Innovative Suspension Structures: The Role of Straight Elements Under Asymmetric Loads

Abstract

Suspension structures, known for their excellent properties, have been widely used to cover medium and large spans. Their efficiency lies in their ability to primarily withstand permanent and variable loads through tension. Consequently, suspension roof structures typically adopt a parabolic shape, which remains in equilibrium under symmetric loads. However, when subjected to asymmetric loads, such structures experience significant kinematic displacements. To reduce these displacements, suspension systems with bending stiffness, commonly referred to as “rigid” cables, are employed. Such elements increase the sustainability of the suspension system compared with conventional spiral ropes. Although previous studies have analyzed the behavior of such systems under symmetric loads, this article examines the performance of an innovative cable–strut system composed of straight “rigid” elements under asymmetric loads. The behavior of three different types of suspension structures under asymmetric loads is analyzed. A non-linear analysis of forces and displacements is conducted in this system, assessing the impact of bending stiffness on the structural response. The results indicate that the proposed two-level suspension system performs more effectively under asymmetric loads than both conventional parabolic suspension structures and suspension systems comprising two straight “rigid” elements. It was found that the total forces and stresses in the “rigid” upper chord elements of the two-level system are the lowest among all the systems considered. Therefore, this system is particularly suitable for covering medium- and large-span roofs, especially when subjected to relatively large asymmetric loads.

1. Introduction

With the increasing demand for covering large building roof spans, various steel structures have been employed, including suspension structures, which have proven to be highly successful [1,2]. In particular, this type of load-bearing structure allows for long spans, such as those seen on bridges [3,4]. Due to their efficiency and excellent architectural esthetics, suspension structures have been widely used for medium- and long-span roofs for several decades [5,6,7,8]. Given their relatively low structural mass, they are also classified as lightweight structures [9]. The efficiency of these structures lies in their ability to primarily withstand permanent and temporary loads through tensile (axial) forces [10,11]. Suspension roof structures come in various shapes, ranging from parabolically contoured single-curvature plates to spatial (double-curvature) configurations [12,13,14]. Among the suspension structures that can be used for both bridges and roofs, the two-span suspended structure with its upper chord designed as string deserves special mention, as this improves the working conditions of the deck [15]. The vast majority of these structures are designed using spiral ropes [16,17], which require pretensioning to control and reduce displacements [1,18]. However, these cables are susceptible to corrosion [19,20]. The shape of a suspension roof is most commonly a parabolic shape, as this closely approximates the equilibrium shape of the load-bearing structures (cables) under a distributed symmetric load [6,7,21]. However, this design has drawbacks, particularly with regard to asymmetric temporary loads (e.g., snow and wind), which can cause significant kinematic displacements in suspension cables [22,23]. Additionally, wind loads can induce considerable upward displacements in such structures [24]. It is also important to note that the modulus of elasticity of spiral ropes is lower than that of conventional structural steel, significantly increasing their deformability [17,25]. To mitigate this issue, heavy decking panels are often used in conjunction with prestressing, which increases the axial forces of suspension structures and requires massive anchoring systems [1,26].

To avoid such negative consequences, bending stiffness suspension load-bearing structures (EI ≠ 0) have been introduced in recent decades to stabilize the shape of suspension roofs [27,28,29]. These structures, also known as “rigid cables” or inverted arches, experience both tensile and bending stresses under load. Such “rigid cables” have also been proposed for suspension bridges [30,31]. Such “rigid cables” are more sustainable compared with spiral cables. Their use enables the application of lightweight partition structures on roofs, reducing tensile forces in cables and consequently decreasing the mass of anchoring systems. A key advantage of “rigid” suspension structures is that they do not require pretensioning or complex anchoring connections. Their cross-sections are typically made from conventional rolled or welded steel profiles, simplifying both manufacturing and installation [32,33]. Regarding these rigid suspension structure joints, we would like to note that the joints of the structure analyzed in this article are designed and manufactured in a similar way to the joints of conventional steel structures, using nodal plates and welding.

Additionally, corrosion poses a lesser threat to these structures compared with traditional spiral ropes. However, “rigid” cables are still predominantly designed in a parabolic shape, similar to suspension roofs using spiral ropes [27,28,29]. This curvature presents challenges not only in operation, but also in manufacturing. In response, new suspension structures composed of straight elements with bending stiffness have been introduced for roofing applications [34,35]. This allows the construction of flat roofs, which are advantageous from an operational standpoint. However, such structures have a major drawback: Their shape is not in equilibrium with evenly distributed symmetric loads. As a result, symmetric loads cause significant displacements and bending moments in suspended straight elements [34,35].

In terms of structural arrangement, tensegrity structures [36] are structurally similar to the suspension systems discussed in this article. In tensegrity structures, the bars are subjected to compressive forces, while the cables carry tensile forces. However, under external compressive loads, the overall stability of the structural system becomes a critical issue for tensegrity structures [37,38]. It is important to note that in the suspension systems considered here, which are composed of rigid elements and subjected to uniformly distributed external loads applied to the upper chord, tensile forces or a combination of tensile and bending forces occur, except in the struts. Therefore, for these suspension structures, the problem of overall stability can be addressed by applying appropriate structural constraints.

Regarding the analysis of suspension roof structures, a substantial number of studies have focused on their static behavior [39,40,41,42], etc. Most of these studies refer to the behavior of flexible cables, and some also address bending stresses in such cables [3,43,44]. However, the research on “rigid cables” is relatively limited and focusses primarily on parabolic forms [45,46]. The existing studies on suspension roof structures with bending stiffness predominantly analyze symmetrical loading, despite the well-known issue of high deformability under asymmetrical loading. There is a noticeable lack of research dedicated to the detailed analysis of asymmetrically loaded suspension structures with bending stiffness.

To improve suspension roof structures, it is crucial not only to analyze their compositional parameters but also to explore new forms that improve the operational conditions of the roof covering. Therefore, the development of an innovative suspension support system consisting of straight “rigid” elements is particularly relevant. One such system is an innovative suspension load-bearing roof structure composed of straight “rigid” elements [47].

This article examines the behavior of new suspension roof systems composed of straight “rigid” elements under asymmetric loading conditions. A methodology is presented to calculate stresses and displacements in such asymmetrically loaded suspension systems, which incorporates geometric nonlinearity. Through numerical experiments, this study identifies the behavioral characteristics of these systems and assesses the influence of their straight “rigid” elements on displacement. An analysis of these suspension systems and the stress variations under asymmetric loading is also presented, considering the ratio between asymmetric and symmetric loads.

