Modeling of suspended weightlessness on the cables of the beam system, by changing the tension forces

Толық мәтін

Introduction

The problem of optimal and rational design of structures is relevant in aviation and aerospace engineering [1; 2]. Review, classification and design analysis of solar panels for spacecraft are considered in [3], where solar batteries made of rigid panels, flexible substrate solar panels, inflatable solar arrays, self-expanding solar panels and other structures.

The development of modern flexible solar cell carrier is given in [4]. In [5], the dynamic aspects of weight-free systems for large-sized transformable elements of spacecraft during deployment are considered. Numerous copyright certificates are known for the development of zero-gravity simulation stands, for example [6]. Schemes for cable weight-lifting of structures are common, affecting three-dimensional, two-dimensional and one-dimensional objects, for example, for antenna reflectors, each single-link spoke is considered as a beam, hinged at one end on a fixed base and pulled up by a cable [7].

In [8], a method for calculating the weightlessness of large-sized transformable elements of spacecraft during ground tests is considered using the example of a beam rigidly clamped at the end and suspended on cables. Deformation is not taken into account.

Thus, the weightless elements, in most cases, are considered to be infinitely rigid in bending and are pulled up by cables at the center of gravity so that secondary reactions do not occur at the articulation points.

For purposes of [3–5], we can conclude that it is necessary to master the issues of modeling the weightlessness of a structure, taking into account its deformation. This raises the problem of regulating stresses, deformations and deflections by additional tension (pre-tension) of certain parts of structures, in particular, tension by cables. The regulators will be the tension forces of the cables; the values of which should be determined from the condition that the sum of the squared deflections of the elastic line of the beam (beams) was minimal. As a result, the problem is reduced to a nonlinear programming problem.

The weightlessness factor, as an imitation of weightlessness, is considered as zeroing out any force parameter (reaction or torque). For example, the beam is rigidly clamped and the values of the forces in the cables are found. It is necessary to deweight so that there is no reaction in the support. To do this, we consider the calculation of a beam that has the ability to move in the direction of reaction with a zero angle of rotation. The calculated forces in the cables will ensure the weightlessness of the reaction. For example, we need to eliminate the moment, then we solve the problem with a hinged-fixed support. Comparing the angle of rotation found here with the bending moment under rigid clamping will allow us to select the spring stiffness to ensure equivalence.

The work considers linearly deformable systems. The cables have infinitely high tensile rigidity. The elastic line of the beam contains an infinite number of points, so a discrete problem with a finite number of points on the elastic line is analyzed. A nonlinear programming problem may have local minima. A test is performed to determine whether there is a single optimal plan depending on the choice of variables entered into the basis.

 

1. Formulation of the problem of minimizing deflections

Let us consider a multi-span beam, consisting of three beams of total length , as a geometrically variable system (Fig. 1, a). To turn this system of beams into a statically determinate one, it is necessary to impose three additional constraints, since this system has three degrees of freedom. Let the beam (Fig. 1, b) be acted upon by active concentrated forces $P_1,{Р}_2\mathrm{,...,}{P}_n$, distributed loads $q_1,q_2\mathrm{,...,}{q}_k.$, and moment ${М}_{А}$. To balance these loads, we will apply de-weighting forces $N_1,N_2,\mathrm{...,}{N}_S$ to ensure the equilibrium of system. If the number of secondary forces $N_1,N_2,\mathrm{...,}{N}_S$ is less than the number of degrees of freedom of the beam, then the beam remains a geometrically variable system. If the number of secondary forces $N_1,N_2,\mathrm{...,}{N}_S$ is equal to the number of degrees of freedom of the beam, then the forces are calculated from the equilibrium equations (we assume that the forces are distributed correctly). In this case, there is nothing to optimize; the equilibrium of the beam and the equilibrium of its parts are satisfied. If the number of secondary forces $N_1,N_2,\mathrm{...,}{N}_S$ is greater than the number of degrees of freedom of the beam, then it is possible to vary the forces, achieving the required weight-loss parameters, for example, eliminating reactions in joints, reducing deflections at given points, regulating internal forces and stresses. The beam, which initially does not have a sufficient number of support connections, must be in equilibrium under the influence of secondary forces (naturally, there should be no signs of instantaneous variability).

The problem of modeling the deformation of beams suspended on cables with the condition of minimizing the sum of squares of its deflections by varying the tension forces leads to a nonlinear programming problem [9; 10]. Let us express the objective function F in terms of deflections at n points:

 

$F(R_A,{\theta }_A,M_A,N_1,N_2,\mathrm{...,}{N}_S)=\sum _{k=1}^n{\left[\left.v\left(z_k\right)\right]\right.}^2\to \min$, $z_k=k(L/n)$, $k=\mathrm{1,}\mathrm{2,}\mathrm{3,...,}n$.             (1)

Where

$v\left(z_k\right)=f(R_A,{\theta }_A,M_A,N_1,N_2,\mathrm{...,}{N}_S)$,                                              (2)

 

the deflection function at a point with a coordinate $z_k$ with unknown parameters $R_A,{\theta }_A,M_A,N_1,N_2,\mathrm{...,}{N}_S$;  $k$– point number (the distance between points is the same);  ${\theta }_A$– angle of rotation; $R_A$  reaction at point A of the beam;  $M_A$– bending moment at point A;   $N_1,N_2,\mathrm{...,}{N}_S$– the desired forces.

