# Analisis estructural

Páginas: 9 (2118 palabras) Publicado: 12 de enero de 2011
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 como las reacciones de todas las barras del pórtico de la figura. Calcular la expresión de la curva de momentos flectores de la barra DE y el ángulo girado en E aplicando el método de superposición. Larelación entre los momentos de inercia de las barras es: I1 = 2 ⋅ I2 = 3 ⋅ I3

2 T/m A 5T 3 D C I1 I1 I2 1 T/m E I1 F I2 I1 B

4

I3

I2

I3

1.25 T/m

G

H 2

I

1.5
(*)

4

4

1º . Determinamos los coeficientes elásticos (βi, Ki y ri).
− NUDO A K AD =
(*)

4 ⋅ E ⋅ I2 4 ⋅ E ⋅ I1 2 = = ⋅ E ⋅ I1 2⋅3 l 3

La barra IF está sometida a una carga uniforme de succión, devalor 1.25 T/m

1

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural

K AB =

4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1 = = E ⋅ I1 l 4 1 β AB = 2 1 β AD = 2 K AD 23 rAD = = = 0 .4 K AD + K AB 2 3 + 1 K AB 1 rAB = = = 0 .6 K AD + K AB 2 3 + 1

− NUDO B K BA = 4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1 = = E ⋅ I1 l 4 4 ⋅ E ⋅ I2 4 ⋅ E ⋅ I1 2 K BE = = = ⋅ E ⋅ I1 l 2⋅3 3 1 β BA = 2 1 βBE = 2 K BA 1 rBA = = = 0 .6 K BA + K BE 1 + 2 3 K BE 23 rBE = = = 0 .4 K BA + K BE 1 + 2 3

− NUDO D 4 ⋅ E ⋅ I2 4 ⋅ E ⋅ I1 2 = = ⋅ E ⋅ I1 l 2⋅3 3 1 β DA = 2 K DC = 0 β DC = 0 4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1 K DE = = = E ⋅ I1 l 4 K DA =

2

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural

1 2 3 ⋅ E ⋅ I3 3 ⋅ E ⋅ I1 1 K DG = = = ⋅ E ⋅ I1 l 4⋅3 4 βDG = 0 β DE = rDA = rDC K DA =0 K DA 23 = = 0.35 + K DC + K DE + K DG 2 3 + 0 + 1 + 1 4

rDE = rDG

K DE 1 = = 0.52 K DA + K DC + K DE + K DG 2 3 + 0 + 1 + 1 4 K DG 14 = = = 0.13 K DA + K DC + K DE + K DG 2 3 + 0 + 1 + 1 4

− NUDO E K ED = 4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1 = = E ⋅ I1 l 4 1 β ED = 2 1 β EB = 2 4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1 K EF = = = E ⋅ I1 l 4 1 β EF = 2 4 ⋅ E ⋅ I2 4 ⋅ E ⋅ I1 1 K EH = = = ⋅ E ⋅I1 l 2⋅4 2 1 β EH = 2 K ED 1 rED = = = 0.315 K ED + K EB + K EF + K EH 1 + 2 3 + 1 + 1 2 K EB 23 rEB = = = 0.21 K ED + K EB + K EF + K EH 1 + 2 3 + 1 + 1 2 K EF 1 rEF = = = 0.315 K ED + K EB + K EF + K EH 1 + 2 3 + 1 + 1 2 K EH 12 rEH = = = 0.16 K ED + K EB + K EF + K EH 1 + 2 3 + 1 + 1 2

3

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra de Ingeniería Rural− NUDO F 4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1 = = E ⋅ I1 l 4 1 β FE = 2 3 ⋅ E ⋅ I3 3 ⋅ E ⋅ I1 1 K FI = = = ⋅ E ⋅ I1 l 4 ⋅3 4 β FI = 0 K FE 1 rFE = = = 0.80 K FE + K FI 1 + 1 4 K FI 14 rFI = = = 0.20 K FE + K FI 1 + 1 4 K FE =

2º . Calculamos los momentos y pares de empotramiento.
2 T/m A 4 B m A = +2.67 T ⋅ m mB = −2.67 T ⋅ m M A = MB = − q ⋅ l2 2 ⋅ 42 =− = −2.67 T ⋅ m 12 12

1 T/m MD = − D 2m 2m E

q ⋅c3 12 ⋅ l 2

 12 ⋅ a ⋅ b 2   ⋅ l − 3 ⋅ b +   c2    12 ⋅ a 2 ⋅ b   ⋅ l − 3 ⋅ a +   c2  

q ⋅ c3 ME = − 12 ⋅ l 2

q=1 a=1 b=3 c=2 l=4

1⋅ 2 3 MD = − 12 ⋅ 4 2 1⋅ 2 3 12 ⋅ 4 2

 12 ⋅ 1⋅ 3 2  4 − 3⋅3 +  = −0.92 ⋅ 22     12 ⋅ 12 ⋅ 3   = −0.42 ⋅  4 − 3 ⋅1+   22  

ME = −

4

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

Cátedra deIngeniería Rural

mD = 0.92 T ⋅ m mE = −0.42 T ⋅ m

5T

MC = 0 MD = −P ⋅ l = −5 ⋅ 1.5 = −7.5 T ⋅ m D 1.5 m MI = 0 1.25 T/m MF = − mi = 0 m F = − 2 .5 T ⋅ m q ⋅ l2 1.25 ⋅ 4 2 =− = − 2 .5 T ⋅ m 8 8 mD = −7.5 T ⋅ m

C

I 4m

F

Pares de empotramiento
2.67 A -2.67 B

C

-7.5 0.92 D

-0.42 E

-2.5

F

G

H

I

5

Cátedra de Ingeniería Rural

3º . Cross: Transmisiones.
+ 0.79 +0.02 -0.06 +0.08 -0.16 +0.18 -0.80 +1.15 -2.29 +2.67 0.60 -0.04 +0.02 -0.11 +0.09 -0.54 +0.19 -1.53 +1.15 0.4 - 0.77 + 4.62 - 1.49 +0.03 -0.03 +0.15 -0.08 +0.37 -0.40 +2.29 -1.15 -2.67 0.60 +0.02 -0.02 +0.10 -0.17 +0.25 -0.22 +1.53 0.40 + 1.49 + 0.55 +0.03 -0.06 +0.14 -0.49 +0.29 -0.65 +1.71 -0.42 0.315 - 0.61...

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