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ECIV 720 A Advanced Structural
Mechanics and Analysis
Lecture 13 & 14: Quadrilateral Isoparametric ElementsStiffness MatrixNumerical IntegrationForce VectorsModeling Issues
Two Dimensional – Plane StressThin Planar Bodies subjected to in plane loading
Tvuu
Tyx fff
Tyx TTT
tdAdV
Tiyixi PPP
Stress and Strain
Txyyx σ
Txyyx ε
Stress Strain Relationship
T
y
u
y
u
y
v
x
u
ε
FEM Solution: Area Triangulation
Area is Discretized into Triangular Shapes
Constant Strain Triangle
For Every Triangular Element
•Strain-Displacement Matrix B
122131132332
211332
123123
000
000
2
1
yxyxyx
xxx
yyy
AB
1 2
3q6
q5
q4
q3
q2
q1
vu
Constant strain
11 LN
22 LN
133 LN
Nqu ,
Bqε
FEM Solution: Quadrilateral Mesh
Area is Discretized into Quadrilateral Shapes
FEM Solution: Quadrilateral Elements
X
Y
1 (x1,y1) 2 (x2,y2)
4 (x4,y4)
q8
q7
q4
q3
q1
q2
8 Degrees of Freedom
3 (x3,y3)
q6
q5v
uP (x,y)
FEM Solution: Objective
• Use Finite Elements to Compute Approximate Solution At Nodes
• Interpolate u and v at any point from Nodal values q1,q2,…q8
q8
q7
q4
q3
q1
q2
vu
q6
q5
To this end…
•Intrinsic Coordinate System•Shape Functions of 4-node quadrilateral
•From 1-D
•Direct •Jacobian of Transformation•Strain-Displacement Matrix•Stiffness Matrix
Intrinsic Coordinate System
12
112
xxxx
Map ElementDefine Transformation
x1x
x2
1=-1 2=1
Recall 1-D
Intrinsic Coordinate System
1 (-1,-1) 2 (1,-1)
4 (-1,1)
Map ElementDefine Transformation
Parent
3 (1,1)
4
1
2
3
Lagrange Shape Functions
1=-1 2=1
12
11N 1
2
12N
For 1-D
N1()N1=1
baN 1
111 N
011 N
2-D Lagrange Shape Functions
12
11,1f
(1,)(1,1)
1 (-1,-1)
What is the lowest order
polynomial
f1(,) along side =-1
that satisfies
f1(-1,-1) =1 & f1(1,-1)=0?
2-D Lagrange Shape Functions
12
1,12f
What is the lowest order
polynomial
f2(,) along side =-1
that satisfies
f2(-1,-1) =1 & f2(-1,1)=0?
(1,)(1,1)
1 (-1,-1)
12
1,12f
f2=0
f2=1
Construction of Lagrange Shape Functions
(1,)(1,1)
1 (-1,-1)
12
11,1f
f1=1
f1=0
What is the lowest order Polynomial F1(,) that satisfiesF1(-1,-1) =1 & F1(1,-1)=F1(1,1)=F1(-1,1)=0?
