# Modeling and Analysis of a Large-Scale Double-Level Guyed Mast for Membrane Antennas

Authors: Bingyan Li; Yuxuan Liu; Rongqiang Liu; Hongwei Guo; Qiang Cong; Qiuhong Lin; Shangling Qiao
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

This paper proposes a double-level guyed membrane antenna for stiffness improvement of a large-scale tri-prism deployable mast using the collapsible tubular mast (CTM). Initially, the construction of the antenna and the modeling of the CTM boom are illustrated. Afterwards, the central mast with isosceles triangular cross section is mathematically equivalent to a continuum beam, in which the equations of motion and the constitutive relations are derived. Based on the equivalent central beam, the double-level guyed mast for the membrane antenna is modeled as a 2(3-SPS-S) mechanism, and then velocity Jacobian matrices and stiffness matrices of SPS branches are constructed. Additionally, the total stiffness matrix of the equivalent mechanism is derived with the principle of virtual work and then evaluated as an accurate approach. Finally, with the aim to improve the static stiffness of the double-level guyed mast, the numerical analysis using the Genetic Algorithm (GA) is carried out for optimizing the distribution of guys in terms of anchor positions and attachment heights.


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## Body

## 1. Introduction

The increasing multiplicity and complexity of spacecrafts generates the need for large-scale and lightweight antennas. As a thin foldable material, the membrane is particularly suitable for space missions and commonly used to construct antenna reflectors. The application potential of a membrane antenna is expected to be fulfilled by a stable and deployable support structure. To address this issue, the Institute of Structural Mechanics of the German Aerospace Center developed the ultra-lightweight deployable boom and proved its applicability using a 20 m× 20 m demonstrator [1]. Since then, several deployable composite boom concepts have been developed, such as the sheath-based rollable lenticular-shaped and low-friction (SHEARLESS) boom, the triangular rollable and collapsible (TRAC) boom, X boom, and the collapsible tubular mast (CTM) boom [2]. The analysis object in this research is constructed with the CTM boom, which is the most widely used deployable boom. The characteristics of the CTM boom have been investigated numerically and experimentally. Bai et al. analyzed the geometrical and mechanical properties of the CTM boom in fold deformation and assessed the in-plane strain and interlaminar shear stress on the bonding interface [3]. Chu and Lei studied the mechanical design, dynamic analysis, and mechanical behavior of the CTM boom [4]. Firth and Pankow exerted buckling tests on the CTM boom supported by a conformal root and found that the bending and torsion are linear with the load before buckling point [5]. Hu et al. showed the linear displacement-load curve before local buckling in the axial direction of the CTM boom through axial compression experiments [6].Two parallel CTM booms are able to support the membrane in an antenna, but the stiffness is hard to meet the standard for large-scaled antennas. Therefore, three parallel CTM booms are used to construct a CTM truss which has a triangular cross section, thus improving the stiffness of the membrane antenna. Murphey proposed methods for weight-optimal design of booms and for the rational judgment of boom performance [7, 8]. Eiden et al. presented a continuous-longeron space mast which can change from a helical coiled state into a straight deployed configuration [9]. However, for the large-scaled membrane antenna, the space thermal environment will inevitably induce deformation to the supporting structure, affecting the shape accuracy of the membrane [10]. Therefore, it is crucial to reduce the deformation by improving the static stiffness. For this purpose, it is in need to find a commonly accepted methodology to design the guyed mast which consists of a tall mast laterally supported at several levels along its height by sets of inclined pretensioned guys spaced around the mast [11]. Xiao et al. employed four tension cables to improve the stiffness of a membrane antenna supported by two compliant booms and established an equivalent 4-SPS-S parallel mechanism of the membrane antenna frame [12]. Páez and Sensale modeled the mast as a continuous equivalent beam-column on nonlinear elastic supports in guyed masts, which used the stability functions based on the Timoshenko beam-column [13]. The nonlinear behaviors of guyed masts have been widely studied. Belevičius et al. simplified the nonlinear behavior of the guyed mast by idealizing the nonlinear guys as approximate boundary conditions to optimize the mast structure for self-weight and wind loading [14]. Heydari et al. divided the guyed mast into linear and nonlinear parts and analyzed them separately and optimized the sizes of members, initial cable tensions, the positions of anchor on the ground, and tie level of cables on the mast [15]. Ballaben and Rosales established a nonlinear 3D finite element formulation to study the dynamic properties and the dynamic response of a 3D guyed mast [16]. Besides, Alshurafa et al. developed several nonlinear finite element models to explore the effects of parameters on the strength and stiffness of the guyed masts [17]. Some optimizations and parametric analyses have been carried out as well. Belevičius et al. optimized the double-level guyed mast to obtain the minimum weight design using the Genetic Algorithm and simulated annealing algorithm [18, 19] and revealed the dependence of the mast mass on the number of guys’ levels [20]. Moreover, Elena Parnás et al. discussed the influence of asymmetrical anchors on behavior of the guyed mast under wind loads [21].To reduce the computational effort in the pre-design stages, a mast is usually modeled as a continuous equivalent beam-column. The continuum modeling offers a practical and efficient approach for the analysis and design of large-scaled structures, thus having a wide range of applications. Considering local effects in each repeating mast element, Noor et al. developed continuum models for large beamlike lattice trusses subjected to static, thermal, and dynamic loadings [22, 23]. Salehian et al. derived the governing partial differential equations of vibration of a repeated-type structure into a Timoshenko equation [24]. Zhang et al. evaluated the equivalent bending stiffness of a hybrid deck-truss structure using the equivalent continuum beam method based on the homogenization concept and shearing equivalence principle [25]. In addition, Liu et al. established an equivalent dynamic model for the prestressed space antenna truss with the static condensation method on the bases of the energy equivalence principle and Timoshenko beam theory [26].For stiffness analysis, the guyed mast is more complex than the mast without guys to be analyzed. The analysis of the guyed mast used to mainly depend on the nonlinear finite element method, which takes much computational effort. Therefore, it is necessary to establish the equivalent mechanism of the guyed mast. Xiao et al. employed four tension cables to improve the stiffness of a membrane antenna in which the one-guy-level mast is modeled as a 4-SPS-S parallel mechanism [12]. Dai and Ding established the equilibrium between the twist deflection, the supporting wrenches, and the external wrench on a three-legged compliant platform and for the first time developed an analogy between the feeder and the parallel mechanisms [27]. Based on the principle of virtue work and the determined forces, Lu and Hu proposed an equivalent mechanism approach for establishing the stiffness matrices of parallel kinematic machines to solve the elastic deformation [28].This paper proposes a new modeling and analysis methodology for guyed masts, especially for double-level guyed masts with isosceles triangular cross section. Initially, the design of the membrane antenna and the characteristics of the CTM boom are illustrated. Afterwards, the central mast is mathematically modeled as a continuous equivalent beam in consideration of the material and geometric parameters of each element. Then, the double-level guyed mast is modeled as a 2(3-SPS-S) mechanism, and the total stiffness matrix of the equivalent mechanism is derived. Finally, the influence of anchor positions and attachment heights of guys on the stiffness performance is revealed, and the optimal distribution of guys is proposed based on the Genetic Algorithm (GA).


