Lecture 2.pdf

CENG6504: Finite Element Methods
Introduction to Stiffness
(Displacement) Method
Dr. Tesfaye Alemu
1

Learning Objectives
• To define the stiffness matrix
• To derive the stiffness matrix for a spring element
• To demonstrate how to assemble stiffness matrices into
a global stiffness matrix
• To illustrate the concept of direct stiffness method to
obtain the global stiffness matrix and solve a spring
assemblage problem
• To describe and apply the different kinds of boundary
conditions relevant for spring assemblages
• To show how the potential energy approach can be used
to both derive the stiffness matrix for a spring and solve
a spring assemblage problem
2

The Stiffness (Displacement) Method
This section introduces some of the basic concepts on which
the direct stiffness method is based.
The linear spring is simple and an instructive tool to illustrate
the basic concepts.
The steps to develop a finite element model for a linear spring
follow our general 8 step procedure.
1. Discretize and Select Element Types - Linear spring
elements
2. Select a Displacement Function - Assume a variation
of the displacements over each element.
3. Define the Strain/Displacement and Stress/Strain
Relationships - use elementary concepts of equilibrium
and compatibility.
3

4. Derive the Element Stiffness Matrix and Equations -
Define the stiffness matrix for an element and then
consider the derivation of the stiffness matrix for a linear-
elastic spring element.
5. Assemble the Element Equations to Obtain the Global
or Total Equations and Introduce Boundary
Conditions - We then show how the total stiffness matrix
for the problem can be obtained by superimposing the
stiffness matrices of the individual elements in a direct
manner.
The term direct stiffness method evolved in reference to
this method.
4

6. Solve for the Unknown Degrees of Freedom (or
Generalized Displacements) - Solve for the nodal
displacements.
7. Solve for the Element Strains and Stresses - The
reactions and internal forces association with the bar
element.
8. Interpret the Results
5

1. Select Element Type - Consider the linear spring shown
below. The spring is of length L and is subjected to a
nodal tensile force, T directed along the x-axis.
1x
f 2x
f
Note: Assumed sign conventions 6

2. Select a Displacement Function - A displacement
function u(x) is assumed.
 
1 2
u a a x
In general, the number of coefficients in the displacement
function is equal to the total number of degrees of freedom
associated with the element. We can write the
displacement function in matrix forms as:
  1
1 x 2
2 2 x 1
1
a
u x
a
 
  
 
7

We can express u as a function of the nodal displacements
ui by evaluating u at each node and solving for a1 and a2.
  
1 1
( 0)
u x u a
   
2 2 1
( )
u x L u a L a
Solving for a2:
2 1
2
u u
a
L


Substituting a1 and a2 into u gives:

 
 
 
 
2 1
1
u u
u x u
L





Boundary Conditions
   
  
   
   
1 2
1
x x
u u
L L 8

In matrix form:
Or in another form:
 
 
 
   
 
 
 
   
1
2
1
u
x x
u
u
L L
 
 
  
 
1
1 2
2
u
u N N
u
Where N1 and N2 are defined as:
The functions Ni are called interpolation functions
because they describe how the assumed displacement
function varies over the domain of the element. In this case
the interpolation functions are linear.
  
1 2
1
x x
N N
L L
9

1
2
1
u
x x
u
u
L L
 
 
 
   
 
 
 
   
  1
1 2
2
u
u N N
u
 
  
 
2
x
N
L

1 2
u a a x
 
1 1
x
N
L
 
 
1 2 1
N N
10

3. Define the Strain/Displacement and Stress/Strain
Relationships - Tensile forces produce a total elongation
(deformation)  of the spring. For linear springs, the force
T and the displacement u are related by Hooke’s law:
T k

where deformation of the spring  is given as:
  
( ) (0)
u L u 2 1
u u
  
1x
f 2x
f
T T
 
1x
f T

2x
f T
11

4. Step 4 - Derive the Element Stiffness Matrix and
Equations - We can now derive the spring element
stiffness matrix as follows:
Rewrite the forces in terms of the nodal displacements:
 
