Alanoca Quenta a. Freddy[1]

Anlisis estructural Avanzado

Facultad De Ingeniera Escuela Profesional De Ingeniera Civil

UNIVERSIDAD PRIVADA ANTENOR ORREGO

ANALISIS ESTRUCTURAL AVANZADO




A Temtica:

I. introduccin

II. comparacin de mtodos de solucin matricial

III. mtodo de rigidez:

1. introduccin

2. mtodo de la deflexin de la pendiente teora y aplicaciones.

3. Mtodo de rigidez por deflexin de teora y aplicaciones

4. Mtodo de rigidez directo con matrices [A] teora y problemas

5. Mtodo de rigidez directo con cosenos directos teora y problema

6. Mtodo de la condensacin esttica

7. Mtodo de rigidez para vigas-brazo rgido teora de aplicaciones

8. Mtodo de rigidez para prtico-placa

9. Mtodo de rigidez 3-D teora y aplicaciones




X Viga

Ray Rby

Y

F2

F

Rax

1. VIGA 1:

Ecuaciones (EQ)

M =0 FX =0

F =0 3EQ = FY =0

MZ =0

EN 3-D

FX =0

F =0 FY =0

FZ =0

MX =0

M =0 MY =0

MZ =0

HIPERESTATICIDAD DE LA ESTRUCTURA EXTERNAMENTE (GHE)

< 0 inestable (hiposttico)

GHE = NR NEQ = 0 isosttica

> 0 hiperesttica

Apoyo fijo apoyo mvil

3D




M M M

Ma

X

Y

Ray Rby

Rax

Rcy

Ry

Rx

Ry Ry

Rx Rx

NR =nmero de reacciones

NEQ = nmero de ecuaciones

De la VIGA 1 el GHE: GHE = 3 3 = 0 ______ isosttica.

2. VIGA CONTINUA

NR = 5 NEQ = 3

GHE = 5 3 = 2 hiperesttica de 2do grado externamente.

3. PORTICO

NR = 9 NEQ = 3

GHE = 9 3 = 6 hiperesttica de 6to grado

- Grado de hiperestaticidad total ( GHT )

- Grado de hiperestaticidad interna ( GHI )




- grado de hiperestaticidad interna ( GHI )

- nmero de barras ( NB )

- numero de reacciones ( NR )

- numero de nudos ( NN )

GHT = GHI + GHE

GHE = NR NEQ

GHT = 3 NB + NR 3 NN

GHI = GHT GHE

DE LA VIGA 2

GHE = 2do grado

GHT = 3 (2) + 5 3 (3) = 2do grado

GHI = GHT GHE

GHI = 2 2 = 0

DEL PORTICO 3

GHE = 9 3 = 6to

GHT = 3 (10) + 9 3 (9) = 12

GHI = GHT GHE

GHI = 12 6 = 6do grado




4. ARMADURA (estructura especial, total son 6 fuerzas.)

GHT = GHE + GHI

GHE = 0

GHT = NB + NR 2 NN

GHT = 20 + 3 2(10) = 3

5. ARMADURA 2

GHE = 3er

GHT = 3(12) + 6 3(10) = 12

GHI= 9no

3 D

X3

X3

X2

X2

X1

X1

Rotula

Rotula




Z

X

Y

1. 3-D

NEQ = 6 (3 D)

Fx = 0 Fy = 0 Fz = 0 Mx = 0 My = 0 Mz = 0

NR = 24

GHE = NR NEQ = 24 6 = 18vo

GHT = 6NB + NR 6n (3 D)

GHT = 6(8) + 24 6(8) = 24

GHI = GHT GHE = 24 18 = 6to

2.

GHE = 5 6 = -1 hipostatico (inestable)

GHT = 6(8) + 5 6(8) = 5to

GHI = 5 (-1) = 6to




3. ARMADURA 3 - D

GHE = 9 6 = 3er grado

GHT = GHE + GHI

GHT = NB + NR 3m ARM 3 D

GHT = 20 + 9 3(8) = 5

GHI = 5 3 = 2do grado

HIPERESTATICIDAD CINEMATICA ( # G.D.L.)

3 DESPLAZAMIENTOS

a y b rotacin

b traslacin

3 G.D.L (CINEMATICA)

A

Y

X




HAY 6 G.D.L

SI EA =

METODO DE LA FLEXION DE LA PENDIENTE

Ecuaciones de la deflexin de la pendiente:

= + + + = + + +




Desplazamientos de:

Rotacin:

Traslacin:

EJEMPLO 1:

Resolver:

Solucin:

Paso 1:

Paso 2:

M0ab = - M0ba = (P x L)/ 8 = (4 x 6) / 8 = 3 T-m

M0bc = - M0cb = (W x L2)/ 12 = (3 x 52) / 12 = 6.25 T-m

3 T-m -3 T-m 6.25 T-m




Paso 3:

= 0 (I) = 0 (II)

+ = 0 = 0 Paso 4:

Mba = M0ba + 2EI / 6 2b + 0 + 0 = -3 + (4EI / 6) b Mbc = M0bc + 2EI / 5 2b + c + 0 = 6.25 + (4EI / 5) b + (2EI / 5) c Mcb = M0cb + 2EI / 5 2c + b + 0 = -6.25 + (4EI / 5) c + (2EI / 5) b

(a) Y (b) en I -3 + (4EI / 6) a + 6.25 + (4EI / 5) b + (2EI / 5) c = 0 1.47EI b + 0.4EI c = -3.25 (I) (c) En II

-6.25 + (4EI / 5) c + (2EI / 5) b = 0 0.4EI b + 0.8EI c = 6.25 (II)

1.47 0.4 b -3.25 /EI

0.4 0.8 c 6.25/EI

b = -5.02/EI c = 10.33/EI




Mba = -3 + (4EI / 6) (-5.02/EI) = -6.35 T-m Mbc = 6.25 + (4EI / 5) (-5.02/EI) + (2EI / 5) (10.33/EI) = 6.35 T-m Mab = 3 + (2EI / 6) (-5.02/EI) = 1.33 T-m

Diagrama de momento flector:

EJEMPLO 2:




* Cuando es empotramiento no se considera giro y el momento es cero

Mba = M0ba + (2EI / 3) 2b + 0 + 3/Lba = 0 + (4EI / 3) b + (2EI / 3) Mbc = M0bc + ( 2EI / 5) 2b + c + 0 = 4.17 + ( 4EI / 5 ) b + ( 2EI / 5 ) c

Mcb = M0cb + (2EI / 5) 2c + b + 0 = -4.17 + (4EI / 5) c + (2EI / 5) b

Mcd = M0cd + (2EI / 3) 2c + 0 + 3/Lcd = 0 + ( 4EI / 3 ) c + ( 2EI / 3 )

(a) Y (b) en I

0 + (4EI / 3) b + (2EI / 3) + 4.17 + (4EI / 5) b + (2EI / 5) c = 0




2.13 EI b + 0.4 EI c + 0.67 EI = -4.17 (I) (c) Y (d) en II

-4.17 + (4EI / 5) c + (2EI / 5) b + 0 + (4EI / 3) c + (2EI / 3) = 0

0.4EI b + 2.13 EI c + 0.67 EI = 4.17 (II)

Mab = M0ab + (2EI / 3) 0+ b + 3/Lab = 0 + (2EI / 3) b + (2EI / 3)

Mdc = M0dc + (2EI / 3) 0 + c + 3/Lab = 0 + ( 2EI / 3 ) c + ( 2EI / 3 ) Mab + Mba + Mdc + Mcd = 15

(e), (a), (f) Y (d) en III 0 + (2EI / 3) b + (2EI / 3) + 0 + (4EI / 3) b + (2EI / 3) + 0 + ( 2EI / 3 ) c + ( 2EI / 3 ) + 0 + ( 4EI / 3 ) c + ( 2EI / 3 ) = 0




2 EI b + 2 EI c + 2.67 EI = 15

2.13 0.4 0.67 b -4.17/EI

0.4 2.13 0.67 c 4.17/EI 2 2 2.67 15 /EI

b = -4.88/EI c = -0.061/EI = 9.31/EI

Mba = (4EI / 3) (-4.88/EI) + (2EI / 3) (9.31/EI) = -0.3 T-m Mbc = 4.17 + ( 4EI / 5 ) (-4.88/EI ) + ( 2EI / 5 ) (-0.061/EI ) = 0.24 T-m Mcb = -4.17 + (4EI / 5) (-0.061/EI) + (2EI / 5) (-4.88/EI) = -3.19 T-m Mcd = ( 4EI / 3 ) (-0.061/EI ) + ( 2EI / 3 ) ( 9.31/EI ) = 6.14 T-m

Mab = (2EI / 3) (-4.88/EI) + (2EI / 3) (9.31/EI) = 2.95 T-m

Mdc = ( 2EI / 3 ) (-0.061/EI ) + ( 2EI / 3 ) ( 9.31/EI ) = 6.14 T-m




METODO MODIFICADO DE LA FLEXION DE LA PENDIENTE

= ( ) + +

Ejercicio 1:

-Se condensa solo en los extremos, cuando esta empotrado no se condensa.