## 2. Structural Characteristics of Membrane Antenna

In case of lateral deformation, a double-level guyed mast shown in Figure1 is proposed to strengthen the structural static stiffness and improve the shape accuracy of the membrane. With one additional level of guys to the traditional single-level guyed mast, the constraints for the mast are increased, thus enhancing the static stiffness. In the membrane antenna, a large-scaled strip membrane is attached to the deployable central mast which has an isosceles triangular cross section. There are two lateral battens of equal length and one bottom batten in the section frame. The cross section of each batten is tube-shaped, and the outer diameter and thickness of each batten tube are denoted as db and tb, respectively. Three conformal connections fixed onto the three CTM booms in the deployment process are located at each vertex connecting neighbor battens. Three battens and three conformal connections construct an integrated frame to support the CTM booms. The width and height of the triangular frame are denoted as b0 and h0, respectively. Three deployment mechanisms forming an isosceles triangle are fixed on the base, and each of them can deploy a CTM boom from the base outwards. A continuous CTM boom is divided into a number of longerons by the frames, and the central mast is divided into several repeating units as well. There are two diagonal cables distributed on each rectangular face of the mast unit. When the central mast is fully deployed, it is supported by two levels of guy cables. Each of the three outer guys is anchored on the base and attached to the corresponding vertex of the outermost triangular frame. Meanwhile, each of the three inner guys is anchored on the base as well and attached to the corresponding vertex of a triangular frame between the base and the outermost frame.Figure 1
Double-level guyed mast for the membrane antenna.As shown in Figure2, the cross section of a CTM boom is symmetric, and a quarter of its inner profile is illustrated by line A′B′C′D′. The thickness of the CTM boom is denoted as t. The radii of arcs A′B′ and B′C′ are both denoted as Rl. Straight line C′D′ indicates the bonding plane of the top and bottom parts and has a width of q.Figure 2
Cross section of the CTM boom.Thex’-direction of a CTM boom is along its axis and the cross section is on the y'z'-coordinate plane, which is shown in Figure 2. The moment of area about the y'-axis, denoted as Iy1, and that about the z’-axis, denoted as Iz1, are, respectively, expressed as(1)Iy1=4∫03/2RltRlRl2−y2dy+4∫3/2Rl3RltRlRl−Rl2−y−3Rl22Rl2−y−3Rl2dy,(2)Iz1=4∫03/2RltRly2Rl2−y2dy+4∫3/2Rl3RltRly2Rl2−3Rl−y2dy+4∫3Rl3Rl+qty2dy.Calculating equations (1) and (2) yields(3)Iy1=8π3−33tRl3,Iz1=16π3−93tRl3+43t3Rl+q3.The calculation method referred in [29] is used to examine the equivalent torsional moment of area for the CTM boom, and the following result is obtained:(4)J1=18πRl3t.Besides, the cross-sectional area of the CTM boom, denoted asA1, is given by(5)A1=83πtRl+4qt.


