Civil Engineering (CE) Exam > Civil Engineering (CE) Notes > Structural Analysis > The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 2
The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 2 | Structural Analysis - Civil Engineering (CE) PDF Download
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Page 1
411
3
33
0000
01 0 1
15 10 2 10
0000 3 x 10 10
010 1
k
-
? ?
? ?
-
×××
? ?
?? =
??
? ? ×
? ?
-
? ?
(7)
The global stiffness matrix is of the order 8 8 × ,assembling the three member
stiffness matrices, one gets
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
k (8)
Writing the load displacement equation for the truss
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640
(9)
In the present case, the displacements
1
u and
2
u are not known. All other
displacements are zero. Also 0
2 1
= = p p (as no joint loads are applied).Thus,
Page 2
411
3
33
0000
01 0 1
15 10 2 10
0000 3 x 10 10
010 1
k
-
? ?
? ?
-
×××
? ?
?? =
??
? ? ×
? ?
-
? ?
(7)
The global stiffness matrix is of the order 8 8 × ,assembling the three member
stiffness matrices, one gets
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
k (8)
Writing the load displacement equation for the truss
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640
(9)
In the present case, the displacements
1
u and
2
u are not known. All other
displacements are zero. Also 0
2 1
= = p p (as no joint loads are applied).Thus,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640 (10)
Thus unknown displacements are
1
1
3
2
147.14 47.14 0 1
1
(150.82 )
47.14 147.14 0 1 10
u
u
-
- ?????? ??
=-
?? ? ? ? ?
??
-
???? ?? ??
(11)
4
1
4
2
7.763 10
7.763 10
um
um
-
-
=×
=×
Now reactions are calculated as
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
1
1
0
0
640
0
0
0
0
0
0
0 0 0 0 0 0
0 100 0 0 0 0
0 0 14 . 47 14 . 47 0 0
0 0 14 . 47 14 . 47 0 0
0 0 0 0 100 0
0 0 0 0 0 0
0 0
0 100
14 . 47 14 . 47
14 . 47 14 . 47
100 0
0 0
10
2
1 3
8
7
6
5
4
3
u
u
p
p
p
p
p
p
3
4
5
6
7
8
0
77.63
77.63
77.63
77.63
0
p
p
p
p
p
p
??
??
??
??
-
??
??
??
??
?? ? ?
=
?? ? ?
?? ? ?
?? ? ?
-
?? ? ?
??
????
??
kN (12)
Page 3
411
3
33
0000
01 0 1
15 10 2 10
0000 3 x 10 10
010 1
k
-
? ?
? ?
-
×××
? ?
?? =
??
? ? ×
? ?
-
? ?
(7)
The global stiffness matrix is of the order 8 8 × ,assembling the three member
stiffness matrices, one gets
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
k (8)
Writing the load displacement equation for the truss
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640
(9)
In the present case, the displacements
1
u and
2
u are not known. All other
displacements are zero. Also 0
2 1
= = p p (as no joint loads are applied).Thus,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640 (10)
Thus unknown displacements are
1
1
3
2
147.14 47.14 0 1
1
(150.82 )
47.14 147.14 0 1 10
u
u
-
- ?????? ??
=-
?? ? ? ? ?
??
-
???? ?? ??
(11)
4
1
4
2
7.763 10
7.763 10
um
um
-
-
=×
=×
Now reactions are calculated as
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
1
1
0
0
640
0
0
0
0
0
0
0 0 0 0 0 0
0 100 0 0 0 0
0 0 14 . 47 14 . 47 0 0
0 0 14 . 47 14 . 47 0 0
0 0 0 0 100 0
0 0 0 0 0 0
0 0
0 100
14 . 47 14 . 47
14 . 47 14 . 47
100 0
0 0
10
2
1 3
8
7
6
5
4
3
u
u
p
p
p
p
p
p
3
4
5
6
7
8
0
77.63
77.63
77.63
77.63
0
p
p
p
p
p
p
??
??
??
??
-
??
??
??
??
?? ? ?
=
?? ? ?
?? ? ?
?? ? ?
-
?? ? ?
??
????
??
kN (12)
The support reactions are shown in Fig.26.2c.The member forces can be easily
calculated from reactions. The member end forces can also be calculated by
using equation (26.10a) and (26.10b). For example, for member (1),
o
0 = ?
100 10
3 '
2
× = p [] 0 1 0 1 - 4
4
0
0
7.763 10
7.763 10
-
-
? ?
? ?
? ?
? ?
×
? ?
? ?
×
??
(13)
= 77.763 kN. Thus the member (1) is in tension.
Member (2)
o
45 = ?