4.5T-m

1.8T-m 2.7T-m

6.3T-m

6.3T-m

7.2T-m




Mba = M0ba + (2EI / 4) 2b + 0 + 3/Lba = -2 + (4EI / 4) b + (6EI / 16) Mbc = M0bc + ( 2EI / 6) 2b + c + 0 = 1.8 + ( 4EI / 6 ) b + ( 2EI / 6 ) c


Mcd = M0 cd - (M0dc/2) + (3EI / Ldc) c + 0 = 6.3 - (-7.2/ 2) + (3EI/ 6) c

(a) Y (b) en I

-2 + (4EI / 4) b + (6EI / 16) + 1.8 + (4EI / 6) b + (2EI / 6) c = 0




1.67 EI b + 0.33 EI c + 0.38 EI = 0.2 (I)

(c) Y (d) en II

-2.7 + (4EI / 6) c + (2EI / 6) b + 6.3 - (-7.2/ 2) + (3EI/ 6) c = 0

0.33EI b + 1.17 EI c + 0 EI = -7.2 (II)

(III)

Remplazando en (III):

Mab = M0ab + (2EI / 4) 0 + b + 3/Lab = 2 + (2EI / 4) b + (6EI / 16)

(e) Y (a) en III

-2 + (4EI / 4) b + (6EI / 16) + 2 + (2EI / 4) b + (6EI / 16) = 8




1.5EI b + 0 EI c + 0.75 EI = 8 (III)

1.67 0.33 0.38 b 0.2/ EI

0.33 1.17 0 c = -7.2 /EI

1.5 0 0.75 8 /EI

b = -2.14/EI c = -5.55/EI = 14.75/EI

Mba = -2 + (4EI / 4) (-2.14/EI) + (6EI / 16) (14.75/EI) = 1.39 T-m Mbc = 1.8 + (4EI / 6) (-2.14/EI) + (2EI / 6) (-5.55/EI) = -1.48 T-m

Mcb = -2.7 + (4EI / 6) (-5.55/EI) + (2EI / 6) (-2.14/EI) = -7.12 T-m

Mcd = 6.3 - (-7.2/ 2) + (3EI/ 6) (-5.55/EI) = 7.12 T-m

Mab = 2 + (2EI / 4) (-2.14/EI) + (6EI / 16) (14.75/EI) = 6.46 T-m Diagrama de momento flector:

C




Ejercicio 2:

Solucin:

4.44 T-m 2.22 T-m 3.75 T-m 2.5 T-m




Mba = M0ba - (M0ab/2) + (3EI / Lab) b + 0 = -2.22 - (4.44/ 2) + (3EI/ 6) b

Mbc = M0bc - (M0cb/2) + (3EI / Lbc) b + 0 = 2.5 - (-3.75/ 2) + (3EI/ 5) b

(a) Y (b) en I

-2.22 - (4.44/ 2) + (3EI/ 6) b + 2.5 - (-3.75/ 2) + (3EI/ 5) b = 0

1.1 EI b = 0.065 (I)

b = 0.059/EI

Remplazando b en (a) y (b):

Mba = -2.22 - (4.44/ 2) + (3EI/ 6) (0.059/EI) = - 4.41 T-m

Mbc = 2.5 - (-3.75/ 2) + (3EI/ 5) (0.059/EI) = 4.41 T-m Diagrama de momento flector:




Ejercicio 3:

Paso 1:

Condensar giro a




paso2:

Paso3:




Mba = M0ba -(M0ab/2)+ (3EI /Lba) b + /Lba = 0+0+ (3EI/ 3.5)b+(3EI/12.25)

Mbc = M0bc + (2EI / 5) 2b + c + 0 = 4.17 + (4EI / 5) b + (2EI / 5) c


Mce = M0ce + (2EI /3.5) 2c + 0 + 3/Lce =1.47+ (4EI /3.5) c + (6EI /12.25)

Mcd = M0cd + (2EI /5) 2c + d + 0 = 0 + (4EI / 5) c + (2EI / 5) d

Mdc = M0dc + (2EI /5) 2d + c + 0 = 0 + (4EI / 5) d + (2EI / 5) c

Mdf = M0df + (2EI /3.5) 2d + 0 +3/Ldf = 0 + (4EI /3.5) d + (6EI/12.25)

Remplazando:

(a) Y (b) en I

0 + 0 + (3EI/ 3.5)b + (3EI/12.25 ) + 4.17 + ( 4EI / 5 ) b + ( 2EI / 5 ) c = 0

1.66 EI b + 0.4 EI c + 0 EI d + 0.24 EI = -4.17 (I) (c), (d) y (e) en II

-4.17 + (4EI / 5) c + (2EI / 5) b +1.47+ (4EI /3.5) c + (6EI /12.25) +




0 + (4EI / 5) c + (2EI / 5) d = 0

0.4 EI b + 2.74 EI c + 0.4 EI d + 0.49 EI = 2.7 (II) (f) Y (g) en III

0 + (4EI / 5) d + (2EI / 5) c + 0 + (4EI /3.5) d + (6EI/12.25) = 0

0 EI b + 0.4 EI c + 1.94 EI d + 0.49 EI = 4 (III)

Para hallar la otra ecuacin:

+ + + 3 - 3 3.5 = 0

+ + = 3.5 IV

Ha x 3.5 = 0 He x 3.5 + Mec + Mce 3 x 1.5 = 0




Ha = 0 He = 4.5 - Mec - Mce

He x 3.5 + Mfd + Mdf 3.5 x 2.3 = 4

Hf = 12.05 - Mfd Mdf

Remplazando Ha, He y Hf en IV:

4.5 - Mec - Mce + 12.05 - Mfd - Mdf =12.25

Mec + Mce + Mfd + Mdf = 4.3 IV

Mec = M

0ec + (2EI /3.5) 0 + c + 3/Lec = -1.10+ (2EI /3.5) c + (6EI /12.25)

Mfd = M0fd + (2EI /3.5) 0 + d +3/Lfd = -1.23 + (2EI /3.5) d + (6EI/12.25)

(d), (g), (h) y (i) en IV

1.47+ (4EI /3.5) c + (6EI /12.25) + 0 + (4EI /3.5) d + (6EI/12.25) +

-1.10+ (2EI /3.5) c + (6EI /12.25) + -1.23 + (2EI /3.5) d +(6EI/12.25) = 4.3

0 EI b + 1.71 EI c + 1.71 EI d + 1.96 EI = 5.16 (IV) 1.66 0.4 0 0.24 b -4.17/ EI

0.4 2.74 0.4 0.49 c 2.7/ EI

0 0.4 1.94 0.49 d 4/ EI

0 1.71 1.71 1.96 5.16/ EI




b = -2.79/EI c = 1.11/EI d = 1.81/EI = 0.08/EI

Remplazando b, c, d y :

Mba = (3EI/ 3.5) (-2.79/EI) + (3EI/12.25) (0.08/EI) = - 2.37 T-m

Mbc = 4.17 + (4EI / 5) (-2.79/EI) + (2EI / 5) (1.11/EI) = 2.38 T-m

Mcb = -4.17 + (4EI / 5) (1.11/EI) + (2EI / 5) (-2.79/EI) = - 4.39 T-m

Mce = 1.47+ (4EI /3.5) (1.11/EI) + (6EI /12.25) (0.08/EI) = 2.78 T-m

Mcd = 0 + (4EI / 5) (1.11/EI) + (2EI / 5) (1.81/EI) = 1.61 T-m

Mdc = 0 + (4EI / 5) (1.81/EI) + (2EI / 5) (1.11/EI) = 1.89 T-m

Mdf = 0 + (4EI /3.5) (1.81/EI) + (6EI/12.25) (0.08/EI) = 2.11 T-m

Mec = -1.10+ (2EI /3.5) (1.11/EI) + (6EI /12.25) (0.08/EI) = -0.43 T-m

Mfd = -1.23 + (2EI /3.5) (1.81/EI) + (6EI/12.25) (0.08/EI ) = -0.16 T-m

Mab = 0




Diagrama de momento flector

= 0

= 0

= 0

= 0

= 0




EA = axial = 0

Ejercicio 4:

