Chap-1 Preliminary Concepts and Linear Finite Elements.pptx

1
Prof. Samirsinh Parmar
Asst. Professor, Dept. of Civil Engg.
Dharmasinh Desai University, Nadiad, Gujarat, INDIA
Mail: samirddu@gmail.com
CHAPTER- 1

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Table of Contents
1.1. INTRODUCTION
1.2. VECTOR AND TENSOR CALCULUS
1.3. STRESS AND STRAIN
1.4. MECHANICS OF CONTINUOUS BODIES
1.5. FINITE ELEMENT METHODS

3
VECTOR AND TENSOR
CALCULUS
1.2

4
Vector and Tensor
• Vector: Collection of scalars
• Cartesian vector: Euclidean vector defined using Cartesian
coordinates
– 2D, 3D Cartesian vectors
– Using basis vectors: e1 = {1, 0, 0}T, e2 = {0, 1, 0}T, e3 = {0, 0, 1}T
 
   
 
   
   
 
1
1
2
2
3
u
u
, or u
u
u
u u
  
1 1 2 2 3 3
u u u
u e e e

5
Index Notation and Summation Rule
• Index notation: Any vector or matrix can be expressed in
terms of its indices
• Einstein summation convention
– In this case, k is a dummy variable (can be j or i)
– The same index cannot appear more than twice
• Basis representation of a vector
– Let ek be the basis of vector space V
– Then, any vector in V can be represented by



3
k k k k
k 1
a b a b

 

N
k k k k
k 1
w w
w e e
   
   
   
   
   
   
1 11 12 13
i 2 ij 21 22 23
3 31 32 33
v A A A
[v ] v [A ] A A A
v A A A
v A

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Index Notation and Summation Rule cont.
• Examples
– Matrix multiplication:
– Trace operator:
– Dot product:
– Cross product:
– Contraction: double dot product
   
j k j k ijk j k i
u v ( ) e u v
u v e e e


 



ijk
0 unless i, j,k are distinct
e 1 if (i, j,k) is an even permutation
1 if (i, j,k) is an odd permutation
 
  

3 3
ij ij ij ij
i 1 j 1
J : A B A B
A B
Permutation
symbol
  
   
    
ij ik kj
11 22 33 kk
1 1 2 2 3 3 k k
C A B
tr( ) A A A A
u v u v u v u v
C A B
A
u v

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Cartesian Vector
• Cartesian Vectors
• Dot product
– Kronecker delta function
– Equivalent to change index j to i, or vice versa
• How to obtain Cartesian components of a vector
   

1 1 2 2 3 3 i i
j j
u u u u
v
u e e e e
v e
       
i i j j i j i j i j ij i i
(u ) (v ) uv ( ) uv uv
u v e e e e


  


ij
1 if i j
0 if i j
X1
X2
X3
e1 e2
e3
u
v
     
i i j j j ij i
(v ) v v
e v e e Projection

8
Direct tensor notation Tensor component notation Matrix notation
Notation Used Here
  
a b   i i
ab   T
a b
 
A a b 
ij i j
A ab  T
A ab
 
b A a 
i ij j
b A a 
b Aa
 
b a A 
j i ij
b aA 
T T
b a A

9
Tensor and Rank
• Tensor
– A tensor is an extension of scalar, vector, and matrix
(multidimensional array in a given basis)
– A tensor is independent of any chosen frame of reference
– Tensor field: a tensor-valued function associated with each point
in geometric space
• Rank of Tensor
– No. of indices required to write down the components of tensor
– Scalar (rank 0), vector (rank 1), matrix (rank 2), etc
– Every tensor can be expressed as a linear combination of rank 1
tensors
– Rank 1 tensor v: vi
– Rank 2 tensor A: Aij
– Rank 4 tensor C: Cijkl

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Tensor Operations
• Basic rules for tensors
• Tensor (dyadic) product: increase rank
    
i j i j ij i j
uv A uv
A u v e e
   
   
    
( ) ( )
( ) ( )
( )( ) ( )
u v w u v w
w u v v w u
u v w x v w u x   
u v v u

  
    
 
( ) ( )
( )
( ) ( ) ( )
TS R T SR
T S R TS TR
TS T S T S
1T T1 T
Different notations
 
TS T S
Identity tensor
 ij
[ ]
1

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Tensor Operations cont.
• Symmetric and skew tensors
– Symmetric
– Skew
– Every tensor can be uniquely decomposed by symmetric and
skew tensors
– Note: W has zero diagonal components and Wij = - Wji
• Properties – Let A be a symmetric tensor
 
 
T
1
2
T
1
2
( )
( )
S T T
W T T

 
T
T
S S
W W
 
T S W


: 0
: :
A W
A T A S

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Example
• Displacement gradient can be considered a tensor (rank 2)
  
  
  
  
  
  
 
 

   
  
   

 
 
 
 
1 1 1
1 2 3
2 2 2
1 2 3
3 3 3
1 2 3
u u u
X X X
u u u
X X X
u u u
X X X
u
u
X

   
    

   
    
  
 
    
 
 
 
 
   
 
 
 
 
 
3
1 1 2 1
1 2 1 3 1
3
1 2 2 2
2 1 2 3 2
3 3 3
1 2
3 1 3 2 3
u
u u u u
1 1
X 2 X X 2 X X
u
u u u u
1 1
2 X X X 2 X X
u u u
u u
1 1
2 X X 2 X X X
( ) ( )
sym( ) ( ) ( )
( ) ( )
u