## 3. Equivalent Mathematical Model of Central Mast

### 3.1. Unit of Central Mast

To offer a foundation for efficient analysis and design of the double-level guyed mast, the equivalent beam model of the central mast is established. The simplified repeating unit of the central mast shown in Figure3 consists of two bottom battens, four lateral battens, three longerons, two bottom diagonal cables, and four lateral diagonal cables. The six joints in the unit are denoted by numbers from 1 to 6. The origin of the coordinate system is located at the centroid of the bottom triangular frame in the unit. Three types of angles in this unit are defined as follows: the angle between bottom batten and bottom diagonal cable is α, the angle between bottom batten and lateral batten is β, and the angle between lateral batten and lateral diagonal cable is θ.Figure 3
Repeating unit of the central mast.If the length of each element in the unit is denoted with its two joint numbers, such thatl12 indicates the length of the batten between joints 1 and 2, there will be five different lengths, respectively, denoted as l1, l2, l3, l4, and l5 in this unit, and the following equality relations can be concluded:(6)l14=l25=l36=l1,l12=l13=l45=l46=l2,l23=l56=l3,l15=l24=l34=l16=l4,l35=l26=l5.Furthermore, the relations among parameters of lengths and angles can be described by(7)l1=l2tanθ=l3tanα=l4sinθ=l5sinα.The displacement components of the repeating mast unit are expressed as follows.(8)ux,y,z=u0x−yϕz0x+zϕy0x,vx,y,z=v0x+yεy0x+z−ϕx0x+12γyz0x,wx,y,z=w0x+zεz0x+yϕx0x+12γyz0x,where u0x, v0x, and w0x are the displacement components at y = z = 0; ϕx0x, ϕy0x, and ϕz0x are the rotation components; εy0x and εz0x are the extensional strains in the y and z directions; and γyz0x is the shearing strain in the plane of the cross section. Based on equation (8), the displacement components of an arbitrary point of the repeating unit are expanded in a Taylor series in the x direction, thus obtaining equation (9) if the derivative terms of strain are ignored:(9)ux,y,z≈u0+xεx0−yϕz0+xκy0+zϕy0+xκz0,vx,y,z≈v0+xϕz0+γxy0+yεy0+z−ϕx0−xκx0+12γyz0+12x2κy0,wx,y,z≈w0+xγxy0−ϕy0+zεz0+yϕx0+xκx0+12γyz0−12x2κz0,where κx0 is the twisting curvature and κy0 and κz0 are the bending curvatures. Then, the following equations can be obtained:(10)κy0=∂2v∂x2,κz0=−∂2w∂x2,κx0=∂ϕx∂x.Based on the kinematic assumptions, the strain components have a linear variation in they and z directions as follows:(11)εx=εx0−yκy0+zκz0,εy=εy0,εz=εz0,γxy=γxy0+y∂εy0+z−κx0+12∂γyz0,γxz=γxz0+z∂εz0+yκx0+12∂γyz0,γyz=γyz0,where εx0 is the extensional strain of the centerline; κx0 is the twist in the x direction; κy0 and κz0 are the curvature changes in the y and z directions; and γxy0, γxz0, and γyz0 are the transverse shear strains.The axial strain of the batten, longeron, and cable elements in the repeating unit can be expressed as follows:(12)εk=εxklk2+εykmk2+εzknk2+γyzkmknk+γxzklknk+γxyklkmk,where εk is the axial strain of the kth element; lk, mk, and nk are the direction cosines of the axes of the kth element; and εxk, εyk, εzk, γyzk, γxzk, and γxyk are the strain components of the kth element. As a batten belongs to two neighbor units, only one bottom batten and two lateral elements should be included in the number of elements. Accordingly, there are 12 elements in a repeating unit. The strain state of each batten, longeron, and cable element is represented by the strain of the middle point in each corresponding element. If an element connects joints i and j, the position vector of its middle point will be denoted as pij. Hence, the position matrix P of the whole mast unit can be expressed as follows:(13)P=p25p36p14p23p12p13p24p15p16p34p26p35=l12−l1cα2sα−l1sβcθ3sθl12l1cα2sα−l1sβcθ3sθl1202l1sβcθ3sθ00−l1sβcθ3sθ0−l1cα4sαl1sβcθ6sθ0l1cα4sαl1sβcθ6sθl12−l1cα4sαl1sβcθ6sθl12−l1cα4sαl1sβcθ6sθl12l1cα4sαl1sβcθ6sθl12l1cα4sαl1sβcθ6sθl120−l1sβcθ3sθl120−l1sβcθ3sθ,where s represents sin and c represents cos. For the kth element, its direction vector can be expressed by lk2,mk2,nk2,mknk,lknk,lkmk. Therefore, the corresponding direction matrix Q to P can be established with the direction vectors of all elements in the mast unit, which is expressed as follows:(14)Q=1000001000001000000100000c2βs2βsβcβ000c2βs2β−sβcβ00s2θc2θc2βc2θs2βc2θcβsβsθcθsβsθcθcβs2θc2θc2βc2θs2βc2θcβsβ−sθcθsβ−sθcθcβs2θc2θc2βc2θs2β−c2θcβsβ−sθcθsβsθcθcβs2θc2θc2βc2θs2β−c2θcβsβsθcθsβ−sθcθcβs2αc2α000sαcαs2αc2α000−sαcα.With the obtained position matrix and direction matrix, equation (12) is transformed into the following expression:(15)εk=IkΑΕ0+PΚ0QTIkT,where Ik, Α, Ε0, and Κ0 are shown in the Appendix. In this calculation, the first 3 elements are longerons, the next 3 elements are battens, and the last 6 elements are cables.


### 3.2. Equivalent Beam

The strain energy of each longeron and batten includes the axial strain energy, the torsion strain energy, and the bending strain energy, whereas that of each cable only includes the axial strain energy. Therefore, the total strain energy of the repeating central mast unit can be expressed as(16)UM=∑k=1312∫l1E1A1εk2+E1Iy1∂2v∂x22+E1Iz1∂2w∂x22+G1J1∂ϕx∂x2dx+∑k=4612∫l1E2A2εk2+E2Iy2∂2v∂x22+E2Iz2∂2w∂x22+G2J2∂ϕx∂x2dx+∑k=71212∫l1E3A3εk2dx,where E1A1, E2A2, and E3A3 denote the axial stiffnesses of the longeron, batten, and beam, respectively; E1Iy1, E1Iz1, E2Iy2, and E2Iz2 denote the bending stiffnesses of the longeron and batten in y and z directions, respectively; and G1J1 and G2J2 denote the torsion stiffnesses of the longeron and batten, respectively. Based on the energy equivalence principle, the total strain energy of the repeating mast unit is equal to that of an anisotropic Timoshenko beam model which is expressed as(17)UB=12∫l1εTDεdx,where the strain vector ε and the elasticity matrix D of the equivalent beam are, respectively, written as(18)ε=εx0κy0κz0κx0γxz0γxy0T,D=EA¯η12η13η14η15η16EI¯zη23η24η25η26EI¯yη34η35η36GJ¯η45η46SYMGA¯xzη56GA¯xy,where EA¯, EI¯z, EI¯y, GJ¯, GA¯xz, and GA¯xy denote the extension, bending, twisting, and shearing stiffnesses of the equivalent beam, and the rest of the matrix elements are the coupling coefficients of the stiffnesses. By comparing the corresponding coefficients in the results obtained from equations (16) and (17), the stiffness of the equivalent beam can be solved as(19)EA¯=3A1E1+2A3E3sin4α+4A3E3sin4θ,EI¯y=3E1Iy1+19A3E3l12cos2θsin2θsin2β+23A1E1l12cot2θsin2β+29A3E3l12cot2θsin4αsin2β,EI¯z=3E1Iz1+12A1E1l12cot2α+14A3E3l12cot2αsin4θ,GJ¯=3G1J1+19A3E3l12cos2βcos4θsin2β+29A3E3l12cos2αcot2θsin2αsin2β+13A3E3l12cosβcos3θcotαsin2βsinθ+14A3E3l12cos2θcot2αsin2βsin2θ,GA¯xz=4A3E3cos2θsin2βsin2θ,GA¯xy=2A3E3cos2αsin2α+4A3E3cos2βcos2θsin2θ.In this section, the equivalent model of the central mast has been established and will be applied in the static stiffness computation for the equivalent mechanism of the double-level guyed mast.


