Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads
Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads are derived by using a novel double finite integral transform method. Since only the basic elasticity equations for orthotropic thin plates are used, the method presented in this paper eliminates t...
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| Zitieren: | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads / Rui Li, Yang Zhong, Bin Tian, Jian Du // Прикладная механика. — 2011. — Т. 47, № 1. — С. 131-143. — Бібліогр.: 30 назв. — англ. |
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| author_facet | Rui Li Yang Zhong Bin Tian Jian Du |
| citation_txt | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads / Rui Li, Yang Zhong, Bin Tian, Jian Du // Прикладная механика. — 2011. — Т. 47, № 1. — С. 131-143. — Бібліогр.: 30 назв. — англ. |
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| description | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads are derived by using a novel double finite integral transform method. Since only the basic elasticity equations for orthotropic thin plates are used, the method presented in this paper eliminates the need to predetermine the deformation function and is hence completely rational thus more accurate than conventional semi-inverse methods, which presents a breakthrough in solving plate bending problems as they have long been bottlenecks in the history of elasticity. Numerical results are presented to demonstrate the validity and accuracy of the approach as compared with those previously reported in the literature.
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2011 П РИКЛАДНАЯ МЕХАНИКА Том 47, № 1
ISSN0032–8243. Прикл. механика, 2011, 47, № 1 131
R u i L i
1
, Y a n g Z h o n g
1
, B i n T i a n
1
, J i a n D u
2
EXACT BENDING SOLUTIONS OF ORTHOTROPIC RECTANGULAR
CANTILEVER THIN PLATES SUBJECTED TO ARBITRARY LOADS
1School of Civil Engineering, Dalian University of Technology, Dalian 116024,
Liaoning Province, PR China; e-mail: lirui19851985@yahoo.cn;
2Institute of Advanced Manufacturing Technology, Dalian University of Technology,
Dalian 116024, Liaoning Province, PR China
Abstract. Exact bending solutions of orthotropic rectangular cantilever thin plates sub-
jected to arbitrary loads are derived by using a novel double finite integral transform
method. Since only the basic elasticity equations for orthotropic thin plates are used, the
method presented in this paper eliminates the need to predetermine the deformation function
and is hence completely rational thus more accurate than conventional semi-inverse meth-
ods, which presents a breakthrough in solving plate bending problems as they have long
been bottlenecks in the history of elasticity. Numerical results are presented to demonstrate
the validity and accuracy of the approach as compared with those previously reported in the
literature.
Key words: Rectangular cantilever thin plate; exact solution; finite integral transform
1. Introduction.
Orthotropic rectangular thin plates are widely used in various engineering applications
such as decks of contemporary steel bridges, corrugated plates and reinforced concrete slabs
stiffened by orthogonal ribs. The bending of orthotropic, especially isotropic rectangular
thin plates with various combinations of boundary conditions has been investigated for
many years by different authors. It is well known that explicit analytic solutions of
orthotropic or isotropic rectangular thin plates are available only for the cases with two op-
posite sides simply supported (i.e. Navier’s solution, Levy’s solution, etc) while it is, so far,
difficult to get the solutions which exactly satisfy both the partial differential equation and
other boundary conditions of a plate. Accordingly, various methods have been studied. One
of the most commonly used methods for exact bending solutions of isotropic plates is the
superposition method, which could be extended to orthotropic plates [1 – 5]. Meanwhile, the
technique of Fourier series expansion is another procedure for solving complex structures
[6] as well as accurate bending solutions of plates [7]. Besides, a number of numerical
methods have been utilized by many researchers to analyze problems of plates and shells [8,
9] such as the finite difference method [10 – 13], finite element method [14, 15], finite strip
method [16], integral equation method [17], method of discrete singular convolution [18],
method of differential quadrature [19], differential quadrature element method [20], mesh-
less method [21] and spline element method [22].
Cantilever thin plate is an important structural element while its bending has been one
of the most difficult problems in the theory of elastic thin plate. Some approximate methods
have been utilized for the problem of isotropy. The method of finite difference was firstly
used to solve a cantilever plate with concentrated edge load by Holl [10]. The problem is
also solved by Barton [11], Macneal [12], Livesly and Birchall [23] separately with the
same method. Besides, some other approximate analysis of the bending of a rectangular
132
cantilever plate by uniform normal pressure was presented by Nash [13]. The generalized
variational principle was applied to rectangular thin plates by Shu and Shih [24], and the
principle was then used by Plass et al. [25] for deflection and vibration problems of cantile-
ver plates. Leissa and Niedenfuhr [26] obtained the solution for uniformly loaded cantile-
vered square plates using the technique of point matching and the Rayleigh – Ritz method.
