Civil Engineering Reference
In-Depth Information
300 lb
(0, 40)
(40, 40)
500 lb
2
2
3
1
1
(0, 0)
Figure 3.5
Two-element truss with
external loading.
■
Solution
The no
dal coordi
nates are such that
1
=
/
4
and
2
=
0
and the element lengths are
L
1
=
√
40
2
≈
56
.
57 in
.
,
L
2
=
40 in
.
The characteristic element stiffnesses are then
+
40
2
A
1
E
1
L
1
1
.
5(10)(10
6
)
56
.
57
=
2
.
65(10
5
) lb/in
.
k
1
=
=
1
.
5(10)(10
6
)
40
A
2
E
2
L
2
=
3
.
75(10
5
) lb/in
.
k
2
=
=
As the element orientation angles and numbering scheme are the same as in Example 3.1,
we use the result of that example to write the global stiffness matrix as
1
.
325
1
.
325
0
0
−
1
.
325
−
1
.
325
1
.
325
1
.
325
0
0
−
1
.
325
−
1
.
325
0
0
3
.
75
0
−
3
.
75
0
[
K
]
=
10
5
lb/in.
0
0
0
0
0
0
−
1
.
325
−
1
.
325
−
3
.
75
0
5
.
075
1
.
325
−
1
.
325
−
1
.
325
0
0
1
.
325
1
.
325
Incorporating the displacement constraints
U
1
=
U
2
=
U
3
=
U
4
=
0
, the global equilib-
rium equations are
1
.
325
1
.
325
0
0
−
1
.
325
−
1
.
325
0
0
0
0
U
5
U
6
F
1
F
2
F
3
F
4
500
300
1
.
325
1
.
325
0
0
−
1
.
325
−
1
.
325
0
0
3
.
75
0
−
3
.
75
0
10
5
=
0
0
0
0
0
0
−
1
.
325
−
1
.
325
−
3
.
75
0
5
.
075
1
.
325
−
1
.
325
−
1
.
325
0
0
1
.
325
1
.
325
