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Chapter 3 - Development of Truss Equations
Development of Truss Equations
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 1.
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
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.
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Development of Truss Equations
Development of Truss Equations
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Chapter 3 - Development of Truss Equations
Development of Truss Equations
Development of Truss Equations
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Development of Truss Equations
Development of Truss Equations
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Consider the derivation of the stiffness matrix for the linearelastic, constant cross-sectional area (prismatic) bar
element show below.
This application is directly applicable to the solution of pinconnected truss problems.
Stiffness Matrix for a Bar Element
Consider the derivation of the stiffness matrix for the linearelastic, constant cross-sectional area (prismatic) bar
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.
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Stiffness Matrix for a Bar Element
Consider the derivation of the stiffness matrix for the linearelastic, constant cross-sectional area (prismatic) bar
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.
Stiffness Matrix for a Bar Element
The strain-displacement relationship is:
  E

du
dx
From equilibrium of forces, assuming no distributed loads
acting on the bar, we get:
A x  T  constant
Combining the above equations gives:
AE
du
 T  constant
dx
Taking the derivative of the above equation with respect to the
local coordinate x gives:
d 
du 
AE
0

dx 
dx 
CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
The following assumptions are considered in deriving the bar
element stiffness matrix:
1. The bar cannot sustain shear force:
f1y  f2 y  0
2. Any effect of transverse displacement is ignored.
3. Hooke’s law applies; stress is related to strain:
 x  E x
Stiffness Matrix for a Bar Element
Step 1 - Select Element Type
We will consider the linear bar element shown below.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Step 2 - Select a Displacement Function
A linear displacement function u is assumed: u  a1  a2 x
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:

x
u   1  
 L 
 u  u1 
u  2
 x  u1
 L 
x   u1 
 
L  u2 
Stiffness Matrix for a Bar Element
Step 2 - Select a Displacement Function
Or in another form:
u  N1
u 
N2   1 
u2 
where N1 and N2 are the interpolation functions gives as:
N1  1 
x
L
N2 
x
L
The linear displacement
function û plotted over the
length of the bar element is
shown below.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Step 3 - Define the Strain/Displacement
and Stress/Strain Relationships
The stress-displacement relationship is:
x 
du u2  u1

dx
L
Step 4 - Derive the Element Stiffness Matrix and Equations
We can now derive the element stiffness matrix as follows:
T  A x
Substituting the stress-displacement relationship into the
above equation gives:
 u  u1 
T  AE  2

 L 
Stiffness Matrix for a Bar Element
Step 4 - Derive the Element Stiffness Matrix and Equations
The nodal force sign convention, defined in element figure, is:
f1x  T
therefore,
f2 x  T
u u 
f1x  AE  1 2 
 L 
 u  u1 
f2 x  AE  2

 L 
Writing the above equations in matrix form gives:
 f1x  AE  1 1  u1 
 
 
L  1 1 u2 
f2 x 
Notice that AE/L for a bar element is analogous to the spring
constant k for a spring element.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Step 5 - 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 ( e )
e 1
n
F  F    f ( e )
e 1
Where k and f are the element stiffness and force matrices
expressed in global coordinates.
Stiffness Matrix for a Bar Element
Step 6 - Solve for the Nodal Displacements
Solve the displacements by imposing the boundary conditions
and solving the following set of equations:
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:
x 
du u2  u1

dx
L
 x  E x
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 1 - Bar Problem
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.
Determine: (a) the global stiffness matrix, (b) the displacement
of nodes 2 and 3, and (c) the reactions at nodes 1 and 4.
Stiffness Matrix for a Bar Element
Example 1 - Bar Problem
For elements 1 and 2:
k
(1)
k
(2)

1  30  106  
1
2
2 node numbers for element 1
3 node numbers for element 2
1 1 lb
6  1 1 lb
10

in
 1 1
 1 1 in




30
For element 3:
3
k
(3)

4 node numbers for element 3
 2  15  106  
30
1 1 lb
6  1 1 lb
 1 1 in  10  1 1 in




As before, the numbers above the matrices indicate the
displacements associated with the matrix.
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Stiffness Matrix for a Bar Element
Example 1 - Bar Problem
Assembling the global stiffness matrix by the direct stiffness
methods gives:
E1
E2
E3
 1 1 0 0 

2 1 0 
6  1

K  10
 0 1 2 1


 0 0 1 1
Relating global nodal forces related to global nodal
displacements gives:
 F1x 
 1
F 

 2x 
6  1

10
 
 0
F3 x 

F4 x 
 0
1 0 0   u1 
2 1 0  u2 
 
1 2 1 u3 

0 1 1 u4 
Stiffness Matrix for a Bar Element
Example 1 – Bar Problem
The boundary conditions are: u1  u4  0
 F1x 
 1
F 

 2x 
6  1

10
 
 0
F3 x 

F4 x 
 0
1 0
0   u1 
2 1 0  u2 
 
1 2 1 u3 

0 1 1 u4 
Applying the boundary conditions and the known forces
(F2x = 3000 lb.) gives:
u
3000 
6  2 1  2 

  10 
 

 0 
 1 2  u3 
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 1 – Bar Problem
Solving for u2 and u3 gives:
u2  0.002 in
u3  0.001 in
The global nodal forces are calculated as:
 F1x 
 1
F 

