Lecture 3.pdf

CENG6504: Finite Element Methods
Truss Member
Dr. Tesfaye Alemu
1

Learning Objectives
• To derive the stiffness matrix for a bar element.
• To illustrate how to solve a bar assemblage by the direct
stiffness method.
• To introduce guidelines for selecting displacement
functions.
• To describe the concept of transformation of vectors in
two different coordinate systems in the plane.
• To derive the stiffness matrix for a bar arbitrarily oriented
in the plane.
• To demonstrate how to compute stress for a bar in the
plane.
• To show how to solve a plane truss problem.
• To develop the transformation matrix in three-
dimensional space and show how to use it to derive the
stiffness matrix for a bar arbitrarily oriented in space.
• To demonstrate the solution of space trusses.
2

Having set forth the foundation on which the direct stiffness
method is based, we will now derive the stiffness matrix for
a linear-elastic bar (or truss) element using the general
steps outlined in Chapter 2.
We will include the introduction of both a local coordinate
system, chosen with the element in mind, and a global or
reference coordinate system, chosen to be convenient (for
numerical purposes) with respect to the overall structure.
We will also discuss the transformation of a vector from the
local coordinate system to the global coordinate system,
using the concept of transformation matrices to express
the stiffness matrix of an arbitrarily oriented bar element in
terms of the global system.
Development of Truss Equations
3

Next we will describe how to handle inclined, or skewed,
supports.
We will then extend the stiffness method to include space
trusses.
We will develop the transformation matrix in three-dimensional
space and analyze a space truss.
We will then use the principle of minimum potential energy
and apply it to the bar element equations.
Finally, we will apply Galerkin’s residual method to derive the
bar element equations.
4

5

6

7

8

9

10

11

12

Stiffness Matrix for a Bar Element
Consider the derivation of the stiffness matrix for the linear-
elastic, constant cross-sectional area (prismatic) bar
element show below.
This application is directly applicable to the solution of pin-
connected truss problems.
13

element show below.
where T is the tensile force directed along the axis at
nodes 1 and 2, x is the local coordinate system directed
along the length of the bar.
14

element show below.
The bar element has a constant cross-section A, an initial
length L, and modulus of elasticity E.
The nodal degrees of freedom are the local axial
displacements u1 and u2 at the ends of the bar.
15

The strain-displacement relationship is:
du
E
dx
  
 
From equilibrium of forces, assuming no distributed loads
acting on the bar, we get:
constant
x
A T
  
Combining the above equations gives:
constant
du
AE T
dx
 
Taking the derivative of the above equation with respect to the
local coordinate x gives:
0
d du
AE
dx dx
 

 
 
16

The following assumptions are considered in deriving the bar
element stiffness matrix:
1. The bar cannot sustain shear force: 1 2 0
y y
f f
 
2. Any effect of transverse displacement is ignored.
3. Hooke’s law applies; stress is related to strain: x x
E
 

17

Step 1 - Select Element Type
We will consider the linear bar element shown below.
The bar element has a constant cross-section A, an initial
length L, and modulus of elasticity E.
18

Step 2 - Select a Displacement Function
1 2
u a a x
 
A linear displacement function u is assumed:
The number of coefficients in the displacement function, ai, is
equal to the total number of degrees of freedom associated
with the element.
Applying the boundary conditions and solving for the unknown
coefficients gives:
2 1
1
u u
u x u
L

 
 
 
 
1
2
1
u
x x
u
u
L L
 
 
 
   
 
 
 
   
19

Step 2 - Select a Displacement Function
Or in another form:
where N1 and N2 are the interpolation functions gives as:
  1
1 2
2
u
u N N
u
 
  
 
1 2
1
x x
N N
L L
  
The linear displacement
function plotted over the
length of the bar element is
shown here.
u
20

Step 3 - Define the Strain/Displacement
and Stress/Strain Relationships
The stress-displacement relationship is:
2 1
x
u u
du
dx L


 
Step 4 - Derive the Element Stiffness Matrix and Equations
We can now derive the element stiffness matrix as follows:
x
T A

Substituting the stress-displacement relationship into the
above equation gives:
2 1
u u
T AE
L

 
  
  21

The nodal force sign convention, defined in element figure, is:
Step 4 - Derive the Element Stiffness Matrix and Equations
therefore,
Writing the above equations in matrix form gives:
1 2
x x
f T f T
  
1 2 2 1
1 2
x x
u u u u
f AE f AE
L L
 
   
 
   
   
1 1
2 2
1 1
1 1
x
x
f u
AE
f u
L

   
 

   
 

 
   
Notice that AE/L for a bar element is analogous to the spring
constant k for a spring element. 22

The global stiffness matrix and the global force vector are
assembled using the nodal force equilibrium equations, and
force/deformation and compatibility equations.
Step 5 - Assemble the Element Equations
and Introduce Boundary Conditions
   
( ) ( )
1 1
n n
e e
e e
K F
 
   
 
K k F f
Where k and f are the element stiffness and force matrices
expressed in global coordinates.
23

Solve the displacements by imposing the boundary conditions
and solving the following set of equations:
Step 6 - Solve for the Nodal Displacements

F Ku
Step 7 - Solve for the Element Forces
Once the displacements are found, the stress and strain in
each element may be calculated from:
2 1
x x x
u u
du
E
dx L
  

  
24

Consider the following three-bar system shown below. Assume
for elements 1 and 2: A = 1 in2 and E = 30 (106) psi and for
element 3: A = 2 in2 and E = 15 (106) psi.
Example 1 - Bar Problem
Determine: (a) the global stiffness matrix, (b) the displacement
of nodes 2 and 3, and (c) the reactions at nodes 1 and 4.
25

For elements 1 and 2:
For element 3:
  
6
(1) (2) 6
1 30 10 1 1 1 1
10
30 1 1 1 1
lb lb
in in
  
   
  
   
 
   
k k
1 2 node numbers for element 1
  
6
(3) 6
2 15 10 1 1 1 1
10
30 1 1 1 1
lb lb
in in
  
   
 
   
 
   
k
As before, the numbers above the matrices indicate the
displacements associated with the matrix.
26

Assembling the global stiffness matrix by the direct stiffness
methods gives:
Relating global nodal forces related to global nodal
displacements gives:
6
1 1 0 0
1 2 1 0
10
0 1 2 1
0 0 1 1

 
 
 
 

 
 
 

 
K
1 1
2 2
6
3 3
4 4
1 1 0 0
1 2 1 0
10
0 1 2 1
0 0 1 1
x
x
x
x
F u
F u
F u
F u

   
 
   
 
 
   
 

   
 
 
   
 
   

 
   
E1 E 2 E 3
27


     
     
 
   
 

   
 
 
   
 
   

   
 
1
2 2
6
3 3
4
1 1 0 0 0
1 2 1 0
10
0 1 2 1
0 0 1 1 0
x
x
x
x
F
F u
F u
F
The boundary conditions are:
1 4 0
u u
 
