Post on 25-Dec-2015
description
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:
Momento de empotramiento.
M°BC=−M °CB=−W L2
12−PL8
=−136
M°CB=136
Remplazando en la ecuación de MOHR.
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
MBC=2 E ( I ) (2θB+θC−3(0))−136
MBC=2 E ( I ) (2θB+θC )−136
M ji=2 E K ji (2θ j+θi−3∆ ji)+M ° ji
MCB=2E ( I ) (2θC+θB−3(0))+ 136
MCB=2E ( I ) (2θC+θB )+ 136
TRAMO CD:
Momento de empotramiento.
M°CD=−M °DC=−W L2
12=−32
M°DC=32
Remplazando en la ecuación de MOHR.
Dónde: θD=0
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
MCD=2 E( 2 I3 ) (2θC+θD−3 (0 ) )−32
MCD=4 E( I3 ) (2θC )−32
M ji=2 E K ji (2θ j+θi−3∆ ji)+M ° ji
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:
Momento de empotramiento.
M°12=−M °21=−W L2
12=−32
M°21=32
Remplazando en la ecuación de MOHR.
Dónde: θ1=0
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 12=2E ( I3 ) (2θ1+θ2−3(0))−32
M 12=2E ( I3 ) (2θ1+θ2 )−32
M 12=23
(θ2 ) EI−32
M ji=2 E K ji (2θ j+θi−3∆ ji)+M ° ji
M 21=2E ( I3 ) (2θ2+θ1−3(0))+32
M 21=43
(θ2 ) EI+ 32
Tramo 23:
Momento de empotramiento.
M°12=−M °21=0
Reemplazando en la ecuación de MOHR
Dónde: θ3=0
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 23=2 E( I4 ) (2θ2+θ3−3 (0))
M 23=θ2 EI
M ji=2 E K ji (2θ j+θi−3∆ ji)+M ° ji
M 32=2E ( I4 ) (2θ3+θ2−3(0))
M 32=θ2EI
2
TRAMO 24:
Momento de empotramiento.
M°24=−M° 42=−W L2
12−PL8
=−316
M° 42=316
Remplazando en la ecuación de MOHR.
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 24=2 E( I4 ) (2θ2+θ4−3(0))−316
M 24=θ2EI+θ4EI
2−316
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 42=θ4EI +θ2EI
2+ 316
Tramo 45:
Momento de empotramiento.
M° 45=−M°54=0
Reemplazando en la ecuación de MOHR
Dónde: θ5=0
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 45=2 E( I4 ) (2θ4+θ5−3(0))
M 45=θ4EI
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 54=2 E( I4 ) (2θ5+θ4−3(0))
M 54=θ4 EI
2
Tramo 45:
Momento de empotramiento.
M°67=−M°76=0
Reemplazando en la ecuación de MOHR.
Dónde: θ7=0
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 67=2 E( I4 ) (2θ6+θ7−3(0))
M 67=θ6 EI
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 76=2 E( I4 ) (2θ7+θ6−3(0))
M 76=θ6 EI
2
TRAMO 46:
Momento de empotramiento.
M° 46=−W L2
12− Pa2b
L2=−1057
150
M°64=W L2
12+ Pab
2
L2=¿ 913
150
Remplazando en la ecuación de MOHR.
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 46=2 E( I5 ) (2θ4+θ6−3 (0))−1057150
M 46=4 θ4 EI
5+2θ6EI
5−1057150
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 64=2 E( I5 ) (2θ6+θ4−3 (0))+ 913150
M 64=4θ6EI
5+2θ4 EI
5+ 913150
Tramo 68:
Momento de empotramiento.
M°68=−W L2
12=−94
M° 86=0
Remplazando en la ecuación de MOHR.
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
M 68=2 E( I3 ) (2θ6+θ8−3(0))−94
M 68=4θ6EI
3+2θ8EI
3−94
M ij=2E K ij (2θi+θ j−3∆ij )+M °ij
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
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
Cálculo de los momentos y cortantes:
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
Cálculo de los momentos y cortantes:
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