CC BY
29Yu.Ya. Tyukalov,
Vyatka State University, Kirov, Russia
Key words: finite element method; moments approximation; bending plates; buildings; constructions
doi: 10.18720/MCE.82.16
Finite element models in stresses for bending plates
Конечно элементные модели в напряжениях для изгибаемых пластин
Д-р техн. наук, профессор Ю.Я. Тюкалов,
Вятский государственный университет, г. Киров, Россия
Ключевые слова: метод конечных элементов; аппроксимация моментов; изгибаемые пластины; здания; конструкции
Abstract. Finite element models for plate bending problems are constructed on the basis of approximations of moments fields. The bending and twisting moments are approximated in the finite element area by piecewise constant functions. The solution is based on the functional of the additional energy. Algebraic equations of equilibrium of nodes of grid of finite elements are formed using the principle of possible displacements and are included in the functional with the help of Lagrange multipliers. The necessary expressions for rectangular and triangular finite elements are obtained. Calculations of square clamped and hinged-supported plates on the action of uniformly distributed load are performed. Comparison of the obtained results with the results by the finite element method calculations in displacements is presented. It is shown that the presented method of calculating bent plates by the finite element method in stresses has the property of convergence from above. The displacements obtained by this method converge to the exact values from above, while the values of the moments is determined with reserve. When the grid of finite elements is crushed, the difference of the two solutions, in stresses and in displacements, decreases monotonically and the accuracy of the obtained results can be estimated from the value of this difference.
Аннотация. Конечно-элементные модели для задач изгиба пластин построены на основе аппроксимаций полей моментов. Изгибающие и крутящие моменты аппроксимируются по области конечных элементов кусочно-постоянными функциями. Решение строится на основе функционала дополнительной энергии. Алгебраические уравнения равновесия узлов сетки конечных элементов формируются при помощи принципа возможных перемещений и включаются в функционал при помощи множителей Лагранжа. Получены необходимые выражения для прямоугольных и треугольных конечных элементов. Выполнены расчеты квадратных защемленных и шарнирно-опертых плит на действие равномерно распределенной нагрузки. Приведено сравнение полученных результатов с результатами расчетов по методу конечных элементов в перемещениях. Показано, что представленный метод расчета изгибаемых пластин методом конечных элементов в напряжениях обладает свойством сходимости сверху - перемещения, полученные данным методом сходятся к точным значениям сверху, при этом величины моментов определяются в запас прочности. При измельчении сетки конечных элементов разность двух решений, в напряжениях и в перемещениях, монотонно уменьшается и по величине этой разности можно оценивать точность полученных результатов.
1. Introduction
Bending plates are widely used in construction of various buildings and structures. They are important constructive elements. From their strength very often depends the security and reliability of the entire structure. A more accurate definition of the internal forces, arising in bent plates, from various types of loads and impacts, remains an actual task now. Therefore, a lot of scientific articles and books are devoted to the construction of various finite elements modeling the bending of plates [1, 2], but the problem remains topical.
When finite element method is used based on the displacements fields approximation, it is difficult to select suitable functions for the approximations [3-6]. As is known, when we solve problems of bending of plates by the finite element method, the polynomials, used to approximate the displacements, must have higher order, in comparison with the plane problem of the theory of elasticity. It is difficult to ensure the continuity of the stress fields at the node points and along the boundaries of the finite elements [2]. Stresses
and forces are usually determined for the central points of finite elements. In [7, 8], displacement fields along the boundaries of finite elements are accorded by means of additional equations introduced by means of penalty functions or Lagrange multipliers.
In paper [9], the Ritz method is used to solve the problem of nonlinear bending of elliptic plates, and comparison is made with the solutions of other authors. The construction of finite elements for bent plates is possible based on the application of the equations of the three-dimensional theory of elasticity [10, 11]. This approach allows to obtain good results when analyzing thick plates, when it is necessary to consider the shear strains.
A few papers are devoted to the use of mixed and hybrid methods for the construction of finite elements [12-15]. With this approach, the approximations of displacement fields and stress (force) fields are used, which leads to an increase in the total number of unknowns. In addition, the obtained solutions do not possess the properties of the lower or upper boundary. In [16], the displacement function is chosen as polynomial of the fourth degree to solve the problem of bending of clamped rectangular plate. The results of calculations of deflections, bending moments and shearing forces for square plates of different thicknesses are presented and compared with the results of other authors.
The finite element models for bending the plate according to the Kirchhoff theory can be constructed since approximations only for forces [17-19]. In this case, the solution is built on the basis functional of the additional energy. Algebraic equilibrium equations for nodes are forming using the principle of possible displacements and they are including in the functional as penalty functions. It is shown that in the case of using piecewise-constant functions to approximate the nodal moments, the solution has property of upper boundary.
In [22], a comparative analysis of different types of finite elements is carried out for the calculation of bent plates. More than ten types of finite elements are developed since the Lagrange principle, and they have a different number of degrees of freedom in the form of nodal displacements and angles of rotation. Also presented many finite elements developed since mixed variational principles, when both nodal forces and displacements are used as unknowns. In book consider only two finite elements developed since the method of forces having 9 and 16 degrees of freedom. Obviously, finite elements based on the method of forces are not developed enough.
To solve the plates bending problems, incompatible elements of the displacement method building based on the Lagrange functional are widely used. When formulating such elements, additional conditions are introduced for matching the slope angles of the normal along the boundaries of finite elements [24, 25], or independent approximations are used for displacements and angles of rotation [26, 27, 30], and then additional conditions are used to harmonize the introduced approximations. Also, various hybrid [27, 28], equilibrium [33] and mixed formulations [35] of the finite element method are used for the solution.
The present paper is aimed at constructing the solution of the problem of bending of Kirchhoff plates by the finite element method based on approximation of forces and using only the principles of minimum of additional energy and possible displacements.
2. Methods
The solution of plate bending problems in stresses can be obtained using the additional energy functional [1, 2]:
nc — U*+V*
-2f>
[M]T[E]-1{M}dn
f{T]T{K}dS
^ mm,
(1)
U* - additional energy of the strains, V*- potential of boundary forces corresponding to the specified displacements [1]; {A} - vector given displacements of nodes; {T} - vector boundary forces; S - boundary
surface, on which the displacement nodes are given; D - subject area; {M} — { My
M
xy
- vector of
bending and torsional moments; [£] 1 - matrix of flexibility of the material:
[E]-1—^
L J E^t3
1
0 0
1
0 0 2(1 +p).
