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"AÑO DE LA LUCHA CONTRA LA CORRUPCION Y LA IMPUNIDAD"

METODO DE LA RIGIDEZ DIRECTO

CURSO :

ANALISIS ESTRUCTURAL II

DOCENTE : ALAVA

ING. MG. HORACIO SORIANO

INTEGRANTES:

-INOCENTE PEREZ, SAMAEL -REATEGUI TINOCO, LAURA -REYNA ESPINOZA, ELMER -FLORES RAMIREZ, NIELS -FASANANDO CUEVA, VIVIANA -POQUIOMA MAMANI, CARLOS

2019

1. CALCULAR DFC Y DMF

2Tn/m 2EI

C 2m

B

EI A

6Tn/m

3m

EI

D

4m

I. GRADOS DE LIBERTAD

3°

1° 2° B

C

A 4° D

5m

4Tn/m

GDL 1 1° B

C

A

D

1.5 EI=6EI L^2

B

12EI EI L^3=1.5

6EI =0.24 EI L^2

12EI L^3=0.096EI

C

1.5EI= 6EI L^2

K11 K21 K13 K14

=(1.5+0.096)EI= 1.596EI = 1.5EI =0.24EI =0.24EI

A -12EI

EI L^3=1.5

D 6EI L^2=0.24 EI

12EI L^3=-0.096EI

GDL 2 2 EI= 4EI L

2° B

B

C

C 2 EI= 4EI L

3/2 EI=6EI L^2

B

K12 K22 K32 K42

A

D 1 EI= 2EI L

A -1.5 EI= -6EI L^2

= 1.5EI = (2+2)EI=4EI =1EI =0

2EI=EI L

GDL 3 3° B

1EI= 2EI L

C

2EI= 4EI L

B

C

4EI =0.8 EI L

C

0.4EI= 2EI L

D

6EI = 0.24EI L^2

A

D K13 K23 K33 K34

=0.24EI =0 =(2+0.8)EI=2.8EI =0.4EI

GDL 4 0.4EI= 2EI L

B

C

C

-6EI= -0.24EI L^2

K14 K24 K34 K44

A 0.8EI= 4EI L

4°

2. EMSAMBLAMOS LA MATRIZ DE RIGIDEZ

1

K

=

1.596 1.5 0.24 0.24

2

3

4

1.5 4 1 0

0.24 1 2.8 0.4

0.24 0 0.4 0.8

1 2 3 4

D 6EI = 0.24EI L^2

= 0.24EI =0 =0.4EI =0.8EI

-6EI = -0.24EI L^2

3. FUERZAS INTERNAS BARRA AB : Momento de empotramiento perfecto -0.533 Tn.m

B

WL^2 =(4)(2^2) =0.8 Tn.m 20 20

1.20 Tn

2m

WL^2 =(4)(2^2) =0.533 Tn.m 30 30

4Tn/m A

2.8 Tn

0.8 Tn.m

Cortantes B 2/3

+ MA=0

FY=0

1/3

0.8+0.533 4(2) (0.667) +2RBA=0 2 RBA=1.20 Tn

-4(2) +1.2+RAB=0 2 RAB=2.80 Tn

A

FAB=

-1.20 Tn -0.533Tn.m 0 Tn 0 Tn.m

1 2 0 0

BARRA BC :

2Tn/m

8/3 Tn.m

4 Tn

-8/3 Tn.m

FBC=

C

B 4m

Momento de empotramiento perfecto WL^2 =(2)(4^2) =2.667 Tn.m 12 12

4 Tn

0 2 3 0

Cortantes FY=0

+ MB=0 -8/3+8/3 -4(2) +4RCB=0 RCB=4 Tn

-WL^2 =(-2)(4^2) =-2.667 Tn.m 12 12

0 Tn 8/3Tn.m -8/3 Tn 0 Tn.m

-4(2) +4RBC=0 RBC=4 Tn

BARRA CD : 4.5 Tn

C

5 Tn.m

Momento de empotramiento perfecto

EI

WL^2 =(6)(5^2) =-7.5 Tn.m 20 20

5m

6Tn/m

WL^2 =(6)(5^2) =5 Tn.m 30 30 D 10.5 Tn -7.5 Tn.m

1/3

C Cortantes + MC=0

FY=0

5-7.5- 6(5) (10/3) +2RDC=0 2 RDC=10.5 Tn

-6(5) +10.5+RCD=0 2 RCD=4.50 Tn

2/3

6Tn/m

D 4.5 Tn 0Tn.m 5 Tn -7.5 Tn.m

FCD=

1 0 3 4

FUERZA EN LOS NUDOS

1 2 3 4

0 0 0 0

fn=

4. MATRIZ FUERZA

F = fn

F=

0 0 0 0

-

-

fb

3.3 32/15 7/3 -7.5

=

3.3 -32/15 -7/3 7.5

1 2 3 4

5. DESPLAZAMIENTOS

-1

=

u

K

F

-1

u

1.596 1.5 0.24 0.24

=

1.5 4 1 0

0.24 1 2.8 0.4

0.24 0 0.4 0.8

-3.3 -32/15 -7/3 7.5

*

-5.7314878

= 2.3618869 -2.483624 12.58626

m rad rad rad

6.CALCULO DE MOMENTOS FINALES

Mf =

0.8 -0.533 2.666 -2.666 5 -7.5

+

1.5 1.5 0 0 0.24 0.24

1 2 2 1 0 0

Me +

Ke

0 0 1 2 0.8 0.4

0 0 0 0 0.4 0.8

u

*

-5.7314878 2.3618869 -2.483624 12.58626

=

-5.44 -4.41 4.41 -6.27 6.27 0

Tn.m Tn.m Tn.m Tn.m Tn.m Tn.m

7. GRAFICOS

DFC 3.53Tn 6.25 Tn 6.25Tn

4.47 Tn

2.25 Tn 8.75Tn

DMF -4.41 Tn.m 6.27Tn.m

-4.41 Tn.m

B C

5.44Tn.m

-6.27 Tn.m