The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

Structural Analysis

GATE : The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

 Page 1


 
                                                         
4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
 
 
  
()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-
 
 
  kN 424 . 0 4239 . 0 ? =      (7) 
 
Example 26.3 
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the 
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume 
AE=300KN for all members and 
75000
1
= a per 
o
C. 
 
 
 
Page 2


 
                                                         
4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
 
 
  
()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-
 
 
  kN 424 . 0 4239 . 0 ? =      (7) 
 
Example 26.3 
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the 
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume 
AE=300KN for all members and 
75000
1
= a per 
o
C. 
 
 
 
                                                         
 
 
Solution 
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints 
and members are numbered as shown in Fig.26.4b. In the given problem 
4 3 2 1
, , , u u u u and 
5
u represent unconstrained degrees of freedom. Due to support 
conditions, 0
8 7 6
= = = u u u . 
 
The temperature of the member (2) is raised by 
o
50 C.Thus, 
 
m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
     (1) 
 
The forces are developed in member (2), as it was prevented from expansion.  
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p
 
Page 3


 
                                                         
4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
 
 
  
()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-
 
 
  kN 424 . 0 4239 . 0 ? =      (7) 
 
Example 26.3 
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the 
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume 
AE=300KN for all members and 
75000
1
= a per 
o
C. 
 
 
 
                                                         
 
 
Solution 
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints 
and members are numbered as shown in Fig.26.4b. In the given problem 
4 3 2 1
, , , u u u u and 
5
u represent unconstrained degrees of freedom. Due to support 
conditions, 0
8 7 6
= = = u u u . 
 
The temperature of the member (2) is raised by 
o
50 C.Thus, 
 
m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
     (1) 
 
The forces are developed in member (2), as it was prevented from expansion.  
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p
 
                                                         
             
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
1
0
1
0
     (2) 
 
The displacement of the member (5) was raised by C
o
40 . Thus, 
 
m T L L
3 5
10 771 . 3 40 2 5
000 , 75
1
-
× = × × = ? = ? a
 
 
The forces developed in member (5) as it was not allowed to expand is 
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
707 . 0
707 . 0
707 . 0
707 . 0
10 771 . 3 300
3
8
7
6
5
t
t
t
t
p
p
p
p
 
           
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1
1
1
1
8 . 0
      (3) 
 
The global force vector due to thermal load is 
 
()
()
()
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
      (4) 
 
Writing the load-displacement relation for the entire truss is given below. 
 
Page 4


 
                                                         
4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
 
 
  
()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-
 
 
  kN 424 . 0 4239 . 0 ? =      (7) 
 
Example 26.3 
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the 
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume 
AE=300KN for all members and 
75000
1
= a per 
o
C. 
 
 
 
                                                         
 
 
Solution 
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints 
and members are numbered as shown in Fig.26.4b. In the given problem 
4 3 2 1
, , , u u u u and 
5
u represent unconstrained degrees of freedom. Due to support 
conditions, 0
8 7 6
= = = u u u . 
 
The temperature of the member (2) is raised by 
o
50 C.Thus, 
 
m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
     (1) 
 
The forces are developed in member (2), as it was prevented from expansion.  
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p
 
                                                         
             
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
1
0
1
0
     (2) 
 
The displacement of the member (5) was raised by C
o
40 . Thus, 
 
m T L L
3 5
10 771 . 3 40 2 5
000 , 75
1
-
× = × × = ? = ? a
 
 
The forces developed in member (5) as it was not allowed to expand is 
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
707 . 0
707 . 0
707 . 0
707 . 0
10 771 . 3 300
3
8
7
6
5
t
t
t
t
p
p
p
p
 
           
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1
1
1
1
8 . 0
      (3) 
 
The global force vector due to thermal load is 
 
()
()
()
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
      (4) 
 
Writing the load-displacement relation for the entire truss is given below. 
 
                                                         
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 2 . 0 0
071 . 0 271 . 0 0 2 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 2 . 0 0 071 . 0 071 . 0
0 2 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 2 . 0 0 129 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
2 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 20 . 0 071 . 0 271 . 0
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
AE
p
p
p
p
p
p
p
p
(5) 
 
In the above problem 0
8 7 6 5 4 3 2 1
= = = = = = = = p p p p p p p p and 
0
8 7 6
= = = u u u . 
 
Thus solving for unknown displacements, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
- -
-
- -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8 . 0
0
0
8 . 1
8 . 0
0
0
0
0
0
271 . 0 0 0 071 . 0 071 . 0
0 129 . 0 071 . 0 0 0
0 071 . 0 271 . 0 0 20 . 0
071 . 0 0 0 271 . 0 071 . 0
071 . 0 0 2 . 0 071 . 0 271 . 0
1
5
4
3
2
1
AE
u
u
u
u
u
  (5) 
 
Solving equation (5), the unknown displacements are calculated as 
 
m u
u m u m u m u
0013 . 0
0 , 0005 . 0 , 0020 . 0 , 0013 . 0
5
4 3 2 1
- =
= - = = =
    (6) 
 
Now, reactions are computed as, 
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- - -
=
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
0 071 . 0 071 . 0 2 . 0 0
2 . 0 071 . 0 071 . 0 0 0
071 . 0 2 . 0 0 071 . 0 071 . 0
5
4
3
2
1
8
7
6
u
u
u
u
u
p
p
p
  (7) 
 
All reactions are zero as truss is externally determinate and hence change in 
temperature does not induce any reaction. Now member forces are calculated by 
using equation (26.10b) 
 
Member (1): L=5m,
o
0 = ? 
 