Paso 1:

paso2:

Momentos del tramo ab:

= 0.44 T-m = -0.66 T-m

Condensar giro d




Momentos del tramo bc

Momentos del tramo cd

1.11T-m 1.11T-m

0.45T-m 0.66T-m

M0bc = 1.11 T-m + 0.45T-m = 1.56 T-m

1.56T-m 1.77T-m

-2.23T-m 2.23T-m

0.44T-m -0.66T-m

2.67T-m 2.89T-m

0.84T/m

M0cb = -1.11 T-m - 0.66T-m = -1.77 T-m

M0cd = 2.23 T-m + 0.44T-m = 2.67 T-m

M0dc = -2.23 T-m - 0.66T-m = -2.89 T-m




Paso3:

Mba = M0ba + (2EI / 4) 2b + 0 + 0 = -0.66 + (4EI / 4) b

Mbc = M0bc + (2EI / 4) 2b + c + 0 = 1.56 + (4EI / 4) b + (2EI / 4) c

Mcb = M0cb + (2EI /4) 2c + c + 0 = -1.77+ (4EI /4) c + (2EI /4) b

Mcd = M0cd - (M0dc/2) + (3EI /Ldc) c + 0 = 2.67 - (-2.89/2) + (3EI/ 4) c

Remplazando:

(a) Y (b) en I

-0.66 + (4EI / 4) b + 1.56 + (4EI / 4) b + (2EI / 4) c = 0

2EI b + 0.5EI c = -0.90 (I)

(c) Y (d) en II

-1.77+ (4EI /4) c + (2EI /4) b + 2.67 - (-2.89/2) + (3EI/ 4) c = 0

0.5EI b + 1.75EI c = -2.35 (II)




2 0.50 b -0.90/ EI

0.5 1.75 c -2.35/ EI

b = -0.12/EI c = -1.31/EI

Remplazando b y c:

Mba = -0.66 + (4EI / 4) (-0.12/EI) = - 0.78 T-m

Mbc = 1.56 + (4EI / 4) (-0.12/EI) + (2EI / 4) (-1.31/EI) = 0.78 T-m

Mcb = -1.77+ (4EI /4) (-1.31/EI) + (2EI /4) (-0.12/EI) = - 3.14 T-m

Mcd = 2.67 - (-2.89/2) + (3EI/ 4) (-1.31/EI) = 3.14 T-m

Mdc = 0





M0db =-2.4 T-m M0bd =1.6 T-m

Ejercicio 5:

Solucin:

3m

2.5m

4m

3T/m

3T/m

4m 3m -1.5 T-m

M0ab =1.5 T-m M0ba =-1.5 T-m




Mba = M0ba -(M

0ab / 2)+ (3EI /Lab) b + 0 = -1.5 - (1.5 / 2) + (3EI/ 3) b

Mbc = M0bc + (2EI / 2.5) 2b+c+0 = 0 + (4EI / 2.5) b + (2EI / 2.5) c

Mbd = M0bd - (M0db/2) + (3EI /Ldb) b + 0 = 1.6 - (-2.4/2) + (3EI/ 4) b

Mcb = M0cb+ (2EI /2.5) 2c + b+0 = 0 + (4EI / 2.5) c + (2EI / 2.5) b

Remplazando:

(a), (b) y (c) en (I)

-1.5-(1.5/2) + (3EI/ 3) b + (4EI/ 2.5) b + (2EI/ 2.5) c + 1.6-(-2.4/ 2) + (3EI/ 4) b = 0

3.35 EI b + 0.8 EI c + = - 3.55 (I)




(d) En (II)

(4EI / 2.5) c + (2EI/2.5) b+ = 0

0.8 EI b + 1.6 EI c = 0 (II)

3.35 0.8 b -3.55/ EI

0.8 1.6 c 0

b = -1.20/EI c = -0.60/EI

Mba = -1.5 - (1.5 / 2) + (3EI/ 3) (-1.20/EI) = -3.45 T-m

Mbc = 0 + (4EI / 2.5) (-1.20/EI) + (2EI / 2.5) (-0.60/EI) = -2.4 T-m

Mbd = 1.6 - (-2.4/2) + (3EI/ 4) (-1.20/EI) = 1.9 T-m

Mcb = 0 + (4EI / 2.5) (-0.60/EI) + (2EI / 2.5) (-1.20/EI) = 1.92 T-m





MATRIZ DE RIGIDEZ POR DEFINICION

EJEMPLO:




D1 y D2 SON DE ROTACION Y D3 DE TRASLACION

VECTOR DE DESPLAZAMIENTO GLOBALES DE LA ESTRUCTURA

EJEMPLO:

{} = EJEMPLO:

{} =

LEY DE HOOKE GENERALIZADA: {} = { } K = MATRIZ DE RIGIDEZ GLOBAL DE LA ESTRUCTURA

EJEMPLO: {} = { } SI D1 = 1 , D2 = D3 = 0




K11 K12 K13K21 K22 K23K31 K32 K33 D1 = 1D2 = 0D3 = 0 = K11K21K31 = Q1Q2Q3

SI D2 = 1 , D1 = D3 = 0


FUERZAS EXTERNAS UNITARIAS




SI D3 = 1 , D1 = D2 = 0


CALCULO DE LOS COEFICIENTES DE RIGIDEZ:

EJEMPLO #1:

D1 = 1 , D2 = D3 = 0




K11 = Mbc + Mba

- Hallar Mbc

Mbc = M0bc + (2EIV / LV) 2D1 + D2 + 0

Mbc = 0 + (2EIV / LV) 2 (1) + (0) + 0

Mbc = 4EIV / LV

- Hallar Mba

Mba = M0ba + (2EIC / h) 2D1 + D2+ (3D3/h)

Mba = 0+ (2EIC / h) 2 (1) + 0 + (3x0/h)

Mba = 4EIC / h

Remplazando:




K11 = 4EIV / LV + 4EIC / h

K21 = Mcb + Mcd

- Hallar Mcb

Mcb = M0cb + (2EIV / LV) 2D2 + D1 + 0

Mcb = 0 + (2EIV / LV) 2 (0) + (1) + 0

Mcb = 2EIV / LV

- Hallar Mcd

Mcd = M0cd + (2EIC / h) 2D2 + D1+ (3D3/h)

Mcd = 0+ (2EIC / h) 2 (0) + 0 + (3x0/h)

Mcd = 0

Remplazando:

K21 = 2EIV / LV

Vba x h - Mab - Mba = 0

= + = + = Vba = 6EIC/h2

F(x) = 0




K31 Vba = 0

K31 = 6EIC/h2

D2 = 1 , D1 = D3 = 0

K12 = Mbc + Mba K22 = Mcb + Mcd K32 Vcd = 0

K12 = 2EIV / LV K22 = 4EIV / LV + 4EIC / h K32 = 6EIC/h2

D3 = 1 , D1 = D2 = 0




K13 = Mbc + Mba

- Hallar Mbc

Mbc = M0bc + (2EIV / LV) 2D1 + D2 + 0

Mbc = 0 + (2EIV / LV) 2 (0) + (0) + 0

Mbc = 0

- Hallar Mba

Mba = M0ba + (2EIC / h) 2D1 + D2+ (3D3/h)

Mba = 0+ (2EIC / h) 2 (0) + 0 + (3x1/h)

Mba = 6EIC / h2

Remplazando:

K13 = 6EIC / h2




K23 = Mcb + Mcd

- Hallar Mcb

Mcb = M0cb + (2EIV / LV) 2D2 + D1 + 0

Mcb = 0 + (2EIV / LV) 2 (0) + (0) + 0

Mcb = 0

- Hallar Mcd

Mcd = M0cd + (2EIC / h) 2D2 + D1+ (3D3/h)

Mcd = 0+ (2EIC / h) 2 (0) + 0 + (3x1/h)

Mcd = 6EIC / h2

Remplazando:

K23 = 6EIC / h2

Vba x h - Mab - Mba = 0

- Hallar Mab

Mab = M0ab + (2EIC / h) 2D2 + D1+ (3D3/h)

Mac = 0+ (2EIC / h) 2 (0) + 0 + (3x1/h)

Mab = 6EIC / h2

= + = + = Vba = 12EIC/h3 y Vcd = 12EIC/h3




F(x) = 0

K33 Vba Vcd = 0

K33 = 12EIC/h3 + 12EIC/h3 = 24EIC/h3

4

+ 4

2

622

4

+ 4

626

26

224

3

K21 = K12

K31 = K13

K32 = K23

EJEMPLO #2:

Hallar K de la estructura mostrada:




D1 = 1 , D2 = D3 = 0

K11 = 4EIV / LV + 4EIC / h K21 = 2EIV / LV K31 = 6EIC/h2

K11 = 4EIV / 5 + 4EIC / 3 K21 = 2EIV / 5 K31 = 6EIC/9

D2 = 1 , D1 = D3 = 0




Hallar:

= + + = + . + .