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Рис. 1. Модель (расчетная схема) балки с тремя шарнирами: а – балка как геометрически изменяемая система; б – балка с действующими на нее нагрузками

Fig. 1. Model (design scheme) of a beam with three hinges: a – a beam as a geometrically variable system; b – a beam with loads acting on it

 

As restrictions, equilibrium equations are added to (1): $\sum у=0$– the sum of the projections of forces on the y-axis;  $\sum m_i=0$– the sum of the moments of all forces relative to the point i. The system of restrictions has the form of equalities. From the compiled system of restrictions, we select basic variables, for example, $R_A,N_1,N_2$, which we substitute into (1). Now the objective function (1) will contain only free variables, i.e. parameters :${\theta }_A,M_A,\mathrm{...,}{N}_S$

$F({\theta }_A,M_A,\mathrm{...,}{N}_S)=\sum _{k=1}^n{\left[\left.v\left(z_k\right)\right]\right.}^2$,                                                 (3)

 

which allows solving the optimization problem without restrictions [11]. Partial derivatives (3) with respect to free parameters (the parameters are often called coordinates):

 

$\frac{\partial F}{\partial {\theta }_A}=0$, $\frac{\partial F}{\partial M_A}=0$, … $\frac{\partial F}{\partial N_S}=0$,                                               (4)

 

give a system of linear algebraic equations for the required parameters that determine the optimal plan for calculating deflections (2) at discrete points.

Derivatives of the deflection function (2)

 

$\theta \left(z\right)=dv\left(z\right)/dz$; $M\left(z\right)=-EJ{d}^{2}v\left(z\right)/d{z}^2$; $Q\left(z\right)=dM\left(z\right)/dz$; $q\left(z\right)=dQ\left(z\right)/dz$            (5)

 

give functions of rotation angle, bending moment, shear force and load $q=q\left(z\right)$.

The basic solution of nonlinear optimization problems may have local minima, so the work will show convergence to the optimal plan for various combinations of basis and free variables. To calculate deflections (2), the method of initial parameters was used as a direct method of integrating a fourth order differential equation with discontinuous functions [12]. Derivatives of functions are calculated numerically [13].

Beams can be not only geometrically variable, as shown in Fig. 1, but also without supports, i.e. movable, which contradicts the concepts of kinematic analysis of structural mechanics [14]. However, the balance of the beams is achieved by tensioning the cables (Fig. 2). In Fig. 2, a we will show the design diagram of a beam without support connections with an active moment M. Here, equilibrium is ensured by a pair of equal forces. Let us present a design diagram of a composite beam consisting of four beams (Fig. 2, b). Having imagined the floor diagram of beams, we can see that the main beam is statically indeterminate; the secondary beam, statically determinate, rests on it. Both right beams, the third and fourth one, are supported by forces $N_4-N_7$.

                       

 

Рис. 2. Расчетные схемы балок: а – опорных связей нет – равновесие поддерживается силами ; б – составная балка – равновесие поддерживается реакциями опор и силами   

Fig. 2. Design schemes of beams: a – there are no support links – the equilibrium is maintained by forces; b – composite beam – the balance is maintained by the reactions of the supports and forces  

                                   

It should be noted that to calculate deflections (2) you can use the variational-difference method in the form of the finite difference method and FEM. However, the first method requires introducing additional contour nodes at all hinge joints, which complicates programming, and the second method requires defining deflections and rotation angles at all discrete points (we shall call them nodes), which increases the dimension of the problem. The method of initial parameters taking into account intermediate hinges requires calculating only deflections at discrete points, and at the joint nodes of beams the deflection and increment of the rotation angle are calculated.

Let us consider examples of beam calculations that show the unity of the optimal solution for a non-linear programming problem depending on the choice of variables introduced into the basis, and the convergence of solutions from the assigned number of points at which the deflection is calculated.

2. Optimization of deflections of the beam hinged-fixed and suspended on two cables

We consider an I-beam hinged at one end and supported by two cables (Fig. 3). The cables have infinitely high tensile rigidity. In Fig. 2 their action on the beam is shown by the forces and N₁ и N₂.