 2x 
6  1

10
 
 0
F3 x 

F4 x 
 0
1 0 0   0  2000 
2 1 0  0.002   3000 


 lbs
1 2 1  0.001  0 

0 1 1  0  1000 
Stiffness Matrix for a Bar Element
Selecting Approximation Functions for Displacements
Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement
function.
1. Common approximation functions are usually
polynomials.
2. The approximation function should be continuous within
the bar element.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Selecting Approximation Functions for Displacements
Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement
function.
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.
Stiffness Matrix for a Bar Element
Selecting Approximation Functions for Displacements
Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement
function.
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.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Selecting Approximation Functions for Displacements
Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement
function.
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.
Stiffness Matrix for a Bar Element
Selecting Approximation Functions for Displacements
The interpolation function must allow for a rigid-body
displacement, that means the function must be capable of
yielding a constant value.
a1  u1  u2
Consider the follow situation: u  a1
Therefore:
u  N1u1  N2u2   N1  N2  a1
Since u = a1 then: u  a1   N1  N2  a1
This means that:
N1  N2  1
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.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
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.
Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
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.
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Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
Let’s consider that d does not coincident 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.
Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
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:
d  u1i  v1 j  u1i  v1 j
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Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
Consider the following diagram:
Using vector addition: a  b  i
Using the law of cosines, we get: | a || i | cos 
Similarly: | b || i | sin
| a | cos 
| b | sin
Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
Consider the following diagram:
The vector a is in the i direction and b is in the j direction,
therefore:
a | a | i   cos  i
b | b |   j    sin   j 
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Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
Consider the following diagram:
The vector i can be rewritten as: i  cos  i  sin j
The vector j can be rewritten as: j  sin i  cos  j
Therefore, the displacement vector is:
u1i  v1 j  u1  cos  i  sin j   v1  sin i  cos  j   u1i  v1 j
Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
Consider the following diagram:
Combining like coefficients of the local unit vectors gives:
u1 cos  v1 sin  u1
u1 sin  v1 cos  v1
u1   C S  u1 
 
 
v1   S C  v1 
C  cos 
S  sin
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Transformation of Vectors in Two Dimensions
The previous equation relates the global displacement d to
the d local displacements
 C S
The matrix is called the transformation matrix. 

 S C 
The figure below shows u expressed in terms of the global
coordinates x and y.
u   Cu  Sv
Stiffness Matrix for a Bar Element
Example 2 - Bar Element Problem
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.
Using the following expression we just derived, we get:
u   Cu  Sv
u2  cos 60o (0.1)  sin 60o (0.2)  0.223 in
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Stiffness Matrix for a Bar Element
Global Stiffness Matrix
We will now use the transformation relationship developed
above to obtain the global stiffness matrix for a bar element.
Stiffness Matrix for a Bar Element
Global Stiffness Matrix
We known that for a bar element in local coordinates we have:
 f1x  AE  1 1  u1 
 
 
L  1 1 u2 
f2x 
f   kd
We want to relate the global element forces f to the global
displacements d for a bar element with an arbitrary
orientation.
 f1x 
 u1 
f 
 
 1y 
 v1 

k
 
 
f2 x 
u2 
f2 y 
v 2 
f = kd
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Stiffness Matrix for a Bar Element
Global Stiffness Matrix
Using the relationship between local and global components,
we can develop the global stiffness matrix.
We already know the transformation relationships:
u1  u1 cos   v1 sin
u2  u2 cos   v 2 sin
Combining both expressions for the two local degrees-offreedom, in matrix form, we get:
 u1  C S
 
u2   0 0
 u1 
0 0   v1 
 
C S  u2 
v 2 
d = T *d
C S 0 0 
T*  

 0 0 C S
Stiffness Matrix for a Bar Element
Global Stiffness Matrix
A similar expression for the force transformation can be
developed.
 f1x 
 
 f1x  C S 0 0   f1y 
f  T*f
 
 

f2x   0 0 C S  f2 x 
f2 y 
Substituting the global force expression into element force
equation gives:
T * f  kd
f  = kd 
Substituting the transformation between local and global
displacements gives:
 T * f  kT *d
d = T *d
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Stiffness Matrix for a Bar Element
Global Stiffness Matrix
The matrix T* is not a square matrix so we cannot invert it.
Let’s expand the relationship between local and global
displacement.
 u1   C
 v    S
 1 
 
u2   0
v 2   0
0   u1 
0   v1 
 
0 C S  u2 

0 S C  v 2 
S
C
0
0
d = Td
where T is:
C
 S
T
0

0
S 0 0
C 0 0

0 C S

0 S C 
Stiffness Matrix for a Bar Element
Global Stiffness Matrix
We can write a similar expression for the relationship between
local and global forces.
 f1x   C
f   
S
 1y 
  
f
 2x   0

f2y   0
0   f1x 
 
C 0 0   f1y 
 
0 C S  f2 x 

0 S C  f2 y 
S
0
f  = Tf
Therefore our original local coordinate force-displacement
expression
 f1x  AE  1 1  u1 
 
 
L  1 1 u2 
f2x 
f  = kd
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Stiffness Matrix for a Bar Element
Global Stiffness Matrix
May be expanded:
 f1x 
1
f  

 1y  AE  0
  
L  1
f2 x 

f2y 
0
0 1 0   u1 
0 0 0  v1 
 
0 1 0  u2 

0 0 0  v 2 
The global force-displacement equations are:
f   kd

Tf  kTd
Multiply both side by T -1 we get: f  T -1kTd
where T-1 is the inverse of T. It can be shown that: T 1  T T
Stiffness Matrix for a Bar Element
Global Stiffness Matrix
The global force-displacement equations become: f = T TkTd
Where the global stiffness matrix k is: k  T TkT
Expanding the above transformation gives:
 C 2 CS C 2 CS 

2
2
CS S 
AE  CS S
k
L  C 2 CS C 2
CS 


2
S2 
CS
 CS S
We can assemble the total stiffness matrix by using the above
element stiffness matrix and the direct stiffness method.
n
K  K    k ( e )
e 1
n
F  F    f ( e )
e 1
F  Kd
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Chapter 3 - Development of Truss Equations
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Stiffness Matrix for a Bar Element
Global Stiffness Matrix
Local forces can be computed as:
 f1x 
1
f  

 1y  AE  0
  
L  1
f2 x 

f2y 
0
0 1 0   u1 
1



0 0 0  v1  AE  0

  
0 1 0  u2 
L  1


0 0 0  v 2 
0
0 1 0   C
0 0 0   S

0 1 0  0

0 0 0  0
0   u1 
C 0 0   v1 
 
0 C S  u2 

0 S C  v 2 
S
0
 f1x 
 Cu1  Sv1  Cu2  Sv 2 
f  


0
 1y  AE 

  
L  Cu1  Sv1  Cu2  Sv 2 
f2 x 


0


f2y 
Stiffness Matrix for a Bar Element
Example 3 - Bar Element Problem
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.
 C 2 CS C 2 CS 

