#### Transcript MANE 4240 & CIVL 4240 Introduction to Finite Elements

```MANE 4240 & CIVL 4240
Introduction to Finite Elements
Prof. Suvranu De
Finite element formulation for
1D elasticity using the
Rayleigh-Ritz Principle
Lecture notes, Logan 3.10
Summary:
• Stiffness matrix and nodal load vectors for 1D elasticity
problem
y
F
A(x) = cross section at x
b(x) = body force distribution
(force per unit length)
x E(x) = Young’s modulus
x
x=L
x=0
Potential energy of the axially loaded bar corresponding to the
exact solution u(x)
2
L
1
 du 
(u)   EA  dx   bu dx  Fu(x  L)
0
2 0
 dx 
L
Potential energy of the bar corresponding to an admissible
displacement w(x)
2
L
1
 dw 
(w)   EA
 dx  0 bw dx  Fw(x  L)
0
2
 dx 
L
Finite element idea:
Step 1: Divide the truss into finite elements connected to each
other through special points (“nodes”)
1
El #1
4
3
2
El #2
El #3
Total potential energy=sum of potential energies of the elements
2
L
1
 dw 
(w)   EA
dx   bw dx  Fw(x  L)

0
2 0
 dx 
L
x1=0
x2
El #1
El #2
x4=L
x3
El #3
Total potential energy
2
L
1
 dw 
(w)   EA
dx   bw dx  Fw(x  L)

0
2 0
 dx 
L
Potential energy of element 1:
2
x2
1
 dw 
1 (w)   EA
 dx  x bw dx
x
1
2 1
 dx 
x2
Potential energy of element 2:
2
x3
1
 dw 
 2 (w)   EA
 dx  x bw dx
x
2
2 2
 dx 
x3
x1=0
x2
El #1
El #2
x4
x3
El #3
Potential energy of element 3:
2
x4
1
 dw 
 3 (w)   EA
dx   bw dx  Fw(x  L)

x3
2 x3
 dx 
x4
Total potential energy=sum of potential energies of the elements
(w)  1 (w)   2 (w)   3 (w)
Step 2: Describe the behavior of each element
In the “direct stiffness” approach, we derived the stiffness matrix
of each element directly (See lecture on Springs/Trusses).
Now, we will first approximate the displacement inside each
element and then show you a systematic way of deriving the
stiffness matrix (sections 2.2 and 3.1 of Logan).
TASK 1: APPROXIMATE THE DISPLACEMENT WITHIN
EACH ELEMENT
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN
EACH ELEMENT
TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH
ELEMENT (this class) USING THE RAYLEIGH-RITZ
PRINCIPLE
Summary
Inside an element, the three most important approximations in
terms of the nodal displacements (d) are:
Displacement approximation in terms of shape functions
(1)
w(x)  N d
Strain approximation in terms of strain-displacement matrix
(2)
ε(x) B d
Stress approximation in terms of strain-displacement matrix and
Young’s modulus
  EB d
(3)
The shape functions for a 1D linear element
1
1
x2 - x
N1 (x) 
x 2  x1
x - x1
N 2 (x) 
x 2  x1
x1
El #1
x2
x
Within the element, the displacement approximation is
x2 - x
x - x1
w(x) 
d1x 
d 2x
x 2  x1
x 2  x1
For a linear element
Displacement approximation in terms of shape functions
 x 2 - x x - x1  d1x 
w(x)  
 d 
 x 2  x1 x 2  x1   2x 
Strain approximation
d1x 
dw
1
 1 1  
ε

dx x 2  x1
d 2x 
Stress approximation
d1x 
E
 1 1  
  Eε 
x 2  x1
d 2x 
x1=0
x2
x3
x4=L
El #1
El #2 El #3
For the entire bar, the displacement approximation is
w(x)  w (1) (x) w (2) (x) w (3) (x)
Where w(i)(x) is the displacement approximation within element (i).
Let use set d1x=0. Then, can you seen that the above approximation
does satisfy the two conditions of being an admissible function on
the entire bar, i.e.,
(1) w(x  0)  0
dw
(2)
exist s
dx
TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH
ELEMENT USING THE RAYLEIGH-RITZ PRINCIPLE
Potential energy of element 1:
x2
1 x2
1 (w)     Adx   bw dx
x1
2 x1
Lets plug in the approximation
w(x)  N d
1 T
1 (d)  d
2
  EB d
ε(x) B d

x2
x1

B EB Adx d  d
T
T

x2
x1
T
N
b dx

Lets see what the matrix

x2
x1
T
is for a 1D linear element
Recall that
B
Hence
1
 1 1 
x 2  x1
1  1
1
 1 1 
B EB
 E
x 2  x1  1  x 2  x1
 1
E
E

 1 1  
2 
x 2  x 1   1 
x 2  x 1 2
T
 1  1
 1 1 



x2
x1
1
T
 x 2  x1 
2
 1 1 x2
 1 1  x1 AEdx 



x2
x1
AEdx
 x  x 
1
2
1
2
 1 1
 1 1 


Now, if we assume E and A are constant

x2
x1
T

x2
x1
AEdx
 x  x 
1
2

1
2
 1 1 AE(x 2  x1 )  1 1
 1 1  

2 

1
1
x

x

  2 1 

AE  1 1
 x 2  x1  1 1 
Remembering that (x2-x1) is the length of the element, this is the
stiffness matrix we had derived directly before using the direct
stiffness approach!!
