Transcript Document
ECIV 720 A
Advanced Structural Mechanics
and Analysis
Lecture 13 & 14:
Quadrilateral Isoparametric Elements
Stiffness Matrix
Numerical Integration
Force Vectors
Modeling Issues
Two Dimensional – Plane Stress
Thin Planar Bodies subjected to in plane loading
u u v
T
T T
P P
f fx
x
i
ix
T
fy
T
T
y
Piy
dV tdA
T
Stress and Strain
σ x y xy
ε x y xy
T
T
Stress Strain Relationship
u
ε
x
v
y
u u
y y
T
FEM Solution: Area Triangulation
Area is Discretized into Triangular Shapes
Constant Strain Triangle
For Every Triangular Element
3
q6
N1 L1 x
q5
x
1
N 2 L2 h
v
q2
u
q1
q4
2
q3
h
ux ,h Nq
ε Bq
N3 L3 1 x h
•Strain-Displacement Matrix B
y23
1
B
0
2A
x32
0
x32
y23
y31
0
x13
0
x13
y31
y12
0
x21
0
x21
y12
Constant strain
FEM Solution: Quadrilateral Mesh
Area is Discretized into Quadrilateral Shapes
FEM Solution: Quadrilateral Elements
q8
q6
q7
4 (x4,y4)
q5
v
q1
Y
3 (x3,y3)
u
P (x,y)
q4
q2
X
1 (x1,y1)
q3
2 (x2,y2)
8 Degrees of Freedom
FEM Solution: Objective
q8
q6
q7
q5
v
q1
q2
u q
4
• Use Finite Elements to Compute
Approximate Solution At Nodes
• Interpolate u and v at any point
from Nodal values q1,q2,…q8
q3
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
x1
x
Recal
l 1-D
x2
Map Element
Define Transformation
2
x x1 1
x
x2 x1
x1=-1
x
x2=1
1
Intrinsic Coordinate System
4
h
2
x
h
4 (-1,1)
3 (1,1)
3
Map Element
Define Transformation
x
1 (-1,-1)
2 (1,-1)
Parent
Lagrange Shape Functions
For 1-D
N1 x a bx
N1=1
N1 1 1
N1(x)
x1=-1
x
1
N1 x 1 x
2
x2=1
N1 1 0
1
N 2 x 1 x
2
2-D Lagrange Shape Functions
1 (-1,-1)
What is the lowest order
polynomial
f1(x,1) along side h=-1
that satisfies
f1(-1,-1) =1 & f1(1,-1)=0?
h
x
(1,h)
(1,1)
1
f1 x ,1 1 x
2
2-D Lagrange Shape Functions
What is the lowest order
polynomial
f2(1,h) along side x1=-1
that satisfies
f2(-1,-1) =1 & f2(-1,1)=0?
1 (-1,-1)
h
x
(1,h)
(1,1)
1
f 2 x1 ,h 1 h
2
Construction of Lagrange Shape Functions
f1=1
f2=1
1
f1 x ,1 1 x
2
1
f 2 1,h 1 h
2
1 (-1,-1)
f2=0
h
f1=0
x
(1,h)
(1,1)
Bi-Linear Surface
What is the lowest order Polynomial F1(x,h) that
satisfies
F1(-1,-1) =1 & F1(1,-1)=F1(1,1)=F1(-1,1)=0?