The results demonstrate that the novel two-level cable–strut suspension roof system is more effective than conventional curved suspension roof structures under asymmetric loading, particularly in terms of displacement and stress criteria.

2. Analytical Solutions of Asymmetrically Loaded Suspension System with Stiffness in Bending Elements

2.1. Calculation of Suspension Structure with Two Straight Elements

Before discussing the innovative suspension system, we will examine its prototype, which is composed of two stiff elements in bending straight elements (EI ≠ 0), see Figure 1, SK-1 structure. These elements with bending stiffness allow us to create a flat roof surface. This would not be possible to create if the elements of the suspension structure were absolutely flexible (EI = 0). Such a suspension structure of straight “rigid” bars under the influence of asymmetric loads experiences such displacements: the central node (hinge) “went up”, and the straight “rigid” bars incurve (see Figure 2a). The analysis and calculation of this symmetrically loaded straight element suspension structure are presented in [47].

Analyzing the following case of asymmetric loading, when the suspension structure consists of two straight bars, the temporary asymmetric load s is distributed over half of its span length, i.e., through one of the straight elements “1–2” (see Figure 2b). Under the action of loads, the straight tension elements “1–2” and “2–3” will deflect with displacements ${∆f}_{\mathrm{1,2}}$ and ${∆f}_{\mathrm{2,3}}$, and the central node “2” will move not only with a vertical displacement ${∆f}_c$, but also a horizontal displacement ${∆h}_2$. The tensile forces $H_1$ and $H_2$ and corresponding bending moments $M_{1\left(x\right)}$ and $M_{2\left(x\right)}$ will appear in these elements. In the case of asymmetric loading, the calculation of this suspension structure becomes more complicated compared with symmetric loading, because we have one additional unknown—the horizontal displacement of the central node ${∆h}_2$. It is, therefore, necessary to analyze both sides of the structure separately: one subjected to combined $(g+s)$ loads and the other subjected to only the $g$ load. In this case (in the case of asymmetric loading), there are already two mutually unequal unknowns—the fictitious displacement ${∆f}_{fic,1}$ and ${∆f}_{fic,2}$ of both straight elements. It should be noted that the effect of asymmetric load on suspension structures can be estimated by applying the ratio of loads γ = s/g. The ratio γ indicates the influence of asymmetric loading on the deformation of the structure. The higher the value of the ratio γ, the more dominant the displacements of kinematic origin become.

According to the value of the load ratio γ, it is possible to write the dependency of the fictitious displacements of both straight tension elements from the equilibrium condition in the central node “2” (see Figure 2b):

$$∆f_{fic,1}=∆f_{fic,2}\left(1+\gamma \right).$$

(1)

It can be seen from Formula (1) that the difference in the fictitious displacements of the two tension bars in the loaded part of the structure $∆f_{fic,1}$ and the unloaded part $∆f_{fic,2}$ is increasing when the values of the load ratio γ increasing.

The equation of equilibrium of the true tension “rigid” element (bar) “1–2” can be written as follows (see Figure 2). The actual “1” element displacement $w_1\left(x\right)$ will be equal to:

$$w_1\left(x\right)=∆f_{fic,1}\left[{\displaystyle \frac{8}{{\left(k_{1}l\right)}^2}}\left(ch{k}_{1}x+{\displaystyle \frac{\left(1-ch{k}_{1}l\right)sh{k}_{1}x}{sh{k}_{1}l}}-1\right)+{\displaystyle \frac{4x}{l}}-{\displaystyle \frac{4x^2}{l^2}}\right]+∆f_c{\displaystyle \frac{x}{l}},$$

(2)

here $k_{1}l=\sqrt{{\displaystyle \frac{H_1{l}^2}{EI{cos}^3{\phi }_1}}}$—slenderness parameter of element “1–2”; $∆f_c$—vertical displacement of the central node (see Figure 2b); $ch{k}_{1}x=0.5(e^{k_{1}x}+e^{-k_{1}x}$)—hyperbolic cosine; and $sh{k}_{1}x=0.5(e^{k_{1}x}-e^{-k_{1}x}$)—hyperbolic sine.

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”:

$$∆f_{fic,1}=\sqrt[3]{{\displaystyle \frac{3{\left(g+s\right)l}^4}{64EA·\Phi \left(k_{1}l\right)·{\beta }_1{cos}^5{\phi }_1}}},$$

(3)

$$\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\Phi \left({kl}_1\right)=1-{\displaystyle \frac{24}{{\left(k_{1}l\right)}^2}}\left(1-{\displaystyle \frac{2}{k_{1}l}}th{\displaystyle \frac{k_{1}l}{2}}\right)+{\displaystyle \frac{6\left(sh{kl}_1-k_{1}l\right)}{{\left(k_{1}l\right)}^3{ch}^2\left(k_{1}l/2\right)}};$$

(4)

where $s_0={\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$cos{\phi }_0$}\right.}$—initial length of element “1–2”; $∆s_{el,1}={\displaystyle \raisebox{1ex}{$H_{1}l$}\!\left/ \!\raisebox{-1ex}{$EAcos{\phi }_{0}cos{\phi }_1$}\right.}$—elastic elongation of element “1–2”; and $s_1=l+0.5{\int }_0^l{\left[w_1^{\prime }\left(x\right)\right]}^{2}dx$—length of element “1–2” after deformation:

$${\beta }_1=1+{\displaystyle \frac{3}{8}}{\displaystyle \frac{l}{{∆f_{fic,1}}^{2}cos{\phi }_1\Phi \left(k_{1}l\right)}}\left({\displaystyle \frac{l}{cos{\phi }_1}}-{\displaystyle \frac{l}{cos{\phi }_0}}\right).$$

(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 $\Phi \left(k_{1}l\right)$ 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:

$$∆f_{fic,2}=\sqrt[3]{{\displaystyle \frac{3g{l}^4}{64EA·\Phi \left(k_{2}l\right)·{\beta }_2{cos}^5{\phi }_2}}},$$

(6)

$$\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}:\Phi \left(k_{2}l\right)=1-{\displaystyle \frac{24}{{\left(k_{2}l\right)}^2}}\left(1-{\displaystyle \frac{2}{k_{2}l}}th{\displaystyle \frac{k_{2}l}{2}}\right)+{\displaystyle \frac{6\left(sh{k}_{2}l-k_{2}l\right)}{{\left(k_{2}l\right)}^3{ch}^2\left(k_{2}l/2\right)}};$$

