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# The Direct Stiffness Method: Plane Frames - 2 GATE Notes | EduRev

## GATE : The Direct Stiffness Method: Plane Frames - 2 GATE Notes | EduRev

``` 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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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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?
?
?
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?
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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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-
-
=
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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,

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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-
-
=
?
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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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## Structural Analysis

30 videos|122 docs|28 tests

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