2
2
CS S 
AE  CS S
k
L  C 2 CS C 2
CS 


2
S2 
CS
 CS S
Therefore:
C  cos30o 
3
2
S  sin30o 
1
2
CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 3 - Bar Element Problem
The global elemental stiffness matrix is:
 3

4

3
(2 in 2 ) 30  106 psi 
4

k
60 in
 3
4

 3
 
4


3
4
1
4
3

1
4
4
3
4
3

4
3
4
3
4

4

1 
4  lb
in
3 
4 

1 
4 
 3
Simplifying the global elemental stiffness matrix is:
 0.750
 0.433
k  106 
 0.750

 0.433
0.433
0.250
0.433
0.250
0.750 0.433 
0.433 0.250  lb
 in
0.750
0.433 

0.433
0.250 
Stiffness Matrix for a Bar Element
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:
 f1x  AE  1 1  u1 
 
 
L  1 1 u2 
f2x 
The force-displacement equation for f2x is:
f2x 
AE
 1
L
u 
1  1 
u2 
The stress in terms of global displacement is:
 u1 
 
C S 0 0  v1  E
E
   1 1 
     Cu1  Sv1  Cu2  Sv 2 
L
 0 0 C S  u2  L
v 2 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 4 - Bar Element Problem
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.

210  106
2
 1

3
1
3
(0)  (0.5) 
(0.75) KN m
  (0.25) 
4
2
4
 2

  81.32  103 kN m  81.32 MPa
2
Stiffness Matrix for a Bar Element
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.
27/77
CIVL 8/7117
Chapter 3 - Development of Truss Equations
28/77
Stiffness Matrix for a Bar Element
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.
Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
Element Node 1
1
1
2
1
3
1
Node 2
2
3
4

90o
45o
0o
C
0
0.707
1
S
1
0.707
0
CIVL 8/7117
Chapter 3 - Development of Truss Equations
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Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
The global elemental stiffness matrix are:
 C 2 CS C 2 CS 

2
2
CS S 
AE  CS S
k
L  C 2 CS C 2
CS 


2
S2 
CS
 CS S
u1
element 1:
C 0
S 1 
k (1)
v1
0

(2 in 2 )(30  106 psi ) 0

0
120 in

0
u1
element 2:
C
2
2
S
2
2

k
(2)
C 1 S  0

k (3)
v1
1

(2 in 2 )(30  106 psi )  0

 1
120 in

0
v2
u3
v3
1 1 1
1 1 1 lb
 in
1 1 1 

1 1 1 
1

(2 in 2 )(30  106 psi )  1

 1
240 2 in

 1
u1
element 3:
u2
0 0 0
1 0 1 lb
 in
0 0 0

1 0 1 
v1
u4
v4
0 1 0 
0 0 0  lb
 in
0 1 0

0 0 0
Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
The total global stiffness matrix is:
 1.354
 0.354
 0
0
K  5  105  0.354

 0.354
1
 0

0.354
1.354
0
0
0
1
0.354
0.354
0.354
0.354
1
0
0
1
0
0
0
1
0
0
0
0
0
0
0.354
0.354
0
0
0
0
0.354
0.354
0.354
0.354
0
0
0
0
0
0
0
0
0
0
0
0
1
0


0
0  lb
in
0

0

0

0
0
0
The total global force-displacement equations are:
 0 
 1.354
10, 000
 0.354
 F2 x 
 0
 F2 y 
5
0
 F3 x   5  10  0.354

 F3 y 
 0.354
 F 
1
 0
 F4 x 

 4y 
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
 u1 
 v 1 
 0 
  0 
 0 
  00 
 0 
 
element 1
element 2
element 3
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Chapter 3 - Development of Truss Equations
30/77
Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
Applying the boundary conditions for the truss, the above
equations reduce to:
 0 
 1.354
10, 000
 0.354
 F2 x 
 0
 F2 y 
5
0
 F3 x   5  10  0.354

 F3 y 
 0.354
 F 
1
 0
 F4 x 

 4y 
 u1 
 v 1 
 0 
0  0 
 
0
 0 
0
 0
0 0 
 0
0 
 
0.354
0
0
0.354
0.354
1
0
1.354
0
0
0
1
0
0.354
0
0.354
0
0
0
0
0
1
0
1
0
0
0
0.354
0.354
0
0
0
0
0.354
0.354
0.354
0.354
0
0
0
0
0
0
0
0
0
0
0
0
1
0
Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
Applying the boundary conditions for the truss, the above
equations reduce to:
u
0


5 1.354 0.354   1 

  5  10 



10,000 
0.354 1.354  v1 
Solving the equations gives:
u1  0.414  10 2 in
v1  1.59  10 2 in
The stress in an element is:

E
 Cu1  Sv1  Cu2  Sv 2 
L
CIVL 8/7117
Chapter 3 - Development of Truss Equations
31/77
Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
Element Node 1
1
1
2
1
3
1
Node 2
2
3
4

90o
45o
0o
C
0
0.707
1
S
1
0.707
0
30  106
 v1   3,965 psi
120
element 1
 (1) 
element 2
 (2)  
element 3
 (3) 
30  106
(0.707)u1  (0.707)v1   1,471 psi
120
30  106
 u1   1,035 psi
120
Stiffness Matrix for a Bar Element
Example 5 - Plane Truss Problem
Let’s check equilibrium at node 1:
F
 (1,471 psi )(2 in 2 )(0.707)  (1,035 psi )(2 in 2 )  0
F
 (3,965 psi )(2 in 2 )  (1,471 psi )(2 in 2 )(0.707)  10,000  0
x
y
CIVL 8/7117
Chapter 3 - Development of Truss Equations
32/77
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
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
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Develop the element stiffness matrices and system equations
for the plane truss below.
Member
Node 1
Node 2
1
2
3
1
2
1
2
3
3
Elemental
Stiffness
k
k
k

