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

Learning Objectives
Development of Beam Equations
To review the basic concepts of beam bending
To derive the stiffness matrix for a beam element
To demonstrate beam analysis using the direct stiffness
method
2
To compare the finite element solution to an exact solution
for a beam
To show how the potential energy method can be used to
derive the beam element equations
To show how the potential energy method can be used to
derive the beam element equations

Development of Beam Equations
In this section, we will develop the stiffness matrix for a beam
element, the most common of all structural elements.
The beam element is considered to be straight and to have
constant cross-sectional area.
3

Development of Beam Equations
We will derive the beam element stiffness matrix by using the
principles of simple beam theory.
The degrees of freedom associated with a node of a beam
element are a transverse displacement and a rotation.
4

Development of Beam Equations
We will discuss procedures for handling distributed loading
and concentrated nodal loading.
We will include the nodal shear forces and bending moments
and the resulting shear force and bending moment diagrams
as part of the total solution.
5

Development of Beam Equations
We will develop the beam bending element equations using
the potential energy approach.
Finally, the Galerkin residual method is applied to derive the
beam element equations
6

Beam Stiffness
Consider the beam element shown below.
The beam is of length L with axial local coordinate x and
transverse local coordinate y.
The local transverse nodal displacements are given by vi and
the rotations by ϕi. The local nodal forces are given by fiy and
the bending moments by mi.
7

3. Forces are positive in the positive y direction.
Beam Stiffness
At all nodes, the following sign conventions are used on the
element level:
1. Moments are positive in the counterclockwise direction.
2. Rotations are positive in the counterclockwise direction.
4. Displacements are positive in the positive y direction.
8

At all nodes, the following sign conventions are used on the
global level:
1. Bending moments m are positive if they cause the beam
to bend concave up.
2. Shear forces V are positive is the cause the beam to
rotate clockwise.
Beam Stiffness
9

Beam Stiffness
(+) Bending Moment
(-) Bending Moment
10

Beam Stiffness
(+) Shear Force
(-) Shear Force
11

Beam Stiffness
The differential equation governing simple linear-elastic beam
behavior can be derived as follows. Consider the beam
shown below.
12

Beam Stiffness
The differential equation governing simple linear-elastic beam
behavior can be derived as follows. Consider the beam
shown below.
Write the equations of equilibrium for the differential element:
0
right side
M

 


0
y
F

 

( )
w x dx
2
dx
 
 
 
  ( )
2
dx
M M dM Vdx w x dx
 
       
 
( ) ( )
V V dV w x dx
    0
13

Beam Stiffness
From force and moment equilibrium of a differential beam
element, we get:
0 0
right side
M Vdx dM
     

0 0
y
F wdx dV
    

d dM
w
dx dx
 
   
 
or
dM
V
dx

or
dV
w
dx
 
14

Beam Stiffness
The curvature  of the beam is related to the moment by:
1 M
EI


 
where  is the radius of the deflected curve, v is the
transverse displacement function in the y direction, E is the
modulus of elasticity, and I is the principle moment of inertia
about y direction, as shown below.
15

Beam Stiffness
The curvature,  for small slopes is given as:
dv
dx
 
2
2
d v
dx
 
Therefore:
2 2
2 2
d v M d v
M EI
dx EI dx
  
Substituting the moment expression into the moment-load
equations gives:
 
2 2
2 2
d d v
EI w x
dx dx
 
 
 
 
For constant values of EI, the above equation reduces to:
 
4
4
d v
EI w x
dx
 
 
 
 
16

Beam Stiffness
Step 1 - Select Element Type
We will consider the linear-elastic beam element shown below.
17

Beam Stiffness
Step 2 - Select a Displacement Function
Assume the transverse displacement function v is:
3 2
1 2 3 4
v a x a x a x a
   
The number of coefficients in the displacement function ai is
equal to the total number of degrees of freedom associated
with the element (displacement and rotation at each node).
The boundary conditions are:
1
( 0)
v x v
 
1
0
x
dv
dx


 2
x L
dv
dx



2
( )
v x L v
 
18

Beam Stiffness
Step 2 - Select a Displacement Function
Applying the boundary conditions and solving for the unknown
coefficients gives:
Solving these equations for a1, a2, a3, and a4 gives:
1 4
(0)
v v a
  3 2
2 1 2 3 4
( )
v L v a L a L a L a
    
1 3
(0)
dv
a
dx

  2
2 1 2 3
( )
3 2
dv L
a L a L a
dx

   
    3
1 2 1 2
3 2
2 1
v v v x
L L
 
 
   
 
 
    2
1 2 1 2 1 1
2
3 1
2
v v x x v
L L
  
 
      
 
 
19

Beam Stiffness
Step 2 - Select a Displacement Function
In matrix form the above equations are:
where
 
[ ]
v N d

   
1
1
1 2 3 4
2
2
[ ]
v
d N N N N N
v


 
 
 
 
 
 
 
 
   
3 2 3 3 2 2 3
1 2
3 3
1 1
2 3 2
N x x L L N x L x L xL
L L
     
   
3 2 3 2 2
3 4
3 3
1 1
2 3
N x x L N x L x L
L L
    
20

Beam Stiffness
Step 2 - Select a Displacement Function
N1, N2, N3, and N4 are called the interpolation functions for a
beam element.
-0.200
0.000
0.200
0.400
0.600
0.800
1.000
0.00 1.00
N1
-0.200
0.000
0.200
0.400
0.600
0.800
1.000
0.00 1.00
N2
-0.200
0.000
0.200
0.400
0.600
0.800
1.000
0.00 1.00
N3
-0.200
0.000
0.200
0.400
0.600
0.800
1.000
0.00 1.00
N4
21

Beam Stiffness
Step 3 - Define the Strain/Displacement
and Stress/Strain Relationships
The stress-displacement relationship is:
We can relate the axial displacement to the transverse
displacement by considering the beam element shown
below:
where u is the axial displacement function.
 