## 3.1. Unit of Central Mast

To offer a foundation for efficient analysis and design of the double-level guyed mast, the equivalent beam model of the central mast is established. The simplified repeating unit of the central mast shown in Figure3 consists of two bottom battens, four lateral battens, three longerons, two bottom diagonal cables, and four lateral diagonal cables. The six joints in the unit are denoted by numbers from 1 to 6. The origin of the coordinate system is located at the centroid of the bottom triangular frame in the unit. Three types of angles in this unit are defined as follows: the angle between bottom batten and bottom diagonal cable is α, the angle between bottom batten and lateral batten is β, and the angle between lateral batten and lateral diagonal cable is θ.Figure 3
Repeating unit of the central mast.If the length of each element in the unit is denoted with its two joint numbers, such thatl12 indicates the length of the batten between joints 1 and 2, there will be five different lengths, respectively, denoted as l1, l2, l3, l4, and l5 in this unit, and the following equality relations can be concluded:(6)l14=l25=l36=l1,l12=l13=l45=l46=l2,l23=l56=l3,l15=l24=l34=l16=l4,l35=l26=l5.Furthermore, the relations among parameters of lengths and angles can be described by(7)l1=l2tanθ=l3tanα=l4sinθ=l5sinα.The displacement components of the repeating mast unit are expressed as follows.(8)ux,y,z=u0x−yϕz0x+zϕy0x,vx,y,z=v0x+yεy0x+z−ϕx0x+12γyz0x,wx,y,z=w0x+zεz0x+yϕx0x+12γyz0x,where u0x, v0x, and w0x are the displacement components at y = z = 0; ϕx0x, ϕy0x, and ϕz0x are the rotation components; εy0x and εz0x are the extensional strains in the y and z directions; and γyz0x is the shearing strain in the plane of the cross section. Based on equation (8), the displacement components of an arbitrary point of the repeating unit are expanded in a Taylor series in the x direction, thus obtaining equation (9) if the derivative terms of strain are ignored:(9)ux,y,z≈u0+xεx0−yϕz0+xκy0+zϕy0+xκz0,vx,y,z≈v0+xϕz0+γxy0+yεy0+z−ϕx0−xκx0+12γyz0+12x2κy0,wx,y,z≈w0+xγxy0−ϕy0+zεz0+yϕx0+xκx0+12γyz0−12x2κz0,where κx0 is the twisting curvature and κy0 and κz0 are the bending curvatures. Then, the following equations can be obtained:(10)κy0=∂2v∂x2,κz0=−∂2w∂x2,κx0=∂ϕx∂x.Based on the kinematic assumptions, the strain components have a linear variation in they and z directions as follows:(11)εx=εx0−yκy0+zκz0,εy=εy0,εz=εz0,γxy=γxy0+y∂εy0+z−κx0+12∂γyz0,γxz=γxz0+z∂εz0+yκx0+12∂γyz0,γyz=γyz0,where εx0 is the extensional strain of the centerline; κx0 is the twist in the x direction; κy0 and κz0 are the curvature changes in the y and z directions; and γxy0, γxz0, and γyz0 are the transverse shear strains.The axial strain of the batten, longeron, and cable elements in the repeating unit can be expressed as follows:(12)εk=εxklk2+εykmk2+εzknk2+γyzkmknk+γxzklknk+γxyklkmk,where εk is the axial strain of the kth element; lk, mk, and nk are the direction cosines of the axes of the kth element; and εxk, εyk, εzk, γyzk, γxzk, and γxyk are the strain components of the kth element. As a batten belongs to two neighbor units, only one bottom batten and two lateral elements should be included in the number of elements. Accordingly, there are 12 elements in a repeating unit. The strain state of each batten, longeron, and cable element is represented by the strain of the middle point in each corresponding element. If an element connects joints i and j, the position vector of its middle point will be denoted as pij. Hence, the position matrix P of the whole mast unit can be expressed as follows:(13)P=p25p36p14p23p12p13p24p15p16p34p26p35=l12−l1cα2sα−l1sβcθ3sθl12l1cα2sα−l1sβcθ3sθl1202l1sβcθ3sθ00−l1sβcθ3sθ0−l1cα4sαl1sβcθ6sθ0l1cα4sαl1sβcθ6sθl12−l1cα4sαl1sβcθ6sθl12−l1cα4sαl1sβcθ6sθl12l1cα4sαl1sβcθ6sθl12l1cα4sαl1sβcθ6sθl120−l1sβcθ3sθl120−l1sβcθ3sθ,where s represents sin and c represents cos. For the kth element, its direction vector can be expressed by lk2,mk2,nk2,mknk,lknk,lkmk. Therefore, the corresponding direction matrix Q to P can be established with the direction vectors of all elements in the mast unit, which is expressed as follows:(14)Q=1000001000001000000100000c2βs2βsβcβ000c2βs2β−sβcβ00s2θc2θc2βc2θs2βc2θcβsβsθcθsβsθcθcβs2θc2θc2βc2θs2βc2θcβsβ−sθcθsβ−sθcθcβs2θc2θc2βc2θs2β−c2θcβsβ−sθcθsβsθcθcβs2θc2θc2βc2θs2β−c2θcβsβsθcθsβ−sθcθcβs2αc2α000sαcαs2αc2α000−sαcα.With the obtained position matrix and direction matrix, equation (12) is transformed into the following expression:(15)εk=IkΑΕ0+PΚ0QTIkT,where Ik, Α, Ε0, and Κ0 are shown in the Appendix. In this calculation, the first 3 elements are longerons, the next 3 elements are battens, and the last 6 elements are cables.


