Vibration Properties of a Concrete Structure with Symmetries Used in Civil Engineering

: The paper aims to study a concrete structure, currently used in civil engineering, which has certain symmetries. This type of problem is common in engineering practice, especially in civil engineering. There are many reasons why structures with identical elements or certain symmetries are used in industry, related to economic considerations, shortening the design time, for constructive, simplicity, cost or logistical reasons. There are many reasons why the presence of symmetries has beneﬁts for designers, builders, and beneﬁciaries. In the end, the result of these beneﬁts materializes through short execution times and reduced costs. The paper studies the eigenvalue and eigenmode properties of vibration for components of the constructions’ structure, often encountered in current practice. The identiﬁcation of such properties allows the simpliﬁcation and easing of the effort necessary for the dynamic analysis of such a structure.


Introduction
Frequently encountered in the design and construction of structures used in civil engineering, symmetries allow, in many cases, the simplification of calculations and dynamic analysis of such a structure. The direct consequence would be the shortening of the design and execution time and, of course, the decrease of the costs generated by these stages. The smaller information provided by a repetitive or symmetrical structure can help ease the computational effort. In the case of a static calculation, methods of approaching this problem are presented in the Strength of Materials courses. In the dynamic case, considering the elastic elements and studying the vibrations, although certain properties have long been observed [1], a systematic study of the problem has not yet been done. A case in which the symmetries introduced by two identical motors and their effect on vibrations were considered was studied [2]. The use of identical systems was applied to the design of a centrifugal pendulum vibration absorber system [3]. Circular symmetry and its induced properties have been reported previously [4,5]. Other particular cases have been studied [6] using a finite difference method and [7] for continuous systems. In the following, we will study the case of a mechanical system consisting of four trusses, two of which are identical. The transverse and torsional vibrations of such a system that are strongly coupled for the chosen case will be studied.
In the field of engineering, not only in civil engineering, but also in other fields such as the machine or machinery manufacturing industry, the automotive industry, and the aerospace industry, there are products, parts of products, machines, and components that contain identical, repetitive elements, which have, in their composition, parts that show symmetries of different types.

Model and Free Vibration Response
We have 2 coplanar reinforced concrete beams with different properties. AD beam is a main beam, considered double clamped at both ends, with the length L 1 + L 2 . CB beam is a secondary beam, considered simply supported at both ends (nodes C and B), with length L 3 + L 3 . The beams are made monolithically, so at point O of the intersection, a rigid knot is created. The secondary beam is arranged symmetrically to the main beam. The 2 beams have the Young's moduli E 1 and E 2 . The 2 beams have different sectional properties: for beam AD, we have the moment of inertia Iz1 and the area of section A 1 , and for beam Symmetry 2021, 13, 656 3 of 18 CB, we have the moment of inertia I z2 and the area of section A 2 , with the property that I z1 > I z2 and A 1 > A 2 . The whole structure is presented in Figure 1.