   
1 2 1
x
T f k u u
 
  
2 2 1
x
T f k u u
We can write the last two force-displacement relationships
in matrix form as:

   
 

   
 

 
   
1 1
2 2
x
x
f u
k k
f u
k k
 
  
1 1 2
x
f k u u
 
   
2 1 2
x
f k u u
12

This formulation is valid as long as the spring deforms
along the x axis. The coefficient matrix of the above
equation is called the local stiffness matrix k:

 
  

 
k k
k k
k
5. Step 4 - Assemble the Element Equations
and Introduce Boundary Conditions
The global stiffness matrix and the global force vector
are assembled using the nodal force equilibrium equations,
and force/deformation and compatibility equations.
 

   ( )
1
N
e
e
K
K k  

   ( )
1
N
e
e
F
F f
where k and f are the element stiffness and force matrices
expressed in global coordinates. 13

6. Step 6 - Solve for the Nodal Displacements
Solve the displacements by imposing the boundary
conditions and solving the following set of equations:
    

F K d
7. Step 7 - Solve for the Element Forces
Once the displacements are found, the forces in each
element may be calculated from:
T k

 
F Kd
 
 
2 1
k u u
14

The Stiffness Method – Spring Example 1
Consider the following two-spring system shown below:
where the element axis x coincides with the global axis x.
For element 1: 1 1
1 1
3 3
1 1
x
x
f u
k k
f u
k k

   
 

   
 

 
   
For element 2: 3 2 2 3
2 2 2 2
x
x
f k k u
f k k u

     

   
 

     
15

We can write the nodal equilibrium equation at each node as:
Both continuity and compatibility require that both elements
remain connected at node 3.
(1) (2)
3 3
u u

(1)
1 1
x x
F f
 (2)
2 2
x x
F f

(1) (2)
3 3 3
x x x
F f f
 
Element number
16

In matrix form the above equations are:
Therefore the force-displacement equations for this spring
system are:
1 1 1 1 3
x
F k u k u
  2 2 3 2 2
x
F k u k u
  
   
    
3 1 1 1 3 2 3 2 2
x
F k u k u k u k u

F Kd
1 1 1 1
2 2 2 2
3 1 2 1 2 3
0
0
x
x
x
F k k u
F k k u
F k k k k u

     
   
 
 
   
 
   
 
  
     
where F is the global nodal force vector, d is called the
global nodal displacement vector, and K is called the
global stiffness matrix.
Element 1 Element 2
17

The elemental stiffness matrices may be written for each
element.
Assembling the Total Stiffness Matrix by Superposition
Consider the spring system defined in the last example:
For element 1: For element 2:

 
  

 
1 3
1
1 1
(1)
3
1 1
u u
u
k k
u
k k
k

 
  

 
3 2
2 2 3
(2)
2 2 2
u u
k k u
k k u
k
18

For element 2:
Write the stiffness matrix in global format for element 1 as
follows:
(1)
1 1
(1)
1 2 2
(1)
3 3
1 0 1
0 0 0
1 0 1
x
x
x
u f
k u f
u f
 

   
 
 
  
   
 
   
 

     
(1)
1 1
(1)
2 2 2
(1)
3 3
0 0 0
0 1 1
0 1 1
x
x
x
u f
k u f
u f
 
   
 
 
 
 
   
 
   
 

     
19

The above equations give:
Apply the force equilibrium equations at each node.

     
   
 
 
   
 
   
 
  
     
1 1 1 1
2 2 2 2
1 2 1 2 3 3
0
0
x
x
x
k k u F
k k u F
k k k k u F
(1)
1 1
(2)
2 2
(1) (2)
3 3 3
0
0
x x
x x
x x x
f F
f F
f f F
     
     
 
     
     
   
 
20

To avoid the expansion of the each elemental stiffness
matrix, we can use a more direct, shortcut form of the
stiffness matrix.