= .

= . =

= . = . F(x) = 0

K33 Vcd = 0

K33 = Vcd

K12 = 2EIV / LV K22 = 4EIV / LV + 3EIC / h K32 = 3EIC/h2

K12 = 2EIV / 5 K22 = 4EIV / 5 + 3EIC / 2.5 K32 = 3EIC/2.52




D3 = 1 , D1 = D2 = 0 Hallar: = + +

= + . + . = .

= . =

= . = .




K33 Vba Vcd = 0 K33 = 12EIC/h3 + 3EIC/h3 K13 = 6EIC / h2 K23 = 3EIC / h2 K33 = 12EIC/h3 + 3EIC/h3 K13 = 6EIC / 32 K23 = 3EIC / 2.52 K33 = 24EIC/33 + 3EIC/2.53 43 + 45 25 63225 45 + 32.5 32.52632 32.52 1233 + 32.53

EJEMPLO #3: RESOLVER POR EL METODO DE RIGIDEZ POR DEFINICION: E = 2 x 106 Ton/m2 I = 0.30 x (0.55)3/12

= 1. G.D.L = 2




2. D1 = 1 y D2 = 0 K11 - Mab =0 K11 = Mab K11 =4EI/5

K21 - Mba - Mbc =0 K21 = Mba + Mbc K21 =2EI/5 D2 = 1 y D1 = 0




K12 - Mab =0 K12 = Mab K12 =2EI/5

K22 - Mba - Mbc =0 K22 = Mba + Mbc K22 =4EI/5 + 4EI/6 Hallar EI:

IE = 0.30 x 0.55312 x(2x106) = 8318.75 Ton m2 K {D} = {Q} 4EI5 2EI52EI5 45 + 46

1

2

= 06 6655

3327.5 3327.5

12200.8 1

2

= 06 D1 = 2.85 x 10-4 D2 = -5.69 x 10-4




Por otro metodo, condensando:

D1 =1

K11 - Mba - Mbc =0 K22 = Mba + Mbc

Hallar el Mab

Mba = M0ba M0ab2 + 3EILab D1 + 0Lab Mba = 0 02 + 3EI5 1 + 05




= K11 =3EI/5 + 4EI/6

35 + 46 1 = 6 10537.051 = 6 1 = 5.69 104

EJEMPLO #4:

E = 2 x 106 T/m2

Solucin:




D1 = 1 , D2 = D3 = D4 = 0

K11 - Mab =0 K11 = Mab K21 - Mba =0 K21 = Mba

K11 =4EI/4 K21 =2EI/4




K41 - Mcb =0 K41 = Mcb K31 Mdb =0 K21 =Mdb

K41 =0 K31 =0

D2 = 1 , D1 = D3 = D4 = 0




K12 - Mab =0 K12 = Mab K22 - Mba -Mbc -Mbd =0

K12 =2EI/4 K22 =4EI/4+4EI/3 +4EI/3.5

K42 Mcb =0 K42 = Mcb K32 Mdb =0 K32 =Mdb

K42 =2EI/3 K32 =2EI/3.5

D3 = 1 , D1 = D2 = D4 = 0




K13 - Mab =0 K13 = Mab K23 - Mbd =0 K23=Mbd

K13 =0 K23 =2EI/3.5


K43 =0 K33 =4EI/3.5

D4 = 1 , D1 = D2 = D3 = 0




K14 - Mab =0 K14 = Mab K24 - Mbc =0 K24=Mbc

K14 =0 K24 =2EI/3


K44 =4EI/3 K34 =0

/2EI/400 2EI/4

+

. + 2EI/3.52EI/3 02EI/3.5

/. 0 02EI/30

/ D1D2D3D4 =

0800

IE = 0.25 x 0.40312 x(2x106) = 2666.67 Ton m2

. 1333.3400 1333.34. 1523.811777.78

01523.81. 0

01777.780.

D1D2D3D4 = 0800

D1 = -5.75 x 10-4 D2 = 1.15 x 10-3 D3 = -5.75 x 10-4 D4 = -5.75 x 10-4




METODO DE RIGIDEZ DIRECTA (Con matrices de transformacin A)

LEY DE HOOKE GENERALIZADA:

{} = K {D}.. (I)

Dnde:

{} mx1 = vector de cargas globales de la estructura

{D} mx1 = vector de desplazamiento globales de la estructura

{K} mxm = matriz de rigidez global de la estructura

Dnde: m = # G.D.L

DEFINIR:

{d} e = Ae {D}. (II)

{d} e = desplazamiento locales del elemento

Ae = matriz de compatibilidad o transformacin del elemento.

Ejemplo: EA =




Solucin:

nicamente por flexin

{de} e = vector desplazamiento del elemento en coordenadas locales.




{q}e = Ke {d}e -----------------------(III)

{q} e = vector de cargas del elemento

D1 = 1 D2 = 1




D3 = 1 D4 = 1

=

0000 0000

0001 0000




{d} e = Ae {D}. (II)

Ejemplo:

= 0000 0100

0001 0000

= 00

Si

{} = {}=

() PRINCIPIO DE TRABAJO VIRTUAL:

Wext = Wint

{} . {} = {} . {}=

{} . {} = {( {}) } . {}=

{} . {} = {( {}) } . {}= {} = {{}) . {} PRINCIPO DE LA CONTRAGRADIENCIA

=

PASOS:




1. Definir los grados de libertad G.D.L {D} m , m = # G.D.L. 2. Generar las matrices de compatibilidad o matrices de transformacin de C/elemento; Ae.

3. Generar la matrices de rigidez en coordenadas locales de C/elemento; Ke. 4. Proceso de ensamblaje, obtencin de la matriz de rigidez global de la estructura,

KG.

{} = {} {} 5. Generar el vector de cargas globales de la estructura {}.

{} = {} + {} {} = {} 6. Resolver {} = KG {D} --------------OBTENER {D}

7. Hallar {q}e = Ke Ae {D} - {q}eeq

8. Hallar {d}e = Ae {D} y D.M.F y D.F.C

{} = KTOTAL {D}

DONDE:

{} = =

, = #. . SI: solo por flexin.




d1 = 1 , d2 = d3 = d4 = 0

d2 = 1 , d1 = d3 = d4 = 0

d3 = 1 , d1 = d2 = d4 = 0

d4 = 1 , d1 = d2 = d3 = 0




=

12EI/36EI/212EI/36EI/2

6EI/24EI/L6EI/22EI/L

12EI/36EI/212EI/36EI/2

6EI/22EI/L6EI/24EI/L

Ejemplo#1:

Resolver: E = 2x 106 T/m2 , EA =

Solucin:

Paso 1:

G.D.L = 2

Paso 2: D1 = 1 , D2 = 0




=

0001 0000 42

Paso 2: D2 = 1 , D1 = 0

=

0100 0001 42

Paso 3:

{} = =

= + = . .

