Civil Engineering (CE) Exam > Civil Engineering (CE) Notes > Structural Analysis > The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3
The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 | Structural Analysis - Civil Engineering (CE) PDF Download
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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
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FAQs on The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the direct stiffness method in truss analysis? |  |
Ans. The direct stiffness method is a numerical technique used in structural analysis to determine the displacements, forces, and reactions in a truss structure. It involves dividing the truss into smaller elements, calculating the stiffness of each element, and assembling them to form the overall stiffness matrix. By applying appropriate boundary conditions and solving the resulting equations, the method provides a solution for the structural response.
| 2. How do temperature changes affect truss analysis using the direct stiffness method? | |
Ans. Temperature changes can cause thermal expansion or contraction in truss structures, leading to changes in the lengths of truss members. These changes introduce additional forces and displacements, which need to be considered in truss analysis using the direct stiffness method. By incorporating the thermal effects into the stiffness matrix, the method can account for the influence of temperature changes on the structural response.
| 3. What are fabrication errors in truss analysis, and how do they impact the direct stiffness method? | |
Ans. Fabrication errors refer to deviations or inaccuracies in the dimensions, alignment, or material properties of truss members during the manufacturing process. These errors can introduce variations in the stiffness and geometry of the truss structure, affecting its overall behavior. In truss analysis using the direct stiffness method, fabrication errors can be accounted for by adjusting the stiffness matrix to incorporate the actual properties of the fabricated truss.
| 4. Is the direct stiffness method suitable for analyzing complex truss structures? | |
Ans. Yes, the direct stiffness method is suitable for analyzing complex truss structures. It is a versatile and widely used technique that can handle trusses with varying member properties, non-uniform cross-sections, and different types of support conditions. By dividing the truss into smaller elements and assembling them using the stiffness matrix, the method can accurately analyze complex truss structures and provide detailed information about their internal forces and displacements.
| 5. Can the direct stiffness method account for other factors such as material nonlinearity or geometric nonlinearities in truss analysis? | |
Ans. The direct stiffness method, in its basic form, is primarily suited for linear elastic truss analysis. It assumes that the truss members behave linearly and follow Hooke's law. Therefore, it does not directly account for material nonlinearity or geometric nonlinearities. However, advanced versions of the direct stiffness method, such as the finite element method, can incorporate nonlinearities by using nonlinear material models and geometric formulations. These advanced techniques enable the analysis of trusses under nonlinear conditions.
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