COMPUTATIONAL METHODS IN ENGINEERING AND SCIENCE
EPMESC X, Aug. 21-23, 2006, Sanya, Hainan, China
©2006 Tsinghua University Press & Springer
New FDM for Plane Elasticity in Polar Coordinates
B. Q. Zhu 1*, J. S. Zhuo 2, J. F. Zhou 1
1
2
College of Mechanical & Electrical Engineering, Hohai University, Changzhou, 213022 China
College of Civil Engineering, Hohai University, Nanjing, 210098 China
Email: [email protected]
Abstract
Circular sector domain is the typical area of plane elasticity in polar coordinates for solution. It can be changed to
rectangle area with suitable variable substitution. With radial coordinate or circumferential coordinate treated as
“time” and dual variables introduced, two symplectic systems of different forms can be established for plane
elasticity in polar coordinates by using Hellinger-Reissner generalized variational principle. In this paper, FDM is
introduced into radial symplectic system of elasticity. The idea of FDM in Lagrange system is adopted. After meshes
and nodes being set, the surrounding area of each node is used as integrating area. From Hamilton’s dual equations,
difference equations for each node are established by using integral interpolation method. Combining the difference
equations of all nodes, whole difference format is constructed. The problems of thick-walled cylinder and curved
beam are calculated by programming. The numerical results show that the proposed FDM is effective. Another
numerical method is provided for plane elasticity in polar coordinates. Elasticity problems are always solved with
one kind of variables in Lagrange system. Traditional FDM includes stress difference method and displacement
difference method. These difference methods have some particular limits because they are restricted by the system
itself. With dual variables for solution, the new FDM can treat complex boundary conditions better. Displacements
and stresses can be obtained directly. The format of FDM with stress boundary condition is discussed in this paper.
The problem with displacement boundary condition can be treated similarly.
Key words: elasticity, plane problem in polar coordinates, radial symplectic system, integral interpolation method,
new FDM
INTRODUCTION
Plane problems in polar coordinates are always solved with one kind of variables in Lagrange system. Traditional
FDM includes stress difference method and displacement difference method. These methods have some particular
limits because they are restricted by the system itself. With radial coordinate or circumferential coordinate treated as
“time” and dual variables introduced, two symplectic systems of different forms can be established for plane
elasticity in polar coordinates by using Hellinger-Reissner generalized variational principle[1-4]. An analytical
method is given to plane elasticity in circular sector domain. The method is applied to the problems of elastic curved
beam, elastic wedge, stress singularity of fracture mechanics, etc. [3-8], and its superiority is shown to that in
Lagrange system. In circumferential symplectic system, a new solution of elasticity in polar coordinates is obtained
[9]. By installing analytical element into FEM program system, semi-analytical FEM is introduced into symplectic
system, and it can be used for stress intensity factor solution of plane crack, etc [10-13]. In this paper, the FDM is
introduced into radial symplectic system. A new difference format is constructed, and another numerical method is
provided for plane elasticity in polar coordinates.
HAMILTON’S DUAL EQUATIONS IN RADIAL SYMPLECTIC SYSTEM
Circular sector domain shown in Fig.1 is the typical area of plane elasticity in polar coordinates for solution. With
the variable substitution ξ = ln ρ (i.e. ρ = eξ ), it can be changed to the rectangle area as shown in Fig.2, in which
ξ 1 = ln R1 , ξ 2 = ln R2 . Via new variables ( S ρ = ρσ ρ , Sϕ = ρσ ϕ , S ρϕ = ρτ ρϕ ) introduced and radial coordinate ξ
⎯ 696 ⎯
treated as “time”, Hellinger-Reissner generalized variational principle can be expressed as follows [2]:
α2
ξ2
−
1
δ∫
α ∫ξ
∂uϕ
∂u ρ
⎧
1
2 ⎫
) + S ρϕ (u&ϕ − uϕ +
)−
[ S ρ2 + Sϕ2 − 2ν S ρ Sϕ + 2(1 + ν ) S ρϕ
]⎬ d ξ d ϕ = 0
⎨ S ρ u& ρ + Sϕ (u ρ +
2E
∂ϕ
∂ϕ
⎩
⎭
(1)
in which, the dot represents the derivative of a variable about ξ .
