Calculus Through Transfer-Matrix Method of Continuous Circular Plates for Applications to Chemical Reactors

Abstract

This paper presents an original approach through Transfer-Matrix Method applied for the calculus of the continuous circular plate embedded at the exterior circumference, charged with asymmetrical uniform load on the entire upper surface of the plate. Continuous circular plates are elements often found in practice, in the machine building, aeronautics, chemical industries (the bottoms of chemical reactors), and in petrochemical, mechanical, robotic, medical, military, nuclear, and aerospace industries. The calculus of continuous circular plates is a special problem both from the point of view of the theory of elasticity and from the point of view of the mathematical approach. The results obtained with Transfer-Matrix Method were compared and validated with those obtained from classical analytical calculation, the Theory of Elasticity. Transfer-Matrix Method is an elegant method and relatively easy to program. In future research, we want to validate our results with those given by the Finite Elements Method and those measured experimentally.

1. Introduction

Continuous circular plates are elements often found in practice, in the machine building industry—the flat bottoms of various cylindrical tanks or the bottoms of reactors, flat covers for cylinders, flanges, sealing devices, pistons from pumps or from the cylinders of internal combustion engines—in aeronautics, and in the chemical industry—the bottoms of chemical reactors—and in petrochemical, mechanical, robotic, medical, military, nuclear, and aerospace industries.

The calculus of continuous circular plates is a special problem both from the point of view of the theory of elasticity and from the point of view of the mathematical approach. This work is the beginning for broader research regarding circular plates. For the calculation of the bottom of a chemical reactor (but not only this) it is important, both in the design and in the verification of existing ones, to have the possibility to find the optimal shape that satisfies the conditions of material resistance, temperature (most of the time the temperature is very high), resistance to the agents with which it comes into contact (corrosion resistance), etc. TMM for iterative calculations offers the possibility of being programmed and of giving numerical results very quickly. If coupled with a program for optimizing the constructive form, the best variant for the studied part can be reached. This possibility of obtaining a quick solution for the necessary optimal form of the part motivated us to start this study, with this article being a first step. Our main contribution, which is also a novelty, is the approach of this study for the bottoms of reactors, which can be modeled as solid circular plates embedded on the outer contour and loaded with a load uniformly distributed over the entire upper surface of the plate. As with any theoretical approach, there are also gaps. That is why working, simplifying hypotheses have been specified, both from the point of view of elasticity theory and from the point of view of mathematics (differential calculus, integrals, functions, operators, etc.) In the future, as our study advances, we hope to be able to apply this algorithm to real parts and give exact solutions.

This paper presents a calculus of a continuous circular plate embedded on an exterior contour and charged with a uniformly distributed load over all of the upper surface of the plate. The calculus presented is based on a mathematical apparatus provided by the Transfer-Matrix Method (TMM), and the results are compared with those obtained by analytical calculation based on the theory of elasticity. The TMM is based on the theory of Dirac’s and Heaviside’s functions and operators.

For our studies, we define a transfer-matrix, that we should integrate in a general manner for the differential equation, which gives the deformation of some plates loaded with an exterior density q(x,y), written by means of the Dirac’s and Heaviside’s functions and operators. After this, we can express the boundary conditions for the displacement calculus, for efforts calculus or for the stress’s calculus in any plate points.

The basics of TMM calculation are given in [1]. The classic Strength of Materials formula is presented in [2,3]. A review about the past, the present, and the future of multibody system transfer matrix method is shown in [4]. Calculus of long rectangular plates embedded in long borders with uniform vertical load on a line parallel to the long borders with TMM is given in [5], and the TMM for calculus of long cylinder tube with industrial applications is presented in [6]. Theoretical aspects about the time-harmonic analysis of acoustic pulsation in gas pipeline systems using the Finite Element Transfer Matrix Method (FETMM) can be found in [7]. The classical theory of plates is given in [8]. Determination of plane stress–strain states of the plates based on the three-dimensional theory of elasticity is presented in [9]. An analysis of the hypotheses used when constructing the theory of beams and plates is given in [10]. The plates calculus by the series representation of the deflection function is presented in [11]. Static analysis of an orthotropic plate is given in [12]. In [13], an analysis of homogeneous and non-homogeneous plates is shown. Approximate analytical solutions in the analysis of thin elastic plates are given in [14]. In [15], some aspects of implementation of boundary elements method in plate theory are shown. Convergence analysis of finite element approach to classical approach for analysis of plates in bending is presented in [16]. Ref. [17] presents application of numerical methods in solving a phenomenon of the theory of thin plates. Review of a few selected theories of plates in bending is given in [18]. The analytical solutions of the mechanical answer of thin orthotropic plates can be found in [19]. Theories and analyses of beams and axisymmetric circular plates are given in [20]. TMM for multi-body systems dynamics modeling with applications for rescue robots is presented in [21], and with applications for piezoelectric stack actuators in [22]. Dynamics modeling and simulation for controlled flexible manipulator can be shown in [23]. Application of discrete-time TMM in gas-turbine dynamics calculation is given in [24]. Research on TMM of multibody system modeling of high-pressure compressor rotor of gas turbine can be found in [25]. Dynamic property analysis on vibratory shaft of road rollers via TMM is given in [26], and dynamic characteristics analysis of spindle on TMM is given in [27]. TMM for determination of the natural vibration characteristics of elastically coupled launch vehicle boosters is presented in [28]. Extended discrete-time transfer matrix approach to modeling and decentralized control of lattice-based structures can be shown in [29]. Hybrid method of discrete time TMM for multibody system dynamics and FEM is presented in [30]. Dynamics analysis and fuzzy anti-swing control design of overhead crane system based on Riccati discrete time TMM is given in [31]. A new form of TMM for multibody systems is presented in [32]. Theory and applications of TMM for Multibody Systems can be found in [33], and reduced multibody system TMM in [34]. Some improvements to the TMM for simulation of nonlinear multibody systems are presented in [35]. A new version of the Riccati TMM for multibody systems consisting of chain and branch bodies can be found in [36]. Modeling and control of magnetorheological 6-DOF Stewart platform based on multibody systems TMM is presented in [37]. Application of the TMM to control problems is given in [38]. Ref. [39] presents visualized simulation and design method of mechanical system dynamics based on TMM for multibody systems. Ref. [40] gives an automatic TMM of multibody system. Ref. [41] presents the calculus through the TMM of a beam with intermediate support with applications in dental restorations. Applications of the calculus by the TMM for long rectangular plates under uniform vertical loads are given in [42]. Axisymmetric free vibration of functionally graded piezoelectric circular plates can be shown in [43]. Analytical thermal analysis of radially functionally graded circular plates with coating or undercoating under transverse and radial temperature distributions is presented in [44]. A comparative study about modal parameters estimation of circular plates manufactured by FDM technique using vibrometry can be shown in [45]. TMM for the determination of the natural vibration characteristics of realistic thrusting launch vehicle is presented in [46].