## 3.2. Equivalent Beam

The strain energy of each longeron and batten includes the axial strain energy, the torsion strain energy, and the bending strain energy, whereas that of each cable only includes the axial strain energy. Therefore, the total strain energy of the repeating central mast unit can be expressed as(16)UM=∑k=1312∫l1E1A1εk2+E1Iy1∂2v∂x22+E1Iz1∂2w∂x22+G1J1∂ϕx∂x2dx+∑k=4612∫l1E2A2εk2+E2Iy2∂2v∂x22+E2Iz2∂2w∂x22+G2J2∂ϕx∂x2dx+∑k=71212∫l1E3A3εk2dx,where E1A1, E2A2, and E3A3 denote the axial stiffnesses of the longeron, batten, and beam, respectively; E1Iy1, E1Iz1, E2Iy2, and E2Iz2 denote the bending stiffnesses of the longeron and batten in y and z directions, respectively; and G1J1 and G2J2 denote the torsion stiffnesses of the longeron and batten, respectively. Based on the energy equivalence principle, the total strain energy of the repeating mast unit is equal to that of an anisotropic Timoshenko beam model which is expressed as(17)UB=12∫l1εTDεdx,where the strain vector ε and the elasticity matrix D of the equivalent beam are, respectively, written as(18)ε=εx0κy0κz0κx0γxz0γxy0T,D=EA¯η12η13η14η15η16EI¯zη23η24η25η26EI¯yη34η35η36GJ¯η45η46SYMGA¯xzη56GA¯xy,where EA¯, EI¯z, EI¯y, GJ¯, GA¯xz, and GA¯xy denote the extension, bending, twisting, and shearing stiffnesses of the equivalent beam, and the rest of the matrix elements are the coupling coefficients of the stiffnesses. By comparing the corresponding coefficients in the results obtained from equations (16) and (17), the stiffness of the equivalent beam can be solved as(19)EA¯=3A1E1+2A3E3sin4α+4A3E3sin4θ,EI¯y=3E1Iy1+19A3E3l12cos2θsin2θsin2β+23A1E1l12cot2θsin2β+29A3E3l12cot2θsin4αsin2β,EI¯z=3E1Iz1+12A1E1l12cot2α+14A3E3l12cot2αsin4θ,GJ¯=3G1J1+19A3E3l12cos2βcos4θsin2β+29A3E3l12cos2αcot2θsin2αsin2β+13A3E3l12cosβcos3θcotαsin2βsinθ+14A3E3l12cos2θcot2αsin2βsin2θ,GA¯xz=4A3E3cos2θsin2βsin2θ,GA¯xy=2A3E3cos2αsin2α+4A3E3cos2βcos2θsin2θ.In this section, the equivalent model of the central mast has been established and will be applied in the static stiffness computation for the equivalent mechanism of the double-level guyed mast.


## 4. Equivalent Mechanism of Double-Level Guyed Mast

### 4.1. Stiffness of SPS Branches

Since the central mast is modeled as an equivalent beam, it is shown as the central beamO0O2 perpendicular to the base in Figure 4. The triangular frames which the outer and the inner guys are attached to are, respectively, represented as the outer platform and the inner platform. The inner and the outer platforms link with the central beam through spherical joints O1 and O2, respectively. The replacement of the tension cable with a SPS kinematic branch is proposed by [12]. The SPS branch consists of two spherical joints and a prismatic joint, which performs the same movement with that of a cable. Therefore, the double-level guyed mast is modeled as a 2(3-SPS-S) mechanism in Figure 4. Coordinate frames O0−x¯0y¯0z¯0, O1−x¯1y¯1z¯1, and O2−x¯2y¯2z¯2 are located at O0, O1, and O2 on the central beam, respectively. Besides, z¯0, z¯1, and z¯2 denote the axial direction of the beam. In a SPS mechanism, the anchors on the base, the actuating prismatic joints, and the attachment points to the platform are, respectively, denoted as Aij, Bij, and Ciji=1,2,j=1,2,3 in which i = 1 means the SPS mechanism represents an inner guy and i = 2 means it represents an outer guy.Figure 4
Equivalent 2(3-SPS-S) mechanism of the double-level guyed mast.The relations between the velocities of the actuating prismatic joints and the platform can be expressed by the velocity Jacobian matrixJi1i=1,2 as follows:(20)Ji1=ri1Tei1×ri1Tri2Tei2×ri2Tri3Tei3×ri3T3×6,where eij=Cij−Oi and rij=Cij−Aij/Cij−Aijj=1,2,3. If the actuating force and the elastic deformation of the SPS branch are, respectively, denoted as Fiaj and dlij, the stiffness matrix Kia of the SPS branches can be established as(21)Fia1Fia2Fia3=Kiadli1dli2dli3,where Kia is a 3×3 symmetric stiffness matrix as follows:(22)Kia=Killi1−1000Killi2−1000Killi3−1,where kil is the one-meter stiffness of the guy and lij is the measured length of the guy. The state of each guy is denoted as Cisjk′, in which k' indicates the times of the iteration. When a guy is tensioned, its corresponding Cisjk′ is 1; otherwise is 0. If sik′=diagCis1k′Cis2k′Cis3k′ expresses the state of each level of guys, the stiffness matrix in the k'th iteration can be written as(23)K¯ia=Kiasik′.