R1
O
=0
R2
Figure 1: Circular sector domain in polar coordinates
2
1
Figure 2: Rectangle area after variable substitution
Performing variation of Eq. (1) to S ϕ , it follows that
S ϕ = E (u ρ +
∂uϕ
∂ϕ
) + νS ρ
(2)
Substituting Eq. (2) into Eq. (1), Hamiltonian mixed energy variational principle is given
∂uϕ 2
∂u ρ
∂uϕ
1
)
− u ϕ ) + E (u ρ +
S ρ u& ρ + S ρϕ u&ϕ + νS ρ (u ρ +
) + S ρϕ (
⎨
−
∂ϕ
2
∂ϕ
∂ϕ
1
⎩
1
2 ⎫
−
[(1 − ν 2 ) S ρ2 + 2(1 + ν ) S ρϕ
]⎬ d ξ d ϕ = 0
2E
⎭
α
δ∫
ξ2 ⎧
α ∫ξ
(3)
Expanding Eq. (3), Hamilton’s dual equations in radial symplectic system are obtained
u& ρ = − ν u ρ − ν
u& ϕ = −
∂u ρ
∂ϕ
S& ρ = Eu
ρ
∂uϕ
∂ϕ
+
1−ν
E
2
Sρ
2 (1 + ν )
S ρϕ
E
∂uϕ
∂ S ρϕ
+ E
+ νS ρ −
∂ϕ
∂ϕ
+ uϕ +
∂ 2uϕ
∂S ρ
∂u ρ
− S ρϕ
− E
−ν
S& ρϕ = − E
2
∂ϕ
∂ϕ
∂ϕ
⎫
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎭
(4)
FORMAT OF NEW FDM
The basic idea of the new FDM in radial symplectic system is similar to that in Lagrange system. After meshes and
nodes being set in the area for solution, the surrounding area of each node is used as integrating area. Difference
⎯ 697 ⎯
equations for each node are established by using integral interpolation method. The following discussion is focused
on the problem with stress boundary condition.
1. Difference equations for inner node Uniform meshes are set in rectangle area ( ξ1 ≤ ξ ≤ ξ 2 , ϕ1 ≤ ϕ ≤ ϕ 2 ). For
convenience, the inner node is marked P0. Its adjacent nodes are P1, P2, P3, and P4. hξ and hϕ are, respectively, the
steps in direction ξ and direction ϕ (as shown in Fig.3). The surrounding area (hatched part in Fig.3) of P0 is used
as integrating area D.
2
2
3
3
3
2
0
4
1
1
1
4
4
Figure 3: Internal node and its surrounding area
By integrating Eq. (4) over the area D and using Green’s formula, it follows that