In addition, Chang [3 – 5] derived exact solutions for the bending of both uniformly loaded
and concentrated loaded isotropic rectangular cantilever plates by using the method of su-
perposition, which involved a skillful superposition of several problems, yet used smart trial
functions.
Integral transform is one of the best approaches to obtain exact solutions of some partial
differential equations in the theory of elasticity [27]. The method has been often adopted to
analyze some structural engineering problems [28]. However, to the authors’ knowledge,
there have been no reports on the analysis for an orthotropic or isotropic cantilever thin plate
using finite integral transform.
In the present paper, a novel double finite integral transform method is adopted to ac-
quire exact bending solutions of orthotropic rectangular cantilever thin plates under arbitrary
loading. Unlike the traditional semi-inverse approaches in classical plate analysis employed
by Timoshenko [1] and others such as Chang [3 – 5], where a trial deflection function has to
be predetermined, the analysis in here is completely rational without any trial functions. The
procedure of solution presented here enables one to attempt exact solutions for more prob-
lems of plates which have to hitherto be analyzed using semi-inverse method or approximate
approaches. It can be not only applied to other combinations of boundary conditions but also
further extended to the problems of moderately thick plates as well as buckling, vibration,
etc., some of which will be reported in future. To verify the accuracy of the approach in this
paper, several cases of a rectangular cantilever thin plate are examined and the results are
presented for an easy comparison with those found in the previous literatures. Excellent
agreement is observed thus confirming the accuracy and applicability of the present method.
2. Integral transform and exact bending solutions for an orthotropic rectangular
cantilever thin plate.
The coordinate system of a thin, orthotropic rectangular cantilever plate under consid-
eration is illustrated in Fig. 1, where 0 x a≤ ≤ and 0 y b≤ ≤ . The governing partial differ-
ential equation for bending of the plate for which the principal directions of orthotropy coin-
cide with the x and y axes [1, 29] is
4 4 4
4 2 2 4
2x y
W W W
D H D q
x x y y
∂ ∂ ∂
+ + =
∂ ∂ ∂ ∂
, (1)
where W is the transverse deflection of plate midplane, q is the distributed transverse load,
xD and yD are the flexural rigidities about the y and x axes,
1
2 xyH D D= + is called the
effective torsional rigidity in which
1 2 1x yD D Dν ν= = is defined in terms of the reduced
Fig. 1. An orthotropic rectangular cantilever thin plate
133
Poisson's ratios
1
ν and
2
ν , respectively.
The internal forces of the plate are
2 2
12 2x x
W W
M D D
x y
∂ ∂
= − +
∂ ∂
; (2)
2 2
12 2y y
W W
M D D
y x
∂ ∂
= − +
∂ ∂
; (3)
2
2xy xy
W
M D
x y
∂
= −
∂ ∂
; (4)
2 2
2 2x x
W W
Q D H
x x y
∂ ∂ ∂
= − +
∂ ∂ ∂
; (5)
2 2
2 2y y
W W
Q D H
y y x
∂ ∂ ∂
= − +
∂ ∂ ∂
; (6)
xy
x x
M
V Q
y
∂
= +
∂
; (7)
xy
y y
M
V Q
x
∂
= +
∂
, (8)
where xM , yM , xyM , xQ , yQ , xV and yV are the bending moments, torsional moment,
shear forces and total shear forces, respectively.
The boundary conditions of the plate can be expressed as
0
0xW = = ; (9)
0
0
x
W
x =
∂
=
∂
; (10)
0
0; 0; 0x y yx a y y b
M M M
= = =
= = = ; (11, a – c)
0
0; 0; 0x y yx a y y b
V V V
= = =
= = = ; (12, a – c)
2 2
, 0 ,
2 0; 2 0xy xy
x a y x a y b
W W
D D
x y x y
= = = =
∂ ∂
− = − =
∂ ∂ ∂ ∂
. (13, a, b)
In the particular case of isotropy, we have
1 2
ν ν ν= = , x yD D H D= = = ,
1
D Dν= and
(1 ) / 2xyD Dν= − , where D is the flexural rigidity and ν is Poisson’s ratio. Hence the
above equations can reduce to those of an isotropic plate.
