Método de Cross y deflexion - ejercicios

1) Resolver la viga que se muestra en la figura, usando el método de las deformaciones angulares. E= conocido

Solución:

1-. Rigideces relativa:

K AB=K BA=I4

; K BC=KCB=2 I2

=I ; KCD=K DC=2 I3

2-. Analizamos los tramos para los empotramientos.

Tramo AB:

Momento de empotramiento.

M°AB=−M°BA=−W L2

12=−83

M°BA=83

Remplazando en la ecuación de MOHR.

M ij=2E K ij (2θi+θ j−3∆ij )+M °ij

M AB=2 E( I4 ) (2θA+θB−3(0))−83

M AB=E ( I2 )(2θA+θB )−83

M ji=2 E K ji (2θ j+θi−3∆ ji)+M ° ji

MBA=2E ( I4 ) (2θB+θA−3 (0))+ 83

MBA=E( I2 ) (2θB+θ A )+ 83

TRAMO BC:


M°BC=−M °CB=−W L2

12−PL8

=−136

M°CB=136



MBC=2 E ( I ) (2θB+θC−3(0))−136

MBC=2 E ( I ) (2θB+θC )−136


MCB=2E ( I ) (2θC+θB−3(0))+ 136

MCB=2E ( I ) (2θC+θB )+ 136

TRAMO CD:


M°CD=−M °DC=−W L2

12=−32

M°DC=32


Dónde: θD=0


MCD=2 E( 2 I3 ) (2θC+θD−3 (0 ) )−32

MCD=4 E( I3 ) (2θC )−32


MDC=2E ( 2 I3 ) (2θD+θC−3 (0))+32

MDC=4 E ( I3 ) (θC )+ 32

3-.Ecuaciones de equilibrio.

∑M A= 0; M AB=0⇉E ( I2 )(2θ A+θB )−83=O……(1)

∑M B= 0; MBA+M BC=0⇉E( I2 ) (2θB+θA )+ 83+2 E ( I ) (2θB+θC )−13

6=O

⇉E ( I2 )(θ A )+5 E ( I )θB+2 E ( I )θC+12=O……(2)

∑MC= 0; MCB+MCD=0⇉2 E ( I ) (2θC+θB )+ 136

+4 E ( I3 ) (2θC )−32=O

⇉2 E ( I ) (θB )+20 E ( I )θC+23=O……(3)

De (1) obtenemos:

θA=83−θB2……(4)

De (3) obtenemos:

θC=−110

−3θB10

……(5)

Reemplazamos (4) y (5) en (2):

E( I2 ) (θA )+5 E ( I )θB+2E (I )θC+12=O

θA12+5θB+2θC=

−12

( 83−θB2

) 12+5θB+2(

−110

−3θB10

)=−12

θB=−98249 EI

≅−0.393EI

De (4) obtenemos:

θA=83−θB2

θA=713249EI

≅ 2.863EI

De (5) obtenemos:

θC=−110

−3θB10

θC=3

166 EI≅ 0.018

EI

4-. Calculamos los momentos.

M AB=E ( I2 )(2 713249 EI

+ −98249EI )−83=0

MBA=E( I2 )(2 −98249 EI

+ 713249 EI )+ 83=3.705T

MBC=2 E ( I )(2 −98249 EI

+ 3166 EI )−136 =−3.705T

MCB=2E ( I )(2 3166 EI

+ −983249 EI )+ 136 =1.452T

MCD=4 E( I3 )(2 3166 EI )−32=−1.452T

MDC=4 E ( I3 )( 3166 EI )+ 32=1.53T

5-. Calculamos las cortantes.

De la ecuación Qij=V ij−1l

(M ij+M ji)

QAB=2 ( 42 )−14 (0+3.705 )=3.08

QBA=−2( 42 )−14 (3.705+0 )=−4.93

QBC=2( 22 )+62−12 (−3.705+1.452 )=6.13

QCB=−2( 22 )−62−12

(1.452−3.71 )=−3.87

QCD=2( 32 )−13 (−1.452+1.53 )=2.97

QDC=−2( 32 )−13 (1.53−1.452 )=3.03

Método de Cross

Nº1:

a) Cálculo de las rigideces:

KAB=0

KBA=( 34 ) I4=3 I16KBC=2 I

2=I

KCB=2 I2

=I

KCD=2 I3

KDC=2 I3

b) Cálculo de los factores de distribución

∑ d1=1

Nudo B:

KAB+KBC=19 I16

Nudo C:

KCB+KCD=I+ 2 I3

=5 I3

dA=0 dD=1

dBA=

3 I1619 I16

=0.158 dBA= I

19 I16

=0.842

dCB= I5 I3

=0.6 dCD=

2 I35 I3

=0.4

c) Momentos de empotramiento perfecto

MAB=0

MBA=2x 42

8=4

MBC=−2x 22

12−6 x 2

8=−2.17

MCB=2.17

MCD=−2 x32

12=−1.5

MDC=1.5

LOS MOMENTOS: LAS CORTANTES: Qij=v−1l(M ij+M ji)

MAB=0 QAB=4−14

(0+3.71 )=3.07

MBA=3.71 QBA=−4− 14

(0+3.71 )=−4.93

MBC=−3.71 QBC=5−12

(−3.71+1.35 )=6.18

MCB=1.35 QCB=−5−12

(−3.71+1.35 )=−3.82

MCD=−1.35 QCD=3−13

(−1.35+1.58 )=2.95

MDC= 1.50 QDC=−3−13

(−1.35+1.58 )=−3.05

2)

a) Cálculo de las rigideces:

K12=I4 K34=

34 (1.5 I4 )=9 I32

K21=I4

K43=0

K23=I4

K56=I4

K32=I4

K65=I4

K35=I3

K53=I3

b) Cálculo de factores de distribución :

∑ d1=1

d1=d6=1d4=0

Nudo 2:

K21+K23=I2

Nudo 3:

K32+K34+K35=83 I96

Nudo 5:

K53+K56=7 I12

d1=1

d21=

I4I2

=0.5 d32=

I483 I86

=0.289 d53=

I37 I12

=0.571

d23=

I4I2

=0.5 d35=

I383 I96

=0.386 d56=

I47 I12

=0.429

d34=

9 I3283 I96

=0.325

c) Momentos de empotramiento perfecto

M 12=0

M 21=0

M 43=0

M 34=0

M 56=0

M 65=0

−M 23=M 32=4 x 42

12=5.33

M 35=−4 x 32

12−8 x 1x 2

2

32=−6.56

M 53=¿ 4 x 32

12+ 8 x2 x1

2

32=4.78

Cálculo de los momentos y cortantes:

Qij=v−1l(M ij+M ji)

M 12=1.29 Q12=0−14

(1.29+2.57 )=−0.97

M 21=2.57 Q21=−0.97

M 23=−2.57 Q23=8−14

(−2.57+7.02 )=6.89

M 32=7.12 Q32=−8−14

(−2.57+7.02 )=−9.12

M 34=0.45 Q34=0−14

(0.45+0 )=−0.11

M 43=0 Q43=−0.11

M 35=−7.46 Q35=4 x 32

+ 8 x 23

−13

(−7.46+2.16 )=13.1

M 53=2.16 Q53=−4 x 32

−8 x13

−13

(−7.46+2.16 )=−6.9

M 56=−2.16 Q56=0−14

(−2.16−1.077 )=0.81

M 64=−1.077 Q65=0.81

2) Resolver la estructura que se muestra en la figura usando el método de las Deformaciones Angulares, E es el módulo de elasticidad del material.

Solución:

1-. Rigideces relativa:

K12=K21=I3

; K24=K 42=K23=K32=I4

; K46=K64=I5

K45=K 54=I4

; K68=K 86=I3

; K67=K76=I3

2-. Analizamos los tramos para los empotramientos.

Tramo 12:


M°12=−M °21=−W L2

12=−32

M°21=32


Dónde: θ1=0


M 12=2E ( I3 ) (2θ1+θ2−3(0))−32

M 12=2E ( I3 ) (2θ1+θ2 )−32

M 12=23

(θ2 ) EI−32


M 21=2E ( I3 ) (2θ2+θ1−3(0))+32

M 21=43

(θ2 ) EI+ 32

Tramo 23:


M°12=−M °21=0

Reemplazando en la ecuación de MOHR

Dónde: θ3=0


M 23=2 E( I4 ) (2θ2+θ3−3 (0))

M 23=θ2 EI


M 32=2E ( I4 ) (2θ3+θ2−3(0))

M 32=θ2EI

2

TRAMO 24:


M°24=−M° 42=−W L2

12−PL8

=−316

M° 42=316



M 24=2 E( I4 ) (2θ2+θ4−3(0))−316

M 24=θ2EI+θ4EI

2−316


M 42=θ4EI +θ2EI

2+ 316

Tramo 45:


M° 45=−M°54=0

Reemplazando en la ecuación de MOHR

Dónde: θ5=0


M 45=2 E( I4 ) (2θ4+θ5−3(0))

M 45=θ4EI


M 54=2 E( I4 ) (2θ5+θ4−3(0))

M 54=θ4 EI

2

Tramo 45:


M°67=−M°76=0

Reemplazando en la ecuación de MOHR.

Dónde: θ7=0


M 67=2 E( I4 ) (2θ6+θ7−3(0))

M 67=θ6 EI


M 76=2 E( I4 ) (2θ7+θ6−3(0))

M 76=θ6 EI

2

TRAMO 46:


M° 46=−W L2

12− Pa2b

L2=−1057

150

M°64=W L2

12+ Pab

2

L2=¿ 913

150



M 46=2 E( I5 ) (2θ4+θ6−3 (0))−1057150

M 46=4 θ4 EI

5+2θ6EI

5−1057150


M 64=2 E( I5 ) (2θ6+θ4−3 (0))+ 913150

M 64=4θ6EI

5+2θ4 EI

5+ 913150

Tramo 68:


M°68=−W L2

12=−94

M° 86=0



M 68=2 E( I3 ) (2θ6+θ8−3(0))−94

M 68=4θ6EI

3+2θ8EI

3−94


M 86=2 E( I3 ) (2θ8+θ6−3(0))

M 86=¿4θ8EI

3+2θ6 EI

3

3-. Ecuaciones de equilibrio.

∑M 2= 0; M 21+M 24+M 23=0

43

(θ2 ) EI+ 32+θ2 EI+

θ4 EI

2−316

+θ2EI=O……(1)

∑M 4=0; M 42+M 45+M 46=0

θ4EI +θ2EI

2+ 316

+θ4 EI+4θ4 EI

5+2θ6 EI

5−1057150

=O…… (2)

∑M 6=0; M 64+M 67+M 68=0

4θ6EI

5+2θ4 EI

5+ 913150

+θ6EI +4θ6EI

3+2θ8EI

3−94=O…….(3)

∑M 8=0; M 21+M 24+M 23=0

4θ8EI

3+2θ6EI

3=O……(4)

Resolviendo

θ2=0.994

θ4=0.703

θ6=−1.47

θ8=0.735

4-.Calculamos los momentos.

M 12=23

(θ2 ) EI−32=23

(0.994 )−32=−0.837

M 21=43

(θ2 ) EI+ 32=43

(0.994 )+ 32=2.825

M 23=θ2 EI=0.994

M 32=12

(θ2 ) EI=12

(0.994 )=0.497

M 24=θ2EI+θ4EI

2−316

=0.994+ 0.7032

−316

=−3.821

M 42=θ4EI +θ2EI

2+ 316

=0.703+ 0.9942

+ 316

=6.367

M 45=θ4EI=0.703

M 54=12

(θ4 ) EI=12

(0.703 )=0.352

M 67=θ6 EI=−1.47

M 76=12

(θ6 ) EI=12

(−1.47 )=−0.735

M 46=4 θ4 EI

5+2θ6EI

5−1057150

= 45

(0.703 )+25

(−1.47 )−1057150

=−7.072

M 64=4θ6EI

5+2θ4 EI

5+ 913150

=45

(−1.47 )+ 25

(0.703 )+ 913150

=5.192

M 68=4θ6EI

3+2θ8EI

3−94=4 (−1.47 )

3+2 (0.735 )3

− 94=−3.72

M 86=4θ8EI

3+2θ6 EI

3=4 (0.735)

3−2(−1.47)

3=0

5-. Calculamos las cortantes.

Qij=V ij−1l

(M ij+M ji)

Q12=2( 32 )−13 (−0.837+2.825 )=2.337

Q21=−2( 32 )−14 (2.825−0.837 )=−3.663

Q23=0−14

(0.994+0.497 )=−0.373

Q32=0−14

(0.497+0.994 )=−0.373

Q24=2 ( 42 )+2.5−14 (−3.821+6.367 )=5.864

Q42=−2( 42 )−14 (6.367−3.821 )=−7.137

Q45=0−14

(0.703+0.352 )=−0.264

Q54=0−14

(0.703+0.352 )=−0.264

Q67=0−14

(−1.47−0 .735 )=0.551

Q76=0−14

(−1.47−0 .735 )=0.551

Q46=2(52 )+4 ( 35 )−15 (−7.072+5.192 )=7.776

Q64=−2(52 )−4( 25 )−15 (5.192−7.072 )=−6.224

Q68=2( 32 )−13 (−3.72+0 )=4.24

Q86=2( 32 )−13 (−3.72 )=−1.76

DIAGRAMAS

D.F.C

D.M.F.

Nº1: CROSS

d) Cálculo de las rigideces:

K12=I3

K46=I5

K21=I3

K64=I5

K23=I4

K67=I4

K24=I4

K76=I4

K42=I4

K68=I4

K45=I4

K86=0

e) Cálculo de factores de distribución :

∑ d1=1

d1=d3=d5=d7=1d8=0

Nudo 2:

K21+K23+K24=10 I12

Nudo 4:

K42+K45+K46=7 I10

Nudo 6:

K64+K67+K68=7 I10

d1=1

d21=

I310 I12

=0.4 d42=

I47 I10

=0.357 d64=

I57 I10

=0.286

d23=

I410 I12

=0.3 d45=

I47 I10

=0.357 d67=

I47 I10

=0.357

d24=

I410 I12

=0.3 d46=

I57 I10

=0.286 d68=

I47 I10

=0.357

f) Momentos de empotramiento perfecto

M 12=−2x 32

12=−1.5

M 21=1.5

M 24=2x 42

12+ 5 x 48

=−5.17

M 42=5.17

M 46=−2 x52

12−4 x 2x 3

2

52=−7.05

M 64=2x 52

12+ 4 x3 x2

2

52=6.09

M 68=−2x 32

8=−2.25

M 86=0


Qij=v−1l(M ij+M ji)

M 12=−0.84 Q12=2( 32 )−13 (−0.837+2.825 )=2.337

M 21=2.83 Q21=−2( 32 )−14 (2.825−0.837 )=−3.663

M 23=0.99 Q23=0−14

(0.994+0.497 )=−0.373

M 32=0.496 Q32=0−14

(0.497+0.994 )=−0.373

M 24=−3.83 Q24=2 ( 42 )+2.5−14 (−3.821+6.367 )=5.864

M 42=6.37 Q42=−2( 42 )−14 (6.367−3.821 )=−7.137

M 45=0.70 Q45=0−14

(0.703+0.352 )=−0.264

M 54=0.35 Q54=0−14

(0.703+0.352 )=−0.264

M 46=−7.08 Q46=2(52 )+4 ( 35 )−15 (−7.072+5.192 )=7.776

M 64=5.19 Q64=−2(52 )−4( 25 )−15 (5.192−7.072 )=−6.224

M 67=−1.48 Q67=0−14

(−1.47−0 .735 )=0.551

M 76=0.74 Q76=0−14

(−1.47−0 .735 )=0.551

M 68=−3.73 Q68=2( 32 )−13 (−3.72+0 )=4.24

M 86=0 Q86=2( 32 )−13 (−3.72 )=−1.76

DFC

DMF

Nº2:

Grados de libertad: 8

θ2 , θ3 ,θ4 ,θ5 ,θ7 , θ8 ,△3=△ 4=△7=△m

△2=△5=△8=△n

θ1=θ6 ,=θ9=0

M 12=2EK (2θ1+θ2−3 R )±M 12

→

M 21=2EK (2θ2+θ1−3 R )±M 21

→

Miembro 12:

M 21

→

=M 21

→

=0

K= I4=0.25 I R=

△n

4=0.25△n

M 12=2x 0.25 EI (θ2−0.75△ n)

M 21=2x 0.25 EI (2θ2−0.75△n )

Miembro 23:

M 23

→

=M 32

→

=0

K= I4=0.25 I R=

△m

4=0.25△m

M 12=2x 0.25 EI (2θ2+θ3−0.75△m )

M 21=2x 0.25 EI (2θ3+θ2−0.75△m )

Miembro 25:

M 25

→

=M 52

→

=0

K= I4=0.25 I R=0

M 25=2 x0.25 EI (2θ2+θ5 )

M 52=2x 0.25 EI (2θ5+θ2 )

Miembro 34:

−M 34

→

=M 43

→

=2 x42

12=2.67

K= I4=0.25 I R=0

M 34=2 x0.25 EI (2θ3+θ4 )−2.67

M 43=2 x0.25 EI (2θ4+θ3 )+2.67

Miembro 65:

M 65

→

=M 56

→

=0

K= I4=0.25 I R=

△n

4=0.25△n

M 65=2 x0.25 EI (θ5−0.75△n )

M 56=2 x0.25 EI (2θ5−0.75△n )

Miembro 54:

M 54

→

=M 45

→

=0

K= I4=0.25 I R=

△m

4=0.25△m

M 54=2 x0.25 EI (2θ5+θ4−0.75△m )

M 45=2 x0.25 EI (2θ4+θ5−0.75△m )

Miembro 58:

M 34

→

=M 43

→

=0

K= I4=0.25 I R=0

M 58=2 x0.25 EI (2θ5+θ8 )

M 85=2 x0.25 EI (2θ8+θ5 )

Miembro 47:

−M 47

→

=M 74

→

=2 x52

12=4.17

K= I5=0.2 I R=0

M 47=2 x 0.2EI (2θ4+θ7 )−4.17

M 74=2 x0.2 EI (2θ7+θ4 )+4.17

Miembro 98:

M 98

→

=M 89

→

=0

K= I4=0.25 I R=

△n

4=0.25△n

M 65=2 x0.25 EI (θ8−0.75△n )

M 56=2 x0.25 EI (2θ8−0.75△ n )

Miembro 87:

M 87

→

=M 78

→

=0

K= I4=0.25 I R=

△m

4=0.25△m

M 87=2 x0.25 EI (2θ8+θ7−0.75△m )

M 78=2 x0.25 EI (2θ7+θ8−0.75△m )

Ecuaciones de equilibrio en los nudos:

M 21+M 25+M 23=0

3θ2+0.5θ3+0.5θ5−0.375△ n−0.375△m=0

M 32+M 34=0

0.5θ2+2θ3+0.5θ4−0.375△m−2.67=0

M 43+M 47+M 45=0

0.5θ3+2.8θ4+0.5θ5+0.4θ7−0.375△m−1.5=0

M 74+M 78=0

0.4θ4+1.8θ7+0.5θ8−0.375△m+4.17=0

M 87+M 89+M 85=0

0.5θ5+0.5θ7+3θ8−0.375△n−0.375△m=0

M 52+M 54+M 58+M 56=0

0.5θ2+0.5θ4+4θ5+0.5θ8−0.375△n−0.375△m=0

Ecuaciones de equilibrio de fuerzas horizontales:

M32

4+M 45

4+M 78

4−4=0

0.5θ2+θ3+θ4+0.5θ5+θ7+0.5θ8−1.125△m−16=0

M21

4+M 56

4+M 89

4−5=0

θ2+θ5+θ8−1.125△n−20=0

Resolviendo las ecuaciones:

3θ2+0.5θ3+0.5θ5−0.375△ n−0.375△m=0

0.5θ2+2θ3+0.5θ4−0.375△m−2.67=0

0.5θ3+2.8θ4+0.5θ5+0.4θ7−0.375△m−1.5=0

0.4θ4+1.8θ7+0.5θ8−0.375△m+4.17=0

0.5θ5+0.5θ7+3θ8−0.375△n−0.375△m=0

0.5θ2+0.5θ4+4θ5+0.5θ8−0.375△n−0.375△m=0

0.5θ2+θ3+θ4+0.5θ5+θ7+0.5θ8−1.125△m−16=0

θ2+θ5+θ8−1.125△n−20=0

θ2=−7 θ8=−6.28

θ3=−2.46 △n=−33.35

θ4=−1.53 △m=−31.62

θ5=−4.24

θ7=−6.82


M 12=2x 0.25 EI (2θ2+θ3−0.75△m ) Qij=v−1l(M ij+M ji)

M 12=8.97 Q12=0−14

(8.97+5.47 )=−3.61

M 21=5.47 Q21=−3.61

M 23=3.63 Q23=0−14

(3.63+5.90 )=−2.38

M 32=5.90 Q32=−2.38

M 34=−5.90 Q34=4−14

(−5.90 )=5.48

M 43=0 Q43=−4−14

(−5.90 )=−2.53

M 47=−8.12 Q47=5−15

(−8.12−1.90 )=7

M 74=−1.90 Q74=−5−15

(−8.12−1.90 )=−3

M 78=1.90 Q78=0−14

(1.90+2.17 )=−1.02

M 87=2.17 Q87=0−14

(1.90+2.17 )=6.18

M 25=−9.12 Q25=0−14

(−9.12−7.74 )=4.22

M 52=−7.74 Q52=4.22

M 45=8.21 Q45=0−14

(8.21+6.85 )=−3.777

M 54=6.85 Q54=−3.777

M 56=8.27 Q56=0−14

(8.27+10.39 )=−4.67

M 65=10.39 Q65=−4.67

M 58=−7.38 Q58=0−15

(−7.38−8.4 )=3.16

M 85=−8.4 Q85=3.16

M 89=6.23 Q89=0−14

(6.23+9.37 )=−3.9

M 98=9.37 Q98=−3.9

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Método de Cross y deflexion - ejercicios