Civil Engineering (CE) Exam > Civil Engineering (CE) Notes > Structural Analysis > The Direct Stiffness Method: Beams - 4
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Page 1
We know that 0
8 7 6 5 4 3
= = = = = = u u u u u u . Thus solving for unknowns and
, yields
1
u
2
u
?
?
?
?
?
?
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=
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?
? -
2
1
0 . 2 5 . 0
5 . 0 0 . 2
33 . 2
5
u
u
EI
zz
(8)
?
?
?
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?
? -
?
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-
-
=
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?
333 . 2
5
0 . 2 5 . 0
5 . 0 0 . 2
75 . 3
1
2
1
zz
EI
u
u
(9)
?
?
?
?
?
?
?
?
?
? -
=
909 . 1
977 . 2
1
zz
EI
Thus displacements are,
zz zz
EI
u
EI
u
909 . 1
and
977 . 2
2 1
=
-
= (10)
The unknown joint loads are given by,
?
?
?
?
?
?
?
?
?
? -
?
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-
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=
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?
?
909 . 1
977 . 2
1
375 . 0 0
5 . 0 0
0 375 . 0
0 5 . 0
0 375 . 0
375 . 0 0
8
7
6
5
4
3
zz
zz
EI
EI
p
p
p
p
p
p
(11)
?
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?
?
?
?
?
-
-
-
=
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
Page 2
We know that 0
8 7 6 5 4 3
= = = = = = u u u u u u . Thus solving for unknowns and
, yields
1
u
2
u
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
? -
2
1
0 . 2 5 . 0
5 . 0 0 . 2
33 . 2
5
u
u
EI
zz
(8)
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
333 . 2
5
0 . 2 5 . 0
5 . 0 0 . 2
75 . 3
1
2
1
zz
EI
u
u
(9)
?
?
?
?
?
?
?
?
?
? -
=
909 . 1
977 . 2
1
zz
EI
Thus displacements are,
zz zz
EI
u
EI
u
909 . 1
and
977 . 2
2 1
=
-
= (10)
The unknown joint loads are given by,
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
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-
-
=
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
977 . 2
1
375 . 0 0
5 . 0 0
0 375 . 0
0 5 . 0
0 375 . 0
375 . 0 0
8
7
6
5
4
3
zz
zz
EI
EI
p
p
p
p
p
p
(11)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
The actual reactions at the supports are calculated as,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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-
-
-
=
?
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-
-
-
+
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=
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+
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=
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?
284 . 3
715 . 1
116 . 1
489 . 1
116 . 10
716 . 5
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
4
67 . 2
0
0
9
5
8
7
6
5
4
3
8
7
6
5
4
3
8
7
6
5
4
3
p
p
p
p
p
p
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
(12)
Member end actions for element 1
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
-
-
-
?
?
?
?
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-
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-
- - -
-
-
+
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=
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?
?
?
977 . 2
116 . 1
488 . 1
116 . 1
977 . 2
0
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
0
0
0
0
4
3
2
1
zz
zz
EI
q
q
q
q
? ?
? ?
? ?
0 0 EI
(13)
Member end actions for element 2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
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?
?
?
?
-
- - -
-
-
+ =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
0
977 . 2
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
58 . 4
4 . 5
98 . 2
6 . 4
(14)
Page 3
We know that 0
8 7 6 5 4 3
= = = = = = u u u u u u . Thus solving for unknowns and
, yields
1
u
2
u
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
? -
2
1
0 . 2 5 . 0
5 . 0 0 . 2
33 . 2
5
u
u
EI
zz
(8)
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
333 . 2
5
0 . 2 5 . 0
5 . 0 0 . 2
75 . 3
1
2
1
zz
EI
u
u
(9)
?
?
?
?
?
?
?
?
?
? -
=
909 . 1
977 . 2
1
zz
EI
Thus displacements are,
zz zz
EI
u
EI
u
909 . 1
and
977 . 2
2 1
=
-
= (10)
The unknown joint loads are given by,
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
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?
?
?
?
?
?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
977 . 2
1
375 . 0 0
5 . 0 0
0 375 . 0
0 5 . 0
0 375 . 0
375 . 0 0
8
7
6
5
4
3
zz
zz
EI
EI
p
p
p
p
p
p
(11)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
The actual reactions at the supports are calculated as,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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-
-
-
=
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-
-
+
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=
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+
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=
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?
284 . 3
715 . 1
116 . 1
489 . 1
116 . 10
716 . 5
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
4
67 . 2
0
0
9
5
8
7
6
5
4
3
8
7
6
5
4
3
8
7
6
5
4
3
p
p
p
p
p
p
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
(12)
Member end actions for element 1
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
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?
-
- - -
-
-
+
?
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?
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=
?
?
?
?
?
?
?
?
?
?
?
?
?
977 . 2
116 . 1
488 . 1
116 . 1
977 . 2
0
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
0
0
0
0
4
3
2
1
zz
zz
EI
q
q
q
q
? ?
? ?
? ?
