Civil Engineering Reference
In-Depth Information
U
5
2
(0, 0,
30)
2
U
4
U
6
Y
2
Z
X
U
2
U
11
1
1
4
1
(40, 0, 0)
U
8
U
10
U
1
(0, 0, 30)
4
U
3
U
12
3
3
(0,
30, 0)
U
7
3
U
9
(a)
(b)
Figure 3.8
(a) A three-element, 3-D truss. (b) Numbering scheme.
■
Solution
First, note that the 3-D truss with four nodes has 12 possible displacements. However,
since nodes 1-3 are fixed, nine of the possible displacements are known to be zero. There-
fore, we need assemble only a portion of the system stiffness matrix to solve for the three
unknown displacements. Utilizing the numbering scheme shown in Figure 3.8b and the
element-to-global displacement correspondence table (Table 3.6), we need consider only
the equations
=
K
10, 10
K
10, 11
K
10, 12
U
10
U
11
U
12
0
−
K
11, 10
K
11, 11
K
11, 12
5000
0
K
12, 10
K
12, 11
K
12, 12
Prior to assembling the terms required in the system stiffness matrix, the individual
element stiffness matrices must be transformed to the global coordinates as follows.
Element 1
1
50
[(40
(1)
=
−
0)
I
+
(0
−
0)
J
+
(0
−
30)
K
]
=
0
.
8
I
−
0
.
6
K
Hence,
c
x
=
.
=
=−
.
6
, and Equation 3.59 gives
0
8,
c
y
0,
c
z
0
0
.
64
−
0
.
48
−
0
.
64
0
.
48
0
0
0
0
0
0
0
0
K
(1)
=
−
0
.
48
0
0
.
36
0
.
48
0
−
0
.
36
3(10
5
)
lb/ln.
−
0
.
64
0
0
.
48
0
.
64
0
−
0
.
48
0
0
0
0
0
0
0
.
48
−
0
−
0
.
36
−
0
.
48
0
0
.
36
