Finite Element Analysis of Plane...
57
Chapter 3 Finite Element Analysis of Plane Elasticity 1
Transcript of Finite Element Analysis of Plane...
Chapter 3
Finite Element Analysis of
Plane Elasticity
1
1- Basic Equations of Solid Mechanics (3-D) 1.1- Stress and strain 1.2- Strain-displacement equations (Kinematics equations) 1.3- Linear Constitutive Equations 1.4 Potential Energy for a Linear Elastic Body (general for
2- Dimensional Specializations of Elasticity
2.1- Plane Strain 2.2- Plane Stress 2.3- Axisymmetric Problem
3- Plane Elasticity
3.1- Strain Energy in Plane Elasticity 3.2- Kinematic Relations for FE analysis of Plane Stress and
Strain Problems 3.3- Initial Stresses and Strains 4- Plane Stress Rectangular Element
4.1- Stiffness Matrix 4.2- Load Vector 4.3- Strains and Stresses in Rectangular Element 5- Plane Stress Triangular Elements 5.1- Constant Stress Triangular Element (C.S.T) 5.2- Linear Stress Triangular Element (L.S.T.) 5.3- Quadratic Stress Triangular Element (QST)
5.4- Boundary with Springs
6- Natural Coordinates and Shape Functions "C0" 6.1- Line and Rectangular Elements 6.2- Triangular elements 7- Curved, Isoparametric Elements 7.1- Geometric Conformability of elements 7.2- Constant Derivative Condition
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8- Quadrilateral Iso parametric Elements 8.1- Four node Quadrilateral Element 8.2- Eight Node Quadrilateral Element (Plane Elasticity) 8.2.1- Element Properties (Evaluation of stiffness Matrix)
8.2.2- Stiffness Matrix 8.2.3- Comment on det[J] 8.2.4- Body Forces 8.2.5- Specific Boundary Stress (stress boundary condition)
9- Numerical Integration (Gauss Quadrature) 9.1- Line Integration
9.2- Gauss Quadrature Numerical Integration Over a Rectangle
9.3- Comment on Numerical Integration 9.4- Choice of Quadrature Rule, Instabilities 9.5- Stress Calculation and Gauss Points
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1. Basic Equations of Solid Mechanics (3-D) 1.1- Stress and strain State of stress in a volume (3D) zxyzxyzyx
T τττσσσσ =
=zyx σσσ ,, Normal components of stress =zxyzxy τττ ,, Components of shear stress
Strain zxyzxyzyx
T γγγεεεε =
1.2- Strain-displacement equations (Kinematics equations) u , v and w are displacements in x,y and z directions, respectively.
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛
∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
+∂∂
=222
21
xw
xv
xu
xu
xε
Retaining only the first order term and neglecting second order:
zw
yv
xu
zyx ∂∂
=∂∂
=∂∂
= εεε .
Only for small deformation, that each derivative is much smaller than unity
yw
xw
yv
xv
yu
xu
yu
xv
xy ∂∂
∂∂
+∂∂
∂∂
+∂∂
∂∂
+∂∂
+∂∂
=γ
.
xw
zu
zv
yw
yu
xv
zxyzxy ∂∂
+∂∂
=∂∂
+∂∂
=∂∂
+∂∂
= γγγ
for large deformation higher terms must be retained “ geometric nonlinearity ” 1.3- Linear Constitutive Equations The stress tensor and strain tensor are related. These relations depend on the nature of the material and are called constitutive laws. We shall be concerned in most of this text with linear elastic behavior, wherein each stress components is linearly related, in the general case, to all the strains by equations of the form( and vice versa):
klijklij C ετ = C’s are at most functions of position. This law is called generalized Hook’s law.
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Since τij and εkl are second-order tensor fields and symmetric, Cijkl must be symmetric in ij and kj and also fourth-order tensor field. We may assume that the material is homogeneous( same composition throughout) so Cijkl is constant for a given reference. Hook’s law in one dimensional space can be written as: DoneE −= )( εσ The generalized Hook’s law in matrix notation can be written as.
[ ] [ ] σε
εσDCD
==−3
[ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
666261
262221
161211
.......
..............
ccccccccc
C
Starting with 81 (34) terms for Cijkl due to symmetry only 21 terms are independent for Linear elastic, anisotropic and homogeneous material. Therefore, [C] and [D] are symmetric with 21 experimental evaluation of elastic constant. For linear orthotropic, [C] becomes:
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
66
55
44
33
2322
131211
..
.........
cc
ccccccc
9 constants
The stress-strain equations for orthotropic materials may be written in terms of the young’s moduli and poisson’s ratios as dollows:
zx
zxzx
yz
yzyz
xy
xyxy
zz
yy
yzx
x
xzz
zz
zyy
yx
x
xyy
zz
zxy
y
yxx
xx
GGG
EEE
EEE
EEE
τγ
τγ
τγ
σσυ
συ
ε
συ
σσυ
ε
συ
συ
σε
===
+−−=
−+−=
−−=
1
1
1
5
There are 12 material parameters which only 9 are independent. This is because of the followings:
xz
x
zx
z
zy
z
yz
y
yx
y
xy
x EEEEEEυυυυυυ
===
Linear Isotropic Elasticity Isotropic material are those that have point of symmetry, that is, every plane is a plane of symmetry of material behavior. This property requires that mechanical properties of a material at a point are not dependent on direction. Thus a stress such as τxx must be related to all the strains εij for reference xyz exactly as the stresses τx’x’ is related to all the strains ε’ij for a reference x’y’z’ rotated relative to xyz. Accordingly, Cijkl must have the same components for all references. A tensor such as Cijkl whose components are invariant wrt a rotation of axes is called isotropic tensor. Only two independent elastic constants are necessary to represent the behavior for linear, isotropic and elastic material, the stress-strain relation in this case can be written as:
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+
+
+
−
−−
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
zx
yz
xy
z
y
x
zx
yz
xy
z
y
x
E
ESymmetric
E
E
EE
EEE
τττσσσ
υ
υ
υ
υ
υυ
γγγεεε
)1(2
0)1(2
00)1(2
0001
0001
0001
or in terms of stress components:
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−−
−−
−+=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
zx
yz
xy
z
y
x
zx
yz
xy
z
y
x
Symmetric
E
γγγεεε
υ
υ
υυ
υυυυυ
υυ
τττσσσ
221
0221
00221
000100010001
)21)(1(
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It is well to know that a material may be both isotropic and inhomogeneous or, conversely, anisotropic and homogeneous. These two characteristics are independent. 1.4 Potential Energy for a Linear Elastic Body (general form) The potential energy can be written as:
1
1
),,(
)(
dSwTvTuTdVwfvfufwvudu
tractionsurfaceandforcesbodyLoadsAppliedtheofEnergyPotentialWEnergyStrainU
WU
S
zyx
V Vzyx ∫∫∫∫∫ ∫∫∫ ⎟
⎠⎞
⎜⎝⎛ ++−⎟
⎠⎞
⎜⎝⎛ ++−=
==
−=
−−−−−−
π
π
S1 is surface of the body on which surface tractions are prescribed. d(u,v,w) is strain energy per unit volume (strain energy density). The last two integrals represent the work done by the constant external forces, that is, the body forces and surface tractions
. A bar at the top of a letter indicates that the quantity is specified.
zyx fandff−−−
,,
zyx TandTT−−−
,
⎭⎬⎫
⎩⎨⎧=
⎭⎬⎫
⎩⎨⎧
⎭⎬⎫
⎩⎨⎧=
⎭⎬⎫
⎩⎨⎧
=
⎭⎬⎫
⎩⎨⎧−⎟
⎠
⎞⎜⎝
⎛⎭⎬⎫
⎩⎨⎧−=
=−==
−−−−
−−−−
−−
∫∫∫∫∫
zyx
T
zyx
T
T
S
T
V
TT
TT
TTTT
ffff
wvuuwhere
dSTudVfuC
energystraindlE
dlDdVCdVwvudu
1
2
1
2][21
)21
21:1(][
21
21),,(
εεπ
σεσεεσε
For springs on the boundary we can write for a 2-D case:
springscoupledofeffectsvu
vududu
uududutoaddedbemustboundarytheonsprings
uududu
jiij
T
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+=
+=
+=
2221
1211
21
21
][21
αααα
α
α
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2-Dimensional Specializations of Elasticity Some times due to geometry and loading configurations a 3-D problem may reduce to problem in one or 2-D. 2.1- Plane Strain In this case, strain normal to the plane is zero. Long body whose geometry and loading do not vary significantly in the longitudinal direction are taken as plain strain cases, such as dams, retaining walls, etc. In plain stress problems, we may consider only a slice of unit thickness. If f(x,y) is variable at a cross section some distance away from the ends, we may assume w=0 (displacement along the z direction. Then
)(0 yxzzxyzz then σσνσγγε +==== nonzero strains are εx , εy and εz. Then we can write:
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−
−+=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
xy
y
x
xy
y
x E
γεε
ννν
νν
νντσσ
22100
0101
)21)(1(
Earth dam
2.2- Plane Stress In this case, stress normal to the plane is zero such that problem can be characterized by very small dimensions in the z direction for example thin plate loaded in its plane. No loading are applied on the surface of the plate, then:
)(1
0 yxzzzxyz then εεν
νεσττ +−
====
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xyyx and τσσ , are averaged over the thickness and independent of z.
