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Page 1
Instructional Objectives
After reading this chapter the student will be able to
1. Derive member stiffness matrix of a beam element.
2. Assemble member stiffness matrices to obtain the global stiffness matrix for a
beam.
3. Write the global load-displacement relation for the beam.
4. Impose boundary conditions on the load-displacement relation of the beam.
5. Analyse continuous beams by the direct stiffness method.
28.1 Introduction
In the last lesson, the procedure to analyse beams by direct stiffness method has
been discussed. No numerical problems are given in that lesson. In this lesson,
few continuous beam problems are solved numerically by direct stiffness method.
Example 28.1
Analyse the continuous beam shown in Fig. 28.1a. Assume that the supports are
unyielding. Also assume that EI is constant for all members.
The numbering of joints and members are shown in Fig. 28.1b. The possible
global degrees of freedom are shown in the figure. Numbers are put for the
unconstrained degrees of freedom first and then that for constrained
displacements.
Page 2
Instructional Objectives
After reading this chapter the student will be able to
1. Derive member stiffness matrix of a beam element.
2. Assemble member stiffness matrices to obtain the global stiffness matrix for a
beam.
3. Write the global load-displacement relation for the beam.
4. Impose boundary conditions on the load-displacement relation of the beam.
5. Analyse continuous beams by the direct stiffness method.
28.1 Introduction
In the last lesson, the procedure to analyse beams by direct stiffness method has
been discussed. No numerical problems are given in that lesson. In this lesson,
few continuous beam problems are solved numerically by direct stiffness method.
Example 28.1
Analyse the continuous beam shown in Fig. 28.1a. Assume that the supports are
unyielding. Also assume that EI is constant for all members.
The numbering of joints and members are shown in Fig. 28.1b. The possible
global degrees of freedom are shown in the figure. Numbers are put for the
unconstrained degrees of freedom first and then that for constrained
displacements.
The given continuous beam is divided into three beam elements Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. In the given problem some displacements
are zero, i.e., 0
8 7 6 5 4 3
= = = = = = u u u u u u from support conditions.
In the case of beams, it is not required to transform member stiffness matrix from
local co-ordinate system to global co-ordinate system, as the two co-ordinate
system are parallel to each other.
First construct the member stiffness matrix for each member. This may be done
from the fundamentals. However, one could use directly the equation (27.1)
given in the previous lesson and reproduced below for the sake convenience.
Page 3
Instructional Objectives
After reading this chapter the student will be able to
1. Derive member stiffness matrix of a beam element.
2. Assemble member stiffness matrices to obtain the global stiffness matrix for a
beam.
3. Write the global load-displacement relation for the beam.
4. Impose boundary conditions on the load-displacement relation of the beam.
5. Analyse continuous beams by the direct stiffness method.
28.1 Introduction
In the last lesson, the procedure to analyse beams by direct stiffness method has
been discussed. No numerical problems are given in that lesson. In this lesson,
few continuous beam problems are solved numerically by direct stiffness method.
Example 28.1
Analyse the continuous beam shown in Fig. 28.1a. Assume that the supports are
unyielding. Also assume that EI is constant for all members.
The numbering of joints and members are shown in Fig. 28.1b. The possible
global degrees of freedom are shown in the figure. Numbers are put for the
unconstrained degrees of freedom first and then that for constrained
displacements.
The given continuous beam is divided into three beam elements Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. In the given problem some displacements
are zero, i.e., 0
8 7 6 5 4 3
= = = = = = u u u u u u from support conditions.
In the case of beams, it is not required to transform member stiffness matrix from
local co-ordinate system to global co-ordinate system, as the two co-ordinate
system are parallel to each other.
First construct the member stiffness matrix for each member. This may be done
from the fundamentals. However, one could use directly the equation (27.1)
given in the previous lesson and reproduced below for the sake convenience.
