# Innovative Suspension Structures: The Role of Straight Elements Under Asymmetric Loads
## 1 Introduction
### Analysis of Asymmetrically Loaded Innovative Single-Level Suspension System
#### 2.2.2 Calculation of the Asymmetrically Loaded Lower Chord  
In an asymmetrical loading of the system, the elements of the lower chord will deform as shown in Figure 5. The node \"3\" (central hinge) will shift not only vertically by the value \(\Delta v_{3}\) but also horizontally by the value \(\Delta h_{3}\) towards the side of the structure loaded with the asymmetric load. In this case, the displacement of nodes \"4\" and \"7\" will be \(\Delta f_{4}\) and \(\Delta f_{7}\). It is obvious that the forces of the struts \(F_{4}\) (left side) and \(F_{7}\) (right side) will not be equal as a result of the different loads acting on the upper chord parts (left and right). The left and right sides of the lower chord can be considered separate suspension cables (structures) consisting of two straight elements \"1-4\", \"3-4\" and \"3-7\", \"6-7\", with a common connection at node \"3\" (see Figure 5). Therefore, as in the calculation of the upper chord, we will analyze the separate parts of the lower chord. The calculation scheme for the left of the lower chord is presented in Figure 6a. The lower chord of the two element suspension structure is acted upon by a strut (its axial force), which transfers the concentrated load from the upper chord. The tensile force  
Figure 5: Deformed scheme of an asymmetrically loaded lower chord. Figure 6: Deformed scheme of the left (**a**) and right (**b**) side of the lower chord. of this suspension structure at the central node \"3\" will be indicated by \(H_{3}\). When loading with a concentrated force \(F_{4}\) onto a scheme of the lower part of the chord, it becomes obvious that the axial forces of elements \"1-4\" and \"3-4\" will be equal. The average tensile force of this suspended structure is calculated according to:  
\[H_{m,ur}=0.125F_{4}(l-\Delta{h_{3}})/(f_{2A}+\Delta{f_{4}}), \tag{35}\]  
where \(f_{2A}\)--the height of the strut (see Figure 6). Then the tensile force \(H_{3r}\) of the lower chord can be calculated as follows:  
\[H_{3r}=H_{mur}+\Delta{H_{r}}\, \tag{36}\]  
where \(\Delta{H_{r}}=F_{4}sin{\varphi_{mr}}/2\). The formula for calculating the axial forces in elements \"1-4\" and \"3-4\" will be obtained by:  
\[N_{1,4}=N_{3,4}=\frac{H_{mr}}{cos{\varphi_{mr}}}\left[1+4\frac{\left(f_{2A}+ \Delta\ f_{4}\right)^{2}cos{\varphi_{mr}}}{\left(l-\Delta{h_{3}}\right)^{2}} \right]^{0.5}. \tag{37}\]  
To avoid errors in simplified calculations of suspension structures, we write the following deformation consistency equation:  
\[2\Delta{s_{el,r}}\cdot s_{0,r}=s_{1,r}^{2}-s_{0,r}^{2}. \tag{38}\]  
Then  
\[2\Delta{s_{el,r}}\cdot s_{0,r}=\frac{F_{2A}\ l^{2}\left(l-\Delta{h_{3}}\right) \ \propto^{1,5}}{2\ \ EA\ \left(f_{2A}+\Delta\ f_{2 mr}\right)cos^{2}\ \varphi_{0}\ cos{\varphi_{mr}}}\ ; \tag{39}\]  
\[s_{1,r}^{2}-s_{0,r}^{2}=\left(l-\Delta{h_{3}}\right)^{2}-l^{2}+\left(f_{0}+ \Delta{f_{c}}\right)^{2}-f_{0}^{2}+4\left[\left(f_{2A}+\Delta{f_{2 mr}}\right) ^{2}-f_{2,4}^{2}\right], \tag{40}\]  
where \(\propto^{1,5}=1+4f_{2,4}(\frac{cos{\varphi_{0}}}{l})^{2}\). The concentrated force acting on the lower chord element at node \"4\" is determined from the following Equation:  
\[F_{4}=\frac{8(f_{2A}+\Delta f_{2 mr})Ecos^{2}{\varphi_{0}cos{\varphi_{mr}}}}{ \left[l^{2}\left(l-\Delta{h_{3}}\right)\ \propto^{1,5}\right.}\left[\frac{\left(l-\Delta{h_{3}}\right)^{2}-l^{2}}{4}- \frac{\left(f_{0}+\Delta{f_{c}}\right)^{2}-f_{0}^{2}}{4}+\left(f_{2A}+\Delta {f_{2 mr}}\right)^{2}-f_{2A}^{2}\right]. \tag{41}\]  
From expression (41) it is clearly seen that the axial force in the strut \(F_{2A}\) is variable and depends not only on the displacement of the node \"2\" \(\Delta\ f_{2 mr}\), but also on the displacement of the central node \"3\" \(\Delta f_{c}\). Under the assumption that strut \"2-4\" does not deform, equality of the displacements of the upper and lower chords \(\Delta f_{2 mr}=\Delta f_{4}\) will be obtained.