(2)
where: E - modulus of elasticity of the material; J is the Poisson ratio; t is the thickness of the plate.
In accordance with the principle of minimum additional energy, the moments functions Mx,My,Mxy, must satisfy the differential equations of equilibrium. The system of differential equations of equilibrium for bent plates, under the action of distributed load q, has the following form [2]:
-Q:
-х -y -M -M
xy
дх
-y
-Mv -M
xy
- y
- х
-Qx = 0,
- Qy = o.
(3)
Eliminating the transverse forces Qx and Qy from (3), we obtain the known differential equation of equilibrium of second-order:
d2M, dx2
+ о д2мху
+ ■
d2Mv
дхду ду
+ q = 0.
(4)
Since, in the general case, it is practically impossible to select the basis functions for Mx, My, Mx y, which satisfy the differential equation (4), it is suggested to introduce additional stress functions of Southwell [2]. Then the moments are expressed in terms of the derivatives of the introduced stress functions. This approach leads to additional difficulties in specifying loads and boundary conditions and has not received wide practical spread.
In [17-19] another approach is proposed for the solution. The subject area is divided into rectangular or triangular finite elements. The fields of moments in the region of the finite element can be approximated by linear or piecewise constant functions (Figure 1). Linear approximating functions (Figure 1a) ensure the continuity of the moment fields across the entire subject area. Piecewise-constant functions (Figure 1b) satisfy the differential equation of equilibrium (4) in the absence of a distributed load and are continuous along the boundaries of finite elements but have discontinuities inside the region of the finite element.
{M3} {M3}
/л A
Ш\/ H/{M:} {Mj^e-1/{м2}
{M2} {M4}
a)
b)
Figure 1. Variants of the approximation of moments in the region of the finite element: a) the moments vary linearly; b) the moments are piecewise-constant
For simplicity, we assume that the given node displacements are absent. Then, using for moments any variant of approximating functions (Figure 1), the expression for the functional (1) can be written in the following matrix form:
Пс = 1{M}T[D]{M} ^ win,
(5)
where{M} is a vector of unknown stresses for the whole system; [D] is the matrix of flexibility for the entire system.
z
Figure 2. Possible displacement of the node 13 and adjacent finite elements
Using the principle of possible displacements (Figure 2), for all non-supported nodes of the system, we will get algebraic equations of equilibrium of forces along the vertical axis Z:
[Ci}T{Mi] + Pi = 0, iESz
(6)
where(Mj) is the vector of unknown node moments of all finite elements adjacent to node i; Sz is the set of nodes, that have non-fixed displacement along the vertical axis Z; Pt is the generalized force corresponding to the potential of external loads for possible displacements of node i along the Z axis; [Ci] is vector containing coefficients on unknown node moments in the equilibrium equation of node i along the Z axis. Algebraic equilibrium equations (6) ensure the equilibrium of the moment fields in discrete sense. Unknown parameters are only the moments in nodes of finite elements grid. In [17-19], the solution of this problem was considered using the penalty functions method, which makes it possible to obtain the functional (7):
n = \{M}T[D]{M} + ZieBs a ({Ci}T{Mi} + P).
(7)
a is the penalty parameter, recommendations for the choice of which are given in [17-19]. Equating the derivatives (7) along the unknown nodal moments to zero, we obtain a system of linear algebraic equations. The matrix of non-zero coefficients of this system of equations will have ribbon structure for any variant of the moments approximation.
In this paper, to minimize the functional (5), in the presence of constraints in the form of system of algebraic equations (6), we will use the method of Lagrange multipliers:
nc = 1{M}T[D]{M} + IlieHzwi({Ci}T{Mi} + Pt) ^ min,
(8)
Wi - vertical displacement of node i. In this solution, additional unknowns appear in the form of node displacements. But we must accent that the approximations of the displacement field in region of finite element are not used in the getting (8).
We represent the expression (8) in form more convenient for constructing the solution: nc = l[M]T[D][M] + [w]t([F] - [L][M]) ^ min,
(9)
where {w} is the global vector of unknown nodal displacements for the whole system; {F} is vector whose elements are equal to the works of external forces on the corresponding unit displacements of the nodes; [L] is the matrix of "equilibrium", whose rows are formed from the corresponding vectors {C¿}. If we equate the derivatives nc c along the vector {M} to zero, we obtain the equations of consistency of deformations which expressed in the forces:
[D]{M} - [L]T{w} = 0.
Derivatives nc along the vector [w] are the system of equations of equilibrium of nodes
[F] - [L][M] = 0.
Combining (10) and (11), we obtain the following system of linear algebraic equations:
[D] -[L]T]({M}
[L] [0] J ({w}
0
-{F}}-
(10)
(11)
(12)
Expressing the vector [M] from the first matrix equation and substituting it into the second, we obtain
[K] = [L][D]-1[L]T, (13)
[K][w] = [F], (14)
[M] = [D]-1[L]T[w]. (15)
Thus, solving the system of algebraic equations (14), we obtain the values of the node displacements [w], and then the vector of moments [M] from (15).
If linear functions are used to approximate the moments (Figure 1a), then the matrix [K] will be filled and the solution of the system of linear algebraic equations (14) or the direct solution of the system of equations (12) will require large computational costs. Therefore, below, we will consider variant of the approximation of moments in the region of finite element by piecewise constant functions (Figure 1b). In this case, the matrix [D] is block-diagonal, and the matrix [K] will have ribbon structure of non-zero elements.
Will obtain the necessary expressions for the elements of the matrices [D], [L] and the vector [F], which enter (9), when rectangular and triangular finite elements are used to discredit the subject area (Figure 3).
Уз
Yi Y2
Xi Хз Хг
Figure 3. Triangular and rectangular finite elements
М,-
l ,x
We introduce the following notations: {M J = { Miy } - vector of moments in node i in local
М
М^
l,Xy;
coordinate system X1OY1 ; [Мь} = { Мiy
М
i,xy.
- vector of moments at node i in the global coordinate
system XOY. Mi x - bending moment at node i, directed along the Xaxis; Miy is the bending moment at node i, directed along the Y axis; Mixy is the torque at node i.