Model and Free Vibration Response
We have 2 coplanar reinforced concrete beams with different properties. AD beam is a main beam, considered double clamped at both ends, with the length L1 + L2. CB beam is a secondary beam, considered simply supported at both ends (nodes C and B), with length L3 + L3. The beams are made monolithically, so at point O of the intersection, a rigid knot is created. The secondary beam is arranged symmetrically to the main beam. The 2 beams have the Young's moduli E1 and E2. The 2 beams have different sectional properties: for beam AD, we have the moment of inertia Iz1 and the area of section A1, and for beam CB, we have the moment of inertia Iz2 and the area of section A2, with the property that Iz1 > Iz2 and A1 > A2. The whole structure is presented in Figure 1. Thus, the model of the mechanical system considered ( Figure 2) consists of two identical trusses OB and OC rigidly fixed, perpendicular to a third bar AOD. The trusses can have transverse vibrations in a direction perpendicular to the ABDC plane and torsional vibrations. The trusses are clamped in the points B and C. In A and D, the AOD trusses are clamped, so the displacement, slope, and torsion angle at these points are zero.  We will study a continuous concrete truss, homogeneous, with constant section. If there are no forces distributed or concentrated along the length of the truss, the vibrations of this are described by the well-known Equations [26]: We have 2 coplanar reinforced concrete beams with different properties. AD beam is a main beam, considered double clamped at both ends, with the length L1 + L2. CB beam is a secondary beam, considered simply supported at both ends (nodes C and B), with length L3 + L3. The beams are made monolithically, so at point O of the intersection, a rigid knot is created. The secondary beam is arranged symmetrically to the main beam. The 2 beams have the Young's moduli E1 and E2. The 2 beams have different sectional properties: for beam AD, we have the moment of inertia Iz1 and the area of section A1, and for beam CB, we have the moment of inertia Iz2 and the area of section A2, with the property that Iz1 > Iz2 and A1 > A2. The whole structure is presented in Figure 1.  We will study a continuous concrete truss, homogeneous, with constant section. If there are no forces distributed or concentrated along the length of the truss, the vibrations of this are described by the well-known Equations [26]: We will study a continuous concrete truss, homogeneous, with constant section. If there are no forces distributed or concentrated along the length of the truss, the vibrations of this are described by the well-known Equations [26]: The notations used in Equation (1) are the following: v-is the truss deflection, A-s the cross section, ρ is the mass density, E-Young's modulus, and I z is the second area moment of inertia with respect to the z axis and x is the ordinate of the point having the deflection v. To solve Equation (1), we look for a solution of the form [27][28][29]: v(x, t) = Φ(x) sin(pt + θ). (2) Equation (2) must check Equation (1) at any time and, imposing this condition, we obtain: Denote: Then, Equation (1) becomes: In Equation (5), Φ represents the function, depending on the abscissa x, which gives us the deformed bar (eigenmode) corresponding to its eigenpulsation p. The solution is: The constants C 1 , C 2 , C 3 , C 4 are determined considering the boundary conditions for this problem.
In the following, we will use Equation (5) for the domains defined by the four trusses, obtaining, in this way, four differential equations of the fourth order, corresponding to the frames AO, OD, OB, OC (see Appendix A).
The study of torsional vibrations for a straight bar, unloaded over the length, leads to the second order differential equation: where ϕ is the angle of torsion of the cross section being at the distance x from the end of the truss, J = ρI p is the polar moment of inertia, I p is the polar second moment of the area, and G is shear's modulus. The solution Equation (7) is sought in the form: Equation (8) must verify Equation (7), from which we obtain: where the notation was made: The solution is: Denoted by M b is the bending moment of a bar in section x, T the shear force that appears in the cross section and with M t the torque. For the studied system in the paper, the boundary conditions are: (a) For the AO truss, the end A is clamped, so: v AO (0, t) = 0; v AO (0, t) = 0; ϕ(0, t) = 0; (b) For the OD truss, the end D is supported, so: v OD (L 1 , t) = 0; v OD (L 1 , t) = 0; ϕ(0, t) = 0; (c) For the OB truss, the end B is supported so: Imposing these conditions, for the considered four bars, 12 boundary conditions are obtained. The two moments and the shear force are expressed by the known relationships in the mechanics of the deformable solid [30][31][32][33]: Using the Equation (12), the solutions presented in Appendixes, and the boundary conditions (a), (b), (c), and (d) result in 12 linear equations involving the unknown constants (Appendix A).
The continuity of the elastic system at point O leads to the following conditions: we have once that the displacements in C for all trusses: . These mean nine conditions presented in the Appendix B, from which nine linear equations result.
Three more conditions are needed to obtain the constants in the written differential equations. They are obtained by considering the equilibrium of an infinitesimal element of the mass containing the point O.
The sum of the four shear forces in O must be zero, so: A similar equilibrium relation is obtained for bending and torsional moments: Writing these three conditions results in three linear equations (Appendix C). The unknowns are: (19) or: In such way, a homogeneous linear system with 24 equations with 24 unknowns was obtained. In order to have other solutions besides the trivial solution zero, the determinant of the system must be zero. Putting this condition, the obtained eigenfrequencies of the system can be determined from the obtained equation.
The continuous models used in our studies are excellent for a classical analysis of such systems. The boundary conditions, written for these models, ultimately lead to a linear system of homogeneous equations. For this system to have a solution other than the trivial solution, zero, it is necessary that the system determinant be equal to zero. Imposing this condition leads to writing the characteristic equation that will provide, by solving, its eigenfrequencies of vibration. Once these values are known, then, we can determine your eigenmodes for these frequencies.
The matrix of the system can be written, in a concise form: where the matrices A ij are presented in Appendix C.
The condition: offers the eigenvalues of the system of differential equations.