(7)

$${\beta }_2=1+{\displaystyle \frac{3}{8}}{\displaystyle \frac{l}{{∆f_{fic,2}}^{2}cos{\phi }_1\Phi \left(k_{2}l\right)}}\left({\displaystyle \frac{l}{cos{\phi }_2}}-{\displaystyle \frac{l}{cos{\phi }_0}}\right).$$

(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 ∆f_(fic,1), ∆f_(fic,2), ∆f_c, ∆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 ∆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:

$$∆h_c=∆h_2={\displaystyle \frac{[{\left(1+\gamma \right)}^2-1]}{\left[{\left(1+\gamma \right)}^2+1\right]cos{\phi }_m}}\left[\left({\displaystyle \frac{l}{cos{\phi }_0}}-{\displaystyle \frac{l}{cos{\phi }_m}}\right)+{\displaystyle \frac{g\left(1+0.5\gamma \right)l^3}{2\left(f_0-∆f_c\right)EA·cos{\phi }_{0}cos{\phi }_m}}\right].$$

(9)

Formula (9) shows that the horizontal displacement of the central node ∆h_c depends not only on the load ratio γ and the vertical displacement of the central node “2” ∆f_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 $∆f_{fic,m}$ can be used, which is calculated as follows:

$$∆f_{fic,m}=0.25\left(f_0-∆f_c\right){\displaystyle \frac{(1+\gamma )}{\left(1+0.5\gamma \right)}}cos{\phi }_m,$$

(10)

where ${\phi }_m={0.5(\phi }_1+{\phi }_3)$—the average angle of straight suspension elements after deformation.

Then the following additional equation can be written:

$$H_m={\displaystyle \frac{g\left(1+0.5\gamma \right)L^2}{8(f_0-∆f_c)}},$$

(11)

where $L=2l$—total span of the structure.

It should be noted that at each iteration step, when calculating the values of fictitious displacements $∆f_{fic,1}$ and $∆f_{fic,2}$, the tensile forces H₁ and H₂ of the straight elements are also determined by:

$$H_1={\displaystyle \frac{\left(g+s\right)l^2}{8∆f_{fic,1}}},$$

(12)

$$H_2={\displaystyle \frac{g{l}^2}{8∆f_{fic,2}}}.$$

(13)

When the fictitious displacements $∆f_{fic1}$ and $∆f_{fic2}$ of elements “1–2” and “2–3” are obtained, then the true displacement values of elements “1–2” and “2–3” in the middle of their span can be calculated as follows:

$$∆f_1=∆f_{fic1}\left[{\displaystyle \frac{8}{{\left(k_{1}l\right)}^2}}\left(ch{(0.5k}_{1}l)+{\displaystyle \frac{\left(1-ch{k}_{1}l\right)sh{(0.5k}_{1}l)}{sh{k}_{1}l}}-1\right)+1\right]+0.5∆f_c,$$

(14)

$$∆f_2=∆f_{fic2}\left[{\displaystyle \frac{8}{{\left(k_{2}l\right)}^2}}\left(ch{(0.5k}_{2}l)+{\displaystyle \frac{\left(1-ch{k}_{2}l\right)sh{(0.5k}_{2}l)}{sh{k}_{2}l}}-1\right)+1\right]+0.5∆f_c.$$

(15)

It is necessary to mention that asymmetrical loading compared with symmetrical loading in straight tension elements will cause lower axial forces but higher displacements and bending moments. It is clear that the straight elements reduce the horizontal displacement of the central node $∆h_c$.

2.2. Analysis of Asymmetrically Loaded Innovative Single-Level Suspension System

The analysis of the calculation formulas for a suspension structure with two straight elements shows that under an asymmetric load, its elements experience large displacements and bending moments (element “1–2”). To stabilize the initial shape of the structure, i.e., to reduce its displacements, a new innovative suspension system was developed, in which the straight suspended elements are “supported” by a strut which is supported by a lower chord (see Figure 3).

The upper chord, which has a bending stiffness (EI ≠ 0), is load-bearing, and the lower chord is stabilizing (see Figure 3). This chord is most often designed from flexible bar elements (EI → 0). It is also necessary to note that the “rigid” elements of the upper chord (“1–2”, “2–3” and “3–5”, “5–6”) are hinged and connected to each other (see Figure 3).

To ensure the overall stability of the suspension system, bracing is installed at the struts (see Figure 3). This restricts the ability of the struts and bottom chord to deform out of the plane of the suspension system.

2.2.1. Calculation of Asymmetrically Loaded Upper Chord Elements

It is necessary to note that under asymmetric loads, the calculation of this suspension system becomes more complicated than in the case of symmetric loads [47]. Analogous to the two element suspension structure due to the effect of the asymmetrical variable load s, an additional unknown appears—the horizontal displacement of the central hinge (node “3”) $∆h_{3r}$. For this reason, the suspension system considered is divided into two parts: the left one, loaded with loads g + s, and the right one, which is loaded only by the permanent load g. The calculation scheme for the upper chord on the left side is presented in Figure 4. Under the action of asymmetric loading, the horizontal displacement will occur not only in node “3” but also in node “2” of the strut and this chord connection. In addition, node “2”, will shift perpendicular to the initial axis by the value $∆f_{2r}$ (see Figure 4).

Under the action of loads, the displacements of the elements of the upper chord “1–2” and “2–3” are $∆f_{1r}$ and $∆f_{2r}$ and the deflection angles of this elements are ${\phi }_1$ and ${\phi }_2$ (see Figure 4). When reviewing the behavior of the upper chord elements “1–2” and “2–3” it can be stated that their calculation is analogous to the calculation of the elements of a two element suspension structure.

The analysis of the equilibrium of node “2” can be obtained using the following dependence:

$$H_{1,r}^{\mathrm{*}}\cong H_{2,r}^{\mathrm{*}}+{(g+s)}_h·l_{1pr},$$

(16)

where $H_{1,r}^{*}$ and $H_{2,r}^{*}-$ projection of tensile forces of elements “1–2” and “2–3” on line”1–3”; ${(g+s)}_h$—projection of loads g + s on line “1–3”; and $l_{1pr}-$ distance between nodes “1” and “2” (see Figure 4).