We want the modeling of continuous circular plates to be a starting point in the calculation of circular bottoms from a chemical reactor. That is an important possibility to calculate a continuous circular plate embedded on the exterior contour and charged with a uniformly distributed load over all upper surface of the plate, with the opportunity to program this algorithm and to obtain a simple code of this calculus.

2. Materials and Methods

2.1. Materials

Due to the fact that the continuous circular plates take place in the chemical reactor, the material from which it is made must have special characteristics. Chemical reactors are made of refractory stainless steel. This is a very resistant material and combines both the properties of stainless steel—which are highly resistant to electrochemical corrosion—and refractoriness—the property that allows it to maintain its mechanical properties at high temperatures. Refractory stainless steel is resistant to high temperatures, the action of external loads, plastic deformation, and/or rupture, for a long time. Therefore, the calculus of the resistance of the component elements of a chemical reactor is very important, in occurrence, for the circular plate.

2.2. Methods: Algorithm for Continuous Circular Plates Calculus Charged with Axisymmetric Load on the Entire Upper Surface of the Plate

The chemical reactors are generally cylindrical or tubular, and they have a base (bottom) in the shape of a continuous circular plate, which can be considered as embedded on the exterior circumference, on which a load acts asymmetrically uniform distributed over the entire upper surface of the plate. The analytical calculus method will be presented next and then, the TMM approach to calculus of a continuous plate embedded on the exterior circumference and loaded with a uniformly distributed axisymmetric force will be presented.

2.2.1. Calculus Premises

For calculus of continuous circular plates with axisymmetric load, (Figure 1), it is necessary to consider several working hypotheses, [1]:

• The external forces act perpendicular to the mean plane of the circular plate;
• Under the action of these external forces, the plate deforms and curves;
• The curvature of the plate occurs in two planes, giving rise to an elastic surface with double curvature;
• For the law of variation in the arrow, given in Cartesian coordinates, w(x,y) characterizes the shape of this elastic surface;
• The circular plate has thickness h;
• It is assumed that the numerical values of the function w(x,y) are very small in relation to the thickness h of the plate.

Several simplifying working hypotheses are introduced into the theory of continuous circular plate bending as follows:

• The aligned points that are on the normal to the mean surface before the stress remain aligned on the same normal to the deformed surface after the stress;
• The normal stresses in sections parallel to the midplane are negligible compared to the bending stresses;
• It is assumed that there is no crushing between the overlapping layers of the plate;
• There may be local crushing when a concentrated force acts on the plate;
• The sign convention for the arrow is that the positive arrow is to up;
• The sign convention for the angle is as follows: the positive angle is counterclockwise;
• When the arrow w decreases, the angle ω is negative and ${\displaystyle \frac{dw}{dr}}$ is negative too.

2.2.2. Fundamental Differential Equations of a Continuous Circular Plate Charged with Axisymmetric Load

It is considered a continuous circular plate loaded with an exterior density charge q(x, y). The general equation for the deformation of any plate is as (1), [1] as follows:

$$\Delta \Delta w\left(x,y\right)={\displaystyle \frac{q\left(x,y\right)}{A}},$$

(1)

where

$$\Delta \Delta ={\displaystyle \frac{{\partial }^4}{\partial x^4}}+{\displaystyle \frac{2{\partial }^4}{\partial x^{2 }\partial y^2}}+{\displaystyle \frac{{\partial }^4}{\partial y^4}},$$

(2)

and

$$A={\displaystyle \frac{E{h}^3}{12\left(1-{\nu }^2\right)}},$$

(3)

with the following notations:

-

A is the bending stiffness of the circular plate;

-

E is the Young modulus or the longitudinal modulus of elasticity;

-

h is the constant thickness of the circular plate;

-

ν is the Poisson’s coefficient.