Figure 4. Separation of a triangular finite element into regions with constant moments: a) an arbitrary triangle; b) right-angled triangle
Since the bending moments and torques are piecewise constant, the expression for the additional deformation energy in the global coordinate system can be written in the form of simple sum:
>
U* =1l?=1Ai{Miym-1{Mil
Ai=?l=i\As+ri'T
As
S = lAi ■
(16) (17)
where m is the total number of nodes; niR is the number of rectangular elements connected to node i; niT -the number of triangular elements connected to node i; As is the area of finite element. We introduce the notation for the flexibility matrix of "neighborhoods" of the node i; Af is the area of the part of the triangular finite element with constant bending moments (Figure 4a). This point can be defined as the point of intersection of perpendiculars, drawn from the middle of the sides (Figure 4a).
For a rectangular finite element, the separation of an element into regions with constant moments is uniquely - into four equal regions. For a triangular element, each side must be equally divided, but there must also be a point inside the element in which the three areas, related to the nodes, intersect. If the largest angle of the triangle is greater than 90 degrees, then such a point lies outside the triangle. In this case, the triangle is divided into regions by lines passing through the middle of the sides - these lines will be parallel to the sides of the triangle. Such a division of a triangular element into regions with constant
moments makes it possible to obtain more accurate results, than in the case of division simply into three
-
equal parts - -As.
In Figure 4a, the point O is the center of a circle described about the triangle. OA, OB, OC -perpendiculars, directed from the middle of the corresponding sides of the triangle. Denote the lengths of the sides of the triangle - 1i2,123, l3i. From the schemes in Figure 4a we get:
W2I23I31
R =
4A
Ai^flR2-'^}2-^
(18)
a%=^2 \r2-'^ + ^3 ¡r2 -
i2
l23
4
4
4
l23
T
As3 = '-f \R2
L23
4
+ f ¡R2
L3i
4
It is obvious that for a right-angled triangle (Figure 4b) - A\= A2 = -AS,A3 =-As. For an
4 2
-
equilateral triangle - A- = A2 = A3 = - As. If one of the angles of triangle is greater than 90°, then, as
-well as for a right-angled triangle, we take: A- = A\ =-As,A3 = -As. Besides, you do not need to use too elongated triangles. Better such triangles divided into two more correct triangles.
We introduce the notation for the flexibility matrix of "neighborhoods" of node and for the global flexibility matrix for the entire system, which consists of matrices of flexibility for all nodes of the system:
[Di]=Ai[E]-i,[D] =
[D-]
[Dm]
It is obvious that the matrix [D] is easily invertible analytically.
(19)
[D]-i =
[D-]-i
[■Dm]-
(20)
Consider the derivation of the equilibrium equations for possible displacement of node of rectangular finite element in the local coordinate system XiOYi (Figure 5). For rectangular finite element, we introduce also the local coordinate system ^cq, connected with its center (Figure 5) and enter the basic functions, which are expressed in normalized local coordinates in the following form:
2
2
2
2
(1 + £iO(1 + ViV) 2x 2y Nt(x,y) =---------= —,v =-r,i = 1,2,3,4. (21)
lK JJ 4 ^ a 1 b
The index denotes the local node number of the finite element; x, y - coordinates along the axes Xx and Yt, respectively; are the local normalized coordinates of the node i taking the values 1 or-1. The nodes are numbered counterclockwise, beginning with the lower-left node.
a
a) b)
Figure 5. a - possible displacement of the node 1 and deformation of the rectangular finite element; b - the red line shows the graph of the change in the angle of rotation of the sides of the finite element from the possible movement of the node 1
The displacements of the points of the finite element after the possible displacement of the node ( = 1,2,3,4) can be expressed in the following form:
(1 + ^Od + ViV)
Sw —
4
(22)
Possible displacements along each coordinate axis vary linearly, so the curvatures Skx and 8ky along the axes Xx and Yt are respectively zero, and the torsional curvature Skxy is constant.
Skx —
d2(Sw) dx2
— 0, Skv —
d2(Sw) dy2
— 0,Skxy —
d2(Sw) = dxdy a • b'
(23)
Thus, with possible displacement of node i, only internal torques Mxy and normal bending moments
Mn along the boundaries of finite elements re at the corresponding rotation angles Sßn will work. Then the work of internal forces for the finite element k on possible displacements of node i can be expressed in the following form:
= §0 Mxy 5kxydxdy + HYe Mn 8pnds.
For the boundaries of element directed along the Xx axis, the bending moment My and the corresponding angle of rotation of the finite element along the axis Yt - 8(py are normal. For boundaries directed along the axis Yt, respectively Mx and 5(x are normal (Figure 5b).
(24)
d(Sw) 1 + VVi) d(Sw) m(1 + ffi)
S(Px—^- —-—-,S(y —
dx
2 a
d y
2 b
(25)
In Figure 5b, the red line shows the graph of the change in the angle of rotation along the boundaries of the finite element after possible displacement of the node 1.
Considering that the moments are constant in each quarter of the finite element, the first integral in expression (24) is calculated simply:
ra rb
S0 S0 Mxy Skxydxdy — = i Mxy,jSkxy — 2 = 1 Mxy,j-
im^i M
(26)
The work of the bending moments Mx for the angles of rotation of the sides 1-4 and 2-3 of the finite element (Figure 4b) is calculated by integrals
с г0 ii(i+vm)
^^ ъ г0
reMxS(xds —
ъ77 С Г1 ii(1+Vni)
dTl-zMx^j$Q
2 2a ' 2 ->0 2a
dy —
In expression (27), the coefficients ^ — £4, which are to the right of the moments, consider the directions of the moments Mx and of the rotation angle Spx. On the side 1-4, the moments Mx and the rotation Spx are directed equally, so the first two terms in (27) will be positive, since ^ = = — 1. On the side 2-3, the moments and the rotation are directed in the opposite direction, so the third and fourth terms will be negative, since ^ = = 1. Calculating the integrals in (27), we can obtain the following compact expression:
b
My
freMxS(pxds = —btilUSj^+^Mx,. Similarly, one can obtain an expression for the operation of the moments M
"4 ^ fi , MA 77
y
y,j-
fT. My 8cpyds = -±Vi ZU Vj (l+-f)M Summing the expressions (26), (28), and (29), we obtain
tut = -±ji ZU fj (1 + MXJ- ^ ZU Vj (1 +
m
2 ¿-¡j = 1 lv'xy,j-
fift\ M
y,
(28)
(29)
(30)
Substituting i to be equal from 1 to 4 in (30), we obtain expressions for the operation of internal forces (moments) for possible displacements of nodes from the first to the fourth.