⎫
⎪
⎪
D
D
∂D
∂D
⎪
2(1 + ν )
⎪
S ρϕ d ξ d ϕ = 0
− uϕ d ϕ + u ρ d ξ − uϕ d ξ d ϕ −
E
⎪
D
D
∂D
∂D
⎬
− S ρ d ϕ − Eu ρ d ξ d ϕ − Eu ϕ d ξ − νS ρ d ξ d ϕ + S ρϕ d ξ = 0 ⎪
⎪
D
D
∂D
∂D
∂D
⎪
∂u ϕ
⎪
d ξ + νS ρ d ξ + S ρϕ d ξ d ϕ = 0 ⎪
− S ρϕ d ϕ + Eu ρ d ξ + E
ϕ
∂
D
∂D
∂D
∂D
∂D
⎭
∫
−
uρ d ϕ +
∫∫νu ρ d ξ d ϕ + ∫ νuϕ d ξ − ∫∫
∫
∫
∫∫
∫
∫∫
∫
∫
1 −ν 2
Sρ dξ dϕ = 0
E
∫∫
∫
∫∫
∫
∫
∫
(5)
∫∫
Considering a general function g (ϕ , ξ ) and with g i representing the function value at Pi (i = 0,1,L 4) , following
expressions are deduced from integral midpoint formula and difference formula
−
∫ g dϕ ≈ (g
N2
− g N 4 ) hϕ ≈
∂D
∫∫
g2 − g4
hϕ ,
2
∫ g dξ ≈ (g
N1
− g N3 )hξ ≈
∂D
g1 − g 3
hξ
2
⎡ g − g0 g0 − g3 ⎤
hξ
⎡ ∂g
⎤
∂g
∂g
d ξ ≈ ⎢( ) N1 − ( ) N3 ⎥ hξ ≈ ⎢ 1
−
⎥ hξ = [ g1 + g 3 − 2 g 0 ]
hϕ ⎥⎦
hϕ
∂ϕ
∂ϕ ⎦
⎢⎣ hϕ
⎣ ∂ϕ
∂D
∫
g d ξ d ϕ ≈ g 0 hξ hϕ ,
D
Substituting these into Eq. (5) and simplifying them, the difference equations for inner node are established
uρ2 − uρ4
2hξ
uϕ 2 − uϕ 4
2hξ
Sρ2 − Sρ4
2hξ
+ νu ρ 0 + ν
+
u ρ1 − u ρ 3
2hϕ
− Eu ρ 0 − E
S ρϕ 2 − S ρϕ 4
2hξ
uϕ1 − uϕ 3
+E
−
2hϕ
− uϕ 0 −
2(1 + ν )
S ρϕ 0 = 0
E
uϕ 1 − uϕ 3
2hϕ
u ρ1 − u ρ 3
2hϕ
1 −ν 2
Sρ0 = 0
E
+E
− νS ρ 0 +
S ρϕ 1 − S ρϕ 3
2hϕ
uϕ 1 + uϕ 3 − 2uϕ 0
2
hϕ
+ν
=0
S ρ1 − S ρ 3
2hϕ
+ S ρϕ 0
⎯ 698 ⎯
⎫
⎪
⎪
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪
= 0⎪
⎪⎭
(6)
2. Difference equations for the node on boundaries ϕ = ±α On boundaries ϕ = ±α , the node and its surrounding
area are shown in Fig. 4.
2
2
2
3
0
1
1
4
2
3
2
3
1
2
0
3
4
4
4
4
1
4
(a) ϕ = −α
(b) ϕ = α
Figure 4: Node on boundaries ϕ = ±α and its surrounding area
On boundary ϕ = −α , the integrals of general function g (ϕ , ξ ) are
− ∫ g d ϕ ≈ ( g N2 − g N4 )
hϕ
∂D
∫ g dξ ≈ (g N
2
g2 − g4
hϕ
4
g1 − g 0
hξ
2
− g 0 )hξ ≈
1
≈