Construction of Lagrange Shape Functions
f1(x,-1)
f2(-1,h)
hP
xP
h
P
x
(1,h)
(1,1)
F1 N1 x P ,hP f1 x P ,1 f 2 1,hP
1
1 x P 1 hP
4
Construction of Lagrange Shape Functions
F1(x,h)
1 (-1,-1)
For Every Point (x,h)
h
x
(1,1)
1
F1 N1 x ,h f1 x ,1 f 2 1,h 1 x 1 h
4
Construction of Lagrange Shape Functions
1
f1 x ,1 1 x
2
F2(x,h)
h
2 (1,-1)
x
(1,h)
(1,1)
1
f 2 1,h 1 h
2
Bi-Linear Surface
1
F2 x ,h N 2 f1 x ,1 f 2 1,h 1 x 1 h
4
Construction of Lagrange Shape Functions
F3(x,h)
1
f1 x ,1 1 x
2
h
x
3 (1,1)
1
f 2 1,h 1 h
2
Bi-Linear Surface
1
F3 x ,h N 3 f1 x ,1 f 2 1,h 1 x 1 h
4
Construction of Lagrange Shape Functions
F4(x,h)
1
f 2 1,h 1 h
2
4 (-1,1)
h
x
1
f1 x ,1 1 x
2
Bi-Linear Surface
1
F4 x ,h N 4 f1 x ,1 f 2 1,h 1 x 1 h
4
In Summary
h
4 (-1,1)
1
N1 1 x 1 h
4
3 (1,1)
x
1 (-1,-1)
2 (1,-1)
1
N 2 1 x 1 h
4
1
N 3 1 x 1 h
4
1
N 4 1 x 1 h
4
Shape Functions in a Direct Way
h
4 (-1,1)
3 (1,1)
Assume Bi-Linear
Variation
x
Complete Polynomial
Ni x ,h a bx ch dxh
1 (-1,-1)
2 (1,-1)
i=1,2,3,4
With conditions
1 if
N i x j ,h j
0 if
i j
i j
j=1,2,3,4
Shape Functions in a Direct Way
Each, i, provides 4 Equations with 4 unknowns
e.g. i=1
N1 x ,h a bx ch dxh
N1 1,1 a b c d 1
N1 1,1 a b c d 0
N1 1,1 a b c d 0
N1 1,1 a b c d 0
1 if
N i x j ,h j
0 if
i j
i j
1
N1 1 x 1 h
4
Field Variables in Discrete Form
q8
q7
q6
q5
v
q1
q2
u q
4
q3
1
N1 1 x 1 h
4
1
N 2 1 x 1 h
4
1
N 3 1 x 1 h
4
1
N 4 1 x 1 h
4
Geometry
x N1 x1 N 2 x2 N 3 x3 N 4 x4
y N1 y1 N 2 y2 N 3 y3 N 4 y4
Displacement
u N1q1 N 2q3 N 3q5 N 4q7
v N1q2 N 2 q4 N 3q6 N 4q8
Field Variables in Discrete Form
Geometry
x N1
y 0
0
N1
N2
0
0
N2
N3
0
0
N3
N4
0
x1
y
0 1
N 4
x4
y4
N4
0
q1
0
N4
q8
Displacement
u N1
v 0
0
N1
N2
0
0
N2
N3
0
0
N3
1
Intrinsic Coordinate System
4
h
2
x
h
4 (-1,1)
3 (1,1)
3
x
Map Element
Define Jacobian
J
1 (-1,-1)
2 (1,-1)
Parent
Transformation
The Jacobian from (x,y) to (x,h) may be obtained as:
Consider any function f(x,y)
defined on area of element
h
4 (-1,1)
f f x f y
x x x y x
3 (1,1)
x
1 (-1,-1)
2 (1,-1)
f f x f y
h x h y h
Transformation
f f x f y
x x x y x
f x
x x
f x
h h
f f x f y
h x h y h
4 (-1,1)
h
3 (1,1)
x
1 (-1,-1)
2 (1,-1)
y f
x x
f
y
h y
f
f
x
x
f J f
h
y
x
x
J
x
h
Jacobian of Transformation
y
x N1 x1 N 2 x2 N 3 x3 N 4 x4
x
y
y N1 y1 N 2 y2 N 3 y3 N 4 y4
h
x
N i
N1
N 2
N 3
N 4
xi
x1
x2
x3
x4
x i 1 x
x
x
x
x
4
4
x
N i
xi
h i 1 h
4
y
N i
yi
x i 1 x
y
N i
yi
h i 1 h
4