To calculate the deformations and stresses in all points of the continuous circular plate, the differential Equation (1) must be integrated. The integration of this equation depends on the contour of the integration domain. The theoretical solution is complete for circular plates with axisymmetric loading.

It is noted (as in Figure 2) as follows:

-

w(r)—the arrow of the continuous circular plate at a distance r from the plate axis;

-

ω(r)—the rotation angle by which the normal is rotated.

Between w(r) and ω(r) there is the following relation (4):

$$\omega ={\displaystyle \frac{dw}{dr}}.$$

(4)

It is considered an axial section of the continuous circular plate, as in Figure 2a.

Two normal of the continuous circular plate are considered; the first located at a distance r and the second normal at a distance (r + dr).

It is noted as follows:

$$BC=dr,$$

(5)

where a segment at z-coordinate before loading. After deformation, segment BC reaches position B′C′. It can be written (6) as follows:

$${B\prime }^{C\prime }-BC=z\left(\omega +d\omega \right)-z\omega ,$$

(6)

or as follows (7):

$${B\prime }^{C\prime }-BC=zd\omega .$$

(7)

The specific radial deformation ε_r is as follows (8):

$${\epsilon }_r=z{\displaystyle \frac{d\omega }{dr}}.$$

(8)

Point C′ exists on a circle of radius (r + z ω). The specific tangential deformation ε_t is as follows (9):

$${\epsilon }_t={\displaystyle \frac{2\pi z\omega }{2\pi r}},$$

(9)

or as follows (10):

$${\epsilon }_t=z{\displaystyle \frac{\omega }{r}}.$$

(10)

The specific radial deformation ε_r and the specific tangential deformation ε_t can be written in terms of radial stress σ_r and tangential stress σ_t (from the theory of elasticity as in [2,3]) as follows (11):

$$\left\{\begin{array}{c}{\epsilon }_r={\displaystyle \frac{1}{E}}\left({\sigma }_r-\nu {\sigma }_t\right)\\ {\epsilon }_t={\displaystyle \frac{1}{E}}\left({\sigma }_t-\nu {\sigma }_r\right)\end{array}\right..$$

(11)

The radial stress σ_r and the tangential stress σ_t can be written in terms of the specific radial strain ε_r and the specific tangential strain ε_t as in (12) as follows:

$$\left\{\begin{array}{c}{\sigma }_r={\displaystyle \frac{E}{1-{\nu }^2}}\left({\epsilon }_r+\nu {\epsilon }_t\right)\\ {\sigma }_t={\displaystyle \frac{E}{1-{\nu }^2}}\left({\epsilon }_t+\nu {\epsilon }_r\right)\end{array}\right..$$

(12)

With (7) and (10), relations (12) can be written as follows (13):

$$\left\{\begin{array}{c}{\sigma }_r={\displaystyle \frac{Ez}{1-{\nu }^2}}\left({\displaystyle \frac{d\omega }{dr}}+\nu {\displaystyle \frac{\omega }{r}}\right)\\ {\sigma }_t={\displaystyle \frac{Ez}{1-{\nu }^2}}\left({\displaystyle \frac{\omega }{r}}+\nu {\displaystyle \frac{d\omega }{dr}}\right)\end{array}\right..$$

(13)

It is considered a sectoral element cut from a continuous circular plate as in Figure 3b.

It is noted with M_r, the radial moment, and with M_t, the tangential moment, per unit of length along a radial axis, applied to the faces of this sectoral element.

It can be written the resultant of these moments the two faces as follows (14):

$$\left\{\begin{array}{c}{M}_r·r d\phi =r d\phi {\int }_{-{\displaystyle \frac{h}{2}}}^{+{\displaystyle \frac{h}{2}}}{\sigma }_r zdz\\ M_t dr=dr {\int }_{-{\displaystyle \frac{h}{2}}}^{+{\displaystyle \frac{h}{2}}}{\sigma }_t zdz\end{array}\right.,$$

(14)

or as follows (15):

$$\left\{\begin{array}{c}{M}_r·r d\phi ={\displaystyle \frac{E{h}^3}{12(1-{\nu }^2)}}\left({\displaystyle \frac{d\omega }{dr}}+\nu {\displaystyle \frac{\omega }{r}}\right)·r d\phi \\ M_t dr={\displaystyle \frac{E{h}^3}{12(1-{\nu }^2)}}\left({\displaystyle \frac{\omega }{r}}+\nu {\displaystyle \frac{d\omega }{dr}}\right)·dr\end{array}\right..$$

(15)

With (3), relations (15) can be written as follows (16):

$$\left\{\begin{array}{c}{M}_r=A\left({\displaystyle \frac{d\omega }{dr}}+\nu {\displaystyle \frac{\omega }{r}}\right)\\ M_t=A\left({\displaystyle \frac{\omega }{r}}+\nu {\displaystyle \frac{d\omega }{dr}}\right)\end{array}\right..$$