It can be seen from Equation (16) that due to the effect of the load ${(g+s)}_h$ the tensile forces in elements “1–2” and “2–3” will not be equal to each other. It is necessary to note that the tensile forces $H_{1,r}^{*}$ and $H_{2,r}^{*}$ are calculated as follows:

$$H_{1,r}^{*}=H_1/\cos {\phi }_{1r},$$

(17)

$$H_{2,r}^{*}=H_2/\cos {\phi }_{2r},$$

(18)

where $H_1={\displaystyle \frac{(g+s)l_{pr}^2}{8∆f_{fic,\mathrm{r}1}}}$; $H_2={\displaystyle \frac{(g+s)l_{pr}^2}{8∆f_{fic,\mathrm{r}2}}}$.

Taking into account Equation (16), the dependence between the fictitious displacements of the upper chord elements “1–2” and “2–3” can be written as follows:

$$∆f_{fic,\mathrm{r}2}={∆f}_{fic,\mathrm{r}1}\left(1+8∆f_{fic,\mathrm{r}2}·tg{\phi }_m/l_{pr}\right).$$

(19)

Using, as in the case of symmetric loading, the average fictitious displacement $∆f_{fic,m}$ and the average tensile force $H_m$ can be obtained as follows:

$${∆f}_{fic,mr}^2\cong 0.5\left({∆f}_{fic,r1}^2+{∆f}_{fic,r2}^2\right),$$

(20)

$$H_{mr}\cong 0.5\left(H_2+H_1\right).$$

(21)

Using the deformation compatibility, the formula for calculating the average fictitious displacement of the upper chord elements can be obtained as follows:

$$∆f_{fic,mr}=\sqrt[3]{{\displaystyle \frac{3(g+s)l^4}{64EA·{\Phi }_{mr}\left(k_{mr}{l}_{pr}\right)·{{\mathrm{ɳ}}_{mr}·cos}^2{\phi }_0}}},$$

(22)

where

$${\mathrm{ɳ}}_{mr}=1+{\displaystyle \frac{3·l_{pr}}{8{{\Phi }_{mr}\left(k_{mr}{l}_{pr}\right)(∆f_{fic,mr})}^{2}cos{\phi }_m}}\left({\displaystyle \frac{l_{pr}}{cos{\phi }_m}}-{\displaystyle \frac{l_{pr}}{cos{\phi }_0}}\right).$$

(23)

It should be mentioned that the calculation of the average tensile forces and fictitious displacements ($H_{mr}$ and $∆f_{fic,mr}$) of the upper chord elements was carried out in a manner similar to the two straight elements of the structure using the following equilibrium equation:

$$H_{mr}={\displaystyle \frac{(g+s)l_{pr}^2}{8∆f_{fic,mr}}}.$$

(24)

When the values of the average tensile force $H_m$ and the average fictitious displacement $∆f_{fic,m}$ are obtained by the iterative procedure, then the fictitious displacements of the upper chord elements “1–2” and “2–3” can be calculated as follows:

$${∆f}_{fic,1}=∆f_{fic,mr}{\left({\displaystyle \frac{2}{1+{\mu }^2}}\right)}^{0.5},$$

(25)

$${∆f}_{fic,2}=∆f_{fic,mr}{\left({\displaystyle \frac{2}{1+1/{\mu }^2}}\right)}^{0.5},$$

(26)

where ${\mu }_r={\displaystyle \frac{H_{mr}+0.5(g+s)l_{pr}}{H_{mr}-0.5(g+s)l_{pr}}}$.

When calculating the suspension system, it is important to know the forces in the strut “2–4”. The force $F_{2r}$ (axial force of strut “2–4”) acting on node “2” is calculated according to Equation (27):

$$F_{2\mathrm{r}}=N_{c2}=p{l}_{pr}\cos {\phi }_m-\left(H_1+H_2\right)∆f_{2r}·cos{\phi }_m/l_{pr}.$$

(27)

After rearranging:

$$N_{c2}=p{l}_{lr}\cos {\phi }_m\left(1-0.25∆f_{2r}/∆f_{fic,m}\right).$$

(28)

From Expressions (27) and (28), it is obvious that the axial force of the strut has a variable magnitude and depends both on the displacement of node “2” $∆f_{2r}$ and on the average fictitious displacement of the upper chord elements $∆f_{fic,mr}$ and, at the same time, on the fictitious displacements of elements “1–2” and “2–3” ${∆f}_{fic,1}$ and ${∆f}_{fic,2}$.

The left and right sides of the upper chord are connected by a common node “3” and its displacements $∆v_3$ and $∆h_3$.

The right side of the upper chord calculation scheme is presented in Figure 4b.

It should be noted that the calculation sequence for the right side of the upper chord is analogous to the calculation sequence for the left side, and it is loaded only with a permanent load g (see Figure 4b). Therefore, the final calculation expressions without detailed derivation will be presented. The formula for calculating the average fictitious displacement of the elements of the right side of the upper chord element is as follows:

$$∆f_{fic,ml}=\sqrt[3]{{\displaystyle \frac{3g{l}_{pl}^4}{64EA·{\Phi }_{ml}\left(k_{ml}{l}_{pl}\right)·{{\mathrm{ɳ}}_{ml}·cos}^2{\phi }_0}}}, \mathrm{here}$$

(29)

$${\mathrm{ɳ}}_{ml}=1+{\displaystyle \frac{3·l_{pl}}{8{{\Phi }_{ml}\left(k_{ml}{l}_{pl}\right)(∆f_{fic,ml})}^{2}cos{\phi }_m}}\left({\displaystyle \frac{l_{pl}}{cos{\phi }_m}}-{\displaystyle \frac{l_{pl}}{cos{\phi }_0}}\right).$$

(30)

The average tensile force will be equal to the following:

$$H_{ml}={\displaystyle \frac{g{l}_{plr}^2}{8∆f_{fic,ml}}}.$$

(31)

Then the fictitious displacements of the upper chord elements “3–5” and “5–6” are calculated as follows:

$${∆f}_{fic,3}=∆f_{fic,ml}{\left({\displaystyle \frac{2}{1+{\mu }^2}}\right)}^{\mathrm{0,5}},$$

(32)

$${∆f}_{fic,4}=∆f_{fic,ml}{\left({\displaystyle \frac{2}{1+1/{\mu }^2}}\right)}^{\mathrm{0,5}},$$

(33)

where ${\mu }_l={\displaystyle \frac{H_{ml}+0.5g{l}_{pl}}{H_{ml}-0.5g{l}_{pl}}}$.