281 . 94 10
3 '
2
× = p [-0.707 -0.707 0.707 0.707]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
×
×
-
-
3
3
10 2942 . 3
10 2942 . 3
0
0
kN. 78 . 109
2
- = ' p
Thus member (2) is in compression
Page 4
411
3
33
0000
01 0 1
15 10 2 10
0000 3 x 10 10
010 1
k
-
? ?
? ?
-
×××
? ?
?? =
??
? ? ×
? ?
-
? ?
(7)
The global stiffness matrix is of the order 8 8 × ,assembling the three member
stiffness matrices, one gets
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
k (8)
Writing the load displacement equation for the truss
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640
(9)
In the present case, the displacements
1
u and
2
u are not known. All other
displacements are zero. Also 0
2 1
= = p p (as no joint loads are applied).Thus,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640 (10)
Thus unknown displacements are
1
1
3
2
147.14 47.14 0 1
1
(150.82 )
47.14 147.14 0 1 10
u
u
-
- ?????? ??
=-
?? ? ? ? ?
??
-
???? ?? ??
(11)
4
1
4
2
7.763 10
7.763 10
um
um
-
-
=×
=×
Now reactions are calculated as
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
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?
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?
+
?
?
?
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?
?
?
?
?
?
?
?
?
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?
?
?
?
?
?
?
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
1
1
0
0
640
0
0
0
0
0
0
0 0 0 0 0 0
0 100 0 0 0 0
0 0 14 . 47 14 . 47 0 0
0 0 14 . 47 14 . 47 0 0
0 0 0 0 100 0
0 0 0 0 0 0
0 0
0 100
14 . 47 14 . 47
14 . 47 14 . 47
100 0
0 0
10
2
1 3
8
7
6
5
4
3
u
u
p
p
p
p
p
p
3
4
5
6
7
8
0
77.63
77.63
77.63
77.63
0
p
p
p
p
p
p
??
??
??
??
-
??
??
??
??
?? ? ?
=
?? ? ?
?? ? ?
?? ? ?
-
?? ? ?
??
????
??
kN (12)
The support reactions are shown in Fig.26.2c.The member forces can be easily
calculated from reactions. The member end forces can also be calculated by
using equation (26.10a) and (26.10b). For example, for member (1),
o
0 = ?
100 10
3 '
2
× = p [] 0 1 0 1 - 4
4
0
0
7.763 10
7.763 10
-
-
? ?
? ?
? ?
? ?
×
? ?
? ?
×
??
(13)
= 77.763 kN. Thus the member (1) is in tension.
Member (2)
o
45 = ?
281 . 94 10
3 '
2
× = p [-0.707 -0.707 0.707 0.707]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
×
×
-
-
3
3
10 2942 . 3
10 2942 . 3
0
0
kN. 78 . 109
2
- = ' p
Thus member (2) is in compression
Example 26.2
Analyze the truss shown in Fig.26.3a, if the member BC is made 0.01m too short
before placing it in the truss. Assume AE=300 kN for all members.
Page 5
411
3
33
0000
01 0 1
15 10 2 10
0000 3 x 10 10
010 1
k
-
? ?
? ?
-
×××
? ?
?? =
??
? ? ×
? ?
-
? ?
(7)
The global stiffness matrix is of the order 8 8 × ,assembling the three member
stiffness matrices, one gets
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
k (8)
Writing the load displacement equation for the truss
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
10
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640
(9)
In the present case, the displacements
1
u and
2
u are not known. All other
displacements are zero. Also 0
2 1
= = p p (as no joint loads are applied).Thus,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
- - -
- - -
=
0 0 0 0 0 0 0 0
0 100 0 0 0 0 0 100
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 14 . 47 14 . 47 0 0 14 . 47 14 . 47
0 0 0 0 100 0 100 0
0 0 0 0 0 0 0 0
0 0 14 . 47 14 . 47 100 0 14 . 147 14 . 47
0 100 14 . 47 14 . 47 0 0 14 . 47 14 . 147
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
0
0
1
1
0
0
1
1
640 (10)
Thus unknown displacements are
1
1
3
2
147.14 47.14 0 1
1
(150.82 )
47.14 147.14 0 1 10
u
u
-
- ?????? ??
=-
?? ? ? ? ?
??
-
???? ?? ??
(11)
4
1
4
2
7.763 10
7.763 10
um
um
-
-
=×
=×
Now reactions are calculated as
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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+
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?
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?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
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?
?
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
?
?
?
?
?
?
?
?
?
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
1
1
0
0
640
0
0
0
0
0
0
0 0 0 0 0 0
0 100 0 0 0 0
0 0 14 . 47 14 . 47 0 0
0 0 14 . 47 14 . 47 0 0
0 0 0 0 100 0
0 0 0 0 0 0
0 0
0 100
14 . 47 14 . 47
14 . 47 14 . 47
100 0
0 0
10
2
1 3
8
7
6
5
4
3
u
u
p
p
p
p
p
p
3
4
5
6
7
8
0
77.63
77.63
77.63
77.63
0
p
p
p
p
p
p
??