Jacobian of Transformation
N i
1
1
1 xix 1 hih xi 1 hih
x x 4
4
N i
1
1
1 xix 1 hih 1 xix hi
h h 4
4
Jacobian of Transformation
4
x
N i
1 4
J 11
xi xi 1 hih xi
x i 1 x
4 i 1
1
1 h x1 1 h x2 1 h x3 1 h x4
4
4
x
N i
1 4
J 21
xi hi 1 xix xi
h i 1 h
4 i 1
1
1 x x1 1 x x2 1 x x3 1 x x4
4
Jacobian of Transformation
4
y
N i
1 4
J 21
yi xi 1 hih yi
x i 1 x
4 i 1
1
1 h y1 1 h y2 1 h y3 1 h y4
4
4
y
N i
1 4
J 22
yi hi 1 xix yi
h i 1 h
4 i 1
1
1 x y1 1 x y2 1 x y3 1 x y4
4
In Summary
1
h
4 (-1,1)
4
h
2
x
x
1 (-1,-1)
3
x
x
J
x
h
3 (1,1)
y
x J 11
y J 21
h
dA dxdy det J dx dh
2 (1,-1)
J 12
J 22
Without Proof
Jacobian of Transformation
4
x
N i
1 4
J 11
xi xi 1 hih xi
x i 1 x
4 i 1
1
1 h x1 1 h x2 1 h x3 1 h x4
4
4
x
N i
1 4
J 21
xi hi 1 xix xi
h i 1 h
4 i 1
1
1 x x1 1 x x2 1 x x3 1 x x4
4
Jacobian of Transformation
4
y
N i
1 4
J 21
yi xi 1 hih yi
x i 1 x
4 i 1
1
1 h y1 1 h y2 1 h y3 1 h y4
4
4
y
N i
1 4
J 22
yi hi 1 xix yi
h i 1 h
4 i 1
1
1 x y1 1 x y2 1 x y3 1 x y4
4
In Summary
f
x J 11
f
J 21
h
f
J 12 x
f
J 22
y
Or the inverse
f
x J 11
f
J 21
y
f
1
J 12 x
1
f
J 22 det J
h
J 22
J
21
f
J 12 x
f
J 11
h
In Summary
h
4 (-1,1)
3 (1,1)
x
1 (-1,-1)
2 (1,-1)
x N1 x1 N 2 x2 N 3 x3 N 4 x4
1
N1 1 x 1 h
4
1
N 2 1 x 1 h
4
1
N 3 1 x 1 h
4
1
N 4 1 x 1 h
4
u N1q1 N 2q3 N 3q5 N 4q7
y N1 y1 N 2 y2 N 3 y3 N 4 y4 v N1q2 N 2 q4 N 3q6 N 4q8
Strain Tensor from Nodal Values of
Displacements
u
x
v
ε
y
u v
y x
Strain Tensor from Nodal Values of
Displacements
Thus we need to evaluate
u
x
u
y
and
v
x
v
y
Strain Tensor from Nodal Values of
Displacements
Recall that
f
x
1
f
det J
y
J 22
J
21
f
J 12 x
J 11 f
h
Consequently,
f=v
f=u
u
x
1
u
det J
y
J 22
J
21
u
J 12 x
J 11 u
h
v
x
1
v
det J
y
J 22
J
21
v
J 12 x
v
J 11
h
Strain Tensor from Nodal Values of
Displacements
u
x
1
v
ε
y det J
u v
y x
J 22 J 12
0
0
J 21 J 11
A
0
J 21
J 22
u
x
0 u
h
J 11
v
J 12 x
v
h
Strain Tensor from Nodal Values of
Displacements
Next we need to evaluate
u
x
u N1
u(x ,h )
v 0
u
h
0
N1
N2
0
v
x
0
N2
T
v
h
N3
0
0
N3
N4
0
q1
0
Nq
N4
q8
Strain Tensor from Nodal Values of
Displacements
u N1
u(x ,h )
v 0
0
N1
N2
0
0
N2
N3
0
0
N3
N4
0
q1
0
Nq
N4
q8
4
N i
N 3
u
N1
N 2
N 4
q2i 1
q1
q3
q5
q7
x i 1 x
x
x
x
x
4
N i
u
q2i 1
h i 1 h
4
N i
v
q2 i
x i 1 x
4
N i
v
q2 i
h i 1 h
Strain Tensor from Nodal Values of
Displacements
u
x
1 h
0
0
1 h
u
0
1 x
0
h 1 1 x
v
1 h
1 h
0
4 0
x
1 x
0
1 x
0
v
h
u
x
q1
q
u
2
h
v G
x
q8
v
h
1 h