∂D
1
∫∫ g d ξ d ϕ ≈ 2 g 0 hξ hθ
D
Since the stress boundary condition is under discussion, therefore
∫ S ρϕ d ξ ≈ (S ρϕ )
−
N1 hξ
∂D
∫
Eu ρ d ξ +
∂D
∫
E
∂D
≈ ( Eu ρ + E
≈ (E
∂uϕ
2
C3
C4
S ρϕ |ϕ = −α d ξ ≈
+E
S ρϕ 1 + S ρϕ 0
2
hξ −
∫
C3
C4
S ρϕ |ϕ =−α d ξ
∫
d ξ + νS ρ d ξ
∂ϕ
∂D
∫
+ νS ρ ) N1 hξ −
∂ϕ
u ρ1 + u ρ 0
∂uϕ
∫
u ϕ 1 − uϕ 0
hϕ
C3
C4
+ν
( Eu ρ + E
S ρ1 + S ρ 0
2
∂uϕ
+ νS ρ ) |ϕ =−α d ξ
∂ϕ
)hξ −
∫
C3
C4
Sϕ |ϕ =−α d ξ
Substituting these into Eq. (5) and simplifying them, the difference equations for the node on boundary ϕ = −α take
the final form
uρ 2 − uρ 4
2hξ
uϕ 2 − u ϕ 4
2hξ
Sρ2 − Sρ4
2hξ
+ νu ρ 0 + ν
+
u ρ1 − u ρ 0
hϕ
+E
−
hϕ
− uϕ 0 −
− Eu ρ 0 − E
S ρϕ 2 − S ρϕ 4
2hξ
uϕ 1 − uϕ 0
hϕ
2(1 + ν )
S ρϕ 0 = 0
E
uϕ 1 − uϕ 0
hϕ
u ρ1 + u ρ 0
1 −ν 2
Sρ0 = 0
E
+E
− νS ρ 0 +
S ρϕ1 + S ρϕ 0
2(uϕ1 − uϕ 0 )
2
hϕ
hϕ
+ν
2
−
hϕ hξ
S ρ1 + S ρ 0
hϕ
−
C3
∫C S ρϕ |ϕ =−α d ξ = 0
4
2
hϕ hξ
⎯ 699 ⎯
C3
∫C Sϕ |ϕ =−α d ξ + S ρϕ 0
4
⎫
⎪
⎪
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪
= 0⎪
⎪⎭
(7)
For that on boundary ϕ = α , the difference equations take the similar form as follows.
uρ 2 − uρ 4
2hξ
uϕ 2 − uϕ 4
2hξ
Sρ2 − Sρ4
2hξ
uϕ 0 − uϕ 3
+ νu ρ 0 + ν
+
hϕ
u ρ1 − u ρ 3
2hϕ
− uϕ 0 −
2hξ
hϕ
uρ 0 + uρ 3
−E
hϕ
1 −ν 2
Sρ0 = 0
E
2(1 + ν )
S ρϕ 0 = 0
E
uϕ 0 − uϕ 3
− Eu ρ 0 − E
S ρϕ 2 − S ρϕ 4
−
−E
− νS ρ 0 −
S ρϕ 0 + S ρϕ 3
hϕ
2(uϕ 0 − uϕ 3 )
2
hϕ
2
+
hϕ hξ
S ρ1 + S ρ 0
−ν
+
hϕ
C2
∫C
S ρϕ |ϕ =α d ξ = 0
1
2
hϕ hξ
C2
∫C
Sϕ |ϕ =α d ξ + S ρϕ 0
1
⎫
⎪
⎪
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪
= 0⎪
⎪⎭
(8)
3. Difference equations for the node on boundaries ξ = ξ1 , ξ 2 On boundaries ξ = ξ1 , ξ 2 , the node and its
surrounding area are shown in Fig. 5.
3
2
2
3
3
3
2
3
1
4
1
1
0
1
1
4
1
4
2
0
3
4
(a) ξ = ξ1
(b) ξ = ξ 2
