The Direct Stiffness Method: Truss Analysis - 4 GATE Notes | EduRev

Structural Analysis

GATE : The Direct Stiffness Method: Truss Analysis - 4 GATE Notes | EduRev

 Page 1


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Transform member stiffness matrix from local to global co-ordinate system. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix. 
3. Analyse plane truss by the direct stiffness matrix. 
4. Analyse plane truss supported on inclined roller supports. 
 
 
25.1 Introduction  
In the previous lesson, the direct stiffness method as applied to trusses was 
discussed. The transformation of force and displacement from local co-ordinate 
system to global co-ordinate system were accomplished by single transformation 
matrix. Also assembly of the member stiffness matrices was discussed. In this 
lesson few plane trusses are analysed using the direct stiffness method. Also the 
problem of inclined support will be discussed. 
 
Example 25.1  
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be 
constant for all the members. 
 
 
 
Page 2


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Transform member stiffness matrix from local to global co-ordinate system. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix. 
3. Analyse plane truss by the direct stiffness matrix. 
4. Analyse plane truss supported on inclined roller supports. 
 
 
25.1 Introduction  
In the previous lesson, the direct stiffness method as applied to trusses was 
discussed. The transformation of force and displacement from local co-ordinate 
system to global co-ordinate system were accomplished by single transformation 
matrix. Also assembly of the member stiffness matrices was discussed. In this 
lesson few plane trusses are analysed using the direct stiffness method. Also the 
problem of inclined support will be discussed. 
 
Example 25.1  
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be 
constant for all the members. 
 
 
 
                                                         
 
 
The numbering of joints and members are shown in Fig. 25.1b. Also, the possible 
displacements (degrees of freedom) at each node are indicated. Here lower 
numbers are used to indicate unconstrained degrees of freedom and higher 
numbers are used for constrained degrees of freedom. Thus displacements 6,7 
and 8 are zero due to boundary conditions.  
 
First write down stiffness matrix of each member in global co-ordinate system 
and assemble them to obtain global stiffness matrix. 
 
Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 
 
 []
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1) 
 
Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 
 
Page 3


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Transform member stiffness matrix from local to global co-ordinate system. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix. 
3. Analyse plane truss by the direct stiffness matrix. 
4. Analyse plane truss supported on inclined roller supports. 
 
 
25.1 Introduction  
In the previous lesson, the direct stiffness method as applied to trusses was 
discussed. The transformation of force and displacement from local co-ordinate 
system to global co-ordinate system were accomplished by single transformation 
matrix. Also assembly of the member stiffness matrices was discussed. In this 
lesson few plane trusses are analysed using the direct stiffness method. Also the 
problem of inclined support will be discussed. 
 
Example 25.1  
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be 
constant for all the members. 
 
 
 
                                                         
 
 
The numbering of joints and members are shown in Fig. 25.1b. Also, the possible 
displacements (degrees of freedom) at each node are indicated. Here lower 
numbers are used to indicate unconstrained degrees of freedom and higher 
numbers are used for constrained degrees of freedom. Thus displacements 6,7 
and 8 are zero due to boundary conditions.  
 
First write down stiffness matrix of each member in global co-ordinate system 
and assemble them to obtain global stiffness matrix. 
 
Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 
 
 []
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1) 
 
Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 
 
                                                         
 []
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 4
2
EA
k       (2) 
 
Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
3
EA
k   (3) 
 
Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2 
 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
4
EA
k      (4) 
 
Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
5
EA
k      (5) 
 
 
The assembled global stiffness matrix of the truss is of the order 8 8 × . Now 
assemble the global stiffness matrix. Note that the element 
1
11
k of the member 
stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix. 
On the member stiffness matrix the corresponding global degrees of freedom are 
indicated to facilitate assembling. Thus,  
 
Page 4


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Transform member stiffness matrix from local to global co-ordinate system. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix. 
3. Analyse plane truss by the direct stiffness matrix. 
4. Analyse plane truss supported on inclined roller supports. 
 
 
25.1 Introduction  
In the previous lesson, the direct stiffness method as applied to trusses was 
discussed. The transformation of force and displacement from local co-ordinate 
system to global co-ordinate system were accomplished by single transformation 
matrix. Also assembly of the member stiffness matrices was discussed. In this 
lesson few plane trusses are analysed using the direct stiffness method. Also the 
problem of inclined support will be discussed. 
 
Example 25.1  
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be 
constant for all the members. 
 
 
 
                                                         
 
 
The numbering of joints and members are shown in Fig. 25.1b. Also, the possible 
displacements (degrees of freedom) at each node are indicated. Here lower 
numbers are used to indicate unconstrained degrees of freedom and higher 
numbers are used for constrained degrees of freedom. Thus displacements 6,7 
and 8 are zero due to boundary conditions.  
 
First write down stiffness matrix of each member in global co-ordinate system 
and assemble them to obtain global stiffness matrix. 
 
Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 
 
 []
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1) 
 
Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 
 
                                                         
 []
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 4
2
EA
k       (2) 
 
Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
3
EA
k   (3) 
 
Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2 
 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
4
EA
k      (4) 
 
Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
5
EA
k      (5) 
 
 
The assembled global stiffness matrix of the truss is of the order 8 8 × . Now 
assemble the global stiffness matrix. Note that the element 
1
11
k of the member 
stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix. 
On the member stiffness matrix the corresponding global degrees of freedom are 
indicated to facilitate assembling. Thus,  
 
                                                         
[]
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
8 7 6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
= EA K
  
 
           (6)  
 
Writing the load-displacement relation for the truss, yields 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
           (7) 
 
The displacements 
1
u to 
5
u are unknown. The displacements 0
8 7 6
= = = u u u .  
 
Also  0
5 3 2 1
= = = = p p p p . But 
4
10 kN p = - . 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
5
4
3
2
1
487 . 0 0 433 . 0 094 . 0 054 . 0
0 25 . 0 0 25 . 0 0
433 . 0 0 866 . 0 0 0
094 . 0 25 . 0 0 575 . 0 0
054 . 0 0 0 0 108 . 0
0
10
0
0
0
u
u
u
u
u
EA
       (8)  
 
Solving which, the unknown displacements are evaluated. Thus,  
 
Page 5


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Transform member stiffness matrix from local to global co-ordinate system. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix. 
3. Analyse plane truss by the direct stiffness matrix. 
4. Analyse plane truss supported on inclined roller supports. 
 
 
25.1 Introduction  
In the previous lesson, the direct stiffness method as applied to trusses was 
discussed. The transformation of force and displacement from local co-ordinate 
system to global co-ordinate system were accomplished by single transformation 
matrix. Also assembly of the member stiffness matrices was discussed. In this 
lesson few plane trusses are analysed using the direct stiffness method. Also the 
problem of inclined support will be discussed. 
 
Example 25.1  
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be 
constant for all the members. 
 
 
 
                                                         
 
 
The numbering of joints and members are shown in Fig. 25.1b. Also, the possible 
displacements (degrees of freedom) at each node are indicated. Here lower 
numbers are used to indicate unconstrained degrees of freedom and higher 
numbers are used for constrained degrees of freedom. Thus displacements 6,7 
and 8 are zero due to boundary conditions.  
 
First write down stiffness matrix of each member in global co-ordinate system 
and assemble them to obtain global stiffness matrix. 
 
Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 
 
 []
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1) 
 
Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 
 
                                                         
 []
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 4
2
EA
k       (2) 
 
Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
3
EA
k   (3) 
 
Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2 
 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
4
EA
k      (4) 
 
Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3 
          
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
5
EA
k      (5) 
 
 
The assembled global stiffness matrix of the truss is of the order 8 8 × . Now 
assemble the global stiffness matrix. Note that the element 
1
11
k of the member 
stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix. 
On the member stiffness matrix the corresponding global degrees of freedom are 
indicated to facilitate assembling. Thus,  
 
                                                         
[]
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
8 7 6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
= EA K
  
 
           (6)  
 
Writing the load-displacement relation for the truss, yields 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
           (7) 
 
The displacements 
1
u to 
5
u are unknown. The displacements 0
8 7 6
= = = u u u .  
 
Also  0
5 3 2 1
= = = = p p p p . But 
4
10 kN p = - . 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
5
4
3
2
1
487 . 0 0 433 . 0 094 . 0 054 . 0
0 25 . 0 0 25 . 0 0
433 . 0 0 866 . 0 0 0
094 . 0 25 . 0 0 575 . 0 0
054 . 0 0 0 0 108 . 0
0
10
0
0
0
u
u
u
u
u
EA
       (8)  
 
Solving which, the unknown displacements are evaluated. Thus,  
 
                                                         
AE
u
AE
u
AE
u
AE
u
AE
u
334 . 13
;
642 . 74
;
668 . 6
;
64 . 34
;
668 . 6
5 4 3 2 1
=
-
= =
-
= =  (9) 
Now reactions are evaluated from equation, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
=
?
?
?
?
?
?
?
?
?
?
334 . 13
642 . 74
668 . 6
64 . 34
668 . 6
1
0 0 0 162 . 0 094 . 0
0 0 433 . 0 094 . 0 054 . 0
094 . 0 0 0 162 . 0 094 . 0
8
7
6
EA
EA
p
p
p
 (10) 
 
 
Thus, 
 
67 8
5.00 kN ; 0 ; 5.00 kN pp p == = .    (11) 
 
Now calculate individual member forces. 
 
Member 1:  m L m l 619 . 4 ; 866 . 0 ; 50 . 0 = = = . 
 
{} []
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
2
1
8
7
1
619 . 4
'
u
u
u
u
m l m l
AE
p 
 
{} []
1
6.667
1
' 0.5 0.866 5.77 kN
34.64 4.619
AE
p
AE
??
=- - =
??
-
??
  (12) 
 
Member 2:  m L m l 0 . 4 ; 0 . 1 ; 0 = = = . 
  
{}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
2
1
4
3
1
0 . 4
'
u
u
u
u
m l m l
AE
p 
 
{} []
1
74.642
1
' 1 1 10.0 kN
34.64 4.619
AE
p
AE
-??
=- =-
??
-
??
   (13) 
 
Member 3:  m L m l 619 . 4 ; 866 . 0 ; 50 . 0 = = - = . 
 
 
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