To solve the partial differential equation Eq. (1), a double finite integral transform ap-
proach is utilized. Since ( ),W x y , defined within a rectangular domain 0 x a≤ ≤ and
0 y b≤ ≤ , is a function of the independent variables x and y , we define a double finite
integral transform by the equation
134
( ) ( )
0 0
, sin cos 1, 3, 5, ; 0, 1, 2...
2
a b
m
mn nW W x y x ydxdy m n
α
β= = =∫ ∫ � . (14)
The inversion formula can be represented as
( )
1,3,5, 0,1,2,
4
, sin cos
2
m
mn n n
m n
W x y W x y
ab
α
δ β
∞ ∞
= =
= ∑ ∑
� �
, (15)
where
m
m
a
π
α = ; n
n
b
π
β = and
1 2, 0;
1, 1,2,3, .
n
if n
if n
δ
=
=
= �
The double integral transforms of higher-order partial derivatives of W appeared in Eq.
(1) are derived respectively as follows:
4
4
0 0
sin cos
2
a b
m
n
W
x ydxdy
x
α
β
∂
=
∂∫ ∫
( ) ( )
2 3 43 21 1
2 2
3 2 0
0 0
1 1 cos
2 4 8 16
b m m
m m m m
n mnx
x ax a x
W W W
W ydy W
xx x
α α α α
β
− −
=
== =
∂ ∂ ∂
= − + − − − +
∂∂ ∂
∫ =
( ) ( )
2 43 21 1
2 2
3 2
0 0
1 1 cos
2 4 16
b m m
m m m
n mn
x ax a x
W W W
ydy W
xx x
α α α
β
− −
== =
∂ ∂ ∂
= − + − − +
∂∂ ∂
∫ ;
(16)
4
4
0 0
sin sin
2
a b
m
n
W
x ydxdy
y
α
β
∂
=
∂∫ ∫
( ) ( )
3 3
2 2 4
3 3
0 00
1 1 sin
2
a
n n m
n n n mn
y b yy b y
W W W W
xdx W
y yy y
α
β β β
= == =
∂ ∂ ∂ ∂
= − − − − + +
∂ ∂∂ ∂
∫ ;
(17)
4
2 2
0 0
sin cos
2
a b
m
n
W
x ydxdy
x y
α
β
∂
=
∂ ∂∫ ∫
( ) ( ) ( )
2 22 21 1
2 2
0, 0, 0, , 0
1 1 1
2 2 4
m m
n n m m m n
mn
x y b x yx a y b x a y
W W W W
W
x y x y y y
α α α β− −
+
= = = == = = =
∂ ∂ ∂ ∂
= − − − + − − + −
∂ ∂ ∂ ∂ ∂ ∂
( ) ( )
2 1
2
2
0
0 00
1 sin 1 cos
4 2 2
a b m
nm m m
n nx
x ay b y
W W W
xdx W ydy
y y x
α α α
β β
−
=
== =
∂ ∂ ∂
− − − − − + =
∂ ∂ ∂
∫ ∫
( ) ( )
2 2 2 1
2
2
0 00
1 sin 1 cos ,
4 4 2
a bm
nm n m m
mn n n
x ay b y
W W W
W xdx ydy
y y x
α β α α
β β
−
== =
∂ ∂ ∂
= − − − − −
∂ ∂ ∂
∫ ∫
(18)
in which three boundary conditions, i.e. Eqs. (9) and (13, a, b), have been imposed to sim-
plify the expressions.
By performing over Eq. (1) the double finite integral transform and the substitution of
Eqs. (16) – (18), we have
135
( ) ( )
3 3 31
2
3 3 3
0 0 0
1 cos 1 sin sin
2 2
b a am
n m m
x n y y
x a y b y b
W W W
D ydy D xdx D xdx
x y y
α α
β
−
= = =
∂ ∂ ∂
− + − − − ∂ ∂ ∂
∫ ∫ ∫
( )
2 2
2 2
0 0 0
1 sin sin
2 2 2 2
a a
n m m m m
y n y n
y b y
H HW W
D xdx D xdx
y y
α α α α
β β
= =
∂ ∂
− − + + + − ∂ ∂
∫ ∫
( )
2 21
2
2
2
0 0 0
1 2 cos cos
4 2
b bm
x m x m
n n n
x a x
D DW W
H ydy ydy
x x
α α
β β β
−
= =
∂ ∂
− − + + + ∂ ∂
∫ ∫
(19)
4 2 2
4
16 2
x m m n
y n mn mn
D H
D W q
α α β
β
+ + + =
,
where ( )
0 0
, sin cos
2
a b
m
mn nq q x y x ydxdy
α
β= ∫ ∫ represents the transform of the load function
( ),q x y .