,
x
du
x y
dx
 
22

Beam Stiffness
Step 3 - Define the Strain/Displacement
and Stress/Strain Relationships
dv
u y
dx
 
23

Beam Stiffness
Step 3 - Define the Strain/Displacement
and Stress/Strain Relationships
One of the basic assumptions in simple beam theory is that
planes remain planar after deformation, therefore:
Moments and shears are related to the transverse
displacement as:
 
,
x
du
x y
dx
 
 
2
2
d v
m x EI
dx
 
  
 
 
3
3
d v
V x EI
dx
 
  
 
2
2
d v
y
dx
 
   
 
24

Beam Stiffness
Step 4 - Derive the Element Stiffness Matrix
and Equations
Use beam theory sign convention for shear force and bending
moment.
V+ V+
M+
M+
25

Beam Stiffness
Step 4 - Derive the Element Stiffness Matrix
and Equations
Using beam theory sign convention for shear force and
bending moment we obtain the following equations:
 
3
1 1 1 2 2
3 3
0
12 6 12 6
y
x
d v EI
f V EI v L v L
dx L
 

     
 
3
2 1 1 2 2
3 3
12 6 12 6
y
x L
d v EI
f V EI v L v L
dx L
 

        
 
2
2 2
1 1 1 2 2
2 3
0
6 4 6 2
x
d v EI
m m EI Lv L Lv L
dx L
 

       
 
2
2 2
2 1 1 2 2
2 3
6 2 6 4
x L
d v EI
m m EI Lv L Lv L
dx L
 

     
26

Beam Stiffness
Step 4 - Derive the Element Stiffness Matrix
and Equations
In matrix form the above equations are:
1 1
2 2
1 1
3
2 2
2 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
y
y
f v
L L
m L L L L
EI
f v
L L L
m L L L L



   
 
   
 

   
 

   
 
  
   
 
   

 
   
where the stiffness matrix is:
2 2
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
L L
L L L L
EI
L L L
L L L L

 
 

 

 
  
 

 
k
1 1
1 1
2 2
2 2
y
y
f v
m
k
f v
m


   
   
   

   
   
   
   
27

Beam Stiffness
Step 5 - Assemble the Element Equations
and Introduce Boundary Conditions
Assume the EI to be constant throughout the beam. A force of
1,000 lb and moment of 1,000 lb-ft are applied to the mid-
point of the beam.
Consider a beam modeled by two beam elements
28

Beam Stiffness
Step 5 - Assemble the Element Equations
and Introduce Boundary Conditions
The beam element stiffness matrices are:
1 1 2 2
2 2
(1)
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v v
L L
L L L L
EI
L L L
L L L L
 

 
 

 

 
  
 

 
k
2 2 3 3
2 2
(2)
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v v
L L
L L L L
EI
L L L
L L L L
 

 
 

 

 
  
 

 
k
29

1 1
2 2
1 1
2 2
3 2 2 2 2
2 2
3 3
2 2
3
3
12 6 12 6 0 0
6 4 6 2 0 0
12 6 12 12 6 6 12 6
6 2 6 6 4 4 6 2
0 0 12 6 12 6
0 0 6 2 6 4
y
y
y
F v
L L
M L L L L
F v
L L L L
EI
M L L L L L L L L L
F v
L L
L L L L
M



   

 
   
 

   
 
   
 
     
 

   
 
   
   
 
   
 
  
   
 

     
 
Beam Stiffness
Step 5 - Assemble the Element Equations
and Introduce Boundary Conditions
In this example, the local coordinates coincide with the global
coordinates of the whole beam (therefore there is no
transformation required for this problem).
The total stiffness matrix can be assembled as:
Element 1 Element 2
30

Beam Stiffness
Step 5 - Assemble the Element Equations
and Introduce Boundary Conditions
The boundary conditions are: 1 1 3 0
v v

  
1 1
2 2
1 1
2 2
3 2 2 2 2
2 2
3 3
2 2
3
3
12 6 12 6 0 0
6 4 6 2 0 0
12 6 12 12 6 6 12 6
6 2 6 6 4 4 6 2
0 0 12 6 12 6
0 0 6 2 6 4
y
y
y
F v
L L
M L L L L
F v
L L L L
EI
M L L L L L L L L L
F v
L L
L L L L
M



   

 
   
 

   
 
   
 
     
 

   
 
   
   
 
   
 
  
   
 

     
 
2
2
3
0
0
0
v


 
 
 
 
 
 
 
 
 
31

Beam Stiffness
Step 5 - Assemble the Element Equations
and Introduce Boundary Conditions
By applying the boundary conditions the beam equations
reduce to:
2
2 2
2
3
2 2
3
1,000 24 0 6
1,000 0 8 2
0 6 2 4
lb L v
EI
lbft L L
L
L L L



     
   
 

   
 
   
 
     
32

Beam Stiffness
Step 6 - Solve for the Unknown Degrees of Freedom
Solving the above equations gives:
2 2 3
3 2 2 2
875 375 125 625 125 125
12 4
L L L L L L
v in rad rad
EI EI EI
 
    
  
Step 7 - Solve for the Element Strains and Stresses
 
2
2
d v
m x EI
dx
 
  
 
1
2
1
2
2
2
v
d N
EI
v
dx


 
 
  
  
 
  
 
 
The second derivative of N is linear; therefore m(x) is linear.
33

Beam Stiffness
Step 6 - Solve for the Unknown Degrees of Freedom
Solving the above equations gives:
2 2 3
3 2 2 2
875 375 125 625 125 125
12 4
L L L L L L
v in rad rad
EI EI EI
 
  
     