  
   

  
   
 
 
   
 
 
 
 
    
 
 
   
 
 
3
1 2 1
2 1 3 1
3
1 2 2
2 1 3 2
3 3
1 2
3 1 3 2
u
u u u
1 1
2 X X 2 X X
u
u u u
1 1
2 X X 2 X X
u u
u u
1 1
2 X X 2 X X
0 ( ) ( )
skew( ) ( ) 0 ( )
( ) ( ) 0
u
Strain tensor
Rotation
tensor

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Contraction and Trace
• Contraction of rank-2 tensors
– contraction operator reduces four ranks from the sum of ranks of
two tensors
• magnitude (or, norm) of a rank-2 tensor
• Constitutive relation between stress and strain
• Trace: part of contraction
– In tensor notation
     
ij ij 11 11 12 12 32 32 33 33
: a b a b a b a b a b
a b
 :
a a a
   
  ij ijkl kl
: , D
D
   
ii 11 22 33
tr( ) A A A A
A
 
tr( ) : :
A A 1 1 A

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Orthogonal Tensor
• In two different coord.
• Direction cosines
• Change basis
  * *
i i j j
u u
u e e
  
*
i j ij
e e  
*
i ij j
e e
 
 
* *
j j i i
*
i ij j
u u
u
u e e
e
  *
j ij i
u u
 T *
u u
e1
*
e2
*
e3
*
e1
e2
e3
We can also show
 
 
* *
j ij i
e e u u
  
    
T * T T
( ) ( )
u u u u
   
   
T T
det( ) 1
1
Orthogonal tensor


 
1 T
   
 
* T *
ij ik kl jl
, T T
T T
Rank-2 tensor transformation

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Permutation
• The permutation symbol has three indices, but it is not a
tensor
• the permutation is zero when any of two indices have the
same value
• Identity
• vector product


 



ijk
1 if ijk are an even permutation : 123, 231, 312
e 1 if ijk are an odd permutation : 132, 213, 321
0 otherwise
     
ijk lmk il jm im jl
e e
  i ijk j k
e u v
u v e

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Dual Vector
• For any skew tensor W and a vector u
– Wu and u are orthogonal
• Let
• Then,
      
T
0
u Wu u W u u Wu
 
ij ijk k
W e w

   
 
 
     
 
 
 
  
   
12 13 23
12 23 13
13 23 12
0 W W W
W 0 W W
W W 0 W
W w
  
ij j ijk k j ikj k j
W u e w u e w u
 
Wu w u
Dual vector of skew tensor W
  1
i ijk jk
2
w e W

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Vector and Tensor Calculus
• Gradient
– Gradient is considered a vector
– We will often use a simplified notation:
• Laplace operator
• Gradient of a scalar field f(X): vector
( )
 
X
f
i
i
X
 
  
 
e
X
,
i
i j
j
v
v
X



2
i j
i j j j
X X X X
 
 
   
       
 
   
   
   
e e
( ) i
i
X
f
f

 

X e

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Vector and Tensor Calculus
• Gradient of a Tensor Field (increase rank by 1)
• Divergence (decrease rank by 1)
– Ex)
• Curl
j
i j j i j
i i
X X
f
f


       
 
e e e e
f f
  i
i j j
i i
X X
f
f

 

    
 
 
 
e e
f
,
jk j k

   e

,
i ijk k j
e v
  
v e

19
Integral Theorems
• Divergence Theorem
• Gradient Theorem
• Stokes Theorem
• Reynolds Transport Theorem
d d
 
     
 
A n A
d d
 
    
 
A n A
( )d d
c
c

     
 
n v r v

c
r
d
d d ( ) d
dt t
  

     

  
A
A n v A
n: unit outward normal vector

20
Integration-by-Parts
• u(x) and v(x) are continuously differentiable functions
• 1D
• 2D, 3D
• For a vector field v(x)
• Green’s identity
b b
b
a
a a
u(x)v (x)dx u(x)v(x) u (x)v(x)dx
 
 
 
 
 
i
i i
u v
vd uvn d u d
x x
  
 
    
 
  
u d u( )d u d
  
         
  
v v n v
2
u vd u v d u vd
  
          
  
n

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Example: Divergence Theorem
• S: unit sphere (x2 + y2 + z2 = 1), F = 2xi + y2j + z2k
• Integrate d
S
S

 F n
S
dS d
2 (1 y z)d
2 d 2 yd 2 zd
2 d
8
3


  

    
   
     
 


 

  

F n F

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Surface Traction (Stress)
• Surface traction (Stress)
– The entire body is in equilibrium with
external forces (f1 ~ f6)
– The imaginary cut body is in equilibrium due
to external forces (f1, f2, f3) and internal
forces
– Internal force acting at a point P
on a plane whose unit normal is n:
– The surface traction depends on the unit
normal direction n.
– Surface traction will change as n changes.
– unit = force per unit area (pressure)
f1
f2
f3
f4
f6
f5
y
z
F
n
f1
f2
f3
P
A
x
( )
A 0
lim
A
 



n F
t
( )
1 1 2 2 3 3
t t t
  
n
t e e e

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Cartesian Stress Components
• Surface traction changes according to the direction of
the surface.
• Impossible to store stress information for all directions.
• Let’s store surface traction parallel to the three
coordinate directions.
• Surface traction in other directions can be calculated
from them.
• Consider the x-face of an infinitesimal cube
11 12
13
x y
z
x
y
z
F
(x) (x) (x)
(x)
1 2 3
1 2 3
t t t
 