Applying the boundary conditions and the known forces
(F2x = 3,000 lb) gives:
2
6
3
3,000 2 1
10
0 1 2
u
u
  
   

   
 

     
28

Solving for u2 and u3 gives: 

2
3
0.002
0.001
u in
u in
The global nodal forces are calculated as:
1
2 6
3
4
1 1 0 0 0 2,000
1 2 1 0 0.002 3,000
10
0 1 2 1 0.001 0
0 0 1 1 0 1,000
x
x
x
x
F
F
lb
F
F
 
       
       
 
     
 
 
     
 
 
     
 
     
 
     
 
29

Consider the following guidelines, as they relate to the one-
dimensional bar element, when selecting a displacement
function.
Selecting Approximation Functions for Displacements
1. Common approximation functions are usually
polynomials.
2. The approximation function should be continuous within
the bar element.
30

3. The approximating function should provide interelement
continuity for all degrees of freedom at each node for
discrete line elements, and along common boundary lines
and surfaces for two- and three-dimensional elements.
function.
31

For the bar element, we must ensure that nodes common
to two or more elements remain common to these
elements upon deformation and thus prevent overlaps or
voids between elements.
The linear function is then called a conforming (or
compatible) function for the bar element because it
ensures both the satisfaction of continuity between
adjacent elements and of continuity within the element.
function.
32

4. The approximation function should allow for rigid-body
displacement and for a state of constant strain within the
element.
Completeness of a function is necessary for
convergence to the exact answer, for instance, for
displacements and stresses.
function.
33

The interpolation function must allow for a rigid-body
displacement, that means the function must be capable of
yielding a constant value.
Consider the follow situation:
Therefore:
1 1 1 2
u a a u u
  
 
1 1 2 2 1 2 1
u N u N u N N a
   
Since u = a1 then:
This means that: 1 2 1
N N
 
 
1 1 2 1
u a N N a
  
The displacement interpolation function must add to unity at
every point within the element so the it will yield a constant
value when a rigid-body displacement occurs. 34

In many problems it is convenient to introduce both local and
global (or reference) coordinates.
Local coordinates are always chosen to conveniently
represent the individual element.
Global coordinates are chosen to be convenient for the whole
structure.
Transformation of Vectors in Two Dimensions
35

Given the nodal displacement of an element, represented by
the vector d in the figure below, we want to relate the
components of this vector in one coordinate system to
components in another.
36

Let’s consider that d does not coincide with either the local or
global axes. In this case, we want to relate global
displacement components to local ones. In so doing, we will
develop a transformation matrix that will subsequently be
used to develop the global stiffness matrix for a bar element.
37

We define the angle  to be positive when measured
counterclockwise from x to x’. We can express vector
displacement d in both global and local coordinates by:
1 1 1 1
u v u v
   
   
d i j i j
38

The vector a is in the direction and b is in the direction,
therefore:

i 
j
      
| | cos | | sin
 
   
     
a a i i b b j j
40

The vector i can be rewritten as: cos sin
 
 
 
i i j
The vector j can be rewritten as: sin cos
 
 
 
j i j
Therefore, the displacement vector is:
   
1 1 1 1 1 1
cos sin sin cos
u v u v u v
   
       
      
i j i j i j i j41

Combining like coefficients of the local unit vectors gives:
1 1 1
cos sin
u v u
  
 
1 1 1
sin cos
u v v
  
  
cos
sin
C
S




1 1
1 1
u u
C S
v v
S C

   
 

   
 
 
 
   
42

The matrix is called the transformation matrix:
The previous equation relates the global displacement d to
the d local displacements
C S
S C
 
 

 
The figure below shows u expressed in terms of the global
coordinates x and y.
u Cu Sv
  
43

Example 2 - Bar Element Problem
Using the following expression we just derived, we get:
The global nodal displacement at node 2 is u2 = 0.1 in and
v2 = 0.2 in for the bar element shown below. Determine the
local displacement.
2 cos60 (0.1) sin60 (0.2)
o o
u  
u Cu Sv
  
0.223 in

44

Global Stiffness Matrix
We will now use the transformation relationship developed
above to obtain the global stiffness matrix for a bar element.
45

We want to relate the global element forces f to the global
displacements d for a bar element with an arbitrary
orientation.
We known that for a bar element in local coordinates we have:
1 1
2 2
1 1
1 1
x
x
f u
AE
f u
L
 

   
 

   
 
 

 
   
1 1
1 1
2 2
2 2
x
y
x
y
f u
f v
k
f u
f v
   
   
   

   
   
   
 
 
  

f k d
f = kd
46

Combining both expressions for the two local degrees-of-
freedom, in matrix form, we get:
Using the relationship between local and global components,
we can develop the global stiffness matrix.
We already know the transformation relationships:
1 1 1 2 2 2
cos sin cos sin
u u v u u v
   
 
   
1
1 1
2 2
2
0 0
0 0
u
u v
C S
u u
C S
v
 
 

     

   
 
  
   
 
 
 *
d = T d
* 0 0
0 0
C S
C S
 
  
 
T
47

Substituting the global force expression into element force
equation gives:
A similar expression for the force transformation can be
developed.
1
1
1 *
2
2
2
0 0
0 0
x
y
x
x
x
y
f
f
f C S
f
f C S
f
 
 

     

 
   
 
  
   
 
 
f T f
  
f = k d

 
* *
T f k T d
 
 
*
T f k d
 *
d = T d
Substituting the transformation between local and global
displacements gives:
48

The matrix T* is not a square matrix so we cannot invert it.
Let’s expand the relationship between local and global
displacement.
where T is:
1 1
1 1
2 2
2 2
0 0
0 0
0 0
0 0
u u
C S
v v
S C
u u
C S
v v
S C

   
 
   
 
 
   
 

   
  
   
 
   
 
 
   
0 0
0 0
0 0
0 0
C S
S C
C S
S C
 
 

 

 
 

 
T

d = Td
49

We can write a similar expression for the relationship between
local and global forces.
Therefore our original local coordinate force-displacement
expression
1 1
1 1
2 2
2 2
0 0
0 0
0 0
0 0
x x
y y
x x
y y
f f
C S
f f
S C
f f
C S
f f
S C

   
 
   
 
 
   
 

   
  
   
 
   
 
 
   
1 1
2 2
1 1
1 1
x
x
f u
AE
f u
L
 

   
 

   
 
 

 
   

f = Tf
  
f = k d
50

May be expanded:
The global force-displacement equations are:
1 1
1 1
2 2
2 2
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
x
y
x
y
f u
f v
AE
f u
L
f v
 
    
 
   
 
 
   
 

   
 
 

   
 
   
 
   
 
  

f k d
Multiply both side by T -1 we get: 
 -1
f T k Td
where T-1 is the inverse of T. It can be shown that: 