1 x
0
0
0
0
1 h
1 x
1 h
1 x
0
0
q1
q2
0
1 h
1 x
q8
0
Strain Tensor from Nodal Values of
Displacements
= AG q
= B q
Both A and G are linear functions of x and h
Stresses
x
E
y
2
1
xy
0 x
1
1
0 y
1
0 0
xy
2
σ Dε
= B q
σ DBq
1
4
Element Stiffness Matrix ke
h
2
x
3
1 T
U e ε σDdV
2 Ve
dV tdA
= B qe
Ue
1 T T
q e B DBtdA q e
le
2
1 T
T
q e t B DBdAq e
2 A
= D B qe
ke
8x8 matrix
Element Stiffness Matrix ke
Furthermore
dV tdA t det Jdxdh
and
k e t B DBdA
T
A
1 1
t B DB det Jdxdh
T
1 1
Numerical
Integration
Integration of Stiffness Matrix
1 1
k e t B DB det Jdxdh
T
1 1
T
B (8x3)
D (3x3)
B (3x8)
ke (8x8)
Integration of Stiffness Matrix
Each term kij in ke is expressed as
3 T 3
kij t Bim Dml Blj det J x ,h dxdh
l 1
1 1 m 1
1 1
1 1
t g x ,h dxdh
1 1
Linear Shape Functions is each Direction
Numerical Integration
Integrals
Definite
Indefinite
1 3
x
dx
x C
3
2
2
2
x
dx
0
8
3
10
8
x2
6
4
2
0
0
0.5
1
1.5
2
2.5
3
Numerical Integration
Definite integrals can be computed numerically
b
f x dx w f x
i
a
i
i
Objective:
• Determine points xi
• Determine coefficients wi
10
8
x2
6
4
2
0
0
0.5
1
1.5
2
2.5
3
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)
a=x1 x2 x3 x4
x5
x6
x7=b
a=x1 x2 x3 x4
x5
x6
x7=b
Upper Sum
U(f;xi)
Numerical Integration
It can be shown that
b
L f ; xi f x dx wi f xi U f ; xi
a
i
b
lim L f ; xi f x dx wi f xi lim U f ; xi
i
a
i
i
Mid-Point Rule
1
A1 x2 x1 f x2 x1
2
1
A6 x7 x6 f x6 x7
2
a=x1 x2 x3 x4
1
x1 x2
2
1
x2 x3
2
b
a
x5
x6
1
x4 x5
2
x7=b
1
x6 x7
2
n 1
1
f x dx wi f xi xi 1 xi f xi 1 xi
2
i
i 1
Mid Point Rule
n 1
1
f x dx wi f xi xi1 xi f xi1 xi
b
i
a
i 1
2
Simple to comprehend and implement
Large number of intervals is required for accuracy
…
a=x1
b=xn
Trapezoid Rule
A1
A6
1
x2 x1 f x2 f x1
2
a=x1 x2 x3 x4
b
a
x5
x6
1
x7 x6 f x7 f x6
2
x7=b
1 n1
f x dx wi f xi xi 1 xi f xi 1 f xi
2 i 1
i
Trapezoid Rule
b
a
1 n1
f x dx wi f xi xi 1 xi f xi 1 f xi
2 i 1
i
Simple to comprehend and implement
Exact for polynomials f(x) = ax+b
Large number of intervals is required for accuracy
(Less than midpoint rule)
Simpson’s Rule
a2h
Step 1
f x dx
a
a
a+2h
h
a2h
a
h
h
f x dx f a 4 f a h f a 2h
3
Simpson’s Rule
Step 2
Ii
h
xi
xi1
xi
Ii+1
h
xi+1/2
…
xi+1
h
f x dx f a 4 f a h f a 2h I i
3
b
n 1
f x dx I
a
i 0
i
Quadratures
Objective
b
f xdx w f x w f x w f x w f x
i
a
i
1
1
2
2
n
n
i
Where do such formulae come from?
Theory of Interpolation….