After single finite cosine and sine transforms over Eqs. (12, a) and (12, b, c) respec-
tively, we obtain
( )
3 3
3 2
0
2 cos 0
b
x xy n
x a
W W
D H D y dy
x x y
β
=
∂ ∂
+ + =
∂ ∂ ∂
∫ ;
( )
3 3
3 2
0 0
2 sin 0
2
a
m
y xy
y
W W
D H D x dx
y x y
α
=
∂ ∂
+ + =
∂ ∂ ∂
∫ ; (20, a – c)
( )
3 3
3 2
0
2 sin 0
2
a
m
y xy
y b
W W
D H D x dx
y x y
α
=
∂ ∂
+ + =
∂ ∂ ∂
∫ ,
which can be written as
( )
3
2
3
0 0
cos 2 cos ;
b b
x n xy n n
x ax a
W W
D ydy H D ydy
xx
β β β
==
∂ ∂
= + ∂∂
∫ ∫
( ) 23
3
0 0 00
2
sin sin ;
2 4 2
a a
xy mm m
y
yy
H DW W
D xdx xdx
yy
αα α
==
+∂ ∂
=
∂∂
∫ ∫ (21, a – c)
( ) 23
3
0 0
2
sin sin
2 4 2
a a
xy mm m
y
y by b
H DW W
D xdx xdx
yy
αα α
==
+∂ ∂
=
∂∂
∫ ∫ .
Substituting Eqs. (21, a – c) into Eq. (19) yields
( )
( ) 2
2
0
2
1 sin
4 2
a
xy mn m
y n
y b
H D W
D xdx
y
α α
β
=
− ∂
− − + +
∂
∫
( ) 2
2
0 0
2
sin
4 2
a
xy m m
y n
y
H D W
D xdx
y
α α
β
=
− ∂
+ + −
∂
∫
136
( ) ( )
2 21
2
2
2
0 0 0
1 2 cos cos
4 2
b bm
x m x m
xy n n n
x a x
D DW W
H D ydy ydy
x x
α α
β β β
−
= =
∂ ∂
− − + − + + ∂ ∂
∫ ∫
4 2 2
4
16 2
x m m n
y n mn mn
D H
D W q
α α β
β
+ + + =
. (22)
Let
0
0 0
sin ; sin ;
2 2
a
a
m m
m m
y b y
W W
I xdx J xdx
y y
α α
= =
∂ ∂
= =
∂ ∂
∫ ∫
2
20
0 0
cos ; cos
b
b
n n n n
x a x
W W
K ydy L ydy
x x
β β
= =
∂ ∂
= = ∂ ∂
∫ ∫
(23, a, d)
Accordingly Eq. (22) is expressed in terms of unknown constants mI , mJ , nK and nL as
4 2 2
4
1
16 2
mn
x m m n
y n
W
D H
D
α α β
β
= ×
+ +
( )
( ) ( )
( ) ( )
2 2
2 2
21
2
2
2 2
1
4 4
1 2 .
4 2
xy m xy mn
mn y n m y n m
m
x m x m
xy n n n
H D H D
q D I D J
D D
H D K L
α α
β β
α α
β
−
− −
× + − + − + +
+ − + − −
(24)
Via substitution of Eq. (24) into Eq.(15), one can get the expression of ( ),W x y with
mI , mJ , nK and nL for 1,3,5,m = � and 0,1,2n = � . Eq. (24) can meet the boundary
conditions described by Eqs. (9), (12, a – c) and (13, a, b), as indicated above.
By substituting Eq. (15) into the remaining boundary conditions represented by Eqs.
