CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
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.
The Stiffness (Displacement) Method
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 linearelastic 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.
1/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness (Displacement) Method
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
The Stiffness (Displacement) Method
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 axis.
f1x
Note: Assumed sign conventions
f2x
2/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness (Displacement) Method
2. Select a Displacement Function - A displacement
function is assumed.
u  a1  a2 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:
a 
u  1 x   1 
a2 
The Stiffness (Displacement) Method
We can express u as a function of the nodal displacements
ui by evaluating u at each node and solving for a1 and a2.


 Boundary Conditions
u( x  L )  u2  a2L  a1 

Solving for a2:
u( x  0)  u1  a1
a2 
u2  u1
L
Substituting a1 and a2 into u gives:
 u  u1 
u  2
 x  u1
 L 
 x
x
  1   u1    u2
 L
L
3/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
4/32
The Stiffness (Displacement) Method
In matrix form:
Or in another form:
x   u1 
 
L  u2 

x
u   1  
 L 
u  N1
u 
N2   1 
u2 
Where N1 and N2 are defined as:
x
N1  1 
L
N2 
x
L
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.
The Stiffness (Displacement) Method
u  a1  a2 x

x
u   1  
 L 
u 
N2   1 
u2 
u  N1
N1  1 
N2 
x
L
x   u1 
 
L  u2 
x
L
N1  N2  1
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
5/32
The Stiffness (Displacement) Method
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:
f1x
f2 x
T
T
T  k
where deformation of the spring  is given as:
  u(L )  u(0)
  u2  u1
f1x  T
f2 x  T
The Stiffness (Displacement) Method
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:
T  f1x  k  u2  u1   f1x  k  u1  u2 
T  f2 x  k  u2  u1 
 f2 x  k  u1  u2 
We can write the last two force-displacement relationships
in matrix form as:
 f1x   k k   u1 
 
 
f2 x   k k  u2 
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness (Displacement) Method
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.
N
K  K    k
N
F  F    f ( e )
(e )
e 1
e 1
where k and f are the element stiffness and force matrices
expressed in global coordinates.
The Stiffness (Displacement) Method
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
 F  Kd
7. Step 7 - Solve for the Element Forces
Once the displacements are found, the forces in each
element may be calculated from:
T  k  k  u2  u1 
6/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
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:
 f1x   k1 k1   u1 
 
 
f3 x   k1 k1  u3 
For element 2:
f3 x   k 2 k 2  u3 
 
 
f2 x   k 2 k 2  u2 
The Stiffness Method – Spring Example 1
Both continuity and compatibility require that both elements
remain connected at node 3.
Element number
u3(1)  u3(2)
We can write the nodal equilibrium equation at each node as:
F1x  f1x (1)
F3 x  f3 x (1)  f3 x (2)
F2 x  f2 x (2)
7/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
8/32
The Stiffness Method – Spring Example 1
Therefore the force-displacement equations for this spring
system are:
F1x  k1u1  k1u3
Element 1
F2 x  k 2u3  k2u2
Element 2
F3 x   k1u1  k1u3    k2u3  k 2u2 
In matrix form the above equations are:
0
 F1x   k1
  
k2
F2 x    0
F   k k
2
 3x   1
k1   u1 
 
k 2  u2 

k1  k 2  u3 
F  Kd
where F is the global nodal force vector, d is called the
global nodal displacement vector, and K is called the
global stiffness matrix.
The Stiffness Method – Spring Example 1
Assembling the Total Stiffness Matrix by Superposition
Consider the spring system defined in the last example:
The elemental stiffness matrices may be written for each
element.
For element 1:
For element 2:
u1
u3
 k k1  u1
k (1)   1

 k1 k1  u3
u3
u2
 k  k 2  u3
k (2)   2

 k 2 k 2  u2
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 1
Write the stiffness matrix in global format for element 1 as
follows:
(1)
 1 0 1  u1  f1x 
   
k1  0 0 0  u2   f2(1)x 


   (1) 
 1 0 1  u3  f3 x 
For element 2:
(1)
0 0 0   u1  f1x 
   
k 2 0 1 1 u2   f2(1)x 


0 1 1  u3  f3(1)x 
The Stiffness Method – Spring Example 1
Apply the force equilibrium equations at each node.
f1(1)
  0   F1x 
x
   (2)   
 0   f2 x   F2 x 
f (1)  f (2)  F 
 3x   3x   3x 
The above equations give:
0
 k1
 0
k2