Figure 5: Node on boundaries ξ = ξ 1 , ξ 2 and its surrounding area
On boundary ϕ = −α , the integrals of general function g (ϕ , ξ ) are
∫ g dϕ ≈ ( g N
−
2
− g 0 ) hϕ ≈
∂D
⎡ ∂g
∂g
∫ ∂ϕ d ξ ≈ ⎢⎣( ∂ϕ ) N
1
−(
∂D
g2 − g0
hϕ
2
∫ g dξ ≈ (g N
1
− g N3 )
∂D
hξ
2
≈
g1 − g 3
hξ
4
1
∫∫ g d ξ d ϕ ≈ 2 g 0 hξ hϕ
D
hξ
⎤ hξ ⎡ g1 − g 0 g 0 − g 3 ⎤ hξ
∂g
≈⎢
−
= [ g1 + g 3 − 2 g 0 ]
) N3 ⎥
⎥
∂ϕ ⎦ 2 ⎢⎣ hϕ
hϕ ⎦⎥ 2
2hϕ
But
C1
C1
Sρ0 + Sρ2
C4
C4
2
− ∫ S ρ d ϕ ≈ − ∫ S ρ |ξ =ξ1 d ϕ + ( S ρ ) N 2 hϕ = − ∫ S ρ |ξ =ξ1 d ϕ +
∂D
C1
C1
S ρϕ 0 + S ρϕ 2
C4
C4
2
− ∫ S ρϕ d ϕ ≈ − ∫ S ρϕ |ξ =ξ1 d ϕ + ( S ρϕ ) N 2 hϕ = − ∫ S ρϕ |ξ =ξ1 d ϕ +
∂D
hϕ
hϕ
So the difference equations for the node on boundary ξ = ξ1 are
uρ2 − uρ0
hξ
uϕ 2 − uϕ 0
hξ
Sρ0 + Sρ2
hξ
+ νu ρ 0 + ν
+
u ρ1 − u ρ 3
2hϕ
− Eu ρ 0 − E
S ρϕ 0 + S ρϕ 2
hξ
uϕ 1 − u ϕ 3
+E
−
2hϕ
− uϕ 0 −
2(1 + ν )
S ρϕ 0 = 0
E
u ϕ 1 − uϕ 3
2hϕ
u ρ1 − u ρ 3
2hϕ
1 −ν 2
Sρ0 = 0
E
+E
− νS ρ 0 +
S ρϕ1 − S ρϕ 3
2hϕ
2
−
hϕ hξ
C1
∫C S ρ |ξ =ξ
4
1
dϕ = 0
S ρ1 − S ρ 3
1
2
+ S ρϕ 0 −
(uϕ1 + uϕ 3 − 2uϕ 0 ) + ν
2
2hϕ
hξ hϕ
hϕ
⎯ 700 ⎯
C1
∫C S ρϕ |ξ =ξ
4
1
⎫
⎪
⎪
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪
d ϕ = 0⎪
⎪⎭
(9)
Similarly, difference equations for that on boundary ξ = ξ 2 are
uρ0 − uρ 4
hξ
uϕ 0 − uϕ 4
hξ
+ νu ρ 0 + ν
+
S ρϕ 0 + S ρϕ 4
−
uϕ 1 − uϕ 3
− Eu ρ 0 − E
hξ
2hϕ
u ρ1 − u ρ 3
+E
+E
2hϕ
⎫
⎪
⎪
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪
d ϕ = 0⎪
⎪⎭
1 −ν 2
Sρ0 = 0
E
2(1 + ν )
S ρϕ 0 = 0
E
− uϕ 0 −
2hϕ
hξ
−
2hϕ
u ρ1 − u ρ 3
Sρ0 + Sρ4
−
uϕ1 − uϕ 3
S ρϕ1 − S ρϕ 3
− νS ρ 0 +
2
+
hϕ hξ
2hϕ
C2
∫C
3
S ρ |ξ =ξ 2 d ϕ = 0
S ρ1 − S ρ 3
1
2
(uϕ1 + uϕ 3 − 2uϕ 0 ) + ν
+ S ρϕ 0 +
2
2hϕ
hξ hϕ
hϕ
C2
∫C
3
S ρϕ |ξ =ξ 2
(10)
4. Difference equations for the node at corners At four corners, the node and its surrounding area are shown in
Fig. 6.