(10) and (11, a – c) observing the differentiation procedure of trigonometric series [7, 30],
we obtain
0,1,2, 1,3,5,
2
cos 0;m
n mn n
n m
W y
ab
α
δ β
∞ ∞
= =
=∑ ∑
� �
(25, a – d)
( ) ( ) ( )
21 1
2
2 2
1 1 1
0,1,2, 1,3,5,
4
1 1 1 cos 0;
4
m m
n x m
n m m x n n mn n
n m
D
D I D J D K D W y
ab
α
δ β β
∞ ∞ − −
= =
− − − + − − + =
∑ ∑
� �
( ) ( )
21
2 12
1
1,3,5, 0,1,2,
4
1 1 sin 0;
4 2
m
n m m
n y m y m n y n mn
m n
D
D I D J D K D W x
ab
α α
δ β
∞ ∞ −
= =
− − + − − + =
∑ ∑
� �
( ) ( ) ( )
21
2 1
2
1
1,3,5, 0,1,2,
4
1 1 1 sin 0.
4 2
m
n n nm m
n y m y m n y n mn
m n
D
D I D J D K D W x
ab
α α
δ β
∞ ∞ −
+
= =
− − + − − + − =
∑ ∑
� �
137
Multiplying Eqs. (25, a – b) by cos n ydyβ and following by integration from 0 to b yields
Eqs. (26, a – b). Multiplying Eqs. (25, c – d) by sin
2
m xdx
α
and following by integration
from 0 to a yields Eqs. (26, c – d). They are given as follows:
( )
1,3,5,
0 0, 1, 2 ;m mn
m
W nα
∞
=
= =∑
�
�
( ) ( ) ( ) ( )
21 1
2
2 2
1 1 1
1,3,5,
1 1 1 0 0, 1, 2
4
m m
n x m
m m x n n mn
m
D
D I D J D K D W n
α
β
∞ − −
=
− − − + − − + = =
∑
�
�
( ) ( ) ( )
21
2 1
2
1
0,1,2,
1 1 0 1, 3, 5,
4
m
n m
n y m y m n y n mn
n
D
D I D J D K D W m
α
δ β
∞ −
=
− − + − − + = =
∑
�
�
( ) ( ) ( ) ( )
21
2 1
2
1
0,1,2,
1 1 1 0 1, 3, 5,
4
m
n n nm
n y m y m n y n mn
n
D
D I D J D K D W m
α
δ β
∞ −
+
=
− − + − − + − = =
∑
�
�
(26, a – d)
Substituting Eq. (24) into Eqs. (26, a –d), we arrive finally at
( ) ( )
1
2
1,3,5, 1,3,5, 1,3,5,
1 1
m
n
m mn mn m m mn mn m m mn mn n
m m m
C R I C R J C T Kα α α
∞ ∞ ∞ −
= = =
− − + − −∑ ∑ ∑
� � �
( )2
1,3,5, 1,3,5,
0, 1, 2, ;
2
x
m mn n m mn mn
m m
D
C L C q nα α
∞ ∞
= =
− = − =∑ ∑
� �
�
(27)
( ) ( ) ( )
1 1
2 2
1 1
1,3,5, 1,3,5,
1 1 1
m m
n
mn mn mn m mn mn mn m
m m
D C E R I D C E R J
∞ ∞− −
= =
− − − − − − +
∑ ∑
� �
( ) ( )
1
1
2
1,3,5, 1,3,5,
1 1
2
m
m x
x mn mn mn n m mn mn n
m m
D
D C E T K C E Lα
∞ ∞−
−
= =
+ − − − + =
∑ ∑
� �
(28)
( )
1,3,5,
0, 1, 2mn mn mn
m
C E q n
∞
=
= =∑
�
� ;
( ) ( ) ( )
0,1,2, 0,1,2,
1
n
n y mn mn mn m n y mn mn mn m
n n
D C F R I D C F R Jδ δ
∞ ∞
= =
− − − − +∑ ∑
� �
( ) ( )
1
2
1
0,1,2, 0,1,2,
1
2
m
x
n mn mn mn n n m mn mn n
n n
D
D C F T K C F Lδ δ α
∞ ∞−
= =
+ − − + =∑ ∑
� �
(29)
0,1,2,
mn mn mn
n
C F q
∞
=
= ∑
�
( )1, 3, 5n = � ;
138
( ) ( ) ( )
0,1,2, 0,1,2,
1
n
n y mn mn mn m n y mn mn mn m
n n
D C F R I D C F R Jδ δ
∞ ∞
= =
− − − − +∑ ∑
� �
( ) ( ) ( ) ( )
1
2
1
0,1,2, 0,1,2,
1 1 1
2
m
n nx m
n mn mn mn n n mn mn n
n n
D
D C F T K C F L
α
δ δ
∞ ∞−
= =
+ − − − + − =∑ ∑
� �
(30)
( )
0,1,2,
1,3,5, ,mn mn mn
n
C F q m
∞
=
= =∑
�
�
where
( )
2 21
2 2 1
2
14 2 2
4
; 1 ; ;
4 4
16 2
m
m x m m
mn mn n mn y n
x m m n
y n
D D
C E D F D
D H
D
α α α
β β
α α β
β
−
= = + − = +
+ +
( )
( )
2 2
2 2
2
; 2 .