The axial force in strut “5–7” will be equal to the following:

$$N_{c5}=F_{5l}=g{l}_{pl}\cos {\phi }_m\left(1-0.25∆f_{5l}/∆f_{fic,ml}\right).$$

(34)

It is important to note that with equal values of the horizontal displacement of the left and right sides $∆h_{3r}=∆h_{3l}$ the vertical displacements of their elements “2–3” and “3–5” will not be equal. The axial forces of these elements and struts $N_{c2}$ and $N_{c5}$ will also differ. These differences will be greater with the higher value of the load ratio γ = s/g.

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 $∆v_3$ but also horizontally by the value $∆h_3$ towards the side of the structure loaded with the asymmetric load. In this case, the displacement of nodes “4” and “7” will be $∆f_4$ and $∆f_7$. It is obvious that the forces of the struts F₄ (left side) and F₇ (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 of this suspension structure at the central node “3” will be indicated by H₃. When loading with a concentrated force F₄ 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:

$$H_{m,\mathrm{u}r}={0.125F}_4(l-∆h_3)/\left(f_{\mathrm{2,4}}+∆f_4\right),$$

(35)

where $f_{\mathrm{2,4}}$—the height of the strut (see Figure 6).

Then the tensile force $H_{3r}$ of the lower chord can be calculated as follows:

$$H_{3r}=H_{mur}+∆H_r,$$

(36)

where $∆H_r=F_{4}sin{\phi }_{mr}$/2.

The formula for calculating the axial forces in elements “1–4” and “3–4” will be obtained by:

$$N_{\mathrm{1,4}}=N_{\mathrm{3,4}}={\displaystyle \frac{H_{mr}}{cos{\phi }_{mr}}}{\left[1+4{\displaystyle \frac{{\left(f_{\mathrm{2,4}}+∆f_4\right)}^{2}cos{\phi }_{mr}}{{(l-∆h_3)}^2}}\right]}^{0.5}.$$

(37)

To avoid errors in simplified calculations of suspension structures, we write the following deformation consistency equation:

$$2∆s_{el,r}·s_{0,r}=s_{1,r}^2-s_{0,r}^2.$$

(38)

Then

$$2∆s_{el,r}·s_{0,r}={\displaystyle \frac{F_{\mathrm{2,4}}{l}^2\left(l-∆h_3\right){\propto }^{\mathrm{1,5}}}{2EA\left(f_{\mathrm{2,4}}+∆f_{2mr}\right){cos}^2{\phi }_{0}cos{\phi }_{mr}}};$$

(39)

$$s_{1,r}^2-s_{0,r}^2={\left(l-∆h_3\right)}^2-l^2+{\left(f_0+∆f_c\right)}^2-f_0^2+4\left[{\left(f_{\mathrm{2,4}}+∆f_{2mr}\right)}^2-f_{\mathrm{2,4}}^2\right],$$

(40)

where ${\propto }^{\mathrm{1,5}}=1+4f_{\mathrm{2,4}}{\left({\displaystyle \frac{cos{\phi }_0}{l}}\right)}^2$.

The concentrated force acting on the lower chord element at node “4” is determined from the following Equation:

$$F_4={\displaystyle \frac{8\left(f_{\mathrm{2,4}}+∆f_{2mr}\right)EA{cos}^2{\phi }_{0}cos{\phi }_{mr}}{l^2(l-{∆h_3)\propto }^{\mathrm{1,5}}}}\left[{\displaystyle \frac{{(l-∆h_3)}^2-l^2}{4}}-{\displaystyle \frac{{\left(f_0+∆f_c\right)}^2-{f_0}^2}{4}}+{\left(f_{\mathrm{2,4}}+∆f_{2mr}\right)}^2-f_{\mathrm{2,4}}^2\right].$$

(41)

From expression (41) it is clearly seen that the axial force in the strut $F_{\mathrm{2,4}}$ is variable and depends not only on the displacement of the node “2” $∆f_{2mr}$, but also on the displacement of the central node “3” $∆f_c$. Under the assumption that strut “2–4” does not deform, equality of the displacements of the upper and lower chords $∆f_{2mr}=∆f_4$ will be obtained.

To achieve greater calculation accuracy, the displacement of node “4” is conditionally divided into elastic $∆f_{4r,el}$ and kinematic $∆f_{4r,k}$ as follows:

$$∆f_{4r}=∆f_{4r,el}+∆f_{4r,k},$$

(42)

where $∆f_{4r,k}=\sqrt{{f_{\mathrm{2,4}}}^2+0.25\left[l^2-{\left(l-∆h_3\right)}^2-{\left(f_0+∆f_c\right)}^2-f_0^2\right]}$.

Taking into account (42), the following formula for calculating the concentrated force is obtained as follows:

$$N_{cr}=F_{\mathrm{2,4}}={\displaystyle \frac{8\left(f_{\mathrm{2,4}k}+∆f_{4r,el}\right)EA{cos}^2{\phi }_{0}cos{\phi }_{mr}}{l^2(l-{∆h_3)\propto }^{\mathrm{1,5}}}}\left({∆f_{4r,el}}^2+2f_{\mathrm{2,4}k}∆f_{4r,el}\right).$$

(43)

The advantage of Formula (43) is that the left part of the lower chord can be calculated as a suspended structure with fixed supports and having an initial sag equal to $f_{\mathrm{2,4},k}$.= $f_{\mathrm{2,4}}+∆f_{4r,k}$. According to this expression, the axial force of the strut F_2,4 depends only on the elastic displacement $∆f_{4,el}$. Therefore, taking into account the displacements of the upper chord, an approximate expression can be written to calculate this elastic displacement as follows:

$$∆f_{4r,el}={\displaystyle \frac{b_r\left(1-∆f_{4,k}/∆f_{fic,mr}\right)}{1+b_r\frac{\left(1-∆f_{4,k}/∆f_{fic,mr}\right)}{f_{\mathrm{2,4},k}}+\mathrm{0,25}{b}_r/∆f_{fic,mr}}},$$

(44)

where $b_r={\displaystyle \frac{g(1+\gamma )l^2{{(l-∆h_3)}^2\propto }^{\mathrm{1,5}}}{32EA{f}_{\mathrm{2,4},k}^2{cos}^2{\phi }_0}}.$

By applying Formula (44), it is possible to simplify the iterative calculation and know the value of the fictitious displacement $∆f_{fic,mr}$ of the upper chord to calculate approximate $∆f_{4,el}$ without an iteration procedure. The error of Formula (44) does not exceed 1–2%.