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⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
xy
y
x
xy
y
x E
γεε
νν
ν
ντσσ
2100
0101
1 2 y y
x z Plane stress: thin plate with in-plane loading 2.3- Axisymmetric Problem In this case, axisymmetric solids subjected to axially symmetric loading. Due to symmetry, stress components are independent of angular coordinates. Hence, all derivatives wrt θ vanish and component
zrzrv θθθθ ττγγ ,,,, become zeros. Then nonzero components relation can be written as:
z, w
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
−−
−−
−+=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
∂∂
+∂∂
=∂∂
==∂∂
=
rz
z
r
rz
z
r
rzzr
symmetry
E
rw
zu
zw
ru
ru
γεεε
νν
ννννν
νντσσσ
γεεε
θθ
θ
221
010101
)21)(1(
θ
u,r Cylinder under axisymmetric loading
3. Plane Elasticity In a domain of Ω for i=1,2,3 and j=1,2,3 we have:
EquationsKinematicsuu
EquationsveConstitutiE
EquationsmEquilibriuf
ijjiij
klijklij
ijij
)(21
,,
,
+=
=
=
ε
ετ
τ
These three set of equations have to be satisfied within the two dimensional domain of Ω. Further, these are also subjected to some boundary conditions. T yy
τnn υ τns
τxx Ω τxy
τyy
x x Following three sets of boundary conditions occur commonly in plane elasticity.
a) Homogeneous boundary conditions
ui
Tjij
Sonu
Son
0
0
=
=υτ
jυ =components of unit outward normal ST=Stress free part of the boundary Su=Part of the boundary where displacements are specified to be zero
b) Mixed homogeneous boundary conditions
Miijjij
ui
Tjij
SonuSonu
Son
00
0
=+=
=
αυτ
υτ
ijα =Constants such as spring constant SM=Part of the boundary where mixed conditions are specified (e.g. springs or elastic foundation etc.)
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c) Non homogeneous boundary conditions (Mixed)
Miiijjij
uii
Tijij
SonCu
Sonuu
SonT
−
−
−
=+
=
=
αυτ
υτ
Where quantities with the bar on the top are specified. Let us consider the following condition for now:
Miijjij
ui
Tijij
SonuSonu
SonT
00
=+=
=−
αυτ
υτ
Potential energy expression is then given by (t=thickness):
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+−Ω−Ω= ∫∫∫∫
−
ΩΩ
dSuudSuTdufdtMT S
iiS
iiiiijij αετπ21
21
equation 1
where the last two integrals are the boundary integrals (line integrals). 3.1- Strain Energy in Plane Elasticity Previously, the strain energy had been written either in tensor notation or matrix notation. In x, y coordinates, strain energy can be written as:
( )
( )dxdytU
thentconsistif
dVU
EnergyStrainU
xyxyyyyyxxxx
xyxyyyyyxxxxV
εγετετ
εγετετ
++=
++=
=
∫∫∫
∫∫∫
Ω21
:tan21
Substituting for stresses from constitutive equations of plane stress:
dxdyEtU xyyyxxyyxx ⎟⎠⎞
⎜⎝⎛ −
+++−
= ∫∫∫Ω
2222 2
12)1(2
1 γνενεεεν
Substituting for strains from kinematic relations:
( ) dxdyvuuuuuEtU xyyxyx ⎟⎠⎞
⎜⎝⎛ +
−+++
−= ∫∫∫
Ω
2222 2
12)1(2
1 ννν
11
In plane stress formulation, if we change ν with ν/(1-ν) and E with E/(1-ν2) then the constitutive equations would be for plane strain. You can compare the different boundary conditions of above with the boundary conditions obtained in assignment no.1 problem 4. 3.2- Kinematic Relations for FE Analysis of Plane Stress and Strain
Problems
In simplified notation:
[ ] uLvu
xy
y
x
xy
yy
xx
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∂∂
∂∂
∂∂
∂∂
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
εγεε
0
0
[ ] ετ D= Where [D] is the elasticity matrix obtained previously. [D] for plane stress and strain problems are different. [L] is the linear operator matrix. Next consider approximations for u(x,y) and v(x,y). Suppose those are given by (within an element):
( ) ][][:][]][[][
]][][[][]][[
][
000000
),(),(
3
3
2
2
1
1
321
321
332211
332211
DDnoteDLD
LDDL
u
vuvuvu
vu
u
vvvyxvuuuyxu
TTTeT
e
e
ee
e
===
==
=
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎦
⎤⎢⎣
⎡=
⎭⎬⎫
⎩⎨⎧
=
++=++=
φδετ
δφετ
δφε
δφ
φφφφφφ
φφφφφφ
Substituting above equations into the strain energy, we get strain energy within an element:
12
( )[ ] [ ]
yandxoftindependenis
dBDBtU
LB
dLDLtU
e
eeTTee
eeTTee
e
e
]][[][2
]][[][
]][[][]][[2
δ
δδ
φ
δφφδ
Ω=
=
Ω=
∫∫
∫∫
Ω
Ω
Assuming homogeneous boundary conditions i.e. the last two integrals in equation 1are zero. The body force term then yields:
0][
][]][[][
][]][[][2
..
][
=⎭⎬⎫
⎩⎨⎧
−
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛Ω
⎭⎬⎫
⎩⎨⎧
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛Ω=
Ω⎭⎬⎫
⎩⎨⎧
−Ω=
=Ω⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
=Ω⎭⎬⎫
⎩⎨⎧
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
=⎭⎬⎫
⎩⎨⎧
Ω⎭⎬⎫
⎩⎨⎧
=Ω⎭⎬⎫
⎩⎨⎧
−
−
ΩΩ
−
ΩΩ
−
−
Ω
−
Ω
−
−−
−
Ω
−
Ω
∫∫∫
∫∫∫
∫∫
∫∫
eee
eTeeTTee
eTTeeeTTee
ee
y
xTTeeT
y
x
eTeT
FK
dftdBDBt
dftdBDBt
forcesbodyofPEorforcesbodybydoneworkei
Wdf
ftdfut
f
ffwhere
dfutduft
ee
ee
ee
ee
δ
φδδδδπ
φδδδπ
φδ
13
3.3- Initial Stresses and Strains We showed that stresses are given by: [ ] ετ D= In general, material within the element boundaries may be subjected to initial strain 0ε such as may be due to temperature changes, shrinkage etc.. In addition, the body may be stresses by some known stresses 0τ (residual stresses, …), such stresses, for instance could be measured but the prediction of which, without the full knowledge of the material history is impossible. When both, initial stresses and strains are taken into account, then the stresses due to externally applied loads are given by: [ ] ( ) [ ]
( ) ( ) ( ) eTeTe
eT
e
dtdDtU
stressgcauStrain
dtU
bygiveniselementjththeenergyinstrainthenowDD
D
ee
e
Ω−+Ω−−=
−==⎭⎬⎫
⎩⎨⎧
Ω⎭⎬⎫
⎩⎨⎧
=
+−=+−=
∫∫
∫
ΩΩ
−
−
Ω
0000
0
00
00
][21
sin
21
:][
τεεεεεε
εεε
τε
τεεττεετ
Expanding the above equation yields:
( )
eTeT
eTTTTe
dtdt
dDDDDtU
ee
e
Ω−Ω
+Ω+−−=
∫∫
∫
ΩΩ
Ω
000
0000 ][][][][21
τετε
εεεεεεεε
The first term on the LHS is the same as that we obtained before: eT
e dDtUe
Ω= ∫Ω
εε ][2
14
The second and third terms are equal:
( )
( ) eTT
e
TTT
eeT
eeT
eT
e
eeTTeTeT
eTeT
eTeTeTe
dDBtF
Hence
Df
where
dBftdLftF
dLDtdDtdt
dtdt
dDtdDtdDtU
e
ee
eee
ee
eee
Ω−=⎭⎬⎫
⎩⎨⎧
−=⎭⎬⎫
⎩⎨⎧
Ω⎭⎬⎫
⎩⎨⎧
=Ω⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
=
Ω−=Ω−Ω=
Ω−Ω
+Ω+Ω−Ω=
∫
∫∫
∫∫∫
∫∫
∫∫∫
Ω
−
−
−
Ω
−
Ω
−
ΩΩΩ
ΩΩ
ΩΩΩ
00
00
0000
00
*
0
00
*
0
][][
:
][
][]][[
]][[][][*
][2
][][21
ετ
ετ
δδφδ
δφετεεετ
τετε
εεεεεε
44 344 21
444 3444 21
Now if we rewrite the potential energy expression ,
Te
ee
eTe
eT
eeTee
FfK
YieldsMinimizing
tsConsFfK
−⎭⎬⎫
⎩⎨⎧
=
+−⎭⎬⎫
⎩⎨⎧
+=
−
−
δ
δδδδπ
][
:
tan][21
Initial strains and stresses (know priori) contribute to the load vector.