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
k
z z z z
z z z z
z z z z
z z z z
4 6 2 6
6 12 6 12
2 6 4 6
6 12 6 12
2 2
2 3 2 3
2 2
2 3 2 3
(1)
The degrees of freedom of a typical beam member are shown in Fig. 28.1c.
Here equation (1) is used to generate element stiffness matrix.
Member 1: , node points 1-2. m L 4 =
The member stiffness matrix for all the members are the same, as the length
and flexural rigidity of all members is the same.
[]
1
3
5
6
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
'
1 3 5 6 . .
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(2)
On the member stiffness matrix, the corresponding global degrees of freedom
are indicated to facilitate assembling.
Member 2: , node points 2-3. m L 4 =
[]
2
4
1
3
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
2 4 1 3 . .
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(3)
Member 3: , node points 3-4. m L 4 =
Page 4
Instructional Objectives
After reading this chapter the student will be able to
1. Derive member stiffness matrix of a beam element.
2. Assemble member stiffness matrices to obtain the global stiffness matrix for a
beam.
3. Write the global load-displacement relation for the beam.
4. Impose boundary conditions on the load-displacement relation of the beam.
5. Analyse continuous beams by the direct stiffness method.
28.1 Introduction
In the last lesson, the procedure to analyse beams by direct stiffness method has
been discussed. No numerical problems are given in that lesson. In this lesson,
few continuous beam problems are solved numerically by direct stiffness method.
Example 28.1
Analyse the continuous beam shown in Fig. 28.1a. Assume that the supports are
unyielding. Also assume that EI is constant for all members.
The numbering of joints and members are shown in Fig. 28.1b. The possible
global degrees of freedom are shown in the figure. Numbers are put for the
unconstrained degrees of freedom first and then that for constrained
displacements.
The given continuous beam is divided into three beam elements Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. In the given problem some displacements
are zero, i.e., 0
8 7 6 5 4 3
= = = = = = u u u u u u from support conditions.
In the case of beams, it is not required to transform member stiffness matrix from
local co-ordinate system to global co-ordinate system, as the two co-ordinate
system are parallel to each other.
First construct the member stiffness matrix for each member. This may be done
from the fundamentals. However, one could use directly the equation (27.1)
given in the previous lesson and reproduced below for the sake convenience.
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
k
z z z z
z z z z
z z z z
z z z z
4 6 2 6
6 12 6 12
2 6 4 6
6 12 6 12
2 2
2 3 2 3
2 2
2 3 2 3
(1)
The degrees of freedom of a typical beam member are shown in Fig. 28.1c.
Here equation (1) is used to generate element stiffness matrix.
Member 1: , node points 1-2. m L 4 =
The member stiffness matrix for all the members are the same, as the length
and flexural rigidity of all members is the same.
[]
1
3
5
6
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
'
1 3 5 6 . .
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(2)
On the member stiffness matrix, the corresponding global degrees of freedom
are indicated to facilitate assembling.
Member 2: , node points 2-3. m L 4 =
[]
2
4
1
3
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
2 4 1 3 . .
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(3)
Member 3: , node points 3-4. m L 4 =
[]
7
8
2
4
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
7 8 2 4 . .
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(4)
The assembled global stiffness matrix of the continuous beam is of the
order . The assembled global stiffness matrix may be written as, 8 8 ×
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
-
-
-
- - -
- - -
-
-
=
1875 . 0 375 . 0 0 0 1875 . 0 0 375 . 0 0
375 . 0 0 . 1 0 0 375 . 0 0 5 . 0 0
0 0 1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
0 0 375 . 0 0 . 1 0 375 . 0 0 5 . 0
1875 . 0 375 . 0 0 0 375 . 0 1875 . 0 0 375 . 0
0 0 1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
375 . 0 5 . 0 0 0 0 375 . 0 0 . 2 5 . 0
0 0 375 . 0 5 . 0 375 . 0 0 . 0 5 . 0 0 . 2
zz
EI K
(5)
Now it is required to replace the given members loads by equivalent joint loads.
