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Element 4:
1
4 p 0 ,
p b =
p a =
0
7
80
T
1
80
3
80
1
120
P 40
=
p 0
(22)
P 0
The complete set of equations KV
=
0 would appear as
12
6
12
6
w a
θ
64
6
2
a
12
6
12
+
12
6
6
12
6
w
b
626
64
+
4
6
2
θ
b
EI
12
6
12
+
12
6
6
12
6
w
c
6
2
6
64
+
4
6
2
θ
3
c
12
6
12
+
12
6
6
12
6
w
d
6
2
6
64
+
0
6
2
θ
d
12
6
12
6
w
e
θ
6
2
6
4
e
K
V
37
/
80
6
/
80
33
/
80
+
27
/
80
17
/
240
13
/
240
23
/
80
+
17
/
80
p 0
4
=
0
(23)
1
/
20
1
/
30
13
/
80
+
7
/
80
7
/
240
1
/
80
3
/
80
1
/
120
P 0
=
0
For this beam, the displacement boundary conditions are on the left end,
w a = θ a =
0, and
on the right end,
.
The reduced simultaneous equations of (23) are readily found to have the solution with
=
w
=
0
e
L
/
4
.
] T
V
=
[
w b
θ b w c
θ c w d
θ d θ e
4
p 0
5333 ] T
=
[0
.
3063
0
.
4229
0
.
6000
0
.
1000
0
.
4688
0
.
3437
0
.
(24)
EI
5.3.13
Hinges and Other Indeterminate Nodal Conditions
The conditions, e.g., hinges or supports, illustrated in Fig. 5.26, which constrain one response
variable (usually the value is zero) while generating a discontinuity (a reaction) in the
complementary variable, occur frequently in structural models.
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