15
4- Plane Stress Rectangular Element We have four corner nodes and it is logical to have u and v displacements as dof at each node. Hence, there are 8 dof per elements. y, η v4 v3 3 u44 η=y/a
1 2 v2
v1
u3
u1
b ξ=x/a x, ξ u2 a Let the generalized forces at the ith node be Ui and Vi. Assume the polynomials (within the element) for u and v:
bilinearLinear
hgfevdcbau
ξηηξηξξηηξηξ
+++=+++=
43421),(),(
The ξη terms are bilinear, because for constant η, u and v are linear in ξ, and for constant ξ, u and v are linear in η. Check for convergence can be done by following requirements:
a) Rigid body modes Two translation u )tan( tConseva == One rotation )tan( tConsfcvu
−=∂∂
−∂∂
ξη
b) Constant strain This implies that u and v should be arbitrary linear functions i.e. gyfxevcybxau ++=++=
c) Continuity Both displacements u and v are required to be continuous between elements. "continuity of disp. And its derivative to the order of (n-1) where n is the highest derivatives in the PE function".
d) Spatial Isotrophy Inclusion of xy instead of x2 or y2 (ξη instead of ξ2 or η2) satisfies this requirement.
Requirements a, b, c and d are all satisfied by the equation of the displacements. Note along the edges u and v are linear. e.g. edge 1-2:
16
ξηξξηξ
fevbau
+=+=
),(),(
Therefore, we have to match displacements at only two nodes (1 and 2) to make u and v continuity along an edge.
mpmp
mpmp
vvvv
uuuu
3241
3241
==
==
m4 3
1 2
1 p
4 3
2 Therefore, displacements equations for u and v satisfy the convergence requirement. Now find constants a to h in displacement equations in terms of generalized degree of freedom (nodal dof) ui's and vi's.
cauudcbauu
bauuauu
+==+++==
+====
)1,0()1,1()0,1()0,0(
4
3
2
1
Solving the above equations for a to h in terms of ui's to get: Similarly for vi's .
ηξφξηφηξφηξφ
φηξφηξ
ηξξηηξηξηξηξξηηξηξηξ
)1()1()1)(1(:
),(),(
)1()1()1)(1(),()1()1()1)(1(),(
4321
4
1
4
1
4321
4321
−==−=−−=
==
−++−+−−=−++−+−−=
∑∑==
where
vvuu
orvvvvvuuuuu
iii
iii
( )( )143)1(),(
021)1(),(
3
2
4
1
=−+−=
=−+−=
ηξξηξ
ηξξηξ
edgealonguuu
edgealonguuummm
ppp
Then matching will make u continuous along the boundary between the two elements. Similarly for v displacement.
mpmp uanduuandu 3241 ,
17
4.1- Stiffness Matrix Substitute u and v in the expression for strain energy:
( )
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=∂∂
=∂∂
=
==
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎭⎬⎫
⎩⎨⎧ ++⎟
⎠⎞
⎜⎝⎛ −
+++
−=
=
=⎟⎠⎞
⎜⎝⎛ +
−+++
−=
∫ ∫
∫∫∫Ω
jj
ii
xi
ii
i
jijijijijiji
jijijijijiji
T
eTxyyxyx
ua
ua
u
jioversummation
ddvv
avu
abuu
b
vuab
vvb
uuaEtU
vuvuvuvu
KdxdyvuuuuuEtU
ξ
φ
ξφ
ξφ
φηφ
φ
ηξφφφφφφν
φφνφφφφ
ν
δ
δδννν
ξη
ξξξηηη
ηξηηξξ
11
4,3,2,14,3,2,1
1212
1
211
)1(21
][21
212
)1(21
2
22
221
0
1
02
44332211
2222
Castigliano's Theorem (1st) i
ei u
UF
∂∂
= Ue is the strain energy of the
element, ui is the generalized displacement and Fi is the generalized force corresponding to the generalized displacement ui. Then the first row of the stiffness matrix is given by:
ratioaspectSabcall
ba
babEtK
egrationafterand
ddub
ua
EabtUK
ddv
abu
b
vab
uaEabt
uU
K
e
e
jjjj
jjjje
jjei
=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −
−+−
=
−−=−−=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −
+−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎭⎬⎫
⎩⎨⎧ +⎟
⎠⎞
⎜⎝⎛ −
++
−=
∂∂
=
∫ ∫
∫ ∫
221,1
11
11121112
1
0
1
0211,1
112
1121
0
1
02
1,
21)1(24
)1(12
int)1()1(
211
)1(2
222
1
22
)1(2
ννν
ξφηφ
ηξφφνφφν
ηξφφφφν
φφνφφ
νδ
ηξ
ηηξξ
ξηηη
ξξξξ
Similarly we can determine the other components, finally:
18
SS
k
SS
kS
SkS
Sk
SS
kkS
Sk
SkkS
Sk
kkkkkkkkkkkkkkk
kkkkkkkkkkk
kkkkkkk
SYMMkkk
EtKe
)1(22
)1(2)1(14)1(12
)1(4)31(23)1(22
)1(234)1(
23)1(24
.
)1(12][
10
987
654
321
321059265
1582754
326592
15427
32105
158
32
1
2
ν
ννν
ννν
ννν
ν
−−=
−−−=−+−=−−−=
−+−=−−=−−=
−+=+=−+=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−
−−−
−
−=
4.2- Load Vector
yetknownnotisFWF
recallVUVUVUVUF
KF
ee
eTe
Te
eee
δ
δ
∂
∂=
=
=
:
][
44332211
In order to determine the load vector for an element we need to know the loading, i.e. body forces, boundary stresses etc.. Consider the case of gravity loading: fy=-γ where γ is the specific weight. The work done by gravity is:
101010104
4
,
4
),(
862
4
1
1
0
1
012
1
0
1
0
AtF
bygivenisloadgravity
FFAtV
WF
similarly
AtdabdtV
WF
dabdvtdAvtW
Teg
eee
ge
ege
iie
g
γ
γ
γηξφγ
ηξφγηξγ
−=
==−=∂
∂=
−=−=∂
∂=
−=−=
∫ ∫
∫ ∫∫∫
19
Next considering an element pth with one edge coinciding with the boundary of the domain as shown on the figure: Suppose both normal and shear stresses are prescribed on this boundary. Therefore, work done by the boundary stresses is:
3 Boundary edge4
20
b
ξ=s/b
1 2
τnn
τns
mannerclockwisecounterisegrationdsuTtW ii
ST
T
int−
∫=
Further suppose the normal and tangential stresses vary along edge 2-3 boundary stresses may then be approximated by a parabolic expression:
b ξ=s/b
4 3
1 2
Boundary node number p3
P2
p1
3
)2()44()231(
)()(
)2()44()231()(
'')(
42)
2(
)0()(
23
22
21
3
1
32
22
12
23213
2
3212
11
2321
ξξϕξξϕξξϕ
ξϕξ
ξξξξξξξ
+−=−=+−=
=
+−+−++−=
++==
++==
==++=
∑=
iii
ii
pp
pppp
sboftermsinscforsolvebcbccbpp
bcbccbpp
cppscsccsp
2
1
Normal and tangential (or shear) stresses can now be approximated by above equation using the stress values at the temporary boundary nodes 1,2 and 3 i.e. let:
332211
332211
nsynsynsy
nnxnnxnnx
pppppp
ττττττ
======
In order to use them in the equation of potential energy, we also need to know u and v distribution along the edge2-3. From before we know it is linear: If any of the other edges have prescribed stresses on it, i.e. more than one edge coincides with the stress boundary, then we can generate another load
vector similar to above vectors, with non zero entry indifferent positions of the load vector. For example, if edge 3-4 also has prescribed stresses, then F1
eT= F2eT =F3
eT =F4eT=0 the new FT
e for 3-4 is then added to FTe for
edge2-3.
( ) ( ) ( ) ( )[ ]
( )
( )
( )
( )
( ) ( ) ( ) ( )[ ]
eT
eg
e
yyxxyyxxTe
T
yyTeT
xxTeT
yyTeT
xxTeT
eTeTeTeT
yyyyxxxxT
yxT
FFF
bygiventhenisvectorloadtotalelementthisforthen
ppppppppbtF
onlyedgeonstressestodue
ppbtv
WF
ppbtuWF
ppbtvWF
ppbtuWF
FFFF
vppvppuppuppbtW
obtainweegrationon
dvpbtdupbtW
vvvuuu
+=
++++=
−
+=∂
∂=
+=∂∂
=
+=∂∂
=
+=∂∂
=
====
+++++++=
+=
+−=+−=
∫∫
:,
002222006
:32
26
26
26
26
02
22226
:,int
)()()()(
)1()()1()(
32322121
323
6
323
5
212
4
212
3
8731
332221332221
1
0
1
0
32
32
ξξξξξξ
ξξξξξξ
21
4.3- Strains and Stresses in Rectangular Element Recall:
[ ]
[ ]
( ) ( )
( ) ( ) ⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−+−+−+−
++−++−=
∂∂
+∂∂
=∂∂
+∂∂
=
−+−++−=∂∂
=∂∂
=
−+−++−=∂∂
=∂∂
=
−++−+−−=−++−+−−=
ηξξηγ
ξη
ε
ηξ
ε
ηξξηηξηξηξηξξηηξηξηξ
43214321
2121
432141
432121
4321
4321
11
1111
)(11
)(11)1()1()1)(1(),()1()1()1)(1(),(
vvvva
uuuub
vva
uubv
au
bxv
yu
vvvvvvb
vby
v
uuuuuua
uax
uvvvvvuuuuu
xy
yy
xx
Stresses are then obtained from equations [ ] ετ D= . [D] can be etermined based on the problem at hand is plane stress or plane strain. Note strain εxx is linear in η and εyy linear in ξ where as γxy is complete linear. Further, these are not continuous across the interelement boundaries. It is convenient to obtain strains at the centroid of the element, i.e. at ξ=η=0.5
[ ]
[ ]
( ) ( )43214321
4321
4321
21
2121
21
vvvva
uuuub
vvvvb
uuuua
xy
yy
xx
−++−+++−−=
++−−=
−++−=
γ
ε
ε
Then proceed to obtain stresses at the centroid from theses strains.
