# Innovative Suspension Structures: The Role of Straight Elements Under Asymmetric Loads
## 1 Introduction
### Calculation of Suspension Structure with Two Straight Elements
By means of the equation for deformation compatibility of the suspended element, we obtain the design formula of the fictitious displacement of the tension \"rigid\" element \"1-2\":  
\[\Delta f_{fc,1}=\sqrt{\frac{3\ (g+s)\ l^{4}}{64\ EA\cdot\Phi(k_{1}l)\cdot\beta_{1 }\ cos^{5}\ \varphi_{1}}}, \tag{3}\]  
\[\text{here}\ \Phi(kl_{1})=1-\frac{24}{(k_{1}l)^{2}}\bigg{(}1-\frac{2}{k_{1}l }th\frac{k_{1}l}{2}\bigg{)}+\frac{6\ (shkl_{1}-k_{1}l)}{(k_{1}l)^{3}ch^{2}(k_{1}l/2)}; \tag{4}\]  
where \(s_{0}=\
icefrac{{1}}{{cos\varphi_{0}}}\)--initial length of element \"1-2\"; \(\Delta s_{el,1}=\
icefrac{{H_{1}l}}{{EA\cos\varphi_{0}\cos\varphi_{1}}}\)--elastic elongation of element \"1-2\"; and \(s_{1}=l+0.5\int_{0}^{l}\big{[}w_{1}^{\prime}(x)\big{]}^{2}dx\)--length of element \"1-2\" after deformation:  
\[\beta_{1}=1+\frac{3}{8}\frac{l}{\Delta f_{fc,1}{}^{2}cos\varphi_{1}\Phi(k_{1} l)}\bigg{(}\frac{l}{cos\varphi_{1}}-\frac{l}{cos\varphi_{0}}\bigg{)}. \tag{5}\]  
It should be noted that Formula (3) is analogous to the formula for calculating the displacement of an absolutely flexible string [48] and the function \(\Phi(k_{1}l)\) evaluates the influence of stiffness \(EI\) on the displacements of the tension element. An expression for calculating the fictitious displacement of the right-side element \"2-3\" loaded only with a permanent symmetrical load g is obtained in the same way (see Figure 2a) as follows:  
\[\Delta f_{fc,2}=\sqrt[3]{\frac{3\ g\ l^{4}}{64\ EA\cdot\Phi(k_{2}l)\cdot\beta _{2}\ cos^{5}\varphi_{2}}}, \tag{6}\]  
\[\text{here}:\ \Phi(k_{2}l)=1-\frac{24}{\big{(}k_{2}l\big{)}^{2}}\bigg{(}1-\frac{2 }{k_{2}l}th\frac{k_{2}l}{2}\bigg{)}+\frac{6\ (shk_{2}l-k_{2}l)}{\big{(}k_{2}l\big{)}^{3}ch^{2}(k_{2}l/2)}; \tag{7}\]  
\[\beta_{2}=1+\frac{3}{8}\frac{l}{\Delta f_{fc,2}{}^{2}cos\varphi_{1}\Phi(k_{2} l)}\bigg{(}\frac{l}{cos\varphi_{2}}-\frac{l}{cos\varphi_{0}}\bigg{)}. \tag{8}\]  
Calculations of all suspension structures consisting of elements \"1-2\" and \"2-3\" are performed in the iteration procedure. Asymmetric loading increases the complexity and sensitivity of structural calculations. The main unknowns of the iterative calculation are \(\Delta\)f_(fic,1), \(\Delta\)f_(fic,2), \(\Delta\)f_c, \(\Delta\)h_c, and H_1 and H_2. Obviously, the iterative calculation process becomes more complicated than in the case of symmetrical loading [47]. When a certain initial value of the displacement \(\Delta\)f_c of the vertical central hinge in the iterative calculation is accepted, you can also calculate the horizontal displacement of this hinge as follows:  
Figure 2:0  
Formula (9) shows that the horizontal displacement of the central node \(\Delta\)h_c depends not only on the load ratio \(\gamma\) and the vertical displacement of the central node \"2\" Af_c, but also on the initial sag of the structure f_0 and the axial stiffness EA of the straight tension and bending bar. This introduces additional challenges compared with symmetric loading in determining the displacements and forces within the structure. To simplify the calculations of the suspension structure, the average fictitious displacement \(\Delta f_{fic,m}\) can be used, which is calculated as follows:  
Figure 2:1  
where \(\varphi_{m}=0.5\) (\(\varphi_{1}+\varphi_{3}\))--the average angle of straight suspension elements after deformation. Then the following additional equation can be written:  
Figure 2:2  
where \(L=2l\)--total span of the structure.