t e e e
+
(x)
11 1 12 2 13 3
    
t e e e
+
Normal
stress
Shear
stress

25
Stress Tensor
– First index is the face and the second index is its direction
– When two indices are the same, normal stress, otherwise shear
stress.
– Continuation for other surfaces.
– Total nine components
– Same stress components are defined for the negative planes.
• Rank-2 Stress Tensor
• Sign convention
11 22
33
12
13
21
23
31
32
x y
z
x
y
z
ij i j
  
e e

11 x
12 y
sgn( ) sgn( ) sgn( F )
sgn( ) sgn( ) sgn( F )
   
   
n
n

26
Symmetry of Stress Tensor
– Stress tensor should be symmetric
9 components 6 components
– Equilibrium of the angular moment
– Similarly for all three directions:
– Let’s use vector notation:
12
21
x
y
12
21
O
l
l
A B
C D
12 21
12 21
M l( ) 0
     
   

11 12 13
ij 12 22 23
13 23 33
[ ]
  
 
 
    
 
 
  
 
11
22
33
12
23
13
{ }

 
 

 
 

  

 
 

 

 

12 21 23 32 13 31
, ,
        
Cartesian components
of stress tensor

27
Stress in Arbitrary Plane
– If Cartesian stress components are known, it is possible to
determine the surface traction acting on any plane.
– Consider a plane whose normal is n.
– Surface area (ABC = A)
– The surface traction
– Force balance
3 1 2
PAB An ; PBC An ; PAC An
     
x
y
z
B
A
C
33
31
32
22
23
21
11
13
12
t(n)
n
P
( ) ( ) ( )
( )
1 2 3
1 2 3
t t t
  
n n n
n
t e e e
( )
1 11 1 21 2 31 3
1
F t A An An An 0
       
 n
( )
11 1 21 2 31 3
1
t n n n
     
n

28
Cauchy’s Lemma
• All three-directions
• Tensor notation
– stress tensor; completely characterize the state of stress at a
point
• Cauchy’s Lemma
– the surface tractions acting on opposite sides of the same surface
are equal in magnitude and opposite in direction
( )
12 1 22 2 32 3
2
( )
13 1 23 2 33 3
3
t n n n
t n n n
     
     
n
n
( )
11 1 21 2 31 3
1
t n n n
     
n
( ) ( )
    
n n
t n t n
 

 
( ) ( )
n n
t t

29
Projected Stresses
• Normal stress
• Shear stress
• Principal stresses
• Mean stress (hydrostatic pressure)
• Stress deviator
ij i j
( ) nn
       
n
n t n n n

2 2
( ) ( ) ( ) ( )
      
n n
n t n n t n n

1 2 3
/ / , ,
   
n
t n
m dev
11 m 12 13
12 11 m 23
13 23 11 m
:
   
    
 
 
     
 
 
    
 
s 1 I
s
 
Which stresses are frame indifferent?
m 11 22 33
1 1
p tr( ) ( )
3 3
        

1
dev 3
  
I I 1 1
1
ijkl ik jl il jk
3
I ( )
     
Rank-4 identity tensor
Rank-4 deviatoric identity tensor

30
Strain
• Strain: a measure of deformation
– Normal strain: change in length of a line segment
– Shear strain: change in angle between two perpendicular line
segments
• Displacement of P = (u, v, w)
• Displacement of Q & R
P(x,y,z) Q
R P'(x+u,y+v,z+w)
Q'
R'
x
y
z
x
y
Q
Q
Q
u
u u x
x
v
v v x
x
w
w w x
x

  


  


  

R
R
R
u
u u y
y
v
v v y
y
w
w w y
y

  


  


  


31
Strain
– Strain is defined as the elongation per unit length
– Tensile (normal) strains in x- and y-directions
– Strain is a dimensionless quantity. Positive for elongation and
negative for compression
P P
x ux
y
uy
x x
11
x 0
y y
22
y 0
u u
lim
x x
u u
lim
y y
 
 
 
  
 
 
  
 
Textbook has different, but
more rigorous derivations

32
Shear Strain
– Shear strain is the tangent of the change in angle between two
originally perpendicular axes
– Shear strain (change of angle)
– Positive when the angle between two positive (or two negative)
faces is reduced and negative when the angle is increased.
– Valid for small deformation
P
ux
uy
q2
q1
/2 – g12
y
1 1
x
2 2
u
~ tan
x
u
~ tan
y

q q 


q q 

y y
x x
12 1 2
x 0 y 0
u u
u u
lim lim
x y x y
   
 
 
g  q  q    
   
y x
12 12
u u
1 1
2 2 x y

 

  g  
 
 
 
 
x
y

33
Strain Tensor
• Strain Tensor
• Cartesian Components
• Vector notation
ij i j
  
e e

11 12 13
ij 12 22 23
13 23 33
[ ]
  
 
 
   
 
 
  
 

11 11
22 22
33 33
12 12
23 23
13 13
{ }
2
2
2
 
   
   
 
   
   
 
 
   
 g
   
   
 g
   
 g
   


34
Volumetric and Deviatoric Strain
• Volumetric strain (from small strain assumption)
• Deviatoric strain
0
V 11 22 33 11 22 33
0
V V
(1 )(1 )(1 ) 1
V