We unite the node moments in the local coordinate system X1OY1 for the finite element k to the
vector {Mk}.
(Mx,i
Mv 1 My
ly,1 mxy,1 Mx,2 My,2 m xy,2 m x,3 my,3 mxy,3 mx,4 my,4 mxy,4j
We also introduce vector combining the values of the work of internal forces for possible displacements of nodes of finite element
isui,)
Mrv7 Mx3
Mvq My
Mx4 My
My
o.
(31)
Kl=
SUk2,z SUk3,z .SUkAlzJ
(32)
Then we can write the following expression in matrix form:
[SUk] = [Lk] {Mk}. Using (30), we obtain expressions for the elements of the matrix [l ].
(33)
r-3b -3a 1 3b - a 1 b a 1 -b 3a 1
8 a 8 b 2 8 a 8 b 2 8 a 8b 2 8 a 8b 2
3b - a -1 -3b -3a -1 -b 3a -1 b a -1
8 a 8 b 2 8 a 8 b 2 8 a 8b 2 8 a 8b 2 (34)
b a 1 -b 3a 1 -3b -3a 1 3b - a 1 . (34)
8 a 8 b 2 8 a 8 b 2 8 a 8b 2 8 a 8b 2
-b 3a -1 b a -1 3 b - a -1 -3b -3a -1
[ 8 a 8 b 2 8 a 8 b 2 8 a 8b 2 8 a 8b 2 -
The node moments {M } in local and [Mk] in global coordinate system are connected by the
matrix of the direction cosines
[ l] =
cos2 a sin2 a
sin2 a cos2 a
Lsin a cos a — sin a cos a
2 sin a cos a 2 sin a cos a cos2 a — sin2 a.
(35)
a is the angle between the Y1 axis and the Y axis (Figure 3). Using (35), we obtain the matrix of the direction cosines for the finite element
[Sk] =
[1]
И
[I]
wJ
(36)
The work of internal forces (33) can be represented in the following form:
V—kl
[SU?] = [Lk][Mk], [Lk] = [L ] [Sk].
(37)
The matrix can be called the local matrix of "equilibrium" of the finite element. From matrices [Lk] for finite elements, in accordance with the numbering of the nodes and elements, global "equilibrium" matrix [L] is formed for the whole system.
The potential of the external concentrated and uniformly distributed loads for possible displacements of the node i along the global coordinate axis is determined by the formula (38).
1 ,
8Vi = Pi+-qkab = Ri. (38)
Pt - the force concentrated at node i; qk - uniformly distributed by element load. The generalized forces Ri are placed in the vector [F] (see (12)).
Consider possible displacement of node of triangular finite element (Figure 6). We can express possible displacements of the nodes of finite element using triangular coordinates:
Swi(x,y) = Li,i = 1,2,3.
1
Li =J (ai + biX + Ciy), ai = Xi+^i+2 - xi+2yi+x,
(39)
(40)
bi = 7i + 1 yi + 2,Ci = Xi+2 Xi+1. A is the area of the triangular element; xi, yi are the coordinates of node i (Figure 3).
Figure 6. Possible displacement of the node 1 of the triangular finite element
Triangular coordinates are the natural coordinates of the triangular region. The function Li takes value of 1 in node i and values of zero in the other two nodes. Since the function Li is linear with respect to x and y coordinates, curvatures Skx and 8ky along the X and Y axes, respectively, and torsional curvature Skxy for a possible node displacement will be zero. In this case, the finite element moves as
rigid whole without deforming, and the work will be performed only by bending moments Mn normal to boundaries of the finite element on the corresponding angles of rotation of the sides of the element (Figure 6).
In Figure 6 lines 1A, AB, AC are perpendicular to the corresponding sides of the triangles 2-3,1-2, 3-1. The angles of rotation of the sides are denoted respectively - (p12, p23, (p31. The values of angles of sides rotation can be easily determined by means of geometric scheme (Figure 6):
1 BB1 cosB31 CC1 cosB12 ....
W23 = -,W12 = - = -— ,W31 =- = -—. (41)
T23 h.23 AB h-23 AC h23
The moments Mn normal to the side of the element are directed oppositely with respect to internal moments and are expressed through the bending and twisting moments in the following form:
Mn = —Mx sin2 atj — My cos2 atj + 2Mxy sin atj cos a^. (42)
In (42) aij is the angle between the side i — j of the element and the Xaxis.
x
j xI . yj—yi cos aij =—-, sin aij =—-. (43)
i
Using (42), (43), we obtain the expressions for the nodal moments directed perpendicular to the sides of the finite element (Figure 5):
Mii- -2 = -Mx,i 2 sin2 ai2 -My,i 2 cos2 ai2 + 2Mxy,i sin ai2 cos ai2,
Mb- -i = -Mx,i sin2 a3i -My,i cos2 a3i + 2Mxy,i sin a3i cos a3i,
Mh .2= - Mx,2 sin2 ai2 - My,2 cos2 ai2 + 2Mxy,2 sin ai2 cos ai2,
M$2. 3= - Mx,2 sin2 a23 - My,2 cos2 a23 + 2Mxy,2 sin a23 cos a23,
Ml2. 3= - Mx,3 sin2 a23 - My,3 cos2 a23 + 2Mxy,3 sin a23 cos a23,
Mh- i= - Mx,3 sin2 a3i - My,3 cos2 a3i + 2Mxy,3 sin a3i cos a3i .
(44)
Considering the piecewise constant approximation of the moments, the expression for the work of internal forces on possible displacement of node 1 (Figure 6) can be written in the following form:
SU£Z = \MJnii-2<Pi2k2+\Mini3-1<P3lhl+\M2nil-2<Pl2k2 1 M 2,2-3 V 23^23 — 11 -Mn,2-3V23k3 +-M33-1W31I31.
Substituting (44) into (45), we obtain the expression for 5U1^Z:
Mx,1
SUlz =TT~(—k2 sin2ai2 cos P31 — hisin2a3i cos fa) +
2h23
112 cos2 ai2 cos P31 — I31 cos2 a3i cos fa) +
(45)
MXy,!
