Metodo de cross

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural

Hallar por el método de Cross los diagramas de momentos flectores y esfuerzos cortantes, así comolas reacciones 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

A

I1

B

I1

C

q2=1 T/mI2

I3

I2

5

D 3 3

E 4

F

1º . Determinamos los coeficientes elásticos (βi, Ki y ri).
Nudo A K AD = 4 ⋅ E ⋅ I2 = 0 .8 ⋅ E ⋅ I 2 5 4 ⋅ E ⋅ I1 K AB = = 1.33 ⋅ E ⋅ I2 6 1 β AB = 2 1 βAD = 2

1

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural

rAD = rAB

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

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 BA rBE = =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 + KCF 2 + 0.8 K CB =

2

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural

rCF =

K CF 0 .8 = = 0.29 K CB + K CF 2 + 0.8

2º . Calculamos losmomentos y pares de empotramiento.
1 T/m D 5 A

q ⋅ l2 1⋅ 5 2 MD = M A = − =− = −2.08 T ⋅ m 12 12 mD = +2.08 T ⋅ m m A = −2.08 T ⋅ m

5T

M A = MB = −
A 3m 3m B

P ⋅l 5⋅6 =− = −3.75 T ⋅ m 88

m A = +3.75 T ⋅ m mB = −3.75 T ⋅ m

2 T/m B 4m C

MB = MC = −

q ⋅ l2 2 ⋅ 42 =− = −2.67 T ⋅ m 12 12

mB = +2.67 T ⋅ m m C = −2.67 T ⋅ m

3

Escuela Universitaria de IngenieríaTécnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural

3º . Cross: Transmisiones.
+ 2.77 +0.02 -0.02 -0.01 +0.09 -0.14 -0.20 +0.32 -1.04 +3.75 0.62 +0.01 +0.05 0.38 -0.12 -0.63 - 2.77 -2.08 -...