The deformed scheme of the right part of the lower chord loaded only with a permanent load g is presented in Figure 6b. It is necessary to note that in this part the central hinge will shift to the side loaded with a variable asymmetric load s. Therefore, without providing detailed derivations, the main expressions are presented to calculate the stresses and displacements of the lower chord. The tensile force at node “3” will be calculated analogously to expression (36) as follows:

$$H_{3l}=H_{mul}+∆H_l,$$

(45)

where $∆H_l=F_{7}sin{\phi }_{ml}$/2.

The average tensile force on the right side is equal to

$$H_{ml}={0.125F}_7(l+∆h_3)/\left(f_{\mathrm{5,7}}+∆f_7\right),$$

(46)

where $F_7$—the force of the strut “5–7” transmitted to node “7”.

The kinematic displacement of node “7” is calculated according to the following equation:

$$∆f_{7,k}=\sqrt{{f_{\mathrm{5,7}}}^2+0.25\left[f_0^2-{\left(f_0+∆f_c\right)}^2-{\left(l+∆h_3\right)}^2-l^2\right]}-f_{\mathrm{5,7}}.$$

(47)

Expression (47) shows that the kinematic displacement of node “7” for certain values $∆f_c$ and $∆h_3$ can also take a negative sign, i.e., can be directed upwards.

The axial force in the strut “5–7” is calculated according to the following formula:

$$F_7={\displaystyle \frac{8\left(f_{\mathrm{5,7}k}+∆f_{7l,el}\right)EA{cos}^2{\phi }_{0}cos{\phi }_{ml}}{l^2(l+{∆h_3)\propto }^{\mathrm{1,5}}}}\left({∆f_{7r,el}}^2+2f_{\mathrm{5,7}k}∆f_{5l,el}\right),$$

(48)

where $f_{\mathrm{5,7}k}=f_{\mathrm{5,7}}+∆f_{7,k}.$

Having the expression for the axial force in the strut, it is possible to obtain a formula to calculate the elastic displacements of node “7” as follows:

$$∆f_{7l,el}={\displaystyle \frac{b_l\left(1-∆f_{7,k}/∆f_{fic,ml}\right)}{1+b_l\frac{\left(1-∆f_{7,k}/∆f_{fic,ml}\right)}{f_{\mathrm{5,7}k}}+0.25b_l/∆f_{fic,ml}}},$$

(49)

where $b_l={\displaystyle \frac{g{l}^2{{(l+∆h_3)}^2\propto }^{\mathrm{1,5}}}{32EA{f}_{\mathrm{5,7},k}^2{cos}^2{\phi }_0}}$.

Notice that the formula for calculating the elastic displacement of the right side of the structure is analogous to the formula for calculating the elastic displacement of the left side. The iterative calculation procedure of the lower chord right side is performed in the same sequence as for the left side. It is only necessary to note that the displacements and forces of these parts (left and right) will differ proportionally to the values of the load ratio γ.

2.2.3. Calculation of the Asymmetrically Loaded Suspension System

The calculation of an asymmetrically loaded suspension system is performed by combining the expressions for the calculation of the upper and lower chords of the left and right sides of the system obtained in Section 2.2.1 and Section 2.2.2. The main unknowns of this iterative calculation procedure are the vertical displacement $∆f_c=∆v_3$ and the horizontal displacement $∆h_3$ of the central hinge (node “3”). Other dependent unknowns are the displacements of nodes “4” ($∆f_4$) and “7” ($∆f_7$) of the lower chord of the suspension system and the fictitious displacements of the straight elements of the upper chord ${∆f}_{fic,1}$, ${∆f}_{fic,2}$ (part loaded with asymmetrical load $g+s$) and ${∆f}_{fic,3}$, ${∆f}_{fic,4}$ (part loaded with $g$), as well as the tensile forces in the upper and lower chord elements: H₁, H₂, and $H_{3r}$. An important unknown is the force in the struts F₄ and F₇.

The iterative calculation can be divided into a few steps. Having accepted the initial values of the vertical $∆f_c=∆v_3$ and horizontal $∆h_3$ displacements of the central hinge (node “3”), it is possible to calculate the left and right parts of the suspension system separately. First, the displacements of the nodes of the left $∆f_4$, (according to (42)) and right parts $∆f_7$ (according to (47) and (49)). Then, the left part fictitious average displacement is calculated $∆f_{fic,mr}$ (see (23) and (24)) and fictitious displacements ${∆f}_{fic,1}$ and ${∆f}_{fic,2}$ (see (25) and (26)) and the medium force of their elements $H_{mr}$ (see (24)) are determined. After that, when ${∆f}_{fic,1}$ and ${∆f}_{fic,2}$ are known, the tensile forces $H_{1r}^{*}$ and $H_{2r}^{*}$ (see (17) and (18)) and the forces in the strut “2–4” are determined (see (28)). Then, the calculation sequence for the right side of the upper chord is analogous to the calculation sequence for the left side. The fictitious displacements of this part are calculated $∆f_{fic,ml}$ according to (29) and ${∆f}_{fic,3}$ and ${∆f}_{fic,4}$ according to (32) and (33), and axial forces $H_{ml}$ and $F_{5l}$. according to (46) and (48). After that, the equilibrium conditions of the central node are verified as follows:

$$H_{2r}+H_{3r}=H_{2l}+H_{3l}=\mathrm{Ʃ}H={\displaystyle \frac{p{L}^2}{8\left(f_0+∆f_c\right)}}.$$

(50)

If the equilibrium conditions of the central node are satisfied, then the actual displacements and bending moments of the upper chord elements of both parts of the system can be calculated.

It should be noted that the presented analytical solutions are performed for these reasons. The obtained analytical solutions allow the determination of certain rational parameters of the examined structure without a large number of iterations. Calculating rational parameters using an FEM software would require a large number of convergence iterations.

3. Numerical Analysis of Suspension System Under Asymmetric Loads

To analyze the behavior of suspension systems under asymmetric loads and validate the accuracy of the analytical solutions presented in Section 2, several structural configurations were examined, as illustrated in Figure 1. These configurations include SK-1, a structure with two straight rigid elements; SK-2, a single-level suspension system; SK-3, a two-level suspension system; and SK-4, a conventional parabolic system.

In all cases, the length of the structure is assumed to be l = 36.00 m with a sag of f₀ = 3.60 m. The additional dimensions include h = 4.30 m, l₁ = 2.25 m, l₂ = 4.5 m, and h₁ = 1.00 m.