Initially, the beam is geometrically variable. Let us add balancing forces and create an objective function

 

         $F(R_A,{\theta }_A,N_1,N_2,q)=\sum _{k=1}^n{\left[\left.v\left(z_k\right)\right]\right.}^2\to \min$, $n=24$ .                                    (6)

 

Let us write the deflection function based on the initial parameters method

 

$\frac{R_A{z}^3}{3!}+\frac{N_1{(z-L/3)}^3}{3!}H(z-L/3)+\frac{N_2{(z-2L/3)}^3}{3!}H(z-2L/3)-\frac{q{z}^4}{4!}$.     (7)

 

Where  $v\left(z\right)$– deflection at the point $z$; ${\theta }_{А}$– angle of rotation at the beginning of the beam (initial parameter at point A); $R_A$– reaction at point A; $N_1$– force in the first cable;  $N_2$– force in the second cable (required parameters); q– uniformly distributed linear load; E– Young’s modulus; J– axial moment of inertia of the beam cross-section; H()– Heaviside function.        

Beam length L = 6 m; rolled I-beam profile, hinged-fixed fastening at a point A; Young's modulus of the material E = 2 · 10¹¹ Pa; axial moment of inertia in the bending plane J = 200 · 10^–8 m⁴; linear weight of the beam q = 100 n/m. 

 

2.1. Checking solutions by changing basic variables

Let us consider solutions related to the peculiarities of beam bending as a discrete problem with a finite number of points at which deflections are calculated. We will search for the global minimum of the objective function.

We check that for different basic variables introduced into the objective function, there must be a single optimal solution.

 

Рис. 3. Балка, прикрепленная шарниром и поддерживаемая двумя тросами

Fig. 3. A beam attached by a hinge and supported by two cables

 

As restrictions, we add equilibrium equations to the objective function (6)

 

$N_1(L/3)+N_2(2L/3)-q{L}^2/3=0$;                                                 (8)

 

                                                         $R_A+N_1+N_2-qL=0.$            (9)

Let us consider three options for applying basic variables.

 

Variant 1. We introduce variables into the basis N₁ and N₂. To do this, we obtain the forces from (8) and (9)

 

$N_1=qL/2-2R_A$,                                                            (10)

 

$N_2=qL/2+R_A$,                                                            (11)

 

which, substituted into (6), give the following nonlinear programming problem:

 

                                                   $F=F(R_A,{\theta }_A,q)\to \min$  (12)

without a system of restrictions.

Derivatives with respect to free variables

 

$\partial F/\partial R_A=0$, $\partial F/\partial {\theta }_A=0$  

 

allow you to calculate the angle of rotation ${\theta }_A$ and reaction $R_A$. All parameters for calculating function (7) are defined. 

 

Variant 2. Let us take N₁ and R_A as basic variables. Thereafter from (8) and (9) we derive the following

 

$N_1=3qL/2-2N_2$,                                                            (13)

 

$R_A=N_2-qL/2$,                                                             (14)

 

which, when substituted into (6), give the search problem

 

$F=F(N_2,{\theta }_A,q)\to \min$.                                                    (15)

 

Derivatives with respect to free variables

 

$\partial F/\partial N_2=0$, $\partial F/\partial {\theta }_A=0$     

 

make it possible to calculate the angle of rotation ${\theta }_A$ and the tension force $N_2$. We calculate function (7).

 

Variant 3. As basic variables we take $N_2$ and $R_A$. Thereafter from (8) and (9) we obtain

 

$N_2=3qL/4-N_1/2$,                                                          (16)

 

$R_A=qL/4-N_1/2$,                                                           (17)

 

which, substituted into (6), give the objective function

$F=F(N_1,{\theta }_A,q)\to \min$.                                                      (18)

 

Derivatives with respect to variables $N_1$ and ${\theta }_A$

 

$\partial F/\partial N_1=0$, $\partial F/\partial {\theta }_A=0$     

 

allow you to calculate the angle of rotation ${\theta }_A$ and  tension force $N_1$. Function (7) is defined

In all three variants of the basic variables, we obtained exactly the same required parameters $R_A,{\theta }_A,N_1,N_2$. The found parameters give solutions (diagrams), which we present in Fig. 4: diagram of deflections (Fig. 4, a); diagram of rotation angles (Fig. 4, b); diagram of bending moments (Fig. 4, c); diagram of transverse forces (Fig. 4, d). Fourth derivative of a function deflection $v\left(z\right)$ along the  coordinate gives a diagram $q\left(z\right)=-100\text{ кн/м}=\text{const}$, which is a check (Fig. 4, e).

 

Рис. 4. Функции деформационных и внутренних силовых факторов: а – прогиб; б – эпюра углов поворота; в – эпюра изгибающего момента; г – эпюра поперечных сил; д – эпюра нагрузки q = –100 н/м

Fig. 4. Functions of deformation and internal force factors: a – deflection; b – plot of rotation angles; с – the bending moment plot; d – the transverse forces plot; e – the load plot q = –100 N/м

 

Thus, changing the basis did not affect the solution of the problem. In all three variants, calculating the programming parameters came to the single optimal solution.