Then why is it necessary to go through this complicated procedure??
1. Easy to handle nonuniform E and A
2. Easy to handle distributed loads
For nonuniform E and A, i.e. E(x) and A(x), the stiffness matrix of
the linear element will NOT be
EA  1 1
 x 2  x1   1 1 
But it will ALWAYS be
x2
k   B EB Adx
x1
T
Now lets go back to





1 T  x2 T

T  x2
T
1 (d)  d   B EB Adx  d  d   N b dx 
x1
2  x1



k
f


b


1 T
T
 d k d  d fb
2
Element stiffness matrix
x2
k   B EB Adx
T
x1
Element nodal load vector due to distributed body force
x2
f b   N b dx
x1
T
Apply Rayleigh-Ritz principle for the 1D linear element
Π1 (d )

 0
d1x

Π1 (d)
0
 
Π1 (d )
d
 0

d 2x
Recall from linear algebra (Lecture notes on Linear Algebra)
1 T
T
 1 (d )  d k d  d f b
2
 1 (d)

 kd  fb
d
Hence
Π1 (d)
0
d
 kd  fb
Exactly the same equation that we had before, except that the
stiffness matrix and nodal force vectors are more general
Recap of the properties of the element stiffness matrix
x2
k   B EB Adx
T
x1
1. The stiffness matrix is singular and is therefore non-invertible
2. The stiffness matrix is symmetric
3. Sum of any row (or column) of
the stiffness matrix is zero!
k
11
Why?
Sum of any row (or column) of the stiffness matrix is zero
Consider a rigid body motion of the element
2
1
d1x=1
1
d
1
d2x=1
Element strain ε  0  B d
k d
x1
T
x1
0 
 
0 
T

  B E  Bd  Adx
x2
 k11 k12  1 0
kd
  

k 21 k 22  1 0
 k11  k12  0 and k 21  k 22  0

x2
x2
f b   N b dx
T
x1
b(x)
1
2
f1x  
x2
x1
f 2x  
x2
x1
x2
x1
x1
f b   N b dx  
d2x
d1x
x2
 N 1 ( x) 

 b dx
 N 2 ( x) 
 x2 N ( x) b dx
 f1x 
 x1 1
    x2
 f 2x 
x N 2 ( x) b dx
 1
N 1 ( x ) b dx
N 2 ( x ) b dx





b(x) /unit length
2
1
Replaced by
d1x
d2x
d1x
1 f
1x
2
f2x
d2x
consistent manner
e.g., if b=1
x 2  x1
f1x   N 1 ( x) b dx   N 1 ( x) dx 
x1
x1
2
x2
x2
x 2  x1
f 2 x   N 2 ( x) b dx   N 2 ( x) dx 
x1
x1
2
Divide the total force into two equal halves and lump them at the
nodes
What happens if b(x)=x?
x2
x2
Summary: For each element
Displacement approximation in terms of shape functions
w(x)  N d
Strain approximation in terms of strain-displacement matrix
ε(x) B d
Stress approximation
  EB d
Element stiffness matrix
x2
k   B EB Adx
T
x1
x2
f b   N b dx
x1
T
What happens for element #3?
2
x4
1
 dw 
 3 (w)   EA
 dx  x bw dx  Fw(x  L)
x
3
2 3
 dx 
x4
For element 3
 x 4 - x x - x 3  d 3x 
w(x)  
 
 x 4  x 3 x 4  x 3  d 4x 
 w(x  L)  d 4x
The discretized form of the potential energy
1 T
3 (d)  d
2

x4
x3

B EB Adx d  d
T
T

x4
x3
T
N

b dx  Fd 4x
What happens for element #3?
Now apply Rayleigh-Ritz principle
Π 3 (d)
0
d
0 
 k d  fb   
F
Hence there is an extra load term on the right hand side due to the
concentrated force F applied to the right end of the bar.
NOTE that whenever you have a concentrated load at ANY
node, that load should be applied as an extra right hand side
term.