Properties of the Eigenvalues and Eigenmodes
Let us now consider one of the identical trusses OB and OC. The truss is clamped in O and supported in B (or C). Equations (1) and (7) are also valid for the OB bar (OC), with the boundary conditions: For point O: and for the point B(C): The solution is: for transversal vibrations and: with the imposed boundary conditions for this case: Using conditions expressed by Equations (28)-(31), it is now possible to determine the constants C BO 1 , C BO 2 , C BO 3 , C BO 4 , D BO 1 , D BO 2 from the linear homogenous system: where [A 11 ] is the matrix determined by Equation (A76). If the condition of the existence of non-zero solutions is set: It is now possible to obtain the eigenvalues of the truss OB (or OC).
The following theorems will be proved in the following: The eigenvalues for the OB truss, clamped at one end and supported at the other, are also eigenvalues for the entire mechanical system.

Proof.
We must show that det (A) = 0 implies det (S) = 0. In reference [34], this property is proved in a more general case. It turns out that the property is valid in our case.
It follows that the eigenvalues of a single truss in Appendix D, clamped at one end and supported at the other, are also eigenvalues of the composed system, clamped in A and D and with the ends B and C supported.
Using the matrix done by Equation (A76), and obtaining the eigenvalues for this matrix, the eigenmodes of deformations are obtained using Equation (A85). The following two theorems will be proved: Theorem 2. For eigenvalues that are common to the whole mechanical system ( Figure 3) and to the subsystem in Figure A1 (see Theorem 1), eigenvectors are of the form: (the existence of common eigenvalues is proved by theorem T1).
Symmetry 2021, 13, x FOR PEER REVIEW 7 If the condition of the existence of non-zero solutions is set: It is now possible to obtain the eigenvalues of the truss OB (or OC).
The following theorems will be proved in the following: The eigenvalues for the OB truss, clamped at one end and supported at the othe also eigenvalues for the entire mechanical system.
Proof. We must show that det (A) = 0 implies det (S) = 0. In reference [34], this proper proved in a more general case. It turns out that the property is valid in our case. □ It follows that the eigenvalues of a single truss in Appendix D, clamped at one and supported at the other, are also eigenvalues of the composed system, clamped and D and with the ends B and C supported.
Using the matrix done by Equation (A76), and obtaining the eigenvalues for this trix, the eigenmodes of deformations are obtained using Equation (A85). The follow two theorems will be proved: Theorem 2. For eigenvalues that are common to the whole mechanical system (Figure 3) a the subsystem in Figure A1   Proof. For the eigenvalues obtained from Equation (33), the following system mus solved: Condition (36) implies that a vector can be found, such that: = 0 Suppose that we determined this vector. Equation (35) becomes, after perform Proof. For the eigenvalues obtained from Equation (33), the following system must be solved: Suppose that we determined this vector. Equation (35) becomes, after performing some simple calculations: If we take into account Equation (37), the system of Equations (38)-(40) becomes: From Equation (41), because, in general, detA 12 = 0, it follows immediately: and introducing that in Equation (43), we obtain Φ OB = −Φ OC , a relation which verifies also Equation (42), if we take into account Equation (37). If Φ OB = Φ 1 is denoted, it results in Equation (34).

Theorem 3.
For the other eigenvalues of the system, the eigenvectors are of the form: Proof. For the eigenvalues calculated, the system of Equation (35) must be solved, with detA 11 = 0 or: Subtracting (47) from (46), we get: A block diagram representing the stage of the analysis is presented in Figure 4.  For the eigenfrequencies of the system that are the same with the eigenvalues of a single clamped truss at one end and supported at the other, the eigenmodes are skew For the eigenfrequencies of the system that are the same with the eigenvalues of a single clamped truss at one end and supported at the other, the eigenmodes are skew symmetric, the two identical trusses vibrate in counterphase, and the third truss AOD rests. For the other eigenfrequencies, the identical trusses have identical eigenmodes.