0 ( 3)
22
2
2
3
3
44
(a) ϕ = −α , ξ = ξ1
3
3
1
4
4
00 ( 11)
44
1
2
3
11
33
1
1
4
22
33
1
0 ( 4)
22
33
2
2
3
1
4
4
N2
(b) ϕ = α , ξ = ξ1
− g N4 )
hϕ
hξ
≈
∂D
∫ g dξ ≈ (g
− g N3 )
N1
2
∂D
1
∫∫ g d ξ d ϕ ≈ 4 g
2
≈
(c) ϕ = −α , ξ = ξ 2
(d) ϕ = α , ξ = ξ 2
g2 − g0
hϕ
4
g1 − g 0
hξ
4
0 hξ hϕ
D
But
− ∫ S ρ d ϕ ≈ (S ρ ) N2
∂D
hϕ
2
hϕ
− ∫ S ρϕ d ϕ ≈ ( S ρϕ ) N 2
2
∂D
hξ
∫ S ρϕ d ξ ≈ (S ρϕ ) N
2
1
∂D
C1
Sρ0 + Sρ2
C4
4
− ∫ S ρ |ξ =ξ1 d ϕ =
C4
C1
S ρϕ 0 + S ρϕ 2
4
S ρϕ 1 + S ρϕ 0
C4
4
∂uϕ
∂uϕ
≈ (E
∂D
u ρ1 + u ρ 0
4
∂D
+E
uϕ1 − uϕ 0
2hϕ
+ν
S ρ1 + S ρ 0
4
C1
hϕ − ∫ S ρϕ |ξ =ξ1 d ϕ
C4
C3
− ∫ S ρϕ |ϕ =−α d ξ =
∫ Eu ρ d ξ + ∫ E ∂ϕ d ξ + ∫νS ρ d ξ ≈ ( Eu ρ + E ∂ϕ
∂D
C1
hϕ − ∫ S ρ |ξ =ξ1 d ϕ
C4
− ∫ S ρϕ |ξ =ξ1 d ϕ =
C3
hξ − ∫ S ρϕ |ϕ =−α d ξ
C4
+ νS ρ ) N1
hξ
2
1
4
For the node at ϕ = −α , ξ = ξ1 , the integrals of general function g (ϕ , ξ ) are
∫ g dϕ ≈ (g
0 ( 2)
1
4
Figure 6: Node at corners and its surrounding area
−
2
C3
∂uϕ
C4
∂ϕ
− ∫ ( Eu ρ + E
C3
)hξ − ∫ Sϕ |ϕ =−α d ξ
C4
So the difference equations for the node are
⎯ 701 ⎯
+ νS ρ ) |ϕ =−α d ξ
uρ 2 − uρ 0
hξ
uϕ 2 − uϕ 0
hξ
Sρ 2 + Sρ 0
hξ
+ν uρ 0 +ν
+
hϕ
uρ1 − uρ 0
− uϕ 0 −
hϕ
− Euρ 0 − E
S ρϕ 2 + S ρϕ 0
hξ
uϕ1 − uϕ 0
+E
1 −ν 2
Sρ 0 = 0
E
2(1 + ν )
S ρϕ 0 = 0
E
uϕ1 − uϕ 0
uρ1 + uρ 0
hϕ
−
hϕ
+E
−ν S ρ 0 +
S ρϕ1 + S ρϕ 0
hϕ
4
−
hξ hϕ
(∫
C3
C4
)
C1
S ρϕ |ϕ =−α d ξ + ∫ S ρ |ξ =ξ1 d ϕ = 0
C4
Sρ1 + Sρ 0
2
4
(uϕ1 − uϕ 0 ) + ν
+ S ρϕ 0 −
2
hϕ
hξ hϕ
hϕ
(∫
C3
C4
C1
Sϕ |ϕ =−α d ξ + ∫ S ρϕ |ξ =ξ1
C4
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
d ϕ = 0⎪
⎪⎭
(11)
)
The equations for other corners can be established similarly and won’t be listed here anymore.
Combining the difference equations of all nodes, whole difference format is constructed. The numerical solution can
be obtained for the problem of circular sector domain in polar coordinates.
EXAMPLES
To verify the effectiveness of the new FDM, calculations are implemented by programming. Two Examples are
given here.
1. Example 1 Thick-walled cylinder subjected to inner pressure p=120MPa (Fig. 7). The inner radius is a=50mm,
and external radius b=100mm. The properties of materials are E=200GPa, and ν = 0.3 .
a
p
b
Figure 7: Thick-walled cylinder subjected to inner pressure
This is an Axis-symmetric problem. Taken out from the cylinder by two radial sections whose angle is equal to 10º, a
block can be used for calculation. 10 meshes are set here in direction ξ , i.e. hξ = 0.06931 . The numerical results of
calculation are listed in Table 1.