Page 5


 
                                                         
4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
 
 
  
()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-
 
 
  kN 424 . 0 4239 . 0 ? =      (7) 
 
Example 26.3 
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the 
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume 
AE=300KN for all members and 
75000
1
= a per 
o
C. 
 
 
 
                                                         
 
 
Solution 
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints 
and members are numbered as shown in Fig.26.4b. In the given problem 
4 3 2 1
, , , u u u u and 
5
u represent unconstrained degrees of freedom. Due to support 
conditions, 0
8 7 6
= = = u u u . 
 
The temperature of the member (2) is raised by 
o
50 C.Thus, 
 
m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
     (1) 
 
The forces are developed in member (2), as it was prevented from expansion.  
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p
 
                                                         
             
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
1
0
1
0
     (2) 
 
The displacement of the member (5) was raised by C
o
40 . Thus, 
 
m T L L
3 5
10 771 . 3 40 2 5
000 , 75
1
-
× = × × = ? = ? a
 
 
The forces developed in member (5) as it was not allowed to expand is 
 
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
707 . 0
707 . 0
707 . 0
707 . 0
10 771 . 3 300
3
8
7
6
5
t
t
t
t
p
p
p
p
 
           
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1
1
1
1
8 . 0
      (3) 
 
The global force vector due to thermal load is 
 
()
()
()
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
      (4) 
 
Writing the load-displacement relation for the entire truss is given below. 
 
                                                         
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 2 . 0 0
071 . 0 271 . 0 0 2 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 2 . 0 0 071 . 0 071 . 0
0 2 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 2 . 0 0 129 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
2 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 20 . 0 071 . 0 271 . 0
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
AE
p
p
p
p
p
p
p
p
(5) 
 
In the above problem 0
8 7 6 5 4 3 2 1
= = = = = = = = p p p p p p p p and 
0
8 7 6
= = = u u u . 
 
Thus solving for unknown displacements, 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
- -
-
- -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8 . 0
0
0
8 . 1
8 . 0
0
0
0
0
0
271 . 0 0 0 071 . 0 071 . 0
0 129 . 0 071 . 0 0 0
0 071 . 0 271 . 0 0 20 . 0
071 . 0 0 0 271 . 0 071 . 0
071 . 0 0 2 . 0 071 . 0 271 . 0
1
5
4
3
2
1
AE
u
u
u
u
u
  (5) 
 
Solving equation (5), the unknown displacements are calculated as 
 
m u
u m u m u m u
0013 . 0
0 , 0005 . 0 , 0020 . 0 , 0013 . 0
5
4 3 2 1
- =
= - = = =
    (6) 
 
Now, reactions are computed as, 
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- - -
=
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
0 071 . 0 071 . 0 2 . 0 0
2 . 0 071 . 0 071 . 0 0 0
071 . 0 2 . 0 0 071 . 0 071 . 0
5
4
3
2
1
8
7
6
u
u
u
u
u
p
p
p
  (7) 
 
All reactions are zero as truss is externally determinate and hence change in 
temperature does not induce any reaction. Now member forces are calculated by 
using equation (26.10b) 
 
Member (1): L=5m,
o
0 = ? 
 
                                                         
5
'
2
AE
p = [-1 0 1 0]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
u
u
u
u
     (8) 
 
=
'
2
p 0.1080 Kn 
 
Member 2:  L=5m,
o
90 = ? ,nodal points 4-1 
 
5
'
2
AE
p = [0 -1 0 1]
5
2
1
8
7
10 771 . 3 300
-
× × -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
u
u
u
u
   (9) 
 
      =0.1087 kN 
 
Member (3): L=5m,
o
0 = ? ,nodal points 3-4 
 
5
300
'
2
= p [-1 0 1 0]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
u
u
u
u
       (10) 
 
   =0.0780kN 
 
Member (4): , 5 , 90 m L = =
o
? nodal points 3-2 
 
5
300
'
2
= p [0 -1 0 1]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
4
3
6
5
u
u
u
u
=0      (11) 
 
Member (5): 2 5 , 45 = = L
o
? ,nodal points 3-1 
 
2 5
300
'
2
= p [-0.707 -0.707 0.707 0.707]
3
2
1
6
5
10 333 . 3 300
-
× × -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
u
u
u
u
   (12) 
 
      =-0.8619 kN 
 
Member (6) : 2 5 , 135 = = L
o
? ,nodal points 4-2 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Summary

,

Free

,

pdf

,

ppt

,

Previous Year Questions with Solutions

,

Viva Questions

,

study material

,

Extra Questions

,

practice quizzes

,

Objective type Questions

,

The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

,

Important questions

,

past year papers

,

The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

,

Semester Notes

,

shortcuts and tricks

,

Sample Paper

,

mock tests for examination

,

video lectures

,

The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

,

Exam

,

MCQs

;