() =

= = 347.221041.67347.221041.67

1041.674166.671041.672083.37

347.221041.67347.221041.67

1041.672083.371041.674166.67




= 00 00 00 10 347.221041.67347.221041.67 1041.674166.67

1041.672083.37 347.22

1041.67347.221041.67

1041.672083.371041.674166.67

0001 0000

00 10 00 01 347.221041.67347.221041.67 1041.674166.67

1041.672083.37 347.22

1041.67347.221041.67

1041.672083.371041.674166.67

0100 0001

= 4166.670 00 4166.672083.32 2083.324166.67 8333.332083.32 2083.324166.67

Paso 4: {}

= = = + {} + {}

{q}1 = q1q2q3q4 = 9 Tn9 T m9 Tn

9 T m {q}1eq = 9 Tn

9 T m9 Tn+9 T m




{q}2 = q1q2q3q4 = 2.5 Tn3.75 T m2.5 Tn

3.75 T m {q}1eq = 2.5 Tn

3.75 T m2.5 Tn+3.75 T m

12 = 00 + 0 0 00 0 0 10 999+9 + 0 1 00 0 0 01 2.5

3.752.5+3.75

12 = 00 + +90 + 3.75+3.75 = . .

{} = KTOTAL {D} KTOTAL {D} = {}

8333.33 2083.332083.33 4166.67 D1D2 = . . D1D2 = . . {q} 1 = K1 A1 {D} - {q}1eq




q1 = Viq2 = Miq3 = Vjq4 = Mj

= 347.221041.67347.221041.67 1041.674166.67

1041.672083.33 347.22

1041.67347.221041.67

1041.672083.331041.674166.67

0001 0000 . .

999+9 =

. . .

. {q} 2 = K2 A2 {D} - {q}2eq


= 347.221041.67347.221041.67 1041.674166.67

1041.672083.33 347.22

1041.67347.221041.67

1041.672083.331041.674166.67

0100 0001 . .

2.53.752.5+3.75 =

. . .

Diagrama de momento:

Ejemplo#2:




Paso 1:

D1 = 1 , D2 = D3 = D4 =0 D2 = 1 , D1 = D3 =D4 = 0

Paso 2:

=

0001 0000

0000 0010 =

0000 0001

0000 0010

D3 = 1 , D1 = D2 = D4 =0 D4 = 1 , D1 = D2 =D3 = 0




=

0100 0001

0000 0000 =

0000 0100

0001 0000

Paso 3:

{} = =

= + + + = . .

() = .

= = = = 1185.21777.81185.21777.8 1777.83555.6

1777.81777.8 1185.21777.81185.21777.8

1777.81777.81777.83555.6

= 0000 0000

0001

1000 1185.21777.8

1185.21777.8 1777.83555.6

1777.81777.8 1185.21777.81185.21777.8

1777.81777.81777.83555.6

0001 0000

0000 00

10




0000 0000

0001

0100 1185.21777.8

1185.21777.8 1777.83555.6

1777.81777.8 1185.21777.81185.21777.8

1777.81777.81777.83555.6

0000 0001

0000 00

10 0000

1000 0000

0100 1185.21777.8

1185.21777.8 1777.83555.6

1777.81777.8 1185.21777.81185.21777.8

1777.81777.81777.83555.6

0100 0001

0000 0000

0000 0100

0000 0010

1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

0000 0100

0001 0000

= 7111.21777.801777.8 1777.810666.81777.81777.8

01777.83555.60 1777.81777.802370.4

Paso 4: {}

= = = = = + + + +




{q}1 = q1 = 2.5 Tnq2 = 1.87 T mq3 = 2.5Tnq4 = 1.87 T m , {q}1eq = q1 = 2.5 Tnq2 = 1.87 T mq3 = 2.5 Tnq4 = +1.87 T m

{q}3 = q1 = 3 Tnq2 = 1.5 T mq3 = 3Tnq4 = 1.5 T m , {q}3eq = q1 = 3 Tnq2 = 1.5 T mq3 = 3 Tnq4 = +1.5 T m

1234

= 0800 + 0000

0000 000

1 1000

2.51.872.5+1.87 +

0000 0000

0001

0100 0000 +

0000 1000

0000 0100

31.523+1.5 +

0000 0100

0000 0010

0000 = 123

4

= 0800 + 1.87002.5 +

0000 + 1.51.500 +

0000 = . .

.





. . .

. . . .

. .

. . .

D1D2D34 = 0.379.502.5

= . . .

{q} 1 = K1 A1 {D} - {q}1eq


= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

0001 0000

0000 00

10

. . .

2.51.872.5+1.87 =

. .

. .

{q} 2 = K2 A2 {D} - {q}2eq


= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

. . .

0000 = . . .

.

{q} 3 = K3 A3 {D} - {q}3eq

= = = =

= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

. . .

31.53+1.5 =

. . . .

{q} 4 = K4 A4 {D} - {q}4eq

= = = =

= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

. . .

0000 = . .

.





Ejemplo#3: el mismo que el #2 pero darle solucin con el metodo de la condensacin:




Paso 1:

D1 = 1 , D2 = D3 =0 D2 = 1 , D1 = D3 = 0




=

0001 0000

0010 =

0000 0001

0010

Paso 2:

D3 = 1 , D1 = D2 =0

= = + + +

= = + + + + = = = , =

=

0000 010

1/2 0000

Paso 3:

{} = =

= + + +




= . .

() = .

= = = = 1185.21777.81185.21777.8 1777.83555.6

1777.81777.8 1185.21777.81185.21777.8

1777.81777.81777.83555.6

= 000 000 001 100

1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

0001 0000

0010

000 000 001 010 1185.21777.8

1185.21777.8 1777.83555.6

1777.81777.8 1185.21777.81185.21777.8

1777.81777.81777.83555.6

0000 0001

0010 000 100 000 010

1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

0100 0001

0000 000 010 000 01/20

1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

0000 010

1/2 0000

= 7111.21777.81777.8 1777.88444.61777.8 1777.81777.82370.4 3x3 Paso 4: {}




= = = = = + + + +

{q}1 = q1 = 2.5 Tnq2 = 1.87 T mq3 = 2.5Tnq4 = 1.87 T m , {q}1eq = q1 = 2.5 Tnq2 = 1.87 T mq3 = 2.5 Tnq4 = +1.87 T m

{q}3 = q1 = 3 Tnq2 = 1.5 T mq3 = 3Tnq4 = 1.5 T m , {q}3eq = q1 = 3 Tnq2 = 1.5 T mq3 = 3 Tnq4 = +1.5 T m

123

= 080 + 000 000 001 100 2.51.872.5+1.87 + 000 000 001 010 0000 +




000 100 000 010 31.53+1.5 + 000 010 000 0

1/20 0000 =

123

= 080 + 1.8702.5 + 000 + 1.51.50 + 000 = . . .


7111.21777.81777.8 1777.88444.61777.8 1777.81777.82370.4 D1D2D3 = . . .

= . . .

{q} 1 = K1 A1 {D} - {q}1eq

= = = =

= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

. . .

2.51.872.51.87 =

. . .

. {q} 2 = K2 A2 {D} - {q}2eq

= = = =

= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

. . .

0000 = . .

. .

{q} 3 = K3 A3 {D} - {q}3eq

= = = =

= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

. . .

31.53+1.5 =

. . . .




{q} 4 = K4 A4 {D} - {q}4eq

= = = =

= 1185.21777.81185.21777.8

1777.83555.61777.81777.8

1185.21777.81185.21777.8

1777.81777.81777.83555.6

/

. . .

= . . .


METODO DE CONDENSACION ESTATICA

Sea por ejemplo:




K11K21K31 K12K22K32

K13K23K33

D1D2D3

=

GENERALIZANDO:

{}{} = {}{}

{} + {} = {} (1)

{} + {} = {F} .... (2)




{} + {} = {}

T {} = - T {} I {} = - -1 {} {} = - -1 {} ..... (3)

{} = T {} ..... (4)

DONDE:

T = - -1 . (5)

Remplazando (3) en (2) tenemos:

(- -1 {}) + {} = {F}

{F} = - -1 {} {F} = L {}

L = rigidez lateral.

{F} = L {} (6)

SIENDO:

L = - -1

Ejemplo #1: hallar la rigidez lateral de la estructura mostrada y graficar D.M.F.