Step3:Assembly exactly as you had done before, assemble the
global stiffness matrix and global load vector and solve the
resulting set of equations by properly taking into account the
displacement boundary conditions
Problem:
6”
12”
E=30x106 psi
r=0.2836 lb/in3
Thickness of plate, t=1”
24”
P=100lb
3”
x
Model the plate as 2 finite elements and
(1)Write the expression for element stiffness
matrix and body force vectors
(2)Assemble the global stiffness matrix and
(3)Solve for the unknown displacements
(4)Evaluate the stress in each element
(5)Evaluate the reaction in each support
Solution (1)
Node-element connectivity chart
Finite element model
1
12”
El #1
2
El #2 P=100lb
3
x

12
0
12”
Element # Node 1
Node 2
1
1
2
2
2
3
Stiffness matrix of El #1
k 
(1)
12
0
12
12
0
0
E
(12) 2
T

12
0
A( x)dx   t (6  0.125 x)dx  t  (6  0.125 x)dx  63 in3
k 
(1)
1
 1 1
E
6  1
63

13.125

10


 1 1 
2 

1
1
(12)





 1 1
A( x)dx 


1
1


Stiffness matrix of El #2
k

24
12
(2)

24
12
E
(12)2
T

24
24
12

 1 1
A( x)dx 


1
1


6”
x
12”
24
A( x)dx   t (6  0.125 x)dx  t  (6  0.125 x)dx  45 in 3
k
12
(2)
6 - 0.125x
4.5”
12
1
 1 1
E
6 1

45

9.375

10


 1 1 
2 

1
1
(12)




x
Now compute the element load vector due to distributed body
force (weight)
x2
fb   N b dx
x1
T
For element #1
fb
(1)
12
12
  N b dx   N
T
0
0
12
T
 r A  dx
 r  N A dx
T
0
 r
12
0
 N1(1) ( x) 
 (1)  t (6  0.125 x) dx
 N 2 ( x) 
A( x )
33
 0.2836   lb
30 
9.3588

 lb
 8.508 
1
N1(1) ( x)
12”
N2(1) ( x)
12  x
(1)
N1 ( x ) 
12
x
(1)
N 2 ( x) 
12
2
El #1
x
Superscript in parenthesis indicates
element number
For element #2
fb
(2)
24
24
  N b dx   N
T
12
12
24
T
 r A  dx
 r  N A dx
T
12
 r
24
12
 N 2(2) ( x) 
 (2)  t (6  0.125 x) dx
 N3 ( x) 
A( x )
24 
 0.2836   lb
 21
6.8064 

 lb
5.9556 
( 2)
2
N ( x)
N3(2) ( x)
1
El #1
2
El #2
3
x
24  x
N ( x) 
12
x  12
N 3( 2 ) ( x) 
12
( 2)
2
12”
12”
Solution (2) Assemble the system equations
0 
 13.125 13.125
K  106   13.125
22.5
9.375 
 0
9.375 9.375 
f  fb  f concentrated load
 9.3588 


fb  8.508  6.8064  lb
 5.9556 


 0 
 
 100  lb
 0 
 
 9.3588 


 f  115.3144  lb
 5.9556 


Solution (3)
Hence we need to solve
0   d1x  9.3588  R1 
 13.125 13.125
  

106   13.125
22.5
9.375 d 2x    115.3144 
 0
9.375 9.375  d 3x   5.9556 
R1 is the reaction at node 1.
Notice that since the boundary condition at x=0 (d1x=0) has not been
taken into account, the system matrix is not invertible.
Incorporating the boundary condition d1x=0 we need to solve the
following set of equations
 22.5 9.375 d 2x  115.3144
10  
 


d

9.375
9.375
5.9556

  3x  

6
Solve to obtain
d 2 x  0.92396 105 
in
 
5 
d
 3 x  0.98749 10 
Solution (4) Stress in elements
Notice that since we are using linear elements, the stress within
each element is constant.
In element #1
 (1)  EB (1) d (1)
 d1x 
E

1 1  

x2  x1
d 2x 
30 106

d 2x
d1x  0
12
 23.099 psi
In element #2
 (2)  EB (2) d (2)
d 2x 
E

1 1  

x3  x2
d 3x 
30 106

 d3x -d 2x 
12
 1.5882 psi
Solution (5) Reaction at support
Go back to the first line of the global equilibrium equations…
0   d1x  9.3588  R1 
 13.125 13.125
  

106   13.125
22.5
9.375 d 2x    115.3144 
 0
9.375 9.375  d 3x   5.9556 
 R1  130.6288 lb (The –ve sign indicates that the force is in the –ve x-direction)
R1
6”
Check
12”
The reaction at the wall from force
equilibrium in the x-direction
24”
R1  P  
24
x 0
r A( x) dx
24
P=100lb
3”
x
 100  r t  (6  0.125 x) dx
x 0
 130.6288 lb
Problem: Can you solve for the displacement and stresses
analytically?
Check out
uanal
 4.727 109 x 2  9.487 107 x
for 0  x  12

9 2
7
6
4.727 10 x  2.0797 10 x  8.89 10 for 12  x  24
Stress
 ( x)anal
duanal
6 duanal
E
 30 10
dx
dx
Comparison of
displacement solutions
-4
1.2
x 10
Analytical solution
Displacement (in)
1
0.8
0.6
Finite element solution
0.4
0.2
0
0
5
10
15
x (in)
20
Notice:
1. Slope discontinuity at x=12 (why?)
2. The finite element solution does not produce the exact
solution even at the nodes
3. We may improve the solution by
(1) Increasing the number of elements
linear)
Comparison of stress solutions
30
25
Stress (psi)
20
15
Finite element solution
Analytical solutions
10
5
0
-5
0
5
10
15
20
x (in)
The analytical as well as the finite element stresses are
discontinuous across the elements
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