# Equivalent Axial Stiffness of Horizontal Stays
## 1 Introduction
From Equations (2) and (3) it follows  
\[E_{t,eq}(\sigma_{0})=\frac{d\sigma_{0}}{d\varepsilon}=\frac{d\sigma_{0}}{d \varepsilon_{\varepsilon}+d\varepsilon_{f}}=\;\frac{E}{1+\frac{E}{E^{*}( \sigma_{0})}}, \tag{4}\]  
Hypothesizing a parabolic deformed configuration, and assuming the cable sag \(d\) much smaller than the chord \(a\), i.e., \(d<0.1\;a\), the length \(L\) of an inextensible cable can be approximated by  
\[L\cong a+\frac{8\;d^{2}}{3\;a}. \tag{5}\]  
Recalling that the cable sag depends on \(N_{0}\) through the equilibrium equation,  
\[d=\frac{p\;a^{2}}{8\;N_{0}}, \tag{6}\]  
the contribution to the chord variation due to the modification of the cable configuration induced by \(dN_{0}\cdot da^{*}\) can be expressed by  
\[da^{*}=-\frac{d(L-a)}{dN_{0}}dN_{0}\equiv-\frac{16}{3\;a}\frac{d}{a}\frac{d}{dN _{0}}\bigg{(}\frac{p\;a^{2}}{8\;N_{0}}\bigg{)}dN_{0}=\frac{p^{2}\;a^{3}}{12\;N_ {0}^{3}}dN_{0}. \tag{7}\]Finally, setting \(da^{\star}=a\;d\varepsilon_{f}\) and simplifying, Equation (7) provides  
\[\frac{1}{E_{f}}=\frac{d\varepsilon_{f}}{d\sigma_{0}}=\frac{\gamma^{2}\;a^{2}}{12 \;\sigma_{0}^{3}}. \tag{8}\]  
Remarking \(E_{f}=E^{\star}(\sigma_{0})\), [PERSON] deduces thus the equivalent modulus, expressed by Equation (1), simply substituting Equation (8) in Equation (4). When a large stress variation needs to be taken into account in the cable, it is common practice to consider, rather than the tangent modulus, the equivalent secant elastic modulus, \(E_{s,eq}\). Let \(\sigma_{01}\) be the normal stresses at the initial point of the given loading process and \(\sigma_{02}\) the normal stress at the final point of the given loading process; \(\Delta\sigma_{0}=\sigma_{02}-\sigma_{01}\), \(E_{s,eq}\) can be expressed by  
\[E_{s,eq}=\frac{\Delta\sigma_{0}}{\Delta\varepsilon}=\frac{E}{1+\frac{(\gamma \;a)^{2}}{24\;\sigma_{01}^{3}}\;\frac{1+\overline{q}}{\overline{q}^{2}}E}, \tag{9}\]  
where \(\overline{\sigma}=\sigma_{02}/\sigma_{01}\). According to [PERSON] [3], Equation (9) is obtained using the Dischinger approach and considering the chord length variation, \(\Delta a^{\star}\), resulting from a large variation, \(\Delta N_{0}=N_{02}-N_{01}\), of the horizontal component of the tensile force. By integrating Equation (7), after some elementary passages, it results  
\[ \begin{array}{rl}\Delta a^{\star}=\int\limits_{N_{01}}^{N_{02}}\frac{p^{2}\; a^{3}}{12\;N_{0}^{3}}dN_{0}&=\frac{p^{2}\;a^{3}}{24}\left(\frac{1}{N_{01}^{2}}- \frac{1}{N_{02}^{2}}\right)=\frac{\gamma^{2}\;a^{3}}{24}\left(\frac{1}{\sigma _{01}^{2}}-\frac{1}{\sigma_{02}^{2}}\right)\\ &=\frac{\gamma^{2}\;a^{3}}{24}\left(\frac{\sigma_{02}^{2}-\sigma_{01}^{2}}{ \sigma_{01}^{2}}\right)=\frac{\gamma^{2}\;a^{3}}{24\;\sigma_{01}^{3}}\;\frac{1 +\overline{q}}{\overline{q}^{2}}\Delta\sigma,\end{array}  \tag{10}\]  
and, consequently,  
\[\frac{\Delta\sigma_{0}}{\Delta\varepsilon_{f}}=\frac{1}{E_{s}^{\star}(\sigma _{01},\sigma_{02})}=\frac{\gamma^{2}\;a^{2}}{24\;\sigma_{01}^{3}}\;\frac{1+ \overline{\sigma}}{\overline{\sigma}^{2}}. \tag{11}\]  
As a matter of fact, these solutions are satisfactory only when the cable highly stressed and, therefore, the cable sag is small with respect to its chord; consequently, several theoretical and numerical studies have been devoted to refine the solutions, aiming to enlarge their field of application [4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15]. Among them, particularly remarkable appear the approach adopted by [PERSON] [11], who introduced the characteristic parameter of a suspended cable, \(\lambda^{2}\), including both the shape and the deformational properties of the cable. The abovementioned parameter \(\lambda^{2}\) is defined by  
\[\lambda^{2}=\left(\frac{\gamma\;a}{\sigma_{0}}\right)^{2}\frac{a\;E}{\sigma_{0 }\;L_{e}}, \tag{12}\]  
where \(L_{e}\),  
\[L_{e}=\int\limits_{0}^{a}\left(\frac{ds}{dx}\right)^{3}dx, \tag{13}\]  
is a relevant property of the cable, usually known as virtual length.