1 T
T T

 
Tf k Td
51

The global force-displacement equations become:
Where the global stiffness matrix k is:
Expanding the above transformation gives:
We can assemble the total stiffness matrix by using the above
element stiffness matrix and the direct stiffness method.
= 
T
f T k Td

 T
k T k T
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
L CS CS
C C
S S
CS CS
  

 


 

 


 


 
k
   
( ) ( )
1 1
n n
e e
e e
K F
 
   
 
K k F f 
F Kd
52

Local forces can be computed as:
1 1
1 1
2 2
2 2
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
x
y
x
y
f u
f v
AE
f u
L
f v
 
    
 
   
 
 
   
 

   
 
 

   
 
   
 
   
 
1 1 1 2 2
1
2 1 1 2 2
2
0
0
x
y
x
y
f Cu Sv Cu Sv
f AE
f L Cu Sv Cu Sv
f

    
 
   

   

 
  
   
 
 
 
  
 
1
1
2
2
1 0 1 0 0 0
0 0 0 0 0 0
1 0 1 0 0 0
0 0 0 0 0 0
u
C S
v
S C
AE
u
C S
L
v
S C
  
   
 
   
  
   
  
   
  
     

     
53

Therefore:
For the bar element shown below, evaluate the global stiffness
matrix. Assume the cross-sectional area is 2 in2, the length
is 60 in, and the E is 30 x 106 psi.
3
cos30
2
o
C  
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
L CS CS
C C
S S
CS CS
  

 


 

 


 


 
k
1
sin30
2
o
S  
54

Simplifying the global elemental stiffness matrix is:
The global elemental stiffness matrix is:
 
2 6
3 3
3 3
4 4 4 4
3 3
1 1
(2 ) 30 10 4 4 4 4
60 3 3
3 3
4 4 4 4
3 3
1 1
4 4 4 4
in psi
in
 
 
 
 
 
 

 

 
 
 
 
 
 
 
k
6
0.750 0.433 0.750 0.433
0.433 0.250 0.433 0.250
10
0.750 0.433 0.750 0.433
0.433 0.250 0.433 0.250
lb
in
 
 
 
 
 

 
 
 
 
 
k
55

Computation of Stress for a Bar in the x-y Plane
For a bar element the local forces are related to the local
displacements by:
1 1
2 2
1 1
1 1
x
x
f u
AE
f u
L
 

   
 

   
 
 

 
   
The force-displacement equation for is:
2x
f 
  1
2
2
1 1
x
u
AE
f
u
L

 
    

 
The stress in terms of global displacement is:
 
1
1
2
2
0 0
1 1
0 0
u
v
C S
E
u
L C S
v

 
 
   
   
 
   
 
 
 
1 1 2 2
E
Cu Sv Cu Sv
L
    
56

For the bar element shown below, determine the axial stress.
Assume the cross-sectional area is 4 x 10-4 m2, the length is
2 m, and the E is 210 GPa.
The global displacements are known as
u1 = 0.25 mm, v1 = 0, u2 = 0.5 mm,
and v2 = 0.75 mm.
 
1 1 2 2
E
Cu Sv Cu Sv
L
     
6
210 10 1 3 1 3
(0.25) (0) (0.5) (0.75)
2 2 4 2 4
KN
m

 

    
 
 
2
3
81.32 10 kN
   81.32 MPa
 57

Solution of a Plane Truss
We will now illustrate the use of equations developed above
along with the direct stiffness method to solve the following
plane truss example problems.
A plane truss is a structure composed of bar elements all lying
in a common plane that are connected together by
frictionless pins.
The plane truss also must have loads acting only in the
common plane.
58

Example 5 - Plane Truss Problem
The plane truss shown below is composed of three bars
subjected to a downward force of 10 kips at node 1. Assume
the cross-sectional area A = 2 in2 and E is 30 x 106 psi for all
elements.
Determine the x and y displacement at node 1 and stresses in
each element.
59

Element Node 1 Node 2  C S
1 1 2 90o 0 1
2 1 3 45o 0.707 0.707
3 1 4 0o 1 0
60

The global elemental stiffness matrix are:
0 1
C S
 
element 1:
1 0
C S
 
element 2:
element 3:
2 2
2 2
C S
 
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
L CS CS
C C
S S
CS CS
  

 


 

 


 


 
k
1 1 2 2
2 6
(1)
0 0 0 0
0 1 0 1
(2 )(30 10 )
0 0 0 0
120
0 1 0 1
lb
in
u v u v
in psi
in
 
 

  
 
 
 

 
k
1 1 3 3
2 6
(2)
1 1 1 1
1 1 1 1
(2 )(30 10 )
1 1 1 1
240 2
1 1 1 1
lb
in
u v u v
in psi
in
 
 
 
 
  
 
 
 
 
 
 
k
1 1 4 4
2 6
(3)
1 0 1 0
0 0 0 0
(2 )(30 10 )
1 0 1 0
120
0 0 0 0
lb
in
u v u v
in psi
in

 
 
  
 
 

 
k
61

The total global stiffness matrix is:
5
1.354 0.354 0 0 0.354 0.354 1 0
0.354 1.354 0 1 0.354 0.354 0 0
0 0 0 0 0 0 0 0
0 1 0 1 0 0 0 0
0.354 0.354 0 0 0.354 0.354 0 0
0.354 0.354 0 0 0.354 0.354 0 0
1 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0
5 10 lb
in
  
  

 
 

 
 
 
 
 
 
 
 
 
K
The total global force-displacement equations are:
5
0
1.354 0.354 0 0 0.354 0.354 1 0
10, 000
0.354 1.354 0 1 0.354 0.354 0 0
2 0 0 0 0 0 0 0 0
2 0 1 0 1 0 0 0 0
0.354 0.354 0 0 0.354 0.354 0 0
3
0.354 0.354 0 0 0.354 0.354 0 0
3
1 0 0 0 0 0 1 0
4
0 0 0 0 0 0 0 0
4
5 10
F x
F y
F x
F y
F x
F y

  

  

 
 

   
  
  
  

 

 

 

   
 
1
1
0
0
0
0
0
0
u
v
 
 

 

 
  
  
  
  
 
element 1
element 2
element 3
62

5
0
1.354 0.354 0 0 0.354 0.354 1 0
10, 000
0.354 1.354 0 1 0.354 0.354 0 0
2 0 0 0 0 0 0 0 0
2 0 1 0 1 0 0 0 0
0.354 0.354 0 0 0.354 0.354 0 0
3
0.354 0.354 0 0 0.354 0.354 0 0
3
1 0 0 0 0 0 1 0
4
0 0 0 0 0 0 0 0
4
5 10
F x
F y
F x
F y
F x
F y

  

  

 
 

   
  
  
  

 

 

 

   
 
1
1
0
0
0
0
0
0
u
v
 
 

 

 
  
  
  
  
 
Applying the boundary conditions for the truss, the above
equations reduce to:
63

Applying the boundary conditions for the truss, the above
1
5
1
0 1.354 0.354
5 10
10,000 0.354 1.354
u
v
 