(16)

It can be noted at an equivalent moment M_eq as follows (17):

$$M_{eq} ={\displaystyle \frac{M_r+M_t}{A(1+\nu )}},$$

(17)

or it can be written as follows (18):

$$M_{eq}={\displaystyle \frac{\omega }{r}}+\nu {\displaystyle \frac{d\omega }{dr}},$$

(18)

or as follows (19):

$$M_{eq}={\displaystyle \frac{1}{r}}{\displaystyle \frac{d\omega }{dr}}+\nu {\displaystyle \frac{d^2\omega }{d{r}^2}},$$

(19)

or as follows (20):

$$M_{eq}={\displaystyle \frac{1}{r}}{\displaystyle \frac{d}{dr}}(r\omega ).$$

(20)

M_eq (equivalent to a moment) is not a bending moment, but it is proportional to the bending moment.

It can be written as the equilibrium equation of the sectoral element of force projections on a vertical direction, where T is the cutting force per unit of length on a circumference of radius r, as follows (21):

(T + dT) (r + dr) dφ − T r dφ + q(r) r dr dφ = 0,

(21)

where we have as follows (22):

$${\displaystyle \frac{d}{dr}}\left(rT\right)=-q\left(r\right)·r.$$

(22)

It can be written as the moment equilibrium equation of the sectoral element, summing all the moments in rapport to an axis tangent to an arc of a circle with radius r, in the midplane, as follows (23):

$$(M_r+\mathrm{d}{M}_r) (\mathrm{r}+\mathrm{dr}) \mathrm{d}\mathsf{\phi }-M_r \mathrm{r} \mathrm{d}\mathsf{\phi }+\mathrm{q}(\mathrm{r}) \mathrm{r} \mathrm{dr} \mathrm{d}\mathsf{\phi }{\displaystyle \frac{dr}{2}}-M_t \mathrm{dr} \mathrm{d}\mathsf{\phi } (\mathrm{T}+\mathrm{dT}) (\mathrm{r}+\mathrm{dr})\mathrm{dr} \mathrm{d}\mathsf{\phi }=0,$$

(23)

Neglecting higher order terms, we obtain the following (24):

$$M_t-{\displaystyle \frac{d}{dr}}\left(r{M}_r\right)=r T.$$

(24)

Substitute M_r and M_t in terms of M_eq it can be obtain the following (25):

$${\displaystyle \frac{rT}{A}}={\displaystyle \frac{\omega }{r}}-r{\displaystyle \frac{d^2\omega }{d{r}^2}}-{\displaystyle \frac{d\omega }{dr}},$$

(25)

or as follows (26):

$${\displaystyle \frac{rT}{A}}=-r{\displaystyle \frac{d{M}_{eq}}{dr }}.$$

(26)

It is noted as follows (27):

$$T_{eq}={\displaystyle \frac{r·T}{A}},$$

(27)

where T_eq is an equivalent cutting force.

The following differential relations (28)–(31) are obtained as follows:

$${\displaystyle \frac{d{T}_{eq}}{dr}}=-{\displaystyle \frac{r}{A}} q\left(r\right),$$

(28)

$${\displaystyle \frac{d{M}_{eq}}{dr}}=-{\displaystyle \frac{T_{eq}}{r}},$$

(29)

$${\displaystyle \frac{d\left(r\omega \right)}{dr}}=r·M_{eq},$$

(30)

$${\displaystyle \frac{dw}{dr}}=\omega .$$

(31)

In relations (28)–(31) we have two physical quantities—arrow w and rotation angle ω and M_eq—quantity proportional to bending moment and T_eq, related to shear force through relation (27). The relations (28)–(31) and it will be obtained, respectively, as follows:

-

relation (28) gives (32):

$$T_{eq}=T_{eqR}-q_1\left(r\right),$$

(32)

with as follows (33):

$$q_1\left(r\right)= {\int }_R^r{\displaystyle \frac{\rho }{A}}q\left(\rho \right)d\rho ,$$

(33)

-

relation (29) gives as follows (34):

$$M_{eq}=M_{eqR}-T_{eqR}log{\displaystyle \frac{r}{R}}+q_2\left(r\right),$$

(34)

with as follows (35):

$$q_2\left(r\right)= {\int }_R^r{q}_1\left(\rho \right){\displaystyle \frac{d\rho }{\rho }}.$$

(35)

The following can be seen (36):

$${\displaystyle \frac{d\left(r\omega \right)}{dr}}=r·M_{eqR}-T_{eqR} r log{\displaystyle \frac{r}{R}}+r q_2\left(r\right),$$

(36)

where (37) is as follows:

$$\omega = {\displaystyle \frac{R}{r}}{\omega }_R+{\displaystyle \frac{M_{eqR}}{2}}\left(r-{\displaystyle \frac{R^2}{r}}\right)-{\displaystyle \frac{T_{eqR}}{2}}\left( r log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{R^2}{2r}}-{\displaystyle \frac{r}{2}}\right)+{\displaystyle \frac{q_3\left(r\right)}{r}},$$

(37)

with (38) as follows:

$$q_3\left(r\right)= {\int }_R^r{\rho q}_2\left(\rho \right)d\rho .$$

(38)

