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
ux ,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 DBdAq 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   xi1  xi f  xi1  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 n1
f x dx   wi f xi    xi 1  xi  f xi 1   f xi 
2 i 1
i
Trapezoid Rule
b

a
1 n1
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
a2h
Step 1
 f x dx
a
a
a+2h
h
a2h

a
h
h
f x dx   f a   4 f a  h   f a  2h 
3
Simpson’s Rule
Step 2
Ii
h
xi
xi1

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 xdx   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   px    li x  f xi  li(x): cardinal functions
i 1
Recall Shape Functions
Quadratures
b
b
b
n
n
 f xdx   pxdx   f x  l xdx   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   px 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
 qx x dx  0
k
k  0,1,..., n - 1
a
Let x1,x2,…,xn be the roots of q(x). Then
b
 f xdx   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 εφ 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 εφ 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 T23   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 T23  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…