   
 
   
 

     
Solving the equations gives: 2
1
2
1
0.414 10
1.59 10
u in
v in


 
  
The stress in an element is: 1 1 2 2
E
Cu Sv Cu Sv
L
  
    
 
where is the local node number
i
64

1 1 2 90o 0 1
2 1 3 45o 0.707 0.707
3 1 4 0o 1 0
element 1  
6
(1)
1
30 10
3,965
120
v psi


  
1 1 2 2
E
Cu Sv Cu Sv
L
  
    
 
element 2  
6
(2)
1 1
30 10
(0.707) (0.707) 1,471
120
u v psi


   
65

element 3  
6
(3)
1
30 10
1,035
120
u psi


   
1 1 2 90o 0 1
2 1 3 45o 0.707 0.707
3 1 4 0o 1 0
1 1 2 2
E
Cu Sv Cu Sv
L
  
    
 
66

Let’s check equilibrium at node 1:
(2)
1 cos(45 )
x x
F f
 

(2)
1 sin(45 )
y x
F f
 

(3)
1x
f

(1)
1x
f
 10,000lb

67

Let’s check equilibrium at node 1:
2 2
(1,471 )(2 )(0.707) (1,035 )(2 ) 0
x
F psi in psi in
  

2 2
(3,965 )(2 ) (1,471 )(2 )(0.707) 10,000 0
y
F psi in psi in
   

68

Develop the element stiffness matrices and system equations
for the plane truss below.
Assume the stiffness of each element is constant. Use the
numbering scheme indicated. Solve the equations for the
displacements and compute the member forces. All elements
have a constant value of AE/L
69

Develop the element stiffness matrices and system equations
for the plane truss below.
Member Node 1 Node 2 Elemental
Stiffness

1 1 2 k 0
2 2 3 k 3/4
3 1 3 k /2
70

Compute the elemental stiffness matrix for each element. The
general form of the matrix is:
Stiffness

1 1 2 k 0
2 2 3 k 3/4
3 1 3 k /2
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
k
L CS CS
C C
S S
CS CS

 

 


 

 


 


 
71

For element 1:
Stiffness

1 1 2 k 0
2 2 3 k 3/4
3 1 3 k /2
1 1 2 2
1
1
(1)
2
2
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
u v u v
u
v
k
u
v

 
 
 

 

 
 
k
72

2 2 3 3
2
2
(2)
3
3
1 1 1 1
1 1 1 1
2 1 1 1 1
1 1 1 1
u v u v
u
v
k
u
v
 
 
 
 
 

 
 
 
 
 
k
For element 2:
Stiffness

1 1 2 k 0
2 2 3 k 3/4
3 1 3 k /2 73

1 1 3 3
1
1
(3)
3
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
u v u v
u
v
k
u
v
 
 

 

 
 

 
k
For element 3:
Stiffness

1 1 2 k 0
2 2 3 k 3/4
3 1 3 k /2
74


 
 

 
 
  
  
 
 
 
 
 
  
 
1 1 2 2 3 3
1
1
2
2
3
3
2 0 2 0 0 0
0 2 0 0 0 2
2 0 3 1 1 1
2 0 0 1 1 1 1
0 0 1 1 1 1
0 2 1 1 1 3
u v u v u v
u
v
u
k
v
u
v
K
Assemble the global stiffness matrix by superimposing the
elemental global matrices.
element 1
element 2
element 3
75

The unconstrained (no boundary conditions satisfied)
equations are: 1
1
1
1
1
2
2
2
3
3
3
3
2 0 2 0 0 0
0 2 0 0 0 2
2 0 3 1 1 1
2 0 0 1 1 1 1
0 0 1 1 1 1
0 2 1 1 1 3
x
y
x
y
F
u
F
v
P
u
k
P
v
F
u
F
v
 
  
 
 
 
 
  
 
 
 
 
 
      

   
 

     
 
   
 
 
   
 
      
     
The displacement at nodes 1 and 3 are zero in both directions.
Applying these conditions to the system equations gives:
 

   
 
   
  
   
 
   
    

   
 

     
 
   
 
 
   
 
    
     
1
1
1
2
2
2
3
3
2 0 2 0 0 0 0
0 2 0 0 0 2 0
2 0 3 1 1 1
2 0 0 1 1 1 1
0 0 1 1 1 1 0
0 2 1 1 1 3 0
x
y
x
y
F
F
P
u
k
P
v
F
F 76

Applying the boundary conditions to the system equations
gives:
Solving this set of equations is fairly easy. The solution is:
2 1
2 2
3 1
2 1 1
u P
k
v P
    
 

   
  

     
1 2
2
P P
u
k

 1 2
2
3
P P
v
k


77

Member (element) 1:
Using the force-displacement relationship the force in each
member may be computed.
 
1 2
x
f k Cu
 
1 1 1 2 2
1
2 1 1 2 2
2
0
0
x
y
x
y
f Cu Sv Cu Sv
f
k
f Cu Sv Cu Sv
f
    
 
   
   

 
 
   
 
 
   
 
1 0
C S
 
2 0
y
f 
1 0
y
f 
1 2
P P
k
k

 
 
 
 
 
1 2
P P
  
 
2 2
x
f k Cu
 1 2
P P
k
k

 
  
 
1 2
P P
 
78

Member (element) 2:
 
2 2 2
x
f k Cu Sv
 
2 2 2 3 3
2
3 2 2 3 3
3
0
0
x
y
x
y
f Cu Sv Cu Sv
f
k
f Cu Sv Cu Sv
f
    
 
   
   

 
 
   
 
 
   
 
1 1
2 2
C S
  
 
3 2 2
x
f k Cu Sv
  
1 2 1 2
3
1 1
2 2
P P P P
k
k k
 
 
   
   
  
 
   
   
   
   
 
1 2 1 2
3
1 1
2 2
P P P P
k
k k
 
 
   
   
  
 
   
   
   
   
 
2
2P
 
2
2P

79

Member (element) 3:
1 1
3 3
0 0
0 0
x y
x y
f f
f f
 
 
The solution to this simple problem can be readily checked by
using simple static equilibrium equations.
1 1 1 3 3
1
3 1 1 3 3
3
0
0
x
y
x
y
f Cu Sv Cu Sv
f
k
f Cu Sv Cu Sv
f
    
 
   
   

 
 
   
 
 
   
 
0 1
C S
 
80

Consider the two bar truss shown below.
Determine the displacement in the y direction of node 1 and
the axial force in each element.
Assume E = 210 GPa and A = 6 x 10-4 m2
81

The global elemental stiffness matrix for element 1 is:
(1) 3
cos 0.6
5
  
6 4
(1)
0.36 0.48 0.36 0.48
0.48 0.64 0.48 0.64
210 10 (6 10 )
5 0.36 0.48 0.36 0.48
0.48 0.64 0.48 0.64

 
 