22
5- Plane Stress Triangular Elements 5.1- Constant Stress Triangular Element (C.S.T) Note: cheap element
u2
u3
v2
v3
v1
x
y
u1
3
2 1
Assume linear displacements, i.e.
fyexdyxvcybxayxu
++=++=
),(),(
Here, we have three node triangle with two dof per node- six dof per element. We have six parameters a, b, c, d, e, and f to go with six dof. We can proceed in exactly the same manner as we did for the plane stress rectangle. However, we will follow a more general approach involving transformation matrix. Define x and y as the global coordinates and ξ and η and the local coordinates for a triangular element as shown in the figure. Specify the nodal coordinate as (x1,y1), (x2,y2) and (x3,y4) for nodes 1, 2 and 3, respectively. The calculated a, b, c and θ are:
(x1,y1)
(x2,y2)
(x3,y3)
θ
c
v2 u2
v3 u3
2
3
1
v1 u1
a
b
η y, v
ξ
x, u
23
( ) ( )
( ) ( ) ( )( ) ( )([ ]
( ) ( )
)
( )( ) ( )([ ]
( ) ( )
)
( )( ) ( )( )[ ]121312131313
121312131313
122312322332
212
212
21212
1212
1sincos
1sincos
1sincos
)(sin
)(cos
yyxxxxyyr
xxyyc
yyyyxxxxr
yyxxb
yyyyxxxxr
yyxxa
yyxxrr
yybayy
rxx
baxx
−−−−−=−−−=
−−+−−=−+−=
−−−−−=−−−=
−+−=−
=+−
=
−=
+−
=
θθ
θθ
θθ
θ
θ
Now assume displacement in ξ and η system:
ηξηξ
ηξηξ
654
321
),(
),(
aaav
aaau
++=
++=−
−
now find in terms of nodal displacements 621 ....,, aaa 311 ....,,−−−vvu
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡−
−
=
=
⎭⎬⎫
⎩⎨⎧
=
=
+=+==
+=+==
−=−−=−=
−−−−−−−
−
−−−
−−−
−−−
cc
aa
bb
T
aaaaaaA
vuvuvu
AT
caavcaacuu
aaavaaaauu
baavbaabuu
T
T
010000000101000000010100000001
][
][
),0(
)0,(
)0,(
654321
332211
643313
542212
541211
δ
δ
Note that det[T]=-c2(a+b) ≠0 for practical triangles, hence [T] is nonsingular and can be inverted A=[T]-1
−δ .
24
Strain Energy Calculation
( ) ηξννν
ηξννν
ηξηξ
ξηηξηξ
ddaaaaaaaaEtU
avavauau
ddvuvuvuEtU
Ae
Ae
⎟⎠⎞
⎜⎝⎛ ++
−+++
−=
====
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+
−+++
−=
∫∫
∫∫−−−−
−−−−−−
2553
2362
26
222
6,5,3,2,
2
,,,,2
,2
,2
22
12)1(2
212
)1(2
Integrand consists of constant terms and ∫∫ +=A
cbadd )(21ηξ
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−
−−
−
+=
∂∂
=∂∂
=∂∂
=
==
−
−−
= =∑∑
10000
02
102
100000000
02
102
10000010
000000
)1(2)(][
...
][21
21
2
22
11
6
1
6
1
ν
νν
ννν
νcbaEtk
etcaU
akaU
akeiaU
ak
AkAaakU
ejj
ejj
i
ejij
Tjiij
i je
Now transform to generalized displacement in local axes. Note that ( ) ]][[][
21][
21 11
−−
−−
−−
−
−== δδδδ TkTkU
TTTe from this equation ( ) 11 ]][[][][ −
−−
−= TkTk
T
y, v
Where is the stiffness matrix in local coordinate system. Next transform
from coordinates to u and v in x, y coordinate or global coordinates.
][−k
][−k ηξ ,, invu
−−
Consider rotation between ξ, η and x, y. η, v
θθθθ
θθθθ
cossinsincos
cossinsincos
vuvvuu
orvuvvuu
+−=+=
+=−=
−−
−−−− v
ξ, u v uθ
u x, u Why do we neglect translation between x, y and ξ, η? Because rigid body translation of an element does not contribute to strain energy, hence we neglect it.
25
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡−
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
=
=−
0000
]0[cossinsincos
][][]0[]0[]0[][]0[]0[]0[][
][
][
332211
θθθθ
δδ
δ
rr
rr
R
R
vuvuvuT
Again strain energy is invariant and:
( ) ( )
][]][[][][]][[][][
][21]][[][
21
][][][21][
21
11 RTkTRRkRk
kRkR
RkRkU
TTT
TTT
TTe
−−
−
−
−
−
−
−
−
==
==
==
δδδδ
δδδδ
[k] is the final stiffness matrix in the global coordinate system. Computing Steps 1- Specifying global coordinates of element nodes (x1,y1), (x2,y2) and (x3,y3) 2- Calculate a, b, c and θ 3- Calculate [T] and invert numerically 4-Calculate [ ], multiply [T]
−k -1[R]=[Q]
5- Calculate [k]=[Q]T[ ][Q]
−k
6- Return [k] to the main program Accuracy of Constant Stress Triangles We have u and v linear in x and y. Error in u and v from taylor's series is f(l2) where l is some typical size of an element. Therefore, error in strain(constant) is f(l). Hence, errot in strain energy is given by f(l2), if l=L/N where N is number of elements along a typical length of a problem, then the error in strain energy is given by f(1/N2). Unfortunately, we do not know the constant of proportionality α in the error =α/N2.
26
5.2- Linear Stress Triangular Element (L.S.T.) Linear stress implies, we need quadratic or parabolic displacement field:
265
24321
265
24321
),(
),(
ybxybxbybxbbyxv
yaxyaxayaxaayxu
+++++=
+++++=
Note that six parameters a1,…,a6 and six u1,…..,u6 and similarly, six b1,…,b6 and six v1,…..,v6 are equivalents for 12 dof per element. We have assumed complete polynomial of degree 2. A complete polynomial is invariant regarding rotation and allows order of error to be determined from Taylor expression, with expression in displacement function, we also satisfy the convergence requirement i.e. rigid body modes, constant strains, and spatial isotrophy of the displacement field. y,v
3
27
x,u
1 u2
6 5
4 2
v2u6
v6 What about continuity? Along and edge, u and v will be quadratic function of edge coordinate. For the quadratic shown along the edge, we have 3 parameters α, β and γ and 3 dof u1, u4 and u2. Therefore, equating, these u’s along the edge will satisfy the continuity requirement.
1 4 2
η
ξ
U=α+βξ+γξ2
5.3- Quadratic Stress Triangular Element (QST) Again use the same coordinate system (local) as for CST, i.e. ξ, η (u ).
−−
v
28
1
7 6
5
4
3 2
9
8 10
Centroidal node
y, v
x, u
3
1 (x1,y1)
(x3,y3) (x2,y2)
2 η, v
− ξ, u −
a
b
,
,
For stress to be quadratic within the element, need a polynomial which is at least cubic:
[ ]
[ ]3210210100
0123012010
),(
),(
10
110
10
1
320
219
218
317
21615
214131211
310
29
28
37
265
24321
==
==
+++++++++=
+++++++++=
∑
∑
=+
−
=
−
−
−
nav
mau
aaaaaaaaaayxv
aaaaaaaaaayxu
i
nmi
i
nmi
ii
ii
ηξ
ηξ
ηξηηξξηξηξηξ
ηξηηξξηξηξηξ
To transform the generalized coordinates a1, . . . , a20 into the nodal degrees of freedom, we need to decide on discretization in terms of dof and the number of nodes.
a) At each nodes the dof are u . We require 10 dof in and 10 dof in
for a
−−
v−
u−
v 1, . . . , a20. Then obtain:
[ ]20321
10102211
......
....
:][
aaaaA
vuvuvu
whereAT
T
vu
T
a
cc
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
=
−
−
δ
δ
29
3 24 u4, v4
b) at each corner nodes take u as the dof and u at the centroid. Thus, we have six dof per corner nodes and 18+ 2=20 dof per element.
ηξηξ ,,,, ,,,,,−−−−−−
vvvuu−−
v,
Then obtain:
1u1, u1,ξ, u1,η
[ ]20321
44111111
......
..
:][
aaaaA
vuvvvuuu
whereAT
T
T
b
=
⎥⎦⎤
⎢⎣⎡=
=
−−−−−−−
−
ηξηξδ
δ
Determinant of [Tb]=c14(a+b)14/729 which is nonzero for all practical problems. Regardless of which discretization we use, we always compute [ k ] in generalized coordinates a
−
1, . . . , a20.
][21
:)!2(
!!])([),(
int
212
)1(2
111
210
1
10
1
2,
210
1
10
1
2,
110
1,
2
,,,,2
,2
,2
AkAU
termsallofevaluationAfternmnmbacddnmF
likeegralgeneralGet
ddmmaaddu
mmaaumau
ddvuvuvuEtU
Te
mmnnm
A
nnmm
Aijiji
iA
nnmm
ijiji
i
nm
iii
Ae
jiji
jijiii
−
+++
+−+
==
−
+−+
==
−−
=
−
−−−−−−
=
++−−==
=
==
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +
−+++
−=
∫∫
∫∫∑∑∫∫
∑∑∑
∫∫
ηξηξ
ηξηξηξ
ηξηξ
ηξννν
ξ
ξξ
ξηηξηξ
Now transform to nodal variables (DOF) in local coordinates, then transform to global coordinates i.e. into u, v etc.
