Civil Engineering (CE) Exam > Civil Engineering (CE) Notes > Structural Analysis > The Direct Stiffness Method: Plane Frames - 3
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Page 1
The load vector corresponding to unconstrained degrees of freedom is (vide
30.4d),
{}
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=
6
24
12
6
5
4
p
p
p
p
k
(4)
In the given frame constraint degrees of freedom are .
Eliminating rows and columns corresponding to constrained degrees of freedom
from global stiffness matrix and writing load-displacement relationship for only
unconstrained degree of freedom,
9 8 7 3 2 1
, , , , , u u u u u u
Page 2
The load vector corresponding to unconstrained degrees of freedom is (vide
30.4d),
{}
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=
6
24
12
6
5
4
p
p
p
p
k
(4)
In the given frame constraint degrees of freedom are .
Eliminating rows and columns corresponding to constrained degrees of freedom
from global stiffness matrix and writing load-displacement relationship for only
unconstrained degree of freedom,
9 8 7 3 2 1
, , , , , u u u u u u
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-
-
6
5
4
3
44 . 44 10 44 . 4
10 3 . 1338 0
44 . 4 0 5 . 2001
10
6
24
12
u
u
u
(5)
Solving we get,
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×
×
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3 -
5 -
-6
6
5
4
10 0.13 -
10 1.695 -
10 6.28
u
u
u
(6)
65
45
6.28 10 m., 1.695 10 uu
--
=× =- ×
Let be the support reactions along degrees of freedom
respectively (vide Fig. 30.4e). Support reactions are calculated by
9 8 7 3 2 1
, , , , , R R R R R R
9 , 8 , 7 , 3 , 2 , 1
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9
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3
2
1
9
8
7
3
2
1
33 . 13 10 0
10 5 0
0 0 2000
88 . 8 0 44 . 4
0 3 . 1333
44 . 4 0 48 . 1
10
6 5 4
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
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92 . 25
40 . 25
57 . 12
85 . 16
59 . 22
42 . 11
92 . 1
40 . 1
57 . 12
14 . 1
59 . 22
57 . 0
24
24
0
18
0
12
9
8
7
3
2
1
R
R
R
R
R
R
(7)
Page 3
The load vector corresponding to unconstrained degrees of freedom is (vide
30.4d),
{}
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=
6
24
12
6
5
4
p
p
p
p
k
(4)
In the given frame constraint degrees of freedom are .
Eliminating rows and columns corresponding to constrained degrees of freedom
from global stiffness matrix and writing load-displacement relationship for only
unconstrained degree of freedom,
9 8 7 3 2 1
, , , , , u u u u u u
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-
-
6
5
4
3
44 . 44 10 44 . 4
10 3 . 1338 0
44 . 4 0 5 . 2001
10
6
24
12
u
u
u
(5)
Solving we get,
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×
×
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3 -
5 -
-6
6
5
4
10 0.13 -
10 1.695 -
10 6.28
u
u
u
(6)
65
45
6.28 10 m., 1.695 10 uu
--
=× =- ×
Let be the support reactions along degrees of freedom
respectively (vide Fig. 30.4e). Support reactions are calculated by
9 8 7 3 2 1
, , , , , R R R R R R
9 , 8 , 7 , 3 , 2 , 1
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6
5
4
3
9
8
7
3
2
1
9
8
7
3
2
1
33 . 13 10 0
10 5 0
0 0 2000
88 . 8 0 44 . 4
0 3 . 1333
44 . 4 0 48 . 1
10
6 5 4
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
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92 . 25
40 . 25
57 . 12
85 . 16
59 . 22
42 . 11
92 . 1
40 . 1
57 . 12
14 . 1
59 . 22
57 . 0
24
24
0
18
0
12
9
8
7
3
2
1
R
R
R
R
R
R
(7)
Example 30.2
Analyse the rigid frame shown in Fig 30.5a by direct stiffness matrix method.
Assume and . The flexural
rigidity
54
200 GPa ; 1.33 10 m
ZZ
EI
-
==×
2
01 . 0 m A =
EI and axial rigidity EA are the same for all beams.
Solution:
The plane frame is divided in to three beam elements as shown in Fig. 30.5b.
The numbering of joints and members are also shown in Fig. 30.5b. The possible
degrees of freedom at nodes are also shown in the figure. The origin of the
global co- ordinate system is taken at A (node 1).
Page 4
The load vector corresponding to unconstrained degrees of freedom is (vide
30.4d),
{}
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?
?
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?
-
- =
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?
?
?
?
?
?
?
