The Direct Stiffness Method: Truss Analysis - 7 GATE Notes | EduRev

Structural Analysis

GATE : The Direct Stiffness Method: Truss Analysis - 7 GATE Notes | EduRev

 Page 1


                                                         
 
 
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate 
axes are shown at node 3. At node 2, roller is supported on inclined support. 
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that 
4
u could be 
set to zero. All the possible displacement degrees of freedom are also shown in 
the figure. In the first step calculate member stiffness matrix. 
 
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2 
         12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .     
     
 
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
   (1) 
Page 2


                                                         
 
 
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate 
axes are shown at node 3. At node 2, roller is supported on inclined support. 
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that 
4
u could be 
set to zero. All the possible displacement degrees of freedom are also shown in 
the figure. In the first step calculate member stiffness matrix. 
 
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2 
         12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .     
     
 
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
   (1) 
                                                         
 
 
 
Member 2: . 00 . 4 , 30 , 0
"
m L
x x
= ° = ° = ? ? Nodal points 2-3 
         50 . 0 " ; 866 . 0 " ; 0 ; 1 = = = = m l m l .     
 
 
 
Page 3


                                                         
 
 
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate 
axes are shown at node 3. At node 2, roller is supported on inclined support. 
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that 
4
u could be 
set to zero. All the possible displacement degrees of freedom are also shown in 
the figure. In the first step calculate member stiffness matrix. 
 
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2 
         12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .     
     
 
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
   (1) 
                                                         
 
 
 
Member 2: . 00 . 4 , 30 , 0
"
m L
x x
= ° = ° = ? ? Nodal points 2-3 
         50 . 0 " ; 866 . 0 " ; 0 ; 1 = = = = m l m l .     
 
 
 
                                                         
 
[]
4
3
6
5
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 4
4 3 6 5
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
   (2) 
 
Member 3: . 00 . 3 , 90 m L
x
= ° = ? , 1 ; 0 = = m l Nodal points 3-1 
         .   
      
[]
2
1
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 3
2 1 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (3) 
 
     
For the present problem, the global stiffness matrix is of the order() 6 6 × . The 
global stiffness matrix for the entire truss is. 
 
[]
6
5
4
3
2
1
333 . 0 0 0 0 333 . 0 0
0 25 . 0 125 . 0 217 . 0 0 0
0 125 . 0 065 . 0 132 . 0 014 . 0 . 019 . 0
0 217 . 0 132 . 0 385 . 0 119 . 0 159 . 0
333 . 0 0 014 . 0 119 . 0 405 . 0 096 . 0
0 0 019 . 0 159 . 0 096 . 0 128 . 0
6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
- -
- - - -
-
= EA k
  (4) 
 
Writing load-displacement equation for the truss for unconstrained degrees of 
freedom, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
=
?
?
?
?
?
?
?
?
?
? -
3
2
1
385 . 0 119 . 0 159 . 0
119 . 0 405 . 0 096 . 0
159 . 0 096 . 0 128 . 0
0
5
5
u
u
u
    (5) 
 
Solving , 
 
AE
u
AE
u
AE
u
12 . 33
;
728 . 3
;
408 . 77
3 2 1
= =
-
=
   (6) 
 
Page 4


                                                         
 
 
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate 
axes are shown at node 3. At node 2, roller is supported on inclined support. 
Hence it is required to use nodal co-ordinates " " y x - at node 2 so that 
4
u could be 
set to zero. All the possible displacement degrees of freedom are also shown in 
the figure. In the first step calculate member stiffness matrix. 
 
Member 1: . 00 . 5 , 87 . 6 , 13 . 143
"
m L
x x
= ° = ° = ? ? Nodal points 1-2 
         12 . 0 " ; 993 . 0 " ; 6 . 0 ; 80 . 0 = = = - = m l m l .     
     
 
[]
4
3
2
1
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 5
4 3 2 1
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
   (1) 
                                                         
 
 
 
Member 2: . 00 . 4 , 30 , 0
"
m L
x x
= ° = ° = ? ? Nodal points 2-3 
         50 . 0 " ; 866 . 0 " ; 0 ; 1 = = = = m l m l .     
 
 
 
                                                         
 
[]
4
3
6
5
014 . 0 119 . 0 072 . 0 096 . 0
119 . 0 986 . 0 596 . 0 794 . 0
072 . 0 596 . 0 36 . 0 48 . 0
096 . 0 794 . 0 48 . 0 64 . 0
0 . 4
4 3 6 5
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- - -
-
=
EA
k
   (2) 
 
Member 3: . 00 . 3 , 90 m L
x
= ° = ? , 1 ; 0 = = m l Nodal points 3-1 
         .   
      
[]
2
1
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 3
2 1 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
     (3) 
 
     
For the present problem, the global stiffness matrix is of the order() 6 6 × . The 
global stiffness matrix for the entire truss is. 
 
[]
6
5
4
3
2
1
333 . 0 0 0 0 333 . 0 0
0 25 . 0 125 . 0 217 . 0 0 0
0 125 . 0 065 . 0 132 . 0 014 . 0 . 019 . 0
0 217 . 0 132 . 0 385 . 0 119 . 0 159 . 0
333 . 0 0 014 . 0 119 . 0 405 . 0 096 . 0
0 0 019 . 0 159 . 0 096 . 0 128 . 0
6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
- -
- - - -
-
= EA k
  (4) 
 
Writing load-displacement equation for the truss for unconstrained degrees of 
freedom, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
-
=
?
?
?
?
?
?
?
?
?
? -
3
2
1
385 . 0 119 . 0 159 . 0
119 . 0 405 . 0 096 . 0
159 . 0 096 . 0 128 . 0
0
5
5
u
u
u
    (5) 
 
Solving , 
 
AE
u
AE
u
AE
u
12 . 33
;
728 . 3
;
408 . 77
3 2 1
= =
-
=
   (6) 
 
                                                         
Now reactions are evaluated from the equation 
 
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
-
- =
?
?
?
?
?
?
?
?
?
?
12 . 33
728 . 3
40 . 77
1
0 333 . 0 0
217 . 0 0 0
132 . 0 014 . 0 . 019 . 0
6
5
4
AE
AE
p
p
p
   (6) 
 
45 6
2.85 kN ; 7.19 kN ; 1.24 kN pp p ==- =- 
 
 
Summary 
Sometimes the truss is supported on a roller placed on an oblique plane. In such 
situations, the direct stiffness method as discussed in the previous lesson needs 
to be properly modified to make the displacement perpendicular to the roller 
support as zero. In the present approach, the inclined support is handled in the 
analysis by suitably modifying the member stiffness matrices of all members 
meeting at the inclined support. A few problems are solved to illustrate the 
procedure. 
 
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