 
 
   

 
 
 
 
 
k
Simplifying the above expression gives:
1 1 2 2
(1)
0.36 0.48 0.36 0.48
0.48 0.64 0.48 0.64
25,200
0.36 0.48 0.36 0.48
0.48 0.64 0.48 0.64
u v u v
 
 
 
 
 

 
 
 
 
 
k
(1) 4
sin 0.8
5
  
82

(2)
cos 0
 
6 4
(2)
0 0 0 0
0 1 0 1
(210 10 )(6 10 )
4 0 0 0 0
0 1 0 1


 
 

   

 
 

 
k
1 1 3 3
(2)
0 0 0 0
0 1.25 0 1.25
25,200
0 0 0 0
0 1.25 0 1.25
u v u v
 
 

 

 
 

 
k
(2)
sin 1
 
83

The total global equations are:
The displacement boundary conditions are:
1 1
1 1
2 2
2 2
3 3
3 3
0.36 0.48 0.36 0.48 0 0
0.48 1.89 0.48 0.64 0 1.25
0.36 0.48 0.36 0.48 0 0
25,200
0.48 0.64 0.48 0.64 0 0
0 0 0 0 0 0
0 1.25 0 0 0 1.25
x
y
x
y
x
y
F u
F v
F u
F v
F u
F v
     
 
   
 
  
   
 
   
 
 
   

   
 
 
   
 
   
 
  
 
 
     
 


1 2 2 3 3 0
u u v u v

    
element 1
element 2
84

   
   
     
  
     
     
 
 

   
 
 
   
 
   
 
   
 

     
 
1 1
1 1
2
2
3
3
0.36 0.48 0.36 0.48 0 0
0.48 1.89 0.48 0.64 0 1.25
0.36 0.48 0.36 0.48 0 0 0
25,200
0.48 0.64 0.48 0.64 0 0 0
0 0 0 0 0 0 0
0 1.25 0 0 0 1.25 0
x
y
x
y
x
y
F u
F v
F
F
F
F
The total global equations are:
By applying the boundary conditions the force-displacement
1
25,200(0.48 1.89 )
P v

 

1
2
2
3
3
x
x
y
x
y
F
P
F
F
F
F
 
 
 
 
 
 
 
 
 
 
 
85

Solving the equation gives:
By substituting P = 1,000 kN and  = -0.05 m in the above
equation gives:
The local element forces for element 1 are:
5
1 (2.1 10 ) 0.25
v P 

  
1 0.0337
v m

1
1 1
2 2
2
0.05
0.0337
1 1 0.6 0.8 0 0
25,200
1 1 0 0 0.6 0.8
x
x
u
f v
f u
v
 
 
 


       

   
   

   
   
 
 
The element forces are: 1 2
76.6 76.7
x x
f kN f kN
  
Tension 86

The local element forces for element 2 are:
The element forces are:
1
1 1
3 3
3
0.05
0.0337
1 1 0 1 0 0
31,500
1 1 0 0 0 1
x
x
u
f v
f u
v
 
 
 


       

   
   

   
   
 
 
1 3
1,061 1,061
x x
f kN f kN
  
Compression
87

Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
Let’s derive the transformation matrix for the stiffness matrix for
a bar element in three-dimensional space as shown below:
88

The coordinates at node 1 are x1, y1, and z1, and the coor-
dinates of node 2 are x2, y2, and z2. Also, let x, y, and z be
the angles measured from the global x, y, and z axes,
respectively, to the local axis.
89

The three-dimensional vector representing the bar element is
gives as:
u v w u v w
     
     
d i j k i j k
90

Taking the dot product of the above equation with gives:

i
( ) ( ) ( )
u v w u
   
     
i i j i k i
By the definition of the dot product we get:
2 1
x
x x
C
L


  
i i
where 2 2 2
2 1 2 1 2 1
( ) ( ) ( )
L x x y y z z
     
cos
x x
C 

where Cx, Cy, and Cz are projections of on to i, j, and k,
respectively.

i
2 1
y
y y
C
L


  
j i 2 1
z
z z
C
L


  
k i
cos
y y
C 
 cos
z z
C 

91

Therefore: x y z
u C u C v C w
   
The transformation between local and global displacements is:
  *
d T d
1
1
1 1
2 2
2
2
0
0 0
0 0 0
x y z
y
x z
u
v
u w
C C C
C
u C C u
v
w
 
 
 
  
 
 

   
 

     
 
 
 
0
0 0
0 0 0
x y z
y
x z
C C C
C
C C
 
  
 
*
T
92

The transformation from the local to the global stiffness matrix
is:

 T
k T k T
0
0
0
0 0 0
1 1
0 1 1 0 0 0
0
0
x
y
z x y z
y
x x z
y
z
C
C
C C C C
AE
C
C C C
L
C
C
 
 
 
   

 
    
 

   
 
 
 
 
k
2 2
2 2
2 2
2 2
2 2
2 2
x y x y
x z x z
x x
y y
y z y z
x y x y
y z y z
z z
x z x z
x y x y
x z x z
x x
y y
y z y z
x y x y
y z y z
z z
x z x z
C C C C
C C C C
C C
C C
C C C C
C C C C
C C C C
C C
C C C C
AE
C C C C
C C C C
L C C
C C
C C C C
C C C C
C C C C
C C
C C C C
 
 

 
 

 
 
 

  
 

 
 
 

 
 

 
 
k
93

The global stiffness matrix can be written in a more convenient
form as:
AE
L
 
 

 
  

 
k
2
2
2
x x y x z
x y y y z
x z y z z
C C C C C
C C C C C
C C C C C

 
 
  
 
 
94

Inclined, or Skewed Supports
If a support is inclined, or skewed, at some angle  for the
global x axis, as shown below, the boundary conditions on
the displacements are not in the global x-y directions but in
the x’-y’ directions.
95

Inclined, or Skewed, Supports
We must transform the local boundary condition of v’3 = 0
(in local coordinates) into the global x-y system.
96

Therefore, the relationship between of the components of the
displacement in the local and the global coordinate systems
at node 3 is:
3 3
3 3
' cos sin
' sin cos
u u
v v
 
 
   
 

   
 

 
   
We can rewrite the above expression as:
   
3 3 3
' [ ]
d t d

We can apply this sort of transformation to the entire
displacement vector as:
       
1 1
' [ ] [ ] '
T
d T d or d T d
 
 
3
cos sin
sin cos
t
 
 
 
  

 
97

Where the matrix [T1]T is:
Both the identity matrix [I] and the matrix [t3] are 2 x 2 matrices.
The force vector can be transformed by using the same
transformation.
T
1
3
[ ] [0] [0]
[ ] [0] [ ] [0]
[0] [0] [ ]
I
T I
t
 
 

 
 
 
   
1
' [ ]
f T f

In global coordinates, the force-displacement equations are:
   