The arrow is given by (39) as follows:

$$w=w_R+R{\omega }_{R}log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{M_{eqR}}{2}}\left({\displaystyle \frac{r^2-R^2}{2}}-R^{2}log{\displaystyle \frac{r}{R}}\right)-{\displaystyle \frac{T_{eqR}}{2}}\left({\displaystyle \frac{r^2+R^2}{2}}log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{R^2-r^2}{2}}\right)+q_4\left(r\right),$$

(39)

with (40) as follows:

$$q_4\left(r\right)= {\int }_R^r{q}_3\left(\rho \right){\displaystyle \frac{d\rho }{\rho }}.$$

(40)

2.2.3. Transfer Matrix for a Continuous Circular Plate Charged with Axisymmetrical Load

We have the following relations (41):

$$\left\{\begin{array}{c}w\left(r\right)=w_R+R{\omega }_{R}log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{M_{eqR}}{2}}\left({\displaystyle \frac{r^2-R^2}{2}}-R^{2}log{\displaystyle \frac{r}{R}}\right)-{\displaystyle \frac{T_{eqR}}{2}}\left({\displaystyle \frac{r^2+R^2}{2}}log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{R^2-r^2}{2}}\right)+q_4\left(r\right)\\ \omega \left(r\right)= {\displaystyle \frac{R}{r}}{\omega }_R+{\displaystyle \frac{M_{eqR}}{2}}\left(r-{\displaystyle \frac{R^2}{r}}\right)-{\displaystyle \frac{T_{eqR}}{2}}\left( r log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{R^2}{2r}}-{\displaystyle \frac{r}{2}}\right)+{\displaystyle \frac{q_3\left(r\right)}{r}}\\ M_{eq}\left(r\right)=M_{eqR}-T_{eqR}log{\displaystyle \frac{r}{R}}+ q_2\left(r\right)\\ T_{eq}\left(r\right)=T_{eqR}- q_1\left(r\right)\end{array}.\right.$$

(41)

With expressions (41), it can be written as a matrix relation as follows (42):

$$r{\left\{\begin{array}{c}w\left(r\right)\\ \omega \left(r\right)\\ M_{eq}\left(r\right)\\ T_{eq}\left(r\right)\end{array}\right\}}_r={\left[\begin{array}{cccc}1& R log{\displaystyle \frac{r}{R}}& {\displaystyle \frac{r^2-R^2}{4}}-{\displaystyle \frac{R^2}{2}}log{\displaystyle \frac{r}{R}}& -{\displaystyle \frac{r^2+R^2}{4}}log{\displaystyle \frac{r}{R}}-{\displaystyle \frac{R^2-r^2}{4}}\\ 0& {\displaystyle \frac{R}{r}}& {\displaystyle \frac{r}{2}}-{\displaystyle \frac{R^2}{2r}}& -{\displaystyle \frac{r}{2}}log{\displaystyle \frac{r}{R}}-{\displaystyle \frac{R^2}{2r}}+{\displaystyle \frac{r}{2}}\\ 0& 0& 1& -log{\displaystyle \frac{r}{R}}\\ 0& 0& 0& 1\end{array}\right]}_r{\left\{\begin{array}{c}{w}_R\\ {\omega }_R\\ {M_{eq}}_R\\ {T_{eq }}_R\end{array}\right\}}_R+{\left\{\begin{array}{c}{q}_4\left(r\right)\\ {\displaystyle \frac{q_3\left(r\right)}{r}}\\ q_2\left(r\right)\\ -q_1\left(r\right)\end{array}\right\}}_r .$$

(42)

For the continuous circular plate, it can be defined as a fictitious state vector at the rayon r as follows (43):

$${\left\{V\left(r\right)\right\}}_r^{-1}={\left\{V\right\}}_r^{-1}={\left\{w\left(r\right), \omega \left(r\right),M_{eq}\left(r\right),T_{eq}\left(r\right)\right\}}_r^{-1},$$

(43)

and a vector corresponding to the external loads as follows (44):

$${\left\{Ve\left(r\right)\right\}}_r^{-1}= {\left\{Ve\right\}}_r^{-1}={\left\{q_4\left(r\right), {\displaystyle \frac{q_3\left(r\right)}{r}}, q_2\left(r\right), -q_1\left(r\right)\right\}}_r^{-1}.$$

(44)

The expression of the transfer-matrix [TM]_r from the matrix relation (42) is as follows (45):

$${{\left[TM\right]}_r=\left[\begin{array}{cccc}1& R log{\displaystyle \frac{r}{R}}& {\displaystyle \frac{r^2-R^2}{4}}-{\displaystyle \frac{R^2}{2}}log{\displaystyle \frac{r}{R}}& -{\displaystyle \frac{r^2+R^2}{4}}log{\displaystyle \frac{r}{R}}-{\displaystyle \frac{R^2-r^2}{4}}\\ 0& {\displaystyle \frac{R}{r}}& {\displaystyle \frac{r}{2}}-{\displaystyle \frac{R^2}{2r}}& -{\displaystyle \frac{r}{2}}log{\displaystyle \frac{r}{R}}-{\displaystyle \frac{R^2}{4r}}+{\displaystyle \frac{r}{4}}\\ 0& 0& 1& -log{\displaystyle \frac{r}{R}}\\ 0& 0& 0& 1\end{array}\right]}_r.$$

(45)

Matrix relation (42) can be written more condensed as follows (46):

$${{\left\{V\right\}}_r=\left[TM\right]}_r{\left\{V\right\}}_R+{\left\{V_e\right\}}_r,$$

(46)

with the following notations:

-

[TM]_r is the transfer-matrix of the circular plate at the rayon r;

-

{V}_r is the fictitious state vector at the rayon r;

-

{V}_R is the fictitious state vector at the rayon r = R;

-

{Ve}_r is the vector at the rayon r corresponding to the exterior loads.