5.4- Boundary with Springs Recall
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+−Ω−Ω= ∫∫∫∫
−
ΩΩ
dSuudSuTdufdtMT S
iiS
iiiiijij αετπ21
21
suppose edge 1-2 of the element shown is on an elastic foundation. Foundation modulus of spring stiffness can be approximated by:
ξξξ NNN KKK 21 )1()( +−=
ξ 2
30
u1, v1
1 u2, v2
K1
N
K2N
For CST, v displacement along the edge is given by: ξξξ 21 )1()( vvv +−=
the last term in equation of the PE (on the RHS) is:
ξξξ dvKlU Nl
B )()(21 2
0∫=
Contribution to the nodal forces is then given by:
[ ] [ ]
[ ]
⎭⎬⎫
⎩⎨⎧
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
=⎭⎬⎫
⎩⎨⎧
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡++⎥
⎦
⎤⎢⎣
⎡+=
∂∂
=∂∂
=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡++⎥
⎦
⎤⎢⎣
⎡+=
−+−=∂
∂=
∂∂
=
==∂∂
=
∫
2
1
2121
2121
4
2
221
121
244
221
121
2
21012
2
654321332211
4121212
1212124
4121212
1212124
)1()1()(
vv
KKKK
KKKK
lFF
vKKvKKlvUUF
vKKvKKlF
dvvKlv
UUF
vuvuvuwhereUF
NNNN
NNNN
B
B
NNNNBBB
NNNNB
lBBB
i
BBi
δ
ξξξξξδ
δδδδδδδδ
For tangential springs along edge 1-2: ξξξ TTT KKK 21 )1()( +−=
and: ξξξ 21 )1()( uuu +−=
and:
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−−−−−−−−−−−−−−−−−−−−−−−−−−−
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎭⎬⎫
⎩⎨⎧
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛+
=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
6
5
4
3
2
1
6
5
4
3
2
1
2
1
2121
2121
2
1
:
4121212
1212124
δδδδδδ
YYXX
YYXX
FFFFFF
followsasismatrixstiffnessthetoaddition
uu
KKKK
KKKK
lFF
TTTT
TTTT
B
B
31
6- Natural Coordinates and Shape Functions "C0" Natural Coordinates are dimensionless, homogeneous and independent of size and shape of the elements. 6.1- Line and Rectangular Elements Relation between natural and global coordinates:
)1(21)()1(
21)(
)1(21)1(
21
21
2
1
21
ssNssNXNX
XsXsX
ii
i +=−==
++−=
∑=
Y Constant A and E
S=+1 S=-1 U2 U1 S 2
32
X1
1
X2 X, U
ι
Line element in global and natural coordinate system Here N1 and N2 are functions of the natural coordinates s and are called element shape functions or interpolation functions. These shape functions or interpolation functions can be used in describing the linear displacement field u within the bar or line element.
dXdsUU
dXds
dsdU
dXdU
UUUUUNU
UsUsU
ii
i
2
)1()1(
)1(21)1(
21
12
21
2
1
21
−===
=+=−=
++−=
∑=
ε
Note above equation for U provides only C0 continuity since only U is continuous across the node between two adjacent elements.
ectedasl
UUldX
dsorlXXdsdX
exp
222
12
12
−=
==−
=
ε
Hence, strain-displacement transformation matrix [B] is given by:
[ ]
[ ]
[ ] dsJElAK
elasticityofulusEDWheredXBDBAK
UU
l
lB
Tl
][1111
][
mod][]][[][][
111
111][
1
12
0
2
1
+−⎭⎬⎫
⎩⎨⎧+−
=
==
⎭⎬⎫
⎩⎨⎧
−=
−=
∫
∫+
−
ε
Where [J] is the Jacobian relating the element length in global coordinate system to an element length in natural sys. For one dimensional problem:
⎥⎦
⎤⎢⎣
⎡−
−=⎥
⎦
⎤⎢⎣
⎡−
−=
==
∫+
−1111
21111
][
21
12 l
AEdsllAEK
dslJdsdX
This is the stiffness matrix of the line element. For rectangular elements, the natural coordinates take in the values of E1 on the edges of the rectangle as shown. For four node rectangles, the interpolation functions or shape functions are:
)1)(1(41)1)(1(
41
)1)(1(41)1)(1(
41
43
21
tsNtsN
tsNtsN
+−=++=
−+=−−= t=+1u4, v4 u3, v3
4 3 ts =-1 s=+1s
1 2
u1, v1 u2, v2 t=-1 Recall bilinear polynomials used for u and v in rectangular plane stress element. If the origin of x and y co-ordinates for the element is chosen at the centroid, the shape functions φ1 , .., φ4 are the same as N1, .., N4 except we have s and t coordinates instead of ξ and η (ξ=x/a, η=y/b) We can use natural coordinates to express u and v within the element as:
33
21
21
21
21
4
1
4
1
)1(21)1(
21
)1(21)1(
21
1,21.)1,1()1,1(.)1,1()1,1(
vsvsv
ususu
tedgealongfurtheretcvvvvetcuuuu
vNvuNu ii
iii
i
++−=
++−=
−=−=−+=−−=−+=−−
== ∑∑==
t s
y (0,0,1) (x3,y3)
i.e. u and v vary linearly along edge 1-2, therefore equating the nodal displacements u and v at nodes along an edge provides continuity of the displacements i.e. C0 continuity. 6.2- Triangular elements (x,y)
p(L1, L2, L3) L2=0 L1=0 A1 A2 Use area coordinates L1, L2 and L3
34
x
(1,0,0)(0,1,0)
(x1,y1) (x2,y2)
A3
1321
321
33
22
11
=++−−=
===
LLLtrianngleofAreaA
AA
LA
AL
AA
L
L3=0
The only two of the area coordinates L1, L2 and L3 are independent. Relation between area coordinates and the global coordinates of any point p is given by:
[ ] [ yxZLLLZT
yx
abAabAabA
ALLL
yx
xxyyyxyxxxyyyxyxxxyyyxyx
ALLL
invertingLLL
yyyxxx
yx
TT 1][
1
222
21
1
21
1111
321
3312
2231
1123
3
2
1
12211221
31133113
23322322
3
2
1
3
2
1
321
321
==ℑ=ℑ
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−−−−−−
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
]
Geometric interpolation of the terms in above equation is as follows:
DB
∂
∂
∂
∫
∫i
d
I
s
A
Area O12=A12Area O13=A13
y
L1=11
3
2
h hL1
hdL1
dA
−b
b
L1=0b=b(1-L1)
ifferentiation y chain rule:
)!2(!!!2
:
33)
33
2()1(
)1(
:
2sin
21
,
2sin
21
321
10
21
111
1
01
111
3
1
3
1
3
3
2
2
1
13
1
1
+++=
==−=−=
−==
=∂∂
∂∂
=∂
=∂∂
∂∂
=∂
∂∂
∂∂
+∂
∂∂
∂+
∂∂
∂∂
=∂∂
∂∂
=∂
∫
∫
∑
∑
∑
−
=
=
=
rqprqpA
dALLL
ngeneral
AbhLbhdLLbhLdAL
dLLbhdLhbA
nntegratio
Aa
yL
ceL
aAy
imilarly
Ab
xL
ceL
bAx
LxL
LxL
LxL
LxL
x
rqp
A
L
A
ii
ii
i
ii
ii
i
i
i
i
. Constant Stress Triangles (CST)
Area O23=A23Area Oij=Aij
ya2 a1
3 3 b1b2
2 2 1 b3 1a3
x x A12
A13
35
36
x,u
y,v
1 u2
65
4
3
2
v2 u6
v6
C0 continuity
7 63 2
5
49
8 10y, v u
332211
332211
vLvLvLvuLuLuLu
++=++=
Three dof in u and v for a total of six dof (3 nodes)
This gives a linear approximation for u and v within the element. Or N1=L1, N2=L2 and N3=L3. Note that L1, L2 and L3 are linear function of x and y.
u2
u3
v2
v3
v1
x, u
y, v
u1
3
21
B. Linear Stress Triangles (LST) Six dof in u and v for a total of 12 dof (6 nodes)
iii
iii
vNvuNu
LLNLLNLLNLLN
LLNLLN
∑∑==
==
=−==−=
=−=
3
1
3
1
136333
325222
214111
4)12(4)12(
4)12(
C. Quadratic Stress Triangles (QST) Ten dof in u and v for a total of 20 (10 nodes). Again C0 continuity.
iii
iii
vNvuNu
LLLNLLLNLLLNLLLNLLLNLLLNLLLNLLLN
LLLNLLLN
∑∑==
==
=−=−=−=−=−−=−=−−=
−=−−=
10
1
10
1
321102215
11391314
31383333
33272222
23261111
272/)13(92/)13(92/)13(92/)13(92/)23)(13(2/)13(92/)23)(13(
2/)13(92/)23)(13(
v
37
. Tetrahedrals gles:
DAs we saw in trian
E. Derivation of Stiffness Matrix for Plane Stress CST element Strain energy Ue given by:
dxdyvvbbvubauuaa
vuabvvaauubb
AEtU
UoequationsabovengSubstituti
vbuaAx
vyu
vaAy
v
ubAx
u
vLvLvLvuLuLuLu
thicknesstdxdyEtU yyxxe ⎜⎛ ++= ∫∫ 222
2 εε
jijijiiijiji
jijijijijiji
i jAe
e
iiiii
xy
iii
yy
iii
xx
xyyyxxA
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
++−
+++
−=
+=∂∂
+∂∂
=
=∂∂
=
=∂∂
=
++=++=
=⎟⎠⎞
⎝
−+
−
∑ ∑∫∫
∑
∑
∑
= =
=
=
=
22
1
2
)1(8
int
)(21
21
21
21
)1(2
3
1
3
122
3
1
3
1
3
1
332211
332211
2
ν
ν
ν
γ
ε
ε
γνενεν
)!3(!!!!6:
...