The suspension structures shown in Figure 1 employ cross-sections as specified in

Table 1. Cross-sections of suspension structure elements.

Structural Element	Suspension Structure Scheme (Figure 1)	Cross-Section Label	Cross-Section Type (EN 10365) [49]
Upper Chord	SK-1, SK-2, SK-3	I1	IPE 360
		R1	RHS 250 × 250 × 10
		R2	SHS 400 × 200 × 6.3
		R3	SHS 250 × 150 × 10
		R4	SHS 300 × 100 × 10
		P1	Plate—180 × 40
Lower Chord	SK-2		32 mm
	SK-3		25 mm
Strut	SK-2, SK-3		RHS 60 × 60 × 4

. In the SK-1 structure and the upper chords of the SK-2 and SK-3 systems, six different bending stiffness values were considered, corresponding to the following cross-sections (I1–P1). The cross-sectional areas of the upper chord bars were nearly identical (ranging from $A=72.00-72.70 {\mathrm{c}\mathrm{m}}^2$), ensuring uniform elastic deformations under axial forces. The modulus of elasticity for the steel used was set at E = 200 GPa.

For all structural configurations, both asymmetric and symmetric loads were applied to the rigid upper chord of the structure, as shown in Figure 7. The numerical values of the permanent $g_1$ and variable load $s_1$ are provided in

Table 2. Characteristic values of permanent and variable loads.

$\mathsf{\gamma }$	$\mathbf{Permanent}\mathbf{Load}{\mathit{g}}_1\mathbf{k}\mathbf{N}/\mathbf{m}$	$\mathbf{Variable}\mathbf{Load}{\mathit{s}}_1\mathbf{k}\mathbf{N}/\mathbf{m}$
1.00	5.130	5.130
1.50	4.400	6.600
2.00	3.850	7.700

. To analyze the structural responses under different ratios of variable to permanent loads, three load ratios were considered: γ = 1.0, 1.5, and 2.0. Displacements were calculated using the characteristic values of $g_1$ and $s_1$, while stresses were determined using the design values of these loads.

A numerical analysis was performed using the finite element method (FEM), treating the structures as static geometrically non-linear. A FEM program written in Fortran by the article authors was employed. In presenting the results of the numerical analysis under asymmetric loads, comparative data from symmetric load was obtained.

Figure 1 (SK-1) and Figure 7 illustrate the two-element suspension structure and the applied asymmetric loads. Figure 8 presents the displacements of this two element suspension structure for γ = 1.5. Under asymmetric loading, the largest vertical deflections occur on the left side of the structure, where both permanent and variable loads are applied. Notably, significant upward displacements are observed at the central hinge (point “2”), particularly when the system has low stiffness (e.g., P1 stiffness). To stabilize the structure’s behavior, the use of rigid elements with higher stiffness (e.g., I1 stiffness) is recommended. Furthermore, Figure 8 highlights that suspension structures are significantly more sensitive to asymmetric loads than to symmetric ones, exhibiting larger displacements under asymmetric conditions.

Significant displacements were observed in the two element suspension structure. To reduce these displacements, the stiffness of the rigid bars must be increased. However, this leads to large bending moments, necessitating a new design approach with additional bars to support the main rigid elements.

The accuracy of analytical solutions was evaluated by performing numerical analysis on a suspension structure with two straight rigid elements. The analytical solutions demonstrated errors of no more than 2% for axial forces and approximately 4–5% for displacements and bending moments in asymmetrically loaded suspension structures with different bending stiffness in straight elements.

The design scheme for the single-level suspension system (SK-2) is shown in Figure 1, with cross-sectional characteristics defined earlier in this section.

Comparing results between the single-level suspension system (SK-2) and the two straight bar suspension structure (SK-1), the displacements (see Figure 9) at points “2” and “4” in SK-2 decreased compared with points “a” and “b” in SK-1 under asymmetric loads. Specifically, the vertical displacement at point “2” decreased by a factor of 7.72 to 13.01, depending on stiffness, while at points “a” and “b” of SK-2, the displacements decreased by a factor of 1.31 to 4.84. However, at low stiffness (P1), the displacements remained relatively large (0.432 m). At higher bending stiffness levels (I1 and R1), the vertical displacements at points “a”, “2”, and “b” were similar.

The displacements of the central node “3” in the SK-2 system and node “2” in the SK-1 structure decreased by nearly half. However, at low bending stiffness in the upper chord (P1 and R4), the displacements due to asymmetric loads were larger than those from symmetric loads. At higher stiffness levels (I1, R1, and R2), the vertical displacements changed direction and moved downward. At point “4”, the vertical displacements decreased by a factor of 1.51 to 7.64, but at low stiffness (P1 and R4), they changed direction and moved upward. This indicates that with minimal bending stiffness in the upper chord, the system is less effective against asymmetric loads. However, with medium stiffness (R1, R2, and R3), the single-level suspension system effectively minimizes displacement variations across all points of the upper chord, even under asymmetric loading. The influence of the ratio of permanent to variable loads $\gamma$ on displacement distribution is illustrated in Figure 10. The numerical results indicate that under asymmetric loading, the load ratio γ has minimal impact if the upper chord is sufficiently stiff (I1, R1). Only at very low bending stiffness (P1) does the load ratio significantly affect displacements.

The two-level suspension system (SK-3) is illustrated in Figure 1 with cross-sectional characteristics defined earlier in this section.

This innovative two-level suspension system (SK-3) is more efficient than both the single-level system (SK-2) and two straight element (SK-1) suspension structures in terms of displacements and stress tension forces distribution. The displacements in the two-level suspension system (Figure 11) significantly decreased compared with the single-level system when the upper chord stiffness was set to I1, R1–R4. Under all examined bending stiffnesses, the central hinge “5” and other upper chord nodes shifted downward, even with asymmetric loading. However, when the upper chord had minimal stiffness (P1), the system became inefficient, with central node “5” and nodes “6”, “7”, and “8” shifting upward. These results confirm that a two-level system with a rigid upper chord is significantly more effective than one with a flexible upper chord.

Figure 12 illustrates the influence of the load ratio γ on the displacement distribution. The numerical results for the two-level system, which are similar to those of the suspension single-level system, indicate that under asymmetric loading, the load ratio has minimal effect if the upper chord has sufficient stiffness (I1–R4). Only at very low bending stiffness (P1) does the load ratio significantly impact the displacements.