Table 1 The numerical results of calculation for thick-walled cylinder problem
ρ , mm
Analytical solution
u ρ , mm
σ ρ , MPa
σ ϕ , MPa
u ρ , mm
New FDM
σ ρ , MPa
σ ϕ , MPa
50
0.0590
−120
200
0.0590
−119.53
200.26
53.59
0.0560
−99.29
179.29
0.0559
−99.45
178.85
57.43
0.0533
−81.26
161.26
0.0533
−80.92
161.47
61.56
0.0509
−65.56
145.56
0.0508
−65.69
145.23
65.98
0.0486
−51.90
131.90
0.0487
−51.66
132.07
70.71
0.0467
−40.00
120
0.0466
−40.11
119.75
75.79
0.0449
−29.64
109.64
0.0450
−29.48
109.79
81.23
0.0434
−20.63
100.63
0.0433
−20.72
100.44
87.06
0.0421
−12.78
92.78
0.0421
−12.67
92.90
93.30
0.0409
−5.948
85.95
0.0409
−6.02
85.80
100
0.0400
0
80
0.0400
0.06
80.09
⎯ 702 ⎯
a
b
2. Example 2 Curved beam subjected to equal and opposite forces at two ends (Fig. 8). The inner radius is
a=150mm, and external radius b=200mm. The magnitude of the force F =1 kN. The properties of materials are
E=200GPa, and ν = 0.3 .
O
F
F
Figure 8: Curved beam subjected equal and opposite forces at two ends
Taking symmetry into consideration, only the right half of the beam is calculated. 18 meshes are set in direction ϕ
and 16 in direction ξ , i.e. hϕ = 0.08727 and hξ = 0.01798 . Assuming u ρ |ϕ =90°, ρ =a = 0 , the calculations for
displacements on section ϕ = 0 are listed in Table 2 and the calculations for stresses on section ϕ = 90° in Table 3.
Table 2 The calculations of displacements on section ϕ = 0
Analytical solution
New FDM
ρ , mm
u ρ , mm
uϕ , mm
u ρ , mm
uϕ , mm
150
2.0448
0.9372
1.9807
0.9028
155.49
2.0448
1.0187
1.9807
0.9821
161.19
2.0448
1.1023
1.9807
1.0635
167.09
2.0448
1.1885
1.9808
1.1473
173.21
2.0448
1.2774
1.9810
1.2340
179.55
2.0448
1.3697
1.9812
1.3238
186.12
2.0448
1.4655
1.9815
1.4172
192.94
2.0448
1.5654
1.9818
1.5148
200
2.0448
1.6697
1.9823
1.6170
Table 3 The calculations of stresses on section ϕ = 90°
Analytical solution
New FDM
ρ , mm
σ ρ , MPa
σ ϕ , MPa
σ ρ , MPa
σ ϕ , MPa
150
0
485.98
0. 34
459.65
155.49
14.71
364.21
14.35
343.43
161.19
24.27
248.08
23.50
232.50
167.09
29.23
136.90
28.27
126.27
173.21
30.06
30.06
29.10
24.16
179.55
27.20
-73.00
26.39
−74.33
186.12
21.02
-172.81
20.51
−169.68
192.94
11.85
-269.82
11.74
−262.32
200
0
-364.48
0.38
−352.63
The numerical results of the examples above show that the new FDM is effective.
⎯ 703 ⎯
CONCLUSIONS
A new FDM is introduced into radial symplectic system of elasticity. The format of FDM is constructed, and a new
numerical method is provided for plane elasticity in polar coordinates. With dual variables for solution, the new
FDM can treat complex boundary conditions better. Displacements and stresses can be obtained directly. The format
of new FDM with stress boundary condition is discussed in this paper. The problem with displacement boundary
condition can be treated similarly.
In the process of constructing the format with integral interpolation method, uniform meshes are chosen and
rectangle area is used as the surrounding area of each node. If varied meshes are chosen or polygon area is used,
different format of FDM can be constructed. Whether this kind of format has any conservative property needs to be
studied further, and its well-posedness should be proved later.
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⎯ 704 ⎯