Civil Engineering (CE) Exam > Civil Engineering (CE) Notes > Structural Analysis > The Direct Stiffness Method: Beams - 4
The Direct Stiffness Method: Beams - 4 | Structural Analysis - Civil Engineering (CE) PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
? -
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
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,
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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)
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,
?
?
?
?
?
?
?
?
?
? -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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 =
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,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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,
Read More
|
34 videos|140 docs|31 tests
|
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.
Related Exams
About this Document
|
4.73/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: The Direct Stiffness Method: Beams - 4 for Civil Engineering (CE) 2024 is part of Structural Analysis preparation.
The notes and questions for The Direct Stiffness Method: Beams - 4 have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The Direct Stiffness Method: Beams - 4 covers topics
like and The Direct Stiffness Method: Beams - 4 Example, for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for The Direct Stiffness Method: Beams - 4.
Introduction of The Direct Stiffness Method: Beams - 4 in English is available as part of our Structural Analysis
for Civil Engineering (CE) & The Direct Stiffness Method: Beams - 4 in Hindi for Structural Analysis course.
Download more important topics related with notes, lectures and mock test series for Civil Engineering (CE)
Exam by signing up for free. Civil Engineering (CE): The Direct Stiffness Method: Beams - 4 | Structural Analysis - Civil Engineering (CE)
Description
Full syllabus notes, lecture & questions for The Direct Stiffness Method: Beams - 4 | Structural Analysis - Civil Engineering (CE) - Civil Engineering (CE) | Plus excerises question with solution to help you revise complete syllabus for Structural Analysis | Best notes, free PDF download
Information about The Direct Stiffness Method: Beams - 4
In this doc you can find the meaning of The Direct Stiffness Method: Beams - 4 defined & explained in the simplest way possible. Besides explaining types of
The Direct Stiffness Method: Beams - 4 theory, EduRev gives you an ample number of questions to practice The Direct Stiffness Method: Beams - 4 tests, examples and also practice Civil Engineering (CE)
tests
|
Explore Courses for Civil Engineering (CE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Exam
,study material
,The Direct Stiffness Method: Beams - 4 | Structural Analysis - Civil Engineering (CE)
,past year papers
,Extra Questions
,Sample Paper
,Previous Year Questions with Solutions
,Summary
,Semester Notes
,Viva Questions
,mock tests for examination
,Objective type Questions
,shortcuts and tricks
,practice quizzes
,Important questions
,The Direct Stiffness Method: Beams - 4 | Structural Analysis - Civil Engineering (CE)
,The Direct Stiffness Method: Beams - 4 | Structural Analysis - Civil Engineering (CE)
,ppt
,video lectures
,Free
,MCQs
;
Additional Information about The Direct Stiffness Method: Beams - 4 for Civil Engineering (CE) Preparation
The Direct Stiffness Method: Beams - 4 Free PDF Download
The The Direct Stiffness Method: Beams - 4 is an invaluable resource that delves deep into the core of the Civil Engineering (CE) exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the The Direct Stiffness Method: Beams - 4 now and kickstart your journey towards success in the Civil Engineering (CE) exam.
Importance of The Direct Stiffness Method: Beams - 4
The importance of The Direct Stiffness Method: Beams - 4 cannot be overstated, especially for Civil Engineering (CE) aspirants.
This document holds the key to success in the Civil Engineering (CE) exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
The Direct Stiffness Method: Beams - 4 Notes
The Direct Stiffness Method: Beams - 4 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to The Direct Stiffness Method: Beams - 4.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, The Direct Stiffness Method: Beams - 4 Notes on EduRev are your ultimate resource for success.
The Direct Stiffness Method: Beams - 4 Civil Engineering (CE) Questions
The "The Direct Stiffness Method: Beams - 4 Civil Engineering (CE) Questions" guide is a valuable resource for all aspiring students preparing for the
Civil Engineering (CE) exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study The Direct Stiffness Method: Beams - 4 on the App
Students of Civil Engineering (CE) can study The Direct Stiffness Method: Beams - 4 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the The Direct Stiffness Method: Beams - 4,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of The Direct Stiffness Method: Beams - 4 is prepared as per the latest Civil Engineering (CE) syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup