Metodo De Cross

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 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.La relación entre los momentos de inercia de las barras es:
I1 = 2 ⋅ I2 = 3 ⋅ I3

2 T/m
A

B

I1

5T
I2

3

I2
1 T/m

D

I1

I1

I3

4

F

E

C I1

I2

G

I3

1.25 T/m

I

H
2
4

1.5

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 IFestá sometida a una carga uniforme de succión, de valor 1.25 T/m

1

Cátedra de Ingeniería Rural

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

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

K AB =

− NUDO B
4 ⋅ E ⋅ I1 4 ⋅ E ⋅ I1
=
= E ⋅ I1
l4
4 ⋅ E ⋅ I2 4 ⋅ E ⋅ I1 2
K BE =
=
= ⋅ E ⋅ I1
l
2⋅3
3
1
β BA =
2
1
β BE =
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

K BA =

− 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

Cátedra de Ingeniería Rural

EscuelaUniversitaria de Ingeniería Técnica Agrícola de Ciudad Real

1
2
3 ⋅ E ⋅ I3 3 ⋅ E ⋅ I1 1
K DG =
=
= ⋅ E ⋅ I1
l
4⋅3
4
β DG = 0
β DE =

rDA =
rDC

K DA
=0

K DA
23
=
= 0 .3 5
+ K DC + K DE + K DG 2 3 + 0 + 1 + 1 4

K DE
1
=
= 0 .5 2
K DA + K DC + K DE + K DG 2 3 + 0 + 1 + 1 4
K DG
14
=
=
= 0 .1 3
K DA + K DC + K DE + K DG 2 3 + 0 + 1 + 1 4

rDE =
rDG

− NUDO E4 ⋅ 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 .2 1
K ED + K EB + K EF + K EH 1 + 2 3 + 1 + 1 2
K EF
1
rEF =
=
= 0.315
K ED + K EB + KEF + K EH 1 + 2 3 + 1 + 1 2
K EH
12
rEH =
=
= 0 .1 6
K ED + K EB + K EF + K EH 1 + 2 3 + 1 + 1 2

K ED =

3

Cátedra de Ingeniería Rural

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

− 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 .8 0
K FE + KFI 1 + 1 4
K FI
14
rFI =
=
= 0 .2 0
K FE + K FI 1 + 1 4
K FE =

2º . Calculamos los momentos y pares de empotramiento.
2 T/m

M A = MB = −

q ⋅ l2
2 ⋅ 42
=−
= −2.67 T ⋅ m
12
12

B

A

m A = +2.67 T ⋅ m

4

mB = −2.67 T ⋅ m

1 T/m

q ⋅ c3
12 ⋅ l 2

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

q ⋅ c3
ME = −
12 ⋅ l 2

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

MD = −
E

D
2m

2m

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

1⋅ 2 3
MD = −
12 ⋅ 4 2

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

1⋅ 2 3
12 ⋅ 4 2

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

ME = −

4

Cátedra de Ingeniería Rural

Escuela Universitaria de Ingeniería Técnica Agrícola de Ciudad Real

mD = 0.92 T ⋅ m
mE = −0.42 T ⋅ mMC = 0

5T

MD = −P ⋅ l = −5 ⋅ 1.5 = −7.5 T ⋅ m
D

C

mD = −7.5 T ⋅ m

1.5 m
MI = 0
1.25 T/m
F

I
4m

MF = −

q ⋅ l2
1.25 ⋅ 4 2
=−
= − 2 .5 T ⋅ m
8
8

mi = 0
m F = − 2 .5 T ⋅ m

Pares de empotramiento
2.67
A

-7.5 0.92
D

G

-0.42
E

F

H

-2.5

C

-2.67
B

I

5

Cátedra de Ingeniería Rural

Escuela Universitaria de Ingeniería...