               
V 11 22 33 kk
        
1 1
V ij ij V ij
3 3
e
       
e 1

dev :

e I 
x2
x3
x1
1
e11
e22
e33
1
1
Exercise: Write Idev in matrix-vector notation

35
Stress-Strain Relationship
• Applied Load shape change (strain) stress
• There must be a relation between stress and strain
• Linear Elasticity: Simplest and most commonly used
Proportional
limit
Yield stress
Ultimate
stress
Strain
hardening
Necking
Fracture


Young’s
modulus

36
Generalized Hooke’s Law
• Linear elastic material
– In general, Dijkl has 81 components
– Due to symmetry in ij, Dijkl = Djikl
– Due to symmetry in kl, Dijkl = Dijlk
– from definition of strain energy, Dijkl = Dklij
• Isotropic material (no directional dependence)
– Most general 4-th order isotropic tensor
– Have only two independent coefficients
(Lame’s constants)
21 independent coeff
ijkl ij kl ik jl il jk
ij kl ik jl il jk
D
( )
        
         
ij ijkl kl
: , D
   
D
 
2
    
D 1 1 I

37
Generalized Hooke’s Law cont.
• Stress-strain relation
– Volumetric strain:
– Off-diagonal part:
– Bulk modulus K: relation b/w volumetric stress & strain
– Substitute so that we can separate volumetric part
• Total deform. = volumetric + deviatoric deform.
ij ijkl kl ij kl ik jl il jk kl kk ij ij
D [ ( )] 2
                  
kk 11 22 33 v
        
12 12 12
2
    g  is the shear modulus
1 m jj kk jj jj kk
I 3 2 (3 2 )
             
2
m kk v
3
( ) K
       
Bulk modulus
2
3
K
   

38
• Stress-strain relation cont.
2
ij kk ij ij
3
2
kk ij ij kk ij
3
1
ij kl kl ik jl ij kl kl
3
ij kl dev ijkl kl
(K ) 2
K 2
K 2 [ ]
K 2 (I )
       
       
           
 
     
 
dev
σ K 2 : ε
    
 
 
1 1 I
Deviatoric part
Volumetric part
v
m
σ K 2
σ
    
  
1 e
1 s
dev :

e I 
Deviatoric strain
Important for plasticity; plastic deformation only occurs in deviatoric part
volumetric part is always elastic
dev :

s I 
Deviatoric stress

39
• Vector notation
– The tensor notation is not convenient for computer implementation
– Thus, we use Voigt notation
– Strain (6×1 vector), Stress (6×1 vector), and C (6×6 matrix)
2nd-order tensor vector
4th-order tensor matrix
1,1
11
2,2
22
3,3
33
12 1,2 2,1
23 2,3 3,2
13 1,3 3,1
u
u
u
2 u u
2 u u
2 u u
 

 
 
 
  
 
 
 

 
 
 
 
 
 
 
 
 
 
 
  
  
 

11
22
33
12
23
13

 
 

 
 

 
 

 
 

 

 
D
 
12 + 21 = 212 You don’t need 2 here

40
3D Solid Element cont.
• Elasticity matrix
• Relation b/w
Lame’s constants
and Young’s modulus
and Poisson’s ratio
(3 2 )
, E
2( )
E E
,
(1 )(1 2 ) 2(1 )
    
  
     

   
     
dev
K 2
   
D 1 1 I
2 0 0 0
2 0 0 0
2 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
    
 
 
    
 
 
    
  

 
 

 

 
D
1
1
1
0
0
0
 
 
 
 
  
 
 
 
 
1
2 1 1
3 3 3
1 2 1
3 3 3
1 1 2
3 3 3
dev 1
2
1
2
1
2
0 0 0
0 0 0
0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
 
 
 
 
 
 
 
 

 
 
 
 
 
I

41
Plane Stress
• thin plate–like components parallel to the xy–plane
• the plate is subjected to forces in its plane only
• 13 = 23 = 33 = 0
• 13 = 23 = 0, but 33 ≠ 0
11 11
22 22
2
1
12 12
2
1 0
E
{ } 1 0
1
0 0 (1 )
 
  
   
   
 
    
   
 
 
   
 
   g
   
 


42
Plane Strain
• the strains with a z subscript are all equal to zero
• deformation in the z–direction is constrained, (i.e., u3 = 0)
• 13 = 23 = 33 = 0
• can be used if the structure is infinitely long in the z–
direction
• 13 = 23 = 0
11 11
22 22
1
12 12
2
1 0
E
{ } 1 0
(1 )(1 2 )
0 0
 
    
   
   
 
      
   
 
   
   
 
   g
   
 

 
33 11 22
E
(1 )(1 2 )

    
   

43
MECHANICS OF
CONTINUOUS BODIES
1.4

44
Balance of Linear Momentum
• Balance of linear momentum
• Stress tensor (rank 2):
• Surface traction
• Cauchy’s Lemma
b
d d d
  
      
  
n
f t a
ij i j
  
e e

11 12 13
ij 21 22 23
31 32 33
  
 
 
 
    
   
 
  
 
 
n
t n 

 
n n
t t

    
n n
t n t n
 


n
tn
X1
X2
X3
e1 e2
e3
X
fb: body force
tn: surface traction

45
Balance of Linear Momentum cont
• Balance of linear momentum
– For a static problem
• Balance of angular momentum
b
( )d d d
  