int
]det[
10
4321
13422341
4
3
2
1
4321
4321
4321
++++=
==
=
=
⇑=
≤≤
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎨
⎥⎥⎥
⎦⎢⎢⎢
⎣
=
⎪⎪⎭
⎪⎪⎬
⎪⎪⎩
⎪⎪⎨
∫∫∫ srqpsrqVpdVLLLLnIntegratio
etcVVVVge
ppoatjklverticesbysurroundedlTetrahedraofvolumeVV
scoordinatevolumeVV
L
VVolume
L
LLL
zzzzyyyyxxxx
zyx
srqP
V
pp
ijklpi
ii
i
11111⎪⎧
⎥⎤
⎢⎡⎫⎧ L
The integrand in above eqn. consistes of all constant terms and integration is simply multiplication with area A:
[ ] [ ]
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+++++++++++++++++++++
=
−==
−=
⎥⎦⎤
⎢⎣⎡ −
+−
=∂∂
∂=
∂∂∂
=
⎥⎦⎤
⎢⎣⎡ −
+−
=∂∂
∂=
∂∂∂
=
⎥⎦⎤
⎢⎣⎡ −
+−
=∂∂
∂=
∂∂∂
=
⎥⎦⎤
⎢⎣⎡ −
+−
=∂∂
∂=
∂∂∂
=
⎥⎦⎤
⎢⎣⎡ −
+−
=∂
∂=
∂
∂=
=====
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⎭⎬⎫
⎩⎨⎧ −
+
+⎭⎬⎫
⎩⎨⎧ −
++⎭⎬⎫
⎩⎨⎧ −
+
−=
∑∑= =
3333
33333333
323232322222
2332323222222222
31313131212121211111
313131312121212111111111
2
2121221
2
42
2
24
21
212
11
2
22
2
22
2121221
2
31
2
13
1111211
2
21
2
12
31
2122
1
2
21
2
11
231211654321332211
2
3
1
3
1
][
21
)1(4
)2
1()1(4
)2
1()1(4
)2
1()1(4
)2
1()1(4
)2
1()1(4
..
212
21
21
)1(8
bbaabaabaabbSymmetricbbaaabbabbaabaabaabbbaabaabb
bbaaabbabbaaabbabbaabaabaabbbaabaabbbaabaabb
k
AEtlet
bbaaA
Etvv
UUk
baA
Etvv
UUk
aabbA
Etuu
UUk
baabA
Etvu
UUk
abA
Etu
UUk
thereforeuvueivuvuvu
vubaab
vvbbaauuaabb
A
Et
U
ee
ee
ee
ee
ee
T
jijiji
jijijijijijii j
e
γγφγγγβγγβγγβγ
γγβγγβγγβγγβγγβγ
νγνβν
α
ννδδ
ννδδ
ννδδ
νννδδ
ννδ
δδδδδδδδδδ
νν
νν
ν
to convert plane stress to plane strain replace E by E and ν by ν E=E/(1-ν2) ν=ν/(1-ν)
38
7- Curved, Isoparametric Elements -Elements with curved boundaries are useful for problems with curved edges or surfaces for 3-dimensional problems. -Unless edges of triangular elements are very small, we may introduce error due to approximation of curved boundaries by straight lines. -We end up facing both the mathematical problem as well as shape problem Elements of basic one-, two- or three dimensional types will be mapped into distorted shapes in the manner indicated in the following figures. In both figures s, t or L1, L2 and L3 coordinates can be distorted into curvilinear set when plotted in Cartesian coordinates. Similarly, single straight line can be transformed into a curved line in Cartesian coordinate and flat sheet can be distorted into a three dimensional space. Figure 3 indicates two examples of two dimensional (s, t) element mapped into a three dimensional (x, y and z) space.
Figure 1. Two dimensional mapping of some elements
39
Figure 2. Thrre-dimensional mapping of some elemnts
Figure 3. Flat elements (of parabolic type) mapped into thre dimensions
40
Coordinate transformations are required:
However, to apply the principle of transformations, there must exist a one to one correspondence between Cartesian and curvilinear coordinates i.e. no severe distortion- folding back or for a line element α i.e. no cross over etc.
.
3
2
1
etcLLL
forts
fyx
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
A most convenient method of establishing co-ordinate transformations is to use shape functions discussed in the previous section and already used to represent the variation of unknown quantities or functions. Express x and y for each element as:
...''...''
2211
2211
++=++=
yNyNyxNxNx
when the shape functions of local coordinates s, t or L1, L2 and L3 in two dimensional problems (fig. 1). Here the shape in x-y coordinate system is distorted and triangular or square shapes in local coordinates are called parent elements. In above equations N'i are the shape functions with unit values at the nodes. For curved shapes in x-y, these must be nonlinear functions of s, t or L1, L2 and L3. x1, x2 … and y1, y2 … are nodal coordinates. 7.1- Geometric Conformability of elements This requires that by the shape function transformation, the mapping of the parent element into real object should not leave any gaps or holes, figure4. Theorem 1: If two adjacent elements are generated from parent elements in which the shape functions satisfy continuity requirements, then the distorted elements will be continuous. The theorem above is obvious and follows from C0 continuity implied by shape functions for any function that is approximated. Here, the functions are x and y and if the adjacent elements are given the same coordinates at the common nodes, continuity is implied. Unknown Function within Distorted, Curvilinear Elements and Continuity Requirements
41
So far we have defined the shape of distorted elements by the shape functions N'I for a parent element. Suppose the unknown function to be determined in φ and can be approximated by:
where Ni are the usual shape functions and for now assume these are different from N'i. φi are the nodal dof.
elementsperdofofnumbermN ii
m
i
== ∑=
φφ1
Theorem 2: If the shape functions Ni are such that continuity of φ is preserved in parent co-ordinates (or local coordinates)- then continuity requirements will be satisfied in distorted elements. Proof is same as for theorem 1 The nodal values φI may or may not be associated with the same nodes as used to specify the element geometry.
Subparametric Isoparametric Super parametric
Points at which φI are specified
Nodes used for element geometry, coordinates are specified
42
Isoparametric Ni=N'iSuper-parametric Variation of geometry is more general than the
unknown φ Sub-parametric More nodes to define φ (i.e. more general) than to
define geometry Subparametric elements have been used more often in practice. 7.2- Constant Derivative Condition This means that constant stress should be included in the approximation.
Consider φ in equation . Constant derivative condition requires: ii
m
i
Ni
φφ ∑=
=1
yx 321 αααφ ++=
Where α1, α2 and α3 are generalized paramAt nodes (xi, yi):
Therefore, constant derivative condition will be obtained if equations ii
'∑=
here xi and yi are element nodal coordinates. For isoparametric elements
eters.
321 yx iii ++= αααφ
( )
)(1
:
321
321
iiyyNxxNN
yieldsequationaboveequating
yNxNN
yxNN
iii
iii
ii
iii
iii
ii
iiii
iii
===
++=
++==
∑∑∑
∑∑∑
∑∑
αααφ
αααφφ
are satisfied. Recall co-ordinate transformation in equation of:
iii
iii
yNy
xNx '∑=
wNi=N'I therefore, equations of above are automatically satisfied.
Need 1=∑ N ii
43
Theorem 3: The constant derivative condition will be satisfied for
heorem alid for sub parametric I can be expressed as a
where Cij are constants. It is obvious that for the case of subparametric elements, N'I i and above equation can be easily
es)
all isoparametric elements providing 1=∑ ii
N
The same requirement is necessary and t velements provided the shape function N'linear combination of Ni i.e.:
iiji
i NCN ∑='
are of lower order than Nsatisfied, however not so far superparametric elements. Some numericaltests may have to be performed in order to satisfy above equations. (Perhaps, a patch test should be performed, also an eigenvalyue analysis of the stiffness matrix may reveal the presence of constant stress nod
Figure 4. Compatibility requiremen in real subdivision for transformation
44
8- Quadrilateral Iso parametric Elements
4
3
2
tsNtsNtsN
tfunctions
+−=++=−+=
+≤≤−+≤
coordinate transformation:
yNy ∑=
=4
1
4
displacement approximations:
are the degrees of freedom of node I in global coordinate system
8.1- Four node Quadrilateral Element
Local coordinates in s and t:
4/)1)(1(4/)1)(1(4/)1)(1(
11114/)1)(1(1 stsN ≤−−−=Shape
iii
iii
xNx ∑=
=1
ui and vi x,y.
x, u
3
2 1 u1
v1
(x1, ) (-1,-1)
(x4,y4) (-1,+1) ( 3)
(+1,+1)
(x ,y(1,-1)
u2
u3u4
v2
4 v3y, v v t x3,y4
s
y1
2) 2 (-1,-1) (1,-1)
(+1,+1) (-1,+1)
s
t
Distorted Elem Parent Element
iii
vNv ∑=
=4
1
4
ent
iii
uNu ∑=
=1
45
note x and y are linear function of s and t , the local coordinates straight edges remain straight from parent to distorted element after transformation.
ted for this lement.
