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Page 2
The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and 0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1
Page 3
The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and 0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1
[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(2)
Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4
[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)
Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2
[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(4)
Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1
[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(5)
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2
Page 4
The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and 0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1
[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(2)
Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4
[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)
Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2
[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(4)
Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1
[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(5)
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2
[]
4
3
8
7
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
4 3 8 7
6
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(6)
There are eight possible global degrees of freedom for the truss shown in the
figure. Hence the global stiffness matrix is of the order ( 8 8 × ). On the member
stiffness matrix, the corresponding global degrees of freedom are indicated to
facilitate assembly. Thus the global stiffness matrix is,
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
AE K (7)
The force-displacement relation for the truss is,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
(8)
The displacements
2 1
,u u
and
3
u
are unknowns.
Here,
12 3
5kN ; 10 ; 0 pp p ==- =
and
0
8 7 6 5 4
= = = = = u u u u u
.
Page 5
The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and 0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1
[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(2)
Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4
[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)
Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2
[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(4)
Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1
[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(5)
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2
[]
4
3
8
7
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
4 3 8 7
6
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(6)
There are eight possible global degrees of freedom for the truss shown in the
figure. Hence the global stiffness matrix is of the order ( 8 8 × ). On the member
stiffness matrix, the corresponding global degrees of freedom are indicated to
facilitate assembly. Thus the global stiffness matrix is,
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
AE K (7)
The force-displacement relation for the truss is,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
(8)
The displacements
2 1
,u u
and
3
u
are unknowns.
Here,
12 3
5kN ; 10 ; 0 pp p ==- =
and
0
8 7 6 5 4
= = = = = u u u u u
.
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
0
0
0
0
0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
0
10
5
3
2
1
8
7
6
5
4
u
u
u
EA
p
p
p
p
p
(9)
Thus,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
-
3
2
1
271 . 0 0 20 . 0
0 271 . 0 071 . 0
20 . 0 071 . 0 271 . 0
0
10
5
u
u
u
(10)
Solving which, yields
AE
u
AE
u
AE
u
825 . 53
;
97 . 55
;
855 . 72
3 2 1
=
-
= =
Now reactions are evaluated from the equation,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
3
2
1
8
7
6
5
4
071 . 0 20 . 0 0
071 . 0 0 0
0 071 . 0 071 . 0
0 071 . 0 071 . 0
071 . 0 0 0
u
u
u
p
p
p
p
p
(11)
45 6 7 8
3.80 kN ; 1.19 kN ; 1.19 kN ; 3.8 0 ; 15.00 kN pp p p kNp =- =- =- = =
In the next step evaluate forces in members.
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1
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FAQs on The Direct Stiffness Method: Truss Analysis - 5 - 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 calculating the stiffness matrix for each element. By assembling these stiffness matrices, the overall stiffness matrix of the truss can be obtained, which can then be used to solve for the displacements and internal forces in the truss.
| 2. How does the direct stiffness method work in truss analysis? | |
Ans. The direct stiffness method works by considering each truss element as a spring-like structure. The stiffness matrix for each element is calculated based on its material properties and geometry. These individual stiffness matrices are then combined to form the overall stiffness matrix for the truss. Using the boundary conditions and applied loads, the displacements and internal forces in the truss can be determined by solving a system of equations formed by the 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. Firstly, it provides an accurate representation of the truss behavior, taking into account the flexibility of each element. Secondly, it allows for the analysis of complex truss structures with varying material properties and geometries. Additionally, it is a versatile method that can be extended to analyze other structural systems beyond trusses, such as frames and beams.
| 4. Are there any limitations to the direct stiffness method in truss analysis? | |
Ans. While the direct stiffness method is a powerful tool in truss analysis, it does have some limitations. One limitation is that it assumes linear behavior of the truss elements, which may not hold for large deformations or nonlinear materials. Additionally, the method requires the truss to be statically determinate, meaning that it should have fewer unknowns than the number of equations. If the truss is indeterminate, additional techniques such as the flexibility method or the use of redundant members may be needed.
| 5. Can the direct stiffness method be used for truss analysis in three-dimensional structures? | |
Ans. Yes, the direct stiffness method can be extended to analyze truss structures in three dimensions. In three-dimensional truss analysis, the stiffness matrix for each element is calculated based on its geometry and material properties in three dimensions. The overall stiffness matrix is then formed by assembling these three-dimensional stiffness matrices. By solving the resulting system of equations, the displacements and internal forces in the three-dimensional truss can be determined.
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The Direct Stiffness Method: Truss Analysis - 5 Civil Engineering (CE) Questions
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