3. Results

Efforts and deformations calculus for a continuous circular plate embedded on the exterior circumference and charged with asymmetrical uniform distributed load on the entire upper surface (Figure 4) will be made.

It is considered a continuous circular plate embedded on the exterior circumference and loaded with asymmetrical uniform distributed load on the entire upper surface (as in Figure 4), with the conditions for the embedded circumference (47) as follows:

$$\left\{\begin{array}{c}{w}_R=0\\ {\omega }_R=0\end{array}\right..$$

(47)

For continuous circular plates, if we make (48):

$$r\to 0,$$

(48)

in matrix (45), the terms do not have finite values, and do not have well-defined values.

For continuous circular plates, embedded on the exterior circumference, in general, in the center of the plate, the arrow is different from 0, (49) as follows:

$$w\left(0\right)\ne 0,$$

(49)

but the rotation angle is 0, (50) as follows:

$$\omega \left(0\right)={\omega }_0=0.$$

(50)

The moments M_r and M_t (Figure 3b) must have finite value and well-defined values. So, for continuous circular plates, four conditions will be set as follows:

• two conditions related to the support mode; in this case, we are talking about a continuous circular plate, embedded on the exterior contour for r = R:

  -

  for the arrow, as follows (51):

  $$w_R\left(R\right)=0,$$

  (51)

  -

  for the rotation angle, as follows (52):

  $${\omega }_R\left(R\right)=0$$

  (52)

• two conditions that will be set in the center of the plate, for r = 0, that are as follows:

  -

  for the rotation angle, as (50): ${\omega }_0=\omega \left(0\right)=0;$

  -

  for the equivalent moment $M_{{eq}_0}$, as follows (53):

$$M_{{eq}_0}\left(0\right)=D,$$

(53)

where D is a finite value, and a well-defined value.

Analytical calculus will be made and then with TMM for a continuous circular plate charged with axisymmetric uniformly distributed load over the surface of all circle with radius R (as in Figure 4).

The charge density is as follows (54):

$$q\left(r\right)=-q Y\left(r\right),$$

(54)

where Y is the Heaviside function.

For the four functions q_i, i = 1, 4 it can be written as follows:

-

For $q_1\left(r\right)$, as follows (55):

$$q_1\left(r\right)= {\displaystyle \frac{q}{2A}}\left(R^2-r^2\right),$$

(55)

-

For q₂(r), we have (56) as follows:

$$q_2\left(r\right)={\displaystyle \frac{q}{2A}}\left({\displaystyle \frac{R^2-r^2}{2}}-R^{2}log{\displaystyle \frac{R}{r}}\right),$$

(56)

-

For q₃(r), we have (57) as follows:

$$q_3\left(r\right)={\displaystyle \frac{q}{4A}}\left({\displaystyle \frac{R^4-r^4}{4}}-R^2{r}^{2}log{\displaystyle \frac{R}{r}}\right),$$

(57)

-

For q₄(r), we have (58) as follows:

$$q_4\left(r\right)={\displaystyle \frac{q}{8A}}\left({\displaystyle \frac{5R^4-4r^2{R}^2-r^4}{8}}-{\displaystyle \frac{R^4}{4}}log{\displaystyle \frac{R}{r}}-R^{2}r log{\displaystyle \frac{R}{r}}\right).$$

(58)

For the exterior circumference, we have conditions (51) and (52).

The condition (51) can be written as follows (59):

$$\underset{r\to 0}{\mathit{lim}}\left[-T_{{eq}_R}log{\displaystyle \frac{r}{R}}+q_2\left(r\right)\right]=D,$$

(59)

or as follows (60):

$$\underset{r\to 0}{\mathit{lim}}\left[-T_{{eq}_R}log{\displaystyle \frac{r}{R}}+{\displaystyle \frac{q}{2A}}\left({\displaystyle \frac{R^2-r^2}{2}}-R^{2}log{\displaystyle \frac{R}{r}}\right)\right]=D,$$

(60)

from which it follows (61):

$$T_{{eq}_R}={\displaystyle \frac{q{R}^2}{2A}}.$$

(61)

The condition (52) can be written as follows (62):

$${\omega }_R-{\displaystyle \frac{R}{2}}{M}_{{eq}_R}-{\displaystyle \frac{R}{4}}{T}_{{eq}_R}-{\displaystyle \frac{1}{R}}\underset{r\to 0}{\mathit{lim}}\left[q_3\left(r\right)\right]=0,$$