4 4
xy m x m
mn y n mn xy n
H D D
R D T H D
α α
β β
−
= + = + −
(31, a – e)
Eqs. (27) – (30) are four infinite systems of linear simultaneous equations with respect
to unknown constants mI , mJ , nK and nL ( 1, 3, 5, ; 0, 1, 2m n= =� � ). In practice, a fi-
nite number of terms in each set of equations are considered and the resulting sets of finite
number of simultaneous equations are solved to determine the constants.
The bending moments along the clamped edge 0x = could be obtained conveniently
using the expression
2 2 2
12 2 20
0,1,2
0 0
2
cosx x x x n n nx
nx x
W W W
M D D D D L y
bx y x
δ β
∞
=
== =
∂ ∂ ∂
= − + = − = −
∂ ∂ ∂
∑
�
. (32)
Substituting the constants mI , mJ , nK and nL obtained into Eq. (24) then (15) gives
the bending solutions of an orthotropic rectangular cantilever plate. The results are theoreti-
cally exact solutions when m and n → ∞ while in practice we only take the larger ones to
obtain desired accuracy.
3. Numerical results.
In order to verify the validity of the results derived in the paper, three cases of loading
of an isotropic rectangular cantilever thin plate with different aspect ratios are examined:
(1) a uniform load of intensity q .
(2) a concentrated load at the center of the plate.
(3) a concentrated load at the middle of the free edge x a= .
Comparison with known results of Holl [10], Nash [13] and Chang [3 – 5] including the
transverse deflections and bending moments at specific locations for cases above is pre-
sented in Tables 1 – 3, respectively, which displays excellent agreement.
For future comparison, exact bending solutions of an orthotropic cantilever thin plate
subjected to a concentrated load at the free corner ( ), 0a are obtained and some numerical
results are tabulated in Table 4. In addition, the deflection surface of the case is illustrated in
Fig. 2 for direct viewing.
For sufficient accuracy of the solutions, we take the first 50 terms of mI and mJ and the
first 100 terms of nK and nL in the calculation. It should be noted that the convergence of
the results is not so fast because of the double trigonometric series adopted. However, the
simultaneous equations can be solved without any difficulty using mathematical packages
such as MATLAB; and above all, the value of the present approach lies in its excellent abil-
ity to obtain exact bending solutions of a plate.
139
Table 1. Deflections and bending moments for a uniformly loaded isotropic rectangular can-
tilever thin plate with 0,3ν = .
a b ( )4W qa D
Ref, [3] FEM* Present
0y = 0,12933 0,12708 0,12722
0,125y a= 0,12998 0,12788 0,12797
0, 25y a= 0,13056 0,12851 0,12857
0,375y a= 0,13091 0,12892 0,12895
x a=
0,5y a= 0,13102 0,12905 0,12908
0x = 0 0 0
0,25x a= 0,011949 0,1182 0,011771
0,5x a= 0,044327 0,043221 0,043283
0.75x a= 0,085046 0,083888 0,084085
1
0y =
x a= 0,12933 0,12708 0,12722
Ref, [13] Ref, [3] Present
0y = 0,135 0,12540 0,124303
0,125y b= 0,12691 0,126015
0,25y b= 0,139 0,12784 0,127106
0,375y b= 0,12825 0,127626
1/2
x a=
0,5y b= 0,141 0,12837 0,127770
a b ( )2
xM qa
Ref, [3] FEM Present
0,125y b= -0,51270 -0,50399 -0,51240
0,25y b= -0,53353 -0,52760 -0,52959
0,375y b= -0,53550 -0,53058 -0,53135
1
0x =
0,5y b= -0,53560 -0,53092 -0,53136
Ref, [13] Ref, [3] Present
0,125y b= -0,51074 -0,51542
0,25y b= -0,5047 -0,51386 -0,51606
0,375y b= -0,51451 -0,51424
1/2
0x =
0,5y b= -0,5082 -0,51049 -0,51362
*
The finite elements solutions in this paper are adopted from Chang [3, 5], which were
offered by Mr. Wu Liang-tze of Peking University.