0 0 EI
(13)
Member end actions for element 2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+ =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
0
977 . 2
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
58 . 4
4 . 5
98 . 2
6 . 4
(14)
Member end actions for element 3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
909 . 1
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
67 . 2
0 . 4
67 . 2
0 . 4
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
72 . 1
28 . 3
58 . 4
72 . 4
(15)
Example 28.2
Analyse the continuous beam shown in Fig. 28.2a. Assume that the supports are
unyielding. Assume EI to be constant for all members.
The numbering of joints and members are shown in Fig. 28.2b. The global
degrees of freedom are also shown in the figure.
The given continuous beam is divided into two beam elements. Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. Also it is observed that displacements
from support conditions. 0
6 5 4 3
= = = = u u u u
First construct the member stiffness matrix for each member.
Member 1: , node points 1-2. m L 4 =
Page 4
We know that 0
8 7 6 5 4 3
= = = = = = u u u u u u . Thus solving for unknowns and
, yields
1
u
2
u
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
? -
2
1
0 . 2 5 . 0
5 . 0 0 . 2
33 . 2
5
u
u
EI
zz
(8)
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
333 . 2
5
0 . 2 5 . 0
5 . 0 0 . 2
75 . 3
1
2
1
zz
EI
u
u
(9)
?
?
?
?
?
?
?
?
?
? -
=
909 . 1
977 . 2
1
zz
EI
Thus displacements are,
zz zz
EI
u
EI
u
909 . 1
and
977 . 2
2 1
=
-
= (10)
The unknown joint loads are given by,
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
977 . 2
1
375 . 0 0
5 . 0 0
0 375 . 0
0 5 . 0
0 375 . 0
375 . 0 0
8
7
6
5
4
3
zz
zz
EI
EI
p
p
p
p
p
p
(11)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
The actual reactions at the supports are calculated as,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
?
?
?
?
?
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?
?
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-
-
-
+
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-
=
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+
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=
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?
?
?
?
?
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?
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?
?
?
?
?
?
?
?
?
?
?
?
284 . 3
715 . 1
116 . 1
489 . 1
116 . 10
716 . 5
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
4
67 . 2
0
0
9
5
8
7
6
5
4
3
8
7
6
5
4
3
8
7
6
5
4
3
p
p
p
p
p
p
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
(12)
Member end actions for element 1
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
977 . 2
116 . 1
488 . 1
116 . 1
977 . 2
0
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
0
0
0
0
4
3
2
1
zz
zz
EI
q
q
q
q
? ?
? ?
? ?
0 0 EI
(13)
Member end actions for element 2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+ =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
0
977 . 2
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
58 . 4
4 . 5
98 . 2
6 . 4
(14)
Member end actions for element 3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
909 . 1
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
67 . 2
0 . 4
67 . 2
0 . 4
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
72 . 1
28 . 3
58 . 4
72 . 4
(15)
Example 28.2
Analyse the continuous beam shown in Fig. 28.2a. Assume that the supports are
unyielding. Assume EI to be constant for all members.
The numbering of joints and members are shown in Fig. 28.2b. The global
degrees of freedom are also shown in the figure.
The given continuous beam is divided into two beam elements. Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. Also it is observed that displacements
from support conditions. 0
6 5 4 3
= = = = u u u u
First construct the member stiffness matrix for each member.
Member 1: , node points 1-2. m L 4 =
The member stiffness matrix for all the members are the same, as the length and
flexural rigidity of all members is the same.
(1)
[]
1
3
5
6
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
'
1 3 5 6 . .
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
On the member stiffness matrix, the corresponding global degrees of freedom
are indicated to facilitate assembling.
Member 2: , node points 2-3. m L 4 =
[]
2
4
1
3
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
2 4 1 3 . .
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(2)
The assembled global stiffness matrix of the continuous beam is of order 6 6 × .
The assembled global stiffness matrix may be written as,
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
- - -
-
-
=
1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
375 . 0 0 . 1 0 375 . 0 0 5 . 0
0 0 1875 . 0 1875 . 0 375 . 0 375 . 0
1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
0 0 375 . 0 375 . 0 0 . 1 5 . 0
375 . 0 5 . 0 375 . 0 0 5 . 0 2
zz
EI K
(3)
Now it is required to replace the given members loads by equivalent joint loads.
The equivalent loads for the present case is shown in Fig. 28.2c. The
displacement degrees of freedom are also shown in figure.
Page 5
We know that 0
8 7 6 5 4 3
= = = = = = u u u u u u . Thus solving for unknowns and
, yields
1
u
2
u
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
? -
2
1
0 . 2 5 . 0
5 . 0 0 . 2
33 . 2
5
u
u
EI
zz
(8)
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
333 . 2
5
0 . 2 5 . 0
5 . 0 0 . 2
75 . 3
1
2
1
zz
EI
u
u
(9)
?
?
?
?
?
?
?
?
?