0
3/4
/2
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Chapter 3 - Development of Truss Equations
33/77
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Compute the elemental stiffness matrix for each element. The
general form of the matrix is:
 C 2 CS C 2 CS 

2
2
CS S 
AE  CS S
k
L  C 2 CS C 2
CS 


2
S2 
CS
 CS S
Member
Node 1
Node 2
1
2
3
1
2
1
2
3
3
Elemental
Stiffness
k
k
k

0
3/4
/2
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
For element 1:
u
k (1)
1
 1
 0
k
 1

 0
v
1
u
2
0 1
0 0
0 1
0 0
Member
Node 1
Node 2
1
2
3
1
2
1
2
3
3
v
2
0
0

0

0
u1
v1
u2
v2
Elemental
Stiffness
k
k
k

0
3/4
/2
CIVL 8/7117
Chapter 3 - Development of Truss Equations
34/77
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
For element 2:
u2
k (2)
v2
u3
v3
 1 1 1 1


k  1 1 1 1

2  1 1 1 1


 1 1 1 1
Member
Node 1
Node 2
1
2
3
1
2
1
2
3
3
u2
v2
u3
v3
Elemental
Stiffness
k
k
k

0
3/4
/2
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
For element 3:
u1
k (3)
v1
0 0
0 1
k
0 0

0 1
u3
v3
0 0
0 1

0 0

0 1
Member
Node 1
Node 2
1
2
3
1
2
1
2
3
3
u1
v1
u3
v3
Elemental
Stiffness
k
k
k

0
3/4
/2
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Chapter 3 - Development of Truss Equations
35/77
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Assemble the global stiffness matrix by superimposing the
elemental global matrices.
u1
v1
u2
v2
u3
v3
 2
 0

k  2
K 
2 0
 0

 0
0
2
0
0
2
0
0
0
2
0
3
1
1
1
0
1
1
1
1
0
1
1
1
1
0
2 

1

 1
 1

3
u1
element 1
v1
element 2
u2
element 3
v2
u3
v3
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
The unconstrained (no boundary conditions satisfied)
equations are:
0 2 0 0
0   u1   F1x 
 2
 0

k  2

2 0
 0

 0
2
0
0
0
2

  
0 0 0 2  v1   F1y 
3 1 1
1  u2   P1 

   
1 1 1 1  v 2   P2 
1 1 1 1  u3   F3 x 

  
1 1 1 3   v 3   F3 y 
The displacement at nodes 1 and 3 are zero in both directions.
Applying these conditions to the system equations gives:
 2
 0

k  2

2 0
 0

 0
0
2
2
0
0
3
0
0
2
0
0
0 0
1 1
1 1 1
1 1 1
1 1 1
0   u1   F1x 

  
2  v1   F1y 
1  u2   P1 

   
1  v 2   P2 
1  u3   F3 x 

  
3   v 3   F3 y 
CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Applying the boundary conditions to the system equations
gives:
k  3 1 u2   P1 
 

2  1 1  v 2   P2 
Solving this set of equations is fairly easy. The solution is:
u2 
P1  P2
k
v2 
P1  3P2
k
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Using the force-displacement relationship the force in each
member may be computed.
Member (element) 1:
 P  P2 
fx 1  k   1
 P2  P1
k 

fy 1  0
 P  P2 
fx 2  k  1
  P1  P2
 k 
fy 2  0
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Using the force-displacement relationship the force in each
member may be computed.
Member (element) 2:
 P  P2   1   P1  3P2   1  
fx 2  k   1

 

   2P2
k
2 

 k   2  
 P  P2   1   P1  3P2   1  
fx 3  k  1
 


    2P2
k
2 
  2 
 k  
fy 2  0
fy 3  0
Stiffness Matrix for a Bar Element
Example 6 - Plane Truss Problem
Using the force-displacement relationship the force in each
member may be computed.
Member (element) 3:
fx 1  0
fy 1  0
fx 3  0
fy 3  0
The solution to this simple problem can be readily checked by
using simple static equilibrium equations.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
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
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
The global elemental stiffness matrix for element 1 is:
cos  (1) 
3
 0 .6
5
sin  (1) 
 0.36

210  10 (6  10 )  0.48

 0.36
5

 0.48
6
k (1)
4
 0. 8
5
4
0.36 0.48 
0.48 0.64 
0.48 
0.48 0.36

0.64 
0.64 0.48
0.48
0.64
Simplifying the above expression gives:
u1
k (1)
 0.36
 0.48
 25,200 
 0.36

 0.48
v1
u2
v2
0.36 0.48 
0.64 0.48 0.64 
0.48 
0.48 0.36

0.64 
0.64 0.48
0.48
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
The global elemental stiffness matrix for element 2 is:
cos  ( 2 )  0
k (2)
sin  ( 2 )  1
0

(210  106 )(6  10 4 ) 0

0
4

0
0
1
0
1
0 0 
0 1
0 0

0 1
Simplifying the above expression gives:
u1
k (2)
v1
u3
v3
0
0
0 
0
0 1.25 0 1.25 

 25,200 
0
0
0
0 


0 1.25 0 1.25 
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
The total global equations are:
 F1x 
 0.36
F 
 0.48
1
y
 

 0.36
F2 x 
   25,200 
F
 2y 
 0.48
F3 x 

0
 

F
0

 3 y 
0.48
1.89
0.36
0.48
0.48
0.64
0
0
0.48
0.36
0.48
0
0.64
0
0.48
0
0.64
0
0
0
1.25
0
0
0
The displacement boundary conditions are:
u1  
u2  v 2  u3  v 3  0
0   u1 
 
1.25   v1 
0  u2 
 
0  v 2 
0  u3 
 
1.25  v 3 
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
The total global equations are:
 F1x 
 0.36
 FP 
 0.48
 1y 

 0.36
F2 x 
   25,200 
F
 2y 
 0.48
 F3 x 

0
 

F
0


3
y

 
0.48
1.89
0.36
0.48
0.48
0.64
0
0
0.48
0.36
0.48
0
0.64
0
0.48
0
0.64
0
0
0
1.25
0
0
0
u1 
0  
v 