(62)

or as follows (63):

$$-{\displaystyle \frac{R}{2}}{M}_{{eq}_R}-{\displaystyle \frac{R}{4}}{T}_{{eq}_R}-{\displaystyle \frac{1}{R}}\left({\displaystyle \frac{q{R}^4}{16A}}\right)=0,$$

(63)

from which it follows (64):

$$M_{{eq}_R}=-{\displaystyle \frac{q{R}^2}{8A}}.$$

(64)

The arrow can be written as follows (65):

$$w\left(r\right)=-{\displaystyle \frac{q}{64A}}{\left(R^2-r^2\right)}^2.$$

(65)

For r = 0, in the center of the plate, we have the following (66):

$$\left|w_{max}\right|=\left|w\left(0\right)\right|={\displaystyle \frac{q{R}^4}{64A}},$$

(66)

where there is identical value from the theory of elasticity in the classical Strength of Materials calculus, [3].

For the calculus with TMM calculus of continuous circular plate charged with axisymmetric uniform distributed load over all the surface of the circle (as in Figure 4), it must be neglecting all terms containing the “log” function, the expressions for the following:

-

q₁ (r) from (55) remains the same;

-

q₂ (r) from (56) becomes as (67):

$$q_2\left(r\right)={\displaystyle \frac{q}{2A}}\left({\displaystyle \frac{R^2-r^2}{2}}\right),$$

(67)

or, as follows (68):

$$q_2\left(r\right)= {\displaystyle \frac{q\left(R^2-r^2\right)}{4A}},$$

(68)

-

q₃ (r) from (57) becomes as (69):

$$q_3\left(r\right)={\displaystyle \frac{q}{4A}}\left({\displaystyle \frac{R^4-r^4}{4}}\right),$$

(69)

or, as follows (70):

$$q_3\left(r\right)= {\displaystyle \frac{q\left(R^4-r^4\right)}{16A}},$$

(70)

-

q₄ (r) from (58) becomes as follows (71):

$${ q}_4\left(r\right)={\displaystyle \frac{q}{8A}}\left({\displaystyle \frac{5R^4-4r^2{R}^2-r^4}{8}}\right),$$

(71)

or, as follows (72):

$$q_4\left(r\right)={\displaystyle \frac{q\left(5R^4-4r^2{R}^2-r^4\right)}{64A}}.$$

(72)

For r = 0, in the center of the plate, you can write the following:

-

q₁ (r) from (55), becomes as follows (73):

$$q_1\left(r\right)={\displaystyle \frac{q{R}^2}{2A}},$$

(73)

-

q₂(r) from (68), becomes as follows (74):

$$q_2\left(r\right)={\displaystyle \frac{q{R}^2}{4A}},$$

(74)

-

q₃(r) from (70), becomes as follows (75):

$$q_3\left(r\right)={\displaystyle \frac{q{R}^2}{4A}},$$

(75)

-

q₄(r) from (72), becomes as follows (76):

$${ q}_4\left(r\right)={\displaystyle \frac{5q{R}^4 }{64A}}.$$

(76)

Neglecting all terms of transfer matrix of expression (45) containing the “log” function (are made 0), the expression (45) becomes as follows (77):

$${{ \left[TM\right]}_r=\left[\begin{array}{cccc}1& 0& {\displaystyle \frac{r^2-R^2}{4}}& -{\displaystyle \frac{R^2-r^2}{4}}\\ 0& {\displaystyle \frac{R}{r}}& {\displaystyle \frac{r}{2}}-{\displaystyle \frac{R^2}{2r}}& -{\displaystyle \frac{R^2}{4r}}+{\displaystyle \frac{r}{4}}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}_r,$$

(77)

For a current radius r, it can write the matrix relation (42) (with relations (73)–(77)) as follows (78):

$${{ \left\{\begin{array}{c}w\left(r\right)\\ \omega \left(r\right)\\ M_{eq}\left(r\right)\\ T_{eq}\left(r\right)\end{array}\right\}}_r=\left[\begin{array}{cccc}1& 0& {\displaystyle \frac{r^2-R^2}{4}}& -{\displaystyle \frac{R^2-r^2}{4}}\\ 0& {\displaystyle \frac{R}{r}}& {\displaystyle \frac{r}{2}}-{\displaystyle \frac{R^2}{2r}}& -{\displaystyle \frac{R^2}{4r}}+{\displaystyle \frac{r}{4}}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}_r{\left\{\begin{array}{c}{w}_R\\ {\omega }_R\\ {M_{eq}}_R\\ {T_{eq }}_R\end{array}\right\}}_R+{\left\{\begin{array}{c}{\displaystyle \frac{5q{R}^4 }{64A}}\\ {\displaystyle \frac{1}{r}}{\displaystyle \frac{q{R}^2}{4A}}\\ {\displaystyle \frac{q{R}^2}{4A}}\\ -{\displaystyle \frac{q{R}^2}{2A}}\end{array}\right\}}_r.$$

(78)

For the exterior circumference, for r = R, the matrix [TM]_r becomes as follows (79):

$${{ \left[TM\right]}_r={\left[TM\right]}_R={\left[1\right]}_R=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}_R,$$

(79)

which does not lead to a result that satisfies, from an engineering point of view, the conditions that are mandatory on the embedded circumference (51) and (52), nor in the center of the plate (50) and (53).