2/)1)(1(4/)1)(1)(1(
284
273
2
251
stNtstsN
tsNtstsN
st
tsNtstsN
−−=−++−−=
+−=−−++−=
+
−−=++−−−=
Try the following coordinate transformation:
8
1
1
tsgyNy iii
iii
== ∑=
=
Displacements are approximated by:
vNv ∑=
=8
1
8
8.2- Eight Node Quadrilateral Element (Plane Elasticity) A detailed derivation of isoparammetric elements is presene
x, u
y, v 3
2 1
v1
u2
u3u4
v2
v4 v3
s
4
u1
(-1,-1) (1,-1)
(+1,+1) (-1,+1)
s
t
Global system S and t are curvilinear coordinates here
Local co-ordinates
8
7
6
5
5
3 4
1
7
6
2
8
t
Shape functions in local coordinates:
2/)1)(1(4/)1)(1)(1(
2/)1)(1(4/)1)(1)(1(
2/)1)(1(4/)1)(1)(1( 62 NtstsN −=+−−+−=
),(8
tsfxN == ∑x
),(
iii
uNu ∑=
=1
iii
46
Both equation for displacement and coordinates satisfy the requirements of theorem 1, 2 and 3 in the previous section and hence the convergence criteria. 8.2.1- Element Properties (Evaluation of stiffness Matrix) Displacement
[ ] [ ]
11616313
8
8
2
2
1
1
332211
321
321
882211
8
8
2
2
1
1
87654321
87654321 0⎡=
⎫⎧ Nu
][
.
.
...
...000
...000
]][[][][..][
.
.000000000000000
××× =
⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
=====
⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪
⎨
⎧
⎥⎦
⎤⎢⎣⎭
⎬⎩⎨
δε
γεε
ε
δεδδ
Bvu
vuvu
xN
yN
xN
yN
xN
yN
yN
yN
yN
xN
xN
xN
NLBBvuvuvuvuuNu
vu
vuvu
NNNNNNNNNNNNNNN
v
xy
yy
xx
TT
But Ni are functions of s and t and so are z, y functions of s and t. Therfore, using chain rule:
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
∂∂∂
∂
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
∂∂∂
⎥⎥⎥
⎦⎢⎢⎢
⎣ ∂∂
∂∂
∂∂=
⎪⎪⎭
⎪⎪⎬
⎪⎪⎩
⎪⎪⎨
∂∂∂
∂∂
∂∂
+∂∂
∂∂
=∂
yNx
N
J
yNx
ty
tx
ss
tNs
sy
yN
sx
xN
sN
i
i
i
i
i
iii
][
∂
⎧∂⎤⎡ ∂∂⎫⎧∂
∂∂
∂∂
+∂∂
∂∂
=∂
∂
NyxN
ty
yN
tx
xN
tN
i
iii
47
The matrix [J] is called the Jacobian matrix and can be found explicitely in
rms of local coordinates s and t, and xi and yi using equation 2. The left and side of above eqn can also be evaluated using shape functions.
teh
)1(2/)1(
2/)1()1(
)1(2/)1(
)1()1(
4/)2)(1(4/)2)(1(
4/)2)(1(4/)2)(1(
4/)2)(1(4/)2)(1(
4/)2)(1(4/)2)(1(
:
][][
][
][
..
..
.....
...][
][
828
277
626
255
44
33
22
11
2221
12111
1
88
33
22
11
321
321
82
11
stt
Nt
sN
st
Nts
sN
stt
Nt
sN
st
Nts
sN
stst
Ntst
sN
stst
Ntst
sN
stst
Ntst
sN
stst
Ntst
sN
Next
IIIII
J
tNs
N
J
yNx
NynumericallJinvertNow
yx
yxyxyx
tN
tN
tN
sN
sN
sN
J
yt
Nx
tN
ys
Nx
sN
J
ttss
i
i
i
i
ii
ii
ii
ii
ii
ii
−−=∂
∂−−=
∂∂
−=∂
∂+−=
∂∂
+−=∂
∂−=
∂∂
−−=∂
∂−−=
∂∂
−−=∂
∂−+=
∂∂
++=∂
∂++=
∂∂
−+=∂
∂−−=
∂∂
+−=∂
∂+−=
∂∂
=⎥⎦
⎤⎢⎣
⎡=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
∂∂∂
∂
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
∂∂∂
∂
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∂∂
∂∂
∂∂
∂∂
=
∂∂∂∂
−
−
×
==
88
8
1
8
1
yNy
yNy
xt
Ntxx
sN
sx
ii
ii
ii
i
i
∂=
∂∂=
∂
∂∂
=∂∂
∂∂
=∂∂
==
∑∑
∑∑
48
),(),(]det[
tsyxJ
∂∂
=Also dxdy=det[J] dsdt. Note transform derivatives with respect
x and y to with respect to s and t. ow back to strain
toN ][ δε B= . Matrix [B] which consists of Ni,x and Ni,y, can
ow be determined by using above equations:
1 for both s and t.
BDBtk T det[][][][][ 16333316
1
1
1
11616 ×××
− −
× ∫ ∫=
its of integration are simple when integration is carried out over the parent element in s and t coordinates, any typical term of the integrand matrix of above integration becomes very complicated algebraically and hence, we have to resort to
However, for very
n
]][[][2
1000101
)1(
][
...
...0
...0][
2
22,112,111,122,121
,122,121
,112,111
δετ
εεε
νν
ν
ντττ
τ
ετ
BDD
E
stressplaneforD
stress
NNNINININIwithwithNINIsamesameNINI
B
xy
yy
xx
xy
yy
xx
tsts
ts
ts
s
==
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
=
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+=
⎬,
,
2221
1211
,
,
NN
IIII
NN
ti
i
yi
xi
⎭⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡=
⎭⎬⎫
⎩⎨⎧ ⎫
8.2.2- Stiffness Matrix Stiffness matrix is then given by:
dsdtJdxdy
dxdyBDBtk T
A
]det[
][][][][ 163333161616
=
= ×××× ∫∫
and the limits of integration -1 to +
dsdtJ ]
here the integration ]det[][][][ 16333316 JBDB ××× , the entire expression is same complicated function of s and t. Although the lim
T
numerical integration. simple elements, integration can be carried out exactly.
49
8.2.3- Comment on det[J] A well known condition for one-to-one mapping desired for cured edge elements is that the sign of det[J] should remain unchanged at all points of the domain mapped i.e. s, t plane. Violent distortions may cause alteration
f sign of det[J].
⎩ yp
ork done by p is:
vuvu
fffff
forcesbodyddistributetodueforcesbodyaref
Nu
Tbbbb
Tb
b
TT
=
=
==
=
δ
δδ
δ
for gravity load px=0 and py=-γ (γ= weight per unit volume, constant). From above equations:
1
vectorloadconsistentdsdtJpNtf Tb ∫ ∫
+ +
=
e integrated numerically. For gravity loading ,
further, for constant load,
e. by hand. If p is a function of x nterpolate p from the nodal values in terms of
shape functions Ni i.e. 8
1
where are the values of px(x,y) at the
nodes, similarly for p .
o 8.2.4- Body Forces Suppose there are body forces px and py per unit volume involved in the problem being considered for stress analysis.
⎬⎫
⎨⎧
= xpp
⎭
for plane elasticity, w[ ] dAputdA
pp
vutW T
Ay
x
A
=⎭⎬⎫
⎩⎨⎧
= ∫∫∫∫
[ ][ ]8211
16321
...
...