Step 7 - Solve for the Element Strains and Stresses
 
3
3
d v
V x EI
dx
 
  
 
1
3
1
2
2
2
v
d N
EI
v
dx


 
 
  
  
 
  
 
 
The third derivative of N is a constant; therefore V(x) is
constant.
34

Beam Stiffness
 
2
2
d v
m x EI
dx
 
  
 
1
2
1
2
2
2
v
d N
EI
v
dx


 
 
  
  
 
  
 
 
Step 7 - Solve for the Element Strains and Stresses
 
2 2
1 1 1 2 2
3
6 4 6 2
EI
m Lv L Lv L
L
 
   
 
2 2
2 1 1 2 2
3
6 2 6 4
EI
m Lv L Lv L
L
 
   
Assume L = 120 in, E = 29x106 psi, and I = 100 in4:
Element #1:
3,875lb ft
 
3,562.5lb ft
 
5 4
2 2 3
0.0433 7.758 10 5.586 10
v in rad rad
 
 
      
35

Beam Stiffness
 
2
2
d v
m x EI
dx
 
  
 
1
2
1
2
2
2
v
d N
EI
v
dx


 
 
  
  
 
  
 
 
Step 7 - Solve for the Element Strains and Stresses
 
2 2
2 2 2 3 3
3
6 4 6 2
EI
m Lv L Lv L
L
 
   
 
2 2
3 2 2 3 3
3
6 2 6 4
EI
m Lv L Lv L
L
 
   
Assume L = 120 in, E = 29x106 psi, and I = 100 in4:
Element #2:
2,562.5lb ft
  
0

5 4
2 2 3
0.0433 7.758 10 5.586 10
v in rad rad
 
 
      
36

Beam Stiffness
Step 7 - Solve for the Element Strains and Stresses
Assume L = 120 in, E = 29x106 psi, and I = 100 in4:
Element #1:
743.75lb

743.75lb
 
5 4
2 2 3
0.0433 7.758 10 5.586 10
v in rad rad
 
 
      
 
3
3
d v
V x EI
dx
 
  
 
1
3
1
2
2
2
v
d N
EI
v
dx


 
 
  
  
 
  
 
 
 
1 1 1 2 2
3
12 6 12 6
y
EI
f v L v L
L
 
   
 
2 1 1 2 2
3
12 6 12 6
y
EI
f v L v L
L
 
    
37

Beam Stiffness
Step 7 - Solve for the Element Strains and Stresses
Assume L = 120 in, E = 29x106 psi, and I = 100 in4:
Element #2:
256.25lb
 
256.25lb

5 4
2 2 3
0.0433 7.758 10 5.586 10
v in rad rad
 
 
      
 
3
3
d v
V x EI
dx
 
  
 
1
3
1
2
2
2
v
d N
EI
v
dx


 
 
  
  
 
  
 
 
 
2 2 2 3 3
3
12 6 12 6
y
EI
f v L v L
L
 
   
 
3 2 2 3 3
3
12 6 12 6
y
EI
f v L v L
L
 
    
38

Beam Stiffness
Step 7 - Solve for the Element Strains and Stresses
3,875lb ft
 
3,562.5lb ft

2,562.5lb ft

743.75lb
256.25lb

1,000
F lb
  
1,000
M lb ft
   
39

Beam Stiffness
Example 1 - Beam Problem
Consider the beam shown below. Assume that EI is constant
and the length is 2L (no shear deformation).
The beam element stiffness matrices are:
1 1 2 2
2 2
(1)
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v v
L L
L L L L
EI
L L
L
L L L L
 

 
 

 

 
  
 

 
k
2 2 3 3
2 2
(2)
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v v
L L
L L L L
EI
L L
L
L L L L
 

 
 

 

 
  
 

 
k
40

Beam Stiffness
Example 1 - Beam Problem
The local coordinates coincide with the global coordinates of
the whole beam (therefore there is no transformation required
for this problem).
The total stiffness matrix can be assembled as:
1 1
2 2
1 1
2 2
3 2 2 2
2 2
3 3
2 2
3
3
12 6 12 6 0 0
6 4 6 2 0 0
12 6 24 0 12 6
6 2 0 8 6 2
0 0 12 6 12 6
0 0 6 2 6 4
y
y
y
F v
L L
M L L L L
F v
L L
EI
M L L L L L L
F v
L L
L L L L
M



   

 
   
 

   
 
   
 
  
 

   
 

   
 
   
 
  
   
 

     
 
Element 1 Element 2
41

Beam Stiffness
Example 1 - Beam Problem
The boundary conditions are: 2 3 3 0
v v 
  
1 1
2 2
1 1
2 2
3 2 2 2
2 2
3 3
2 2
3
3
12 6 12 6 0 0
6 4 6 2 0 0
12 6 24 0 12 6
6 2 0 8 6 2
0 0 12 6 12 6
0 0 6 2 6 4
y
y
y
F v
L L
M L L L L
F v
L L
EI
M L L L L L L
F v
L L
L L L L
M



   

 
   
 

   
 
   
 
  
 

   
 

   
 
   
 
  
   
 

     
 
1
1
2
0
0
0
v


 
 
 
 
 
 
 
 
 
42

Beam Stiffness
By applying the boundary conditions the beam equations
reduce to:
1
2 2
1
3
2 2
2
12 6 6
0 6 4 2
0 6 2 8
P L L v
EI
L L L
L
L L L



     
   
 

   
 
   
 
     
Solving the above equations gives:
1 2
1
2
7
3
3
4
1
L
v
PL
EI



 
 
   
   

   
   
   
 