 
  

 
3 2
2 2 3
(2)
2 2 2
u u
k k u
k k u
k
The global stiffness matrix may be constructed by directly
adding terms associated with the degrees of freedom in k(1)
and k(2) into their corresponding locations in the K as
follows: 1 2 3
1 1 1
2 2 2
1 2 1 2 3
0
0
u u u
k k u
k k u
k k k k u

 
 
 
 
 
  
 
K

 
  

 
1 3
1
1 1
(1)
3
1 1
u u
u
k k
u
k k
k
21

Boundary conditions are of two general types:
1. homogeneous boundary conditions (the most common)
occur at locations that are completely prevented from
movement;
2. nonhomogeneous boundary conditions occur where finite
non-zero values of displacement are specified, such as the
settlement of a support.
In order to solve the equations defined by the global
stiffness matrix, we must apply some form of constraints or
supports or the structure will be free to move as a rigid body.
Boundary Conditions
22

Consider the equations we developed for the two-spring
system. We will consider node 1 to be fixed u1 = 0. The
equations describing the elongation of the spring system
become:
1 1 1
2 2 2 2
1 2 1 2 3 3
0 0
0
x
x
x
k k F
k k u F
k k k k u F

     
   
 
 
   
 
   
 
  
     
Expanding the matrix equations gives:
1 1 3
x
F k u
 
2 2 3 2 2
x
F k u k u
  
 
3 2 2 1 2 3
x
F k u k k u
   
2 3
Solve for and
u u




 23

Once we have solved the above equations for the unknown
nodal displacements, we can use the first equation in the
original matrix to find the support reaction.
The second and third equation may be written in matrix form
as:
2 2
2 2
3 3
2 1 2
x
x
u F
k k
u F
k k k
    
 

   
 
 
     
1 1 3
x
F k u
 
For homogeneous boundary conditions, we can delete the
row and column corresponding to the zero-displacement
degrees-of-freedom.
24

Expanding the matrix equations gives:
Let’s again look at the equations we developed for the two-
spring system.
However, this time we will consider a nonhomogeneous
boundary condition at node 1: u1 = .
The equations describing the elongation of the spring
system become:
1 1 1
2 2 2 2
1 2 1 2 3 3
0
0
x
x
x
k k F
k k u F
k k k k u F


     
   
 
 
   
 
   
 
  
     
1 1 1 3
x
F k k u

  2 2 3 2 2
x
F k u k u
  
3 1 1 3 2 3 2 2
x
F k k u k u k u

    
25

By considering the second and third equations because
they have known nodal forces we get:
2 2 3 2 2
x
F k u k u
   3 1 1 3 2 3 2 2
x
F k k u k u k u

    
In matrix form the above equations are:
2 2
2 2
3 3 1
2 1 2
x
x
u F
k k
u F k
k k k 
    
 

   
  
 
     
For nonhomogeneous boundary conditions, we must transfer
the terms from the stiffness matrix to the right-hand-side force
vector before solving for the unknown displacements.
26

Once we have solved the above equations for the unknown
nodal displacements, we can use the first equation in the
original matrix to find the support reaction.
1 1 1 3
x
F k k u

 
27

Consider the following three-spring system:
The elemental stiffness matrices for each element are:

 
  

 
1 3
1
3
(1) 1 1
1000
1 1
k
3 4
3
4
(2) 1 1
2000
1 1

 
  

 
k
4 2
4
2
(3) 1 1
3000
1 1

 
  

 
k
28

Using the concept of superposition (the direct stiffness
method), the global stiffness matrix is:
The global force-displacement equations are:
1000 0 1000 0
0 3000 0 3000
1000 0 3000 2000
0 3000 2000 5000

 
 

 

 
 
 
 
 
K
1 1
2 2
3 3
4 4
1000 0 1000 0
0 3000 0 3000
1000 0 3000 2000
0 3000 2000 5000
x
x
x
x
u F
u F
u F
u F
    