           
  
f a n  
b
[ ( )]d 0

      
 f a

b
( ) 0
     
f a

b b
ij,i j
0 f 0
        
f

b
d d d
  
         
  
n
x f x t x a
T
ij ji
   
 
Divergence Thm

46
Boundary-Valued Problem
• We want to determine the state of a body in equilibrium
• The equilibrium state (solution) of the body must satisfy
– local momentum balance equation
– boundary conditions
• Strong form of BVP
– Given body force fb, and traction t
on the boundary, find u such that
and
• Solution space
h
s
on essential BC
on natural BC
 
  
u 0
t n 
b
0
    
f
 (1)
(2)
(3)


n
t
X1
X2
X3
e1 e2
e3
X
fb
 
2 3 h s
A
D [C ( )] | 0 on , on
         
u u x n t x


47
Boundary-Valued Problem cont.
• How to solve BVP
– To solve the strong form, we want to construct trial solutions that
automatically satisfy a part of BVP and find the solution that
satisfy remaining conditions.
– Statically admissible stress field: satisfy (1) and (3)
– Kinematically admissible displacement field: satisfy (2) and have
piecewise continuous first partial derivative
– Admissible stress field is difficult to construct. Thus, admissible
displacement field is used often

48
Principle of Minimum Potential Energy (PMPE)
• Deformable bodies generate internal forces by
deformation against externally applied forces
• Equilibrium: balance between internal and external forces
• For elastic materials, the concept of force equilibrium can
be extended to energy balance
• Strain energy: stored energy due to deformation
(corresponding to internal force)
• For elastic material, U(u) is only a function of total
displacement u (independent of path)
1
U( ) ( ) : ( )d
2 
 

u u u
  ( ) : ( )

u D u
 
Linear elastic material

49
PMPE cont.
• Work done by applied loads (conservative loads)
• U(u) is a quadratic function of u, while W(u) is a linear
function of u.
• Potential energy
s
b
W( ) d d .
 
     
 
u u f u t
h
g

u1 u2
u3
x1
x2
x3
f s
f b
s
b
( ) U( ) W( )
1
( ) : ( )d d d .
2   
  
       
  
u u u
u u u f u t
 

50
PMPE cont.
• PMPE: for all displacements that satisfy the boundary
conditions, known as kinematically admissible
displacements, those which satisfy the boundary-valued
problem make the total potential energy stationary on DA
• But, the potential energy is well defined in the space of
kinematically admissible displacements
• No need to satisfy traction BC (it is a part of potential)
• Less requirement on continuity
• The solution is called a generalized (natural) solution
 
1 3 h
[H ( )] | 0 on ,
     
u u x
Z
H1: first-order derivatives are integrable

51
Example – Uniaxial Bar
• Strong form
• Integrate twice:
• Apply two BCs:
• PMPE with assumed solution u(x) = c1x + c2
• To satisfy KAD space, u(0) = 0,  u(x) = c1x
• Potential energy:
L
F
x
EAu 0 x [0,L]
u 0 x 0
EAu (L) F x L
  
 
  
1 2
EAu(x) c x c
 
Fx
u(x)
EA

L 2 2
1
0
1
1
U EA(u ) dx EALc
2
W Fu(L) FLc

 
 

1
1 1
d d
(U W) EALc FL 0
dc dc

     1
F Fx
c u(x)
EA EA
  

52
Virtual Displacement Field
• Virtual displacement (Space Z)
– Small arbitrary perturbation (variation) of real displacement
– Let ū be the virtual displacement, then u + ū must be kinematically
admissible, too
– Then, ū must satisfy homogeneous displacement BC
– Space Z only includes homogeneous
essential BCs
• Property of variation
    
u u u u
V Z
 
h
1 3
[H ( )] , 0

   
u u u
Z
In the literature, u is often used instead of ū
d d( )
dx dx

 
 
 
 
u u
0
0
1 d
lim [( ) ( )] ( ) .
d


         
 
u u u u u
  

53
PMPE As a Variation
• Necessary condition for minimum PE
– Stationary condition <--> first variation = 0
• Variation of strain energy
0
0
1 d
( ; ) lim [ ( ) ( )] ( ) 0
d


           
 
u u u u u u u
 Z
for all u
0
d
d 
    
   
  
   
   
   
u u u u
x x x
( ) ( )
  
u u
   :
  D
 
1
2
U( ; ) ( ) : : ( ) ( ) : : ( ) d
( ) : : ( )d
a( , )


   
 
 
 



u u u D u u D u
u D u
u u
   
 
Energy bilinear form

54
PMPE As a Variation cont.
• Variation of work done by applied loads
• Thus, PMPE becomes
– Load form is linear with respect to ū
– Energy form a(u, ū) is symmetric, bilinear w.r.t. u and ū
– Different problems have different a(u, ū) and , but they share
the same property
• How can we satisfy “for all ū ” requirement?
Can we test an infinite number of ū?
s
b
W( ; ) d d ( )
 
       
 
u u u f u t u
a( , ) ( ),
  
u u u u Z
( )
u
( )
u
( ; ) U( ; ) W( ; ) 0
     
u u u u u u
Load linear form

55
Example – Uniaxial Bar
• Assumed displacement u(x) = cx 
– virtual displacement is in the same space with u(x):
• Variation of strain energy
• Variation of applied load
• PMPE
u(x) cx

L L
2
0 0 0
0
L
0
d 1 1
U EA (u u) dx 2EA(u u) u dx
d 2 2
EAu u dx EALcc


 
  