2h
23
jk -^XXLf-i *in2„ _7 ci vi 2,
2h23
2h23
(2112 sin a12 cos a12 cos ß31 + 2l31 sin a31 cos a31 cos ß12) + Mx2
TTr-(-k2 sin2at2 cosß31 + I23 sin2a23) + 2h23
M
(-112 cos2 a-12 cos ß31 + I23 cos2 a23) + (46)
2h
23
Mxy2
—T~ (2l 12 sin ai2 cos ai2 cos ß3i - 2I23 sin a23 cos a23) +
2h23
Mx,3
—X- (—3i sin2a3i cos ßi2 + I23 sin2a23) + 2h23
TT^i-ki cos2 a3i cosßi2 + I23 cos2 a23) +
2h23
Mxy3
—T~ (2hi sin a3i cos a3i cos ßi2 - 2I23 sin a23 cos a23).
2h23
To obtain the expression for the work of internal forces for possible displacement of node 2, we need to perform cyclic replacement of the indices in expression (46): replace index 1 by index 2, index 2 by 3, and index 3 replaced by index 1. Then we obtain the expression for 8U2Z :
Mv
su2,z =ТГ2(-123 sin2a23 cos fa - 1хг sin2 a 12 cosfa) +
2h31
(-123 cos2 а23 cos fa - I12 cos2 а-12 cosfa) +
M.
y,2
2h
31
M.
xy,2
2h
(2123 sin а23 cos a23 cos fa + 2I12 sin ai2 cos ai2 cos fa) +
31
M
2h M
x'3 (-I23 sin2a23 cos P12 + l3isin2a3i) +
31
2h
y 3
' (—23 cos2 a23 cos P12 + I31 cos2 a3i) +
(47)
31
M
xy,3
2h
(2l23 sin a23 cos a23 cos fa - 2I31 sin a3i cos a3i) +
31
Mx,1
(-112 sin2ai2 cos fa + I31 sin2a3i) +
31
2h M
Tryl-(-li2 cos2 ai2 cos fa + I31 cos2 a3i) + 2h
31
Mxy-
2h
(2112 sin a12 cos a12 cos p23 - 2l31 sin a31 cos a31).
31
Performing the cyclic replacement of the indices again, we obtain an expression for SU3kz:
M
sUlz = ^r3(-l3i sin2a3- cos fa - I23 sin2a23 cosfa) +
2h
12
M
2h
y 3
' (-3- cos2 a3- cos fa - I23 cos2 a23 cosfa) +
12
M
xy,3
2h
(2I31 sin a3- cos a3- cos fa + 2I23 sin a23 cos a23 cos fa) +
12
Mx-
(-131 sin2a3- cos fa + 1-2 sin2a-2) +
12
2h M
TTyl-(-l3i cos2 a3- cos fa + 1-2 cos2 a-2) + 2h
(48)
12
Mxy-
2h
(2l31 sin a31 cos a31 cos p23 - 2l12 sin a12 cos a12) +
12
M
x,2
2h M
(-123 sin2a23 cos fa + I12 sin2a12) +
12
2h
y'2
' (-23 cos2 a23 cos fa + I12 cos2 a12) +
12
M
xy,2
2h
(2l23 sin a23 cos a23 cos fa - 21^ sin a12 cos a12).
12
Let us unite the node moments in the global coordinate system XOY for the triangular finite element with the index k into the vector [Mk].
[Mk]T = (Mx,i Myi Mxyi Mx,2 My2 Mxy,2 Mx,3 Myi3 Mxy,s). (49)
We also introduce vector combining the values of the work of internal forces for possible displacements of nodes of finite element:
{suk}=
5U{Z SUk2,z
suk
(50)
'3, zj
Then we can write the matrix expression:
[ sukk] = [Lk][Mk]. (51)
The matrix [Lk] has 3 rows and 9 columns. The elements of the matrix [Lk] are coefficients under the corresponding moments in the expressions (46)-(48). From matrices [Lk] for finite elements, in accordance with the numbering of nodes and elements, global "equilibrium" matrix [ L] is formed for the whole system.
The global equilibrium matrix [L] for the whole system will have "tape-line" structure of non-zero elements. The number of rows of the matrix [L] is equal to the number of unfixed nodes of the system n. Indexes for unknown moments are assigned in accordance with the numbering of nodes. Therefore, the width of the "tape" of non-zero elements of the matrix [L] will be determined by the maximum difference of the node numbers of all finite elements adjoining to the node. For example, for node 13 in Figure 2 the maximum difference of the indexes of nodes, the finite elements adjoining it, mv 13 = 18-7 = 11. Accordingly, the width of the "tape" of non-zero elements for the row of the matrix [L] corresponding to node 13, mL13 = (mp13 + 1)3 = 36. After calculating the width of the "tape" mLii of the matrix [L] for each of the rows, the maximum value mLmax is determined. Thus, for elements of the matrix [L], we can use a rectangular array consisting of mLrnax columns and n lines.
In addition, it is necessary to form an array Ind, consisting of two columns. In the first and second columns, respectively, the minimum and maximum indexes of finite element nodes, for each row of the matrix [L], are stored. For example, for the row corresponding to node 13 in Figure 2, we obtain: Ind[13,1] = 7,Ind[13,2] = 18. The Ind array is used in the construction of the matrix multiplication algorithm for calculating the elements of the matrix [K"] = [L][D]-1[L]T, and, also, for determining the width of the "tape" of non-zero elements of matrix - mK.
The potential of external concentrated and evenly distributed loads for possible displacement of node i along the global coordinate axis is determined by formula (52).
SVi=Pi+1qkAk =Rt. (52)
Pi - the force concentrated at node ; q k - uniformly distributed by element load. The generalized forces Ri are placed in the vector [F] (see (12)).
3. Results and Discussion
According to the program developed in MathCad 14.0, calculations of square and rectangular plates were performed on the action of concentrated force and distributed load (Figure 7). Hinged-supported and clamped plates were calculated in accordance with the Kirchhoff theory. In the calculations, grids of rectangular and triangular finite elements were used. The plate has the following parameters: side length - 6m or 3m, thickness - 1m, modulus of elasticity E = 10000 kN/m2, Poisson's ratio ^ = 0.3, uniformly distributed load q = 10 kN/m2.
As consequence of the symmetry of the problem, the quarter of the plate was calculated, with the corresponding boundary conditions on the sides lying on the symmetry axes. At the nodes lying on the axes of symmetry, the torques were assumed to be zero. The bending moments directed along the plate boundary and moments directed perpendicular to the boundary were assumed to be zero in the support nodes of the hinged plates.