 

2.2. Study of the convergence of solutions depending on the choice of the number of points in which deflections are calculated

A discrete problem with a finite number of points on the elastic line of a beam is considered. Initial data as in paragraph 2.1. The objective function is calculated using formula (6), and deflections are calculated using formula (7).

We check if the objective function is quadratic, the number of nodes on the elastic line of the beam is finite, and the method for calculating deflections is accurate, then convergence of deflections, rotation angles, and bending moments is expected from an increase in the number of discretization nodes.

We perform a numerical experiment for n = 6, n = 12, n = 24. The calculation results (diagrams) are shown in Fig. 5 and in Table 1.

In Fig. 5, a the deflection diagrams are shown. We find the following convergence of deflections. For example, at the end of the beam ($z=L$), the deflection increased from 0.2025 mm on a grid n = 6 to 0.2597 mm on a grid n = 12, which is 28%. Further, on the n = 24 grid, the deflection changed to a value of 0.2935 mm. Compared to a grid of n = 12, this is 13%. The next refinement of the grid should be a difference from the previous one of about 6%. Extrapolation gives the expected deflection of the beam end equal to 0.311 mm.

 

Рис. 5. Оптимальные параметры балки для сеток n=6, n=12 и n=24: а – функции прогибов; б – функции углов поворота; в – функции изгибающих моментов; г – функции поперечных сил; д – заданная распределенная нагрузка (вычислена, как четвертая производная функции прогиба)

Fig. 5. Optimal beam parameters for grids ,  and : a – deflection functions; b – rotation angle functions; c – bending moment functions; d – transverse force functions; e – a given distributed load (calculated as the fourth derivative of the deflection function)

 

We calculate the average values of the sum of squared deflections [15] using the formula

 

${\delta }_n=\sqrt{\sum _{k=1}^n{\left[\left.v\left(z_k\right)\right]\right.}^2/n}$, n = 6, n = 12, n = 24.                                         (19)

 

The values are: ${\delta }_6=\mathrm{0,171}\text{ мм}$, ${\delta }_{12}=\mathrm{0,159}\text{ мм}$, ${\delta }_{24}=\mathrm{0,152}\text{ мм}$. The difference is 7,4 % and 4,3 % respectively.

Fig. 5, b shows diagrams of the rotation angles of the beam sections, which coincided essentially.

The second derivatives of the deflection functions, that are bending moments (Fig. 5, c) calculated on different grids, are practically indistinguishable.

Diagrams of transverse forces (Fig. 5, d) coincide completely and they are represented by one broken line having breaks at the points of application of forces

Fig. 5, d shows the distributed load, calculated as the fourth derivative of the deflection function. This is a solution check. On all three diagrams we got the specified value q = –100 n/m.

To conclude this section, one could say that, based on the change in deflections depending on the grid refinement, grids from n = 12 to n = 24 are sufficient. If the prerogative is the need to calculate stresses, it is enough to assign the grid n = 6

We present the calculated variables in Table 1. Rows 1–3 of the table show the parameters for the grids n=6, n=12, n=24. Lines 4–5 show the relative difference of the desired parameters. The following are written sequentially in the columns: v(n) – deflections of the beam console at the point $z=L$; ${\theta }_{А}\left(n\right)$– initial parameter, angle of rotation of the section; $R_{А}\left(n\right)$– initial parameter, reaction at the beginning of the beam; – calculated parameter, the first force; N₂(n) – calculated parameter, the second force; $\delta \left(n\right)$– root-mean-square deflection of the beam end.     

 

Table 1. Optimal design parameters

Row number	Number of points	v(n)	${\theta }_{А}\left(n\right)$	$R_{А}\left(n\right)$	$N_1\left(n\right)$	$N_2\left(n\right)$	$\delta \left(n\right)$
		m	rad	n	n	n	m
1	n = 6	–0.0002025	–0.00023013	141.956	16.087	441.956	0.00017097
2	n = 12	–0.0002597	–0.00021639	138.464	23.071	438.464	0.00015915
3	n = 24	–0.0002935	–0.00020861	136.454	27.091	436.454	0.00015247
	Percentage difference:
4	n = 12/ n = 6	28 %	6.3 %	2.5 %	43 %	0.8 %	7.4 %
5	n = 24 / n =12	13 %	3.7 %	1.5 %	17.4 %	0.5 %	4.3 %

 

Note that the data in the rows 2 and 3 differ little compared to the parameters in the rows 1 and 2. Rows 4 and 5 can be understood as the rate of convergence of the desired parameters.