Solucin:

D1 = 1 , D2 = D3 = 0

K11 = 4EIV / LV + 4EIC / h

K11 = 4EIV / 7 + 4EIC / 3.5

K21 = 2EIV / LV

K21 = 2EIV / 7

K31 = 6EIC/h2

K31 = 6EIC/12.25

D2 = 1 , D1 = D3 = 0




K22 = 2EIV / LV

K12 = 2EIV / 7

K22 = 4EIV / LV + 4EIC / h

K22 = 4EIV / 7 + 4EIC / 3.5

K32 = 6EIC/h2

K32 = 6EIC/12.25

D3 = 1 , D1 = D2 = 0

K13 = 6EIC/h2

K13 = 6EIC/12.25

K23 = 6EIC/h2

K23 = 6EIC/12.25

K33 = 12EIC/h3

K33 = 12EIC/42.88




4EIC/3.5 + 4EIV/7 4EIV/7 6EIC/12.25 D1 0

4EIV/7 4EIC/3.5 + 4EIV/7 6EIC/12.25 D2 = 0

6EIC/12.25 6EIC/12.25 24EIC/42.88 D3 F

EIV = (2x106)x 0.30 x 0.60312 = 10800 Ton m2 EIC = (2x106)x 0.30 x 0.65312 = 13731.25 Ton m2

21864.3 3085.7 6725.5 D1 0

3085.7 21864.3 6725.5 D2 = 0

6725.5 6725.5 7686.3 D3 7

L = - -1 = . . . . . . . . .

L = 5692.09 T/m2 {F} = L {} {7} = 5692.09 {} = 5692.09/7 = 1.2 x 10-3 m T = - -1




= . . . . . .

= . .

{} = T {}

= . . . = = . .

= = . . COLUMNA:

Msup = M0ba + (2EIC / h) 2D1 + 0+ 3D3 /h

= + . . (. ) + (. ). Msup = 3.05 Tn-m

Minf = M0ab + (2EIC / h) 0 + D1+ 3D3 /h

= + . . (. ) + (. ). Minf = 5.53 Tn-m

VIGA:

MIZ = M0ab + (2EIV / L) 2D1 + D2 + 0

= + {(. ) + (. )} MIZ = -2.99 Tn-m

MDER = M0ba + (2EIV / L) 2D2 + D1+ 0




2.5 Kip/Pie

= + {(. ) + (. )} MIZ = -2.99 Tn-m

DIAGRAMA DE MOMENTOS:

Ejemplo #2: resolver el problema usando el metodo de rigidez directa con matrices de transformacin. A.

18 Kip




E= 29000 KSI , I=1780 plg4

SOLUCIN:

PASO 1:

Armaduras:

Si: d1 = 1 d2 = 1

CABLE

A=1.6plg2

EA=




K11= EA/L K12=-EA/L

K21= -EA/L K22=-EA/L

= / // / Paso 2:

D1 = 1 , D2 = D3 = 0

D2 = 1 , D1 = D3 = 0




D3 = 1 , D1 = D2 =0

= 45 , cos = x/1 , x = cos = cos 45 = 0.707

Paso 3:

{} = =

= + + EI = 29000 x 1780 = 5162x 104 Kip-pie2

EA = 29000 x 1.6 = 46400 Kip




= = 358472.22150833.3358472.22150833.3 2150833.317206666.7

2150833.38603333.3 358472.2

2150833.3358472.22150833.3

2150833.38603333.32150833.317206666.7

= 3866.73866.7 3866.73866.7

= 000 000 001 100 358472.22150833.3

358472.22150833.3 2150833.317206666.7

2150833.38603333.3 358472.2

2150833.3358472.22150833.3

2150833.38603333.32150833.317206666.7

0001 0000

0010

000 100 000 010 358472.22150833.3

358472.22150833.3 2150833.317206666.7

2150833.38603333.3 358472.2

2150833.3358472.22150833.3

2150833.38603333.32150833.317206666.7

0100 0001

0000 000 000.707 3866.73866.7 3866.73866.7 0 0 00 0 0.707

= 34413333.48603333.32150833.3 8603333.317206666.70 2150833.30360405 Paso 4: {}

= = = = + {} + {} + {}




{q}2 = q1q2q3q4 =

15301530 , {q}eq2 =

q1q2q3q4 = 153015+30

123

= 008 + 000 000 001 100 0000 + 000 100 000 010 153015+30 +

000 000.707 0000

=

+

+ +

+

= ++ {} = KTOTAL {D} KTOTAL {D} = {}

34413333.48603333.32150833.3 8603333.317206666.70 2150833.30360405 = ++

= . .

. {q} 1 = K1 A1 {D} - {q}1eq


= 358472.22150833.3358472.22150833.3

2150833.317206666.72150833.38603333.3

358472.22150833.3358472.22150833.3

2150833.38603333.32150833.317206666.7

0001 0000

0010 5.410

64.41065.4105





= . . . .

{q} 2 = K2 A2 {D} - {q}2eq


= 358472.22150833.3358472.22150833.3

2150833.317206666.72150833.38603333.3

358472.22150833.3358472.22150833.3

2150833.38603333.32150833.317206666.7

0100 0001

0000 5.41064.41065.4105

153015+30


= . . . .

{q} 3 = K3 A3 {D} - {q}3eq


= 3866.73866.7 3866.73866.7 0 0 00 0 0.707 5.41064.41065.4105 =

00.100 Diagrama de momento:




METODO DE RIGIDEZ DIRECTA (Cosenos directores)

EJEMPLO:

{D} = DESPLAZ.GLOBALES DE LA ESTRUCTURA EN COORDENADAS GLOBALES

48 G.D.L

{} = VECTOR DE CARGAS GLOBALES DE LA ESTRUCTURA EN COORDENADAS GLOBALES




LEY DE HOOKE GENERALIZADA

{} mx1 = KTOTALmxm {D} mx1 . (I) m= #G.D.L

DEL METODO ANTERIOR;

{} = ATe Ke Ae {D} ..... (II)

ELEMENTO (e)

{d}e =

d1d2d3d4d5d6

6x1

=

{d} e = Ae {D}

{} = A AL {D} .. (III)

Ejes LOCALES

Ejes GLOBALES

Vector de desplazamiento en coordenadas locales/elemento

Se incluye deformaciones axiales.




Dnde:

A=Matriz de cosenos directores. AL = Matriz de localizacin.

d1= d*1 cos + d*2 sen d2= d*1 sen + d*2 cos d2= d*3

d1d2d3d4d5d6 =

cosseno0000

senocos0000 001000

000cosseno0

000senocos0 000001

6x6

d1d2d3d4d5d6

A {d}e = A {d}*e (IV)




Ejemplo:

[AL]1 =

d1d2d3d4d5d6 D1 D2 D3 D4 D5000100

000010 000001

000000 000000

6 x G.D.L 6 x 5

6 x 5 DESPLAZ. DE ELEMENTOS EN COORD. GLOBALES




[A]1 =

010000

100000 001000

000010

000100 000001

FORMULACION DE METODO {d}e = [A]e [AL] e {D} ---------------------------------------- (1) {q}e = [Ke]6x6 {d}e ----------------------------------------------- (2) {} = ATe Ke Ae {D}

{} = ALT AT Ke Ae ALe {D} ----------------------------------- (3) Ke = matriz de rigidez del elemento en coord. Locales.

{} = KTOTAL {D} ----------------------------------------------- (4) {} = {}CONC + [A]eTn

e=1

{q}eeq {} = {}CONC + [AL]Tne=1 [A]T {q}eeq -------------------------------------- (5) {q} e = Ke Ae ALe {D} - {q}eeq --------------------------------- (6)

=90 =0




4 T-m

[K]ARM = d1d2 d1 d2 EA/LEA/L EA/LEA/L

[K]e =

EA/L00EA/L00

012EI/L36EI/L20

12EI/L36EI/L2

06EI/L24EI/L06EI/L22EI/L

EA/L00EA/L00

0

12EI/L36EI/L2012EI/L36EI/L2

06EI/L22EI/L0

6EI/L24EI/L

EJEMPLO N1:

25x45

4 m

2 m

2 m

6 T

25x45

25x45




[AL]1 = d1d2d3d4d5d6

D1 D2 D3 D4 D5 D6000100

000010 000001

000000 000000

000000

[AL]3 = d1d2d3d4d5d6

D1 D2 D3 D4 D5 D6100000

010000 001000

000100 000010

000001 = [I]




[A]1 = [A]2 =

010000

100000 001000

000010

000100 000001

A = 0.25 x 0.45 = 0.1125 m2

E = 2 x 106 T/m2 , L = 4m

I = (0.25 x 0.453) / 12 = 1.89 x 10-3 m4

EI = (2x106)x 0.25 x 0.45312 = 3796.88 Ton m2 EA = (2x106)x(0.25 x 0.45) = 225000 Ton [K]1 = [K]2 = [K]3 =

56250005625000

0711.91423.80711.91423.8

01423.83796.901423.81898.4

56250005625000 0711.91423.80711.9

1423.8 01423.81898.40

1423.83796.9

{KTOTAL} = ALeT AeT Ke ne=1

ALe Ae {KT} = A1T A1TK1A1A1 + 2A2T K22A2 + 3 A3T K333

[] =

. .

