MANE 4240 & CIVL 4240
Introduction to Finite Elements
Introduction to the Stiffness
(Displacement) Method:
Analysis of a system of springs
Prof. Suvranu De
Reading assignment:
Chapter 2: Sections 2.1-2.5 + Lecture notes
Summary:
• Developing the finite element equations for a system of
springs using the “direct stiffness” approach
• Application of boundary conditions
• Physical significance of the stiffness matrix
• Direct assembly of the global stiffness matrix
• Problems
FEM analysis scheme
Step 1: Divide the problem domain into non overlapping regions
(“elements”) connected to each other through special points
(“nodes”)
Step 2: Describe the behavior of each element
Step 3: Describe the behavior of the entire body by putting
together the behavior of each of the elements (this is a process
known as “assembly”)
F1x
F2x
k1
F3x
x
k2
Problem
Analyze the behavior of the system composed of the two springs
loaded by external forces as shown above
Given
F1x , F2x ,F3x are external loads. Positive directions of the forces
are along the positive x-axis
k1 and k2 are the stiffnesses of the two springs
F1x
F2x
k1
F3x
x
k2
Solution
Step 1: In order to analyze the system we break it up into smaller
parts, i.e., “elements” connected to each other through “nodes”
F1x
k1
k2
d1x
Element 1
d2x
F3x
x
3
2
1
Node 1
F2x
Element 2
Unknowns: nodal displacements d1x, d2x, d3x,
d3x
F1x
k1
k2
d1x
Element 1
d2x
F3x
x
3
2
1
Node 1
F2x
Element 2
d3x
Solution
Step 2: Analyze the behavior of a single element (spring)
© 2002 Brooks/Cole Publishing / Thomson Learning™
Two nodes: 1, 2
Nodal displacements: d̂ 1x d̂ 2x
Nodal forces: f̂1x f̂ 2x
Spring constant: k
© 2002 Brooks/Cole Publishing / Thomson Learning™
Local (x̂ , ŷ ,ẑ) and global (x,y,z) coordinate systems
Behavior of a linear spring (recap)
x
F
k
k
1
F
d
k
F = Force in the spring
d = deflection of the spring
k = “stiffness” of the spring
Hooke’s Law
F = kd
d
f̂1x
f̂ 2x
© 2002 Brooks/Cole Publishing / Thomson Learning™
Hooke’s law for our spring element
f̂ 2x  k (d̂ 2x  d̂1x )
Eq (1)
Force equilibrium for our spring element (recap free body diagrams)
f̂1x  f̂ 2x  0
 f̂1x  f̂ 2x  k (d̂ 2x  d̂1x )
Eq (2)
Collect Eq (1) and (2) in matrix form
Element force
vector
f̂1x   k - k  d̂1x 
f̂  k̂ d̂
 
 

f̂ 2x  - k k  d̂ 2x 
 
 
Element nodal 
Element
stiffness
matrix
displacement
vector
f̂
k̂
d̂
Note
T
1. The element stiffness matrix is “symmetric”, i.e. k̂  k̂
2. The element stiffness matrix is singular, i.e.,
det (k̂ )  k 2  k 2  0
The consequence is that the matrix is NOT invertible. It is not
possible to invert it to obtain the displacements. Why?
The spring is not constrained in space and hence it can attain
multiple positions in space for the same nodal forces
e.g.,
f̂1x   2 - 2 1 - 2
 
  

f̂ 2x  - 2 2  2  2 
f̂1x   2 - 2 3 - 2
 
  

f̂ 2x  - 2 2  4  2 
Solution
Step 3: Now that we have been able to describe the behavior of
each spring element, lets try to obtain the behavior of the original
structure by assembly
Split the original structure into component elements
Element 2
Element 1
k1
1
2
k2
2
(1)
(1)
f̂1x(1) d̂1x
f̂ 2x(1) d̂ 2x
(1) 
f̂1x(1)   k1 - k1  d̂1x

 (1)   
 (1) 

f̂ 2x  - k1 k1  d̂ 2x 
 
 
(1 )
f̂
(1 )
k̂
Eq (3)
d̂
(1 )
f̂
(2)
1x
d̂
3
(2)
f̂ 2x(2) d̂ 2x
(2)
1x
(2) 
f̂1x(2)   k 2 - k 2  d̂1x