[ ]
f K d

98

Applying the skewed support transformation to both sides of
the equation gives:
By using the relationship between the local and the global
displacements, the force-displacement equations become:
Therefore the global equations become:
   
1 1
[ ] [ ][ ]
T f T K d
 
   
1 1
' [ ][ ][ ] '
T
f T K T d
 
1 1
1 1
2 2
1 1
2 2
3 3
3 3
[ ][ ][ ]
' '
' '
x
y
x T
y
x
y
F u
F v
F u
T K T
F v
F u
F v
   
   
   
   
 

   
   
   
   
   
 
   
[ ]
f K d

   
1
[ ] '
T
d T d

99

Example 9 – Space Truss Problem
Determine the stiffness matrix
for each element.
Consider the plane truss shown below. Assume E = 210 GPa,
A = 6 x 10-4 m2 for element 1 and 2, and A = (6 x 10-4)m2
for element 3.
2
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
k
L CS CS
C C
S S
CS CS

 

 


 

 


 


 
100

(1)
cos 0
 
1 1 2 2
6 2 4 2
(1)
0 0 0 0
0 1 0 1
(210 10 / )(6 10 )
0 0 0 0
1
0 1 0 1
u v u v
kN m m
m
 
 
 

   

 
 

 
k
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
k
L CS CS
C C
S S
CS CS

 

 


 

 


 


 
(1)
sin 1
 
101

(2)
cos 1
 
2 2 3 3
6 2 4 2
(2)
1 0 1 0
0 0 0 0
(210 10 / )(6 10 )
1 0 1 0
1
0 0 0 0
u v u v
kN m m
m


 
 
   

 

 
 
k
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
k
L CS CS
C C
S S
CS CS

 

 


 

 


 


 
(2)
sin 0
 
102

(3) 2
cos
2
 
1 1 3 3
6 2 4 2
(3)
1 1 1 1
1 1 1 1
(210 10 / )(6 2 10 )
1 1 1 1
2 2
1 1 1 1
u v u v
kN m m
m

 
 
 
 
   

 
 
 
 
 
k
2 2
2 2
2 2
2 2
CS CS
C C
S S
CS CS
AE
k
L CS CS
C C
S S
CS CS

 

 


 

 


 


 
(3) 2
sin
2
 
103

Using the direct stiffness method, the global stiffness matrix is:
































5
.
0
5
.
0
0
0
5
.
0
5
.
0
5
.
0
5
.
1
0
1
5
.
0
5
.
0
0
0
1
0
1
0
0
1
0
1
0
0
5
.
0
5
.
0
1
0
5
.
1
5
.
0
5
.
0
5
.
0
0
0
5
.
0
5
.
0
10
260
,
1 5
m
N
K
We must transform the global displacements into local
coordinates. Therefore the transformation [T1] is:
2 2
2 2
2 2
2 2
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0 
 
 
 
 
  
 
 
 
 
1
3
[ ] [0] [0]
[ ] [0] [ ] [0]
[0] [0] [ ]
I
T I
t
 
 

 
 
 
104

The first step in the matrix transformation to find the product of
[T1][K].
  
1
T K
   5
0.5 0.5 0 0 0.5 0.5
0.5 1.5 0 1 0.5 0.5
0 0 1 0 1 0
0 1 0 1 0 0
0.707 0.707 0.707 0 1.414 0.707
0 0 0.707 0 0.707 0
1,260 10
1
N
T K
m
 
  


  

 
 
 
 
 
 
 
2 2
2 2
2 2
2 2
5
1 0 0 0 0 0
0.5 0.5 0 0 0.5 0.5
0 1 0 0 0 0 0.5 1.5 0 1 0.5 0.5
0 0 1 0 0 0 0 0 1 0 1 0
1,
0 1 0 1 0 0
0 0 0 1 0 0
0.707 0.707 0.707 0 1.414 0.707
0 0 0 0
0 0 0.707 0 0.707 0
0 0 0 0
260 10 N
m

 
  


  

   
   
   
   
     
   
   
   
 
 
1
T  
K
105

The next step in the matrix transformation to find the product of
[T1][K][T1]T.
   
2 2
2 2
2 2
2 2
T 5
1 0 0 0 0 0
0.5 0.5 0 0 0.5 0.5
0 1 0 0 0 0
0.5 1.5 0 1 0.5 0.5
0 0 1 0 0 0
0 0 1 0 1 0
0 1 0 1 0 0 0 0 0 1 0 0
0.707 0.707 0.707 0 1.414 0.707 0 0 0 0
0 0 0.707 0 0.707 0
0 0 0 0
1,260 10
1 1
N
T K
m
T

 
  


  

 
   
   
   
 
  
   
   
   
   

   
T
0.5 0.5 0 0 0.707 0
0.5 1.5 0 1 0.707 0
0 0 1 0 0.707 0.707
0 1 0 1 0 0
0.707 0.707 0.707 0 1.5 0.5
0 0 0.707 0 0.5 0.5
5
1
,260 10
1 1
T K N
T
m

 


   

 
 
 
 
 
 
 
 
T
1
T
  
1
T K
106

The displacement boundary conditions are: 1 1 2 3
' 0
u v v v
   
5
1
1
2
2
3
3
1
1
2
2
3
3
1
,260 10
0.5 0.5 0 0 0.707 0
0.5 1.5 0 1 0.707 0
0 0 1 0 0.707 0.707
0 1 0 1 0 0
0.707 0.707 0.707 0 1.5 0.5
0 0 0.707 0 0.5 0.5
'
'
'
'
x
y
x
y
x
y
N
m
F
F
F
F
F
F
u
v
u
v
u
v
 
   
 
   
 
   
 
   

   
   
 
  
 
  
 
  
 
  
 
 
  
 

 


   







107

By applying the boundary conditions the global force-
displacement equations are:
2
5
3
1 0.707
1,260 10
'
0.707 1.5
u
N
m u
  
 
  
 

   
Solving the equation gives: 2 11.91
u mm

2
3
1,000
' 0
x
x
F kN
F

 
  

 
3
' 5.61
u mm

108

Therefore:
The global nodal forces are calculated as:
1
1
2 2
2
3
3
0.5 0.5 0 0 0.707 0 0
0.5 1.5 0 1 0.707 0 0
0 0 1 0 0.707 0.707 11.91
0 1 0 1 0 0 0
0.707 0.707 0.707 0 1.5 0.5 5.61
0 0 0.707 0 0.5 0.5 0
1,260 10
'
'
x
y
x N
mm
y
x
y
F
F
F
F
F
F

 

   

 
     
     
     
 
 
   
 

   
 
   
 
 
 
 
 
 
1 1
500 500
x y
F kN F kN
   
2 3
0 ' 707
y y
F F kN
 
1 1
1 1
2 2
1 1
2 2
3 3
3 3
[ ][ ][ ]
' '
' '
x
y
x T
y
x
y
F u
F v
F u
T K T
F v
F u
F v
   
   
   
   
 

   
   
   
   
   
 
