The Direct Stiffness Method: Plane Frames - 4 GATE Notes | EduRev

Structural Analysis

GATE : The Direct Stiffness Method: Plane Frames - 4 GATE Notes | EduRev

 Page 2


 
 
 
                                                         
The load vector corresponding to unconstrained degrees of freedom is, 
 
 
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24
24
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24
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9
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5
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p
p
p
p
p
p
p
k
     (5) 
 
In the given frame, constraint (known) degrees of freedom are 
. Eliminating rows and columns corresponding to constrained 
degrees of freedom from global stiffness matrix and writing load displacement 
relationship, 
12 11 10 3 2 1
, , , , , u u u u u u
 
 
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-
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9
8
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5
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3
33 . 5 1 1 33 . 1 1 0
1 5 . 500 0 1 5 . 0 0
1 0 5 . 500 0 0 500
33 . 1 0 . 1 0 33 . 5 0 . 1 0 . 1
0 . 1 5 . 0 0 0 . 1 5 . 500 0
0 0 500 0 . 1 0 5 . 500
10
24
24
0
24
24
10
u
u
u
u
u
u
  (6) 
 
Solving we get, 
 
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3 -
5 -
2 -
3 -
5 -
-2
9
8
7
6
5
4
10 3.85
10 5.65 -
10 1.43
10 8.14 -
10 3.84 -
10 43 . 1
u
u
u
u
u
u
      (7) 
 
                                                         
Page 3


 
 
 
                                                         
The load vector corresponding to unconstrained degrees of freedom is, 
 
 
{}
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=
24
24
0
24
24
10
9
8
7
6
5
4
p
p
p
p
p
p
p
k
     (5) 
 
In the given frame, constraint (known) degrees of freedom are 
. Eliminating rows and columns corresponding to constrained 
degrees of freedom from global stiffness matrix and writing load displacement 
relationship, 
12 11 10 3 2 1
, , , , , u u u u u u
 
 
?
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-
-
-
9
8
7
6
5
4
3
33 . 5 1 1 33 . 1 1 0
1 5 . 500 0 1 5 . 0 0
1 0 5 . 500 0 0 500
33 . 1 0 . 1 0 33 . 5 0 . 1 0 . 1
0 . 1 5 . 0 0 0 . 1 5 . 500 0
0 0 500 0 . 1 0 5 . 500
10
24
24
0
24
24
10
u
u
u
u
u
u
  (6) 
 
Solving we get, 
 
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3 -
5 -
2 -
3 -
5 -
-2
9
8
7
6
5
4
10 3.85
10 5.65 -
10 1.43
10 8.14 -
10 3.84 -
10 43 . 1
u
u
u
u
u
u
      (7) 
 
                                                         
Let be the support reactions along degrees of freedom 
 respectively. Support reactions are calculated by 
12 11 10 3 2 1
, , , , , R R R R R R
12 , 11 , 10 , 3 , 2 , 1
 
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9
8
7
6
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3
12
11
10
3
2
1
12
11
10
3
2
1
33 . 1 0 0 . 1 0 0 0
0 500 0 0 0 0
0 . 1 0 5 . 0 0 0 0
0 0 0 33 . 1 0 0 . 1
0 0 0 0 500 0
0 0 0 0 . 1 0 5 . 0
10
9 8 7 6 5 4
u
u
u
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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42 . 19
28 . 28
99 . 10
43 . 3
71 . 19
99 . 0
42 . 19
28 . 28
99 . 10
43 . 3
71 . 19
99 . 0
0
0
0
0
0
0
12
11
10
3
2
1
R
R
R
R
R
R
     (8) 
 
 
Summary 
In this lesson, the analysis of plane frame by the 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. In the 
end, a few problems are solved to illustrate the methodology. 
 
                                                         
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