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# The Direct Stiffness Method: Plane Frames - 1 GATE Notes | EduRev

## GATE : The Direct Stiffness Method: Plane Frames - 1 GATE Notes | EduRev

``` Page 1

Instructional Objectives
After reading this chapter the student will be able to
1. Derive plane frame member stiffness matrix in local co-ordinate system.
2. Transform plane frame member stiffness matrix from local to global co-
ordinate system.
3. Assemble member stiffness matrices to obtain the global stiffness matrix of
the plane frame.
4. Write the global load-displacement relation for the plane frame.
5. Impose boundary conditions on the load-displacement relation.
6. Analyse plane frames by the direct stiffness matrix method.

30.1 Introduction
In the case of plane frame, all the members lie in the same plane and are
interconnected by rigid joints. The internal stress resultants at a cross-section of
a plane frame member consist of bending moment, shear force and an axial
force.  The significant deformations in the plane frame are only flexural and axial.
In this lesson, the analysis of plane frame by direct stiffness matrix method is
discussed. Initially, the stiffness matrix of the plane frame member is derived in
its local co-ordinate axes and then it is transformed to global co-ordinate system.
In the case of plane frames, members are oriented in different directions and
hence before forming the global stiffness matrix it is necessary to refer all the
member stiffness matrices to the same set of axes. This is achieved by
transformation of forces and displacements to global co-ordinate system.

30.2 Member Stiffness Matrix
Consider a member of a plane frame as shown in Fig. 30.1a in the member co-
ordinate system '.  The global orthogonal set of axes ' ' z y x xyz is also shown in the
figure. The frame lies in the xy plane. The member is assumed to have uniform
flexural rigidity EI and uniform axial rigidityEA for sake of simplicity. The axial
deformation of member will be considered in the analysis. The possible
displacements at each node of the member are: translation in - and -
direction and rotation about - axis.
' x ' y
' z

Page 2

Instructional Objectives
After reading this chapter the student will be able to
1. Derive plane frame member stiffness matrix in local co-ordinate system.
2. Transform plane frame member stiffness matrix from local to global co-
ordinate system.
3. Assemble member stiffness matrices to obtain the global stiffness matrix of
the plane frame.
4. Write the global load-displacement relation for the plane frame.
5. Impose boundary conditions on the load-displacement relation.
6. Analyse plane frames by the direct stiffness matrix method.

30.1 Introduction
In the case of plane frame, all the members lie in the same plane and are
interconnected by rigid joints. The internal stress resultants at a cross-section of
a plane frame member consist of bending moment, shear force and an axial
force.  The significant deformations in the plane frame are only flexural and axial.
In this lesson, the analysis of plane frame by direct stiffness matrix method is
discussed. Initially, the stiffness matrix of the plane frame member is derived in
its local co-ordinate axes and then it is transformed to global co-ordinate system.
In the case of plane frames, members are oriented in different directions and
hence before forming the global stiffness matrix it is necessary to refer all the
member stiffness matrices to the same set of axes. This is achieved by
transformation of forces and displacements to global co-ordinate system.

30.2 Member Stiffness Matrix
Consider a member of a plane frame as shown in Fig. 30.1a in the member co-
ordinate system '.  The global orthogonal set of axes ' ' z y x xyz is also shown in the
figure. The frame lies in the xy plane. The member is assumed to have uniform
flexural rigidity EI and uniform axial rigidityEA for sake of simplicity. The axial
deformation of member will be considered in the analysis. The possible
displacements at each node of the member are: translation in - and -
direction and rotation about - axis.
' x ' y
' z

Thus the frame members have six (6) degrees of freedom and are shown in
Fig.30.1a. The forces acting on the member at end j and  are shown in Fig.
30.1b. The relation between axial displacement and axial forces is derived in
chapter 24. Similarly the relation between shear force, bending moment with
translation along  axis and rotation about  axis are given in lesson 27.
Combining them, we could write the load-displacement relation in the local co-
ordinate axes for the plane frame as shown in Fig 30.1a, b as,
k
' y ' z

Page 3

Instructional Objectives
After reading this chapter the student will be able to
1. Derive plane frame member stiffness matrix in local co-ordinate system.
2. Transform plane frame member stiffness matrix from local to global co-
ordinate system.
