Problema3
Páginas: 4 (810 palabras)
Publicado: 2 de junio de 2015
Cátedra de Ingeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
Hallar por el método de Cross los diagramas de momentos flectores y
esfuerzos cortantes, así como lasreacciones de todas las barras del pórtico de
la figura.
La relación entre los momentos de inercia de las barras es:
I1 = 2 ⋅ I2 = 3 ⋅ I3
P=5 T
q1=2 T/m
q2=1 T/m
A
I1
B
I1
C
I2
I3
I2
D
EF
3
3
5
4
1º . Determinamos los coeficientes elásticos (βi, Ki y ri).
Nudo A
4 ⋅ E ⋅ I2
= 0 .8 ⋅ E ⋅ I 2
5
4 ⋅ E ⋅ I1
K AB =
= 1.33 ⋅ E ⋅ I2
6
1
β AB =
2
1
β AD =
2
K AD =
1
Cátedra deIngeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
K AD
0 .8
=
= 0.38
K AD + K AB 0.8 + 1.33
K AB
1.33
=
=
= 0.62
K AD + K AB 0.8 + 1.33
rAD =
rAB
Nudo B
4 ⋅ E ⋅ I1
=0.67 ⋅ E ⋅ I1
6
4 ⋅ E ⋅ I1
K BC =
= E ⋅ I1
4
K BE = 0
1
β BA =
2
1
β BC =
2
β BE = 0
K BA
0.67
rBA =
=
= 0.40
K BA + K BC + K BE 0.67 + 1 + 0
K BC
1
rBC =
=
= 0.60
K BA + K BC + K BE 0.67 + 1 + 0
K BArBE =
=0
K BA + K BC + K BE
K BA =
Nudo C
4 ⋅ E ⋅ I1
= 2 ⋅ E ⋅ I2
4
4 ⋅ E ⋅ I2
K CF =
= 0 .8 ⋅ E ⋅ I 2
5
1
β CB =
2
1
β CF =
2
K CB
2
rCB =
=
= 0.71
K CB + K CF 2 + 0.8
K CB =
2
Cátedra deIngeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
rCF =
K CF
0 .8
=
= 0.29
K CB + K CF 2 + 0.8
2º . Calculamos los momentos y pares de empotramiento.
q ⋅ l2
1⋅ 5 2
MD =M A = −
=−
= −2.08 T ⋅ m
12
12
1 T/m
A
D
mD = +2.08 T ⋅ m
5
m A = −2.08 T ⋅ m
5T
M A = MB = −
P ⋅l
5⋅6
=−
= −3.75 T ⋅ m
8
8
B
A
m A = +3.75 T ⋅ m
3m
3m
mB = −3.75 T ⋅ m
2 T/m
C
B
MB= MC = −
q ⋅ l2
2 ⋅ 42
=−
= −2.67 T ⋅ m
12
12
4m
mB = +2.67 T ⋅ m
m C = −2.67 T ⋅ m
3
Cátedra de Ingeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
3º .Cross: Transmisiones.
-0.63
-2.08
-0.32
+2.08
-0.03
+0.09
+0.05 +0.08
+0.16
-0.14
-0.27 -0.03
-0.02
-0.20
-0.10 -0.41
-0.21
+0.32
+0.64 +0.78
+1.55
-1.04
-0.52 +0.96
+0.48
+3.75...
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
Hallar por el método de Cross los diagramas de momentos flectores y
esfuerzos cortantes, así como lasreacciones de todas las barras del pórtico de
la figura.
La relación entre los momentos de inercia de las barras es:
I1 = 2 ⋅ I2 = 3 ⋅ I3
P=5 T
q1=2 T/m
q2=1 T/m
A
I1
B
I1
C
I2
I3
I2
D
EF
3
3
5
4
1º . Determinamos los coeficientes elásticos (βi, Ki y ri).
Nudo A
4 ⋅ E ⋅ I2
= 0 .8 ⋅ E ⋅ I 2
5
4 ⋅ E ⋅ I1
K AB =
= 1.33 ⋅ E ⋅ I2
6
1
β AB =
2
1
β AD =
2
K AD =
1
Cátedra deIngeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
K AD
0 .8
=
= 0.38
K AD + K AB 0.8 + 1.33
K AB
1.33
=
=
= 0.62
K AD + K AB 0.8 + 1.33
rAD =
rAB
Nudo B
4 ⋅ E ⋅ I1
=0.67 ⋅ E ⋅ I1
6
4 ⋅ E ⋅ I1
K BC =
= E ⋅ I1
4
K BE = 0
1
β BA =
2
1
β BC =
2
β BE = 0
K BA
0.67
rBA =
=
= 0.40
K BA + K BC + K BE 0.67 + 1 + 0
K BC
1
rBC =
=
= 0.60
K BA + K BC + K BE 0.67 + 1 + 0
K BArBE =
=0
K BA + K BC + K BE
K BA =
Nudo C
4 ⋅ E ⋅ I1
= 2 ⋅ E ⋅ I2
4
4 ⋅ E ⋅ I2
K CF =
= 0 .8 ⋅ E ⋅ I 2
5
1
β CB =
2
1
β CF =
2
K CB
2
rCB =
=
= 0.71
K CB + K CF 2 + 0.8
K CB =
2
Cátedra deIngeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
rCF =
K CF
0 .8
=
= 0.29
K CB + K CF 2 + 0.8
2º . Calculamos los momentos y pares de empotramiento.
q ⋅ l2
1⋅ 5 2
MD =M A = −
=−
= −2.08 T ⋅ m
12
12
1 T/m
A
D
mD = +2.08 T ⋅ m
5
m A = −2.08 T ⋅ m
5T
M A = MB = −
P ⋅l
5⋅6
=−
= −3.75 T ⋅ m
8
8
B
A
m A = +3.75 T ⋅ m
3m
3m
mB = −3.75 T ⋅ m
2 T/m
C
B
MB= MC = −
q ⋅ l2
2 ⋅ 42
=−
= −2.67 T ⋅ m
12
12
4m
mB = +2.67 T ⋅ m
m C = −2.67 T ⋅ m
3
Cátedra de Ingeniería Rural
Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real
3º .Cross: Transmisiones.
-0.63
-2.08
-0.32
+2.08
-0.03
+0.09
+0.05 +0.08
+0.16
-0.14
-0.27 -0.03
-0.02
-0.20
-0.10 -0.41
-0.21
+0.32
+0.64 +0.78
+1.55
-1.04
-0.52 +0.96
+0.48
+3.75...
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