The Direct Stiffness Method: Plane Frames - 3 GATE Notes | EduRev

Structural Analysis

GATE : The Direct Stiffness Method: Plane Frames - 3 GATE Notes | EduRev

 Page 1


 
 
The load vector corresponding to unconstrained degrees of freedom is (vide 
30.4d), 
 
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24
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6
5
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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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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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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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1
9
8
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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
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R
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F
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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) 
                                                         
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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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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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
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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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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), 
 
{}
?
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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, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
×
×
×
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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
=× =× 
                                                         
   
[] [] [ ][]
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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× × - × ×
× - × × - × -
× × -
× × - × ×
× × - × ×
× - ×
=
= T k T k
T
 
           (2) 
 
 
Member 3: ° = = 270 ; 4 ? m L ;  node points 3-4  ; 
21
0
xx
l
L
-
== and 
21
1
yy
m
L
-
==- . 
  
 
                                                         
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