 

3. Suspension by cables of the system of three beams connected by hinges

Let us consider a system of I-beams hinged to each other (Fig. 6). Under gravity conditions, the system has its own linear weight (Fig. 6, a). To simulate weightlessness, we will consider four cases of fixing the system at point A and supporting it with six cables. The system is rigidly pinched at point A (Fig. 6, b); hinged-fixed support (Fig. 6, c), sliding embedding (Fig. 6, d); free edge (Fig. 6, d).

It is necessary to determine the tension forces of the cables so that the sum of the squares of the deflections at given points is minimal, i.e.

$F=\sum _{k=1}^n{\left[\left.v\left(z_k\right)\right]\right.}^2\to \min$, $z_k=k(L/6)$,  $k=\mathrm{1,}\mathrm{2,}\mathrm{3,...,}18$ .                            (20)

In (20) F – the objective function; $v\left(z_k\right)$– desired deflection; $z_k$– point coordinate; $k$– point number; $n=18$– amount of points; the lengths of all beams are the same and equal to $L$.

3.1. Rigid

To calculate deflections restraint using the initial parameters method with rigid clamping of the beam at point A (Fig. 6, b), the deflection function is written as:

$v\left(z\right)={\theta }_1(z-L)H(z-L)+{\theta }_2(z-2L)H(z-2L)+$

$+\frac{1}{6EJ}\left[3M_A{z}^2+R_A{z}^3+N_1{\left(z-\frac{L}{3}\right)}^3\right.H\left(z-\frac{L}{3}\right)+N_2{\left(z-\frac{2L}{3}\right)}^{3}H\left(z-\frac{2L}{3}\right)+N_3{\left(z-\frac{4L}{3}\right)}^{3}H\left(z-\frac{4L}{3}\right)+$

$+N_4{\left(z-\frac{5L}{3}\right)}^{3}H\left(z-\frac{5L}{3}\right)+N_5{\left(z-\frac{7L}{3}\right)}^{3}H\left(z-\frac{7L}{3}\right)+$

$+N_6{\left(z-\frac{8L}{3}\right)}^{3}H\left(z-\frac{8L}{3}\right)-\left.\frac{q{z}^4}{4}\right],z\in \left[\mathrm{0,}3L\right].$ (21)

 

 

Where  ${\theta }_1$– the rotation angle (initial parameter at point B); ${\theta }_2$– angle of rotation (initial parameter at point C); ${М}_{А}$– fixing moment (point A); R_A– reaction acting on the beam at point A; $N_i,(i=1-6)$– forces in the cables (required parameters); $q=\text{const}$– uniformly distributed linear load; E– Young’s modulus; J– axial moment of inertia of the beam cross-section; H()– Heaviside function.      

Formula (21) gives the parameters ${\theta }_1$ and ${\theta }_2$, which are the angles of rotation of the beams adjacent to the intermediate hinges [16], i.e. ${\theta }_i^{}={\theta }_i^{\text{справа}}-{\theta }_i^{\text{слева}}$,$i=\mathrm{1,}2$ .  ${\theta }_i^{\text{справа}}$– the angle of rotation at the point  on the right («справа») of the hinge, and  ${\theta }_i^{\text{слева}}$– the angle of rotation at the point  on the left («слева»)  of the hinge. The rotation angles ${\theta }_i^{}$ are additional unknown variables similar to initial parameters $v_{А}$ and ${\theta }_{А}$.  

To equations (21) we add the equilibrium equations

 

$\Sigma m_{С}^{\text{справа}}=0$, $\Sigma m_B^{\text{справа}}=0$, $\Sigma m_A^{\text{справа}}=0$,                                       (22)

 

representing the set of constraints. From (22) we select the basic variables N₄, N₅ ,N₆

 

$\left\{\begin{array}{c}1\\ 0\\ 0\end{array}\right\}{N}_4+\left\{\begin{array}{c}0\\ 1\\ 0\end{array}\right\}{N}_5+\left\{\begin{array}{c}0\\ 0\\ 1\end{array}\right\}{N}_6=\left\{\begin{array}{c}qL/2\\ 2qL-N_3/3\\ 9qL/2-N_1/3-2N_2/3-4N_3/3\end{array}\right\}$,                         (23)

 

 

which we add to the objective function (20). Next observe that $F=F({М}_{А},{\theta }_1,{\theta }_2,N_1,N_2,N_3)$. The equilibrium equation  $\Sigma у=0$– redundant and serves to check solutions. Partial derivatives F with respect to coordinates ${М}_{А},{\theta }_1,{\theta }_2,N_1,N_2,N_3$ give a system of six equations regarding the moment, two angles of rotation and three forces in the cables (note the symmetry of the matrix of the system of equations). The problem was reduced to a quadratic programming problem without restrictions. The minimum objective function (20) determined the optimal solution.