140
Table 2. Deflections and bending moments for an isotropic square cantilever thin plate
subjected to a concentrated load at center of plate 2, 2x a y b= = , with 0,3ν = .
( )2W Pa D
Ref. [4] Present
0y = 0,10353 0,105632
0,125y a= 0,10577 0,107802
0,25y a= 0,10773 0,109713
0,375y a= 0,10904 0,111051
x a=
0,5y a= 0,10957 0,111537
0x = 0 0
0,25x a= 0,010044 0,010181
0,5x a= 0,037122 0,037648
0,75x a= 0,069947 0,071316
0y =
x a= 0,10353 0,10563
( )xM P
Ref, [4] Present
0,125y b= -0,46448 -0,47254
0,25y b= -0,52839 -0,53571
0,375y b= -0,56979 -0,57703
0x =
0,5y b= -0,58645 -0,59030
Table 3. Deflections and bending moments for an isotropic rectangular cantilever thin plate
subjected to a concentrated load at the middle of the free edge , 2x a y b= = , with 0,3ν = .
a b ( )2W Pa D
Ref. [10] Ref. [5] Present
0y = 0,03015 0,035158 0,034363
0,125y a= 0,04993 0,04823 0,041889
0,25y a= 0,08364 0,078971 0,073564
0,375y a= 0,13594 0,12678 0,12423
1/4 x a=
0,5y a= 0,18773 0,16991 0,16882
Ref. [5] FEM Present
0y = 0,34120 0,32912 0,32908
0,125y b= 0,35006 0,33863 0,33833
0,25y b= 0,35929 0,34812 0,34774
0,375y b= 0,36769 0,35663 0,35628
x a=
0,5y b= 0,37239 0,36101 0,36104
0x = 0 0
0,25x a= 0,02386 0,023158
0,5x a= 0,09940 0,096624
0,75x a= 0,2120 0,20483
1
0y =
x a= 0,3412 0,32908
a b ( )xM P
Ref. [10] Ref, [5] Present
0,125y b= -0,12600 -0,13892 -0,11411
0,25y b= -0,22672 -0,23034 -0,22175
0,375y b= -0,37352 -0,40233 -0,39884
1/4 0x =
0,5y b= -0,49672 -0,51798 -0,51287
Ref. [5] FEM Present
0,125y b= -1,0042 -0,99064 -0,99090
0,25y b= -1,1423 -1,0819 -1,0918
0,375y b= -1,1514 -1,1186 -1,1244
1 0x =
0,5y b= -1,1571 -1,1282 -1,1271
141
Table 4. Deflections and bending moments for an orthotropic rectangular cantilever thin
plate subjected to a concentrated load at the free corner , 0x a y= = ( , 0,5 ,x y xy xD D D D= =
1 2
0,3)ν ν= = .
( )2 ,xW Pa D x a=
a b
0y = 0.25y b= 0.5y b= 0.75y b= y b=
1/4 0,3523 0,1231 0,03580 0,01104 0,004960
1/2 0,3611 0,2282 0,1371 0,08764 0,06242
1 0,4477 0,3843 0,3293 0,2857 0,2507
2 0,7512 0,7214 0,6927 0,6657 0,6403
4 1,447 1,432 1,417 1,403 1,388
( )2 , 0xW Pa D y =
a b
0x = 0.25x a= 0.5x a= 0.75x a= x a=
1/4 0 0,03354 0,1170 0,2289 0,3523
1/2 0 0,03407 0,1194 0,2343 0,3611
1 0 0,03970 0,1439 0,2875 0,4477
2 0 0,06316 0,2354 0,4781 0,7512
4 0 0,1205 0,4504 0,9186 1,447
( )2 ,xW Pa D y b=
a b
0x = 0.25x a= 0.5x a= 0.75x a= x a=
1/4 0 0,0002005 0,001156 0,002854 0,004960
1/2 0 0,002577 0,01479 0,03615 0,06242
1 0 0,01343 0,06605 0,1507 0,2507
2 0 0,04396 0,1862 0,3978 0,6403
4 0 0,1084 0,4227 0,8754 1,388
( ) , 0xM P x =
a b
0.125y b= 0.25y b= 0.375y b= 0.5y b= 0.625y b= 0.75y b= 0.875y b=
1/4 -0,7070 -0,3357 -0,1613 -0,08163 -0,04562 -0,02734 -0,01616
1/2 -1,062 -0,7267 -0,5086 -0,3720 -0,2872 -0,2276 -0,1662
1 -1,542 -1,327 -1,152 -1,026 -0,9227 -0,7983 -0,5912
2 -2,567 -2,480 -2,342 -2,255 -2,158 -1,949 -1,503
4 -4,357 -4,702 -4,733 -4,653 -4,485 -4,171 -3,450
Fig. 2. The deflection surface of an orthotropic rectangular cantilever thin plate subjected to
a concentrated load at the free corner , 0x a y= = ( )1a b = .