 (2)   
 (2) 

f̂ 2x  - k 2 k 2  d̂ 2x 
 
 
( 2)
f̂
( 2)
k̂
Eq (4)
d̂
( 2)
To assemble these two results into a single description of the
response of the entire structure we need to link between the local
and global variables.
Question 1: How do we relate the local (element) displacements
back to the global (structure) displacements?
F1x
k2
F2x
F3x
k1
x
Node 1
3
2
1
d1x
Element 1
d2x
Element 2
(1)
d̂1x
 d1x
d̂
(1)
2x
 d̂
(2)
1x
d̂ (2)
2x  d 3x
 d 2x
Eq (5)
d3x
Hence, equations (3) and (4) may be rewritten as
f̂1x(1)   k1
 (1)   
f̂ 2x  - k1
- k1  d1x 
 

k1  d 2x 
f̂1x(2)   k 2
 (2)   
f̂ 2x  - k 2
- k 2  d 2x 
 

k 2  d 3x 
Or, we may expand the matrices and vectors to obtain
f̂1x(1)   k1  k1 0 d1x 
 (1)  
d 

f̂ 2x   - k1 k1 0  2x 
0   0


d

0
0
3x

  
f̂
(1) e
k̂
(1) e
Eq (6)
d
0  d1x 
 0  0 0
f̂ (2)  
d 

 1x   0 k 2  k 2   2x 
f̂ (2)  0 - k
d 3 x 

k
2x 

2
 
2 
f̂
(2)e
k̂
(2)e
Eq (7)
(1) e
k̂
(1) e
f̂
d
Expanded element stiffness matrix of element 1 (local)
Expanded nodal force vector for element 1 (local)
Nodal load vector for the entire structure (global)
d
Question 2: How do we relate the local (element) nodal forces back
to the global (structure) forces? Draw 5 FBDs
F1x
k2
F2x
F3x
k1
x
1 A
B2
D3
C
d1x
d3x
d2x
2
F1x
f̂1x(1)
f̂ 2x(1)
© 2002 Brooks/Cole Publishing / Thomson Learning™
F2x
3
f̂1x(2)
At node 1 : F1x - f̂1x(1)  0
At node 2 : F2x - f̂ 2x(1)  f̂1x(2)  0
At node 3 : F3x - f̂ 2x(2)  0
(2)
f̂ 2x
F3x
In vector form, the nodal force vector (global)
(1)
 F1x   f̂1x 
   (1) (2) 
F  F2x   f̂ 2x  f̂1x 
F   f̂ (2) 
 3x   2x 
Recall that the expanded element force vectors were
f̂
(1) e
f̂1x(1) 
0 
( 2)e
 (1) 
f̂ (2) 
 f̂ 2x  and f̂
  1x 
0 
f̂ (2) 
 2x 
 
Hence, the global force vector is simply the sum of the expanded
element nodal force vectors F 
1x
  (1) e ( 2) e
F  F2x   f̂  f̂
F 
 3x 
But we know the expressions for the expanded local force vectors
from Eqs (6) and (7)
f̂
(1) e
 k̂
(1)e
 f̂
( 2)e
d and f̂
( 2) e
 k̂
(2)e
d
Hence
F  f̂
(1) e
 k̂
(1)e
d  k̂
(2)e
(1)e
(2)e

d   k̂  k̂ d


FKd
F  Global nodal force vector
d  Global nodal displaceme nt vector
K  Global stiffness matrix
 sum of expanded element stiffness matrices
For our original structure with two springs, the global stiffness
matrix is
0 
 k 1  k 1 0  0 0
K  - k1 k1 0  0 k 2  k 2 
 0
0 0 0 - k 2 k 2 
 
k̂
 k1
 - k1
 0
(1) e
 k1
k1  k 2
- k2
k̂
(2)e
0 
 k 2 
k 2 
NOTE
1. The global stiffness matrix is symmetric
2. The global stiffness matrix is singular
The system equations
FKd
 k1
 F1x   k1
  