3. Assemble member stiffness matrices to obtain the global stiffness matrix of
the plane frame.
4. Write the global load-displacement relation for the plane frame.
5. Impose boundary conditions on the load-displacement relation.
6. Analyse plane frames by the direct stiffness matrix method.

30.1 Introduction
In the case of plane frame, all the members lie in the same plane and are
interconnected by rigid joints. The internal stress resultants at a cross-section of
a plane frame member consist of bending moment, shear force and an axial
force.  The significant deformations in the plane frame are only flexural and axial.
In this lesson, the analysis of plane frame by direct stiffness matrix method is
discussed. Initially, the stiffness matrix of the plane frame member is derived in
its local co-ordinate axes and then it is transformed to global co-ordinate system.
In the case of plane frames, members are oriented in different directions and
hence before forming the global stiffness matrix it is necessary to refer all the
member stiffness matrices to the same set of axes. This is achieved by
transformation of forces and displacements to global co-ordinate system.

30.2 Member Stiffness Matrix
Consider a member of a plane frame as shown in Fig. 30.1a in the member co-
ordinate system '.  The global orthogonal set of axes ' ' z y x xyz is also shown in the
figure. The frame lies in the xy plane. The member is assumed to have uniform
flexural rigidity EI and uniform axial rigidityEA for sake of simplicity. The axial
deformation of member will be considered in the analysis. The possible
displacements at each node of the member are: translation in - and -
direction and rotation about - axis.
' x ' y
' z

Thus the frame members have six (6) degrees of freedom and are shown in
Fig.30.1a. The forces acting on the member at end j and  are shown in Fig.
30.1b. The relation between axial displacement and axial forces is derived in
chapter 24. Similarly the relation between shear force, bending moment with
translation along  axis and rotation about  axis are given in lesson 27.
Combining them, we could write the load-displacement relation in the local co-
ordinate axes for the plane frame as shown in Fig 30.1a, b as,
k
' y ' z

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6
5
4
3
2
1
2 2
2 3 2 3
2 2
2 3 2 3
6
5
4
3
2
1
'
'
'
'
'
'
4 6
0
2 6
0
6 12
0
6 12
0
0 0 0 0
2 6
0
4 6
0
6 12
0
6 12
0
0 0 0 0
'
'
'
'
'
'
u
u
u
u
u
u
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
q
q
q
q
q
q
z z z z
z z z z
z z z z
z z z z
(30.1a)

This may be succinctly written as

{} []{} ' ' ' u k q =      (30.1b)

where  is the member stiffness matrix. The member stiffness matrix can also
be generated by giving unit displacement along each possible displacement
degree of freedom one at a time and calculating resulting restraint actions.
[] ' k

30.3  Transformation from local to global co-ordinate system
30.3.1 Displacement transformation matrix
In plane frame the members are oriented in different directions and hence it is
necessary to transform stiffness matrix of individual members from local to global
co-ordinate system before formulating the global stiffness matrix by assembly. In
Fig. 30.2a the plane frame member is shown in local coordinate axes z y x ' ' ' and
in Fig. 30.2b, the plane frame is shown in global coordinate axesxyz . Two ends
of the plane frame member are identified by j and . Let   and
be respectively displacements of ends
k
3 2 1
' , ' , ' u u u
6 5 4
' , ' , ' u u u j and  of the member in local
coordinate system '. Similarly  and  respectively are
displacements of ends
k
' ' z y x
3 2 1
, , u u u
6 5 4
, , u u u
j and k of the member in global co-ordinate system.

Page 4

Instructional Objectives
After reading this chapter the student will be able to
1. Derive plane frame member stiffness matrix in local co-ordinate system.
2. Transform plane frame member stiffness matrix from local to global co-
ordinate system.
3. Assemble member stiffness matrices to obtain the global stiffness matrix of
the plane frame.
4. Write the global load-displacement relation for the plane frame.
5. Impose boundary conditions on the load-displacement relation.
6. Analyse plane frames by the direct stiffness matrix method.

30.1 Introduction
In the case of plane frame, all the members lie in the same plane and are
interconnected by rigid joints. The internal stress resultants at a cross-section of
a plane frame member consist of bending moment, shear force and an axial
force.  The significant deformations in the plane frame are only flexural and axial.