### 4.2. Stiffness Analysis of 2(3-SPS-S) Mechanism

The load state of the central beam is shown in Figure5. The distance between O0 and O1 is d1, and that between O1 and O2 is d2. The external loads acting on the beam at Oi of the central beam are denoted as Ti, which can be expressed as(24)Ti=FixFiyFizMixMiyMizT.Figure 5
Load state of the central beam.Meanwhile,δi representing the displacement at Oi is expressed as(25)δi=ωixωiyωizθixθiyθizT.The elastic deformations in each level of SPS branches can hence be written as(26)dli1dli2dli3=Ji1δi.Besides, based on the obtained results in Section3, the stiffness parameters of this equivalent central beam in the 2(3-SPS-S) mechanism can be expressed as(27)EA=EA¯,EIx=EI¯z,EIy=EI¯y,GJ=GJ¯.As the central mast has been modeled as an anisotropic beam, the load analysis of the central beam is carried out in different directions. The load states of the central beam inx and y directions are shown in Figure 6. The distance from O0 to any point between O0 and O1 is denoted as s1, while that from O0 to any point between O1 and O2 is denoted as s2. Based on the definition of s1 and s2, their ranges can be written as follows.(28)s1∈0,d1,s2∈d1,d1+d2.Figure 6
Load states of the central beam: (a) inx direction and (b) in y direction.
(a)(b)The forces and moments in different directions for a point positioned withsii=1,2 are denoted as Fxi, Fyi, Fzi, Mxi, Myi, and Mzi while the deflections in x and y directions for this point are denoted as ωxi and ωyi. According to the load states shown in Figure 6, the moments in the beam can be calculated as follows:(29)EIxωy1″s1=Mx1s1=M1x+M2x−F1yd1−s1−F2yd1+d2−s1,EIywx1″s1=My1s1=M1y+M2y+F1xd1−s1+F2xd1+d2−s1,EIxωy2″s2=Mx2s2=M2x−F2yd1+d2−s2,EIywx2″s2=My2s2=M2y+F2xd1+d2−s2.Besides, the forces and moments in thez direction of the beam can be obtained as follows:(30)Fz1s1=F1z+F2z,Mz1s1=M1z+M2z,Fz2s2=F2z,Mz2s2=M2z.As the central beam is a continuous beam constrained by the base, the characteristics of its deflection can offer boundary and continuity conditions for the calculation. The boundary conditions for the central beam are expressed as(31)s1=0:ωx1s1=0,ωx1′s1=0,ωy1s1=0,ωy1′s1=0.Meanwhile, the continuity conditions for the central beam are expressed as(32)s1=s2=d1:ωx1s1=ωx2s2,ωx1′s1=ωx2′s2,ωy1s1=ωy2s2,ωy1′s1=ωy2′s2.The formula of force deformation can be derived from equations (29) and (30) with the condition equations, and it is solved as(33)T1T2=K11K12K21K22−1δ1δ2,where K11, K12, K21, and K22 are shown in the Appendix. Transforming equation (33) yields the following expression:(34)T1T2=K¯11K¯12K¯21K¯22δ1δ2,where K¯11, K¯12, K¯21, and K¯22 are shown in the Appendix. Afterwards, the expressions of T1 and T2 are revealed as(35)T1=K¯11δ1+K¯12δ2,(36)T2=K¯21δ1+K¯22δ2.Based on the principle of virtual work, the following equation exists atO1:(37)δ1TT1+dl11dl12dl13TF1a1F1a2F1a3=0.Substituting equations (21) and (26) into (37) yields(38)T1=−J11TK¯11J11δ1.Substituting equation (38) into (35) leads to(39)δ1=K¯δ2,where(40)K¯=−K¯11+J11TK¯1aJ11−1K¯12.Introducing equation (39) into (36) yields the following expression of T2 with δ2:(41)T2=K¯21K¯+K¯22δ2.Assuming the load applied to the whole 2(3-SPS-S) mechanism on the outer platform is denoted asT, the following equation can be established according to the principle of virtual work:(42)dl21dl22dl23TF2a1F2a2F2a3+δ2TT2=δ2TT.The above equation can be rewritten by introducing equations (21) and (26) into (42) as(43)δ2TJ21TK¯2aJ21δ2+δ2TT2=δ2TT.Substituting the expression ofT2 in equation (41) into the above presents(44)T=Kδ2,where(45)K=K¯21K¯+K¯22+J21TK¯2aJ21.In this section, the total stiffness matrix of the equivalent 2(3-SPS-S) mechanism for the double-level guyed mast is finally obtained, which establishes the relation between the load and deformation at the end of the central mast.


## 4.1. Stiffness of SPS Branches

Since the central mast is modeled as an equivalent beam, it is shown as the central beamO0O2 perpendicular to the base in Figure 4. The triangular frames which the outer and the inner guys are attached to are, respectively, represented as the outer platform and the inner platform. The inner and the outer platforms link with the central beam through spherical joints O1 and O2, respectively. The replacement of the tension cable with a SPS kinematic branch is proposed by [12]. The SPS branch consists of two spherical joints and a prismatic joint, which performs the same movement with that of a cable. Therefore, the double-level guyed mast is modeled as a 2(3-SPS-S) mechanism in Figure 4. Coordinate frames O0−x¯0y¯0z¯0, O1−x¯1y¯1z¯1, and O2−x¯2y¯2z¯2 are located at O0, O1, and O2 on the central beam, respectively. Besides, z¯0, z¯1, and z¯2 denote the axial direction of the beam. In a SPS mechanism, the anchors on the base, the actuating prismatic joints, and the attachment points to the platform are, respectively, denoted as Aij, Bij, and Ciji=1,2,j=1,2,3 in which i = 1 means the SPS mechanism represents an inner guy and i = 2 means it represents an outer guy.Figure 4
Equivalent 2(3-SPS-S) mechanism of the double-level guyed mast.The relations between the velocities of the actuating prismatic joints and the platform can be expressed by the velocity Jacobian matrixJi1i=1,2 as follows:(20)Ji1=ri1Tei1×ri1Tri2Tei2×ri2Tri3Tei3×ri3T3×6,where eij=Cij−Oi and rij=Cij−Aij/Cij−Aijj=1,2,3. If the actuating force and the elastic deformation of the SPS branch are, respectively, denoted as Fiaj and dlij, the stiffness matrix Kia of the SPS branches can be established as(21)Fia1Fia2Fia3=Kiadli1dli2dli3,where Kia is a 3×3 symmetric stiffness matrix as follows:(22)Kia=Killi1−1000Killi2−1000Killi3−1,where kil is the one-meter stiffness of the guy and lij is the measured length of the guy. The state of each guy is denoted as Cisjk′, in which k' indicates the times of the iteration. When a guy is tensioned, its corresponding Cisjk′ is 1; otherwise is 0. If sik′=diagCis1k′Cis2k′Cis3k′ expresses the state of each level of guys, the stiffness matrix in the k'th iteration can be written as(23)K¯ia=Kiasik′.