F2x   - k1 k1  k 2
F   0
- k2
 3x  
imply
0  d1x 
 

 k 2  d 2x 
k 2  d 3x 
F1x  k1d1x  k1d 2x
 F2x  -k1d1x  (k1  k 2 )d 2x  k 2 d 3x
F3x  -k2 d 2x  k 2 d 3x
These are the 3 equilibrium equations at the 3 nodes.
F1x
1 A
k2
F2x
k1
B2
x
D3
C
d1x
F3x
d3x
d2x
2
F1x
f̂1x(1)
f̂ 2x(1)
© 2002 Brooks/Cole Publishing / Thomson Learning™
At node 2 : F2x - f̂
 f̂
(2)
1x
At node 3 : F3x - f̂ 2x(2)  0
f̂1x(2)
(2)
f̂ 2x
F3x
F1x  k1 d1x  d 2x   f̂1x(1)
At node 1 : F1x - f̂1x(1)  0
(1)
2x
F2x
3
0
F2x  -k1d1x  (k1  k 2 )d 2x  k 2 d 3x
  k1 d1x  d 2x   k 2 d 2x  d 3x 
 f̂ 2x(1)  f̂1x(2)
(2)
F3x  -k2 d 2x  d3x   f̂ 2x
Notice that the sum of the forces equal zero, i.e., the structure is in
static equilibrium.
F1x + F2x+ F3x =0
Given the nodal forces, can we solve for the displacements?
To obtain unique values of the displacements, at least one of the
nodal displacements must be specified.
Direct assembly of the global stiffness matrix
Global
F1x
k2
F2x
k1
F3x
3
2
1
d1x
Element 1
d2x
x
Element 2
d3x
Local
1
(1)
f̂1x(1) d̂1x
Element 1
k1
2
2
(1)
f̂ 2x(1) d̂ 2x
f̂
(2)
1x
d̂
(2)
1x
Element 2
k2
3
(2)
f̂ 2x(2) d̂ 2x
Node element connectivity chart : Specifies the global node
number corresponding to the local (element) node numbers
ELEMENT Node 1 Node 2
1
1
2
2
2
3
Local node number
Global node number
Stiffness matrix of element 1
Stiffness matrix of element 2
d2x d3x
d1x d2x
k̂
(1)
 k1 - k1  d1x


- k1 k1  d2x
k̂
( 2)
 k2

- k 2
Global stiffness matrix
d2x
d3x
d1x
- k1
 k1
K  - k1 k1  k 2
 0
- k2
0  d1x
- k 2  d2x
k 2  d3x
Examples: Problems 2.1 and 2.3 of Logan
- k 2  d2x

k 2  d3x
Example 2.1
22
3
4
© 2002 Brooks/Cole Publishing / Thomson Learning™
Compute the global stiffness matrix of the assemblage of
springs shown above
d2x
d3x
d4x
d1x
1000
0
0  d1x
 1000
 1000 1000  2000



K
 0
2000

0
 0
2000
0  d2x
 2000  3000  3000 d3x

3000
3000  d
4x
Example 2.3
3
© 2002 Brooks/Cole Publishing / Thomson Learning™
Compute the global stiffness matrix of the assemblage of
springs shown above
 k1

K  -k1
 0
-k1
k1  k 2  k 3
-  k 2  k3 
0


-  k 2  k 3 
 k 2  k 3  
Imposition of boundary conditions
Consider 2 cases
Case 1: Homogeneous boundary conditions (e.g., d1x=0)
Case 2: Nonhomogeneous boundary conditions (e.g., one of the
nodal displacements is known to be different from zero)
Homogeneous boundary condition at node 1
k2=100N/m
k1=500N/m
1
2
Element 2
Element 1
d1x=0
d2x
F3x=5N
3
d3x
x
System equations
0
0   d1x   F1x 
 500 -500
-500 600 -100  d    0 

  2x   
 0
-100 100   d3 x   5 
Global Stiffness Nodal Nodal
load
disp
matrix
vector vector
Note that F1x is the wall reaction which is to be computed as part
of the solution and hence is an unknown in the above equation
Writing out the equations explicitly
-500d 2x  F1x
Eq(1)
Eq(2)
600d 2 x  100d3 x  0
Eq(3)
100d 2 x  100d3 x  5
Eq(2) and (3) are used to find d2x and d3x by solving
 600 100   d 2 x  0 
 100 100   d   5 

  3x   
 d 2 x   0.01 m 
 

d
0.06
m

 3x  
NOTICE: The matrix in the above equation may be obtained from
the global stiffness matrix by deleting the first row and column
0 
 500 -500
-500 600 -100


 0
-100 100 
 600 100 
 100 100 


Note use Eq(1) to compute F1x =-500d 2x  5N
NOTICE:
1. Take care of homogeneous boundary conditions
by deleting the appropriate rows and columns from the
global stiffness matrix and solving the reduced set of
equations for the unknown nodal displacements.
2. Both displacements and forces CANNOT be known at
the same node. If the displacement at a node is known, the
reaction force at that node is unknown (and vice versa)
Imposition of boundary conditions…contd.
Nonhomogeneous boundary condition: spring 2 is pulled at
node 3 by 0.06 m)
k2=100N/m
k1=500N/m
x
1
3
2
Element 2
Element 1
d3x=0.06m
d1x=0
d2x
System equations
0
0   d1x   F1x 
 500 -500
-500 600 -100  d    0 