. .

. .

. .

.

.

. .

. .

. .




2.67 T-m 2.67 T-m

4 Tn 4 Tn

3 T-m 3 T-m 2 m 2 m

+

. .

. .

. .

. .

.

.

. .

. .

. .

+

. .

. .

. .

. .

.

.

. .

. .

. .

=

[KTOTAL] =56961.901423.85625000

056961.91423.80711.91423.8

1423.81423.87593.801423.81898.4

562500056961.901423.80

711.91423.8056961.91423.8

01423.81898.41423.81423.87593.8

{} = {}CONC + [AL]Tne=1

[A]T {q}eeq

2 T/m

4 m

6 T




{}CONC =

1 = 02 = 03 = 04 = 05 = 06 = 4

{} = {}CONC + [AL]1T [A]1T {q}1eq + [AL]3T [A]3T {q}3eq

{} =

000004 +

000000

000000 000000

100000 010000

001000

010000

100000 001000

000010 000

100 000001

03303+3

+ [I] [I]

04

2.6704+2.67

{} =

000004 +

303000 +

04

2.6704+2.67

=

340.33046.67

{} = KTOTAL {D}

KTOTAL {D} = {}




56961.901423.85625000

056961.91423.80711.91423.8

1423.81423.87593.801423.81898.4

562500056961.901423.80

711.91423.8056961.91423.8

01423.81898.41423.81423.87593.8

=

.

.

D1D2D3D4D5D6 =

2.06x1037.4x105

4.89x1042.01x1031.55x1056.34x104

{q} e = Ke Ae ALe {D} - {q}eeq

{q}1 =

q1 = Axiiq2 = Viq3 = Miq4 = Axijq5 = Vjq6 = Mj

= [Ke]1 [A]1 [AL]1 {D} {q}1eq {q}1 =

56250005625000

0711.91423.80711.91423.8

01423.83796.901423.81898.4

562500056250000

711.91423.80711.91423.8

01423.81898.401423.83796.9

2.06x1037.4x105

4.89x1042.01x1031.55x1056.34x104

03303+3

{q}1 =

q1 = 4.16 Tnq2 = 3.77 Tnq3 = 3.12 T mq4 = 4.16 Tnq5 = 2.23 Tnq6 = 2.88 T m




{q}2 =


= [Ke]2 [A]2 [AL]2 {D} {q}2eq {q}2 =

56250005625000

0711.91423.80711.91423.8

01423.83796.901423.81898.4

562500056250000

711.91423.80711.91423.8

01423.81898.401423.83796.9

2.06x1037.4x105

4.89x1042.01x1031.55x1056.34x104

000000

{q}2 =


{q}3 =


= [Ke]3 [A]3 [AL]3 {D} {q}3eq

{q}3 =

56250005625000

0711.91423.80711.91423.8

01423.83796.901423.81898.4

562500056250000

711.91423.80711.91423.8

01423.81898.401423.83796.9

[] []

2.06x1037.4x105

4.89x1042.01x1031.55x1056.34x104

04

2.6704+2.67

{q}3 =





DIAGRMA DE MOMENTO FLECTOR:

EJEMPLO N2: HALLAR LAS FUERZAS INTERNAS EN LA ARMADURA MOSTRADA




P= 50 Klb

L = 20Pie

A= 8 pulg2 (const)

E = 30000 Ksi (const)




[A]ARM = CosSen00 SenCos00

00CosSen

00SenCos4x4 ARMADURAS:

{d}e = d1d3d3d44 x 4 {q}e = q1q3q3q44 x 4




[A]1 = [A]2 = 0100 1000

0001

00104x4 [A]3 = [A]4 = [I]

[A]5 = 0.60.800 0.80.600

000.60.8

000.80.64x4 [A]6 = 0.60.800

0.80.600

000.60.8

000.80.64x4

[AL]1 = d1d2d3d4 D1 D2 D3 D4

0010 0001

0000 0000 [AL]2 =

d1d2d3d4 D1 D2 D3 D4

0000 0000

0010 0001

[AL]3 = d1d2d3d4 D1 D2 D3 D4

0000 0000

0000 0000 [AL]5 =

d1d2d3d4 D1 D2 D3 D4

0000 0000

0010 0001

[AL]6 = d1d2d3d4 D1 D2 D3 D4

0010 0001

0000 0000 [KARM]1=2 =

1250012500

0000 1250012500

0000




[KARM]3=4 = 1666.701666.70 0000

1666.701666.70 0000 [KARM]5=6 =

1000010000

0000 1000010000

0000

[] =

= 5010025125 {} = + .. +

{K} = 00000000

1000 0100

01001000

0001 00

10 12500

125000000

1250012500 0000

0100

1000000

1 0010

00100001

0000 0000

00000000

0010 0001

01001000

0001 00

10 12500

125000000

1250012500 0000

0100

1000000

1 0010

00000000

0010 0001

00000000

0000 0000

10000100

0010 0001

1666.701666.70

00001666.701666.70

0000 1000

01000010

0001 0000

00000000

0000

10000100

0010 0001

10000100

0010 0001

1666.701666.70

00001666.701666.70

0000 1000

01000010

0001 1000

01000010

0001

00000000

0010 0001

0.60.8000.80.600

000.60.8 00

0.80.6 10000

100000000

1000010000 0000

0.60.800

0.80.600000.6

0.8 000.80.6

00000000

0010 0001




0000

00001000

0100 0.60.800

0.80.600

000.60.8

000.80.6

1000010000

00001000010000

0000 0.60.800

0.80.600

000.60.8

000.80.6

00100001

0000 0000

{} = 2026.74801666.70480189000

1666.702026.7480 004801890

KTOTAL {D} = {}

2026.7480

1666.70480189000

1666.702026.7480 004801890

D1D2D3D4 = 5010025

125

D1D2D3D4 =

0.16 plg9.24 x 102 plg0.13 plg3.24 x 102 plg

{q} 1 = K1 A1 AL1 {D} - {q}1eq

{q}1 = 12500125000000

1250012500 0000

0100

1000000

1 0010

00100001

0000 0000

0.16 9.24 102 0.13 3.24 102 {} {q}1 =

0115.5 0 115.5

{q} 2 = K2 A2 AL2 {D} - {q}2eq

{q}2 = 12500125000000

1250012500 0000

0100

1000000

1 0010

00000000

0010 0001

0.16 9.24 102 0.13 3.24 102 {}




{q}2 =

40.50 40.5 0

{q} 3 = K3 A3 AL3 {D} - {q}3eq

{q}3 = 1666.701666.700000

1666.701666.70 0000

10000100

0010 0001

00000000

0000 0000

0.16 9.24 102 0.13 3.24 102 {} {q}3 =

0000

{q} 4 = K4 A4 AL4 {D} - {q}4eq

{q}4 = 1666.701666.700000

1666.701666.70 0000

10000100

0010 0001

10000100

0010 0001

0.16 9.24 102 0.13 3.24 102 {} {q}4 =

50.00

50.00

{q} 5 = K5 A5 AL5 {D} - {q}5eq

{q}5 = 10000100000000

1000010000 0000

0.60.800

0.80.600000.6

0.8 000.80.6

00000000

0010 0001

0.16 9.24 102 0.13 3.24 102 {} {q}5 =

103.920103.920

{q} 6 = K6 A6 AL6 {D} - {q}6eq




{q}6 = 10000100000000

1000010000 0000

0.60.800

0.80.600

000.60.8

000.80.6

00100001

0000 0000

0.16 9.24 102 0.13 3.24 102 {} {q}6 =

22.080

22.080

EJERCICIO PROPUESTO: Wu = 1.4 CM + 1.7 CV

C1 = 18 Tn , C2 = 10 Tn , C3 = 9 Tn

CM1 = 2.5 T/ml , CM2 = 2 T/ml , CM3 = 1 T/ml

CV1 = 1.5 T/ml , CV2 = 1 T/ml , CV3 = 0.5 T/ml

E = 2 x 106 T/m2

SOLUCION:




D1 = 1 D2 = 1

D3 = 1 D4 = 1

D5 = 1 D6 = 1




D7 = 1 D8 = 1

D9 = 1 D10 = 1

D11 = 1 D1 2 = 1

D13 = 1 D1 4 = 1




D15 = 1

MATRIZ DE RIGIDES TOTAL DE TODO EL PORTICO (KTOTAL):