Because we have an axisymmetric load, in the center of the continuous circular plate, for r = 0, an additional condition for $T_{{eq}_0}$ occurs as follows (80):

$$T_{{eq}_0}=0.$$

(80)

From the matrix relation (78), for r = 0, the first line can be written as follows (81):

$$w\left(0\right)=w_R-{\displaystyle \frac{R^2}{4}}{M_{eq}}_R-{\displaystyle \frac{R^2}{4}}{T}_{{eq}_0}+{\displaystyle \frac{5q{R}^4 }{64A}},$$

(81)

or as follows (82):

$$w_{max}=w_R-{\displaystyle \frac{R^2}{4}}{M_{eq}}_R-{\displaystyle \frac{R^2}{4}}{T}_{{eq}_0}+{\displaystyle \frac{5q{R}^4 }{64A}},$$

(82)

because the maximum arrow is found in the center of the plate. Relation (82) is used to calculate the maximum arrow.

With condition (80), from the last line of matrix relation (78), it can be written as follows (83):

$${0=T}_{{eq}_R} -{\displaystyle \frac{q{R}^2}{2A}},$$

(83)

from which it follows (84):

$$T_{{eq}_R}={\displaystyle \frac{q{R}^2}{2A}},$$

(84)

so the relation (84) is identical to relationship (61).

Because the load is axisymmetric, an additional condition (85) is also imposed for M_eq as follows:

$$M_{{eq}_0}{=-M}_{{eq}_R}.$$

(85)

Taking into account (85), the third line in the matrix relation (78) can be written as follows (86):

$$M_{{eq}_0}{=-M}_{{eq}_R}+{\displaystyle \frac{q{R}^2}{4A}},$$

(86)

from which it follows (87):

$$M_{{eq}_0}={\displaystyle \frac{q{R}^2}{8A}},$$

(87)

and (88) as follows:

$$M_{{eq}_R}=-{\displaystyle \frac{q{R}^2}{8A}}.$$

(88)

Relation (88) is identical to relationship (64).

With relation (82), we can calculate the maximum arrow (89) at the center of the continuous circular plate:

$$w_{max}=-{\displaystyle \frac{R^2}{4}}{M}_{{eq}_R}-{\displaystyle \frac{R^2}{4}}{T}_{{eq}_R}+ {\displaystyle \frac{5q{R}^4 }{64A}},$$

(89)

from where the maximum arrow is as follows (90):

$$w_{max}= {\displaystyle \frac{q{R}^4 }{64A}}.$$

(90)

Relation (90) is identical to relationship (66).

4. Discussion

Through this work, we wanted to approach the problem of calculating a continuous circular plate embedded on the outer contour and loaded with a uniformly distributed axisymmetric load from two perspectives: analytical approach and then through TMM. What is interesting is that TMM offers the opportunity to relatively easily program the calculation algorithm, which then calculates the deflection, rotation, bending moment, shear force, and normal and tangential stress in any section of the plate.

Modeling of one continuous circular plate with TMM consists of discretizing the plate into concentric circular elements (circular concentric rings). Each circular ring has two sides: a left side (for any ring it will be the face corresponding to the current radius r) and a right side (for any ring it will be the face corresponding to the radius (r + dr)). The sides of each ring will be numbered as follows: ring 1 will have the left face numbered 0, and the right face will have the number 1; ring 2 will have the left face in common with the right face of ring 1, so it will have the number 1, and the right face will have the number 2 (common with the right face of ring 2); ring number i will have the left face numbered (I − 1), and the right face will have the number i; this will be performed until the last ring, which will have the number n, its left face will have the number (n − 1), and the right face will have the number n. In addition, face n of ring n corresponds to the outer contour of the solid circular plate, for r = R. Each face of the elementary circular ring is associated with a state vector consisting of four elements: the arrow, the rotation angle, the bending moment (in our case a moment proportional to the bending moment), and an element relative to the cutting force. The state vector corresponding to each face will have the number corresponding to the respective face. We can, thus, write the matrix relation (78), which is the basis for the calculation for the TMM. We can then put the boundary and support conditions and, thus, solve the problem. Through this work, we have shown that TMM leads to the same results as those obtained with classical calculation based on the Theory of Elasticity.

This work represents a first step towards broader research on the modeling of continuous circular plates with TMM. We plan to conceive a computational code based on TMM and attach it to a shape optimization program. Then, we will take a concrete example and validate the results obtained with our program with Finite Elements Method (FEM) modeling and with values that will be determined through experimental measurements.

5. Conclusions

This paper presents an original approach through TMM applied for the calculus of continuous circular plate charged with asymmetrical uniform load on the entire upper surface of the plate. The results obtained with TMM were compared and validated with those obtained from classical analytical calculation, from the Theory of Elasticity.

The main drawback of applying this method in our research is that we have not yet been able to model the circular plate in detail. Because it is an iterative calculus, we hope to be able to design a program based on this TMM calculation algorithm, in which we can eliminate these shortcomings.

TMM is an elegant method and is relatively easy to program. We wish and hope to be able to continue our research on the design and verification of chemical reactors and, thus, to be able to contribute to simplifying the resistance calculations of their component elements.