109

Potential Energy Approach to Derive Bar Element Equations
The differential internal work (strain energy) dU in a one-
dimensional bar element is:
Let’s derive the equations for a bar element using the principle
of minimum 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
   
( )( )( )
x x
dU y z x d
 
   
110

Summing the differential energy over the whole bar gives:
If we let the volume of the element approach zero, then:
x x
dU d dV
 

0
x
x x
V
U d dV

 
 
 
  
 
 
 
For a linear-elastic material (Hooke’s law) as shown below:
x x
E
 

0
x
x x
V
E d dV

 
 
 
  
 
 
 
2
1
2
x
V
E dV

 
1
2
x x
V
U dV
 
 
111

The potential energy of the external forces is:
The internal strain energy statement becomes
1
2
x x
V
U dV
 
 
s
V S
X u dV T u dS f u

     
 
M
b x ix i
i 1
where Xb is the body force (force per unit volume), Tx is the
traction (force per unit area), and fix is the nodal concentrated
force. All of these forces are considered to act in the local x
direction.
112

1. Formulate an expression for the total potential energy.
2. Assume a displacement pattern.
3. Obtain a set of simultaneous equations minimizing the
total potential energy with respect to the displacement
parameters.
Apply the following steps when using the principle of minimum
potential energy to derive the finite element equations.
113

We can approximate the axial displacement as:
Consider the following bar element, as shown below:
2 2
0
L
p x x 1x 1 x
s
V S
A
dx f u f u
2
X u dV T u dS
  
  
 

 
b x
  1
1 2
2
u
u N N
u
 
  
 
1 1
x
N
L
  2
x
N
L

114

where N1 and N2 are the interpolation functions gives as:
Using the stress-strain relationships, the axial strain is:
x
du
dx
 
1
2
1 1
x
u
u
L L

 
 
   
 
   
1 1
B
L L
 
 
 
 
The axial stress-strain relationship is:    
[ ]
x x
D
 

  [ ]{ }
x B d
 
1
1 2
2
u
dN dN
u
dx dx
 
 
  
 
   
115

The total potential energy expressed in matrix form is:
For the one-dimensional stress-strain relationship [D] = [E]
where E is the modulus of elasticity.
Therefore, stress can be related to nodal displacements as:
where {P} represented the concentrated nodal loads.
   
[ ][ ]
x D B d
 
               
0
L
T T T T
p x x
V S
A
dx d P u X dV u T dS
2
  
   
  
b x
116

If we substitute the relationship between and into the
energy equations we get:
û d̂
            
           
0
L
T T
T T
p
T T
T T
s
V S
A
d B D B d dx d P
2
d N X dV d N T dS
  
 

 
b x
In the above expression for potential energy p is a function of
the d, that is: p = p( ).
1,
u 2
u
However, [B] and [D] and the nodal displacements u are not a
function of x.
x




 x





117

where
Integration the energy expression with respect to x gives:
       
[ ] [ ] [ ]
2
T T
T T
p
AL
d B D B d d f
  
       
[ ] [ ]
V S
f P N X dV N X dS
  
 
T T
b b
We can define the surface tractions and body-force matrices
as:
   
[ ] x
S
f N T dS
 
T
s    
[ ]
V
f N X dV
 
T
b b
118

Minimization of p with respect to each nodal displacement
requires that:
For convenience, let’s define the following
1 2
0 0
p p
u u
 
 
 
 
     
*
[ ] [ ] [ ]
T T T
U d B D B d

    1
*
1 2
2
1
1 1
[ ]
1
u
L
U u u E
u
L L
L
 

   
   
 
   
 
   
 
 
 
119

The loading on a bar element is given as:
 
* 2 2
1 1 2 2
2
2
E
U u u u u
L
  
    1 1 2 2
T
x x
d f u f u f
 
Therefore, the minimum potential energy is:
 
1 2 1
1
2 2 0
2
p
x
AE
u u f
u L


   

 
1 2 2
2
2 2 0
2
p
x
AE
u u f
u L


    
 120

The above equations can be written in matrix form as:
The stiffness matrix for a bar element is:
This form of the stiffness matrix obtained from the principle of
minimum potential energy is identical to the stiffness matrix
derived from the equilibrium equations.
 
1 1
2 2
1 1
0
1 1
p x
x
u f
AE
u f
d L

     
 
  
   
 
 
    
 
1 1
1 1
AE
k
L

 
  

 
121

Consider the bar shown below:
The energy equivalent nodal forces due to the distributed load
are:
   
0 [ ] x
S
f N T dS
 
T
   
0
0
L
x
1
f L
f Cx dx
f x
L
 

 
   
 
   
   
 
 

1x
2x
122

The total load is the area under the distributed load curve, or:
The equivalent nodal forces for a linearly varying load are:
f
f
 
 
 
1x
2x
2
1
( )( )
2 2
CL
F L CL
 
1
1
of the total load
3 3
x
F
f   2
2 2
of the total load
3 3
x
F
f  
 
0
L
x
1
L
Cx dx
x
L
 

 
 
  
 
 
 

0
0
L
2 3
L
3
Cx Cx
2 3L
Cx
3L
 

 
 
  
 
 
 
2
2
CL
6
CL
3
 
 
 
  
 
 
 
123

Consider the axially loaded bar shown below. Determine the
axial displacement and axial stress. Let E = 30 x 106 psi,
A = 2 in2, and L = 60 in. Use (a) one and (b) two elements in
the finite element solutions.
124

The one-element solution:
The distributed load can be converted into equivalent nodal
forces using:
   
0 [ ] x
S
F N T dS
 
T
 
0
1x
2x
F
F
F
 
  
 
 
0
10
L
x
1
L x dx
x
L
 

 
 
 
 
 
 
 

125

 
0
10
L
1x
2x
x
1
F L x dx
F x
L
 

 
   
 
   
   
 
 

6,000
12,000
1x
2x
F lb
F lb

   

   

 
 
2 2
2
10 10
- +
2 3
10
-
3
L L
L
 
 
 
  
 
 
 
2
2
10
-
6
10
-
3
L
L
 
 
 
  
 
 
 
126

The element equations are:
(1) 6 1 1
10
1 1

 
  

 
k
1
6
2
6,000
1 1
10
12,000
1 1 0 x
u
R

  
   

   
  

     
1 0.006
u in
 
The second equation gives:
6
1 2
10 ( ) 12,000
x
u R
   2 18,000
x
R lb
 
127

The axial stress-strain relationship is: { } [ ]{ }
x x
D
 

 
{ } [ ]
x E B d
 
1
2
1 1 u
E
u
L L
 
 
   
 
   
6 0 0.006
30 10
60

 
   
 
2 1
u u
E
L

 
  
 
3,000 ( )
psi T

128

The two-element solution:
The distributed load can be converted into equivalent nodal
forces.
For element 1, the total force of the triangular-shaped
distributed load is:
1
(30 .)(300 ) 4,500
2
lb
in
in lb
 