10666.7 1777.8 0 0 1777.8 0 0 0 0 0 0 0 3555.6 -1777.8 0

1777.8 14222.2 1777.8 0 0 1777.8 0 0 0 0 0 0 3555.6 -1777.8 0

0 1777.8 14222.2 1777.8 0 0 1777.8 0 0 0 0 0 3555.6 -1777.8 0

0 0 1777.8 10666.7 0 0 0 1777.8 0 0 0 0 3555.6 -1777.8 0

1777.8 0 0 0 10666.7 1777.8 0 0 1777.8 0 0 0 -1777.8 3555.6 -1777.8

0 1777.8 0 0 1777.8 14222.2 1777.8 0 0 1777.8 0 0 -1777.8 3555.6 -1777.8




0 0 1777.8 0 0 1777.8 14222.2 1777.8 0 0 1777.8 0 -1777.8 3555.6 -1777.8

0 0 0 1777.8 0 0 1777.8 10666.7 0 0 0 1777.8 -1777.8 3555.6 -1777.8

0 0 0 0 1777.8 0 0 0 7111.1 1777.8 0 0 0 -1777.8 1777.8

0 0 0 0 0 1777.8 0 0 1777.8 10666.7 1777.8 0 0 -1777.8 1777.8

0 0 0 0 0 0 1777.8 0 0 1777.8 10666.7 1777.8 0 -1777.8 1777.8

0 0 0 0 0 0 0 1777.8 0 0 1777.8 7111.1 0 -1777.8 1777.8

3555.6 3555.6 3555.6 3555.6 -1777.8 -1777.8 -1777.8 -1777.8 0 0 0 0 9481.5 -4740.7 0

-1777.8 -1777.8 -1777.8 -1777.8 3555.6 3555.6 3555.6 3555.6 -1777.8 -1777.8 -1777.8 -1777.8 -4740.7 9481.5 -4740.7

0 0 0 0 -1777.8 -1777.8 -1777.8 -1777.8 1777.8 1777.8 1777.8 1777.8 0 -4740.7 4740.7

L = rigidez lateral.

L = - -1

EJEMPLO

PLACA

PLACA




EA = 6G.D.L (4 ROT. Y 2 TRASL.)

EJERCICIO #3: HALLAR LOS DESPLAZAMIENTOS LATERALES.

D1 = 1 D2 = 1




K11 = 4EI/6 + 4EI/3 K12 = 2EI/6

K21 = 2EI/6 K22 = 4EI/6 + 4EI/3

K31 = -6EI/9 K32 = -6EI/9

K41 = 6EI/9 K42 = 6EI/9

D3 = 1 D4 = 1

K13 = -6EI/9 K14 = 6EI/6




K23 = -6EI/9 K24 = 6EI/9

K33 = 48EI/27 K34 = -24EI/27

K43 = -24EI/27 K44 = 24EI/27

[K]TOTAL =4EI6 + 4EI32EI6

6EI96EI9

2EI64EI6 + 4EI3

6EI96EI9

6EI9

6EI948EI27

24EI27

6EI96EI9

24EI2724EI27

EI = (2x106)x 0.30 x 0.50312 = 6250 Ton m2 [K]TOTAL =

125002083.34166.64166.6

2083.312500

4166.64166.6

4166.64166.611111.15555.5

4166.64166.6

5555.55555.5

L = rigidez lateral. L = - -1 [KL] = 11111.1 -5555.5-5555.5 5555.5 - -4166.6 -4166.64166.6 4166.6 12500 2083.32083.3 12500 -4166.6 4166.6-4166.6 4166.6 [KL] = 8730.15 3174.63174.6 3174.6 L {} = {F}

8730.15 3174.63174.6 3174.6 D3D4 = 108

D3D4 = 3.24 x 103 m5.76 x 103 m

{} = - -1 {}




D1D2 = 12500 2083.32083.3 125001 4166.6 4166.64166.6 4166.6 3.24 x 103 m5.76 x 103 m

D1 = 1D2 = 2 = 7.2 x 104rad7.2 x 104 rad

MATRIZ DE RIGIDEZ LATERAL PORTICO PLACA




MATRIZ DE RIGIDEZ DE LA VIGA BRAZO RIGIDO




SI LA VIGA TRABAJA SOLO POR FLEXION:

SE TIENE:

Parte flexible:

{qe} = Ke4x4 {e} ________________________________________ (1)

POR COMPATIBILIDAD:

VA = 1 VA = Vi + a x i

A = i

VB = 1 VB = Vj - b x j

B = j




vAAvBB

= 1000 a100

0010 00

b1 viivjj

d = [H] {d} H Flexible rgido

H = Matriz de compatibilidad

VA = 1 x Vi + a x i + 0 x Vj + 0 x j

A = 0 x Vi + 1 x i + 0 x Vj + 0 x j

VB = 0 x Vi + 0 x i + 1 x Vj - b x j

B = 0 x Vi + 0 x i + 0 x Vj + 1 x j [K]BRAZO RIGIDO = [H]4x4T [Ke]4x4 [H]4x4 POR EQUILIBRIO:

Vi = VA

Mi = a x VA + MA




Vj = VB

Mj = -b x VB + MB

Vi = 1 x VA + 0 x MA + 0 x VB + 0 x MB

Mi = 0 x VA + 1 x MA + 0 x VB + 0 x MB

Vj = 0 x VA + 0 x MA + 1 x VB + b x MB

Mj = 0 x VA + 0 x MA - b x VB + 1 x MB

viMivjMj = 1a00

a100 001

b 0001

vAMAvBMB {f} = [H] f HT Flexible rgido

POR LA LEY DE HOOKE : f = [Ke] d () Si remplazamos (3) en (2):

{f} = [H]T [Ke] d () Si remplazamos (1) en (4):

{f} = [H]T [Ke] [H] {d} [K]VBRAZ = [H]4x4T [Ke]4x4 [H]4x4




[K] =

12EIL36EIL2 + 12EIL3 xa

12EIL36EIL2 + 12EIL3 xb

6EIL2 + 12EIL3 x a4EIL + 12EIL2 x a + 12EIL3 x a2

6EIL2 12EIL3 xa2EIL + 6EIL2 (a + b) + 12EIL3 xaxb

12EIL3

6EIL2 12EIL3 x a12EIL3

6EIL2 12EIL3 xb

6EIL2 + 12EIL3 xb2EIL + 6EIL2 (a + b) + 12EIL3 xaxb

6EIL2 12EIL3 xb4EIL + 12EIL2 x b + 12EIL3 x b2

[Ke]VRIGIDO =

EA/L00EA/L00

000000 000000

EA/L00EA/L00

000000 000000

KP = PLACA

[K]P =

EA/L0

0EA/L0

0

012L3 ( EIZ1 + 2g)6L2 ( EIZ1 + 2g)0

12L3 ( EIZ1 + 2g)6L2 ( EIZ1 + 2g)

06L2 ( EIZ1 + 2g)4L (1 + g2)( EIZ1 + 2g)0

6L2 ( EIZ1 + 2g)2L (1 g)( EIZ1 + 2g)

EA/L00

EA/L00

0

12L3 ( EIZ1 + 2g)

6L2 ( EIZ1 + 2g)012L3 ( EIZ1 + 2g)

6L2 ( EIZ1 + 2g)

06L2 ( EIZ1 + 2g)2L (1 g)( EIZ1 + 2g)0

6L2 ( EIZ1 + 2g)4L (1 + g2)( EIZ1 + 2g)




LTOTAL = a + b + L

FACTOR DE FORMA:

f = 1.2 f = 10 / 9 f = 2 f = Area axial / Area alma

PROBLEMA:

1. HALLAR: LA MATRIZ DE RIGIDEZ TOTAL

2. HALLAR: LA MATRIZ DE RIGIDEZ LATERAL

3. HALLAR: DESPLAZAMIENTO LATERAL




30 Tn

4.00 m

8.00m 2.00

.20

E = 2 x 106 Ton/m VIGA

PLACA C A

COLUMNA

30 x 70

30 x 70




a=1.00 8.35

3.65

D1 = 1 D2 = 1

[A]1 = d1d2d3d4 D1 D1 D1

0001 0000

0010

Alanoca Quenta a. Freddy[1]

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