      
 
 
 
  
 
 
 

0
d
W F u(L) u(L) Fu(L) FLc
d 
     
   
   

U W c(EALc FL) 0
       
Fx
u(x) cx
EA
 

56
Principle of Virtual Work
• Instead of solving the strong form directly, we want to
solve the equation with relaxed requirement (weak form)
• Virtual work – Work resulting from real forces acting
through a virtual displacement
• Principle of virtual work – when a system is in equilibrium,
the forces applied to the system will not produce any
virtual work for arbitrary virtual displacements
– Balance of linear momentum is force equilibrium
– Thus, the virtual work can be obtained by multiplying the force
equilibrium equation with a virtual displacement
– If the above virtual work becomes zero for arbitrary ū, then it
satisfies the original equilibrium equation in a weak sense
b
0
    
f

b
W ( ) d

      
 f u


57
Principle of Virtual Work cont
• PVW
– Integration-by-parts
– Divergence Thm
– The boundary is decomposed by
b
ij,i j j
( f )u d 0

      
 u Z
b
ij,i j j j
u d f u d
 
     
 
b
ij j ,i ij j,i j j
( u ) u d f u d
 
 
       
 
 
b
i ij j ij j,i j j
n u d u d f u d
  
        
  
h s
j i ij j
u 0 on and n t on
    
h s
    
S
b
j j ij j,i j j
t u d u d f u d
  
       
  

58
Principle of Virtual Work cont
• Since ij is symmetric
• Weak Form of BVP
Internal virtual work = external virtual work
Starting point of FEM
• Symbolic expression
– Energy form:
– Load form:
s
b
ij ij j j j j
d f u d t u d
  
         
   u Z
a( , ) ( )
  
u u u u Z
a( , ) : d

 

u u  
s
b
( ) d d
 
      
 
u u f u t
ij j,i ij j,i ij ij
u sym(u )
     
j
i
i,j ij
j i
u
u
1
sym(u )
2 X X

 

   
 
 
 
 
[ ]{ } { }

K d F
FE equation

59
Example – Heat Transfer Problem
• Steady-State Differential Equation
• Boundary conditions
• Space of kinematically admissible temperature
• Multiply by virtual temperature, integrate by part, and
apply boundary conditions
Q
domain A
Sq
ST
qn
T = T0
n = {nx, ny}T
y
x
T
T k Q 0
k
y
x y
x

   

    
   

 
 
  
0 T
n x x y y q
T T on S
dT dT
q n k n k on S
dx dy




  


 
1
T
T H ( ) T( ) 0, S
     
x x
Z
q
x y n q
S
T T T T
k k d TQd Tq dS , T
x x y y
 
 
   
      
 
   
 
   Z

60
Example – Beam Problem
• Governing DE
• Boundary conditions for cantilevered beam
• Space of kinematically admissible displacement
• Integrate-by-part twice, and apply BCs
4
4
d v
EI f(x), x [0,L]
dx
 
2 3
2 3
dv d v d v
v(0) (0) (L) (L) 0
dx dx dx
   
f(x)
x L
2 dv
v H [0,L] v(0) (0) 0
dx
 
   
 
 
Z
2 2
L L
2 2
0 0
d v d v
EI dx fv dx, v
dx dx
  
  Z

61
Difference b/w Strong and Weak Solutions
• The solution of the strong form needs to be twice
differentiable
• The solution of the weak form requires the first-order
derivatives are integrable bigger solution space than
that of the strong form
• If the strong form has a solution, it is the solution of the
weak form
• If the strong form does not have a solution, the weak
form may have a solution
F
y
x
T
T k Q 0
k
y
x y
x

   

    
   

 
 
  
x y
T T T T
k k d
x x y y

 
   
 
 
   
 


63
Finite Element Approximation
• Difficult to solve a variational equation analytically
• Approximate solution
– Linear combination of trial functions
– Smoothness & accuracy depend on
the choice of trial functions
– If the approximate solution is expressed in the entire domain, it is
difficult to satisfy kinematically admissible conditions
• Finite element approximation
– Approximate solution in simple sub-domains (elements)
– Simple trial functions (low-order polynomials) within an element
– Kinematically admissible conditions only for elements on the
boundary
n
i i
i 1
u(x) c (x)

 f

Exact solution
Approximate
solution
x
u(x)
Finite elements
Nodes
Piecewise-
linear
approximation

64
Finite Elements
• Types of finite elements
1D 2D 3D
• Variational equation is imposed on each element.
One element
1 0.1 0.2 1
0 0 0.1 0.9
dx dx dx dx
   
   

65
Trial Solution
– Solution within an element is approximated using simple polynomials.
– i-th element is composed of two nodes: xi and xi+1. Since two
unknowns are involved, linear polynomial can be used:
– The unknown coefficients, a0 and a1, will be expressed in terms of
nodal solutions u(xi) and u(xi+1).
1 2 3 n1 n+1
n
1 2 n1 n
xi xi+1
i
l
0 1 i i 1
u(x) a a x, x x x
   

66
Trial Solution cont.
– Substitute two nodal values
– Express a0 and a1 in terms of ui and ui+1. Then, the solution is
approximated by
– Solution for Element e:
– N1(x) and N2(x): Shape Function or Interpolation Function
i i 0 1 i
i 1 i 1 0 1 i 1
u(x ) u a a x
u(x ) u a a x
  
