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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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