In [23] analytical solutions of the plate bending problems for various types of the support of sides for distributed and concentrated loads are presented. The displacements are expressed in the form:
QQ.4 PQ.2 о
w = а— ,w = а—, and moments in the form: M = Bqa2,M = BP. The coefficients forthe variants
D D
Et
3
of the calculation schemes are given in Table 1. D = ^ 2 - bending stiffness of a plate. Table 1. Coefficients from [23] for the determination of analytical values
Sizes of sides Support Load а My1 B Mxl B М3 B М2 B
a = b hinged q 0.00406 0.0479 0.0479 - -
a = b clamped q 0.00126 0.0231 0.0231 -0.0513 -0.0513
a = b clamped P 0.0056 - - -0.1257 -0.1257
b — = 2 a clamped q 0.00254 0.0412 0.0158 -0.0571 -0.0829
Figure 7. Finite element grids for square and rectangular plates: a) 5x5 square grid of finite elements; b) 5x5 grid of isosceles triangular finite elements; c) grid of 5x5 rectangular finite elements; d) grid of 5x5 elongated triangular finite elements
The results of calculations of plates according to the proposed method were compared with the results obtained by the finite elements method in the LIRA-SAPR program and analytical solutions. A comparison of solutions is shown in Figures 8-15 and in Tables 2-16. In the figures, the green line shows the analytical solutions from [23]. In the tables and in the figures, the grid parameters are indicated for the whole plate.
17.6)
17.4
17.2
Figure 8. Hinged-supported square plate - rectangular finite elements (Figure 7a), q = 10 kN/m2: a) displacement of the center of the plate - w; b)the bending moment at the center of the plate is M1. 1 - the solution in stresses; 2 - the solution in displacements (LIRA-SAPR)
50 60
Figure 9. Hinged-supported square plate - triangular finite elements (Figure 7a), q = 10 kN/m2: a) displacement of the center of the plate - w; b) the bending moment at the center of the plate is Mi. 1 - the solution in stresses; 2 - the solution in displacements (LIRA-SAPR)
Table 2. The square plate with hinged supports, divided by the square finite elements (Figure 7a), q = 10 kN/m2.
Grid Solution in stresses Solution in displacements LIRA-SAPR
w, mm Mi, kNm/m mL,ma. mK n w, mm Mi, kNm/m ™K n
10x10 59.342 17.4523 41 15 25 58.097 16.8897 21 85
20x20 57.950 17.2919 71 25 100 57.643 17.1518 36 320
30x30 57.695 17.2625 101 35 225 57.559 17.2003 51 705
40x40 57.606 17.2523 131 45 400 57.530 17.2173 66 1240
50x50 57.565 17.2475 161 55 625 57.516 17.2251 81 1925
60x60 57.542 17.2449 191 65 900 57.509 17.2294 96 2760
Exact 57.458 17.244 57.458 17.244
Error, % 0.15 0.005 0.09 0.08
Table 3. The square plate with hinged supports, divided by the triangular finite elements (Figure 7b), q = 10 kN/m2.
Grid Solution in stresses Solution in displacements LIRA-SAPR
w, mm Mi, kNm/m mL,ma mk n w, mm Mi, kNm/m mK n
10x10 57.653 16.9602 41 15 25 56.910 16.9338 21 85
20x20 57.531 17.1260 71 25 100 57.348 17.1573 36 320
30x30 57.510 17.1777 101 35 225 57.423 17.2002 51 705
40x40 57.502 17.2002 131 45 400 57.456 17.2161 66 1240
50x50 57.499 17.2120 161 55 625 57.469 17.2238 81 1925
60x60 57.497 17.2191 191 65 900 57.476 17.2281 96 2760
Exact 57.458 17.244 57.458 17.244
Error, % 0.07 0.14 0.03 0.09
i7i_i_i_i_i_i" 7i(i_i_i_i_i_i" ]ni_i_i_i_i_i"
10 20 30 40 50 60 ' 10 20 30 40 50 60 10 20 30 40 50 60
a) b) c)
Figure 10. Clamped square plate - rectangular finite elements (Figure 7a), q = 10 kN/m2: a) displacement of the center of the plate - w; b) the bending moment at the center of the plate is M1. c) the bending moment at the clamped side is M2.1 - the solution in stresses; 2 - the solution
in displacements (LIRA-SAPR)
w Ml -M2
Figure 11. Clamped square plate - triangular finite elements (Figure 7b), q = 10 kN/m2: a) displacement of the center of the plate - w; b) the bending moment at the center of the plate is Mi. c) the bending moment at the clamped side is М2. 1 - the solution in stresses; 2 - the solution in displacements (LIRA-SAPR)
Table 4. Clamped square plate - rectangular finite elements (Figure 7a), q = 10 kN/m2.
Grid Solution in stresses Solution in displacements LIRA-SAPR
w, mm М1, kNm/m М2, kNm/m w, mm М1, kNm/m М2, kNm/m
10х10 20.293 8.66832 -17.65748 18.257 7.91934 -11.4736
20х20 18.537 8.36097 -18.25078 17.997 8.16432 -14.7482
30х30 18.193 8.29848 -18.37533 17.947 8.20962 -15.9398
40х40 18.069 8.27586 -18.42045 17.930 8.22546 -16.5550
50х50 18.012 8.26521 -18.44167 17.922 8.23280 -16.9304
60х60 17.980 8.25936 -18.45331 17.917 8.23678 -17.1833
Exact 17.832 8.316 -18.468 17.832 8.316 -18.468
Error, % 0.83 0.68 0.08 0.47 0.95 6.96
Figure 11. Clamped square plate - triangular finite elements (Figure 7b), q = 10 kN/m2: a) displacement of the center of the plate - w; b) the bending moment at the center of the plate is М1. c) the bending moment at the clamped side is М2. 1 - the solution in stresses; 2 - the solution in
displacements (LIRA-SAPR).
Table 5. Clamped square plate - triangular finite elements (Figure 7b), q = 10 kN/m2.
