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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 = = - = .
Read More
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FAQs on The Direct Stiffness Method: Truss Analysis - 4 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Direct Stiffness Method in truss analysis? |  |
Ans. The Direct Stiffness Method is a numerical technique used in truss analysis to determine the displacements, forces, and stresses in a truss structure. It involves creating a stiffness matrix for the truss and solving a system of equations to obtain the unknowns.
| 2. How is the stiffness matrix created in the Direct Stiffness Method for truss analysis? | |
Ans. To create the stiffness matrix, the Direct Stiffness Method considers the individual stiffness of each truss member. These stiffness values are combined using the principle of superposition to form the overall stiffness matrix for the truss structure.
| 3. What are the advantages of using the Direct Stiffness Method in truss analysis? | |
Ans. The Direct Stiffness Method offers several advantages in truss analysis. It provides an efficient and systematic approach to solve complex truss structures, considering geometric and material nonlinearity. Additionally, it allows for the analysis of trusses with different support conditions and enables the determination of internal forces and displacements accurately.
| 4. How does the Direct Stiffness Method handle boundary conditions in truss analysis? | |
Ans. The Direct Stiffness Method incorporates the boundary conditions of a truss structure through the use of support displacement constraints and known forces applied at specific nodes. By considering these constraints, the stiffness matrix is modified to account for the fixed or supported nodes, resulting in an accurate analysis of the truss structure.
| 5. Can the Direct Stiffness Method be applied to analyze trusses with varying cross-sectional areas? | |
Ans. Yes, the Direct Stiffness Method can be applied to analyze trusses with varying cross-sectional areas. By assigning different stiffness values to each member based on their respective cross-sectional areas, the method can accurately account for the variation in stiffness throughout the truss structure. This enables the analysis of trusses with non-uniform member properties.
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Nov 15, 2024 Last updated |
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