 
43

Beam Stiffness
Example 1 - Beam Problem
The positive signs for the rotations indicate that both are in the
counterclockwise direction.
The negative sign on the displacement indicates a deformation
in the -y direction.
7
1 3
2 2
1
2
2 2 2
2
3
2 2
3
12 6 12 6 0 0
6 4 6 2 0 0 3
12 6 24 0 12 6 0
4 6 2 0 8 6 2 1
0 0 12 6 12 6 0
0 0 6 2 6 4 0
L
y
y
y
F L L
M L L L L
F L L
P
M L L L L L L
F L L
L L L L
M

  
   
     

     
     
  
 

   
 

   
 
   
 
  
   
 

     
 
44

Beam Stiffness
Example 1 - Beam Problem
The local nodal forces for element 1:
7
3
1
2 2
1
2
2 2
2
12 6 12 6
6 4 6 2 3 0
4 12 6 12 6 0
6 2 6 4 1
L
y
y
f L L P
m L L L L
P
f L L L P
m L L L L PL
  
       
       

     
 
 
     
 
  
     
 
     
 
     
 
The local nodal forces for element 2:
2
2 2
2
3
2 2
3
12 6 12 6 0 1.5
6 4 6 2 1
4 12 6 12 6 0 1.5
6 2 6 4 0 0.5
y
y
f L L P
m L L L L PL
P
f L L L P
m L L L L PL

       
       

     
 
 
     
 
   
     
 
     

     
 
45

Beam Stiffness
Example 1 - Beam Problem
The free-body diagrams for the each element are shown
below.
Combining the elements gives the forces and moments for the
original beam.
46

Beam Stiffness
Example 1 - Beam Problem
Therefore, the shear force and bending moment diagrams are:
47

Beam Stiffness
Example 2 - Beam Problem
Consider the beam shown below. Assume E = 30 x 106 psi and
I = 500 in4 are constant throughout the beam. Use four
elements of equal length to model the beam.
The beam element stiffness matrices are:
2 2
( )
3
2 2
( 1) ( 1)
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
i
v v
i i i i
L L
L L L L
EI
L L
L
L L L L
 
 

 
 

 

 
  
 

 
k
48

Beam Stiffness
Example 2 - Beam Problem
Using the direct stiffness method, the four beam element
stiffness matrices are superimposed to produce the global
stiffness matrix.
Element 1 Element 2
Element 3
Element 4
49

Beam Stiffness
Example 2 - Beam Problem
The boundary conditions for this problem are:
1 1 3 5 5 0
v v v
 
    
50

Beam Stiffness
Example 2 - Beam Problem
The boundary conditions for this problem are:
After applying the boundary conditions the global beam
equations reduce to:
1 1 3 5 5 0
v v v
 
    
2
2 2
2
2 2 2
3
3
4
2 2
4
24 0 6 0 0
0 8 2 0 0
6 2 8 6 2
0 0 6 24 0
0 0 2 0 8
v
L
L L
EI
L L L L L
L
v
L
L L



 
 
 
 
 
   
 
  
   

   
   
   
10,000
0
0
10,000
0
lb
lb

 
 
 
 
  
 

 
 
 
51

Beam Stiffness
Example 2 - Beam Problem
Substituting L = 120 in, E = 30 x 106 psi, and I = 500 in4 into
the above equations and solving for the unknowns gives:
The global forces and moments can be determined as:
2 4 2 3 4
0.048 0
v v in   
     
1 1
2 2
3 3
4 4
5 5
5 25 ꞏ
10 0
10 0
10 0
5 25 ꞏ
y
y
y
y
y
F kips M kips ft
F kips M
F kips M
F kips M
F kips M kips ft
 
  
 
  
  
52

Beam Stiffness
Example 2 - Beam Problem
The local nodal forces for element 1:
The local nodal forces for element 2:
1
2 2
1
3
2
2 2
2
12 6 12 6 0
6 4 6 2 0
12 6 12 6 0.048
6 2 6 4 0
y
y
f L L
m L L L L
EI
f L L L
m L L L L

     
     

   
 

   
 
   
   
 
   

   
 
2
2 2
2
3
3
2 2
3
12 6 12 6 0.048
6 4 6 2 0
12 6 12 6 0
6 2 6 4 0
y
y
f L L
m L L L L
EI
f L L L
m L L L L
 
     
     

   
 

   
 
  
   
 
   

   
 
5
25 ꞏ
5
25 ꞏ
kips
k ft
kips
k ft
 
 
 
  

 
 
 
5
25 ꞏ
5
25 ꞏ
kips
k ft
kips
k ft

 
 

 
  
 
 

 
53

Beam Stiffness
Example 2 - Beam Problem
The local nodal forces for element 3:
The local nodal forces for element 4:
3
2 2
3
3
4
2 2
4
12 6 12 6 0
6 4 6 2 0
12 6 12 6 0.048
6 2 6 4 0
y
y
f L L
m L L L L
EI
f L L L
m L L L L

     
     

   
 

   
 
   
   
 
   

   
 
4
2 2
4
3
5
2 2
5
12 6 12 6 0.048
6 4 6 2 0
12 6 12 6 0
6 2 6 4 0
y
y
f L L
m L L L L
EI
f L L L
m L L L L
 
     
     

   
 

   
 
  
   
 
   

   
 
5
25 ꞏ
5
25 ꞏ
kips
k ft
kips
k ft
 
 
 
  

 
 
 
5
25 ꞏ
5
25 ꞏ
kips
k ft
kips
k ft

 
 

 
  
 
 

 
54

Beam Stiffness
Example 2 - Beam Problem
Note: Due to symmetry about the vertical plane at node 3, we
could have worked just half the beam, as shown below.
Line of symmetry
55

Beam Stiffness
Example 3 - Beam Problem
Consider the beam shown below. Assume E = 210 GPa and
I = 2 x 10-4 m4 are constant throughout the beam and the
spring constant k = 200 kN/m. Use two beam elements of
equal length and one spring element to model the structure.
56