Grid Solution in stresses Solution in displacements LIRA-SAPR
w, mm М1, kNm/m М2, kNm/m w, mm М1, kNm/m М2, kNm/m
10x10 19.921 8.56379 -17.86275 17.251 7.90791 -12.3842
20x20 18.449 8.30952 -18.34177 17.826 8.17397 -15.2137
30x30 18.154 8.26870 -18.42369 17.875 8.21509 -16.2554
40x40 18.048 8.25636 -18.45019 17.890 8.22875 -16.7944
50x50 17.998 8.25136 -18.46175 17.897 8.23494 -17.1235
60x60 17.970 8.24898 -18.46776 17.900 8.23827 -17.3453
Exact 17.832 8.316 -18.468 17.832 8.316 -18.468
Error, % 0.77 0.81 0.001 0.38 0.93 6.08
Figure 12. Clamped square plate - the action of concentrated force in the center P = 10 kN: a) square finite elements - displacement of the center of the plate w; b) triangular finite elements -displacement of the center of the plate w; c) square finite elements - moments in clamped side М2; d) triangular finite elements - moments in clamped side М2. 1 - solution in stresses;
2 - solution in displacements LIRA-SAPR
Table 6. The clamped square plate. The action of the concentrated force P = 10 kN.
Grid Solution in stresses Solution in displacements LIRA-SAPR
square elements triangular elements square elements triangular elements
w, mm M2, kNm/m w, mm M2, kNm/m w, mm M2, kNm/m w, mm M2, kNm/m
10x10 2.7351 -1.17795 2.5213 -1.20274 2.2548 -0.8901 2.1160 -0.96346
20x20 2.3638 -1.23115 2.3044 -1.24078 2.2115 -1.0659 2.1833 -1.09814
30x30 2.2826 -1.24479 2.2546 -1.24980 2.2138 -1.1262 2.1945 -1.14827
40x40 2.2517 -125012 2.2352 -1.25319 2.2108 -1.1604 2.1994 -1.17446
50x50 2.2365 -1.25273 2.2257 -1.25481 2.2093 -1.1792 2.2017 -1.19056
60x60 2.2279 -1.25420 2.2202 -1.25569 2.2084 -1.1922 2.2030 -1.20686
Exact 2.2015 -1.257 2.2015 -1.257 2.2015 -1.257 2.2015 -1.257
Error, % 1.1 0.22 0.85 0.1 0.31 5.16 0.07 3.99
Figure 13. The displacements of the center of rectangular clamped plate (Figure 7c, 7d) under the action of load q = 10 kN/m2: a) rectangular finite elements; b) triangular finite elements. 1 - solution in stresses; 2 - solution in displacements LIRA-SAPR
Figure 14. Bending moments in rectangular clamped plate under the action of load q = 10 kN/m2 for rectangular grid of finite elements (Figure 7c)
Figure 15. Bending moments in rectangular clamped plate under the action of load q = 10 kN/m2 for triangular grid of finite elements (Figure 7d)
Table 7. The clamped rectangular plate (Figure 7c) - rectangular finite elements. Load action
q = 10 kH/m2
Grid Solution in stresses Solution in displacements LIRA-SAPR
w, mm Мз, kNm/m М2, kNm/m My1, kNm/m Mxl, kNm/m w, mm Мз, kNm/m М2, kNm/m My1, kNm/m Mxl, kNm/m
10х10 2.4658 -7.41651 -3.97326 3.81770 1.42455 2.2747 -5.30885 -2.04139 3.56671 1.39448
20х20 2.2990 -7.45081 -4.73785 3.73492 1.42046 2.2491 -6.34108 -3.33420 3.66967 1.41457
30х30 2.2668 -7.45332 -4.93844 3.71812 1.42127 2.2443 -6.70369 -3.87049 3.68882 1.41906
40х40 2.2554 -7.45663 -5.01715 3.71203 1.42178 2.2426 -6.88858 -4.16096 3.69544 1.42065
50х50 2.2500 -7.45716 -5.05563 3.70917 1.42207 2.2418 -7.00068 -4.34272 3.69850 1.42139
60х60 2.2471 -7.45743 -5.07722 3.70759 1.42225 2.2414 -7.07590 -4.46709 3.70017 1.42180
Exact 2.2467 -7.461 -5.139 3.708 1.422 2.2467 -7.461 -5.139 3.708 1.422
Error, % 0.02 0.05 1.2 0.01 0.02 0.24 5.16 13.07 0.2 0.01
Table 8. The clamped rectangular plate (Figure 7d) - triangular finite elements. Load action q = 10 kH/m2
Grid Solution in stresses Solution in displacements LIRA-SAPR
w, mm Мз, kNm/m M2, kNm/m My1, kNm/m Mx1, kNm/m w, mm Мз, kNm/m М2, kNm/m Му1, kNm/m Мх1, kNm/m
10x10 2.46209 -7.47306 -4.0175 3.88984 1.55522 2.20474 -5.55454 -2.21824 3.56493 1.48789
20x20 2.29638 -7.46733 -4.7710 3.75561 1.45769 2.23251 -6.48844 -3.39414 3.67033 1.45181
30x30 2.26548 -7.46316 -4.9588 3.72829 1.43891 2.23718 -6.80927 -3.91612 3.69106 1.43924
40x40 2.25459 -7.46118 -5.0306 3.71818 1.43215 2.23871 -6.97098 -4.20201 3.69753 1.43348
50x50 2.24952 -7.46013 -5.0652 3.71332 1.42894 2.23938 -7.06830 -4.38058 3.70025 1.43034
60x60 2.24676 -7.45951 -5.0843 3.71060 1.42714 2.23973 -7.13339 -4.50216 3.70160 1.42843
Exact 2.2467 -7.461 -5.139 3.708 1.422 2.2467 -7.461 -5.139 3.708 1.422
Error, % 0.0003 0.02 1.06 0.07 0.14 0.3 4.39 12.4 0.17 0.45
Table 9. Errors in calculating bending moments, expressed in percent
Grid Support Load Solution in stresses Solution in displacements LIRA-SAPR
Мз М2 Му1 Мх1 Мз М2 Му1 Мх1
Figure 7а hinged q - - 0.005 0,005 - - 0,08 0,08
Figure 7b hinged q - - 0.14 0.14 - - 0.09 0.09
Figure 7а clamped q 0.08 0.08 0.68 0.68 6.96 6.96 0.95 0.95
Figure 7b clamped q 0.001 0.001 0.81 0.81 6.08 6.08 0.93 0.93
Figure 7а clamped p 0.22 0.22 - - 5.16 5.16 - -
Figure 7b clamped p 0.1 0.1 - - 3.99 3.99 - -
Figure 7c clamped q 0.05 1.2 0.01 0.02 5.16 13.07 0.2 0.01
Figure 7d clamped q 0.02 1.06 0.07 0.14 4.39 12.4 0.17 0.45
By beginning we estimate the convergence of displacements determined by the proposed calculation method in stresses and by the finite element method in displacements:
- for all the considered variants of calculation schemes and loads, the displacements obtained in the stresses, when crushing the grid, tend to the exact values from above - the proposed solution is more flexible and has the property of convergence of displacements from above;
- the solution in displacements (LIRA-SAPR) tends to the exact value from above if rectangular elements are used, and from below, if triangular elements are used - in one case the solution is more flexible, in the other case it is more rigid.