 k1 k 2
k1   u1   F1x 
   
k 2  u2   F2 x 

k1  k2  u3  F3 x 
9/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 1
To avoid the expansion of the each elemental stiffness
matrix, we can use a more direct, shortcut form of the
stiffness matrix.
u1 u3
u3 u 2
 k k1  u1
k (1)   1

 k1 k1  u3
 k  k 2  u3
k (2)   2

  k 2 k 2  u2
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:
u
u
u
1
2
 k1 0
K   0 k2

 k1 k2
3
k1 
k 2 

k1  k2 
The Stiffness Method – Spring Example 1
Boundary Conditions
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 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.
10/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 1
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:
0
k1   0   F1x 
 k1
   
 0
k2
k 2  u2   F2 x 


 k1 k 2 k1  k 2  u3  F3 x 
Expanding the matrix equations gives:
F1x  k1u3
F2 x  k2u3  k2u2
F3 x  k 2u2   k1  k 2  u3
The Stiffness Method – Spring Example 1
The second and third equation may be written in matrix form
as:
k 2  u2  F2 x 
 k2
 k k  k  u   F 
2 3
 2 1
 3x 
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.
F1x  k1u3
For homogeneous boundary conditions, we can delete the
row and column corresponding to the zero-displacement
degrees-of-freedom.
11/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 1
Let’s again look at the equations we developed for the twospring 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:
0
k1      F1x 
 k1
   
 0
k2
k 2  u2   F2 x 


   
 k1 k 2 k1  k 2  u3  F3 x 
Expanding the matrix equations gives:
F1x  k1  k1u3
F2 x  k2u3  k2u2
F3 x  k1  k1u3  k2u3  k2u2
The Stiffness Method – Spring Example 1
By considering the second and third equations because
they have known nodal forces we get:
F2 x  k2u3  k2u2
F3 x  k1  k1u3  k2u3  k2u2
In matrix form the above equations are:
k 2  u2   F2 x 
 k2
 k k  k  u   F  k  
2 3
1 
 2 1
 3x
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.
12/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
13/32
The Stiffness Method – Spring Example 1
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.
F1x  k1  k1u3
The Stiffness Method – Spring Example 2
Consider the following three-spring system:
The elemental stiffness matrices for each element are:
1
3
3
 1 1
k (1)  1000 

 1 1
4
1
3
2
 1 1
k (3)  3000 

 1 1
4
2
4
 1 1
k (2)  2000 

 1 1
3
4
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 2
Using the concept of superposition (the direct stiffness
method), the global stiffness matrix is:
0
1000
0 
 1000
 0
3000 
3000
0

K
 1000
0
3000 2000 


3000 2000 5000 
 0
The global force-displacement equations are:
0
1000
0   u1   F1x 
 1000
 0
3000  u2  F2 x 
3000
0

    
 1000
0
3000 2000  u3  F3 x 


3000 2000 5000  u4  F4 x 
 0
The Stiffness Method – Spring Example 2
We have homogeneous boundary conditions at
nodes 1 and 2 (u1 = 0 and u2 = 0).
The global force-displacement equations reduce to:
0
1000
0   u1   F1x 
 1000
 0
3000  u2  F2 x 
3000
0

    
 1000
0
3000 2000  u3  F3 x 


3000 2000 5000  u4  F4 x 
 0
14/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 2
Substituting for the known force at node 4 (F4x = 5,000 lb)
gives:


 3000 2000  u3    0 
 2000 5000  u  5,000 

  4 
Solving for u3 and u4 gives:
u3 
10
in
11
u4 
15
in
11
The Stiffness Method – Spring Example 2
To obtain the global forces, substitute the displacement in
the force-displacement equations.
0
1000
0  0 
 F1x   1000
F   0
3000
0
3000   0 
 2x  
 

 
F



 10 11
1000
0
3000
2000
3
x
 


F4 x   0
3000 2000 5000  15 11
Solving for the forces gives:
10,000
45,000
F1x  
lb
F2 x  
lb
11
11
F3 x  0
F4 x 
55,000
lb
11
15/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
16/32
The Stiffness Method – Spring Example 2
Next, use the local element equations to obtain the force in
each spring.
For element 1:  f   1000 1000   0 
1x
 
