+
=
0
FFF
x
A
x
1
0
)
=
0
-
1
0
+++
++
5
+
(
5
F
+
F
555A
y
F
y
=
FF
-
A
x
F
x
=
-
5
-
1
0
-
1
5
-
2
0
-
2
5
-
6
0
+
6
=
0
M
F
A
Σ
=
N
22
.
5
k
F
y
F
F
y
F
+
F
=
4
5
F
y
A
y
+
=
4
5
F
22
.
5
k
N
k
N
=
N
22
.
5
k
A
y
F
A
y
3. Barras nulasen total
4
I-BK-CM-DÑ-E4. Simetría (Forma y carga)Si hay simetría en forma y carga5. DCL de los nodosA
F
A
B
F
A
H
F
A
H
s
e
n
α
c
o
s
α
F
A
H
A
x
++
c
o
s
=
0
FF
α
∑
F
=
0
x
0
G
A
F
A
H
A
B
=
x
-
F
-
A
c
o
s
A
H
α
--
1
7
,
6
77
(
)
=
c
o
s
α
F
A
B
F
A
B
∑F
Y
=
0
A
y
+
G
A
F+
F
A
H
s
e
n
α
=
0
0
=
2
2
.
5
k
N
+
1
0
k
N
+
F
A
H
5
k
N
1
0
k
N
--
(
)
22
.
-
=
F
A
H
n
α
s
e
n
α
s
e
=
F
A
H
=
F
A
B
Nodo A:HI
α
=
4
5
∘
G
F
A
B
C
O
S
α
1
2
4
99
k
N
B
A
0
∑
F
x
==
α
4
5
∘
=
α
4
5
∘
Nodo H:Nodo G:
5
k
N
G
A
H
1
0
k
N
F
0
k
N
G
x
F
0
k
N
G
x
-
1
7
.
6
77
kN
-
1
7
.
6
77
k
N
F
0
N
k
A
x
F
0
k
N
A
x
F
A
y
N
22
.
5
k
F
A
y
N
22
.
5
k
5
∘
4
1
4
,
1
4
2
k
N
F
A
B
S
e
n
a
∑
F
x
=
∑F
=
0
YF
+
F
+
F
=
0
c
o
s
∘
()
4
5
+
F
+
F
0
=
HI
G
HH
B
HB
co
s
∘
()
4
5
HI
+F
0
=
+
2
4
,
7
4
7
c
o
s
∘
(
)
4
5
-
⎛⎝⎞⎠
1
4
,
1
4
2
-
1
4
.
1
4
2
-
HI
F
=
3
1
.
6
4
k
N
H
I
∑F
=
0
Y
F
=
H
B
A
H
+
∘
4
5
S
e
n
∘
(
)
1
7
,
6
77
4
5
S
e
n
∘
(
)
4
5
=
0
-
=
-
s
e
n
4
5
∘
(
)
F
0
-
5
1
2
,
4
9
9
-
H
B
=
-
s
e
n
4
5
∘
(
)
F
-
1
7
,
4
99
0
-
1
7
,
4
99
s
e
n
4
5
(
)
=
H
B
F
HB
H
B
-
5
F
H
B
-
5
F
s
e
n
)
F
=
0
∘
(
4
5
S
e
n
(
)
++
1
4
,
1
4
2
c
o
s
4
5
+
0
=
0
1
0
=
0
1
4
,
1
4
2
c
o
s
4
5
=
0
F
c
o
s
4
5
=
0
+
F
G
H
G
X
=
0
s
e
n
4
5
-
10
+
1
0
=
F
G
H
s
e
n
4
5
F
G
H
=
F
G
H
1
4
,
1
4
2
k
N
2
4
,
7
4
7
k
N
-,
1
2
4
9
9
kN
-
1
7
.
6
77
k
N
BCLMNoDEFGHJkPÑ
FF
y
5
kN
5
kN
5
k
N
5
k
N
1
0
k
N
1
m
1
m
1
m
F
x
F
1
m
1
m
N
22
.
5
k
0
k
N
∼∼∼∼∼∼
∼∼
F
A
y
N
22
.5
k
I
F
A
x
0
k
N
4
5
∘
A
5
k
N
1
0
k
N
1
m
BCLMNoDEFGHJkPÑ
F
F
y
5
kN
5
k
N
5
kN
1
m
1
m
1
m
F
x
F
1
m
1
m
∼∼∼∼∼∼∼∼
F
A
y
I
F
A
x
0
k
N
4
5
∘
A5
k
N
1
m
0
k
N
Método de Nodos1. Verificar Sistema Interno y Externo
r
=
q
=
44
m
+
r
=
2
j
3
5
=
3
4
HiperestáticoEstable3
1
+
4
=
2
1
7
()
Método de secciones1. Verificar Sistema Interno y Externo
r
=
q
=
44
m
+
r
=
2
j
3
5
=
3
4
HiperestáticoEstable
31
+
4
=
2
1
7
()
Determina la fuerza en los elementos J-K, J-C y B-C
∑
F
x
=
0
∑F
y
=
0
2. DC Determinar reacciones de los Apoyos2. DC Determinar reacciones de los Apoyos
∑
F
x
=
0
∑
F
y
=
0
A
x
=
0
=
F
y
A
y
5
k
N
5
k
N
5
kN
2
A
y
=
5
(
7
)
k
N
=
3
5
A
y
-
2
A
y
=
1
7
.
5
k
N
1
7
.
5
k
N
FF
J
D
F
C
B
J
K
∑M
=
0
∑
M
=
0
-
1
7
,
5
1
,
5
(
)
+
5
1
(
)
+F
1
(
)
=
0
=
2
1
,
2
5
C
F
k
N
-
3
2
(
)
+
5
2
(
)
+
5
1
(
)
-
F
(
1
)
=
0
∑
M
=
0
c
J
K
F
=
J
K
9
J
B
C
B
C
4
5
∘
k
N
+
F
co
s
4
5
()
=
0
J
D
F
J
K
+
F
B
C
F
J
K
F
B
C
-
c
o
s
4
5
(
)
=
F
J
D
=
F
J
D
2
1
,
2
5
-
9
c
o
s
4
5
()
=
1
7
,
3
2
4
∑
F
x
=
0