The distribution of axial forces under asymmetric and symmetric loads is shown in Figure 13. The results reveal that the axial forces are nearly identical when the upper chord stiffness ranges from I1 to R4. Therefore, Figure 13 shows the results only when the stiffnesses are I1, R4, and P1. However, with very low bending stiffness, the axial forces increase with the largest differences reaching 9.0% (31.0 kN) under symmetric loading and 37.8% (52.1 kN) under asymmetric loading. Notably, under asymmetric loading, the axial forces are greater than under symmetric loading only on the less loaded side of the structure, closer to the central point “5”. Therefore, it can be concluded that asymmetric loading does not significantly influence axial forces compared with symmetric loading.

Table 3. The ratio of axial forces in the lower chord under symmetric and asymmetric loading for different load factors and stiffness types.

Stiffness	$\mathbf{Load}\mathbf{Factor}\mathit{\gamma }$	$\frac{{\mathit{N}}_{\mathit{a}\mathit{s}\mathit{y}\mathit{m}}}{{\mathit{N}}_{\mathit{s}\mathit{y}\mathit{m}}}$
		Bar 1–6	Bar 6–3	Bar 3–7	Bar 7–5
I1	1.00	1.008	1.008	0.518	0.518
	1.50	1.011	1.011	0.418	0.418
	2.00	1.012	1.012	0.351	0.351
P1	1.00	0.950	0.950	0.532	0.532
	1.50	0.935	0.935	0.444	0.444
	2.00	0.923	0.923	0.364	0.365

$N_{sym}$, $N_{asym}$—axial force under symmetrical and asymmetrical loading.

presents the results of the influence of asymmetric loading on the bars of the lower chord of the SK-2 structure. It can be observed that, for any load factor $\gamma$, the axial forces on the more heavily loaded side of the structure (bars 1–6 and 6–3) under asymmetric loading are very close to those under symmetric loading—especially when the upper chord is stiff (stiffness I1). When the upper chord has low stiffness (P1), the axial forces under asymmetric loading are lower than under symmetric loading, with the reduction becoming more significant as the load factor increases (variable loading dominates). On the less loaded side of the structure (bars 3–7 and 7–5), the axial forces under asymmetric loading are much lower compared with symmetric loading. However, this difference is smaller when the upper chord is stiffer (I1) and becomes larger as the load factor $\gamma$ increases.

Figure 14 shows the distribution of bending moments in the upper chord under asymmetric and symmetric loads. It should be noted that these bending moments are considerably lower than those observed in the single-level system. Additionally, the magnitude of bending moments strongly depends on the bending stiffness of the upper chord. For instance, under asymmetric loading, the bending moment values for I1 and P1 stiffnesses differ by up to 3.05 times, and under symmetric loading, by up to 5.19 times.

It may seem that the most efficient design for minimizing bending moments occurs when the upper chord has a low bending stiffness (P1). However, a more accurate assessment of the suspended system’s efficiency, particularly with a stiff upper bending chord, can be made by comparing the normal stresses of the upper chord cross-section. These stresses are calculated by combining the effects of axial forces and bending moments for various bending stiffness values, as illustrated in Figure 15. It is important to recall that the cross-sectional areas of the examined upper chords varied only slightly. The stress graphs in Figure 15 reveal that the most efficient cross-sections correspond to the highest stiffness values, where the maximum stress is σ = 86.9 MPa. In contrast, the lowest stiffness in the upper chord results in significantly higher stress values, reaching σ = 316.5 MPa.

The effectiveness of suspension systems (SK-2 and SK-3) compared with suspension structures composed only of bending stiffness elements was previously demonstrated in the analysis of the SK-1 structure. However, it is also important to highlight the advantages of these systems over the conventional parabolic suspension system (SK-4) as shown in Figure 1. In the SK-4 structure, the vertical coordinates follow a quadratic parabolic distribution, typical of conventional suspension structures. Figure 16 presents calculation results for the SK-4 structure. Under asymmetric loading, even with the highest bending stiffness (I1), upward displacements occur on the left side of the structure. Additionally, with minimal bending stiffness (P1), which closely resembles a cable system, the upward displacements are significantly larger. It is important to note that, in all stiffness cases, the displacements of the conventional cable suspension structure are two to three times greater than those of the two-level suspension system.

4. Conclusions

This article examines the design and analysis of rigid element suspension structures, with a particular focus on an innovative two-level cable–strut suspension system composed of straight, stiff bending elements subjected to asymmetric loads. Based on the analysis of these structural systems, the following conclusions can be drawn:

• A novel two-level suspension system has been developed, integrating straight, “rigid” elements in the upper chord and a specially shaped lower chord. This configuration stabilizes the initial geometry of the flat roof under both symmetrical and asymmetrical loads, while simultaneously enhancing operational performance.
• The behavior of the lower chord under asymmetric loading was analyzed, leading to a revised calculation method for structural forces and displacements by dividing the displacements into kinematic and elastic. This approach minimizes the volume of iterative calculations required under asymmetric loading conditions. The axial force in the struts of the suspension system was found to be variable, depending on the bending stiffness of the upper chord, and the axial stiffness of the lower chord.
• The Finite Element Method (FEM) analysis demonstrated that increasing the bending stiffness of the upper chord significantly decreases both the total displacements and internal stresses in the asymmetrically loaded system. Increasing the stiffness from P1 to I1 not only reduces deflections of the upper chord elements but also minimizes displacements at the nodes throughout the entire suspension structure. Notably, by selecting an appropriate bending stiffness for the upper chord, it is possible to prevent upward displacements or negative values of displacements in the central node and the unloaded part of the system even at high load ratios γ.
• Under asymmetric loading, the suspended strut system redistributes stresses between the upper and lower chords, effectively stabilizing displacements. The stresses in the loaded and unloaded parts of the upper and lower chords differ significantly. For a load ratio of γ = 1.5, the axial forces in the loaded part of the upper chord are approximately 2.17 times smaller than those in the unloaded part. Conversely, the axial forces in the loaded part of the lower chord are about 2.66 times greater than in the unloaded part. The bending moments in the upper chord also vary substantially: those in the loaded part are approximately 2.5 times greater than in the unloaded part. At maximum stiffness (I1), the total stresses in the loaded upper chord approach those in the unloaded part. Furthermore, the stresses in the most rigid element (I1) are approximately 3.67 times lower than in the least rigid element (P1), confirming the efficiency and rationale of employing rigid elements in this suspension system.