 10 
f3 x   1000 1000   11
The local forces are:
f1x  
10,000
lb
11
f3 x 
10,000
lb
11
A free-body diagram of the spring element 1 is shown below.
The Stiffness Method – Spring Example 2
Next, use the local element equations to obtain the force in
each spring.
For element 2: f3 x   2000 2000  10 11
 
 
f4 x   2000 2000  15 11
The local forces are:
f3 x  
10,000
lb
11
f4 x 
10,000
lb
11
A free-body diagram of the spring element 2 is shown below.
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
17/32
The Stiffness Method – Spring Example 2
Next, use the local element equations to obtain the force in
each spring.
For element 3: f   3000 3000  15 
11
4x
 
 

f2 x   3000 3000   0 
The local forces are:
f4 x 
45,000
lb
11
f2 x  
45,000
lb
11
A free-body diagram of the spring element 3 is shown below.
The Stiffness Method – Spring Example 3
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 1
k (1)  k (2)  k (3)  k ( 4)  200 
 kN / m
 1 1
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 3
Using superposition (the direct stiffness method), the global
stiffness matrix is:
Element 1
Element 2
Element 3
Element 4
0
0
0
 200 200
 200 400 200
0
0


K
0 200 400 200
0


0
0 200 400 200 


0
0
0 200 200 
The Stiffness Method – Spring Example 3
The global force-displacement equations are:
0
0
0   u1   F1x 
 200 200
 200 400 200
0
0  u2  F2 x 

   

0 200 400 200
0  u3   F3 x 


0
0 200 400 200  u4  F4 x 

   

0
0
0 200 200  u5  F5 x 
Applying the boundary conditions (u1 = 0 and u5 = 20 mm)
and the known forces (F2x, F3x, and F4x equal to zero) gives:
0
0   u2   0 
 400 200
 200 400 200
0   u3   0 


 
 0
200 400 200   u4   0 


0
200 200  0.02  F5 x 
 0
18/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 3
Rearranging the first three equations gives:
0  u2  0 
 400 200
 200 400 200  u   0 

 3  
0 200 400  u4  4 

Solving for u2, u3, and u4 gives:
u2  0.005 m
u3  0.01m
u4  0.015 m
Solving for the forces F1x and F5x gives:
F1x  200(0.005)  1.0kN
F5 x  200(0.015)  200(0.02)  1.0kN
The Stiffness Method – Spring Example 3
Next, use the local element equations to obtain the force in
each spring.
For element 1:  f1x   200 200   0 
 


f2 x   200 200  0.005 
f1x  1.0kN
f2 x  1.0kN
For element 2: f2 x   200 200  0.005 
 


f3 x   200 200   0.01 
f2 x  1.0kN
f3 x  1.0kN
19/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 3
Next, use the local element equations to obtain the force in
each spring.
For element 3:
f3 x   200 200   0.01 
 


f4 x   200 200  0.015 
f3 x  1.0kN
For element 4:
f4 x  1.0kN
f4 x   200 200  0.015 
 


f5 x   200 200   0.02 
f4 x  1.0kN
f5 x  1.0kN
The Stiffness Method – Spring Example 4
Consider the following spring system:
The boundary conditions are: u1  u3  u4  0
The compatibility condition at node 2 is:
u2(1)  u2(2)  u2(3)  u2
20/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
21/32
The Stiffness Method – Spring Example 4
Using the direct stiffness method: the elemental stiffness
matrices for each element are:
1
2
2
 k k1 
k (1)   1

 k1 k1 
1
2
2
3
 k k 2 
k (2)   2

 k 2 k 2 
 k k3 
k (3)   3

 k 3 k3 
2
3
Using the concept of superposition (the direct stiffness
method), the global stiffness matrix is:
1
2
3
4


k1
0 
0
 k1
 k k  k 2  k 3 k k 3 
2
K 1 1

0 
k 2
k2
 0
 0
k3
0 k3 


1
2
3
4
The Stiffness Method – Spring Example 4
Applying the boundary conditions (u1 = u3 = u4 = 0) the
stiffness matrix becomes:
1
2
3
4
4