  2x   
 0
-100 100   d3 x   F3 x 
0.06
Note that now F1x and F3x are not known.
Writing out the equations explicitly
-500d 2x  F1x
Eq(1)
600d 2 x  100(0.06)  0
Eq(2)
100d 2 x  100(0.06)  F3 x Eq(3)
Now use only equation (2) to compute d2x
600d 2 x  100(0.06)
 d 2 x  0.01m
Now use Eq(1) and (3) to compute F1x =-5N and F3x=5N
Recap of what we did
Step 1: Divide the problem domain into non overlapping regions
(“elements”) connected to each other through special points
(“nodes”)
Element
Step 2: Describe the behavior of each element ( f̂  k̂ d̂ )
nodal
displacement
vector
Step 3: Describe the behavior of the entire body (by “assembly”).
This consists of the following steps
1. Write the force-displacement relations of each spring in
Global
expanded form
e
f̂  k̂ e d̂
nodal
displacement
vector
Recap of what we did…contd.
2. Relate the local forces of each element to the global forces at
the nodes (use FBDs and force equilibrium).
Finally obtain
F   f̂
e
FKd
Where the global stiffness matrix
K  k
e
Recap of what we did…contd.
Apply boundary conditions by partitioning the matrix and vectors
 K11 K12  d1   F1 
K K  d   F 
22   2   2 
 21
Solve for unknown nodal displacements
K 22 d 2  F2  K 21d1
Compute unknown nodal forces
F1  K11d1  K12 d 2
Physical significance of the stiffness matrix
F1x
k1
F2x
k2
F3x
3
2
1
Element 1
d1x
d2x
In general, we will have a
stiffness matrix of the form
(assume for now that we do not
know k11, k12, etc)
The finite element
force-displacement
relations:
x
Element 2
d3x
 k11 k12
K  k 21 k 22
k 31 k 32
 k11 k12
k
 21 k 22
k 31 k 32
k13 
k 23 
k 33 
k13   d1   F1 
   
k 23  d 2   F2 
k 33  d 3  F3 
Physical significance of the stiffness matrix
The first equation is
k11d1  k12d 2  k13d 3  F1
Force equilibrium
equation at node 1
Columns of the global stiffness matrix
What if d1=1, d2=0, d3=0 ?
F1  k11
F2  k 21
F3  k 31
While nodes 2 and 3 are held fixed
Force along node 1 due to unit displacement at node 1
Force along node 2 due to unit displacement at node 1
Force along node 3 due to unit displacement at node 1
Similarly we obtain the physical significance of the other
entries of the global stiffness matrix
Physical significance of the stiffness matrix
In general
Force at node ‘i’ due to unit displacement at node ‘j’
k ij = keeping
all the other nodes fixed
This is an alternate route to generating the global stiffness matrix
e.g., to determine the first column of the stiffness matrix
F1
k1
F2
k2
d1
Element 1
d2
Find F1=?, F2=?, F3=?
x
3
2
1
F3
Set d1=1, d2=0, d3=0
Element 2
d3
Physical significance of the stiffness matrix
For this special case, Element #2 does not have any contribution.
Look at the free body diagram of Element #1
d̂
d̂ (1)
2x
(1)
1x
f̂1x(1)
k1
x
f̂ 2x(1)
ˆ (1) )  k (0 1)  k
fˆ2x(1)  k1 (dˆ (1)

d
2x
1x
1
1
fˆ1x(1)  fˆ2x(1)  k1
Physical significance of the stiffness matrix
Force equilibrium at node 1
F1
F1 =fˆ1x(1)  k1
f̂1x(1)
Force equilibrium at node 2
F2
F2 =fˆ2x(1)  k1
f̂ 2x(1)
Of course, F3=0
F1 = k1d1 = k1=k11
F2 = -F1 = -k1=k21
F3 = 0 =k31
Physical significance of the stiffness matrix
Hence the first column of the stiffness matrix is
 F1   k1 
   
 F2   k1 
F   0 
 3  
To obtain the second column of the stiffness matrix, calculate the
nodal reactions at nodes 1, 2 and 3 when d1=0, d2=1, d3=0
Check that
 F1   k1 
  

F

k

k
 2  1 2
 F   k 
2 
 3 
Physical significance of the stiffness matrix
To obtain the third column of the stiffness matrix, calculate the
nodal reactions at nodes 1, 2 and 3 when d1=0, d2=0, d3=1
Check that
 F1   0 
   
 F2   k2 
F   k 
 3  2 
Steps in solving a problem
Step 1: Write down the node-element connectivity table
linking local and global displacements
Step 2: Write down the stiffness matrix of each element
Step 3: Assemble the element stiffness matrices to form the
global stiffness matrix for the entire structure using the
node element connectivity table
Step 4: Incorporate appropriate boundary conditions
Step 5: Solve resulting set of reduced equations for the
unknown displacements
Step 6: Compute the unknown nodal forces