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Page 1
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate
axes are shown at node 3. At node 2, roller is supported on inclined support.
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that
4
u could be
set to zero. All the possible displacement degrees of freedom are also shown in
the figure. In the first step calculate member stiffness matrix.
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2
12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
(1)
Page 2
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate
axes are shown at node 3. At node 2, roller is supported on inclined support.
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that
4
u could be
set to zero. All the possible displacement degrees of freedom are also shown in
the figure. In the first step calculate member stiffness matrix.
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2
12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
(1)
Member 2: . 00 . 4 , 30 , 0
"
m L
x x
= ° = ° = ? ? Nodal points 2-3
50 . 0 " ; 866 . 0 " ; 0 ; 1 = = = = m l m l .
Page 3
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate
axes are shown at node 3. At node 2, roller is supported on inclined support.
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that
4
u could be
set to zero. All the possible displacement degrees of freedom are also shown in
the figure. In the first step calculate member stiffness matrix.
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2
12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
(1)
Member 2: . 00 . 4 , 30 , 0
"
m L
x x
= ° = ° = ? ? Nodal points 2-3
50 . 0 " ; 866 . 0 " ; 0 ; 1 = = = = m l m l .
[]
4
3
6
5
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 4
4 3 6 5
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
(2)
Member 3: . 00 . 3 , 90 m L
x
= ° = ? , 1 ; 0 = = m l Nodal points 3-1
.
[]
2
1
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 3
2 1 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)
For the present problem, the global stiffness matrix is of the order() 6 6 × . The
global stiffness matrix for the entire truss is.
[]
6
5
4
3
2
1
333 . 0 0 0 0 333 . 0 0
0 25 . 0 125 . 0 217 . 0 0 0
0 125 . 0 065 . 0 132 . 0 014 . 0 . 019 . 0
0 217 . 0 132 . 0 385 . 0 119 . 0 159 . 0
333 . 0 0 014 . 0 119 . 0 405 . 0 096 . 0
0 0 019 . 0 159 . 0 096 . 0 128 . 0
6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
- -
- - - -
-
= EA k
(4)
Writing load-displacement equation for the truss for unconstrained degrees of
freedom,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
=
?
?
?
?
?
?
?
?
?
? -
3
2
1
385 . 0 119 . 0 159 . 0
119 . 0 405 . 0 096 . 0
159 . 0 096 . 0 128 . 0
0
5
5
u
u
u
(5)
Solving ,
AE
u
AE
u
AE
u
12 . 33
;
728 . 3
;
408 . 77
3 2 1
= =
-
=
(6)
Page 4
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate
axes are shown at node 3. At node 2, roller is supported on inclined support.
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that
4
u could be
set to zero. All the possible displacement degrees of freedom are also shown in
the figure. In the first step calculate member stiffness matrix.
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2
12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
(1)
Member 2: . 00 . 4 , 30 , 0
"
m L
x x
= ° = ° = ? ? Nodal points 2-3
50 . 0 " ; 866 . 0 " ; 0 ; 1 = = = = m l m l .
[]
4
3
6
5
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 4
4 3 6 5
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
(2)
Member 3: . 00 . 3 , 90 m L
x
= ° = ? , 1 ; 0 = = m l Nodal points 3-1
.
[]
2
1
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 3
2 1 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)
For the present problem, the global stiffness matrix is of the order() 6 6 × . The
global stiffness matrix for the entire truss is.
[]
6
5
4
3
2
1
333 . 0 0 0 0 333 . 0 0
0 25 . 0 125 . 0 217 . 0 0 0
0 125 . 0 065 . 0 132 . 0 014 . 0 . 019 . 0
0 217 . 0 132 . 0 385 . 0 119 . 0 159 . 0
333 . 0 0 014 . 0 119 . 0 405 . 0 096 . 0
0 0 019 . 0 159 . 0 096 . 0 128 . 0
6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
- -
- - - -
-
= EA k
(4)
Writing load-displacement equation for the truss for unconstrained degrees of
freedom,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
=
?
?
?
?
?
?
?
?
?
? -
3
2
1
385 . 0 119 . 0 159 . 0
119 . 0 405 . 0 096 . 0
159 . 0 096 . 0 128 . 0
0
5
5
u
u
u
(5)
Solving ,
AE
u
AE
u
AE
u
12 . 33
;
728 . 3
;
408 . 77
3 2 1
= =
-
=
(6)
Now reactions are evaluated from the equation
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
-
- =
?
?
?
?
?
?
?
?
?
?
12 . 33
728 . 3
40 . 77
1
0 333 . 0 0
217 . 0 0 0
132 . 0 014 . 0 . 019 . 0
6
5
4
AE
AE
p
p
p
(6)
45 6
2.85 kN ; 7.19 kN ; 1.24 kN pp p ==- =-
Summary
Sometimes the truss is supported on a roller placed on an oblique plane. In such
situations, the direct stiffness method as discussed in the previous lesson needs
to be properly modified to make the displacement perpendicular to the roller
support as zero. In the present approach, the inclined support is handled in the
analysis by suitably modifying the member stiffness matrices of all members
meeting at the inclined support. A few problems are solved to illustrate the
procedure.
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FAQs on The Direct Stiffness Method: Truss Analysis - 7 - 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 to analyze truss structures. It involves breaking down the truss into individual elements and applying the principles of equilibrium and compatibility to solve for unknown forces and displacements. The method utilizes the stiffness matrix, which relates the forces and displacements of each element, to obtain the overall behavior of the truss.
| 2. How does the direct stiffness method work in truss analysis? | |
Ans. The direct stiffness method works by dividing the truss into smaller elements and representing each element with a stiffness matrix. The stiffness matrix relates the forces and displacements at each node of the truss element. By assembling these stiffness matrices for all elements and considering boundary conditions, a global stiffness matrix is formed. The unknown displacements and forces can then be solved by solving the system of equations formed by the global stiffness matrix.
| 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 allows for the efficient analysis of complex truss structures by breaking them down into smaller, more manageable elements. The method also accounts for the stiffness and flexibility of the truss members, making it applicable to real-world situations. Additionally, the direct stiffness method can handle different types of loading conditions and boundary conditions, providing accurate results for various scenarios.
| 4. What are the limitations of the direct stiffness method in truss analysis? | |
Ans. While the direct stiffness method is a powerful tool for truss analysis, it does have some limitations. One limitation is that it assumes linear behavior of the truss members, which may not hold true in certain cases. Additionally, the method requires a significant amount of computational resources and can become time-consuming for large and complex truss structures. Furthermore, the direct stiffness method may not be suitable for analyzing trusses with large displacements or nonlinear material properties.
| 5. How is the direct stiffness method used in practical applications of truss analysis? | |
Ans. The direct stiffness method is widely used in engineering practice for the analysis of truss structures. It is employed in the design and analysis of bridges, buildings, aerospace structures, and various mechanical systems. The method provides engineers with valuable insights into the behavior and performance of trusses under different loading conditions. By accurately predicting forces, displacements, and stresses, the direct stiffness method helps ensure the structural integrity and safety of truss-based systems.
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