Bi-Linear Surface
(1,)(1,1)
Construction of Lagrange Shape Functions
f2(-1,)f1(-1)
PP
PPPP ffNF
114
1
,11,, 2111
PP
P
Construction of Lagrange Shape Functions
114
1,11,, 2111 ffNF
For Every Point ()
(1,1)
1 (-1,-1)F1(,)
Construction of Lagrange Shape Functions
(1,)(1,1)
2 (1,-1)
12
11,1f
Bi-Linear Surface
114
1,11,, 2122 ffNF
12
1,12f
F2(,)
Construction of Lagrange Shape Functions
3 (1,1)
12
11,1f
Bi-Linear Surface
114
1,11,, 2133 ffNF
12
1,12f
F3(,)
12
1,12f
Construction of Lagrange Shape Functions
4 (-1,1)
Bi-Linear Surface
114
1,11,, 2144 ffNF
F4(,)
12
11,1f
In Summary
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
114
14N
114
13N
114
12N
114
11N
Shape Functions in a Direct Way
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
Assume Bi-Linear
Variation
dcbaN i ,
Complete Polynomial
i=1,2,3,4
With conditions
jiif
jiifN jji 0
1, j=1,2,3,4
Shape Functions in a Direct Way
dcbaN ,1
Each, i, provides 4 Equations with 4 unknowns
e.g. i=1
jiif
jiifN jji 0
1,
11,11 dcbaN
01,11 dcbaN
01,11 dcbaN
01,11 dcbaN 11
4
11N
Field Variables in Discrete Form
114
14N
114
13N
114
12N
114
11N
Geometry
44332211 xNxNxNxNx
44332211 yNyNyNyNy
Displacement
74533211 qNqNqNqNu
84634221 qNqNqNqNv
q8
q7
q4
q3
q1
q2
vu
q6
q5
Field Variables in Discrete Form
4
4
1
1
4321
4321
0000
0000
y
x
y
x
NNNN
NNNN
y
x
Geometry
8
1
4321
4321
0000
0000
q
q
NNNN
NNNN
v
u
Displacement
Intrinsic Coordinate System
1 (-1,-1) 2 (1,-1)
4 (-1,1)
Map Element
Parent
3 (1,1)
4
1
2
3
JDefine Jacobian
Transformation
The Jacobian from (x,y) to (,) may be obtained as:
Consider any function f(x,y)
defined on area of element
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
y
y
fx
x
ff
y
y
fx
x
ff
Transformation
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
y
y
fx
x
ff
y
y
fx
x
ff
y
fx
f
yx
yx
f
f
y
fx
f
f
f
J
Jacobian of Transformation
44332211 xNxNxNxNx
44332211 yNyNyNyNy
44
33
22
11
4
1
xN
xN
xN
xN
xNx
ii
i
4
1ii
i xNx
4
1ii
i yNy
4
1ii
i yNy
yx
yx
J
Jacobian of Transformation
iiii
iN
14
111
4
1
iiiiiN
14
111
4
1
Jacobian of Transformation
4321
4
1
4
111
11114
1
14
1
xxxx
xxNx
J ii
iiii
i
4321
4
1
4
121
11114
1
14
1
xxxx
xxNx
J ii
iiii
i
Jacobian of Transformation
4321
4
1
4
121
11114
1
14
1
yyyy
yyNy
J ii
iiii
i
4321
4
1
4
122
11114
1
14
1
yyyy
yyNy
J ii
iiii
i
In Summary
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
4
1
2
3
Without Proof
2221
1211
JJ
JJyx
yx
J
ddJdxdydA det
Jacobian of Transformation
4321
4
1
4
111
11114
1
14
1
xxxx
xxNx
J ii
iiii
i
4321
4
1
4
121
11114
1
14
1
xxxx
xxNx
J ii
iiii
i
Jacobian of Transformation
4321
4
1
4
121
11114
1
14
1
yyyy
yyNy
J ii
iiii
i
4321
4
1
4
122
11114
1
14
1
yyyy
yyNy
J ii
iiii
i
In Summary
Or the inverse
y
fx
f
JJ
JJf
f
2221
1211
f
f
JJ
JJ
Jf
f
JJ
JJ
y
fx
f
1121
1222
1
2221
1211
det
1
In Summary
114
14N
114
13N
114
12N
114
11N
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
44332211 xNxNxNxNx
44332211 yNyNyNyNy 74533211 qNqNqNqNu
84634221 qNqNqNqNv
Strain Tensor from Nodal Values of Displacements
x
v
y
uy
vx
u
ε