1.25   1 
0  u2 
 
0  v 2 
0  u3 
 
1.25  v 3 
By applying the boundary conditions the force-displacement
equations reduce to:
P  25,200(0.48  1.89v1 )
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
Solving the equation gives:
v1  (2.1 105 )P  0.25
By substituting P = 1,000 kN and  = -0.05 m in the above
equation gives:
v1  0.0337m
The local element forces for element 1 are:
 u1  0.05 
0  v1  0.0337 
 f1x 
 1 1 0.6 0.8 0
   25,200 



u2
0 0.6 0.8  
 1 1  0
f2 x 

v2


The element forces are: f1x  76.6 kN
f2 x  76.7 kN
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 7 - Plane Truss Problem
The local element forces for element 2 are:
 u1  0.05 


 f1x 
 1 1 0 1 0 0  v1  0.0337 
   31,500 





u3
 1 1 0 0 0 1 
f3 x 

v3


The element forces are: f1x  1,061 kN
f3 x  1,061 kN
Stiffness Matrix for a Bar Element
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:
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
The coordinates at node 1 are x1, y1, and z1, and the coordinates 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.
Stiffness Matrix for a Bar Element
Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
The three-dimensional vector representing the bar element is
gives as:
d  ui  vj  wk  u  i  v  j  w  k
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
Taking the dot product of the above equation with i gives:
u(i  i)  v ( j  i)  w (k  i)  u 
By the definition of the dot product we get:
i  i 
where
x2  x1
 Cx
L
j  i 
y 2  y1
 Cy
L
k  i 
z2  z1
 Cz
L
L  ( x2  x1 )2  ( y 2  y1 )2  ( z2  z1 )2
Cx  cos x
Cy  cos y
Cz  cos  z
where Cx, Cy, and Cz are projections of i on to i, j, and k,
respectively.
Stiffness Matrix for a Bar Element
Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
Therefore:
u   Cx u  Cy v  Czw
The transformation between local and global displacements is:
 u1 
v 
 1
 u1  Cx Cy Cz 0 0 0  w1 
 
 
u2   0 0 0 Cx Cy Cz   u2 
v2 
 
w 2 
d  T *d
C C C 0 0 0 
T*   x y z

 0 0 0 C x C y Cz 
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
The transformation from the local to the global stiffness matrix
is:
C 0 
x
k  T TkT
C 0 
 y 
C 0  AE
k z 
 0 Cx  L
 0 Cy 


 0 Cz 
 Cx2

 C xCy
AE  CxCz
k

L  Cx2
 C C
 x y
 CxCz
 1 1 Cx Cy Cz 0 0 0 
 1 1  0 0 0 C C C 
y


x
z
C xC y
Cy2
C xC z
Cy Cz
Cx2
CxCy
C y Cz
CxCy
Cy2
Cy Cz
Cz2
CxCz
Cy Cz
Cz2
CxCz
Cx2
C xC y
C xC z
CxCy
Cy2
Cy Cz
C xCy
Cy2
C y Cz
CxCz 

Cy Cz 
2
Cz 

C xC z 
C y Cz 

Cz2 
Stiffness Matrix for a Bar Element
Transformation Matrix and Stiffness Matrix for
a Bar in Three-Dimensional Space
The global stiffness matrix can be written in a more convenient
form as:
 C x 2 C x C y C x Cz 
AE    


  CxCy Cy 2 Cy Cz 
k


L  

CxCz Cy Cz Cz 2 


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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider the space truss shown below. The modulus of
elasticity, E = 1.2 x 106 psi for all elements. Node 1 is
constrained from movement in the y direction.
To simplify the stiffness matrices
for the three elements, we will
express each element in the
following form:
k
AE
L
 
 

 
 
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider element 1:
L(1)  ( 72)2  (36)2  80.5 in
72
 0.89
80.5
36
Cy 
 0.45
80.5
Cz  0
Cx 
 0.79
   0.40
 0
0.40
0.20
0
0
0

0 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider element 1:
u v w
1
1
1
(0.302 in )(1.2  10 psi )  
k
 
80.5 in

2
6
u v w
2
2
2
  lb
in
 
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider element 2:
L(2)  ( 72)2  (36)2  (72)2  108 in
72
 0.667
108
36
Cy 
 0.33
108
72
Cz 
 0.667
108
Cx 
 0.45
   0.22
 0.45
0.22
0.11
0.45
0.45 
0.45 

0.45 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider element 2:
u v w
1
k
1
(0.729 in )(1.2  10 psi )  
 
108 in

2
6
1
u v w
3
3
3
  lb
in
 
The boundary conditions are:
u2  v 2  w 2  0
u3  v 3  w 3  0
u4  v 4  w 4  0
v1  0
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider element 3: L(3)  ( x 4  x1 )2  ( y 4  y1 )2  ( z4  z1 )2
L(3)  ( 72)2  ( 48)2  86.5 in
72
 0.833
86.5
Cy  0
Cx 
Cz 
48
 0.550
86.5
0.69
   0
0.46
0
0
0
0.46 
0 

0.30 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Consider element 3:
u v w
1
k
1
1
u v w
(0.187 in )(1.2  10 psi )  
 
86.5 in

2
6
4
4
4
  lb
in
 
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
Canceling the rows and the columns associated with the
boundary conditions reduces the global stiffness matrix to:
u1
w1
 9,000 2,450 
K

 2,450 4,450 
The global force-displacement equations are:
 9,000
 2,450

2,450   u1   0 
 

4,450  w1  1,000 
Solving the equation gives:
u1  0.072 in
w1  0.264 in
48/77
CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
It can be shown, that the local forces in an element are:
 fix  AE  Cx
 

L  Cx
f jx 
Cy
Cy
Cz
Cz
Cx
Cx
Cy
Cy
The stress in an element is:

E
 Cx
L
 Cy
 Cz
Cx
 ui 
v 
 i
Cz  w i 
 
Cz   u j 
v j 
 
w j 
 ui 
v 
 i
w i 
Cz   
uj 
v j 
 
w j 
Cy
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
The stress in element 1 is:
 (1) 
1.2  10 6
0.89
80.5
0.45
0
 0.89
 (1)  955 psi
0.45
 0.072 
 0 