n
Let f x px li x f xi li(x): cardinal functions
i 1
Recall Shape Functions
Quadratures
b
b
b
n
n
f xdx pxdx f x l xdx f x w
a
a
i 1
i
i
a
i 1
i
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
i
Example
Establish coefficients of quadrature for the interval
[-2,2] and nodes –1,0,1
f(x)
-2
-1
0
1
2
Cardinal Functions:
x xj
li x
j 1 xi x j
j i
n
1 i n
Example
Cardinal Functions – Lagrange Polynomials:
x x j x 0 x 1 1 2
l1 x
x x
1 0 1 1 2
j 1 xi x j
j i
3
x x j x 1 x 1
2
l2 x
x 1
0 1 0 1
j 1 xi x j
j i
3
x x j x 1 x 0 1 2
l3 x
x x
1 1 1 0 2
j 1 xi x j
j i
3
Example
1 2
1 2
2
f x x x f 1 x 1 f 0 x x f 1
2
2
2
2
2
3
3
f x dx px dx f x l x dx f x w
2
i
i 1
2
i
i 1
2
i
i
2
with
wi li x dx
2
1 2
l1 x x x
2
l2 x x 1
2
1 2
l3 x x x
2
Example
2
2
2
1 2
1
8
2
w1 l1 x dx x x dx x x dx
2
2 2
3
2
2
2
2
4
w2 l2 x dx x 1 dx
3
2
2
2
2
2
2
1 2
1
8
2
w3 l3 x dx x x dx x x dx
2
2 2
3
2
2
Example
2
8
4
8
2 f x dx 3 f 1 3 f 0 3 f 1
f(x)=x2
-2
-1
0
1
2
Quadrature
2
8 4
8 16
2
2 x dx 3 1 3 0 3 1 3
Exact
2
3 2
x
2 x dx 3
2
2
16
3
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
b
qx x dx 0
k
k 0,1,..., n - 1
a
Let x1,x2,…,xn be the roots of q(x). Then
b
f xdx w f x w f x w f x w f x
i
a
i
1
1
2
2
n
n
i
with xi’s as nodes is exact for all polynomials of
degree 2n-1.
Gaussian Quadrature 2-point
W2=1
W1=1
-1
f x dx 1 f
1
f(x)
f(x1)
x1 1 3
1
f(x2)
x2 1 3
1 3
1 3 1 f
1
Gauss Points and Weights
Compare
Let
1
2
x
dx 2 3
f(x)=x2
1
Use 2-points
MidPoint
X1=-1, X2=0, X3=1
1
x dx (0.5) 0.5
2
2
2
1 2
1
X=-0.5
X=0.5
Compare
Let
1
2
x
dx 2 3
f(x)=x2
Trapezoidal
Use 2-points
1
X1=-1, X2=0, X3=1
1
1
1
2
2
2
2
2
x
dx
1
(
1
)
(
0
)
1
(
0
)
(
1
)
1
1
2
2
Compare
Let
1
2
x
dx 2 3
f(x)=x2
Gauss
Use 2-points
x2 x 2 1 3, w2 1
x1 x1 1 3, w1 1
x dx 1
1
2
2
1
2
1 3 1 1 3 2 3
1
x1 1 3
x1 1 3
2-Dimensional Integration
Gaussian Quadrature
1 1
f x ,h dxdh
1 1
wi f x i ,h dh
1
i 1
1
n
w w f x ,h
n
n
j 1 i 1
j
i
i
j
2-D Integration 2-point formula
h
h2 1 3
w1w2 1
w1 1
w2 w2 1
w2 1
x
h1 1 3
w1w1 1
w2 w1 1
x1 1 3
x2 1 3
2-D Integration 2-point formula
h
h1 1 3
x
h1 1 3
1 1
x1 1 3
x2 1 3
f x ,h dxdh
1 1
w1w1 f x1 ,h1 w2 w1 f x2 ,h1 w1w2 f x1 ,h2 w2 w2 f x2 ,h2
Integration of Stiffness Matrix
Stiffness Matrix
k e t B DBdA
T
A
1 1
t B DB det Jdxdh
T
1 1
h
4 (-1,1)
3 (1,1)
x
Integration over
1 (-1,-1)
2 (1,-1)
Integration of Stiffness Matrix
J 22 J12
1
A
0
0
det J
J 21 J11
0
J11
J12
0
J 21
J 22
1 h
0
0
1 h
0
1 x
0
1 1 x
G
1 h
1 h
0
4 0
1 x
0
1 x
0
= AG q
1 h
1 x
0
0
(3x4)
0
1 h
0
1 x
0
0
1 h
0
1 h
1 x
1 x
0
= B q
B (3x8)
(4x8)
Integration of Stiffness Matrix
1 1
k e t B DB det Jdxdh
T
1 1
T
B (8x3)
D (3x3)
B (3x8)
ke (8x8)
Integration of Stiffness Matrix
Each term kij in ke is expressed as
3 T 3
kij t Bim Dml Blj det J x ,h dxdh
l 1
1 1 m 1
1 1
1 1
t g x ,h dxdh
1 1
Linear Shape Functions is each Direction