Beam Stiffness
Example 3 - Beam Problem
The beam element stiffness matrices are:
2 2
(1)
3
2 2
1 1 2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v v
L L
L L L L
EI
L L
L
L L L L
 

 
 

 

 
  
 

 
k
3
2 2
(2)
3
2 2
2 2 3
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v v
L L
L L L L
EI
L L
L
L L L L
 

 
 

 

 
  
 

 
k
The spring element stiffness matrix is:
3 4
(3)
v v
k k
k k

 
  

 
k
3 3 4
(3)
0
0 0 0
0
v v
k k
k k


 
 
 
 
 

 
k
57

Beam Stiffness
Example 3 - Beam Problem
Using the direct stiffness method and superposition gives the
global beam equations.
1 1
2 2
1 1
2 2
2 2 2
2 2
3
3 3
2 2
3 3
4 4
12 6 12 6 0 0 0
6 4 6 2 0 0 0
12 6 24 0 12 6 0
6 2 0 8 6 2 0
0 0 12 6 12 ' 6 '
0 0 6 2 6 4 0
0 0 0 0 ' 0 '
y
y
y
y
F v
L L
M L L L L
F v
L L
EI
M L L L L L
L
F v
L k L k
M L L L L
F v
k k



    
 
   
 

   
 
   
 
  
   
 
 
   
 
   
 
    
  
 

  
 
  
 

   
 
3
'
kL
k
EI




Element 1
Element 2
Element 3
58

Beam Stiffness
Example 3 - Beam Problem
The boundary conditions for this problem are: 1 1 2 4 0
v v v

   
1 1
2 2
1 1
2 2
2 2 2
2 2
3
3 3
2 2
3 3
4 4
12 6 12 6 0 0 0
6 4 6 2 0 0 0
12 6 24 0 12 6 0
6 2 0 8 6 2 0
0 0 12 6 12 ' 6 '
0 0 6 2 6 4 0
0 0 0 0 ' 0 '
y
y
y
y
F v
L L
M L L L L
F v
L L
EI
M L L L L L
L
F v
L k L k
M L L L L
F v
k k



    
 
   
 

   
 
   
 
  
   
 
 
   
 
   
 
    
  
 

  
 
  
 

   
 
3
'
kL
k
EI




2
3
3
0
0
0
0
v


 
 
 
 
 
 
 
 
 
 
 
59

Beam Stiffness
Example 3 - Beam Problem
After applying the boundary conditions the global beam
equations reduce to:
2 2
2 2
3 3
3
2 2
3 3
8 6 2 0
6 12 ' 6
2 6 4 0
y
M L L L
EI
F L k L v P
L
M L L L


 

     
 
     
     
     
 
     
 

     
 
Solving the above
equations gives:
2
2 3 3
3
3 2
3 1
12 7 '
7 1
'
12 7 '
9 1
12 7 '
PL
EI k
PL kL
v k
EI k EI
PL
EI k


 
 

 
 

 
 
 
 
   
 
  
   
 

 
   
   
 

 
 

 
 
 
60

Beam Stiffness
Example 3 - Beam Problem
Substituting L = 3 m, E = 210 GPa, I = 2 x 10-4 m4, and
k = 200 kN/m in the above equations gives:
Substituting the solution back into the global equations gives:
3
2
3
0.0174
0.00249
0.00747
v m
rad
rad


 
 
 
1
2 2
1
2
2 2 2
2 3
3
2 2
3
4
12 6 12 6 0 0 0 0
6 4 6 2 0 0 0 0
12 6 24 0 12 6 0 0
6 2 0 8 6 2 0 0.00249
0 0 12 6 12 ' 6 ' 0.0174
0 0 6 2 6 4 0 0.00747
0 0 0 0 ' 0 '
y
y
y
y
F L L
M L L L L
F L L
EI
M L L L L L rad
L
F L k L k m
M L L L L rad
F k k
  
 
   

   
   
  
   
  
   
   
     
   
 
   
   

 
  0
 
 
 
 
 
 
 
 
 
 
 
61

Beam Stiffness
Example 3 - Beam Problem
Substituting L = 3 m, E = 210 GPa, I = 2 x 10-4 m4, and
k = 200 kN/m in the above equations gives:
Substituting the solution back into the global equations gives:
3
2
3
0.0174
0.00249
0.00747
v m
rad
rad


 
 
 
1
1
2
2
3
3
4
69.9
69.7
116.4
0
50
0
3.5
y
y
y
y
F kN
M kN m
F kN
M
F kN
M
F kN
  
 
   
 
   
   
   

   
   

   
   
   
 
 
62

Beam Stiffness
Example 3 - Beam Problem
The variation of shear force and bending moment is:
69.7kNm
139.5kNm

69.9kN

46.5 kN
63

Beam Stiffness
Distributed Loadings
Beam members can support distributed loading as well as
concentrated nodal loading.
Therefore, we must be able to account for distributed loading.
Consider the fixed-fixed beam subjected to a uniformly
distributed loading w shown the figure below.
The reactions, determined from structural analysis theory, are
called fixed-end reactions.
64

Beam Stiffness
Distributed Loadings
In general, fixed-end reactions are those reactions at the ends
of an element if the ends of the element are assumed to be
fixed (displacements and rotations are zero).
Therefore, guided by the results from structural analysis for the
case of a uniformly distributed load, we replace the load by
concentrated nodal forces and moments tending to have the
same effect on the beam as the actual distributed load.
65

Beam Stiffness
Distributed Loadings
The figures below illustrates the idea of equivalent nodal loads
for a general beam. We can replace the effects of a uniform
load by a set of nodal forces and moments.
66

Beam Stiffness
General Formulation
We can account for the distributed loads or concentrated loads
acting on beam elements by considering the following
formulation for a general structure:
where F0 are the equivalent nodal forces, expressed in
terms of the global-coordinate components.
These forces would yield the same displacements as the
original distributed load.
If we assume that the global nodal forces are not present
(F = 0) then:
0
F = Kd- F
0
F = Kd
67