 
   
 
    
  
   
 
     
     
 
     
Element 2
Element 1
Element 3
29

We have homogeneous boundary conditions at
nodes 1 and 2 (u1 = 0 and u2 = 0).
The global force-displacement equations reduce to:
1 1
2 2
3 3
4 4
1000 0 1000 0
0 3000 0 3000
1000 0 3000 2000
0 3000 2000 5000
x
x
x
x
u F
u F
u F
u F
    
 
   
 
    
  
   
 
     
     
 
     
30

Substituting for the known force at node 4 (F4x = 5,000 lb)
gives:
Solving for u3 and u4 gives:
3
4
3000 2000 0
2000 5000 5,000
u
u
 
    
  
   
 
  
 
 
 
 
3 4
10 15
11 11
u in u in
31

To obtain the global forces, substitute the displacement in
the force-displacement equations.
Solving for the forces gives:
1
2
10
11
3
15
11
4
1000 0 1000 0 0
0 3000 0 3000 0
1000 0 3000 2000
0 3000 2000 5000
x
x
x
x
F
F
F
F

     
     

   
 

   
 
 
   
 
   
 
   
 
   
1 2
10,000 45,000
11 11
x x
F lb F lb
 
3 4
55,000
0
11
x x
F F lb 5,000lb

32

Next, use the local element equations to obtain the force in
each spring.
The local forces are:
For element 1: 
     

   
 

   
 
1
10
11
3
1000 1000 0
1000 1000
x
x
f
f
  
1 3
10,000 10,000
11 11
x x
f lb f lb
A free-body diagram of the spring element 1 is shown below.
33

each spring.
For element 2:

     

   
 

   
 
10
11
3
15
11
4
2000 2000
2000 2000
x
x
f
f
  
3 4
10,000 10,000
11 11
x x
f lb f lb
34

each spring.
For element 3:

     

   
 

   
 
15
11
4
2
3000 3000
3000 3000 0
x
x
f
f
  
4 2
45,000 45,000
11 11
x x
f lb f lb
35

Consider the following four-spring system:
The spring constant k = 200 kN/m and the displacement
 = 20 mm.
Therefore, the elemental stiffness matrices are:

 
     

 
(1) (2) (3) (4) 1 1
200 /
1 1
kN m
k k k k
36

Using superposition (the direct stiffness method), the global
stiffness matrix is:
200 200 0 0 0
200 400 200 0 0
0 200 400 200 0
0 0 200 400 200
0 0 0 200 200

 
 
 
 
  
 
 
 
 
 

 
K
Element 1 Element 2 Element 3 Element 4
37

1 1
2 2
3 3
4 4
5 5
200 200 0 0 0
200 400 200 0 0
0 200 400 200 0
0 0 200 400 200
0 0 0 200 200
x
x
x
x
x
u F
u F
u F
u F
u F
    
 
   
 
     
     

 
     
     
 
     
 
    
     
Applying the boundary conditions (u1 = 0 and u5 = 20 mm)
and the known forces (F2x, F3x, and F4x equal to zero) gives:
2
3
4
5
0
400 200 0 0
0
200 400 200 0
0
0 200 400 200
0 0 200 200 0.02 x
u
u
u
F
  
   
 
   
     
  
   
 
     
     

      38

Rearranging the first three equations gives:
2
3
4
400 200 0 0
200 400 200 0
0 200 400 4
u
u
u

     
   
 
  
   
 
   
 

     
Solving for u2, u3, and u4 gives:
2 3 4
0.005 0.01 0.015
u m u m u m
  
Solving for the forces F1x and F5x gives:
1 200(0.005) 1.0
x
F kN
   
5 200(0.015) 200(0.02) 1.0
x
F kN
   
39

each spring.
For element 1: 
     

   
 

   
 
1
2
200 200 0
200 200 0.005
x
x
f
f
  
1 2
1.0 1.0
x x
f kN f kN
For element 2: 
     