 0.264 
0

 0 
 0 


 0 
The stress in element 2 is:
 (2) 
1.2  106
0.667
108
 (2)  1,423 psi
 0.33
 0.667
 0.667
0.33
 0.072 
 0 


 0.264 
0.667  

 0 
 0 


 0 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 8 – Space Truss Problem
The stress in element 3 is:
 (3) 
1.2  106
0.83
86.5
0
0.55
 0.83
 (3)  2,843 psi
0
 0.072 
 0 


 0.264 
 0.55  

 0 
 0 


 0 
Stiffness Matrix for a Bar Element
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.
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Inclined, or Skewed, Supports
We must transform the local boundary condition of v’3 = 0
(in local coordinates) into the global x-y system.
Stiffness Matrix for a Bar Element
Inclined, or Skewed, Supports
Therefore, the relationship between of the components of the
displacement in the local and the global coordinate systems
at node 3 is:
sin   u3 
u '3   cos 
 
 
v '3    sin  cos   v 3 
We can rewrite the above expression as:
 cos 
t3    sin
d '3   [t3 ]d3 

sin  
cos  
We can apply this sort of transformation to the entire
displacement vector as:
d '  [T1]d
or
d  [T1]T d '
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Inclined, or Skewed, Supports
Where the matrix [T1]T is:
 [I ]
[T1 ]T  [0]

[0]
[0]
[I ]
[0]
[0] 
[0] 

[t3 ]
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.
f '  [T1]f 
In global coordinates, the force-displacement equations are:
f   [K ]d 
Stiffness Matrix for a Bar Element
Inclined, or Skewed, Supports
Applying the skewed support transformation to both sides of
the equation gives:
[T1 ] f   [T1 ][K ] d 
By using the relationship between the local and the global
displacements, the force-displacement equations become:
f '  [T1][K ][T1]T d '
Therefore the global equations become:
 F1x 
F 
 1y 
 F2 x 
T

  [T1 ][K ][T1 ]
 F2 y 
F ' 3 x 


F '3 y 
 u1 
 v 
 1
 u2 
 
 v2 
u '3 
 
v '3 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
Consider the plane truss shown below. Assume E = 210 GPa,
A = 6 x 10-4 m2 for element 1 and 2, and A = 2 (6 x 10-4)m2
for element 3.
Determine the stiffness matrix
for each element.
Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
The global elemental stiffness matrix for element 1 is:
cos  (1)  0
sin (1)  1
u1
k (1)
v1
0 0

(210  106 kN / m 2 )(6  10 4 m 2 ) 0 1

0 0
1m

0 1
u2
v2
0 0
0 1

0 0

0 1
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
The global elemental stiffness matrix for element 2 is:
cos  (2)  1
sin (2)  0
u2
k (2)
1

(210  106 kN / m 2 )(6  104 m 2 )  0

 1
1m

0
v2
0
0
0
0
u3
v3
1 0 
0 0 
1 0

0 0
Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
The global elemental stiffness matrix for element 3 is:
cos (3) 
2
2
sin (3) 
2
2
u1
k (3)
v1
u3
v3
1 1 1
1
6
2
4
2 
1
1 1 1
(210  10 kN / m )(6 2  10 m ) 



1
1 1 1


2 2m


 1 1 1 1
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Chapter 3 - Development of Truss Equations
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Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
Using the direct stiffness method, the global stiffness matrix is:
 0.5
 0.5


0
K  1,260  10 5 N 
m
0

 0.5

 0.5
0.5
0
0  0.5  0.5
1.5
0  1  0.5  0.5

0
1 0
0
1

1
0
1 0
0
1.5
 0.5  1 0
0.5

0
0
0.5
 0.5
0.5
We must transform the global displacements into local
coordinates. Therefore the transformation [T1] is:
 [I ]
[T1 ]  [0]

[0]
[0]
[I ]
[0]
[0] 
[0] 

[t3 ]
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
2
2
 2






2 
2

2
2
0
0
0
0
2
Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
The first step in the matrix transformation to find the product of
[T1][K].
 0.5
 0.5
5N
T1 K   1,260  10 m  00
 0.707
0






0
0.5
0 0
0.5
0.5
1.5
0 1 0.5
0.5
0
1 0
1
0
1
0
1
0
0
0.707 0.707 0 1.414 0.707
0
0.707
0
0.70
The next step in the matrix transformation to find the product of
[T1][K][T1]T.
T1 K T1 
T
 0.5
 0.5
5N
0

 1,260  10
m 0
 0.707
0

0.5
1.5
0
0
0 0.707
1 0.707
0
1
1
0
0 0.707
1
0
0.707
0
0.707
0.707
0
0
1.5
0.5


0.707 
0
0.5

0.5 
0
0
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Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
The displacement boundary conditions are: u1  v1  v 2  v '3  0
 F1x 

0.5



 F1y 
0.5




F
0
 2x 

5

  1,260  10 N m 
F
0
 2y 



 0.707
F
'
 3x 

F ' 

0

 3y 
0.5
0
0 0.707
0   u1


0 1 0.707
0   v1


1 0 0.707 0.707   u2 


1
0 1
0
0   v 2 


0.707 0.707 0
1.5 0.5  u '3 

0 0.707 0
0.5
0.5  v '3 
1.5
0
Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
By applying the boundary conditions the global forcedisplacement equations are:
1,260  105 N
0.707   u2  F2 x  1,000 kN 
 1


m  0.707 1.5  u '3  F '3 x  0

Solving the equation gives:
u2  11.91 mm
u '3  5.61 mm
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Chapter 3 - Development of Truss Equations
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Stiffness Matrix for a Bar Element
Example 9 – Space Truss Problem
 F1x 
F 
 1y 
 F2 x 
T

  [T1 ][K ][T1 ]
 F2 y 
F '3 x 


F '3 y 
The global nodal forces are calculated as:
 F1x 
F 
 0.5
 1y 
 0.5
 F2 x 

0
2N

  1,260  10 mm 
0
F
2
y



F ' 3 x 
 0.707
0



F '3 y 
0.5
1.5
0
0
0 0.707
1 0.707
1
0
1
0
0 0.707
1
0
0.707
0
0.707
0.707
0
0
1.5
0.5
 u1 
 v 
 1
 u2 
 
 v2 
u '3 
 
v '3 
0  0
0  0


0.707  11.91
 0 
0


0.5
  5.61 
0.5   0 
Therefore:
F1x  500 kN
F1y  500 kN
F2 y  0
F '3 y  707 kN
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
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  
The differential internal work (strain energy) dU in a onedimensional bar element is:
dU   x ( y )( z )( x )d  x
CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
If we let the volume of the element approach zero, then:
dU   x d  x dV
Summing the differential energy over the whole bar gives:
 x

U      x d  x  dV

V 
0
For a linear-elastic material (Hooke’s law) as shown below:
 x  E x
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
The internal strain energy statement becomes
1
U    x  x dV
2V
The potential energy of the external forces is:
M
    X bu dV   Tx us dS   fixui
V
S
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.
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Chapter 3 - Development of Truss Equations
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Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Apply the following steps when using the principle of minimum
potential energy to derive the finite element equations.
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.
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Consider the following bar element, as shown below:
L
A
 p    x  x dx  f1x u1  f2 x u2
20
  X bu dV   Txus dS
V
S
We can approximate the axial displacement as:
u  N1
u 
N2   1 
u2 
N1  1 
x
L
N2 
x
L
CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Using the stress-strain relationships, the axial strain is:
x 
du  dN1

dx  dx
dN2   u1 
 
dx  u2 
where N1 and N2 are the interpolation functions gives as:
 1
 x  
 L
1   u1 
 
L  u2 
 x   [B]{d }
 1
B  
 L
1
L 
The axial stress-strain relationship is:  x   [D ]  x 
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Where [D] = [E] for the one-dimensional stress-strain
relationship and E is the modulus of elasticity.
Therefore, stress can be related to nodal displacements as:
 x   [D][B]d 
The total potential energy expressed in matrix form is:
L
A
T
T
T
T
 p    x   x  dx  d  P   u  X b  dV   u Tx  dS
20
V
S
where {P} represented the concentrated nodal loads.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
If we substitute the relationship between û and d̂ into the
energy equations we get:
L
A
T
T
T
T
 p   d  B  D  B d  dx  d  P
20
  d  N 
T
V
T
 X b  dV   d  Ns  Tx  dS
T
T
S
In the above expression for potential energy p is a function of
the d, that is: p = p(u1 , u2 ).
However, [B] and [D] and the nodal displacements u are not a
function of x.
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Integration the energy expression with respect to x gives:
p 
where
AL
T
T
d  [B ]T [D]T [B]d  d  f 
2
f   P   [N ]T  Xb  dV   [N ]T  X b  dS
V
S
We can define the surface tractions and body-force matrices
as:
fs    [N ]T Tx  dS
fb    [N ]T  Xb  dV
S
V
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Minimization of p with respect to each nodal displacement
requires that:
 p
 p
0
0
u1
u2
For convenience, let’s define the following
U   d 
[B ]T [D ]T [B ] d 
U   u
 1
 L 
 1
u 2    [E ]  
 L
 1
 L 
T
*
*
1
1   u1 
 
L  u2 
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
Simplifying the above expression gives:
U* 

E 2
u1  2u1u2  u22
L2

The loading on a bar element is given as:
d f   u1f1x  u2f2 x
T
Therefore, the minimum potential energy is:
 p AE

 2u1  2u2   f1x  0
u1
2L
 p
u2

AE
 2u1  2u2   f2 x  0
2L
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Potential Energy Approach to Derive Bar Element Equations
The above equations can be written in matrix form as:
 p
 d 

AE
L
 1 1  u1   f1x 
 1 1  u    f   0

  2   2x 
The stiffness matrix for a bar element is:
k  
AE  1 1
L  1 1
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.
Stiffness Matrix for a Bar Element
Example 10 - Bar Problem
Consider the bar shown below:
The energy equivalent nodal forces due to the distributed load
are:
 x
L 1 

f
 
f0    [N ]T Tx  dS f0    1x     L Cx dx
f2x  0  x 
S
 L 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 10 - Bar Problem
 Cx 2 Cx 3
 x


1
3L
f1x  L  L 
 2
   
 Cx dx  
L
f2x  0  x 

Cx 3

 L 
3L 0


2
 CL 
 6 
0

  2
 CL 
  3 

The total load is the area under the distributed load curve, or:
L
1
CL2
F  (L )(CL ) 
2
2
The equivalent nodal forces for a linearly varying load are:
f1x 
F 1
 of the total load
3 3
f2 x 
2F 2
 of the total load
3
3
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
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.
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The one-element solution:
The distributed load can be converted into equivalent nodal
forces using:
 x
L 1 
F1x 
 L 
T
F




F

[
N
]
T
dS
  
 10x dx
 0   x 
0
F
x
2x


0


S
 L 
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The one-element solution:
 10L2 10L2   10L2 
 x
1

 2  3   6 
F1x  L  L 
   
 10x dx  


2
10L
10L2 
F2x  0  x 





 L 

  3 
3
F1x   6,000 lb 
 

F2x  12,000 lb 
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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The one-element solution:
 1 1
k (1)  106 

 1 1
The element equations are:
 1 1 u1   6,000 
106 

   
 1 1  0  R2 x  12,000 
u1  0.006 in
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The one-element solution:
The axial stress-strain relationship is: { x }  [D]{ x }
{ x }  E [B ] d 
 1
 E 
 L
1   u1 
 u2  u1 
   E


L  u2 
 L 
 0  0.006 
 30  106 
  3,000 psi (T )
60


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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
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 in.)(300 lb in )  4,500 lb
2
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The two-element solution:
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