The found parameters ${М}_{А},{\theta }_1,{\theta }_2,N_1,N_2,N_3$, substituted into equations (21), give according to formulas (5) the functions of deflections, rotation angles, bending moment and shear force, respectively. Fig. 7–10 show the functions for rigid embedding; solutions for a hinged-fixed edge are considered; diagrams for sliding embedding are shown; functions for the free edge at point A of the beam system are considered.

 

3.2. Hinged-fixed support

For the system of beams with hinged support at point A (Fig. 6, c), the formula for the initial parameters method is as follows:

 $v\left(z\right)={\theta }_{А}z+{\theta }_1(z-L)H(z-L)+{\theta }_2(z-2L)H(z-2L)+$

$+\frac{1}{6EJ}\left[R_A{z}^3+N_1{\left(z-\frac{L}{3}\right)}^3\right.H\left(z-\frac{L}{3}\right)+N_2{\left(z-\frac{2L}{3}\right)}^{3}H\left(z-\frac{2L}{3}\right)+N_3{\left(z-\frac{4L}{3}\right)}^{3}H\left(z-\frac{4L}{3}\right)+$

$+N_4{\left(z-\frac{5L}{3}\right)}^{3}H\left(z-\frac{5L}{3}\right)+N_5{\left(z-\frac{7L}{3}\right)}^{3}H\left(z-\frac{7L}{3}\right)+N_6{\left(z-\frac{8L}{3}\right)}^{3}H\left(z-\frac{8L}{3}\right)-\left.\frac{q{z}^4}{4}\right],z\in \left[\mathrm{0,}3L\right].$ (24)

Where  ${\theta }_{А}$– angle of rotation at point A.

Additional equations (22) allow you to introduce variables N₄, ,N₅ ,N₆ into the basis, as in (23). The reaction for formula (24) is determined from the equilibrium equation $\Sigma у=0$. The functions of deflections, rotation angles, bending moment, and shear force are shown in Fig. 7–10.   

 

3.3. Sliding sealing

We consider the sliding sealing at point A, a fragment of which is shown in Fig. 6, d. The following equilibrium equations are compiled:

 

$\Sigma m_{С}^{\text{справа}}=0$, $\Sigma m_B^{\text{справа}}=0$, $\Sigma у=0$ .                                               (25)

 

We introduce vectors for the variables ,N₄ , N₅, N₆ into the basis, and after calculating the derivatives with respect to free variables from the equilibrium equation $\Sigma m_A^{}=0$, we express the deflection functions M_A.

Now, the deflection function is the following:

$v\left(z\right)=v_A+{\theta }_1(z-L)H(z-L)+{\theta }_2(z-2L)H(z-2L)+$

$+\frac{1}{6EJ}\left[3M_A{z}^2+N_1{\left(z-\frac{L}{3}\right)}^3\right.H\left(z-\frac{L}{3}\right)+N_2{\left(z-\frac{2L}{3}\right)}^{3}H\left(z-\frac{2L}{3}\right)+N_3{\left(z-\frac{4L}{3}\right)}^{3}H\left(z-\frac{4L}{3}\right)+$

$+N_4{\left(z-\frac{5L}{3}\right)}^{3}H\left(z-\frac{5L}{3}\right)+N_5{\left(z-\frac{7L}{3}\right)}^{3}H\left(z-\frac{7L}{3}\right)+N_6{\left(z-\frac{8L}{3}\right)}^{3}H\left(z-\frac{8L}{3}\right)-\left.\frac{q{z}^4}{4}\right],z\in \left[\mathrm{0,}3L\right].$       (26)

 

All solutions are presented in Fig. 7–10.

 

 

3.4. Free edge

In the case of a free edge, there are no connections at point A of the beam (Fig. 6, d), there are 10 unknown parameters. Then observe that $v\left(z\right)=f(v_A,{\theta }_A,{\theta }_1,{\theta }_2,N_1,N_2\mathrm{,...,}{N}_6)$. Each beam disk in a plane has three degrees of freedom. There are 9 of them in total. Each simple hinge takes away two degrees of freedom; 4 in total are eliminated. For a system of three beams, 5 degrees of freedom remain. To turn the beam system under consideration into a geometrically unchangeable one, 5 connections should be added. One degree of freedom is subtracted, since there is no displacement in the direction of the longitudinal force. You can write 4 equilibrium equations, but they are linearly dependent.

We compose three equilibrium equations. The deflection function will look like this:

$v\left(z\right)=v_A+{\theta }_{А}z+{\theta }_1(z-L)H(z-L)+{\theta }_2(z-2L)H(z-2L)+$

$+\frac{1}{6EJ}\left[N_1{\left(z-\frac{L}{3}\right)}^3\right.H\left(z-\frac{L}{3}\right)+N_2{\left(z-\frac{2L}{3}\right)}^{3}H\left(z-\frac{2L}{3}\right)+N_3{\left(z-\frac{4L}{3}\right)}^{3}H\left(z-\frac{4L}{3}\right)+$

$+N_4{\left(z-\frac{5L}{3}\right)}^{3}H\left(z-\frac{5L}{3}\right)+N_5{\left(z-\frac{7L}{3}\right)}^{3}H\left(z-\frac{7L}{3}\right)+N_6{\left(z-\frac{8L}{3}\right)}^{3}H\left(z-\frac{8L}{3}\right)-\left.\frac{q{z}^4}{4}\right],z\in \left[\mathrm{0,}3L\right].$     (27)

We display the solutions in Fig. 7–10.