The equivalent loads for the present case is shown in Fig. 28.1d. The
displacement degrees of freedom are also shown in Fig. 28.1d.
Page 5
Instructional Objectives
After reading this chapter the student will be able to
1. Derive member stiffness matrix of a beam element.
2. Assemble member stiffness matrices to obtain the global stiffness matrix for a
beam.
3. Write the global load-displacement relation for the beam.
4. Impose boundary conditions on the load-displacement relation of the beam.
5. Analyse continuous beams by the direct stiffness method.
28.1 Introduction
In the last lesson, the procedure to analyse beams by direct stiffness method has
been discussed. No numerical problems are given in that lesson. In this lesson,
few continuous beam problems are solved numerically by direct stiffness method.
Example 28.1
Analyse the continuous beam shown in Fig. 28.1a. Assume that the supports are
unyielding. Also assume that EI is constant for all members.
The numbering of joints and members are shown in Fig. 28.1b. The possible
global degrees of freedom are shown in the figure. Numbers are put for the
unconstrained degrees of freedom first and then that for constrained
displacements.
The given continuous beam is divided into three beam elements Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. In the given problem some displacements
are zero, i.e., 0
8 7 6 5 4 3
= = = = = = u u u u u u from support conditions.
In the case of beams, it is not required to transform member stiffness matrix from
local co-ordinate system to global co-ordinate system, as the two co-ordinate
system are parallel to each other.
First construct the member stiffness matrix for each member. This may be done
from the fundamentals. However, one could use directly the equation (27.1)
given in the previous lesson and reproduced below for the sake convenience.
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
k
z z z z
z z z z
z z z z
z z z z
4 6 2 6
6 12 6 12
2 6 4 6
6 12 6 12
2 2
2 3 2 3
2 2
2 3 2 3
(1)
The degrees of freedom of a typical beam member are shown in Fig. 28.1c.
Here equation (1) is used to generate element stiffness matrix.
Member 1: , node points 1-2. m L 4 =
The member stiffness matrix for all the members are the same, as the length
and flexural rigidity of all members is the same.
[]
1
3
5
6
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
'
1 3 5 6 . .
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(2)
On the member stiffness matrix, the corresponding global degrees of freedom
are indicated to facilitate assembling.
Member 2: , node points 2-3. m L 4 =
[]
2
4
1
3
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
2 4 1 3 . .
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(3)
Member 3: , node points 3-4. m L 4 =
[]
7
8
2
4
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
7 8 2 4 . .
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(4)
The assembled global stiffness matrix of the continuous beam is of the
order . The assembled global stiffness matrix may be written as, 8 8 ×
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
-
-
-
- - -
- - -
-
-
=
1875 . 0 375 . 0 0 0 1875 . 0 0 375 . 0 0
375 . 0 0 . 1 0 0 375 . 0 0 5 . 0 0
0 0 1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
0 0 375 . 0 0 . 1 0 375 . 0 0 5 . 0
1875 . 0 375 . 0 0 0 375 . 0 1875 . 0 0 375 . 0
0 0 1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
375 . 0 5 . 0 0 0 0 375 . 0 0 . 2 5 . 0
0 0 375 . 0 5 . 0 375 . 0 0 . 0 5 . 0 0 . 2
zz
EI K
(5)
Now it is required to replace the given members loads by equivalent joint loads.
The equivalent loads for the present case is shown in Fig. 28.1d. The
displacement degrees of freedom are also shown in Fig. 28.1d.
Thus the global load vector corresponding to unconstrained degree of freedom
is,
{}
?
?
?
?
?
?
?
?
?
? -
=
?
?
?
?
?
?
?
?
?
?
=
33 . 2
5
2
1
p
p
p
k
(6)
Writing the load displacement relation for the entire continuous beam,
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- - -
-
-
-
- - -
- - -
-
-
=
?