Conclusions
Buildings and constructions, in general, show different forms of symmetry or are made up of repetitive elements. The existence of these symmetries leads to obtaining advantages related to the calculation, design, and manufacture of the structure. One of the advantages is related to the ease of describing the system by systematizing the information used; then, the properties demonstrated in the paper allow to decrease the time required to perform calculations and all this will allow savings and simplifications in the design process. Then, the existence of identical elements or identical parts can simplify the process of making the structure, by simplifying the labor and effort required to manufacture the structure. In conclusions, the design is simpler, and the realization costs are lower. There are also aesthetic reasons that justify the realization of structures with symmetries. From the point of view of calculation and behavior in static and dynamic cases, symmetries can bring significant advantages. In the strength of materials, symmetries are widely used in the static analysis of structures. However, they can be used for dynamic analysis, so that the vibrations of such structures allow simplification of the calculation and time savings in the design process. The paper has presented several vibration properties of a symmetrical structure made of concrete, used in civil engineering. Such structures are frequently encountered in the construction of buildings and in civil engineering and the knowledge of vibration properties can prove to be an advantage that allows to reduce the time and costs related to the design. We mention that symmetries appearing in all aspects of current life and in engineering applications are common. In consequence, the results obtained can be extended to other situations that may be encountered in practice. Future research could reveal other types of symmetries that will allow the systematization of the results and the proposal of a strategy to approach the design and execution of systems with identical parts or symmetries. Acknowledgments: We want to thank the reviewers who have read the manuscript carefully and have proposed pertinent corrections that have led to an improvement in our manuscript.

Conflicts of Interest:
The authors declare no conflict of interest.

Appendix A
For the truss AO: For the truss OD: For the truss OB: For the truss OC: The following notations are made: Using (A5), the four solutions for the four differential equations of order four (A1)-(A4) are: For torsion, the notation was made: Index 1 corresponds to trusses AO and OD and index 2 to trusses OB and OC. Applying Equation (8) for the four trusses studied, leads us to: For torsional vibrations of the bar AO: for torsional vibrations of the bar OD: for torsional vibrations of the bar OB: for torsional vibrations of the bar OC: The solutions of the four equations (A11)-(A14) are: The solutions contain 24 integration constants that will be determined considering the boundary conditions.
The two moments and the shear force are expressed by the known relationships in the mechanics of the deformable solid [26,27]: However: Truss AO: Truss OD: Truss OB: Truss OC: The following relations will be obtained: which represent a system of 12 equations.

Appendix B
These nine conditions lead to nine relationships: or : 1 in which, together, (A35)-(A46) represent a system of 21 equations. From (A47), we can obtain the equations: From (A48), we can obtain the equations: From (A49), we obtain:

Appendix C
Three more conditions are needed to obtain the constants in the written differential equations.
They are obtained by considering the equilibrium of an infinitesimal element of the mass containing the point O.
The sum of the four shear forces in O must be zero, so: Replacing the expressions of the shear force determined for the four bars in O, we obtain: With the notation: we can write: A similar equilibrium relation is obtained for bending and torsional moments.
Taking into account relations (27)-(43), we obtain: Using (11)- (14) and (23)- (26), the results are: with: A homogeneous linear system was obtained. In order to have other solutions besides the trivial solution zero, the determinant of the system must be zero. Putting this condition, the obtained eigenfrequencies of the system can be determined from the obtained equation. Denote: In the following, we denote: where: In this case, the matrix of the system can be written, in a concise form: (A83) and the system becomes: The condition: det(A) = 0 (A85) offers the eigenvalues of the system of differential Equations (6)- (9) and (19)- (22).

Appendix D
Let us now consider one of the identical trusses OB and OC. The truss is clamped in O and supported in B (or C). Equation (1) is also valid for the OB bar (OC), with the boundary conditions: Symmetry 2021, 13, x FOR PEER REVIEW 16 of 18 In this case, the matrix of the system can be written, in a concise form: (A84) and the system becomes: = 0 (A85) The condition: offers the eigenvalues of the system of differential Equations (6)- (9) and (19)- (22).

Appendix D
Let us now consider one of the identical trusses OB and OC. The truss is clamped in O and supported in B (or C). Equation (1) is also valid for the OB bar (OC), with the boundary conditions: If noted: the solution is: Torsional vibrations are described by Equation (15). By choosing ϕ as: ϕ(x, t) = ψ(x) sin(pt + θ) (A92) and introducing in Equation (15), we obtain: The solution of the differential Equation (A93) will be: ψ(x) = D 1 sin(δ 2 px) + D 2 cos(δ 2 px) (A94)