  


  

1 2
i 1 i
i i 1
(e) (e)
N (x) N (x)
x x x x
u(x) u u
L L


 
 
1 i 2 i 1 i i 1
u(x) N (x)u N (x)u , x x x
 
   

67
Trial Solution cont.
• Observations
– Solution u(x) is interpolated using its nodal values ui and ui+1.
– N1(x) = 1 at node xi, and =0 at node xi+1.
– The solution is approximated by piecewise linear polynomial and its
gradient is constant within an element.
– Stress and strain (derivative) are often averaged at the node.
N1(x) N2(x)
xi xi+1
xi xi+1 xi+2 xi xi+1 xi+2
ui
ui+1
ui+2 du
dx
u

68
1D Finite Elements
• 1D BVP
• Use PVW
• Integration-by-parts
– This variational equation also satisfies at individual element level
2
2
d u
p(x) 0, 0 x 1s
dx
u(0) 0
Boundary conditions
du
(1) 0
dx
   
 


 

2
1
2
0
d u
p u dx 0
dx
 
 
 
 
  
(1)
u H [0,1] u(0) 0
  
Z
Space of kinematically
admissible displacements
1
1 1
0 0
0
du du du
u dx pu dx
dx dx dx
  
 

69
1D Interpolation Functions
• Finite element approximation for one element (e) at a time
• Satisfies interpolation condition
• Interpolation of displacement variation (same with u)
• Derivative of u(x)
(e) (e) (e)
i 1 i 1 2
u (x) uN (x) u N (x)

   
N d
i
(e) (e)
1 2
i 1
u
N N
u
 
   
   
 
d N
(e)
i i
(e)
i 1 i 1
u (x ) u
u (x ) u
 


(e) (e) (e)
i 1 i 1 2
u (x) uN (x) u N (x)

   
N d
(e)
i i (e) (e)
1 2
(e) (e)
i 1 i 1
u u
dN dN
du 1 1
dx dx dx u u
L L
 
   
   
    
   
   
 
     
B d

70
Element-Level Variational Equation
• Approximate variational equation for element (e)
– Must satisfied for all
– If element (e) is not on the boundary, can be arbitrary
• Element-level variational equation
j j
i i
i
x x
(e)T (e)T (e) (e) (e)T (e)T (e)T
x x
i 1
du
(x )
dx
dx p(x)dx
du
(x )
dx 
 

 
     
 
   

 
 
d B B d d N d
(e)
u (x)  Z
(e)
d
j j
i i
i
x x
(e)T (e) (e) (e)T
x x
i 1
du
(x )
dx
dx p(x)dx
du
(x )
dx 
 

 
     
 
   

 
 
B B d N
 
i
(e) (e) (e)
i 1
du
(x )
dx
[ ]{ }
du
(x )
dx 
 

 
   
 

 
k d f
2x2 matrix 2x1 vector

71
Assembly
• Need to derive the element-level equation for all elements
• Consider Elements 1 and 2 (connected at Node 2)
• Assembly
(1) (1)
1
11 12 1 1
2 2
21 22
2
du
(x )
k k u f dx
u f du
k k (x )
dx
 

 
     
 
     
 
   
   

 
(2) (2)
2
2 2
11 12
3 3
21 22
3
du
(x )
u f
k k dx
u f du
k k (x )
dx
 

 
     
 
     
 
   
   

 
(1) (1) (1)
1
11 12 1
1
(1) (1) (2) (2) (1) (2)
2
21 22 11 12 2 2
(2) (2) (2)
3
21 22 3
3
du
(x )
k k 0 f
u dx
k k k k u f f 0
u du
0 k k f (x )
dx
 

     
 
     
   
   
       
       
 
     
   
 
Vanished
unknown term

72
Assembly cont.
• Assembly of NE elements (ND = NE + 1)
• Coefficient matrix [K] is singular; it will become non-
singular after applying boundary conditions
 
 
   
E E
E
D
D
D D
(1) (1) (1)
11 12 1
1
(1) (1) (2) (2) (1) (2)
21 22 11 12 2 2
2
(2) (2) (2) (2) (3)
3
221 22 11 3 3
N (N )
(N )
N N
21 22 N 1 N 1
N N
k k 0 0 f
u
k k k k 0 f f
u
u
0 k k k 0 f f
u f
0 0 0 k k
 

 

 
   
 
   
 
 
   
   
   
 
   
     
     
     
   
 
 
 
D
1
N
N 1
du
(x )
dx
0
0
du
(x )
dx

 
 
 
 
 
 
 
 
 

 
 
[ ]{ } { }

K q F

73
Example
• Use three equal-length elements
• All elements have the same coefficient matrix
• RHS (p(x) = x)
2
2
d u
x 0, 0 x 1 u(0) 0, u(1) 0
dx
     
(e)
(e)
2 2
1 1 3 3
1
, (e 1,2,3)
1 1 3 3
L

 
   
    
   
   
   
k
i 1 i 1
i i
x x
1 i 1
(e)
(e)
x x
2 i
i i 1
(e)
i i 1
N (x) x(x x)
1
{ } p(x) dx dx
N (x) x(x x )
L
x x
3 6
L , (e 1,2,3)
x x
6 3
  



   
 
   

   
 

 
 
 
 
 