Analysis of the obtained results allows us to do the following conclusions about the accuracy of calculating bending moments (Table 9):
- the proposed solution method in stresses makes it possible to obtain the values of the bending moments with greater accuracy than the finite element method in displacements (LIRA-SAPR) for all the considered plate bending problems;
- the greatest error in calculating the moments for solving in stresses is 0.8 %, and for the finite element method in displacements (LIRA-SAPR), the error in calculating the moments in the clamped sides reaches 13 %;
- when using elongated rectangular and triangular elements (Figure 7c, d), the accuracy of calculating the bending moments according to the proposed technique in stresses is reduced a little;
- when calculating by the finite element method in displacements (LIRA-SAPR), the moments are determined for the centers of the finite elements, but for calculation in stresses - for the finite elements nodes;
It is also necessary to note the following computational features of obtaining the solution in stresses:
- additional computational costs associated with matrix multiplication [L] [D]-1[L]T are necessary to form the resolving system of linear equations. Since the matrix [D]-1 is block-diagonal (block size 3x3), and the matrix [L] has form of tape (the width of the tape of non-zero elements mLmax of Tables 2-3), we can use the special algorithm of multiplication of matrices, allowing to reduce computational costs;
- the resolving system of linear equations, in comparison with the solution by the finite element method in displacements, has a smaller number of unknowns n and a smaller width of the tape of non-zero elements mK (Tables 2-3).
The paper [29] compares the results of calculations of clamped plates on the action of a uniformly distributed load for four types of triangular finite elements: BZIC is uncoordinated element of the displacement method [24, 30, 31]; TEC is element with forced compatibility of the inclination angles of the normal [26]; DKT is discrete Kirchhoff theory triangular element [30-31]; HSM is triangular hybrid finite element [33]. In Tables 10-11, the results of calculations for clamped square and rectangular plates, with a ratio of sides of 2 to 1, obtained for action uniformly distributed load, are compared. In Tables 10-11 shows the results of calculations for finite element grid 11 x 11 (for quarter of a plate), which corresponds to an 22 x 22 grid for the whole plate. In the tables, the solutions obtained by the proposed finite element method in stresses is designated FEM-S.
Table 10. A clamped square plate under the action of a uniformly distributed load. A grid of 11x11 for a quarter of the plate
Type of element wD/qa4 Error, % M1 /qa2 Error, % M2/qa2 Error, %
BZIC 0,001277 + 1.3 0,02374 +2.8 -0,05522 +7.6
TEC 0,001269 +0.7 0,02673 + 15.7 0,05455 +6.3
DKT 0.001279 + 1.5 0.02331 +0.91 -0.05212 + 1.6
HSM 0.001269 +0.7 0.02325 +0.65 -0.05068 -1.2
FEM-S 0.001297 +2.9 0.023045 -0.24 -0.05102 -0.55
Exact 0.00126 0.0231 -0.0513
Table 11. A clamped rectangular plate (b / a = 2) under the action of a uniformly distributed load. A grid of 11x11 for a quarter of the plate
Type of element wD/qa4 Error, % Mi,x/qa2 Error, % M1:y/qa2 Error, %
BZIC 0.002583 + 1.7 0.04216 +2.33 0.01495 -5.4
TEC 0.002551 +0.4 0.04666 + 13.2 0.01888 + 19.5
DKT 0.002566 + 1.0 0.04188 + 1.65 0.01650 +4.4
HSM 0.002541 0 0.04193 + 1.77 0.01697 +7.4
FEM-S 0.002585 + 1.8 0.041636 + 1.06 0.016132 +2.1
Exact 0.00254 0.0412 0.0158
Comparison of the values in Tables 10-11 allows us to have the following conclusions:
- rate of convergence of displacements obtained by the proposed method at stresses lower than the rate of convergence of displacements for finite elements TEC, DKT, HSM; for elongated finite elements rate of convergence of displacements, obtained by method in stresses, is near to the rate of convergence of displacements for the BZIC element (Table 11);
- the rate of convergence of the bending moments obtained by the proposed method in the stresses is higher than the rate of convergence of the moments obtained by the method with other considered finite elements;
- the use of elongated finite elements (Table 11) to a lesser degree leads to an increase in the error in calculating the moments by the proposed solution method in stresses, then for other considered finite elements; especially, this applies to bending moments in clamped sides of plate.
Note, that in constructing solution by the proposed technique in stresses, unlike traditional finite element method, we do not form stiffness matrix or flexible matrix for the finite element. A matrix of coefficients of system of linear algebraic equations for the whole system is directly formed. This matrix can be considered a stiffness matrix of the system, since it connects the displacements of nodes with external forces. When constructing the solutions by the proposed method, there is no need to apply any approximating functions for displacements at the finite element domain. For the solution, only approximations of the fields of possible displacements are used, which can be arbitrary, satisfying the kinematic boundary conditions. This approach is general and is applicable to solving various problems of building mechanics: flat and volume problems of the theory of elasticity, shell calculations, calculation of rod systems [20, 21].
4. Conclusion
1. The presented method of calculating the bent plates according to the Kirchhoff theory by the finite element method in stresses has property of convergence from above - the displacements obtained by this method converge to the exact values from above.
2. The solution by method in stresses requires large computational costs when obtaining elements of matrix of the resolving system of linear algebraic equations, but the number of unknowns and the width of the tape of non-zero elements is less than when solving by the finite element method in displacements.
3. When solving by method in stresses, the values of moments is determined directly at nodes of finite element grid, and not at the centers of the finite elements, which allows obtaining more accurate values of the moments, especially in the clamped nodes.
4. The proposed method of solving in stresses is based on the fundamental principles of structural mechanics - the principle of minimum of additional energy and the principle of possible displacements. Constant functions are used to approximate the forces, and linear ones are used for possible displacements, which ensures the convergence of the approximate solution to the exact solution, when we crush the finite element grid.
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Yury Tyukalov,
+7(912)8218977; [email protected]
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Юрий Яковлевич Тюкалов, +7(912)8218977; эл. почта: [email protected]
© Tyukalov, Yu.Ya., 2018
