Civil Engineering Reference
In-Depth Information
k
(5)
23
K
69
=
=
0
k
(5)
24
K
6,10
=
=
0
K
6,11
=
K
6,12
=
0
k
(2)
33
k
(3)
33
k
(4)
33
k
(6)
33
k
(7)
11
K
77
=
+
+
+
+
75)10
5
=
(2
.
65
/
2
+
3
.
75
+
0
+
2
.
65
/
2
+
3
.
k
(2)
34
k
(3)
34
k
(4)
34
k
(6)
34
k
(7)
12
K
78
=
+
+
+
+
0)10
5
=
(2
.
65
/
2
+
0
+
0
−
2
.
65
/
2
+
=
0
k
(6)
13
2)10
5
K
79
=
=−
(2
.
65
/
k
(6)
23
2)10
5
K
7,10
=
=
(2
.
65
/
k
(7)
13
75(10
5
)
K
7,11
=
=−
3
.
k
(7)
14
K
7,12
=
=
0
k
(2)
44
k
(3)
44
k
(4)
44
k
(6)
44
k
(7)
22
K
88
=
+
+
+
+
0)10
5
=
(2
.
65
/
2
+
0
+
3
.
75
+
2
.
65
/
2
+
k
(6)
14
2)10
5
K
89
=
=
(2
.
65
/
k
(6)
24
2)10
5
K
8,10
=
=−
(2
.
65
/
k
(7)
23
K
8,11
=
=
0
k
(7)
24
K
8,12
=
=
0
k
(5)
33
k
(6)
11
k
(8)
11
0)10
5
K
99
=
+
+
=
(3
.
75
+
2
.
65
/
2
+
k
(5)
34
k
(6)
12
k
(8)
12
0)10
5
K
9,10
=
+
+
=
(0
−
2
.
65
/
2
+
k
(8)
13
K
9,11
=
=
0
k
(8)
14
K
9,12
=
=
0
k
(5)
44
k
(6)
22
k
(8)
22
75)10
5
K
10,10
=
+
+
=
(0
+
2
.
65
/
2
+
3
.
k
(8)
23
K
10,11
=
=
0
k
(8)
24
75(10
5
)
K
10,12
=
=−
3
.
k
(7)
33
k
(8)
33
0)10
5
K
11,11
=
+
=
(3
.
75
+
k
(7)
34
k
(8)
34
K
11,12
=
+
=
0
+
0
k
(7)
44
k
(8)
44
75)10
5
K
12,12
=
+
=
(0
+
3
.
Step 6.
Apply the constraints as dictated by the boundary conditions. In this
example, nodes 1 and 2 are fixed so the displacement constraints are
U
1
=
U
2
=
U
3
=
U
4
=
0