Beam Stiffness
General Formulation
We now solve for the displacements, d, given the nodal
forces F0.
Next, substitute the displacements and the equivalent nodal
forces F0 back into the original expression and solve for the
global nodal forces.
0
F = Kd- F
This concept can be applied on a local basis to obtain the local
nodal forces in individual elements of structures as:
0
f = kd- f
68

Beam Stiffness
Example 5 - Load Replacement
Consider the beam shown below; determine the equivalent
nodal forces for the given distributed load.
The work equivalent nodal forces are shown above.
Using the beam stiffness equations:
2
2
1 1
2 2
1 1
3
2 2
2 2
2 2
2
12
2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
y
y
wL
wL
wL
wL
f v
L L
m L L L L
EI
f v
L L L
m L L L L





 
 
 

   
   
   
   

   
 
 
     
 
  
     
 
     

 
   
 
  69

Beam Stiffness
Example 5 - Load Replacement
Apply the boundary conditions:
2
3 2
2
2
12 6
2
6 4
12
wL
v
L
EI
L L L
wL 
 

    
 
 

   
 

   
 
 
 
4
2
3
2
8
6
wL
v EI
wL
EI

 

 
   

   
   

 
 
We can solve for the displacements
1
2 2
1
3
2
2 2
2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v
L L
L L L L
EI
v
L L L
L L L L


  
 
 
 
  
   
 
    
   

   
1 1 0
v 
 
70

Beam Stiffness
Example 5 - Load Replacement
In this case, the method of equivalent nodal forces gives the
exact solution for the displacements and rotations.
To obtain the global nodal forces, we will first define the
product of Kd to be Fe, where Fe is called the effective
global nodal forces. Therefore:
4
3
1
2 2
1
3
8
2
2 2
6
2
12 6 12 6 0
6 4 6 2 0
12 6 12 6
6 2 6 4
e
y
e
e wL
EI
y
e wL
EI
L L
F
L L L L
M EI
L L L
F
L L L L
M
  
   
     

   
 

   
 
   
   
 
   
 
   
 
2
2
2
5
12
2
12
wL
wL
wL
wL
 
 
 
 
 
  
 

 
 
 
  71

Beam Stiffness
Example 5 - Load Replacement
Using the above expression and the fix-end moments in:
0
F = Kd- F
2
2
0
0
wL
wL
 
 
 
 
 
 
  
 
 
 
 
 
 
2 2
1
1
2
2
2 2
2 2
5
12 12
2 2
12 12
y
y
wL wL
F wL wL
M
F wL wL
M
wL wL
   

   
   
     

     
 
 
     
     
 
     
 
   
   
   
0
y
F 

1 0
M




1y
F wL
 
 
2
2 2
wL L
wL
  72

Beam Stiffness
Example 6 - Cantilever Beam
Consider the beam, shown below, determine the vertical
displacement and rotation at the free-end and the nodal
forces, including reactions. Assume EI is constant
throughout the beam.
We will use one element and replace the concentrated load
with the appropriate nodal forces.
73

Beam Stiffness
Example 6 - Cantilever Beam
The beam stiffness equations are:
1 1
2 2
1 1
3
2 2
2 2
2 2
2
8
2
8
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
y
y
P
PL
P
PL
f v
L L
m L L L L
EI
f v
L L L
m L L L L





 
 
 

   
 
 
   
 
  
   
 
 
     
 
  
     
 
     

 
   
 
 
 
Apply the boundary conditions:
1
2 2
1
3
2
2 2
2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v
L L
L L L L
EI
v
L L L
L L L L


  
 
 
 
  
   
 
    
   

   
1 1 0
v 
 
74

Beam Stiffness
Example 6 - Cantilever Beam
The beam stiffness equations become:
To obtain the global nodal forces, we begin by evaluating the
effective nodal forces.
2
3 2
2
12 6
2
6 4
8
P
v
L
EI
PL L L L 
 

    
 
 

   
 

   
 
 
 
3
2
2
2
5
48
8
PL
v EI
PL
EI

 

 
   

   
   

 
 
3
2
1
2 2
1
3 5
48
2
2 2
8
2
12 6 12 6 0
6 4 6 2 0
12 6 12 6
6 2 6 4
e
y
e
e PL
EI
y
e PL
EI
L L
F
L L L L
M EI
L L L
F
L L L L
M
  
   
     

   
 

   
 
   
   
 
   
 
   
 
2
3
8
2
8
P
PL
P
PL
 
 
 
 
 
  
 

 
 
 
 
75

Beam Stiffness
Example 6 - Cantilever Beam
Using the above expression in the following equation, gives:
0
F = Kd- F
2
3
8
2
8
P
PL
P
PL
 
 
 
 
 
  
 

 
 
 
 
e
Kd = F 0
F F
2
0
0
P
PL
 
 
 
 
 
  
 
 
 
 
 
2
8
2
8
P
PL
P
PL
 

 
 
 

 
 
 

 
 
 
 
1
1
2
2
y
y
F
M
F
M
 
 
 
 
 
 
 
F
76

Beam Stiffness
Example 6 - Cantilever Beam
In general, for any structure in which an equivalent nodal
force replacement is made, the actual nodal forces acting on
the structure are determined by first evaluating the effective
nodal forces Fe for the structure and then subtracting off the
equivalent nodal forces F0 for the structure.
Similarly, for any element of a structure in which equivalent
nodal force replacement is made, the actual local nodal
forces acting on the element are determined by first
evaluating the effective local nodal forces for the element
and then subtracting off the equivalent local nodal forces
associated only with the element.
0
f = kd- f
77

