The Direct Stiffness Method: Truss Analysis - 5 GATE Notes | EduRev

Structural Analysis

: The Direct Stiffness Method: Truss Analysis - 5 GATE Notes | EduRev

 Page 2


                                                         
 
 
                                                         
 
The joint and member numbers are indicated in Fig. 25.2b. The possible degree 
of freedom are also shown in Fig. 25.2b. In the given problem 
2 1
,u u and 
3
u 
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due 
to boundary condition. First let us generate stiffness matrix for each of the six 
members in global co-ordinate system. 
 
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1 
 
 
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (1) 
  
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1 
 
Page 3


                                                         
 
 
                                                         
 
The joint and member numbers are indicated in Fig. 25.2b. The possible degree 
of freedom are also shown in Fig. 25.2b. In the given problem 
2 1
,u u and 
3
u 
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due 
to boundary condition. First let us generate stiffness matrix for each of the six 
members in global co-ordinate system. 
 
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1 
 
 
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (1) 
  
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1 
 
                                                         
 
[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (2) 
 
Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4 
 
 
[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (3) 
 
 
Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2 
 
 
[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (4) 
 
Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1 
 
 
[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
    (5) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
Page 4


                                                         
 
 
                                                         
 
The joint and member numbers are indicated in Fig. 25.2b. The possible degree 
of freedom are also shown in Fig. 25.2b. In the given problem 
2 1
,u u and 
3
u 
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due 
to boundary condition. First let us generate stiffness matrix for each of the six 
members in global co-ordinate system. 
 
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1 
 
 
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (1) 
  
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1 
 
                                                         
 
[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (2) 
 
Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4 
 
 
[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (3) 
 
 
Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2 
 
 
[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (4) 
 
Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1 
 
 
[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
    (5) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
                                                         
 
[]
4
3
8
7
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
4 3 8 7
6
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
    (6) 
 
There are eight possible global degrees of freedom for the truss shown in the 
figure. Hence the global stiffness matrix is of the order ( 8 8 × ). On the member 
stiffness matrix, the corresponding global degrees of freedom are indicated to 
facilitate assembly. Thus the global stiffness matrix is, 
 
 
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
AE K     (7) 
 
The force-displacement relation for the truss is, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
 
           (8) 
 
The displacements 
2 1
,u u
and 
3
u
are unknowns. 
Here,
12 3
5kN ; 10 ; 0 pp p ==- =
 and 
0
8 7 6 5 4
= = = = = u u u u u
. 
Page 5


                                                         
 
 
                                                         
 
The joint and member numbers are indicated in Fig. 25.2b. The possible degree 
of freedom are also shown in Fig. 25.2b. In the given problem 
2 1
,u u and 
3
u 
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due 
to boundary condition. First let us generate stiffness matrix for each of the six 
members in global co-ordinate system. 
 
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1 
 
 
[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (1) 
  
Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1 
 
                                                         
 
[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (2) 
 
Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4 
 
 
[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
    (3) 
 
 
Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2 
 
 
[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (4) 
 
Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1 
 
 
[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
    (5) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
                                                         
 
[]
4
3
8
7
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
4 3 8 7
6
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
    (6) 
 
There are eight possible global degrees of freedom for the truss shown in the 
figure. Hence the global stiffness matrix is of the order ( 8 8 × ). On the member 
stiffness matrix, the corresponding global degrees of freedom are indicated to 
facilitate assembly. Thus the global stiffness matrix is, 
 
 
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
AE K     (7) 
 
The force-displacement relation for the truss is, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
 
           (8) 
 
The displacements 
2 1
,u u
and 
3
u
are unknowns. 
Here,
12 3
5kN ; 10 ; 0 pp p ==- =
 and 
0
8 7 6 5 4
= = = = = u u u u u
. 
                                                         
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
0
0
0
0
0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
0
10
5
3
2
1
8
7
6
5
4
u
u
u
EA
p
p
p
p
p
(9) 
 
Thus, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
-
3
2
1
271 . 0 0 20 . 0
0 271 . 0 071 . 0
20 . 0 071 . 0 271 . 0
0
10
5
u
u
u
   (10) 
 
Solving which, yields 
 
AE
u
AE
u
AE
u
825 . 53
;
97 . 55
;
855 . 72
3 2 1
=
-
= = 
 
Now reactions are evaluated from the equation, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
3
2
1
8
7
6
5
4
071 . 0 20 . 0 0
071 . 0 0 0
0 071 . 0 071 . 0
0 071 . 0 071 . 0
071 . 0 0 0
u
u
u
p
p
p
p
p
   (11) 
 
45 6 7 8
3.80 kN ; 1.19 kN ; 1.19 kN ; 3.8 0 ; 15.00 kN pp p p kNp =- =- =- = =
 
In the next step evaluate forces in members. 
 
Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1 
 
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