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Chapter 5
Finite Element Method for
Plate Bending Problems
1
1- Plate Bending (Review of theory) 1.1- Stresses (Isotropic Case) 1.2- Shear Forces 1.3- Equilibrium Equations 1.4- Strain Energy 1.5- Boundary Conditions 1.6- Potential Energy 2- Rectangular Plate Bending Elements 2.1- Non-conforming Rectangular finite element
2.1.1- Stiffness Matrix 2.1.2- Consistent Load Vector 2.1.3- Stresses 2.1.4- Boundary Conditions (Kinematics)
2.2- Note on Continuity 3- Elements for C1 Problems 4- Triangular Elements 5- Nonconformin Triangular Plate Bending Elements 6- Conforming Rectangular Element (16 dof) 7- Alternative Method for Plate Bending Element 8- Triangular Element for Conforming C1 Continuity 8.1-Transformation of Nodal DOF along an Inclined Edge 9- Two-Dimensional Creeping Flow 9.1- Fully Developed Parallel Flow 9.2- Flow Past a Cylinder
2
1- Plate Bending (Review of theory) Linear elastic small deflections Assumptions:
1. Plate is thin (h<<L where L=typical length) 2. Normals perpendicular to the mid-surface remain normal to the
deflected mid-surface. 3. Small deflections (normal to the plate) so that the mid-surface
remains unstretched Also note that τzz<<τxx and τyy
L
z, w
x, u
h
Mid-surface
h/2
z, w
A
B
α
x, u
A' B'
u
α
w
y, v
Deflections: Displacements of B to B' u=-z sin α≅-zα=-z
xw∂∂ similarly, v=-z
yw∂∂
Strains:
yxwz
xv
yu
ywz
yv
xwz
xu
xy
yy
xx
∂∂∂
−=∂∂
+∂∂
=
∂
∂−=
∂∂
=
∂
∂−=
∂∂
=
2
2
2
2
2
2γ
ε
ε
3
1.1- Stresses (Isotropic Case) Plane stress in xy plane
yxwEzE
yw
xwEzE
yw
xwEzE
xyxy
yyxxyy
yyxxxx
∂∂∂
+−=
+=
∂
∂+
∂
∂
−−=+
−=
∂
∂+
∂
∂
−−=+
−=
2
2
2
2
2
22
2
2
2
2
22
1)1(2
)(1
)(1
)(1
)(1
νγ
ντ
νν
ενεν
τ
νν
νεεν
τ
Section properties (resultants stresses as bending and twisting moments)
yxxyxyxy
h
hxy
yyxxyyxxyy
yyxxyyxx
h
hyyxxxx
h
hxx
MwDwEhdzzM
wwDwwEhM
similarly
wwDwwEhdzzwwEdzzM
=−=+
=−=
+−=+−
−=
+−=+−
−=+−
−==
∫
∫∫
−
−−
)1()1(12
(()1(12
,
(()1(12
(1
32
2
))2
3
))2
32
2
2
)2
2
2
νν
τ
ννν
ννν
νν
τ
Where D is the bending rigidity and Mx, My are bending moments per unit length about x and y axes, respectively. Mxy and Myx are the twisting moments per unit length about x and y axes, respectively. 1.2- Shear Forces
dzQdzQ yz
h
hyxz
h
hx ττ ∫∫
−−
==2
2
2
2
Where Qx and Qy are the shear forces per unit length on edges whose normals are x and y axes, respectively.
4
1.3- Equilibrium Equations q(x,y) is the transverse load per unit area: MxyQy Qy
5
dx
dy
x
y
Qx
dxx
QQ x
x ∂∂
+
dyy
QQ y
y ∂
∂+
q(x,y)
y
Qx
Qx
Myy
dxM
M xxxx
∂+
Qy
dyy
MM xy
xy ∂
∂+
dyy
MM yy
yy ∂
∂+
Mxy
dxx
MM xy
xy ∂
∂+
x∂
Mxx
operatorbiharmoniccalledisyyxx
whereyxqwD
yxqwwwD
yxqwwDwDwwD
MandMMforSubstitute
yxqy
Mxy
M
xM
yxqy
Qx
Q
thenxy
M
xM
xQ
xyM
y
My
Q
sinequationngsubstitutiandQandQinatingE
Qy
Mx
M
or
dxdyx
MMdxMdxdyQdydx
xM
MdyMM
Qx
My
Mor
dydxx
MMdyMdxdyQdydx
xM
MdxMM
effectsorderondtheignoringmEquilibriuMoment
yxqy
Qx
Q
givestermscancelling
dxdyyxqdzdyy
QQdydx
xQ
QdxQdyQF
yyyyxxyyxxxx
yyyyxxyyxxyyxxyyxxxx
xyyyxx
yyxyxxyx
xyxxx
xyyyy
yx
xxyxx
xyxyxyx
xxxxxxy
yxyyy
xyxyxyx
xxyyyyx
yx
yy
xxyxz
4
4
22
4
4
444
2
22
2
2
2
2
2
2
2
2
2),(
),()2(
0),()()1(2)(
:,
0),(2),(
:
:lim
0
:
0)()(
0
:
0)()(
:sec
0),(
:
0),()()(
∂
∂+
∂∂
∂+
∂
∂=∇=∇
=++
=++−−−+−
=+∂
∂+
∂∂
∂−
∂
∂=+
∂
∂+
∂∂
∂∂
∂−
∂
∂=
∂∂
∂∂
∂−
∂
∂=
∂
∂
=−∂
∂−
∂∂
=∂
∂+−+−
∂∂
++−=
=+∂
∂+
∂
∂
=∂
∂++−+
∂∂
+−=
=+∂
∂+
∂∂
=+∂
∂++
∂∂
++−−=
∑
∑
∑
ννν
Also encountered in many other engineering problems, e.g. very viscous or creeping incompressible flow, stress analysis using Airy's stress function φ or ∇4φ=0 (compatibility equation) 1.4- Strain Energy This is analogous to beam bending. Energy stored=Elastic work done by moments
( )
[ ]
[ ]dxdywwwwwDU
dxdywDwwDwwwDU
equationsabovefromMMforMngSubstituti
dxdywMwMwMU
wwMMnote
dywdxMdxwdyMdywdxMdxwdyMdU
xyyyxxyyxxA
xyyyxxxxyyxxA
xyyyxx
xyxyyyyyxxxxA
yxxyyxxy
yxyxxyxyyyyyxxxx
222
2
)1(222
)1(2)()(21
:
221
:
)(21)(
21)(
21)(
21
νν
ννν
−+++=
−++++=
+−−=
==
++−−=
∫∫
∫∫
∫∫
x
Mxx wx Mxx wxx+wxxdxw, z dx
6
y=0 x
1.5- Boundary Conditions x=0 x=a
a) Simply Supported Edge y=b 1) along x=0 and x=a edges y w=0 and Mxx=0 But if w=0 along y on x=0 then wy=wyy=0 Therefore wy=wyy=0 Mxx=-D(wxx+νwyy)=0 ∴ wxx=0 2)along y=0 and y=b w=0 and Myy=0 or wyy=0 for w=0 along y=0 and y=b, wx=0 and wxx=0 b) Clamped or built-in edge 1) along x=0 and x=a w=0 and 0==
∂∂ w xxw
2) along y=0 and y=b w=0 and 0==
∂∂ w yyw
c) Free edge
There are no restrictions on displacements- no edge forces. One is tempted to say that along x=a Qx=0, Mxx=0 and Myy=0
This is wrong because only two independent conditions are allowed. According to kirchhoff, should use only two, i.e. Mxx=0 and Txx=0 (effective shear force) where, 0=
∂
∂−=
yM
QT is effective shear force
along the edge.
xyxx
1.6- Potential Energy Potential energy of the plate bent to transverse load q(x,y) is given by:
[ ] qwdxdydxdywwwwwDWU
Axyyyxxyyxx
A∫∫∫∫ −−+++=
−=
222 )1(222
ννπ
π
7
2- Rectangular Plate Bending Elements
y
2.1- Non-conforming Rectangular finite element use deflection and two slopes as generalized displacements at each node i.e. use w, wx, wy as nodal degrees of freedom. This element has wide use application and performs very well.
w3, wx3,wy3 3 4
8
x 2 1
b
a
w1, wx1,wy1 With three dof per nodes, we have 12 dof per element, therefore, require a twelve term polynomial
312
311
310
29
28
37
265
24321),( xyayxayaxyayxaxayaxyaxayaxaayxw +++++++++++=
i.e. complete cubic plus two terms (x3y and xy3) polynomial of above equation satisfies the homogeneous plate equation D∇4w=0, this fact is of little significance in the finite element formulation. Rigid Body Modes W=constant (translation) and also need two rotations. Three rigid body modes required are included through a1+a2x+a3y in the polynomial of above equation. Constant Strain In plate bending, the strains are curvatures and twist i.e. wxx, wyy and wxy. This is provided by the second degree terms i.e. a4x2+a5xy+a6y2 which are also included. Continuity The polynomial in above equation has been chosen carefully and for a very good reason we included x3y and xy3 terms instead of x4 and y4. For constant y, w(x,y) is cubic in x and vice-versa. Now a cubic polynomial in one dimension contains four independent parameters or coefficient which may be specified uniquely by two conditions at each end point (i.e. the end
nodes. This particular feature leads to ensuring displacement continuity between adjacent elements. We will look into it in more detail later. Generalized Displacement The element formulation begins by first solving for generalized displacements from displacement function. This yields the following matrix equation;
⎪⎪⎪⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎪⎪⎪⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
12
11
10
9
8
7
6
5
4
3
2
1
2
32
32
322
3222
33322322
32
2
32
4
4
4
3
3
3
2
2
2
1
1
1
003000200100000000010
00000000133200100
30300101
00000010000000300010000000001000000000100000000000010000000000001
aaaaaaaaaaaa
bbbbb
bbbabababazbababzababzaababbabbaabababa
aaaazaaaa
wwwwwwwwwwww
b
y
x
y
x
y
x
y
x
or w=[T]A where A=<a1,a2,……..,a12> The matrix [T] in equations above can be inverted in order to solve for A as functions of the generalized displacements. Once a1, a2 etc. are substituted back into displacement function, we obtain deflection in terms of the generalized displacements w1, wx1, wy1,….etc. as:
9
Equation 1
[ ]
[ ]
[ ]
[ ] ⎪⎪⎪⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
−−+−−−−−
−−−−−
−−−−
−−+−−−
−−−−−−−−−−
−−−−
=
ηξξξηηξηηξ
ηξξηηξηηξξ
ξηηηξξ
ηηξηηξξξηη
ηξξηηξηξξξη
ηηξηξξ
ηξ
)21)(1()23)(1()1)(1(
)1()21)(1()23(
)1()1(
)21)(1()1()23()1(
)1()1()23)(1()1()23(1
)1()1()1()1(
),(
2
2
2
2
2
2
2
2
2
22
2
2
ab
aa
ba
ab
aa
ba
Ww T
where w is the column vector of nondimensional generalized displacements:
[ ]scoordinateensionalnonare
by
ax
wwawwwawwww xxyxyxT
dim
....// 43222111
−==
=
ηξ
Each row in equation of w(ξ,η) represent a shape function or interpolation function Ni. We may write it as:
.),(),( 131211
12
1etc
awwwwhereNw yxii
i==== ∑
=
δδδδηξηξ
1
N3
1
1
Zero slope
ξ
η η
ξ
N1
Zero slope along here
1001 =
∂∂
== ηξξN
10
11
ow that we have the displacement distribution within the element defined
. Consider for example e joining of two elements A and B together as illustrated in the Figure.
Nby displacement equation, we may ask what continuity will be provided by the element , th
)()()()( 122111022011 xHwxHwxHwxHww xx +++=
A
4 3
21
4 3
21
Bx
y
x
Joining two elements along edges parallel to x
For element A, the displacement along edge 3-4 is obtained by putting η=1 in equation 1.
It may be seen that the same function of ξ occur in these two equations. Therefore, if w1, wx1, w2, wx2 of element B are equated to w4, wx4, w3, wx3, respectively of element A, w will have to be continuous between the elements. Similar arguments are easily made for edges parallel to the y-axis. What about slopes normal to the edges? There is no continuity of slopes normal to the element edges. This can be shown by taking derivatives of w with respect to η and substituting η=1 for element It will be found that
)1()23()1()231()(
:)0(21,)1()23()1()31()(
22
22
2321
23
23
24
324
ξξξξξξξξξ
ηξξξξξξξξξ
−−−+−+−=
=−−−−+−+−=
xB
xxA
awwawww
isedgealongntdisplacemetheBelementforawwawww
+ 1x
2+
A and η=0 for element B.
η∂
way we can make normal slopes continuous by equating w
∂w terms along those edges are cubic and there is no
y2 and wy3 of element B, respectively. Therefore, the element is called non-conforming
y3 and wy4 of element A to w
2.1.1- Stiffness Matrix Calculate the stiffness matrix for the non-conforming plate bending element by substituting equation 1 into expression for strain energy. After, carrying out integration over the area of the element, we obtain the quadratic form in term of generalized displacements (as expected) for strain energy:
][2
WKWU e = Here, [K] is the 12 by 12 stiffness matrix for the element and is given in the following page.
1 T
Note, this matrix has been derived for W as given in equation of δ3=w1/a, δ6=w2/a, δ9=w3/a and δ12=w4/a i.e. in dimensionless displacements. To allow w's to take on dimensionless displacements, the 3rd, 6th, 9th and 12th row should be multiplied by a again. Further, if the degrees of freedom are desired to be arranged as:
[ ]444333222111 yxyxyxyxT wwwwwwwwwwwww =
Then the rows and columns should be rearranged accordingly, e. st
nd rd rdg. 1 and
2 rows should be moved into second and 3 rows. And 3 row should be placed into 1st row, etc, etc., etc.
12
The stiffness matrix for the plate bending element may also be derived llowing the alternative method we discussed for beam element.
igure 1 Stiffness Matrix for 12 parameter Rectangular Element (non-conforming)
foF
10)1(1
5)27(1
10)1(
210)1(
21
5)27(12)
10)1((
10)41(
21
010
)1(230
)1(6
010
)1(30
)1(3
0
30)1(1
1)1(
210
30)1(
61
10)41(
210
15)1(21
101(1
5)27(12
10)1(
10)41(
21
5)27(1
10)1(
210)1(
21 323
νννν
νννννν
−−+
−−+
+−
−−
−
+−
−−−+−
−−−
++
−−+−−
−+
−−−
mmm
mm
mm
mm
mm
mm
mmmm
mm
mmmm
30
)4210
)1(30
)1(3
010
)1(230
)1(6
0
15)1(2
32
10)41(
210
15)1(2
31
10)1(
210
30)1(1
5)27(22
10)41(
10)41(1
5)27(2
10)41(
210)1(1
15)1(2
32
210)41(
215)1(2
30
15)1(2
32
10)1(10
30)1(1
5)27(22
10)41(1
'15
)1(23
22
15)1(22
32
32
22
2
22
3232
2
3
ννννννν
νννν
ν
ν
ννννν
ννννν
νννννν
νννν
ννν
νν
ννν
ν
−+
−+−−
−+
−−−
−−+−
−+−
+−
−−
−+
−+
−−
−
−+
−−
−−
−+
−+
++
−−−
−−
−+
−++
++
+−−
−−−
+−
−−
−+−
+−
−−
−+
−+
−−
−++
++
=−
+
=−
+
mm
mm
mmmm
mm
mmmm
mm
mmmm
mm
mm
mmm
m
mm
mm
mm
mm
mm
mm
mmmm
mm
mmmm
mmm
mm
mm
mm
mm
mm
mm
sRatioPoissonSYMMETRICm
mbamm
m
3
3
−
6
−
5)27(22
10)41(
10)41(1
5)27(2
10)41(
2
15)1(2
32
210)41(
215)1(2
3
15)1(2
32
10)1(105
)27(2210
)41(15
)1(23
2
3232
2
32
mm
mmm
mm
mmm
mmm
mm
mm
mm
mm
mmm
m
CONTINUE
ννννν
νννν
νν
νν
ν
−++
+−−
++
−−−
++
−
−+−
++
−−−
−+
−−−
−++
++−
−+
] 13
2.1.2- Consistent Load Vector Assume uniform pressure q0. Recall from equation of potential energy π, the work done W is given by:
where Ae is the element area, and w is given by equation 1, when equation 1 is substituted into above equation and integrating, the load vector for the element in dimensional form:
When nonconforming elements are used to obtain an approximate solution for some loading, generally we use reasonably large number of elements and can obtain reasonable answer by using lumped load i.e. q0ab/4 at each corner node. However, for very refined elements, we must use consistent
ad vector since much fewer elements are used. In such cases, we may be introducing an undesirable error through lumped loads. 2.1.3- Stresses Bending and twisting moments Define:
0 WpwdxdyqW T
Ae
== ∫∫
⎥⎦⎤
⎢⎣⎡ −−−=
41
242441
242441
242441
2424 babababaabqp e
T
lo
)1(121000101
][
][
1000101
2
3
ννν
νετ
νν
ντ
τ
ε
−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−−
=
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−−
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
EhDDD
D
www
DMMM
momentsandstressesMMM
curvatureandstrainwww
xy
yy
xx
xy
yy
xx
xy
yy
xx
xy
yy
xx
rom the shape functions in equation 1, we can obtain wxx, wyy and wxy. rther, these can be evaluated at various points (xi, yi) or (ξI, ηi) and hence
d at specified points. We must know w before we can compute τ.
FFuMxx, Myy and Mxy can be evaluate
14
2.1.4- Boundary Conditions (Kinematic) Along AB and AD, the plate is
15
Simply supported
Free
Simply supported
B
Clamped
CD
A
y
x
3 4
2 1 b
a w1, wx1,wy1
w3, wx3,wy3
n
simply supported, AB: w=0 and wx=0 AD: w=0 and wy=0
and its normal derivatives or normal slope must be uniquely determined by values along an interface or edge of an element in order to ensure, C1 continuity. Consider edge 3-4 of the rectangular element shown.
l
Along cd, the plate is clamped w=0 and wx=0 and wy=0 Nothing specified on free boundary.
2.2- Note on Continuity Both w
Here, wn=wy, the normal slope. It is desired that w and wy be unique y determined by the values of w and wx and wy at the nodes lying along edge 3-4.
...2 +++= xaxaaw
...2321 ++= xbxbbw
321
∂+
constants ai and bi in each expression st sufficient to determine the expressions by nodal parameters or dof
ssociated with the line. With w and wx as nodal dof at each node i.e. two nodes, we can allow only ur ai (a1, a2 , a 4) or at most cubic variation in x along 3-4.
imilarly only a linear variation can be allowed i.e. two terms (b1 and b2) r wyi. In the me manner, wx can be made continuous along the edge
arallel to the (wx=c1+c2y along 2-3) herefore, alo dge 3-4
y depends on nodal dof of edge 3-4 nd along edge 2-3
x depends on nodal dof of edge 2-3 ifferentiate wy along edge 3-4 wrt x →Wxy
2-3 wrt y →Wyx
xy3-4≠ wyx2-3 ere as for continuous
nctions wxy=wyx (b2≠c2)
ssertion: It is erefore, impossible to use simple polynomials for shape nctions ensuring full co patibility when only w and its slopes are used as
of at nodes. any function satisfying comp ibility are found with the three nodal ariables, they must be such that atifferentia and the cross derivative is not un e. o far we hav applied the arg ent to a rectangular element, we can xtend this for any tw rbitrary directions of interfaces or common edges t node 3 or uadrilaterals). nfortunately, this extension requires continuity of cross derivatives in
tes the continuity requirement of potential energy theorem, also the hysical requirements. If the plate stiffness varies abruptly from element to lement then equality of moments normal to the interface cannot be
maintained.
∂yalong edge 3-4 with the number of jua- fo 3 and a
say axng e
Sfop isT-wa-wDDifferentiate wx along edgeThe first depends on nodal dof of edge 3-4 and the second depends on nodal dof of edge 2-3. At common node 3: wBecause of arbitrary nodal dof at nodes 2 and 4 whfu A thfu mdIf s at
corner nodes they are not continuously vd ble iquS e ume o a
qa (triangularUseveral sets of orthogonal directions, which in fact implies a specification of all second derivatives at a node. This leads to excessive continuity that violape
16
3- Elements for C1 Problems Constructing two-dime sional elements that can be used for problems
φ
1 continuity requires the specification , φx, φy, φxx, φyy, and φxy at the corner
ith sides parallel to the global rs nodes only φ, φx, φy and φxy.
n a oss derivative φxy will be directionally
pted to apply FE techniques to plate-bending
x y xyent.
nrequiring continuity of the field variable as well as its normal derivative φn along element boundaries is far more difficult than constructing elements for Co continuity alone. To preserve C1 continuity, we must be sure that φ and φn are uniquely specified along the element boundaries by the degrees of freedom assigned to the nodes along a particular boundary. The
ifficulties arise from the following principles: d1. The interpolation functions must contain at least some cubic terms
because the three nodal values φ, φx, and φy must be specified at each corner of the element.
2. For non-rectangular elements, C
six nodal values φof at least thenodes. For a rectangular element waxes, we need to specify at the corne
It is sometimes very convenient to specify only φ, φx and φy at corners, but when this is done, it is impossible to have continuous second derivatives at the corner nodes. In ge er l, the crdependent and hence, nonunique at intersections of the sides of the element. Analysts first began to encounter difficulties in formulating elements for C1
roblems when they attempproblems. For such problems, the displacement of the mid plane of the plate for Kirchhoff plate bending theory is the field variable in each element, and interelement continuity of the displacement and its slope is a desirable physical requirement. Also, since the functional for plate bending involves second order derivatives, continuity of slope at element interface is a mathematical requirement because it ensures convergence as element size is reduced. For these reasons, analysts have labored to find elements giving continuity of slope and value. Rectangular Elements Whereas triangulars are the simplest element shapes to establish C0 continuity in two dimensions, rectangles with sides parallel to the global axes are the simplest element shapes of C1 continuity in 2 dimensions. The reason is that the element boundaries meet at right angles, and imposing continuity of the cross derivatives φxy at the corners quarantees continuity of the derivatives that otherwise might be nonunique. A four-node rectangle with φ, φ , φ and φ specified at the corner nodes assigns a 16-dof elem 17
4- Triangular Elements 1For C continuity, by assigning 21 dof to element, we can make a complete
ined by six nodal values, norSlope varies as a quartic function which is uniquely determined by five nodal
ariables, namely φ and φ at each end node plus φ at the midside node. Thbookkfinal mAppar or onvergence in C problems. Experience has indicated that convergence is
quintic polynomial to represent the field variable φ. When φ and all first and second derivatives are specified at the corner nodes. There are only 18 dof, so 3 more are needed to specify the 21-term quintic polynomial.The 3 dof are obtained by specifying the normal derivatives φn at the midside nodes. This element quarantees continuity of φ along element boundaries because, along a boundary where s is the linear coordinate, φ varies in s as aquintic function, which is uniquely determ
mal, φ, φs and φss at each end node. continuity is also assured because the normal slope along each edge
v n nn ne presence of midside nodes is undesirable because they require special
eeping in the coding process, and they increase the bandwidth of the atrix.
antly, C1 continuity is not always a necessary condition f1c
more dependent on the completeness than on the compatibility property of the element. The following table shows a sample of incompatible elements. Any of these elements can be used in the solution of continuum problems involving functionals containing up to second-order derivatives. The analysts may ask, which element should I use to sole my problem? Unfortunately, no general answer can be given because the answer is problem dependent.
18
Some Incompatible Elements for C1 Problems
19
20
1
2
3
b
ac
ξ
η
θ
x
y
5- Nonconform
- we need an element of more general shape - Triangular elements fit curved edges more appropriately than the
rectangular elements - Again consider local coordinates ξ and η. We shall use
transformation matrix to go back to x-y system. - Consider w, wξ, wη as the dof at each node. - A cubic has 10 generalized parameters:
- for the element we have 9 dof but 10 generalized parameters in above equation. Therefore, must delete one of ai (i=1,2,…,10) or add a dof.
Possibilities:
a) use w at centroid as an extra dof -this element doesnot work sometimes and also exhibit poor convergence -Certain orientations may lead to less than a cubic along one of the edges and violates w continuity requirement
b) Throwaway one term- say a5=0 This violates constant curvature or constant strain energy requirement i.e. will not work since wξη=constant not present
c) combine two terms, i.e. equate a =a9 -we get a8(ξ2η + ξη2) which keeps some symmetry. -in general, ruins isotrophy of the polynomial so we expect orientation problems.
in Triangular Plate Bending Elements
310
29
28
37
265
24321 ηξηηξξηξηξηξ aaaaaaaaaaw +++++++++=
8
Recall:
[ ][ ]3111 ...... ηηξ wwwww =
e his h
10.....
a
w ⎫⎧
appens when two of
......010
......01
][
aaa
aaa
bbb
T
t equation is a constraint equation i.e. a8-a9=0 ng it.
c+b-a=0 then det[T]=0 and we cannot invert [T] to
21
110
.. aaA T =
⎭⎩ ×1101010 ][
0AT=⎬⎨ ××
[T] matrix becom s singular sometimes. T). edges are parallel to the global axes (x,y
d) Use area coordinates (Zienkiewics, 9dof triangular element)
-explain lack of full cubic because of only 9 dof. Let us look at (c ) in more detail. [T] matrix
⎪⎪⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−
=
10
9
8
3
2
198
2
.
.
.
11.............
2.
The lasThis is a more elegant way of doiDet[T]=c5(a+b)5(c+b-a) If a=c+b or formulate the element. If this situation is avoided then:
21
[ ]
[ ]
22
a=c
c b=0
y
x
ξ
45
Det [T]=0
η
y
x
3 4
2 1 b
a w1, wx1,wy1,wxy1
w3, wx3,wy3,wxy3
.),(),(cossin0sincos0
001][
][]0[]0[]0[][]0[]0[]0[][
][
][9][
][:
0
1
1
1
1
199102
12
199102110
1
axesyxandbetweenangletheiswhere
RR
RR
R
thenTofcolumnsfirstcontainsT
wTAaswrittenbecanthis
w
ηξθ
θθθθ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
=
⎭⎬⎫
⎩⎨⎧
××
−×××
−
6- Conforming Rectangular Element (16 dof) Nodalis permdof pe ression involving 16 constants could be used. We retain termor its normal slope than cubic along the sides. There are many alternatives as far as choosing the polynomial is concerned. But some of these alternatives may not produce invertible [T] matrix.
][][),(
][1),(
101
2322322
wTp
wTwTηξ
ηξηξηξηξηξηξηξ
=
=
×
][ TA =
w
][][
RwherewRw =
...... 3111 wwwwwtowfromtransformto yyx=
degrees of freedom at each node are w, wx, wy and wxy. Extra dof wxy issible as it does not involve excessive continuity. Thus, we have 16
r element and a polynomial exps which do not produce a higher order variation of w
An alternative is to use Hermitian polynomials. These are one dimensional polynomials and possess certain properties. A Hermitian polynomial Hn
mi(x) is a polynomial of order 2n+1 which gives, where x=xi:
Equation 2
jk
k
k
k
xxwhenormkdx
Hd
and
ntomformkdx
Hd
=≠=
===
0
01
A set of first order Hermitian polynomaols is thus a set of cubics giving shape functions for a line element ij and at the ends, slopes and values of the function are used as nodal degrees of freedom along 1-2
)(1)(
)2(1)(
)32(1)(
)3−
23
x
a
1 2
2(1)(
32
112
22
111
233
102
323101
axxa
x
xaaxxa
x
axxa
x
aaxxxH
−=
−=
−−=
+=
These polynomials are plotted in the following figure. isplacements and slopes at
)()()()()()()()( xyxyxy
y
wyHxHwyHxHwyHxHwyHx +++
The superscript for H has been dropped since all Hmi are 2x1+1=3polynomials (n=1). Further for Hmi(y), just replace x with y and a with b.
hecks . we can show that w(x,y) has three rigid body modes (can be
shown by performing an eigenvalue analysis) . we can also show that w(x,y) has constant strain modes.
3a
H
H
H
3 2 +
2
Note these polynomials provide unit values of done end and zero at the other as was implies by equation 2. assume w(x,y) of the following form:
12013120221102
111014021130212
201121011140201
302022010210101
)()()()()()(
)()()()()()(
)()()()()()(
)()()()()()(),(
yy
yxx
xx
yHxHwyHxHwyHxH
wyHxHwyHxHwyHxH
wyHxHwyHxHwyHxH
wyHxHwyHxHwyHxHyxw
+++
++
+++
+++=
1211312122111211111
4w
H 4xyrd degree
C1
2
3. continuity: look at edge 1-2 of the element: )()()()( 122111022011 xHwxHwxHwxHww xx +++=
a ues wy: only those terms having H11(y) will have non-zero v l()()()( xHwxHwxHwxHww )122111022011 xyxyyyy +++=
24
A4 3
21
4 3
21B
x
y
x
from above two equations, we note w and wy depends on nodal dof at nodes
r edge 1-2. Similarly, we can show that we get the same expressions for w and wy along edge 3-4 except w4 replaces w1, w3 replaces w2, etc. Therefore, equating the nodal variables along edge 1-2 of element A in the figure to nodal variables along edge 3-4 of element B will ensure continuity of w and wy as requitred. In exactly the same manner we can show continuity of w and wx along edges parallel to y axis.
Thus, the plate bending element discussed here is conforming in the sense so that the potential
e as strain energy from below as was shown for the beam problem, i.e. potential energy is bounded above and strain energy is bounded below.
1 and 2 fo
that displacements and normal slopes are continuous energy theorem does apply. We expect monotonic convergence of potential energy as well as strain energy. Potential energy will converge to the exact value from above wher
7- AltThe alternative method for deriving the stiffness matrix and the consistent load v in the previo to any rectangular or triangular elements.
an multiply out these polynomials in eq1 of the
yx +
to cubic terms. Using aylo series approach e r in w is f(h4) where h =typical element
dimension Error in strain f(h2) (strain are second derivatives) Error in strain energy is f(h4) For h=L/N, the strain energy error is f(N-4 where n=number of elements along side of length L Generally for convergence rate study, use square elements. Asymptotic convergence rate is N-4. When w is given in the form of above equation, it is obvious that w(x,y) contains rigid body modes and constant strains. We can write the polynomial in the following form:
ernative Method for Plate Bending Element
ector is presented for the conforming element discussedus section. However, the approach is general enough to apply
Although, we used Hermitian polynomials in deriving the displacement approximation, one cprevious section and obtain the following expression:
3316
3215
2314
313
2212
310
29
28
37
265
24321),(
yxayxayxaxyayxa
ayaxyayxaxayaxyaxayaxaayxw
++++
++++++++++=
n this equation, the polynomial is complete only up
311
IT rro
) a .
[ ][ ]3323213210210100
3231230123012010
),(16
1
=
=
=∑=
i
T
T
nmi
i
n
m
xayxw ii
Let us first obtain the stiffness matrix in terms of ai,s and later transform to obtain [K] in terms of w ,s.
25
[ ]
[ ]
[ ]
[ ]
)1)(1(),(
11
00 ++==
++
∫∫ nmbadxdyyxnmG
nmnm
ab
Note that w
:Define
)1(2
)1(222
22
222
11
1
⎪⎭−
−+++=
=
−+−+
−+++−+
−−
=
∫∫
∑
aya
yxnnmm
dxdywwwwwDU
yxanmw
ji
nnmm
nnmmnnmm
xyyyxxyyxx
nmiii
ixy
jiji
ii
νν
y term has been split into two terms tp preserve symmetry
It is obvious that this integration is not valid when m=-1 or n=-1 and blows up for m≤-1 or n≤-1 at lower limit i.e. x=0 (m=0,1,2,… and n=0,1,2,…). Strain Energy Can be written as:
)1)(1()1)(1(
)1)(1()1)(1(
222
44
00 ⎪
⎪⎪⎬
⎫
⎪
⎪⎪⎨
⎧
+−−+−−
+−−+−−
= −+−+∫∫ dxdyxnmnmnmnm
yxnnnnyxmmmmDU nnmm
ijijjiji
jijijijiab
ejiji
jijijiji
ν
)1(
16
216
1
1
−= −
=
=
∑ yxannw nmiii
iyy
i
ii
)1(
...
][
216
1621
1161616116
4
−=
=
=
−
×××
∑ yxammw
aaaA
ATw
nmiiixx
T
xy
ii
.... 4441111= wwwwwwwww yxxyyxT
A
⎪⎩ jijiν
xx, wyi.e. if we change I with j Ue is still the same.
26
[ ]
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−+−+−+−−+−−
+−++−−++−+−−=
=
0003000200010000000000300020010000000000000100000000000001
966343022001000033232320201003232302302010
100000300200100000000000000100000000000300201000000000000010000000000010000000000000000010000000000000000100000000000000001
][
:][
)2,2()1(2)1)(1()1)(1(
)4,()1)(1(),4()1)(1(
][21
2
2
2
32
222222
2322322322
3232232222
3332233223322322
2
32
2
32
bbbb
bbbbb
baabbababababababaabbaababababaabbababbababababababaabbabababbaabababa
aaaaa
aaaaa
T
computedbetohasmatrixTthenext
nnmmGnnmmnmnmnmnm
nnmmGnnnnnnmmGmmmmDK
AKAU
jijijijiijijjiji
jijijijijijijijiij
e
ννν
-Either we can program the matrix above or determine in a more
l
Where ε is a very small number, ε=10-13, instead of zero. This helps retaining some more accuracy and some times makes the inversion possible especially for triangular elements which may exhibit some orientation preferences. Then:
general form as follows: 4,3,2,1,: nodatheasiyxDefine =
),(),(),(,),(),( 2211
bayxyx
ii
εεεε
===
scoordinate
),(,),(),( 4433 yxbayx =
27
continueJNkyJMkxJNJMJIT
JNkyJMkxJNJITJNkyJMkxJMJIT
)1)(*(*)(*)(**)(*)(),2()(**)(*)1)(*(*)(*)(),1(
−=+−=+JNkyJMkxJIT
jDokI
kDoprogrammedbecanmatrixT
iforkiforkiforkiforkjwhere
iforyxnmT
iforyxnT
iforyxmT
iforyxT
jj
jj
jj
jj
nk
mkjjij
nk
mkjij
nk
mkjij
nk
mkij
59)1)(*(*)(*)1)(*(*)(*)(*)(),3(60
)(**)(*)(**)(),(16,1601)1(*44,159
][16,15,14,134
12,11,10,938,7,6,524,3,2,1116,....,3,2,1
16,12,8,4
15,11,7,3
14,10,6,2
13,9,5,1
11
1
1
−−=+
==+−=
=
=========
==
==
==
==
−−
−
−
The matrix [T] is then inverted and the stiffness matrix is the global oordinates is calculated: c
116161616
116161616 ][][][][ −
××−×× = TkTK
one hbecausmay l . terms like mi+nj-4, etc. For examp =0, n1=0 then m1+m1-4=-4 and n1+n1-4=-4. These are the smallest possible indecises for G(m,n) or [G]. This can be avoided by taking a matrix [F] such that: F(m+5,n+5)=G(m,n) Where [F] has dimensions at least 4 larger than [G] would require.
as to be cautious when computing G(m,n) or [G] matrix. This is e some of the terms (lower order) in the polynomial of equation 1 ead to negative or zero m and n i.ele, m1
4,3,2,14,3,2,10
)1)(1(),()5,5(
,
11
00
===
++===++
++
∫∫jandiforFwhere
nmbadxdyyxnmGnmF
ji
nmnm
ab
28
Load Vector Assume constant load q0/unit area applied to the plate. Therefore, work done is given by:
( )
( ) scoordinateglobalinvectorloadfTf
inmGqf
nmGaqW
fTwfAW
dxdyxaqwdxdyqW
Tiii
iiii
e
TTTe
nmi
i
a b
Ae
ii
11616161
0
16
10
1
16
10
0 00
][
16,..,2,1),(
),(
][
××−
=
−
=
=
==
=
==
==
∑
∑∫ ∫∫∫
Stress Matrix for Obtaining Moments (for an element) Recall:
]][[
][
)1(
01][
][
1000101
][
1
1
113
22
16
3
wTS
wTA
yxnmS
yxnnS
DD
D
www
DMMM
momentsandstressesMMM
curvatureandstrainASwww
jj
jj
ii
nmjjj
nmjjj
nm
xy
yy
xx
xy
yy
xx
xy
yy
xx
xy
yy
xx
−
−
−−
−
=
=
=
−=
⎤⎡ −−=
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−−
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
ε
νετ
νν
ντ
τ
ε
)1(1210001
2EhD−
=⎥⎥⎥
⎦⎢⎢⎢
⎣ −−−
ννν
)1(
)1(
16
216
1
216
1
1
yxannw
yxammw
dxdyxaw
ii
ii
nmiii
iyy
nmiii
ixx
ii
−
=
−
=
=
−=
−=
=
∑
∑
∑
)1( 21
1
yxmmS jj nmjjj
i−
=
−=
11 yxanmw ii nmiiixy
−−=∑
29
where w is the displacement vector for element under consideration and is
⎪⎩
⎪⎨= wTSDMM
xy
yyτ
Note that the matrix [S] is function of x and y and has to be evaluated at the points (xi,yi) where bending moments and twisting moment are desired to be evaluated. The [T]-1 matrices can be stored away e.g. on a file so that these can be used
r determining moments later, i.e. after displacements have been
ioned earlier, the procedure is general and only changes need to be made are integration routine, different data for mi and ni and changing sizes of various matrices. The logic does not change at all.
extracted from the global displacement vector. ⎫⎧M xx
1161
161616333 ][][][ ×−×××=
⎪⎭
⎪⎬
focalculated. As ment
30
8- Triangular Element for Conforming C1 Continuity Using quintic polynomial for the displacement field: Equation 3
]
),(
),(
21
521
420
3219
2318
417
516
415
314
2213
312
411
310
29
287
265
24321
=
++++++++
++++++++++++=
∑
3
[ ][ 00
0123450123401230120101
=
=
5432104321032102101
=
nm yxayx
yaxyayxayxayxaxayaxyayxa
yxaxayaxyayxaxayaxyaxayaxaayxw
ii
required mdent equations to relates a 's to the dof.
ption One: ix dof at corner nodes (1,2,3), i.e. w, wx, wy, wxx, wxy and wyy and one dof t the mid side nodes (4,5,6) i.e. wn
T
Ti
ii
n
m
w
i
There are 21 generalized parameters ai's, therefore either 21 dof areor 21 indepe i
nie
x
y
x
y3
2
1 4
5 6
Option One Option Two
3
1
2
2ˆne
1ˆne
3ˆne
OS
( nn eww ˆ.∇= )
ption two: nly six dof at corner nodes (1,2,3), i.e. w, wx, wy, wxx, wxy and wyy for a tal of 18 dof per element. Additional three equations come from
onstraining the normal slope wn to vary cubically.
a OOtoc
31
x
y
ith edge
s β
li
i(xi,yi)
i
j(xj,yj)
]... nnnyyxyxxyx wwwwwwww
Edge Geometry Consider the ith edge defined by nodes i and j as shown. Let s be the running coordinate along the edge and be the unit outward normal to the ith edge:
[ ]2121 ... T aaaa = ][ aTw =
[ 1 T ww = 65411111
nie
( )
i
nie
i
thiini
iisi
jijii
syandsxedgeithealongalso
jie
jjforijyyxxl
ββ
ββ
sincos:
ˆ
ˆsinˆcosˆ
3,31)()( 22
==
+=
=>+=−+−=
Option One Equation 4
where
i
iji
i
iji
jie
lyy
lxx
ββ
ββ
ˆcosˆsin
sincos
−=
−=
−=
(3,2,121,..,3,2,1cossin 133
13,18 atiandjyxnyxmT n
imiji
ni
mijji
jjjj −+++
−++ ==−= ββ )
][][
6,5,4
cossinˆ.