 

Рис. 6. Система трех балок, соединенных шарнирами: а – система в условиях гравитации; б – моделирование невесомости системы жестким защемление в точке А; в – фрагмент шарнирно-неподвижного опирания системы балок в точке А; г – фрагмент скользящей заделки системы балок в точке А; д – фрагмент свободного края системы балок в точке А

Fig. 6. A system of three beams connected by hinges: a – the system under gravity conditions; b – modeling of the weightlessness of the system by rigid pinching at point A; c – a fragment of the hinge-fixed support of the beam system at point A; d – a fragment of the sliding sealing of the beam system at point A; e – a fragment of the free edge of the beam system at point A

 

In table 2 we present the tension forces of the cables for a system of beams with four types of boundary conditions. We compare the tension forces for a rigidly clamped and hingedly supported beam. On the first beam, the tension forces of the cables are equal: $N_1^{\text{заделка}}=-\mathrm{278,4}\text{ н}$ («заделка» –sealing),$N_1^{\text{шарнир}}=-\mathrm{124,02}\text{ н}$  («шарнир» – hinge) – the difference is 220%;  $N_2^{\text{заделка}}=\mathrm{407,5}\text{ н}$ $, N_2^{\text{шарнир}}=\mathrm{336,07}\text{ н}$– the difference is 21%. On the second beam, the greatest tensile forces are: $N_4^{\text{заделка}}=\mathrm{262,6}\text{ н}$,  $N_4^{\text{шарнир}}=\mathrm{251,9}\text{ н}$– the difference is 4.2%. On the third beam, the tension forces of the cables are almost equal: $N_5^{\text{заделка}}=\mathrm{96,3}\text{ н}$,  $N_5^{\text{шарнир}}=\mathrm{100,7}\text{ н}$– the difference is 4.5 %. In the extreme cable:$N_6^{\text{шарнир}}=\mathrm{174,6}\text{ Н}$ and  $N_6^{\text{заделка}}=\mathrm{176,8}\text{ Н}$– the difference is 0.13 %.  

The type of fastening (hinged or sealing) affects the first beam from fastening. The tension forces of the cables, to a certain extent, level out the deformed and a tense state. In subsequent beams, the tension forces of the cables are practically the same, i.e., they do not depend on the type of fastening of the first beam. The same conclusion can be drawn for beams with sliding sealing.

Models of three beam systems when simulating weightlessness are equivalent to a certain extent. So, for example, if a model of a beam system with hinged support at point A is given a moment (for example, by a mover) equal to  or a spring is set to stiffness $G_{А}=M_A/{\theta }_A=\mathrm{89,9}\text{ нм}/\mathrm{5,78}\cdot {10}^{-5}\text{рад}=\mathrm{1,5}\text{ кн/рад}$, we obtain a model of deformation with rigid pinching. Accordingly, it is necessary to change the tension forces of the cables. 

In table 3, to the cable tension forces from table 2, rotation angles were added at the beginning of the beam system (point A) and additional rotation angles at the hinges, root-mean-square deflections and maximum deflections. When moving from rigid fastening to hinged fastening, then from sliding fastening to the free edge, the rigidity of the system decreases. The root mean square deflections increase:${\delta }_{\text{заделка}}=\mathrm{2,301}\cdot {10}^{-5}\text{ м}$  («заделка» – sealing);${\delta }_{\text{шарнир}}=\mathrm{2,396}\cdot {10}^{-5}\text{ м}$  («шарнир» – hinge);${\delta }_{\text{скользящая}}=\mathrm{2,741}\cdot {10}^{-5}\text{ м}$ («скользящая» – sliding);${\delta }_{\text{свободный\hspace{0.17em}край}}=\mathrm{3,282}\cdot {10}^{-5}\text{\hspace{0.17em}м}$ («свободный край» – free edge), i.e. by 4.1, 14.4 and 19.7%. Maximum deflections in the first three cases occur at the point where the hinge connecting the second and third beams, respectively, is installed, $(\mathrm{4,31}÷\mathrm{4,5}÷\mathrm{5,14})\cdot {10}^{-5}\text{ м}$ and in the middle of the system of beams with a free edge – $\mathrm{6,07}\cdot {10}^{-5}\text{ м}$.