Strain Tensor from Nodal Values of Displacements
Thus we need to evaluate
and
y
ux
u
y
vx
v
Strain Tensor from Nodal Values of Displacements
Recall that
Consequently,
f
f
JJ
JJ
Jy
fx
f
1121
1222
det
1
f=u
u
u
JJ
JJ
Jy
ux
u
1121
1222
det
1
f=v
v
v
JJ
JJ
Jy
vx
v
1121
1222
det
1
Strain Tensor from Nodal Values of Displacements
v
v
u
u
JJJJ
JJ
JJ
J
x
v
y
uy
vx
u
12221121
1121
1222
00
00
det
1ε
A
Strain Tensor from Nodal Values of Displacements
Next we need to evaluate
Nqu
8
1
4321
4321
0000
0000),(
q
q
NNNN
NNNN
v
u
Tvvuu
Strain Tensor from Nodal Values of Displacements
74
53
32
11
4
112 q
Nq
Nq
Nq
Nq
Nu
ii
i
4
112
ii
i qNu
4
12
ii
i qNv
4
12
ii
i qNv
Nqu
8
1
4321
4321
0000
0000),(
q
q
NNNN
NNNN
v
u
Strain Tensor from Nodal Values of Displacements
8
2
1
10101010
10101010
01010101
01010101
4
1
q
q
q
v
v
u
u
8
2
1
q
q
q
v
v
u
u
G
Strain Tensor from Nodal Values of Displacements
= B q
Both A and G are linear functions of and
= AG q
Stresses
xy
y
x
xy
y
xE
2
100
01
01
1 2
Dεσ = B q
DBqσ
Element Stiffness Matrix ke
eV
Te dVU σDε
2
1
= B qe= D B qe
e
A
TTe
el
TTe
e
dAt
tdA
U
e
qDBBq
qDBBq
2
1
2
1
ke
tdAdV
4
1
2
3
8x8 matrix
Element Stiffness Matrix ke
dJdttdAdV det
1
1
1
1
det dJdt
dAt
T
A
Te
DBB
DBBk
Furthermore
and
Numerical
Integration
Integration of Stiffness Matrix
1
1
1
1
det dJdt Te DBBk
B (3x8)
D (3x3)
BT(8x3)
ke (8x8)
Integration of Stiffness Matrix
Each term kij in ke is expressed as
1
1
1
1
1
1
1
1
3
1
3
1
,
,det
ddgt
ddJBDBtkm l
ljmlTimij
Linear Shape Functions is each Direction
0
2
4
6
8
10
0 0.5 1 1.5 2 2.5 3
x2
Numerical Integration
Integrals
Indefinite Definite
Cxdxx 32
3
1
2
0
2
3
8dxx
Numerical Integration
Definite integrals can be computed numerically
b
a iii xfwdxxf
Objective:
• Determine points xi
• Determine coefficients wi
0
2
4
6
8
10
0 0.5 1 1.5 2 2.5 3
x2
Numerical Integration
Depending on choice of wi and xi
Midpoint Rule
Trapezoidal Rule
Simpson's
Gaussian Quadratures
etc
Numerical Integration – Upper & Lower Bounds
Lower Sum
L(f;xi)
Upper Sum
U(f;xi)
a=x1 x2 x3 x4 x5 x6 x7=b
a=x1 x2 x3 x4 x5 x6 x7=b
Numerical Integration
ib
a iiii xfUxfwdxxfxfL ;;
It can be shown that
ii
b
a iiii
ixfUxfwdxxfxfL ;lim;lim
Mid-Point Rule
a=x1 x2 x3 x4 x5 x6 x7=b
212
1xx
322
1xx
542
1xx 762
1xx
1
111 2
1n
iiiii
b
a iii xxfxxxfwdxxf
76676 2
1xxfxxA
12121 2
1xxfxxA
Mid Point Rule
1
111 2
1n
iiiii
b
a iii xxfxxxfwdxxf
Simple to comprehend and implement
Large number of intervals is required for accuracy
a=x1
…
b=xn
Trapezoid Rule
a=x1 x2 x3 x4 x5 x6 x7=b
1
1112
1 n
iiiii
b
a iii xfxfxxxfwdxxf
67676 2
1xfxfxxA
12121 2
1xfxfxxA
Trapezoid Rule
Simple to comprehend and implement
Exact for polynomials f(x) = ax+b
Large number of intervals is required for accuracy
(Less than midpoint rule)
1
1112
1 n
iiiii
b
a iii xfxfxxxfwdxxf
Simpson’s Rule
hafhafafh
dxxfha
a
243
2
Step 1
ha
a
dxxf2
h ha a+2h
h
xi xi+1xi+1/2
h
Simpson’s Rule
h
xi xi+1xi+1/2
h
i
x
x
Ihafhafafh
dxxfi
i
243
1
Step 2
1
0
n
ii
b
a
Idxxf
Ii
Ii+1…
Quadratures
nn
b
a iii xfwxfwxfwxfwdxxf 2211
Objective
Where do such formulae come from?