? -
=
909 . 1
977 . 2
1
zz
EI
Thus displacements are,
zz zz
EI
u
EI
u
909 . 1
and
977 . 2
2 1
=
-
= (10)
The unknown joint loads are given by,
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
977 . 2
1
375 . 0 0
5 . 0 0
0 375 . 0
0 5 . 0
0 375 . 0
375 . 0 0
8
7
6
5
4
3
zz
zz
EI
EI
p
p
p
p
p
p
(11)
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
The actual reactions at the supports are calculated as,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
284 . 3
715 . 1
116 . 1
489 . 1
116 . 10
716 . 5
715 . 0
955 . 0
116 . 1
488 . 1
116 . 1
715 . 0
4
67 . 2
0
0
9
5
8
7
6
5
4
3
8
7
6
5
4
3
8
7
6
5
4
3
p
p
p
p
p
p
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
(12)
Member end actions for element 1
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
977 . 2
116 . 1
488 . 1
116 . 1
977 . 2
0
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
0
0
0
0
4
3
2
1
zz
zz
EI
q
q
q
q
? ?
? ?
? ?
0 0 EI
(13)
Member end actions for element 2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+ =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
909 . 1
0
977 . 2
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
58 . 4
4 . 5
98 . 2
6 . 4
(14)
Member end actions for element 3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
909 . 1
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
67 . 2
0 . 4
67 . 2
0 . 4
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
72 . 1
28 . 3
58 . 4
72 . 4
(15)
Example 28.2
Analyse the continuous beam shown in Fig. 28.2a. Assume that the supports are
unyielding. Assume EI to be constant for all members.
The numbering of joints and members are shown in Fig. 28.2b. The global
degrees of freedom are also shown in the figure.
The given continuous beam is divided into two beam elements. Two degrees of
freedom (one translation and one rotation) are considered at each end of the
member. In the above figure, double headed arrows denote rotations and single
headed arrow represents translations. Also it is observed that displacements
from support conditions. 0
6 5 4 3
= = = = u u u u
First construct the member stiffness matrix for each member.
Member 1: , node points 1-2. m L 4 =
The member stiffness matrix for all the members are the same, as the length and
flexural rigidity of all members is the same.
(1)
[]
1
3
5
6
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
'
1 3 5 6 . .
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
On the member stiffness matrix, the corresponding global degrees of freedom
are indicated to facilitate assembling.
Member 2: , node points 2-3. m L 4 =
[]
2
4
1
3
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
2 4 1 3 . .
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
-
=
zz
EI k
f o d Global
(2)
The assembled global stiffness matrix of the continuous beam is of order 6 6 × .
The assembled global stiffness matrix may be written as,
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
- - -
-
-
=
1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
375 . 0 0 . 1 0 375 . 0 0 5 . 0
0 0 1875 . 0 1875 . 0 375 . 0 375 . 0
1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
0 0 375 . 0 375 . 0 0 . 1 5 . 0
375 . 0 5 . 0 375 . 0 0 5 . 0 2
zz
EI K
(3)
Now it is required to replace the given members loads by equivalent joint loads.
The equivalent loads for the present case is shown in Fig. 28.2c. The
displacement degrees of freedom are also shown in figure.
Thus the global load vector corresponding to unconstrained degree of freedom
is,
{}
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
=
67 . 6
0
2
1
p
p
p
k
(4)
Writing the load displacement relation for the entire continuous beam,
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FAQs on The Direct Stiffness Method: Beams - 4 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Direct Stiffness Method for beams? |  |
Ans. The Direct Stiffness Method is a numerical technique used to analyze the behavior of beams by dividing them into smaller elements. It calculates the stiffness and displacement of each element and then assembles them to obtain the overall response of the beam structure.
| 2. How does the Direct Stiffness Method work? | |
Ans. The Direct Stiffness Method works by considering the beam as a series of interconnected elements. Each element is characterized by its stiffness matrix, which relates the forces and displacements at the element's nodes. The method then solves a system of equations derived from these stiffness matrices to determine the displacements and forces throughout the beam structure.
| 3. What are the advantages of using the Direct Stiffness Method? | |
Ans. The Direct Stiffness Method offers several advantages in beam analysis. It allows for the analysis of complex beam structures, including those with variable cross-sections and supports. The method also provides accurate results, especially when compared to approximate analytical methods. Additionally, it can handle different types of boundary conditions and external loads, making it a versatile tool for structural engineers.
| 4. Are there any limitations to the Direct Stiffness Method? | |
Ans. Yes, the Direct Stiffness Method has some limitations. It requires a significant amount of computation, especially for large-scale structures, which can be time-consuming. The method also assumes linear elastic behavior, which may not accurately represent the material properties of certain beams. Lastly, the Direct Stiffness Method may encounter difficulties in handling beams with significant nonlinearities, such as large deformations or material yielding.
| 5. How is the Direct Stiffness Method applied in practice? | |
Ans. In practice, the Direct Stiffness Method is implemented using software programs specifically designed for structural analysis. Engineers input the geometric and material properties of the beams, along with the boundary conditions and external loads. The software then applies the Direct Stiffness Method to solve the system of equations and generate results such as displacements, reactions, and internal forces. These results can be used to assess the structural integrity and design of the beam system.
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