Ejercicio Resuelto Analisis Estructural Parrillas
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EJERCICIO N° 1
Resuelva completamente la parrilla mostrada, por el método matricial de los desplazamientos:
GJ= 1200 TN-m² EI= 1500 TN-m²
Solución: 1) Se empieza numerando las barras, los nudos y orientando los elementos:
2) Calculamos [k’] las propiedades
BARRA 1 GJ 1200 = =400 L 3 2 EI 2(1500) = =1000 L 3 4 EI 4(1500) = =2000 L 3 6 EI 6(1500) = =1000 L² 3² 12 EI 12(1500) = =666.67 Lᶟ 3ᶟ
BARRA 2 GJ 1200 = =240 L 5 2 EI 2(1500) = =600 L 5 4 EI 4(1500) = =1200 L 5 6 EI 6(1500) = =360 L² 5² 12 EI 12(1500) = =144 Lᶟ 5ᶟ BARRA 3 GJ 1200 = =400 L 3 2 EI 2(1500) = =1000 L 3 4 EI 4(1500) = =2000 L 3 6 EI 6(1500) = =1000 L² 3² 12 EI 12(1500) = =666.67 Lᶟ 3ᶟ
Elaboramos el cuadro de resumen (kN y m) BARRA
L
GJ/L
2EI/L
4EI/L
6EI/L²
12EI/L³
1 2 3
3 5 3
400 240 400
1000 600 1000
2000 1200 2000
1000 360 1000
666.67 144 666.67
3) Hallamos [K] = [B] ⸆. [k’]. [B] BARRA 1
∝ 0
C 1
S 0
[ ]
1 0 0 [R ]= 0 1 0 0 0 1
BARRA 2 ∝ 90
C 0
[
S 1
0 −1 0 [R ]= −1 0 0 0 0 1
BARRA 3
]
∝ 180
C 1
S 0
[
−1 0 0 [R ]= 0 −1 0 0 0 1
]
4) Calculamos [k ' ] de cada barra BARRA 1
[ k ' ]1=¿
[
400 0 0 666 . 67 0 1000 −400 0 0 −666 . 67 0 −1000
0 −400 0 1000 0 −666 .67 2000 0 −1000 0 400 0 −1000 0 666 .67 1000 0 −1000
0 −1000 1000 0 −1000 2000
BARRA 2
[ k ' ]2=¿
BARRA 3
[
240 0 0 −240 0 0
0 144 360 0 −144 −360
0 −240 0 0 360 0 −144 −360 1200 0 −360 600 0 240 0 0 −360 0 144 −360 600 0 −360 1200
]
]
[ k ' ]3=¿
5) Hallamos
[
400 0 0 666 . 67 0 1000 −400 0 0 −666 . 67 0 −1000
[
[B]= [R] [0] [0] [R]
0 −400 0 1000 0 −666 .67 2000 0 −1000 0 400 0 −1000 0 666 .67 1000 0 −1000
]
BARRA 1
[ B ] =¿
BARRA 2
[ B ] =¿
BARRA 3
[ ] 1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
[ ] 0 −1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 0 0 0 1
0 −1000 1000 0 −1000 2000
]
[ B ] =¿
[
−1 0 0 0 0 0 0 −1 0 0 0 0 0 0 1 0 0 0 0 0 0 −1 0 0 0 0 0 0 −1 0 0 0 0 0 0 1
]
6) Hallamos [ k ] BARRA 1
T
[ k ] 1=[ B ] 1∗[ k ' ]1∗[ B ]1=¿
[
400 0 0 −400 0 0 666 . 67 1000 0 −666 .67 0 1000 2000 0 −1000 −400 0 0 400 0 0 −666 .67 −1000 0 666 .67 0 −1000 1000 0 −1000
0 −1000 1000 0 −1000 2000
BARRA 2
[ k ] 2=[ B ] T2∗[ k ' ]2∗[ B ]2=¿
[
144 0 0 240 360 0 −144 0 0 −240 −360 0
360 0 1200 −360 0 600