142
4. Conclusions.
The present paper shows that the bending problem of an orthotropic rectangular cantile-
ver thin plate can be solved accurately by a novel double finite integral transform method.
The main advantage of the approach is it does not require the preselection of a deformation
function, which, however, can scarcely be avoided in the traditional semi-inverse ap-
proaches. Also, the present approach provides an efficient procedure for more accurate re-
sults which should be of both academic and practical importance. Accordingly, it serves as a
completely rational theoretical model for the bending problems of rectangular thin plates.
The present method can be capable of dealing with any other combinations of boundary
conditions while be further extended to the problems of moderately thick plates as well as
buckling, vibration, etc., some of which will be explored in due course.
Acknowledgments This work was supported by the Fundamental Research Funds for
the Central Universities of China (2010) and the Award Young Scholar of Distinction of the
Ministry of Education of China (2010).
Р Е З Ю М Е . За допомогою нового методу подвійного скінченого інтегрального перетворення
отримано точний розв‘язок про згин ортотропної прямокутної консольної пластинки під дією довіль-
ного навантаження. Оскільки використовуються лише основні рівняння пружності для ортотропних
тонких пластинок, то немає необхідності визначати функцію деформації, що є повністю раціональ-
ним і таким чином і більш точним, ніж при застосування звичайних напів-обернених методів. Цей
метод являє собою прорив в розв’язуванні задач про згин, які були довго вузьким місцем в історії
теорії пружності. Числові результати демонструють придатність і точність підходу порівняно з опуб-
лікованими раніше.
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From the Editorial Board: The article corresponds to submitted manuscript.
Поступила 21.09.2009 Утверждена в печать 21.10.2010
|
| id | nasplib_isofts_kiev_ua-123456789-88008 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 0032-8243 |
| language | English |
| last_indexed | 2025-12-07T15:43:37Z |
| publishDate | 2011 |
| publisher | Інститут механіки ім. С.П. Тимошенка НАН України |
| record_format | dspace |
| spelling | Rui Li Yang Zhong Bin Tian Jian Du 2015-11-06T21:17:38Z 2015-11-06T21:17:38Z 2011 Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads / Rui Li, Yang Zhong, Bin Tian, Jian Du // Прикладная механика. — 2011. — Т. 47, № 1. — С. 131-143. — Бібліогр.: 30 назв. — англ. 0032-8243 https://nasplib.isofts.kiev.ua/handle/123456789/88008 Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads are derived by using a novel double finite integral transform method. Since only the basic elasticity equations for orthotropic thin plates are used, the method presented in this paper eliminates the need to predetermine the deformation function and is hence completely rational thus more accurate than conventional semi-inverse methods, which presents a breakthrough in solving plate bending problems as they have long been bottlenecks in the history of elasticity. Numerical results are presented to demonstrate the validity and accuracy of the approach as compared with those previously reported in the literature. en Інститут механіки ім. С.П. Тимошенка НАН України Прикладная механика Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads Точные решения об изгибе ортотропных консольных тонких пластинок при произвольной нагрузке Article published earlier |
| spellingShingle | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads Rui Li Yang Zhong Bin Tian Jian Du |
| title | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads |
| title_alt | Точные решения об изгибе ортотропных консольных тонких пластинок при произвольной нагрузке |
| title_full | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads |
| title_fullStr | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads |
| title_full_unstemmed | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads |
| title_short | Exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads |
| title_sort | exact bending solutions of orthotropic rectangular cantilever thin plates subjected to arbitrary loads |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/88008 |
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