129

Based on equations developed for the equivalent nodal force of
a triangular distributed load, develop in the one-element
problem, the nodal forces are:
(1)
1
(1)
2
1
(4,500)
3
2
(4,500)
3
x
x
f
f
 

 
   

   
   

 
 
1,500
3,000
lb
lb

 
  

 
130

For element 2, the applied force is in two parts: a triangular-
shaped distributed load and a uniform load. The uniform load
is:
(30 )(300 / ) 9,000
in lb in lb
 
(2)
2
(2)
3
1 1
(9,000) (4,500)
2 3
1 2
(9,000) (4,500)
2 3
x
x
f
f
 
 
 
 
 
    

   
 
   
 
 
 
 
 
The nodal forces for element 2 are:
6,000
7,500
lb
lb

 
  

 
131

The final nodal force vector is:
The element stiffness matrices are:
(1)
1 1
(1) (2)
2 2 2
(2)
3 3
x x
x x x
x x
F f
F f f
F f
 
 
 
 
 
   
   
   
(1) (2) 1 1
2
1 1
AE
L
 
 
 

 

k k
1 2
2 3
3
1,500
9,000
7,500
x
R

 
 
 
 
 

 
132

The assembled global stiffness matrix is:
The assembled global force-displacement equations are:
6
1 1 0
2 10 1 2 1
0 1 1

 
 
   
 
 

 
K
1
6
2
3
1 1 0 1,500
2 10 1 2 1 9,000
0 1 1 0 7,500
x
u
u
R
 
     
   
 
    
   
 
   
 
 
     
element 1
element 2
133

After the eliminating the row and column associated with u3x,
we get:
Solving the equation gives:
1
6
2
1 1 1,500
2 10
1 2 9,000
u
u
 
 
   
 
   
 
 
   
 
1
2
0.006
0.00525
u in
u in
 
 
6
2 3
2 10 7,500
x
u R
    3 18,000
x
R
 
Solving the last equation gives:
134

The axial stress-strain relationship is:
1
(1)
2
1 1 x
x
x
d
E
d
L L

 
 
   
 
   
0.006
1 1
30 30 0.00525
E

 
 
   
  
   
750 ( )
psi T

135

The axial stress-strain relationship is:
2
(2)
3
1 1 x
x
x
d
E
d
L L

 
 
   
 
   
0.00525
1 1
30 30 0
E

 
 
   
 
   
5,250 ( )
psi T

136

Comparison of Finite Element Solution to Exact Solution
In order to be able to judge the accuracy of our finite element
models, we will develop an exact solution for the bar element
problem.
The exact solution for the displacement may be obtained by:
where the force P is shown on the following free-body diagram.
0
1
( )
L
u P x dx
AE
 
 
1
( ) (10 )
2
P x x x
 2
5x

137

Therefore:
Applying the boundary conditions:
0
1
( )
L
u P x dx
AE
 
2
1
5
x
o
u x dx
AE
 
3
1
5
( ) 0
3
x
u L C
AE
  
The exact solution for axial displacement is:
 
3 3
5
( )
3
u L x L
AE
 
( )
( )
P x
x
A
 
3
1
5
3
x
C
AE
 
3
1
5
3
L
C
AE
  
2
5x
A

138

A plot of the exact solution for displacement as compared to
several different finite element solutions is shown below.
One element
Two elements
139

A plot of the exact solution for axial stress as compared to
several different finite element solutions is shown below.
One element
Two elements
140

A plot of the exact solution for axial stress at the fixed end
(x = L) as compared to several different finite element
solutions is shown below.
141

Galerkin’s Residual Method and Its Application
to a One-Dimensional Bar
There are a number of weighted residual methods.
However, the Galerkin’s method is more well-known and will be
the only weighted residual method discussed in this course.
In weighted residual methods, a trial or approximate function is
chosen to approximate the independent variable (in our
case, displacement) in a problem defined by a differential
equation.
The trial function will not, in general, satisfy the governing
differential equation.
Therefore, the substitution of the trial function in the differential
equation will create a residual over the entire domain of the
problem.
142

Therefore, the substitution of the trial function in the differential
equation will create a residual over the entire domain of the
problem.
minimum
V
RdV 

In the residual methods, we require that a weighted value of
the residual be a minimum over the entire domain of the
problem.
The weighting function W allows the weighted integral of the
residuals to go to zero.
0
V
RW dV 

143

Using Galerkin’s weighted residual method, we require the
weighting functions to be the interpolation functions Ni.
Therefore:
0 1, 2, ,
i
V
RN dV i n
 
 
144

Example 12 - Bar Element Formulation
Let’s derive the bar element formulation using Galerkin’s
method. The governing differential equation is:
Applying Galerkin’s method we get:
0
d du
AE
dx dx
 

 
 
0
0 1, 2, ,
L
i
d du
AE N dx i n
dx dx
 
 
 
 
 
We now apply integration by parts using the following general
formula:
rds rs sdr
 
 
145

If we assume the following:
then integration by parts gives:
i
r N

d du
ds AE dx
dx dx
 
  
 
0 0
0
L L
i
i
dN
du du
N AE AE dx
dx dx dx
 

0
L
i
d du
AE N dx
dx dx
 

 
 

rds rs sdr
 
 
i
dN
dr dx
dx

du
s AE
dx

146

Recall that:
1 2
1 2
dN dN
du
u u
dx dx dx
 
1
2
1 1 u
du
u
dx L L
 
 
   
 
   
Our original weighted residual expression, with the
approximation for u becomes:
1
2
0
1 1
L
i
u
dN
AE dx
u
dx L L
 
 
  
 
   
 0
L
i
du
N AE
dx

147

Substituting N1 for the weighting function Ni gives:
1
1
2
0
1 1
L
u
dN
AE dx
u
dx L L
 
 
  
 
   

1
2
0
1 1 1
L
u
AE dx
u
L L L
 
   
   
   
     

 
1 2 1x
AE
u u f
L
  
1
0
L
du
N AE
dx 0
x
du
AE
dx 
 0
x x
AE 
 1x
f

0
x x
A 

 
1 2
2
AEL
u u
L
 
1
0
L
du
N AE
dx

148

Substituting N2 for the weighting function Ni gives:
1
2
2
0
1 1
L
u
dN
AE dx
u
dx L L
 
 
  
 
   

1
2
0
1 1 1
L
u
AE dx
u
L L L
 
   
  
   
     

 
1 2 2x
AE
u u f
L
   
2
0
L
du
N AE
dx x L
du
AE
dx 
 x x L
AE 
 2x
f

x x L
A 

 
1 2
2
AEL
u u
L
  
2
0
L
du
N AE
dx

149

Writing the last two equations in matrix form gives:
1 1
2 2
1 1
1 1
x
x
u f
AE
u f
L
    
 

   
 

     
This element formulation is identical to that developed from
equilibrium and the minimum potential energy approach.
150

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