## 4.2. Stiffness Analysis of 2(3-SPS-S) Mechanism

The load state of the central beam is shown in Figure5. The distance between O0 and O1 is d1, and that between O1 and O2 is d2. The external loads acting on the beam at Oi of the central beam are denoted as Ti, which can be expressed as(24)Ti=FixFiyFizMixMiyMizT.Figure 5
Load state of the central beam.Meanwhile,δi representing the displacement at Oi is expressed as(25)δi=ωixωiyωizθixθiyθizT.The elastic deformations in each level of SPS branches can hence be written as(26)dli1dli2dli3=Ji1δi.Besides, based on the obtained results in Section3, the stiffness parameters of this equivalent central beam in the 2(3-SPS-S) mechanism can be expressed as(27)EA=EA¯,EIx=EI¯z,EIy=EI¯y,GJ=GJ¯.As the central mast has been modeled as an anisotropic beam, the load analysis of the central beam is carried out in different directions. The load states of the central beam inx and y directions are shown in Figure 6. The distance from O0 to any point between O0 and O1 is denoted as s1, while that from O0 to any point between O1 and O2 is denoted as s2. Based on the definition of s1 and s2, their ranges can be written as follows.(28)s1∈0,d1,s2∈d1,d1+d2.Figure 6
Load states of the central beam: (a) inx direction and (b) in y direction.
(a)(b)The forces and moments in different directions for a point positioned withsii=1,2 are denoted as Fxi, Fyi, Fzi, Mxi, Myi, and Mzi while the deflections in x and y directions for this point are denoted as ωxi and ωyi. According to the load states shown in Figure 6, the moments in the beam can be calculated as follows:(29)EIxωy1″s1=Mx1s1=M1x+M2x−F1yd1−s1−F2yd1+d2−s1,EIywx1″s1=My1s1=M1y+M2y+F1xd1−s1+F2xd1+d2−s1,EIxωy2″s2=Mx2s2=M2x−F2yd1+d2−s2,EIywx2″s2=My2s2=M2y+F2xd1+d2−s2.Besides, the forces and moments in thez direction of the beam can be obtained as follows:(30)Fz1s1=F1z+F2z,Mz1s1=M1z+M2z,Fz2s2=F2z,Mz2s2=M2z.As the central beam is a continuous beam constrained by the base, the characteristics of its deflection can offer boundary and continuity conditions for the calculation. The boundary conditions for the central beam are expressed as(31)s1=0:ωx1s1=0,ωx1′s1=0,ωy1s1=0,ωy1′s1=0.Meanwhile, the continuity conditions for the central beam are expressed as(32)s1=s2=d1:ωx1s1=ωx2s2,ωx1′s1=ωx2′s2,ωy1s1=ωy2s2,ωy1′s1=ωy2′s2.The formula of force deformation can be derived from equations (29) and (30) with the condition equations, and it is solved as(33)T1T2=K11K12K21K22−1δ1δ2,where K11, K12, K21, and K22 are shown in the Appendix. Transforming equation (33) yields the following expression:(34)T1T2=K¯11K¯12K¯21K¯22δ1δ2,where K¯11, K¯12, K¯21, and K¯22 are shown in the Appendix. Afterwards, the expressions of T1 and T2 are revealed as(35)T1=K¯11δ1+K¯12δ2,(36)T2=K¯21δ1+K¯22δ2.Based on the principle of virtual work, the following equation exists atO1:(37)δ1TT1+dl11dl12dl13TF1a1F1a2F1a3=0.Substituting equations (21) and (26) into (37) yields(38)T1=−J11TK¯11J11δ1.Substituting equation (38) into (35) leads to(39)δ1=K¯δ2,where(40)K¯=−K¯11+J11TK¯1aJ11−1K¯12.Introducing equation (39) into (36) yields the following expression of T2 with δ2:(41)T2=K¯21K¯+K¯22δ2.Assuming the load applied to the whole 2(3-SPS-S) mechanism on the outer platform is denoted asT, the following equation can be established according to the principle of virtual work:(42)dl21dl22dl23TF2a1F2a2F2a3+δ2TT2=δ2TT.The above equation can be rewritten by introducing equations (21) and (26) into (42) as(43)δ2TJ21TK¯2aJ21δ2+δ2TT2=δ2TT.Substituting the expression ofT2 in equation (41) into the above presents(44)T=Kδ2,where(45)K=K¯21K¯+K¯22+J21TK¯2aJ21.In this section, the total stiffness matrix of the equivalent 2(3-SPS-S) mechanism for the double-level guyed mast is finally obtained, which establishes the relation between the load and deformation at the end of the central mast.