Beam Stiffness
Comparison of FE Solution to Exact Solution
We will now compare the finite element solution to the exact
classical beam theory solution for the cantilever beam
shown below.
Both one- and two-element finite element solutions will be
presented and compared to the exact solution obtained by
the direct double-integration method.
Let E = 30 x 106 psi, I = 100 in4, L = 100 in, and uniform load
w = 20 Ib/in.
78

Beam Stiffness
Comparison of FE Solution to Exact Solution
To obtain the solution from classical beam theory, we use the
double-integration method:
( )
M x
y
EI
 
   
 
where the double prime superscript indicates differentiation
twice with respect to x and M is expressed as a function of x
by using a section of the beam as shown:
0
M




0
y
F

 

2 2
( )
2 2
wx wL
M x wLx
    
( )
V x wL wx
  
79

Beam Stiffness
Comparison of FE Solution to Exact Solution
To obtain the solution from classical beam theory, we use the
double-integration method:
( )
M x
y
EI
 
   
 
2 2
2 2
w x L
y Lx dx dx
EI
 
   
 
 

3 2 2
1
6 2 2
w x Lx xL
y C dx
EI
 
    
 
 

4 3 2 2
1 2
24 6 4
w x Lx x L
y C x C
EI
 
     
 
 
4 3 2 2
24 6 4
w x Lx x L
y
EI
 
   
 
 
(0) 0 (0) 0
y y
  
Boundary Conditions
80

Beam Stiffness
Comparison of FE Solution to Exact Solution
Recall the one-element solution to the cantilever beam is:
4
2
3
2
8
6
wL
v EI
wL
EI

 

 
   

   
   

 
 
Using the numerical values for this problem we get:
 
 
 
 
4
6 4
2
3
2
6 4
20 100
8 30 10 100
20 100
6 30 10 100
lb
in
lb
in
in
psi in
v
in
psi in

 

 

 
   

   
   

 

 
 
0.0833
0.00111
in
rad

 
  

 
81

Beam Stiffness
Comparison of FE Solution to Exact Solution
The slope and displacement from the one-element FE
solution identically match the beam theory values evaluated
at x = L.
The reason why these nodal values from the FE solution are
correct is that the element nodal forces were calculated on
the basis of being energy or work equivalent to the
distributed load based on the assumed cubic displacement
field within each beam element.
82

Beam Stiffness
Comparison of FE Solution to Exact Solution
Values of displacement and slope at other locations along the
beam for the FE are obtained by using the assumed cubic
displacement function.
   
3 2 3 2 2
2 2
3 3
1 1
( ) 2 3
v x x x L v x L x L
L L

    
The value of the displacement at the midlength v(x = 50 in) is:
( 50 ) 0.0278
v x in in
  
Using beam theory, the displacement at v(x = 50 in) is:
( 50 ) 0.0295
v x in in
  
83

Beam Stiffness
Comparison of FE Solution to Exact Solution
In general, the displacements evaluated by the FE method
using the cubic function for v are lower than by those of
beam theory except at the nodes.
This is always true for beams subjected to some form of
distributed load that are modeled using the cubic
displacement function.
The exception to this result is at the nodes, where the beam
theory and FE results are identical because of the work-
equivalence concept used to replace the distributed load by
work-equivalent discrete loads at the nodes.
84

Beam Stiffness
Comparison of FE Solution to Exact Solution
The beam theory solution predicts a quartic (fourth-order)
polynomial expression for a beam subjected to uniformly
distributed loading, while the FE solution v(x) assumes a
cubic (third-order) displacement behavior in each beam all
load conditions.
The FE solution predicts a stiffer structure than the actual one.
This is expected, as the FE model forces the beam into
specific modes of displacement and effectively yields a
stiffer model than the actual structure.
However, as more elements are used in the model, the FE
solution converges to the beam theory solution.
85

Beam Stiffness
Comparison of FE Solution to Exact Solution
For the special case of a beam subjected to only nodal
concentrated loads, the beam theory predicts a cubic
displacement behavior.
The FE solution for displacement matches the beam theory
solution for all locations along the beam length, as both v(x)
and y(x) are cubic functions.
86

Beam Stiffness
Comparison of FE Solution to Exact Solution
Under uniformly distributed loading, the beam theory solution
predicts a quadratic moment and a linear shear force in the
beam.
However, the FE solution using the cubic displacement
function predicts a linear bending moment and a constant
shear force within each beam element used in the model.
87

Beam Stiffness
Comparison of FE Solution to Exact Solution
We will now determine the bending moment and shear force
in the present problem based on the FE method.
( )
M x EIy

 
2
2
d
EI
dx

Nd  
2
2
d
EI
dx

N d
  
( )
M x EI
 B d
  2 3 2 2 3 2
6 12 4 6 6 12 2 6
x x x x
L L L L L L L L
 
       
       
       
 
       
 
B
1 1
2 3 2
6 12 4 6
x x
M EI v
L L L L

   
     
   

   

2 2
2 3 2
6 12 2 6
x x
v
L L L L


   
    
    
     88

Beam Stiffness
Comparison of FE Solution to Exact Solution
We will now determine the bending moment and shear force
in the present problem based on the FE method.
Position MFE Mexact
X = 0 -83,333 lb-in -100,000 lb-in
X = 50 in -33,333 lb-in -25,000 lb-in
X = 100 in 16,667 lb-in 0
89

Beam Stiffness
Comparison of FE Solution to Exact Solution
The plots below show the displacement, bending moment ,
and shear force over the beam using beam theory and the
one-element FE solutions.
90

Beam Stiffness
Comparison of FE Solution to Exact Solution
The FE solution for displacement matches the beam theory
solution at the nodes but predicts smaller displacements
(less deflection) at other locations along the beam length.
91