 

Рис. 7. Функции прогибов и силы натяжения тросов для граничных условий слева: а – жесткой заделки; б – шарнирного опирания; в – скользящей заделки; г – свободного края

 Fig. 7. Functions of deflections and tension forces of cables for boundary conditions on the left: a – rigid sealing; b – hinged support; c – sliding sealing; d – free edge

 

Рис. 8. Функции углов поворота и силы натяжения тросов для граничных условий слева: а – жесткой заделки; б – шарнирного опирания; в – скользящей заделки; г – свободного края

Fig. 8. Functions of rotation angles and cable tension force for boundary conditions on the left: a – rigid sealing; b – hinged support; c – sliding sealing; d – free edge

 

Рис. 9. Функции изгибающих моментов и силы натяжения тросов для граничных условий слева: а – жесткой заделки; б – шарнирного опирания; в – скользящей заделки; г – свободного края (Начало)

Fig. 9. Functions of bending moments and tension forces of cables for boundary conditions on the left: a – rigid sealing; b – hinged support; c – sliding sealing; d – free edge (The beginning)

 

Рис. 9. Окончание

Fig. 9. The ending

 

Рис. 10. Функции перерезывающих сил и силы натяжения тросов для граничных условий слева: а – жесткой заделки; б – шарнирного опирания; в – скользящей заделки; г – свободного края (Начало)

Fig. 10. Functions of shearing forces and cable tension forces for boundary conditions on the left: a – rigid sealing; b – hinged support; c – sliding sealing; d – free edge (The beginning)

 

Рис. 10. Окончание

Fig. 10. The ending

 

Table 2. Cable tension forces

Row numbers	Type of fastening at point A	N1	N2	N3	N4	N5	N6
		n	n	n	n	n	n
1	sealing	–278.4	407.5	5.39	262.6	96.3	176.8
2	hinged support	–124.02	336.07	19.9	251.9	100.7	174.6
3	sliding sealing	73.59	259.04	67.58	216.97	115.67	167.17
4	free edge	158.19	133.63	158.19	158.19	133.63	158.19

 

Table 3. Reaction forces and geometric characteristics of the system

Row numbers	Type of fastening at point A	$M_A$	$R_A$	${\theta }_A$	${\theta }_1$	${\theta }_2$	$\delta$	$v\left(z\right)$
		n×m	n	10-5 rad	10-5rad	10-5 rad	10-5 m	10-5 m
1	sealing	–89.9	229.7	0	11,1	14.7	2.301	$v\left(6\text{\hspace{0.17em} м}\right)=\mathrm{4,31}$
2	hinged support	0	110.65	5.78	12.3	15.2	2.396	$v\left(6\text{\hspace{0.17em} м}\right)=\mathrm{4,50}$
3	sliding sealing	–43.79	0	0	15.6	16.7	2.741	$v\left(6\text{\hspace{0.17em} м}\right)=\mathrm{5,14}$
4	free edge	0	0	8.01	–20.1	20.1	3.282	$v\left(\mathrm{4,5}\text{\hspace{0.17em} м}\right)=\mathrm{6,07}$

 

The dimensions of physical quantities are described according to [17].

 

Conclision

In a quadratic programming problem, given different basis variables introduced into the objective function, there is a single optimal solution. Changing the basis does not affect the solution of the problem.

Analysis of the solution to discrete problems (with a finite number of points on the elastic line of beams) showed that grids from n=12 to n=24 points are sufficient. The difference in weighted average deflections is 4.3%.  

The models of three beam systems are equivalent to a certain extent when simulating weightlessness. Any of the considered systems with the presented boundary conditions can be converted into an equivalent one by changing the boundary force factors. For example, if in a model with hinged support we specify a moment or install a spring with a given stiffness, we obtain a deformation model with rigid pinching. Accordingly, it is necessary to adjust the tension forces of the cables

The type of boundary condition has a greater effect on the first beam; the tension forces of the cables level out the deformed and stressed state. In subsequent beams, the tension forces of the cables are practically the same. 

When moving from rigid fastening to hinged fastening, then from sliding fastening to the free edge, the rigidity of the systems decreases. The root-mean-square deflections increase.

Simulating the weightlessness of a system with the condition of minimizing the sum of squared deflections can be useful in preparing physical experiments.

It is possible to generalize the formulation of the problem of regulating the stress-strain state for systems of suspended plates and panels; it is possible to install springs in hinged joints; the forces in the cables can be additionally distributed with weight factors.

Авторлар туралы

Rashid Sabirov

Reshetnev Siberian State University of Science and Technology

Хат алмасуға жауапты Автор.
Email: rashidsab@mail.ru

Ph. D., associate Professor, associate Professor of the Department of technical mechanics

Ресей, Krasnoyarsk

Elena Fisenko

Reshetnev Siberian State University of Science and Technology

Email: rashidsab@mail.ru

senior lecturer

Ресей, Krasnoyarsk

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