Theory of Interpolation….
Let
n
iii xfxlxpxf
1
Recall Shape Functions
li(x): cardinal functions
Quadratures
i
n
ii
b
a
i
n
ii
b
a
b
a
wxfdxxlxfdxxpdxxf
11
It will give correct values for the integral of
every polynomial of degree n-1
Mid-Point Rule is an example of a
quadrature with n=1
Example
Establish coefficients of quadrature for the interval
[-2,2] and nodes –1,0,1
-2 -1 0 1 2
f(x)
ni1 1
n
ijj ji
ji xx
xxxl
Cardinal Functions:
Example
xxxx
xx
xxxl
ijj ji
j
23
11 2
1
11
1
01
0
110
1
10
1 23
12
xxx
xx
xxxl
ijj ji
j
xxxx
xx
xxxl
ijj ji
j
23
13 2
1
01
0
11
1
Cardinal Functions – Lagrange Polynomials:
Example
12
1011
2
1 222 fxxfxfxxxf
ii
iii
i wxfdxxlxfdxxpdxxf
3
1
2
2
3
1
2
2
2
2
xxxl 21 2
1 122 xxl xxxl 2
3 2
1
with
2
2
dxxlw ii
Example
2
2
22
2
22
2
11 3
8
2
1
2
1dxxxdxxxdxxlw
3
41
2
2
22
2
22
dxxdxxlw
3
8
2
1
2
1 2
2
22
2
22
2
33
dxxxdxxxdxxlw
Example
13
80
3
41
3
82
2
fffdxxf
-2 -1 0 1 2
f(x)=x2
3
161
3
80
3
41
3
82
2
2
dxx
Quadrature
3
16
3 2
232
2
2
x
dxx
Exact
Quadratures
So far the placement of nodes has been arbitrary
a=x1 x2 x3 x4 x5 x6 x7=b
They should belong to the interval of integration
Quadrature is accurate for polynomials of degree n-1 (n is the number of nodes)
Gaussian Quadrature
Karl Friedriech Gauss discovered that by a
special placement of nodes the accuracy of the
numerical integration could be greatly increased
Gaussian Quadrature
Theorem on Gaussian nodes
Let q be a polynomial of degree n such that
1-n0,1,...,k 0 dxxxq kb
a
Let x1,x2,…,xn be the roots of q(x). Then
nn
b
a iii xfwxfwxfwxfwdxxf 2211
with xi’s as nodes is exact for all polynomials of
degree 2n-1.
Gaussian Quadrature 2-point
-1 1
f()
311
f()W1=1
312
f()W2=1
3113111
1
ffdxxf
Gauss Points and Weights
Compare
Let f(x)=x2
MidPointMidPoint
1
1
2 32dxx
Use 2-points
X1=-1, X2=0, X3=1
215.0)5.0(1
1
222
dxx
X=-0.5 X=0.5
Compare
Let f(x)=x2
TrapezoidalTrapezoidal
1
1
2 32dxx
Use 2-points
X1=-1, X2=0, X3=1
1)1()0(12
1)0()1(1
2
1 221
1
222
dxx
Compare
Let f(x)=x2
GaussGauss
1
1
2 32dxx
Use 2-points
3231131122
1
1
2
dxx
311 311
1,31 111 wx 1,31 222 wx
2-Dimensional Integration
Gaussian Quadrature
1
1
1
1
, ddf
n
j
n
ijiij fww
1 1
,
1
1 1
, dfwn
iii
2-D Integration 2-point formula
311 312
311
312
11 w
12 w
111 ww 112 ww
122 ww121 ww
2-D Integration 2-point formula
311 312
311
311
1
1
1
1
, ddf
2222212112121111 ,,,, fwwfwwfwwfww
Integration of Stiffness Matrix
1
1
1
1
det dJdt
dAt
T
A
Te
DBB
DBBk
Stiffness Matrix
1 (-1,-1) 2 (1,-1)
4 (-1,1) 3 (1,1)
Integration over
Integration of Stiffness Matrix
12221121
1121
1222
00
00
det
1
JJJJ
JJ
JJ
JA
(3x4)
10101010
10101010
01010101
01010101
4
1G
(4x8)= B q= AG q
B (3x8)
Integration of Stiffness Matrix
1
1
1
1
det dJdt Te DBBk
B (3x8)
D (3x3)
BT(8x3)
ke (8x8)
