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Page 1
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
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6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
Page 2
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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-
-
=
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Page 3
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
-
-
=
?
?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. The origin of the global co-ordinate
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given.
Member 1: ° = = 90 ; 6 ? m L node points 1-2; 0 = l and 1 = m .
Page 4
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. The origin of the global co-ordinate
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given.
Member 1: ° = = 90 ; 6 ? m L node points 1-2; 0 = l and 1 = m .
[ ] [ ] [ ]
1
'
T
kT k T ?? =
??
3333
66
333
1
3333
66
333
1 2 3 4 5 6
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 17.78 10 4.44 10 0 8.88 10
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 8.88 10 4.44 10 0 17.
k
××××
×-×
××× ×
?? =
??
××××
-× ×
×××
3
1
2
3
4
5
6 78 10
??
??
??
??
??
??
??
??
× ??
??
3
(1)
Member 2: ° = = 0 ; 4 ? m L ; node points 2-3 ; 1 = l and 0 = m .
[] [] [ ][ ]
9
8
7
6
5
4
10 66 . 26 10 10 0 10 88 . 8 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
10 88 . 8 10 10 0 10 66 . 26 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
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 6
2
?
?
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?
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?
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?
?
?
× × - × ×
× - × × - × -
× × -
× × - × ×
× × - × ×
× - ×
=
= T k T k
T
(2)
The assembled global stiffness matrix [ ] K is of the order 9 9 × . Carrying out
assembly in the usual manner, we get,
Page 5
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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-
-
=
?
?
?
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?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. The origin of the global co-ordinate
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given.
Member 1: ° = = 90 ; 6 ? m L node points 1-2; 0 = l and 1 = m .
[ ] [ ] [ ]
1
'
T
kT k T ?? =
??
3333
66
333
1
3333
66
333
1 2 3 4 5 6
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 17.78 10 4.44 10 0 8.88 10
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 8.88 10 4.44 10 0 17.
k
××××
×-×
××× ×
?? =
??
××××
-× ×
×××
3
1
2
3
4
5
6 78 10
??
??
??
??
??
??
??
??
× ??
??
3
(1)
Member 2: ° = = 0 ; 4 ? m L ; node points 2-3 ; 1 = l and 0 = m .
[] [] [ ][ ]
9
8
7
6
5
4
10 66 . 26 10 10 0 10 88 . 8 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
10 88 . 8 10 10 0 10 66 . 26 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
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 6
2
?
?
?
?
?
?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
× × - × ×
× - × × - × -
× × -
× × - × ×
× × - × ×
× - ×
=
= T k T k
T
(2)
The assembled global stiffness matrix [ ] K is of the order 9 9 × . Carrying out
assembly in the usual manner, we get,
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
-
- - -
-
- -
- -
- -
-
-
- - -
=
66 . 26 10 0 33 . 13 10 0 0 0 0
10 5 0 10 5 0 0 0 0
0 0 2000 0 0 2000 0 0 0
33 . 13 10 0 44 . 44 10 44 . 4 88 . 8 0 44 . 4
10 5 0 10 3 . 1338 0 0 3 . 1333 0
0 0 2000 44 . 4 0 5 . 2001 44 . 4 0 48 . 1
0 0 0 88 . 8 0 44 . 4 77 . 17 0 44 . 4
0 0 0 0 3 . 1333 0 0 3 . 1333 0
0 0 0 44 . 4 0 48 . 1 44 . 4 0 48 . 1
K (3)
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FAQs on The Direct Stiffness Method: Plane Frames - 2 - Structural Analysis - Civil Engineering (CE)
| 1. What is the direct stiffness method in structural analysis? |  |
Ans. The direct stiffness method is a numerical technique used in structural analysis to determine the displacements, forces, and moments in a structure. It involves representing the structure as an interconnected system of elements, such as beams or trusses, and using the stiffness properties of these elements to calculate the response of the entire structure.
| 2. How is the direct stiffness method applied to plane frames? | |
Ans. In the direct stiffness method for plane frames, the structure is divided into individual members (beams or columns) and nodes (points where members are connected). The stiffness matrix for each member is determined based on its geometry and material properties. Then, the global stiffness matrix for the entire frame is assembled by combining the individual member stiffness matrices. By applying appropriate boundary conditions and solving the resulting system of equations, the displacements and forces in the frame can be determined.
| 3. What are the advantages of using the direct stiffness method in structural analysis? | |
Ans. The direct stiffness method offers several advantages in structural analysis. Firstly, it provides an accurate and efficient way to analyze complex structures, including those with irregular geometries or loadings. Secondly, it allows for easy incorporation of different types of boundary conditions, such as fixed supports or pinned connections. Additionally, it can handle both linear and nonlinear behavior of materials and supports, making it versatile for a wide range of structural problems.
| 4. Are there any limitations or assumptions associated with the direct stiffness method? | |
Ans. Yes, there are certain limitations and assumptions associated with the direct stiffness method. One major assumption is that the structure being analyzed is linearly elastic, meaning it obeys Hooke's law. This assumption may not hold for materials that exhibit significant nonlinear behavior, such as plastic deformation. Additionally, the method assumes that the structure is statically determinate, meaning the number of unknown displacements is equal to the number of equations. If the structure is statically indeterminate, additional techniques, such as the flexibility or stiffness method, need to be used.
| 5. How does the direct stiffness method contribute to the analysis of plane frames in civil engineering? | |
Ans. The direct stiffness method plays a crucial role in the analysis of plane frames in civil engineering. It allows engineers to determine the internal forces and deformations in frame structures, which are essential for assessing their stability, strength, and overall performance. By using this method, engineers can optimize the design of plane frames, ensuring they can withstand the applied loads and meet safety requirements. Additionally, the direct stiffness method facilitates the evaluation of structural modifications or repairs, helping engineers make informed decisions for retrofitting existing frames.
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The Direct Stiffness Method: Plane Frames - 2 Civil Engineering (CE) Questions
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