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?
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?
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? -
8
7
6
5
4
3
2
1
8
7
6
5
4
3
187 0 375 0 0 0 187 0 0 375 0 0
375 0 0 1 0 0 375 0 0 5 0 0
0 0 187 0 375 0 0 187 0 0 375 0
0 0 375 0 0 1 0 375 0 0 5 0
187 0 375 0 0 0 375 0 187 0 0 375 0
0 0 187 0 375 0 187 0 375 0 375 0 0
375 0 5 0 0 0 0 375 0 0 2 5 0
0 0 375 0 5 0 375 0 0 0 5 0 0 2
33 2
5
u
u
u
u
u
u
u
u
. . . .
. . . .
. . . .
. . . .
. . . . .
. . . . .
. . . . .
. . . . . .
EI
p
p
p
p
p
p
.
zz
(7)
where is the joint load vector, {} p { } u is displacement vector.
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FAQs on The Direct Stiffness Method: Beams - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the direct stiffness method for beams? |  |
Ans. The direct stiffness method is a numerical technique used in structural analysis to determine the displacements and internal forces of beams. It involves dividing the beam into smaller elements, calculating the stiffness matrix for each element, and assembling these matrices to form the global stiffness matrix of the entire beam. By solving the equations using the direct stiffness method, the displacements and internal forces can be obtained.
| 2. How is the direct stiffness method applied to beams? | |
Ans. To apply the direct stiffness method to beams, the following steps are typically followed:
1. Divide the beam into smaller elements, such as beam segments or finite elements.
2. Determine the stiffness matrix for each element based on its geometry, material properties, and boundary conditions.
3. Assemble the individual element stiffness matrices to form the global stiffness matrix of the entire beam.
4. Apply appropriate boundary conditions, such as fixing one end of the beam or applying external loads.
5. Solve the system of equations using matrix methods or software to obtain the displacements and internal forces of the beam.
| 3. What are the advantages of using the direct stiffness method for beam analysis? | |
Ans. The direct stiffness method offers several advantages for beam analysis:
1. Flexibility: It can handle complex beam geometries and loadings, making it suitable for a wide range of structural problems.
2. Precision: The method provides accurate results by considering the stiffness of individual elements and their interactions.
3. Efficiency: It reduces the computational effort by dividing the beam into smaller elements and solving equations for each element separately.
4. Compatibility: The direct stiffness method can be integrated with other structural analysis techniques, such as the finite element method, to analyze more complex structures.
5. Versatility: It can be applied to various types of beams, including straight beams, curved beams, and beams with different cross-sectional shapes.
| 4. Are there any limitations or assumptions associated with the direct stiffness method for beam analysis? | |
Ans. Yes, there are some limitations and assumptions associated with the direct stiffness method:
1. Linear behavior: The method assumes that the materials and structural behavior of the beams are linear, neglecting any nonlinear effects.
2. Small deformations: It assumes that the deformations in the beam are small, ensuring that the linear elastic theory is applicable.
3. Homogeneous material: The method assumes that the beam material is homogeneous, with uniform properties throughout.
4. Continuous support conditions: It assumes that the beam is continuously supported and does not consider discontinuities or joint conditions.
5. Static analysis: The direct stiffness method is primarily used for static analysis and may not be suitable for dynamic or time-dependent problems.
| 5. Can the direct stiffness method be used for other structural elements besides beams? | |
Ans. Yes, the direct stiffness method can be applied to analyze other structural elements besides beams. It is a versatile technique that can be extended to analyze frames, trusses, and other types of structures. By dividing these structures into smaller elements and applying the same principles of calculating element stiffness matrices and assembling them into a global stiffness matrix, the direct stiffness method can provide solutions for various types of structural analysis problems.
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The Direct Stiffness Method: Beams - 3 Civil Engineering (CE) Questions
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