:.183,2,1
)1(
)1(
21,..,3,2,13,2,1)1(61
1212121
3
2,5
,4
2,3
1,2
1,1
,
TinvertaTw
nodes
wwewnodessidemid
nodescornertheatdofofcaretakes
yxnnT
yxnmT
yxmmT
yxnT
yxmT
jandiikyxT
i
yxn
ni
mijjjk
imijjjk
nmjjjk
ni
mijjk
mijjk
nmijk
jj
jj
jj
jj
j
jj
×××
−+
+
−+
−+
−+
=
−=
−=
−=
=
−=
=
=
==−+==
ββ
i
ni
j
11 n
ii
−−
iAt
wwn n ∇==∂∂
121
32
Option Two Equation4 (a-f) for corner nodes still apply. These yields 18 eqns and therefore three more equations are still to be accounted for. Note that for a
uintic polynomial, the normal slope along all three edges vary as quartic omial).
Consider only the 5 degree term in equation 3 and denote this partial w(x,y) as wp i.e.:
q(4th degree polyn"additional three equations arise from constraining the normal slope to vary cubically along each edge."
th
24
213 () a
i
−+2520
234
43218
52317
416
521
420
3219
2318
417
516
)sincos5sincos4(sin)sincos3sins
)sincos2sincos3()cossincos4()sincos(
cossinˆ.
sincos:
Sa
aasaw
yw
xw
ewn
wsyandsx
edgeanalongalso
yaxyayxayxayxaxaw
iiiiiiiii
iiiiiiiip
ip
ip
nipp
ii
p
⎥⎥⎦
⎢⎡
−+−
+−+−+=
∂
∂
∂−
∂
∂=∇=
∂
∂==
+++++=
βββββββββ
βββββββββ
ββ
ββ
[…] is the combined coefficient of s4. For wn to be cubic along an edge […] must be set equal to zero and hence yields three more equations, from each edge. Hence,
⎤
19 co2(an ⎢⎣∂
Note the bracked term
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11,18
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33
34
x
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8.2- Transformation of Nodal DOF along an Inclined Edge Before any boundary conditions can be applied along an inclined edge, all first and second derivatives must be transformed to perpendicular and parallel to the edge.
k edge th
s
βk
k
l
nkeskenjnjisis wworiww ,,,, 2,1 λλ
For the first derivatives: iinin woriww ,,, 2,1 nin w,λλ ===
===
here λ are direcwunit outward norm
ni tion cosines of the al and λsi are the
direction cosines of the unit tangential vector . For second derivatives:
[ ] [ ]
nke
ske
[ ][ ]
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,
,
,
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,)(21,2cossinsincos:
][coscossin2sin
cossinsincoscossinsincossin2cos
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90,cossinsincos
××××
×
×
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=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
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⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
−−−==
⎪⎭
⎪⎬
⎫
⎪⎩
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=⎪⎭
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⎫
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⎪⎨
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−−−
=⎪⎭
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===
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βθβθ
θθθθθθθθθθ
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λλλλλλ
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, =⎬⎫
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⎫⎧−
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9- Two-Dimensional Creeping Flow
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35
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4,,
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,
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2
=∇=∂∂
≠
≠
=∂∂
∂∂
=∇∴∂∂
=∇=+−=∇
∇+∇=+−∇=+∇=∇
+−=
+=
==∇=∇=
Ψ∇Ψ∇
==∇=∇=∴≠=
≠−=∂∂
+∂∂
−=∇∴=+=∇
=∴==
∂∂
−=∂∂
=
==∇
==∇
=∇=Ψ+Ψ+Ψ=Ψ∇+Ψ∇
ΩΨ∇+
ΩΨ∇+Ψ∇−Ψ∇=
Ω+Ψ
=
Ω∫∫∫
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d
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ψµδψ
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ψµθθψµ
θµθµθθνµθψθψµψµ
θθ
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µµ
µν
ψψνψδ
τµψνψψψνψν
ψδ
ψψ
δψψν
δψψν
ψνν
δψ
δψδψνδψνδ
δψδψ
2
22
22 =ΨΩΨ∇=Ω∫∫ functionlinestreamdI ν
∇=ΩΨ∇Ψ∇=ΩΩ∫∫∫∫ dI νδνδ
y
2,
22 Ψ∇+Ψ∇−Ψ∇= ∫∫∫∫ nn dsdsI νδψνδψνδΩ
22
eqnfield
ν
:conditionsBoundary
for
9.1- Fully Developed Parallel Flow
36
u0 R=0.5
10
Section 2 Section 1
x
y
Section 2
Ψ=1
Ψ=0
1
x
y
Section 1 twotionon
edgetoponand
edgebottomatandonetiononandyy
sBcyy
V
yyy
U
x
y
y
x
sec0
01
00sec023
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0
)1(6
32
32
=
=Ψ=
=Ψ==Ψ−=
==−=
=
−=∂Ψ∂
=
ψ
ψ
ψψ
ψψψ
9.2- Flow Past a Cylinder Computational domain=20xR away, flow can be assumed uniform Bc,s:
twotionon
edgetoponuanduxalongbottomat
onetiononandyusBc
x
y
x
sec0
100
sec0'
00,
0
=
=Ψ==
=Ψ=
ψ
ψψψ