   
 

   
 
2
3
200 200 0.005
200 200 0.01
x
x
f
f
  
2 3
1.0 1.0
x x
f kN f kN
40

each spring.
For element 3:
For element 4:

     

   
 

   
 
3
4
200 200 0.01
200 200 0.015
x
x
f
f
  
3 4
1.0 1.0
x x
f kN f kN

     

   
 

   
 
4
5
200 200 0.015
200 200 0.02
x
x
f
f
  
4 5
1.0 1.0
x x
f kN f kN
41

Consider the following spring system:
The boundary conditions are: 1 3 4 0
u u u
  
The compatibility condition at node 2 is:
(1) (2) (3)
2 2 2 2
u u u u
  
42

Using the direct stiffness method: the elemental stiffness
matrices for each element are:
Using the concept of superposition (the direct stiffness
method), the global stiffness matrix is:
1 2
1
2
1 1
(1)
1 1
k k
k k

 
  

 
k

 
  

 
2 3
2
3
2 2
(2)
2 2
k k
k k
k
2 4
2
4
3 3
(3)
3 3
k k
k k

 
  

 
k
 

 
 
  
 
  

 
 

 
1 2 3 4
1
2
3
4
1
1
1 2 3 3
1 2
2 2
3 3
0
0
0
0
0 0
k
k
k k k k
k k
k k
k k
K
43

 

 
 
  
 
  

 
 

 
 
1 2 3 4
1
2
3
4
1
1
1 2 3 3
1 2
2 2
3 3
0
0
0
0
0 0
k
k
k k k k
k k
k k
k k
K
Applying the boundary conditions (u1 = u3 = u4 = 0) the
stiffness matrix becomes:
44

Applying the known forces (F2x = P) gives:
Solving the equation gives:
 
1 2 3 2
P k k k u
  
2
1 2 3
P
u
k k k

 
Solving for the forces gives:
1 1 2 3 2 2 4 3 2
x x x
F k u F k u F k u
     
45

Potential Energy Approach to Derive Spring
Element Equations
One of the alternative methods often used to derive the
element equations and the stiffness matrix for an element is
based on the principle of minimum potential energy.
This method has the advantage of being more general than
the methods involving nodal and element equilibrium
equations, along with the stress/strain law for the element.
The principle of minimum potential energy is more adaptable
for the determination of element equations for complicated
elements (those with large numbers of degrees of freedom)
such as the plane stress/strain element, the axisymmetric
stress element, the plate bending element, and the three-
dimensional solid stress element. 46

Total Potential Energy
The total potential energy p is defined as the sum of the
internal strain energy U and the potential energy of the
external forces :
p U
   
Strain energy is the capacity of the internal forces (or
stresses) to do work through deformations (strains) in the
structure.
The potential energy of the external forces  is the capacity
of forces such as body forces, surface traction forces, and
applied nodal forces to do work through the deformation of
the structure.
47

Recall the force-displacement relationship for a linear
spring:
The differential internal work (or strain energy) dU in the
spring is the internal force multiplied by the change in
displacement which the force moves through:
F kx

 
dU Fdx kx dx
 
48

The total strain energy is:
The strain energy is the area under the force-displacement
curve. The potential energy of the external forces is the
work done by the external forces:
  2
0
1
2
x
L
U dU kx dx kx
  
 
Fx
  
49

Therefore, the total potential energy is:
The concept of a stationary value of a function G is shown
below:
2
1
2
p kx Fx
  
The function G is expressed in terms of x.
To find a value of x yielding a stationary value of G(x), we
use differential calculus to differentiate G with respect to x
and set the expression equal to zero.
0
dG
dx

50

We can replace G with the total potential energy p and the
coordinate x with a discrete value di. To minimize p we first
take the variation of p (we will not cover the details of
variational calculus):
The principle states that equilibrium exist when the di define
a structure state such that p = 0 for arbitrary admissible
variations di from the equilibrium state.
1 2
1 2
...
p p p
p n
n
d d d
d d d
  