k1
0 
0
 k1
 k k  k 2  k 3 k k 3 
2
K 1 1

k
0 

k
0

2
2
 0
k 3
0 k3 


1
2
3
4
2
4
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
The Stiffness Method – Spring Example 4
Applying the known forces (F2x = P) gives:
P   k1  k 2  k3  u2
Solving the equation gives:
u2 
P
k1  k 2  k3
Solving for the forces gives:
F1x  k1u2
F3 x  k2u2
F4 x  k3u2
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 threedimensional solid stress element.
22/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
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.
Total Potential Energy
Recall the force-displacement relationship for a linear
spring:
F  kx
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:
dU  Fdx   kx  dx
23/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
24/32
Total Potential Energy
x
The total strain energy is:
U   dU    kx  dx 
L
0
1 2
kx
2
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:
  Fx
Total Potential Energy
Therefore, the total potential energy is:  p 
1 2
kx  Fx
2
The concept of a stationary value of a function G is shown
below:
dG
0
dx
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.
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Total Potential Energy
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):
 p 
 p
d1
 d1 
 p
d 2
 d 2  ... 
 p
d n
 dn
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.
Total Potential Energy
25/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Total Potential Energy
To satisfy p = 0, all coefficients associated with di must be
zero independently, therefore:
 p
d i
0
i  1, 2,, n
or
 p
 d 
0
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.
26/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
27/32
Total Potential Energy – Example 5
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  
U
1 2
kx
2
The variation of p with respect to x is:
  Fx
 p 
 p
x
Since x is arbitrary and might not be zero, then:
Total Potential Energy – Example 5
Using our express for p, we get:
p 
 p
x
1 2
1
kx  Fx  500( lb in ) x 2  1,000lb  x
2
2
 0  500 x  1,000
If we had plotted the total
potential energy function p
for various values of
deformation we would get:
x  2.0 in
x  0
 p
x
0
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Total Potential Energy
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
p 
1
2
k  u2  u1   f1x u1  f2 x u2
2
Expanding the above express gives:
p 


1
k u22  2u1u2  u12  f1x u1  f2 x u2
2
Total Potential Energy
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:
Recall:
 p
d i
Therefore:
 p
u1
 p
u2
0
i  1, 2,, n
or

k
 2u2  2u1   f1x  0
2

k
 2u2  2u1   f2 x  0
2
 p
 d 
0
28/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Total Potential Energy
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:
Therefore:
k  u1  u2   f1x
k  u1  u2   f2x
In matrix form the equations are:
 f1x   k k   u1 
 
 
f2 x   k k  u2 
Total Potential Energy – Example 6
Obtain the total potential energy of the spring system shown
below and find its minimum value.
The potential energy p for element 1 is:
 p(1) 
1
2
k1  u3  u1   f1x u1  f3 x u3
2
The potential energy p for element 2 is:
 p(2) 
1
2
k2  u4  u3   f3 x u3  f4 x u4
2
29/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Total Potential Energy – Example 6
Obtain the total potential energy of the spring system shown
below and find its minimum value.
The potential energy p for element 3 is:
 p(3) 
1
2
k3  u2  u4   f2 x u2  f4 x u4
2
The total potential energy p for the spring system is:
3
 p    p( e )
e 1
Total Potential Energy – Example 6
Minimizing the total potential energy p:
 p
u1
 p
u2
 p
u3
 p
u4
 0  k1u3  k1u1  f1x (1)
 0  k3u2  k3u4  f2 x (3)
 0  k1u3  k1u1  k 2u4  k 2u3  f3 x (1)  f3 x (2)
 0  k 2u4  k 2u3  k3u2  k3u4  f4 x (2)  f4 x (3)
30/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Total Potential Energy – Example 6
In matrix form:  k1 0
0   u1  
k1
f1x

 0 k





k3 u2  
f2 x
0

3

   
(1)
(2) 
 k1 0 k1  k 2 k 2  u3   f3 x  f3 x 


(2)
(3)
 0 k3 k 2 k 2  k3  u4  f4 x  f4 x 
Using the following force equilibrium equations:
f1x (1)  F1x
f2 x (3)  F2 x
f3 x (1)  f3 x (2)  F3 x
f4 x (2)  f4 x (3)  F4 x
Total Potential Energy – Example 6
The global force-displacement equations are:
0
1000
0   u1   F1x 
 1000
 0
3000  u2  F2 x 
3000
0

    
 1000
0
3000 2000  u3  F3 x 


3000 2000 5000  u4  F4 x 
 0
The above equations are identical to those we obtained
through the direct stiffness method.
31/32
CIVL 7/8117
Chapter 2 - The Stiffness Mehtod
Homework Problems
1. Do problems 2.4, 2.10, and 2.22 on pages 66 - 71 in
your textbook “A First Course in the Finite Element
Method” by D. Logan.
End of Chapter 2
32/32