][
][
fdApNtW bA∫∫
)(]det[][1
1 1− −
gain this can b⎭⎬⎫
⎩⎨⎧
−=
γ0
pA
dsdtJNptf Tb ]det[][
1
1
1
1∫ ∫+
−
+
−
=
This integration can be done exactly i.and y, then we may also i
xiii
x pNp ∑=
= xip
y
50
8.2.5- Specific Boundary Stress (stress boundary condition) Boundary work done by given by
iiT dsuTtW = ∫−
stresses is :
T
sT
y
xTT
S
fdspp
ut δ=⎭⎬⎫
⎩⎨⎧
= ∫
irections (((global system) on the boundary. ST is part of boundary on which the stress are prescribed. u is now given by: u=[Ns]δ where [Ns] is the same as [N] except its value on the boundary under consideration. Further, if px and py are functions of x and y on the boundary, again we can use the shape functions Ni to interpolate stress px and py but for Ni on the boundary. For example,
W
TT sonstresstoDueS
where px and py are the stresses in x and y d
consider edge 2-6-3 of the element shown. Stress acting on this boundary is px(x,y) as shown. For edge 2-6-3, s=+1 therefore, only nonzero Ni on s=+1 are:
)1(2
)1()1(2 3
262 ttNtNttN +=−=−−=
x, u
y, v 4 3
2 1
s s 8 6
5
t px37
1
s=+1
px
3
2px2
px1
i 1 are boundary nodes assigned temporarily
51
22
263312
663322
663322
6
6
3
3
2
2
632
632
,
][
000000
dydxdS
FinallypNpNpNp
paspforionapproximat
yNyNyNytionTransformaCoordinatexNxNxNx
Nu
vuvuvu
NNNNNN
vu
xxxx
xx
ss
+=
++=
++=++=
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎦
⎤⎢⎣
⎡=
⎭⎬⎫
⎩⎨⎧
−
−
δ
But x and y are given by above equations i.e. f(t) along 2-6-3. Therefore, dl, the infinitesimal element of length along 2-6-3 is given by:
[ ]12116543
21
1
221
1
21
1
221
1
2
2
3
3
1
1
632
632
263312
21
22
:
][][
][][
][
000000
:),(, y thenyxpstressspecificaalsoisthereifsimilarly
dtttt ∂⎥⎦⎢⎣ ⎠⎝ ∂⎠⎝ ∂
−
362:
ssssssT
s
sat
isTss
sat
isTsTsT
iS
y
X
y
X
y
X
y
x
yyyy
fffffff
and
dtty
txpNNtf
dtty
txpNNtW
pNp
pppppp
NNNNNN
p
p
pNpNpNp
edgealongdxxnotedtyxdldS
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
=
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎦
⎤⎢⎣
⎡=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
++=
−−=∂
⎥⎤
⎢⎡
⎟⎞
⎜⎛ ∂
+⎟⎞
⎜⎛ ∂
==
+=
+
−
+=
+
−
−
−
−
∫
∫
444 3444 21
444 3444 21
δ
This procedure can be repeated for other edges as well. One have to be careful as to s or t =E etc. 9- Numerical Integration (Gauss Quadrature)
52
53
+1ξ1=-0.774597 H1=0.55556
ξ2=0.000 H2=0.88889
ξ1=+.774597 H1=0.55556
f(ξ1)
f(ξ2)
f(ξ3)
n=3 Exact for 5th degree polynomial 2n-1=5
I=0.55556f(-0.774597)+0.88889f(0.000)+0.55556f(+0.774597)
-1
Integration may be done analytically by using closed form formulas from a table of integrals. Alternatively, integration may be done numerically. Gauss quadrature is a commonly used form of numerical integration. Instead of using Newton cotes quadrature method where point at which the function is to be found (i.e. numerical value) are determined a priori (usually equal intervals) we shall use Gauss Quadrature method. The coordinates of sampling points are determined for best accuracy. 9.1- Line Integration
If f(s) is a polynomial, n-point Gauss quadrature yields the exact integral if f(s) is of degree 2n-1 or less. Assume a polynomial expression to estimate I. For n sampling points we have 2n unknown (Hi and si) for which a polynomial of degree 2n-1 can be constructed and integrated exactly. This yields an error of order O(∆2n). Thus, for n=3, a polynomial of degree 5 can be integrated exactly using Gaussian quadrature method. Thus the form f(s)=a+bs is exactly integrated by a one-point rule. The form f(s)=a+bs+cs2 is exactly integrated by a two-point rule, and so on. Use of an excessive number of points, for example, a two-point rule for f(s)=a+bs still yields the exact result.
points are used. Convergence toward exact result may not be monotonic.
)()(1
1
1ii
n
i
sfHdssfI ∑∫=
=
−
==
In above equation, n=number of sampling points often called Gauss points. Hi=Weighting constants si=Coordinates of sampling points For any given n (by choice) Hi and si can be obtained from the tables.
If f(s) is not a polynomial, but (say) the ration of two polynomials, Gauss quadrature yields an approximate result. Accuracy improves as more Gauss
54
s
t
9 points
Rectangle
can use the same tables. Exact for polynomial of ction, s and t.
btain the element , we have to determine
int (s , t )
tf order n over a hexahedron in s 3 points. Three summations and the roduct of three i t factors.
bscissae And Weight Coefficients Of The Gaussian Quadrature Formula
9.2- Gauss Quadrature Numerical Integration over a
),(),(),(111
1
1
1
1
1
1iiji
n
j
n
iii
n
i
tsfHHdttsfHdsdttsfI ∑∑∑∫∫ ∫===
+
−
+
−
+
−
===
Example of n=3, we degree 5 in each direLet us use n=3 for integration of above equation to ostiffness matrix [k]. Thus[B]T[D][B]det[J] at each sample po i i In hree dimensions, Gauss quadrature o volvenp we gh A
9.3- Comment on Numerical Integration
t may introduce additional errors and ducing the error as much as possible. This
Numerical integration, instead of exacfirst attempt may be directed at re
may not be very economical. Therefore, the following should be etermined:
a) The minimum integration requirement permitting convergence f
convergence which would results if exact integration were used. Let p be the order of degree of complete polynomial and m be the order of differentials occurring in the strain energy expression. Providing the integration is exact to the order 2(p-m) or shows an error of O(h2(p-m)+1, or less, then no loss of convergence order will occur. In curvilinear coordinates take a curvilinear dimension h of an element. For C0 problem (m=1), the integration is of the following order: p=1, Linear displacement O(h) p=2, Quadratic displacement O(h3) p=3, Cubic displacement O(h5) With numerical integration , singular stiffness matrix may result for low integration orders making lower order integrations impractical. In general, there should be at least as many integration points as required to yield a number of independent relations equal or greater than the number of overall unknowns. Consider again the plane eight-node isoparametric element discussed previously. Its stiffness matrix integrand [B]T[D][B]det[J] is an 16 by 16 matrix. Because it is a symmetric matrix, only 136 of 256 coefficienta sre different from one another. Each of these coefficients has the form f(s,t) and
. In compu er programming, a
erable com
in
ately integrated ints. Should we use very few points many points to improve the accuracy
d
b) The integration requirements necessary to preserve the rate o
each must be integrated over the element area tp-point integration rule requires p passes through a integration loop. Each pass requires ebaluation of [B] and det[J] at the coordinates of a Gauss point, computation of the product [B]T[D][B]det[J], and multiplication by weight factor. Each pass makes a contribution to [k] which is fully formed when all p passes have been completed. Clearly, there is consid
putation required in this process.
TFor an element of general shape, each coefficient in matrix [B] [D][B]det[J] is the ration of two polynomials in s and t. The polynomial in the denominator comes from J-1 when [J] is inverted det[J] becomes the denominator of every coefficient in J-1 and hence appears the denominator of every coefficient in [B]. Analytical integration of [k] would require the use of cumbersome formulas. Numerical intehration is simpler
ct, s t [k] is only approximbut in general it is not exa o tharegardless of number of integration pofor low computational expense or very
55
of integration? The answer is neither, for reasons explained in the following. 9.4- Choice of Quadrature Rule, Instabilities A FE model is usually inexact, and usually it errs by being too stiff.
auss points sense no strain under a certain deformation mode, the resulting
Overstiffness is usually made worse by using more Gauss points to integrate element stiffness matrix because additional points capture more higher-order terms in [k]. these terms resist some deformation modes that lower-order terms do not, and therefore act to stiffen an element. Accordingly, greater accuracy in the integration of [k] usually produce less accuracy in the FE solution, in addition to requiring more computation. On the other hand, use of too few Gauss points produces an even worse situation known by various names, Instability, spurious singular mode, mechanism, kinematic mode, zero energy mode, and hourglass mode. Instability occurs if one or more deformation modes happen to display zero strain at all Gauss points. One must regard Gauss points as strain sensors. If G
56
a b c d
[k] will have no resistance to that deformation mode. A simple illustration of instabilities are shown. Four-node plane elements are integrated by a one-point Gauss rule. In the lower left element with c a constant, the three instabilities shown have respective forms b)u=cxy v=0 c)u=o v=-cxy d)u=cy(1-x) v=cx(y-1) We easily check that each of these displacement fields produces strains εxx=εyy=γxy=0 at the Gauss point, x=y=0.
a) Undeformed plane 2 by 2 four-node square elements, Gauss points are shown by square B, c, d Instability displacement modes
Nonrectangular elements behave in the same way. Even if the mesh had just enough supports to prevent rigid body motion it could still display these
modes, without strain at the Gauss points, and hence without strain energy. The FE model would have no ressistance to loadings that would activate these modes. The global [k] would be singular re
57
Mesh of four-node square elements with all nodes fixed at the support. Gauss points are shown by square H ode in a single 8-node element ourglass instability displacement mintegrated by Gauss points
gardless of how the tructure is loaded.
ovided by the ess and less with increasing distance from it. node element whose stiffness matrix is integrated with four
rglass instability shown.
often most accurate at Gauss points. It happens that the
s, and common
auss points for stiffness tegration and stress calculation. Stresses at nodes or at other element cations are obtained by extrapolation or interpolation from Gauss point
alues.
sWhen supports are adequate to make [k] nonsingular, there may yet be near-instability that is troublesome. In the figure, all dof are fixed at the support and each element is integrated with one point. Restraint prsupport is felt lA plane eight-Gauss points has the hou There is no way that two adjacent elements can both display this mode while remaining connected. 9.5- Stress Calculation and Gauss Points Calculated stress arelocations of greatest accuracy are apt to be the same Gauss points that were used for integration of the stiffness matrix. In summary, it is common practice to use an order 2 Gauss rule (four points) to integrate [k] of four- and eight-node plane elementpractice to compute strains and stresses at these same points. Similarly, three-dimensional elements often use eight Ginlov