   
  
   
  
51

To satisfy p = 0, all coefficients associated with di must be
zero independently, therefore:
 
0 1, 2, , 0
p p
i
i n or
d d
 
 
  
 

53

Total Potential Energy – Example 5
Consider the following linear-elastic spring system subjected
to a force of 1,000 lb.
Evaluate the potential energy for various displacement
values and show that the minimum potential energy also
corresponds to the equilibrium position of the spring.
54

The total potential energy is defined as the sum of the
internal strain energy U and the potential energy of the
external forces :
p U
   
The variation of p with respect to x is:
Fx
  
2
1
2
U kx

0
p
p x
x

 

 

Since x is arbitrary and might not be zero, then: 0
p
x




55

Using our express for p, we get:
If we had plotted the total
potential energy function p
for various values of
deformation we would get:
 
2 2
1 1
500( ) 1,000
2 2
lb
in
p kx Fx x lb x
    


  

0 500 1,000
p
x
x
 2.0
x in
56

Let’s derive the spring element equations and stiffness
matrix using the principal of minimum potential energy.
Consider the linear spring subjected to nodal forces shown
below:
The total potential energy p
 
2
2 1 1 1 2 2
1
2
p x x
k u u f u f u
    
Expanding the above express gives:
 
2 2
2 1 2 1 1 1 2 2
1
2
2
p x x
k u u u u f u f u
     
57

below:
Recall:
Therefore:
 
0 1, 2, , 0
p p
i
i n or
d d
 
 
  
 

 
2 1 1
1
2 2 0
2
p
x
k
u u f
u


    

 
2 1 2
2
2 2 0
2
p
x
k
u u f
u


   
 58

below:
In matrix form the equations are:
1 1
2 2
x
x
f u
k k
f u
k k

   
 

   
 

 
   
Therefore:  
1 2 1x
k u u f
 
 
1 2 2x
k u u f
  
59

Obtain the total potential energy of the spring system shown
below and find its minimum value.
The potential energy p for element 1 is:
 
2
(1)
1 3 1 1 1 3 3
1
2
p x x
k u u f u f u
    
 
2
(2)
2 4 3 3 3 4 4
1
2
p x x
k u u f u f u
    
60

Obtain the total potential energy of the spring system shown
below and find its minimum value.
The total potential energy p for the spring system is:
 
2
(3)
3 2 4 2 2 4 4
1
2
p x x
k u u f u f u
    
3
( )
1
e
p p
e
 

 
61

Minimizing the total potential energy p:


    

(1)
1 3 1 1 1
1
0
p
x
k u k u f
u


   

(3)
3 2 3 4 2
2
0
p
x
k u k u f
u
(1) (2)
1 3 1 1 2 4 2 3 3 3
3
0
p
x x
k u k u k u k u f f
u


      



      

(2) (3)
2 4 2 3 3 2 3 4 4 4
4
0
p
x x
k u k u k u k u f f
u
62

In matrix form: 1 1 1
1
3 3 2 2
(1) (2)
2
1 2 3 3 3
1
(2) (3)
3 2 3
2 4 4 4
0 0
0
0
0
0
x
x
x x
x x
k u f
k
k k u f
k
k k u f f
k
k k k
k u f f

     
     
    
  
   
 

 
    
     
 
 
   
 
Using the following force equilibrium equations:
(1)
1 1
x x
f F

(3)
2 2
x x
f F

(2) (3)
4 4 4
x x x
f f F
 
(1) (2)
3 3 3
x x x
f f F
 
63

The above equations are identical to those we obtained
through the direct stiffness method.
1 1
2 2
3 3
4 4
1000 0 1000 0
0 3000 0 3000
1000 0 3000 2000
0 3000 2000 5000
x
x
x
x
u F
u F
u F
u F
    
 
   
 
    
  
   
 
     
     
 
     
64

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