−144 0 −360 0 −240 0 −360 0 600 144 0 −360 0 240 0 −360 0 1200
]
]
BARRA 3
[
T
[ k ] 3=[ B ] 3∗[ k ' ] 3∗[ B ]3 =¿
400 0 0 666 . 67 0 −1000 −400 0 0 −666 . 67 0 1000
0 −400 0 −1000 0 −666 .67 2000 00 1000 0 400 0 1000 0 666 .67 1000 0 1000
0 1000 1000 0 1000 2000
7) Hallamos [ K ]
{ Q }=[ K ] { D }
{ } [ ]{ } {Q1 } {Q2 } = {Q3 } {Q4 }
¿ ¿ ¿ ¿
¿ ¿ ¿ ¿
{ D1 } { D2 } { D3 } {D 4 }
[ K ¿ ] =[ K 11 ]1+ [ K 11 ] 2+ [ K 11 ] 3
[ K ¿ ] =¿
[
][
][
400 0 0 144 0 360 400 0 0 + + 0 666 .67 1000 0 240 0 0 666 .67 −1000 0 1000 2000 360 0 1200 0 −1000 2000
[ K ¿ ] =¿
[
944 0 360 0 1573 .34 0 360 0 5200
]
]
]
8) Ensamblaje del vector de cargas de fijación
{Q} = {ℝ} + {F} BARRA 1: Mº Y 12=
−120∗1.72∗1.3 =−50.09 9
120∗1.32∗1.7 Mº Y 21= =38.31 9 Z º 12=
120∗1.7+50.09−38.31 =71.93 3
Zº 21=
120∗1.3−50.09+38.31 =48.07 3
Mº X 12=Mº X 21=0
BARRA 2: Mº X 13=−Mº X 31= Zº 13=Zº 31=
−32∗25 =−66.67 12
32∗5 =80 2
Mº Y 13=Mº Y 31=0
BARRA 3: Mº Y 41=Mº Y 14=0 Zº 41=Zº 14=0 Mº X 41=Mº X 14 =0
9) Cálculo de [ r ' ] BARRA 1:
de cada barra:
{ }
0 −50.09 {r ' 1 } 71.93 [ r ' ] 1= = {r ' 2 } 0 38.31 48.07
{ }
BARRA 2:
{ }
−66.67 0 {r ' 1 } 80 [ r ' ] 2= = {r ' 3 } 66.67 0 80
{ }
En la Barra 2 no coinciden los sistemas de coordenadas, por lo tanto: {r }2=[B]{r ' }2
{ }[
]{ }
{r ' 1 } {r 1 } = [R] [0] {r ' 3 } [0] [ R] {r 3 }
∝=90 º Falta ¿
[ ]{ }
1 0 [ r ] 1= 0 0 1 0
0 0 0 0 −50.09 1 0 ¯ 50.09 0 1 71.93 = 71.93 = {r 1 } 0 {r 2 } ¯1 0 0 38.31 0 0 38.31 48.07 0 1 48.07
{ }
[ ]{ }{ }
1 0 [ r ] 2= 0 ¯1 0 0
0 0 −66.67 −66.67 0 0 −50.09 1 0 0 80 71.93 0 1 80 = = −66.67 −38.31 0 0 66.67 0 0 ¯1 0 0 80 48.07 0 1 80
10) Ensamblamos [R ]
{ }{
R1 {r 1 }1+{r 1 }2 +{r 1 }3 R {r }1 { R }= 2 = R3 {r }2 R4 {r }3
{ R }=
}
{ }{ } { }{ { { {
}{} { } { } { } {
0 −66.67 0 + + 0 −50.09 0 71.93 80 0
−66.67 −50.09 151.93
0 −50.09 71.93
0 −50.09 71.93
−66.67 0 80 −66.67 0 80
=
−66.67 0 80 −66.67 0 80
} } } }
Como {F}, entonces: { ℝ } = {R} - {F}
{} { { { {
−66.67 −50.09 151.93
{ R }={ R }−{F }=
0 −50.09 71.93 −66.67 0 80 −66.67 0 80
11) Hallamos
} } } }
{ }
{Q1 } {Q2 } {Q }= {Q3 } {Q4 }
{} { { { {
} } } }
66.67 50.09 −151.93
{ Q }=−{R }=
0 50.09 −71.93 66.67 0 −80 66.67 0 −80
12) De los valores anteriores tomo : {Q L } {Q L }=[ K ¿ ] +{D L }
Para el problema: {Q1 }=[ K 11 ] +{D1 }