??
??
??
-
??
??
??
??
?? ? ?
=
?? ? ?
?? ? ?
?? ? ?
-
?? ? ?
??
????
??
kN (12)
The support reactions are shown in Fig.26.2c.The member forces can be easily
calculated from reactions. The member end forces can also be calculated by
using equation (26.10a) and (26.10b). For example, for member (1),
o
0 = ?
100 10
3 '
2
× = p [] 0 1 0 1 - 4
4
0
0
7.763 10
7.763 10
-
-
? ?
? ?
? ?
? ?
×
? ?
? ?
×
??
(13)
= 77.763 kN. Thus the member (1) is in tension.
Member (2)
o
45 = ?
281 . 94 10
3 '
2
× = p [-0.707 -0.707 0.707 0.707]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
×
×
-
-
3
3
10 2942 . 3
10 2942 . 3
0
0
kN. 78 . 109
2
- = ' p
Thus member (2) is in compression
Example 26.2
Analyze the truss shown in Fig.26.3a, if the member BC is made 0.01m too short
before placing it in the truss. Assume AE=300 kN for all members.
Solution
A similar truss with different boundary conditions has already been solved in
example 25.1. For the sake of completeness the member of nodes and members
are shown in Fig.26.3b.The displacements
6 5 4 3
, , , u u u u ,
7
u and
8
u are zero due to
boundary conditions. For the present problem the unconstrained degrees of
freedom are
1
u and
2
u .The assembled stiffness matrix is of the order 8 8 × and is
available in example 25.1.
In the given problem the member (2) is short by 0.01m.The forces developed in
member (2) in the global coordinate system due to fabrication error is
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
sin
cos
sin
cos
4
01 . 0
0 2
0 1
0 4
0 3
AE
p
p
p
p
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
75 . 0
0
75 . 0
0
kN (1)
Now force-displacement relations for the truss are
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FAQs on The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 2 - Structural Analysis - Civil Engineering (CE)
| 1. How does temperature changes affect truss analysis using the Direct Stiffness Method? |  |
Ans. Temperature changes can cause thermal expansion or contraction of the truss elements, leading to changes in their lengths. This can result in changes in the internal forces and displacements of the truss members, which need to be considered in the analysis using the Direct Stiffness Method. The temperature effects can be incorporated by modifying the stiffness matrix of the truss elements based on the coefficient of thermal expansion and the temperature change.
| 2. What are fabrication errors and how do they impact truss analysis using the Direct Stiffness Method? | |
Ans. Fabrication errors refer to the deviations or inaccuracies in the physical construction of the truss members compared to their ideal design. These errors can occur due to manufacturing tolerances, welding or assembly imperfections, or any other factors that affect the actual dimensions and alignment of the truss elements. Fabrication errors can lead to variations in the stiffness and geometry of the truss, which can affect the calculated internal forces and displacements using the Direct Stiffness Method.
| 3. How can temperature changes be accounted for in the Direct Stiffness Method analysis of trusses? | |
Ans. Temperature changes can be accounted for in the Direct Stiffness Method analysis of trusses by introducing additional terms in the stiffness matrix of the truss elements. These additional terms represent the thermal expansion or contraction effects and are determined based on the coefficient of thermal expansion and the temperature change. By including these temperature-related terms in the stiffness matrix, the analysis can accurately capture the changes in the truss behavior due to temperature variations.
| 4. What are the potential challenges in incorporating temperature changes and fabrication errors in truss analysis using the Direct Stiffness Method? | |
Ans. Incorporating temperature changes and fabrication errors in truss analysis using the Direct Stiffness Method can pose several challenges. Firstly, accurately determining the coefficient of thermal expansion and the temperature change can be difficult, especially in real-world scenarios. Secondly, quantifying the extent and nature of fabrication errors can be challenging, as they can vary for different truss members. Additionally, modifying the stiffness matrix to account for temperature and fabrication effects can increase the complexity of the analysis, requiring additional computational efforts.
| 5. How can the Direct Stiffness Method be used to analyze trusses with temperature changes and fabrication errors? | |
Ans. The Direct Stiffness Method can be used to analyze trusses with temperature changes and fabrication errors by considering the effects of these factors in the stiffness matrix of the truss elements. By accurately incorporating the thermal expansion or contraction effects and accounting for fabrication errors, the modified stiffness matrix can provide more realistic results for the internal forces and displacements of the truss members. This allows engineers to assess the structural behavior of trusses under various temperature conditions and fabrication tolerances, aiding in the design and evaluation process.
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