 (4,500)
(1)

 f1x   3
  1,500 lb 
 (1)   


f2 x   2 (4,500) 3,000 lb 
 3

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CIVL 8/7117
Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The two-element solution:
For element 2, the applied force is in two parts: a triangularshaped distributed load and a uniform load. The uniform load
is:
(30 in )(300 lb / in )  9,000 lb
The nodal forces for element 2 are:
 1
1

  (9,000)  (4,500) 

f2 x    2
3
 6,000 lb 
 (2)   


f3 x    1 (9,000)  2 (4,500)  7,500 lb 
 
  2
3

(2)
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The two-element solution:
The final nodal force vector is:
 F1x   1,500 
  

F2 x    9,000 
F  R  7,500 
 3x   3x

The element stiffness matrices are:
1
3
k(1)  k(2) 
2
4
2AE  1 1
L  1 1
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The two-element solution:
The assembled global stiffness matrix is:
 1 1 0 
6 
K  2  10 1 2 1


 0 1 1
The assembled global force-displacement equations are:
 1 1 0   u1   1,500 
  

2  106  1 2 1 u2    9,000 


 0 1 1  0  R3 x  7,500 
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The two-element solution:
After the eliminating the row and column associated with d3x,
we get:
 1 1  u1   1,500 
2  106 
 

2 u2  9,000 
 1
Solving the equation gives:
u1  0.006 in
u2  0.00525 in
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 11 - Bar Problem
The two-element solution:
The axial stress-strain relationship is:
 1
 L
 x (1)  E  
 1
 E 
 30
 1
 30
 x (2)  E  
1   d1x 
 
L  d 2 x 
1   0.006 

  750 psi (T )
30  0.00525 
1  0.00525 

  5,250 psi (T )
30  
0

Stiffness Matrix for a Bar Element
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:
L
u
1
P ( x )dx
AE 0
where the force P is shown on the following free-body diagram.
P( x ) 
1
x (10 x )  5 x 2

2
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Comparison of Finite Element Solution to Exact Solution
Therefore:
L
u
x
1
P ( x )dx
AE 0
u
1
5x3
2

 C1
5
x
dx
AE o
3 AE
Applying the boundary conditions:
5x3
u (L )  0 
 C1
3 AE

5L3
C1  
3 AE
The exact solution for axial displacement is:
u (L ) 

5
x 3  L3
3 AE

 (x) 
P( x ) 5x 2

A
A
Stiffness Matrix for a Bar Element
Comparison of Finite Element Solution to Exact Solution
A plot of the exact solution for displacement as compared to
several different finite element solutions is shown below.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Comparison of Finite Element Solution to Exact Solution
A plot of the exact solution for axial stress as compared to
several different finite element solutions is shown below.
Stiffness Matrix for a Bar Element
Comparison of Finite Element Solution to Exact Solution
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.
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
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.
Stiffness Matrix for a Bar Element
Galerkin’s Residual Method and Its Application
to a One-Dimensional Bar
Therefore, the substitution of the trial function in the differential
equation will create a residual over the entire domain of the
problem.
 RdV  minimum
V
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 allows the weighted integral of the
residuals to go to zero.
 RW dV  0
V
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Galerkin’s Residual Method and Its Application
to a One-Dimensional Bar
Using Galerkin’s weighted residual method, we require the
weighting functions to be the interpolation, Ni. Therefore:
 RN
i
dV  0
i  1, 2, , n
V
Stiffness Matrix for a Bar Element
Example 12 - Bar Element Formulation
Let’s derive the bar element formulation using Galerkin’s
method. The governing differential equation is:
d 
du 
AE
0

dx 
dx 
Applying Galerkin’s method we get:
L
d 
du 
 dx  AE dx  N
i
dx  0
i  1, 2, , n
0
We now apply integration by parts using the following general
formula:
 udv  uv   vdu
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 12 - Bar Element Formulation
If we assume the following:
dNi
dx
dx
du
v  AE
dx
u  Ni
dv 
du 
d 
du 
AE
dx

dx 
dx 
then integration by parts gives:
L
du 
du dNi

 Ni AE dx    AE dx dx dx  0

0 0
L
Stiffness Matrix for a Bar Element
Example 12 - Bar Element Formulation
Recall that:
dN2
du dN1

u1 
u2
dx
dx
dxˆ
du  1
 
dx  L
1   u1 
 
L  u2 
Our original weighted residual expression, with the
approximation for u becomes:
L
dN
AE  i
dx
0
 1
 L

1
dx
L 
 u1  
du 
    Ni AE
dx o
u2  
L
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Example 12 - Bar Element Formulation
Substituting N1 for the weighting function Ni gives:
dN1  1
 L
dx

0
L
AE 
 u1  
1
du 
dx

N
AE


1

L 
dx 0
u2  
L
 1  1
AE      
L L
0
1
dx
L 
L
f1x  AE
du
dx
 u1 
du
   AE
dx
u2 
x 0
x 0
AE
u1  u2   f1x
L
Stiffness Matrix for a Bar Element
Example 12 - Bar Element Formulation
Substituting N2 for the weighting function Ni gives:
 1
 L

1
dx
L 
 u1  
du 
    N2 AE
dx 0
u2  
 1  1
AE     
L
L
0 
1
dx
L 
 u1 
du
   AE
dx
u2 
L
dN
AE  2
dx
0
L
f2 x  AE
du
dx
x L
AE
u2  u1   f2x
L
L
AE
L
x L
 1 1  u1   f1x 
 1 1 u   f 

  2   2x 
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Chapter 3 - Development of Truss Equations
Stiffness Matrix for a Bar Element
Problems:
6. Verify the global stiffness matrix for a three-dimensional
bar. Hint: First, expand T* to a 6 x 6 square matrix, then
expand k to 6 x 6 square matrix by adding the appropriate
rows and columns of zeros, and finally, perform the matrix
triple product k = TTk’T.
7. Do problems 3.4, 3.10, 3.12, 3.15a,b, 3.18, 3.23, 3.37,
3.43, 3.48, 3.50, and 3.55 on pages 146 - 165 in your
textbook “A First Course in the Finite Element Method” by
D. Logan.
8. Use SAP2000 and solve problems 3.63 and 3.64.
End of Chapter 3
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