=
6
24
12
6
5
4
p
p
p
p
k
(4)
In the given frame constraint degrees of freedom are .
Eliminating rows and columns corresponding to constrained degrees of freedom
from global stiffness matrix and writing load-displacement relationship for only
unconstrained degree of freedom,
9 8 7 3 2 1
, , , , , u u u u u u
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-
-
6
5
4
3
44 . 44 10 44 . 4
10 3 . 1338 0
44 . 4 0 5 . 2001
10
6
24
12
u
u
u
(5)
Solving we get,
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×
×
×
=
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?
?
?
?
3 -
5 -
-6
6
5
4
10 0.13 -
10 1.695 -
10 6.28
u
u
u
(6)
65
45
6.28 10 m., 1.695 10 uu
--
=× =- ×
Let be the support reactions along degrees of freedom
respectively (vide Fig. 30.4e). Support reactions are calculated by
9 8 7 3 2 1
, , , , , R R R R R R
9 , 8 , 7 , 3 , 2 , 1
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6
5
4
3
9
8
7
3
2
1
9
8
7
3
2
1
33 . 13 10 0
10 5 0
0 0 2000
88 . 8 0 44 . 4
0 3 . 1333
44 . 4 0 48 . 1
10
6 5 4
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
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92 . 25
40 . 25
57 . 12
85 . 16
59 . 22
42 . 11
92 . 1
40 . 1
57 . 12
14 . 1
59 . 22
57 . 0
24
24
0
18
0
12
9
8
7
3
2
1
R
R
R
R
R
R
(7)
Example 30.2
Analyse the rigid frame shown in Fig 30.5a by direct stiffness matrix method.
Assume and . The flexural
rigidity
54
200 GPa ; 1.33 10 m
ZZ
EI
-
==×
2
01 . 0 m A =
EI and axial rigidity EA are the same for all beams.
Solution:
The plane frame is divided in to three beam elements as shown in Fig. 30.5b.
The numbering of joints and members are also shown in Fig. 30.5b. The possible
degrees of freedom at nodes are also shown in the figure. The origin of the
global co- ordinate system is taken at A (node 1).
Now formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. In the present case three degrees of
freedom are considered at each node.
Member 1: ° = = 90 ; 4 ? m L ; node points 1-2 ;
21
0
xx
l
L
-
== and
21
1
yy
m
L
-
== .
The following terms are common for all elements.
52
2
6
5 10 kN/m; 9.998 10 kN
AE EI
L L
=× = ×
23
3
12 4
4.999 10 kN/m; 2.666 10 kN.m
EI EI
L L
=× =×
Page 5
The load vector corresponding to unconstrained degrees of freedom is (vide
30.4d),
{}
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
6
24
12
6
5
4
p
p
p
p
k
(4)
In the given frame constraint degrees of freedom are .
Eliminating rows and columns corresponding to constrained degrees of freedom
from global stiffness matrix and writing load-displacement relationship for only
unconstrained degree of freedom,
9 8 7 3 2 1
, , , , , u u u u u u
?
?
?
?
?
?
?
?
?
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=
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?
?
?
?
-
-
6
5
4
3
44 . 44 10 44 . 4
10 3 . 1338 0
44 . 4 0 5 . 2001
10
6
24
12
u
u
u
(5)
Solving we get,
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?
?
?
?
?
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?
?
?
?
?
?
×
×
×
=
?
?
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?
?
?
?
?
?
?
?
?
?
3 -
5 -
-6
6
5
4
10 0.13 -
10 1.695 -
10 6.28
u
u
u
(6)
65
45
6.28 10 m., 1.695 10 uu
--
=× =- ×
Let be the support reactions along degrees of freedom
respectively (vide Fig. 30.4e). Support reactions are calculated by
9 8 7 3 2 1
, , , , , R R R R R R
9 , 8 , 7 , 3 , 2 , 1
?
?
?
?
?
?
?
?
?
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+
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6
5
4
3
9
8
7
3
2
1
9
8
7
3
2
1
33 . 13 10 0
10 5 0
0 0 2000
88 . 8 0 44 . 4
0 3 . 1333
44 . 4 0 48 . 1
10
6 5 4
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
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92 . 25
40 . 25
57 . 12
85 . 16
59 . 22
42 . 11
92 . 1
40 . 1
57 . 12
14 . 1
59 . 22
57 . 0
24
24
0
18
0
12
9
8
7
3
2
1
R
R
R
R
R
R
(7)
Example 30.2
Analyse the rigid frame shown in Fig 30.5a by direct stiffness matrix method.