 
 
 
f

74
Example cont.
• RHS cont.
• Assembly
• Apply boundary conditions
– Deleting 1st and 4th rows and columns
(3)
(2)
(1)
3
2
1
(3)
(2)
(1)
3
2 4
f
f
f 1 4 7
1 1 1
, ,
54 54 54
2 5 8
f
f
f
 
 
       
     
  
           
     
     
     
1
2
3
4
1
(0)
54
3 3 0 0 2 4
3 3 3 3 0 54 54
0 3 3 3 3 7 5
54 54
0 0 3 3
8
(1)
54
du
dx
u
u
u
u
du
dx
ì ü
ï ï
ï ï
-
ï ï
ï ï
ï ï
ì ü
é ù
- ï ï ï ï
ï ï ï ï
ê ú
ï ï +
ï ï
ï ï
ê ú
- + - ï ï
ï ï ï ï
ê ú =
í ý í ý
ê ú
ï ï ï ï
- + - ï ï ï ï
ê ú +
ï ï ï ï
ï ï ï ï
ê ú
- ï ï ï ï
ë û
î þ ï ï
ï ï
ï ï
+
ï ï
ï ï
î þ
Element 1
Element 2
Element 3
2
3
u
6 3 1
1
9
u
3 6 2
  
   

   
 

   
 
4
2 81
5
3 81
u
u



75
EXAMPLE cont.
• Approximate solution
• Exact solution
– Three element solutions are poor
– Need more elements
4 1
x, 0 x
27 3
4 1 1 1 2
u(x) x , x
81 27 3 3 3
5 5 2 2
x , x 1
81 27 3 3

 


 

    
  
 

  
   
  
   0
0.02
0.04
0.06
0.08
0 0.2 0.4 0.6 0.8 1
x
u(x)
u-approx.
u-exact
 
2
1
u(x) x 1 x
6
 

76
3D Solid Element
• Isoparametric mapping
– Build interpolation functions on the reference element
– Jacobian: mapping relation between physical and reference elem.
• Interpolation and mapping
(a) Finite Element (b) Reference Element
x

z
(1,1,–1)
(1,1,1)
(–1,1,1)
(–1,1,–1)
x1
x2
x3
x4
x5
x6
x7
x8
x2
x1
x3
(1, –1,–1)
(1, –1,1)
(–1, –1,1)
8
I I
I 1
( ) N ( )

 
u u
x x
8
I I
I 1
( ) N ( )

 
x x
x x
I I I I
1
N ( ) (1 )(1 )(1 )
8
  xx    zz
x
Same for mapping
and interpolation

77
• Jacobian matrix
• Derivatives of shape functions
– Jacobian should not be zero anywhere in the element
– Zero or negative Jacobian: mapping is invalid (bad element shape)
8
I
3 3 I
I 1
dN ( )
d
d d


  
x
J x
x
x x
1 1 1
I I I I I I 2 2 2
1 2 3
3 3 3
x x x
N N N N N N x x x
x x x
x x x
  
 
 
x  z
 
        
    

    
x  z    x  z
    
  
 
 
x  z
 
I I
N N
 
 
 
J
x
x
1
I I
N N 
 
 
 
J
x x
J : Jacobian

78
• Displacement-strain relation
8
I I
I 1
( )

 
u B u

I,1
I,2
I,3
I
I,2 I,1
I,3 I,2
I,3 I,1
N 0 0
0 N 0
0 0 N
N N 0
0 N N
N 0 N
 
 
 
 
 

 
 
 
 
 
B
8
I I
I 1
( )

  
u B u
 
i
I,i
i
N
N
x




79
• Transformation of integration domain
• Energy form
• Load form
• Discrete variational equation
1 1 1
1 1 1
d d d d
   
  x  z
    J
8 8
1 1 1
T T T
I I J J
1 1 1
I 1 J 1
a( , ) d d d { } [ ]{ }
  
 
 
 x  z 
 
 
   
u u u B DB J u d k d
8
1 1 1
T b T
I I
1 1 1
I 1
( ) N ( ) d d d { } { }
  

 x  z 
   
u u f J d f
x
T T
h
{ } [ ]{ } { } { }, { }
  
d k d d f d Z

80
Numerical Integration
• For bar and beam, analytical integration is possible
• For plate and solid, analytical integration is difficult, if
not impossible
• Gauss quadrature is most popular in FEM due to simplicity
and accuracy
• 1D Gauss quadrature
– NG: No. of integ. points; xi: integ. point; wi: integ. weight
– xi and wi are chosen so that the integration is exact
for (2NG – 1)-order polynomial
– Works well for smooth function
– Integration domain is [-1, 1]
NG
1
i i
1
i 1
f( )d f( )


x x  w x



81
Numerical Integration cont.
• Multi-dimensions
NG NG
1 1
i j i j
1 1
i 1 j 1
f( , )d d f( , )
 
 
x  x   ww x 

 
NG NG NG
1 1 1
i j k i j k
1 1 1
i 1 j 1 k 1
f( , , )d d d f( , , )
  
  
x  z x  z  ww w x  z

  
NG
Integration
Points (xi)
Weights (wi)
1 0.0 2.0
2 .5773502692 1.0
3
.7745966692
0.0
.5555555556
.8888888889
4
.8611363116
.3399810436
.3478546451
.6521451549
5
.9061798459
.5384693101
0.0
.2369268851
.4786286705
.5688888889
x

x

x

(a) 11
(b) 22 (c) 33

Chap-1 Preliminary Concepts and Linear Finite Elements.pptx

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