2}\right). \tag{14}\]  
For cable-stayed bridges, the parameter \(\lambda^{2}\) is usually smaller than one, but for long stays it can be even bigger than three. For suspension bridges, \(\lambda^{2}\) is usually bigger than 100 [6]. Following Irvine [11], it is possible to estimate the equivalent modulus, \(E_{t,eq1}\), using the simplified expression  
\[E_{t,eq1}=\frac{d\sigma}{d\varepsilon}=\frac{E}{1+\frac{\lambda^{2}}{12}}. \tag{15}\]  
When applicable, Equation (15) is much more precise than Equation (1), although, due to the approximations introduced in deriving it, it leads to some surprisingly bizarre results when \(\sigma_{0}\) is small, \(a\) is great and the virtual length \(L_{e}\) is calculated via exact integration of Equation (13), and not applying the simplified Equation (14). Although arriving at simple and easy to handle formulae, the application of classical methods is subject to severe restrictions, which limit their fields of application, especially for flabby cables. In fact, these approaches do not completely consider the effect of the variation of the cable configuration upon the arrangement of external loads. To allow a general application of the equivalent modulus concept, exact closed form solutions are thus necessary. In a previous paper [16], we suggested an original method, based on the virtual work theorem, to find the equivalent tangent modulus of an inclined stay, taking into account the effect of its self-weight. Assuming a deformed catenary shape of the cable, that method allowed to derive accurate closed-form solutions describing the nonlinear behavior of the cable, in terms of equivalent tangent stiffness. Since deduced considering all relevant aspects of the problem, the formulae proposed in [16] are valid on the whole field of interest, independently on the inclination of the cable, on the stress level, on the sag of the catenary, and on the cable material, so allowing a generalized application of the equivalent stiffness approach. That method provides different expressions, depending on the nature of the relative displacement between the extremities of the cable, i.e., depending on whether both end sections are fixed, or one end section is fixed and the other can moves during the loading process; in other words, depending on whether the cable length is fixed or not during the loading process. Moreover, the solutions are so refined that, in case both cable ends are fixed, they are function of the absolute sign of the vertical component, \(dh\), of the relative displacement between the cable ends themselves, that governs the work done by the self-weight of the cable as a result of that relative displacement. In fact, in [16] (page 1098), where \(k=dh/da\) indicates the ratio between the vertical and the horizontal components of the relative displacement of the cable ends \(A\) and \(B\), and \(\sigma\) the normal stress in the cable whose horizontal component is \(\sigma_{0}\), it is remarked that: \"It must also be emphasized that the apparent modulus can result larger, even significantly, than \(E\), for example when \(kdh>0\) in \(A\), because of the weight of the cable. In this case, for \(\sigma\rightarrow\infty\), the curves \(E_{ft,eqA}=E_{ft,eqA}(\sigma)\) and \(E_{ft,eqB}=E_{ft,eqB}(\sigma)\) approach the horizontal asymptote \(E_{ft,eqA}=E\) from above and from below, respectively. In addition, an in-depth examination of the formula (36.b) induces to infer that negative values of the apparent modulus at the cable end \(B\) cannot be excluded, for example when \(\sigma_{0}\) is small and \(k>0\), and that the region where the apparent modulus is negative raises as the span increases. Negative values of the apparent modulus, clearly unacceptable in design practice, happen when increase of the chord length implies decrease of the stress at the cable end, or vice versa: this is again due to the cable weight.