Beam Stiffness
Comparison of FE Solution to Exact Solution
The bending moment is derived by taking two derivatives on
the displacement function. It then takes more elements to
model the second derivative of the displacement function.
92

Beam Stiffness
Comparison of FE Solution to Exact Solution
The shear force is derived by taking three derivatives on the
displacement function. For the uniformly loaded beam, the
shear force is a constant throughout the singIe-element
model.
93

Beam Stiffness
Comparison of FE Solution to Exact Solution
To improve the FE solution we need to use more elements in
the model (refine the mesh) or use a higher-order element,
such as a fifth-order approximation for the displacement
function.
94

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
Let’s derive the equations for a beam element using the
principle of minimum potential energy.
The procedure for applying the principle of minimum potential
energy is similar to that used for the bar element.
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
   
95

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
The differential internal work (strain energy) dU in a one-
dimensional beam element is: 1
2 x x
V
U dV
 
 
For a single beam element, shown below, subjected to both
distributed and concentrated nodal forces, the potential
energy due to forces (or the work done these forces) is:
2 2
1 1
y iy i i i
i i
S
T v dS P v m
 
    
 

96

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
If the beam element has a constant cross-sectional area A,
then the differential volume of the beam is given as:
The differential element where the surface loading acts is
given as: dS = b dx (where b is the width of the beam
element).
dV dA dx

Therefore the total potential energy is:
 
2
1
2
1
p x x y iy i i i
i
x A x
dA dx T vb dx P v m
   

   

 
97

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
The strain-displacement relationship is:
We can express the strain in terms of nodal displacements
and rotations as:
2
2
x
d v
y
dx
  
 
2 2
3 3 3 3
12 6 6 4 12 6 6 2
x
x L xL L x L xL L
y d
L L L L

 
    
   
 
   
[ ]
x y B d
  
2 2
3 3 3 3
12 6 6 4 12 6 6 2
[ ]
x L xL L x L xL L
B
L L L L
 
    
  
 
98

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
The stress-strain relationship in one-dimension is:
where E is the modulus of elasticity. Therefore:
   
[ ]
x x
E
 

   
[ ][ ]
x y E B d
  
The total potential energy can be written in matrix form as:
         
1
2
T T T
p x x y
x A x
dA dx bT v dx d P
  
  
 
   
[ ]
x y B d
  
99

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
Use the following definition for moment of inertia:
If we define, as a line load (load per unit length) in the
y direction and the substitute the definitions of x and x the
total potential energy can be written in matrix form as:
y
w bT

         
2
0 0
[ ] [ ] [ ]
2
L L
T T T
T T
p
A
E
y d B B d dAdx w d N dx d P
   
 
2
A
I y dA
 
Then the total potential energy expression becomes:
         
0 0
[ ] [ ] [ ]
2
L L
T T T
T T
p
EI
d B B d dx w d N dx d P
   
 
100

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
The nodal forces vector is:
The elemental stiffness matrix is:
Differentiating the total potential energy with respect to the
displacement and rotations (v1, v2, 1 and 2) and equating
each term to zero gives:
   
0 0
[ ] [ ] [ ] 0
L L
T T
EI B B dx d w N dx P
  
 
   
0
[ ]
L
T
f w N dx P
 

 
0
[ ] [ ]
L
T
k EI B B dx
 
101

Beam Stiffness
Potential Energy Approach to Derive
Beam Element Equations
Integrating the previous matrix expression gives:
 
2 2
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
L L
L L L L
EI
k
L L L
L L L L

 
 

 

 
  
 

 
 
0
[ ] [ ]
L
T
k EI B B dx
 
102

Beam Stiffness
Galerkin’s Method to Derive Beam Element Equations
The governing differential equation for a one-dimensional
beam is:
4
4
0
d v
EI w
dx
 
 
 
 
We can define the residual R as:
4
4
0
0 1, 2, 3, and 4
L
i
d v
EI w N dx i
dx
 
 
  
 
 
 
 

If we apply integration by parts twice to the first term we get:
         
, , 0
0 0
L L
L
xxxx i xx i xx i xxx i x xx
EI v N dx EI v N dx EI N v N v
 
  
 
 
where the subscript x indicates a derivative with respect to x 103

Beam Stiffness
Galerkin’s Method to Derive Beam Element Equations
or
Therefore the integration by parts becomes:
Since , then the second derivative of v with respect
to x is:
 
[ ]
v N d

 
2 2
3 3 3 3
12 6 6 4 12 6 6 2
xx
x L xL L x L xL L
v d
L L L L
 
    
  
 
 
[ ]
x x
v B d

       
, ,
0
0 0
[ ] 0
L L
L
i xx i i i x
N EI B dx d N w dx NV N m d
 
   
 
 
1, 2, 3, and 4
i 
104

Beam Stiffness
Galerkin’s Method to Derive Beam Element Equations
The interpolation function in the last two terms can be
evaluated:
The above expression is really four equations (one for each
Ni) and can be written in matrix form as:
  0
0 0
[ ] [ ] [ ] [ ] [ ]
L L
L
T T T T
x
B EI B dx d N w dx N m N V
   
 
   
[ ] 0 [0 1 0 0] [ ] [0 0 0 1]
x x
N x N x L
   
   
[ ] 0 [1 0 0 0] [ ] [0 0 1 0]
N x N x L
   
105

Beam Stiffness
Galerkin’s Method to Derive Beam Element Equations
Therefore, the last two terms of the matrix form of the Galerkin
formulation become (see the figure below):
1 (0) 2 (0)
i V i m
   
3 ( ) 4 ( )
i V L i m L
   
106

Beam Stiffness
Galerkin’s Method to Derive Beam Element Equations
When beam elements are assembled, as shown below:
Two shear forces and two moments form adjoining elements
contribute to the concentrated force and the concentrated
moment at the node common to both elements.
107