Gaussian Quadrature is accurate if
We use 2 Points in each direction
Integration of Stiffness Matrix
h
w1 1
h1 1 3
w2 1
x
h1 1 3
1 1
x1 1 3
x2 1 3
t g x ,h dxdh
1 1
w1w1 g x1 ,h1 w2 w1 g x2 ,h1 w1w2 g x1 ,h2 w2 w2 g x2 ,h2
g x1 ,h1 g x2 ,h1 g x1 ,h2 g x2 ,h2
Element Body Forces
Total Potential
Galerkin
1 T
ε DεtdA
Ae
e 2
u ftdA
T
e
Ae
u Ttdl
T
e
le
u Pi
T
i
i
0
Ve
e
σ εφ dV
T
φ fdV
T
e
Ve
φ TdA
T
e
Ae
φ Pi
T
i
i
Body Forces
Integral of the form
u f det Adxdh
T
Ae
u N1
v 0
x N1
y 0
0
N1
0
N1
N2
0
N2
0
Ae
0
N2
0
N2
N3
0
N3
0
φ f det Adxdh
T
0
N3
0
N3
N4
0
N4
0
q1
0
N4
q8
1
0
N4
8
Body Forces
In both approaches
WP q1
f x N1 det Adxdh
Ae
f y N1 det Adxdh
Ae
q8 f x N 2 det Adxdh
Ae
f
N
det
Ad
x
d
h
y
4
Ae
Linear Shape Functions
Use same quadrature as stiffness maitrx
Element Traction
Total Potential
Galerkin
1 T
ε DεtdA
Ae
e 2
u ftdA
T
e
Ae
u Ttdl
T
e
le
u Pi
T
i
i
0
Ve
e
σ εφ dV
T
φ fdV
T
e
Ve
φ TdA
T
e
Ae
φ Pi
T
i
i
Element Traction
Similarly to triangles, traction is applied
along sides of element
N 0
1
3
v
Ty
h
4
Tx u
x
2
1
1
N 2 1 h
2
1
N 3 1 h
2
N4 0
WP T23 u Ttdl
T
le
Traction
u 0 0 N 2
v 0 0 0
0
N2
N3
0
0
N3
q1
0 0
0 0
q8
For constant traction along side 2-3
WP T23 q1
0
0
Tx
t e l2 3
q8
Ty
2
0
0
Traction
components
along 2-3
h
Stresses
More Accurate at
Integration points
h1 1 3
x
h1 1 3
x1 1 3
σ DBq
x2 1 3
Stresses are
calculated at
any x,h
1 h
1 h 0 1 h 0
0
0
1 h
1 x 0
1 x
0
1 x
0
0
1 1 x
G
1 h 0 1 h 0
1 h
0
1 h
4 0
1 x 0
1 x
1 x
0
1 x
0
0
J 22 J12
1
A
0
0
det J
J 21 J11
0
J 21
J 22
0
J11
J12
Modeling Issues: Nodal Forces
In view of…
1 T
ε DεtdA
Ae
e 2
u ftdA
T
e
Ae
u Ttdl
T
e
le
u Pi
T
i
i
Or virtual potential energy
A node should be
placed at the location
of nodal forces
Modeling Issues: Element Shape
Square : Optimum Shape
Not always possible to use
Rectangles:
Rule of Thumb
Ratio of sides <2
Larger ratios
may be used
with caution
Angular Distortion
Internal Angle < 180o
Modeling Issues: Degenerate Quadrilaterals
Coincident Corner Nodes
4
4
3
x
x
x
x
1
x
x
1
2
x
x
2
3
Integration Bias
Less accurate
Modeling Issues: Degenerate Quadrilaterals
Three nodes collinear
4
Integration Bias
3
x
3
x
x
4
x
1
x
x
x
1
x
2
2
Less accurate
Modeling Issues: Degenerate Quadrilaterals
2 nodes
Use only as necessary to improve representation of
geometry
Do not use in place of triangular elements
A NoNo Situation
y
3
(7,9)
h
(6,4)
x
4
Parent
1
(3,2)
2
J singular
(9,2)
All interior angles < 180
x
Another NoNo Situation
h
x
x, y
not uniquely
defined
Convergence Considerations
For monotonic convergence of solution
Requirements
Elements (mesh) must be compatible
Elements must be complete
Mesh Compatibility
OK
NO NO!
Mesh compatibility - Refinement
Acceptable Transition
However…
Compatibility of displacements OK
Stresses?
Convergence Considerations
We will revisit the issue…