In this lesson, the analysis of plane frame by direct stiffness matrix method is
discussed. Initially, the stiffness matrix of the plane frame member is derived in
its local co-ordinate axes and then it is transformed to global co-ordinate system.
In the case of plane frames, members are oriented in different directions and
hence before forming the global stiffness matrix it is necessary to refer all the
member stiffness matrices to the same set of axes. This is achieved by
transformation of forces and displacements to global co-ordinate system.

30.2 Member Stiffness Matrix
Consider a member of a plane frame as shown in Fig. 30.1a in the member co-
ordinate system '.  The global orthogonal set of axes ' ' z y x xyz is also shown in the
figure. The frame lies in the xy plane. The member is assumed to have uniform
flexural rigidity EI and uniform axial rigidityEA for sake of simplicity. The axial
deformation of member will be considered in the analysis. The possible
displacements at each node of the member are: translation in - and -
direction and rotation about - axis.
' x ' y
' z

Thus the frame members have six (6) degrees of freedom and are shown in
Fig.30.1a. The forces acting on the member at end j and  are shown in Fig.
30.1b. The relation between axial displacement and axial forces is derived in
chapter 24. Similarly the relation between shear force, bending moment with
translation along  axis and rotation about  axis are given in lesson 27.
Combining them, we could write the load-displacement relation in the local co-
ordinate axes for the plane frame as shown in Fig 30.1a, b as,
k
' y ' z

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?
?
?
?
?
?
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6
5
4
3
2
1
2 2
2 3 2 3
2 2
2 3 2 3
6
5
4
3
2
1
'
'
'
'
'
'
4 6
0
2 6
0
6 12
0
6 12
0
0 0 0 0
2 6
0
4 6
0
6 12
0
6 12
0
0 0 0 0
'
'
'
'
'
'
u
u
u
u
u
u
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
q
q
q
q
q
q
z z z z
z z z z
z z z z
z z z z
(30.1a)

This may be succinctly written as

{} []{} ' ' ' u k q =      (30.1b)

where  is the member stiffness matrix. The member stiffness matrix can also
be generated by giving unit displacement along each possible displacement
degree of freedom one at a time and calculating resulting restraint actions.
[] ' k

30.3  Transformation from local to global co-ordinate system
30.3.1 Displacement transformation matrix
In plane frame the members are oriented in different directions and hence it is
necessary to transform stiffness matrix of individual members from local to global
co-ordinate system before formulating the global stiffness matrix by assembly. In
Fig. 30.2a the plane frame member is shown in local coordinate axes z y x ' ' ' and
in Fig. 30.2b, the plane frame is shown in global coordinate axesxyz . Two ends
of the plane frame member are identified by j and . Let   and
be respectively displacements of ends
k
3 2 1
' , ' , ' u u u
6 5 4
' , ' , ' u u u j and  of the member in local
coordinate system '. Similarly  and  respectively are
displacements of ends
k
' ' z y x
3 2 1
, , u u u
6 5 4
, , u u u
j and k of the member in global co-ordinate system.

Let ? be the angle by which the member is inclined to global  x-axis. From
Fig.30.2a and b, one could relate  to  as,
3 2 1
' , ' , ' u u u
3 2 1
, , u u u

? ? sin cos '
2 1 1
u u u + =        (30.2a)

? ? cos sin '
2 1 2
u u u + - =               (30.2b)

3 3
' u u =          (30.2c)

This may be written as,

Page 5

Instructional Objectives
After reading this chapter the student will be able to
1. Derive plane frame member stiffness matrix in local co-ordinate system.
2. Transform plane frame member stiffness matrix from local to global co-
ordinate system.
3. Assemble member stiffness matrices to obtain the global stiffness matrix of
the plane frame.
4. Write the global load-displacement relation for the plane frame.
5. Impose boundary conditions on the load-displacement relation.
6. Analyse plane frames by the direct stiffness matrix method.

30.1 Introduction
In the case of plane frame, all the members lie in the same plane and are
interconnected by rigid joints. The internal stress resultants at a cross-section of
a plane frame member consist of bending moment, shear force and an axial
force.  The significant deformations in the plane frame are only flexural and axial.