Integration of Stiffness Matrix
Each term kij in ke is expressed as
1
1
1
1
1
1
1
1
3
1
3
1
,
,det
ddgt
ddJBDBtkm l
ljmlTimij
Linear Shape Functions is each Direction
Gaussian Quadrature is accurate if
We use 2 Points in each direction
Integration of Stiffness Matrix
311 312
311
311
2222212112121111 ,,,, gwwgwwgwwgww
1
1
1
1
, ddgt
11 w
12 w
22211211 ,,,, gggg
Element Body Forces
ii
Ti
el
T
eA
T
eA
T
e
e
e
tdl
tdA
tdA
Pu
Tu
fu
Dεε2
1
ii
Ti
eA
T
eV
T
eV
T
e
e
e
dA
dV
dV
Pφ
Tφ
fφ
φεσ 0
Total Potential Galerkin
Body Forces
eA
T dAd detfu
Integral of the form
8
1
4321
4321
0000
0000
q
q
NNNN
NNNN
v
u
8
1
4321
4321
0000
0000
NNNN
NNNN
y
x
eA
T dAd detfφ
Body Forces
In both approaches
e
e
e
e
Ay
Ax
Ay
Ax
dAdNf
dAdNf
dAdNf
dAdNf
qqWP
det
det
det
det
4
2
1
1
81
Linear Shape Functions
Use same quadrature as stiffness maitrx
Element Traction
ii
Ti
el
T
eA
T
eA
T
e
e
e
tdl
tdA
tdA
Pu
Tu
fu
Dεε2
1
ii
Ti
eA
T
eV
T
eV
T
e
e
e
dA
dV
dV
Pφ
Tφ
fφ
φεσ 0
Total Potential Galerkin
Element Traction
Similarly to triangles, traction is applied along sides of element
4
2
3
Tx
Ty
u
v
1
4
0
12
1
12
1
0
4
3
2
1
N
N
N
N
el
TT tdlWP Tu
32
Traction
8
1
32
32
000000
000000
q
q
NN
NN
v
u
0
0
0
0
232
8132
y
x
eT T
Tlt
qqWP
For constant traction along side 2-3
Traction
components
along 2-3
Stresses
311 312
311
311
DBqσ
12221121
1121
1222
00
00
det
1
JJJJ
JJ
JJ
JA
10101010
10101010
01010101
01010101
4
1G
More Accurate at
Integration points
Stresses are calculated at any
Modeling Issues: Nodal Forces
ii
Ti
el
T
eA
T
eA
T
e
e
e
tdl
tdA
tdA
Pu
Tu
fu
Dεε2
1
A node should be
placed at the location
of nodal forces
In view of…
Or virtual potential energy
Modeling Issues: Element Shape
Square : Optimum Shape
Not always possible to use
Rectangles:
Rule of Thumb
Ratio of sides <2
Angular Distortion
Internal Angle < 180o
Larger ratios
may be used
with caution
Modeling Issues: Degenerate Quadrilaterals
Coincident Corner Nodes
1
2
3
4
32
1
4
xx
xx
x
xx
x
Integration Bias
Less accurate
Modeling Issues: Degenerate Quadrilaterals
Three nodes collinear
1
2
3
4
xx
xx
1
2
3
4 x
xx
x
Less accurate
Integration Bias
Modeling Issues: Degenerate Quadrilaterals
Use only as necessary to improve representation of geometry
2 nodes
Do not use in place of triangular elements
A NoNo Situation
3
4
1 2
x
y
(3,2) (9,2)
(7,9)
(6,4)
Parent
All interior angles < 180
J singular
Another NoNo Situation
x, y
not uniquely
defined
Convergence Considerations
For monotonic convergence of solution
Elements (mesh) must be compatible
Elements must be complete
Requirements
Mesh Compatibility
OK
NO NO!
Mesh compatibility - Refinement
Acceptable Transition
However…Compatibility of displacements OK
Stresses?
Convergence Considerations
We will revisit the issue…