Assume and . The flexural
rigidity
54
200 GPa ; 1.33 10 m
ZZ
EI
-
==×
2
01 . 0 m A =
EI and axial rigidity EA are the same for all beams.
Solution:
The plane frame is divided in to three beam elements as shown in Fig. 30.5b.
The numbering of joints and members are also shown in Fig. 30.5b. The possible
degrees of freedom at nodes are also shown in the figure. The origin of the
global co- ordinate system is taken at A (node 1).
Now formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. In the present case three degrees of
freedom are considered at each node.
Member 1: ° = = 90 ; 4 ? m L ; node points 1-2 ;
21
0
xx
l
L
-
== and
21
1
yy
m
L
-
== .
The following terms are common for all elements.
52
2
6
5 10 kN/m; 9.998 10 kN
AE EI
L L
=× = ×
23
3
12 4
4.999 10 kN/m; 2.666 10 kN.m
EI EI
L L
=× =×
[] [] [ ][]
6
5
4
3
2
1
10 66 . 2 0 10 1 10 33 . 1 0 10 1
0 10 5 0 0 10 5 0
10 1 0 10 50 . 0 10 1 0 10 50 . 0
10 33 . 1 0 10 1 10 66 . 2 0 10 1
0 10 5 0 0 10 5 0
10 1 0 10 50 . 0 10 1 0 10 50 . 0
6 5 4 3 2 1
'
3 3 3 3
5 5
3 3 3 3
3 3 3 3
5 5
3 3 3 3
1
?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
× × × × -
× × -
× × × × -
× × × × -
× - ×
× - × - × - ×
=
= T k T k
T
(1)
Member 2: ° = = 0 ; 4 ? m L node points 2-3; 1 = l and 0 = m .
[] [] [ ][ ]
9
8
7
6
5
4
10 666 . 2 10 1 0 10 33 . 1 10 1 0
10 1 10 5 . 0 0 10 1 10 5 . 0 0
0 0 10 0 . 5 0 0 10 0 . 5
10 33 . 1 10 1 0 10 666 . 2 10 1 0
10 1 10 5 . 0 0 10 1 10 5 . 0 0
0 0 10 0 . 5 0 0 10 0 . 5
9 8 7 6 5 4
'
3 3 3 3
3 3 3 3
6 6
3 3 3 3
3 3 3 3
6 5
2
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?
× × - × ×
× - × × - × -
× × -
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FAQs on The Direct Stiffness Method: Plane Frames - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the direct stiffness method in plane frames? |  |
Ans. The direct stiffness method is a structural analysis technique used to determine the displacements, forces, and moments in a plane frame structure. It involves creating a stiffness matrix for each element of the frame and assembling them to form the global stiffness matrix. By applying boundary conditions and solving the resulting system of equations, the method allows for the determination of the frame's response to applied loads.
| 2. How is the stiffness matrix of an element determined in the direct stiffness method? | |
Ans. The stiffness matrix of an element in the direct stiffness method is determined by considering the material properties, geometry, and boundary conditions of the element. The stiffness matrix relates the element's forces and moments to its displacements and rotations. It is typically derived using the principle of virtual work or the finite element method, taking into account the element's axial, shear, and flexural behavior.
| 3. What are the advantages of using the direct stiffness method in plane frame analysis? | |
Ans. The direct stiffness method offers several advantages in plane frame analysis. Firstly, it provides an accurate representation of the structural behavior by considering the stiffness and flexibility of each element. Secondly, it allows for the analysis of complex structures with different types of members and varying boundary conditions. Additionally, the method is computationally efficient, making it suitable for large-scale structural analysis. Finally, it can handle non-linear behavior by incorporating appropriate constitutive models into the stiffness matrix formulation.
| 4. What are the limitations of the direct stiffness method in plane frame analysis? | |
Ans. While the direct stiffness method is a powerful tool for plane frame analysis, it does have some limitations. One limitation is that it assumes linear elastic behavior, which may not accurately represent the real-world response of certain materials or structures. Additionally, the method requires the input of accurate and precise material and geometric properties, which can be challenging to obtain in practice. Finally, the direct stiffness method can become computationally intensive for structures with a large number of elements or complex boundary conditions.
| 5. How does the direct stiffness method handle boundary conditions in plane frame analysis? | |
Ans. The direct stiffness method handles boundary conditions in plane frame analysis by incorporating them into the global stiffness matrix. Boundary conditions, such as fixed supports or prescribed displacements, are represented as known values in the displacement vector or as zero entries in the global stiffness matrix. By applying the appropriate boundary conditions, the method allows for the determination of displacements, forces, and moments throughout the frame.
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