The Direct Stiffness Method: Beams - 3 GATE Notes | EduRev

Structural Analysis

GATE : The Direct Stiffness Method: Beams - 3 GATE Notes | EduRev

 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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- - -
-
-
-
- - -
- - -
-
-
=
?
?
?
?
?
?
?
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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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