## 5. Analysis of Static Stiffness of Double-Level Guyed Mast

Based on the theoretical modeling of the double-level guyed mast in Figure4, the static stiffness of the double-level guyed mast can be examined. In a membrane antenna, the bending stiffness of the guyed mast mainly affects the accuracy of a planar membrane. Therefore, the distribution of guys should satisfy the optimal bending stiffness. The ratios of load to deformation in the x and y directions are denoted as the static stiffnesses Kx and Ky, respectively. In this section, the analysis is based on a double-level guyed mast in which some parameters of the central mast have been assigned. The length of each mast unit is 2.5 m, and the total length of the guyed mast is denoted as lm. In the isosceles triangular cross section of the central mast, b0 is 3 m, and h0 is 1 m. Each guy should be tensioned when the structure deforms, and the tension force is set as 100 N. On the base, the anchors of the inner guys form an isosceles triangle whose width and height are denoted as b1 and h1, respectively. Meanwhile, the width and height of the isosceles triangle formed by the anchors of the outer guys are denoted as b2 and h2, respectively. The radius of CTM is denoted as Rl, and the radius of guys is denoted as Rg. Besides, the material and sectional parameters of the mast elements are shown in Table 1.Table 1
Material and sectional parameters of the mast elements.
ItemsBattensDiagonal cablesCTMMaterialCarbon fiberKevlarCompositeElasticity modulusE (GPa)350131114Densityρ (kg/m3)200014351600Poisson’s ratioυ0.300.300.30Considering the size of the base is limited in space,b2 and h2 are set as 9 m and 3 m, respectively, whereas b1, h1, and d1 are set as variables. The ratio of b1 to h1 is set as 3, which is the same as that of b0 to h0. Besides, lm is set as 30 m. Under these conditions, the influences of d1 and h1 on the static stiffnesses Kx and Ky are investigated and shown in Figures 7 and 8, respectively. It is found that increasing h1 can raise both Kx and Ky. Moreover, Kx and Ky will both reach a peak when d1 is around 25 m. When d1 is larger than 25 m, Kx and Ky will both decrease slightly. Therefore, d1 and h1 should be, respectively, set as 25 m and 3 m for the optimal static stiffness of the double-level guyed mast. In this situation, h1 is equal to h2, and b1 is equal to b2.Figure 7
Influence ofd1 and h1 on Kx.Figure 8
Influence ofd1 and h1 on Ky.For the given base in which the width and height are, respectively, set as 9 m and 3 m, the optimal distribution of guys is shown in Figure9. The anchors of the inner guys coincide with those of outer guys. The parameter d2 is 5 m, which means there are two mast units between the inner and the outer platforms.Figure 9
Distribution of guys for optimal static stiffness.If the above optimized configuration is designated as Case 1, other cases can be obtained by changing each parameter of Case 1. Five feature parameters are chosen in a case, includingRg, h1, Rl, lm, and d1. Therefore, the six different cases can be established as follows.Case 1:Rg=2mm,Rl=80mm,h1=3m,lm=30m,d1=25mCase 2:Rg=3mm,Rl=80mm,h1=3m,lm=30m,d1=25mCase 3:Rg=2mm,Rl=100mm,h1=3m,lm=30m,d1=25mCase 4:Rg=2mm,Rl=80mm,h1=4m,lm=30m,d1=25mCase 5:Rg=2mm,Rl=80mm,h1=3m,lm=40m,d1=25mCase 6:Rg=2mm,Rl=80mm,h1=3m,lm=30m,d1=15mTo verify the theoretical model of 2(3-SPS-S) mechanism shown in Figure4, its results are compared with those obtained from finite element analysis (FEA) model. In the FEA model, the ratios of load to deformation in the x and y directions are denoted as Kx∗ and Ky∗, respectively. Three equal loads are applied at the three end nodes of the central mast, respectively. The deformation of the guyed mast is indicated by the average displacement of the three end nodes. The coordinate system used to measure the results is O2−x¯2y¯2z¯2 as shown in Figure 9, which is for both FEA model and theoretical model. In addition, the errors between the results of FEA model and theoretical model are denoted as ξx and ξy, respectively, which are expressed as follows:(46)ξx=Kx−Kx∗Kx∗×100%,ξy=Ky−Ky∗Ky∗×100%.The comparison of the results from FEA model and theoretical model is shown in Table2, and it is found that the errors are acceptable. In particular, the accuracy of the theoretical model in the x direction is higher than that in the y direction. Generally, the equivalent mechanism is reliable enough for the purpose of pre-design.Table 2
Comparison between the results of FEA model and theoretical model.
CasesKx∗ (N/m)Kx (N/m)ξxKy∗ (N/m)Ky (N/m)ξyCase 12.02×1032.05×1031.49%1.17×1041.38×10417.95%Case 22.86×1033.04×1036.29%1.71×1042.04×10419.30%Case 32.27×1032.36×1033.96%1.36×1041.58×10416.18%Case 42.42×1032.59×1037.02%1.52×1041.73×10413.82%Case 58.29×1028.57×1023.38%5.46×1035.76×1035.49%Case 61.96×1031.99×1031.53%1.15×1041.33×10415.65%Further, the Genetic Algorithm (GA) is used to conduct the optimization of the static stiffness of the double-level guyed mast. According to the above comparison,Kx is more accurate than Ky, and Kx performs the same variation regularity with Ky as shown in Figures 7 and 8. Therefore, Kx is chosen in the optimization problem which can be established as follows.(47)Findχ=h1,b1,d1T,MinΨχ=1Kxs.t.b1=3h1,h0≤h1≤h2,0≤d1≤lm.Based on equation (47), the optimization results calculated with GA show h1 is equal to h2, whereas the optimal d1 is related to lm. In addition, η is the ratio of d1 to lm, which is going to be investigated in different parametric conditions. The effects from Rg, Rl, and h2 on the optimal η are studied and presented in Figures 10–12, respectively. When the length of the guyed mast increases, the optimal η decreases in a nonlinear pattern but finally infinitely approaches a constant value. It is revealed in Figure 10 that the larger the radius of guys is, the lower the optimal η will be. Differently, the larger the radius of the CTM is, the higher the optimal η will be obtained, which is depicted in Figure 11. According to Figure 12, it is noticed that increasing h2 can raise the optimal η as well. By comparing Figures 10, 11, and 12, it is demonstrated that Rg, Rl, and h2 influence the optimal η by different means. The variation curves in Figure 10 or 11 tend to be parallel to each other, but those in Figure 12 do not show this characteristic. Besides, when Rl is raised to a certain degree, the increase of the optimal η is limited. Among these three variation conditions, it is found that changing the radius of guys could be the most efficient way to adjust the optimal η.Figure 10
Influence ofRg on the optimal η.Figure 11
Influence ofRl on the optimal η.Figure 12
Influence ofh2 on the optimal η.Given a certain length of the guyed mast, the influence ofRl on the optimal d1 is investigated and presented in Figure 13. The length of the guyed mast is set as 30 m, and the radius of guys varies from 1 mm to 5 mm. When Rg is 1 mm, the increase of Rl hardly influences the optimal d1. Besides, the larger the radius of guys, the more obvious Rl affects the optimal d1. Therefore, the influencing degree of Rl on the optimal d1 depends on the thickness of guys.Figure 13
Influence ofRl on the optimal d1.


## 6. Conclusions

This paper presents the stiffness analysis of a double-level guyed mast for a membrane antenna and the optimal distribution of guys for enhancing static stiffness. Based on the energy equivalence principle, the central mast is mathematically modeled as an anisotropic Timoshenko beam. Afterwards, the double-level guyed mast is modeled as a 2(3-SPS-S) mechanism in which the stiffnesses of the SPS branches and the central double-level loaded mast are derived, respectively. The total stiffness matrix of the equivalent mechanism is obtained according to the principle of virtual work. The comparison between results of FEA model and theoretical model indicates the equivalent mechanism is reliable enough for the purpose of pre-design.The influence of distribution of guys on static stiffness is discussed in detail. The anchors of the inner guys should coincide with those of the outer guys for the optimal static stiffness. The Genetic Algorithm is used to conduct the optimization of the static stiffness, which illustrates the influences fromRg, Rl, and h2 on the optimal η. Among these three variation conditions, changing the radius of guys could be the most efficient way to adjust the optimal η. The analysis and design of the double-level guyed mast provides a reference for the equivalence of a multilevel guyed mast.



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*Source: 3614625-2020-11-25.xml*