In this lesson, the analysis of plane frame by direct stiffness matrix method is
discussed. Initially, the stiffness matrix of the plane frame member is derived in
its local co-ordinate axes and then it is transformed to global co-ordinate system.
In the case of plane frames, members are oriented in different directions and
hence before forming the global stiffness matrix it is necessary to refer all the
member stiffness matrices to the same set of axes. This is achieved by
transformation of forces and displacements to global co-ordinate system.

30.2 Member Stiffness Matrix
Consider a member of a plane frame as shown in Fig. 30.1a in the member co-
ordinate system '.  The global orthogonal set of axes ' ' z y x xyz is also shown in the
figure. The frame lies in the xy plane. The member is assumed to have uniform
flexural rigidity EI and uniform axial rigidityEA for sake of simplicity. The axial
deformation of member will be considered in the analysis. The possible
displacements at each node of the member are: translation in - and -
direction and rotation about - axis.
' x ' y
' z

Thus the frame members have six (6) degrees of freedom and are shown in
Fig.30.1a. The forces acting on the member at end j and  are shown in Fig.
30.1b. The relation between axial displacement and axial forces is derived in
chapter 24. Similarly the relation between shear force, bending moment with
translation along  axis and rotation about  axis are given in lesson 27.
Combining them, we could write the load-displacement relation in the local co-
ordinate axes for the plane frame as shown in Fig 30.1a, b as,
k
' y ' z

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?
?
?
?
?
?
?
?
?
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-
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=
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6
5
4
3
2
1
2 2
2 3 2 3
2 2
2 3 2 3
6
5
4
3
2
1
'
'
'
'
'
'
4 6
0
2 6
0
6 12
0
6 12
0
0 0 0 0
2 6
0
4 6
0
6 12
0
6 12
0
0 0 0 0
'
'
'
'
'
'
u
u
u
u
u
u
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
q
q
q
q
q
q
z z z z
z z z z
z z z z
z z z z
(30.1a)

This may be succinctly written as

{} []{} ' ' ' u k q =      (30.1b)

where  is the member stiffness matrix. The member stiffness matrix can also
be generated by giving unit displacement along each possible displacement
degree of freedom one at a time and calculating resulting restraint actions.
[] ' k

30.3  Transformation from local to global co-ordinate system
30.3.1 Displacement transformation matrix
In plane frame the members are oriented in different directions and hence it is
necessary to transform stiffness matrix of individual members from local to global
co-ordinate system before formulating the global stiffness matrix by assembly. In
Fig. 30.2a the plane frame member is shown in local coordinate axes z y x ' ' ' and
in Fig. 30.2b, the plane frame is shown in global coordinate axesxyz . Two ends
of the plane frame member are identified by j and . Let   and
be respectively displacements of ends
k
3 2 1
' , ' , ' u u u
6 5 4
' , ' , ' u u u j and  of the member in local
coordinate system '. Similarly  and  respectively are
displacements of ends
k
' ' z y x
3 2 1
, , u u u
6 5 4
, , u u u
j and k of the member in global co-ordinate system.

Let ? be the angle by which the member is inclined to global  x-axis. From
Fig.30.2a and b, one could relate  to  as,
3 2 1
' , ' , ' u u u
3 2 1
, , u u u

? ? sin cos '
2 1 1
u u u + =        (30.2a)

? ? cos sin '
2 1 2
u u u + - =               (30.2b)

3 3
' u u =          (30.2c)

This may be written as,

?
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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
'
'
'
'
'
'
u
u
u
u
u
u
l m
m l
l m
m l
u
u
u
u
u
u
(30.3a)

Where, ? cos = l and ? sin = m .

This may be written in compact form as,

{ } [ ] { } u T u = '         (30.3b)

In the above equation, [ is defined as the displacement transformation matrix
and it transforms the six global displacement components to six displacement
components in local co-ordinate axes. Again, if the coordinate of node
] T
j is
and coordinate of node  are (
1 1
,y x ) k ( )
2 2
,y x , then,

L
x x
l
1 2
cos
-
= = ?   and
L
y y
m
1 2
sin
-
= = ? .

Where ()( )
2
1 2
2
1 2
y y x x L - + - =             (30.4)

```
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## Structural Analysis

30 videos|122 docs|28 tests

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