{ } { } {} { } D1 x 66.67 50.09 =[ K 11 ] D1 y −151.93 θ
D1 x 66.67 −1 D 1 y =[ K 11 ] 50.09 −151.93 θ
{ } { } } { }{ D1 x 66.67 −1 D 1 y =❑ 50.09 −151.93 θ
D1 x 0.08398 m. = D1 y 0.03184 m. −0.03503 rad . θ
13) Cálculo de fuerzas de extremo de barras: BARRA 1: {d }=
{ }{ } d1 d2
{ D1 } { D2 }
Hallamos {d } :
{ q ' } =[ k ' ] +{d ' } {d ' }1=[ B ] ⸆ 1 +{d }1
]{ }
[
{d 1 }1 {d ' }1= [R ]⸆ [0 ] [0 ] [R] ⸆ {d 2 }1
[ ]{
1 0 0 0.08398 ' {d 1 }= 0 1 0 0.03184 0 0 1 −0.03503
{
0.08398 ' {d 1 }= 0.03184 −0.03503
}
}
BARRA 2: {d ' }2=[ B ] ⸆ 2 +{d }2
[
]{ }
{d 1 }2 {d ' }2= [R ]⸆ [0 ] [0 ] [R] ⸆ {d 3 }2
[
]{
{
}
0 1 0 0.08398 ' {d 2 }= −1 0 0 0.03184 0 0 1 −0.03503 0.03184 {d ' 1 }= −0.08398 −0.03503
}
BARRA 3: {d ' }3=[ B ] ⸆3 +{d }3
[
]{ }
[
]{
{d 1 }3 {d ' }3= [R ]⸆ [0 ] [0] [R ]⸆ {d 4 }3
−1 0 0 0.08398 {d ' 2 }= 0 −1 0 0.03184 0 0 −1 −0.03503
{
−0.08398 {d ' 1 }= −0.03184 −0.03503
}
}
14) Empleamos: {q ' }n =[ k ' ] n+ {d ' }n
n = 1, 2, 3
BARRA 1:
{ }[
q '1 [ k ' 11 ] [k ' 12] = q ' 2 1 [ k ' 21 ] [k ' 22]
{q ' 1 }1=[ k ' 11 ]1 {d' 1 }1
]{ } 1
d '1 d '2
1
[
400 0 0 {q ' 1 }1= 0 666.67 1000 0 1000 2000
]{ 1
}{ }
0.08398 33.592 = 0.03184 −13.803 −0.03503 1 −38.22
1
{q ' 2 }1=[ k ' 21 ]1 {d ' 1 }1
[
{q ' 2 }1=
−400 0 0 0 −666.67 −1000 0 −1000 1000
]{ 1
}{ }
0.08398 −33.592 = 0.03184 13.803 −0.03503 1 −38.22
BARRA 2:
{ }[
]{ }
q '1 [ k ' 11 ] [k ' 12 ] d ' 1 = q ' 3 2 [ k ' 21 ] [k ' 22 ] 2 d ' 3
2
{q ' 1 }2=[ k ' 11 ]2 {d' 1 }2
[
240 0 0 ' {q 1 }2= 0 144 360 0 360 1200
]{ 2
}{
0.03184 7.6416 = −0.08398 −24.70392 −0.03503 2 −72.2688
}
2
{q ' 3 }2=[ k '21 ]2 {d ' 1 }2
{q ' 3 }2=
[
−240 0 0 0 −144 −360 0 −360 600
]{ 2
0.03184 −7.6416 = −0.08398 24.70392 −0.03503 2 9.2148
}{
}
}{
}
3
2
BARRA 3:
{ }[
q '1 [k ' 11 ] [k ' 12 ] = q ' 4 3 [k ' 21] [k ' 22 ]
]{ } 3
d'1 d'4
3
{q ' 1 }3=[ k '11 ]3 {d '1 }3
[
400 0 0 {q ' 1 }3= 0 666.67 1000 0 1000 2000
]{ 3
−0.08398 −33.592 = −0.03184 −56.2568 −0.03503 3 −101.9
1
{q ' 2 }1=[ k ' 21 ]1 {d ' 1 }1
[
−400 0 0 {q 4 }3 = 0 −666.67 −1000 0 −1000 1000 '
]{ 3
}{ }
0.08398 33.592 0.03184 = 56.2568 −0.03503 3 3.19
GRAFICAS
DIAGRAMA DE CORTANTE
DIAGRAMA DE MOMENTO FLECTOR
3
DIAGRAMA DE TORSION
Resuelva completamente la parrilla mostrada, por el método matricial de los desplazamientos:
GJ= 1200 TN-m² EI= 1500 TN-m²
Solución: 1) Se empieza numerando las barras, los nudos y orientando los elementos:
2) Calculamos [k’] las propiedades
BARRA 1 GJ 1200 = =400 L 3 2 EI 2(1500) = =1000 L 3 4 EI 4(1500) = =2000 L 3 6 EI 6(1500) = =1000 L² 3² 12 EI 12(1500) = =666.67 Lᶟ 3ᶟ
BARRA 2 GJ 1200 = =240 L 5 2 EI 2(1500) = =600 L 5 4 EI 4(1500) = =1200 L 5 6 EI 6(1500) = =360 L² 5² 12 EI 12(1500) = =144 Lᶟ 5ᶟ BARRA 3 GJ 1200 = =400 L 3 2 EI 2(1500) = =1000 L 3 4 EI 4(1500) = =2000 L 3 6 EI 6(1500) = =1000 L² 3² 12 EI 12(1500) = =666.67 Lᶟ 3ᶟ
Elaboramos el cuadro de resumen (kN y m) BARRA
L
GJ/L
2EI/L
4EI/L
6EI/L²
12EI/L³
1 2 3
3 5 3
400 240 400
1000 600 1000
2000 1200 2000
1000 360 1000
666.67 144 666.67
3) Hallamos [K] = [B] ⸆. [k’]. [B] BARRA 1
∝ 0
C 1
S 0
[ ]
1 0 0 [R ]= 0 1 0 0 0 1
BARRA 2 ∝ 90
C 0
[
S 1
0 −1 0 [R ]= −1 0 0 0 0 1
BARRA 3
]
∝ 180
C 1
S 0
[
−1 0 0 [R ]= 0 −1 0 0 0 1
]
4) Calculamos [k ' ] de cada barra BARRA 1
[ k ' ]1=¿
[
400 0 0 666 . 67 0 1000 −400 0 0 −666 . 67 0 −1000
0 −400 0 1000 0 −666 .67 2000 0 −1000 0 400 0 −1000 0 666 .67 1000 0 −1000
0 −1000 1000 0 −1000 2000
BARRA 2
[ k ' ]2=¿
BARRA 3
[
240 0 0 −240 0 0
0 144 360 0 −144 −360
0 −240 0 0 360 0 −144 −360 1200 0 −360 600 0 240 0 0 −360 0 144 −360 600 0 −360 1200
]
]
[ k ' ]3=¿
5) Hallamos
[
400 0 0 666 . 67 0 1000 −400 0 0 −666 . 67 0 −1000
[
[B]= [R] [0] [0] [R]
0 −400 0 1000 0 −666 .67 2000 0 −1000 0 400 0 −1000 0 666 .67 1000 0 −1000
]
BARRA 1
[ B ] =¿
BARRA 2
[ B ] =¿
BARRA 3
[ ] 1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
[ ] 0 −1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 0 0 0 1
0 −1000 1000 0 −1000 2000
]
[ B ] =¿
[
−1 0 0 0 0 0 0 −1 0 0 0 0 0 0 1 0 0 0 0 0 0 −1 0 0 0 0 0 0 −1 0 0 0 0 0 0 1
]
6) Hallamos [ k ] BARRA 1
T
[ k ] 1=[ B ] 1∗[ k ' ]1∗[ B ]1=¿
[
400 0 0 −400 0 0 666 . 67 1000 0 −666 .67 0 1000 2000 0 −1000 −400 0 0 400 0 0 −666 .67 −1000 0 666 .67 0 −1000 1000 0 −1000
0 −1000 1000 0 −1000 2000
BARRA 2
[ k ] 2=[ B ] T2∗[ k ' ]2∗[ B ]2=¿
[
144 0 0 240 360 0 −144 0 0 −240 −360 0
360 0 1200 −360 0 600
−144 0 −360 0 −240 0 −360 0 600 144 0 −360 0 240 0 −360 0 1200
]
]
BARRA 3
[
T
[ k ] 3=[ B ] 3∗[ k ' ] 3∗[ B ]3 =¿
400 0 0 666 . 67 0 −1000 −400 0 0 −666 . 67 0 1000
0 −400 0 −1000 0 −666 .67 2000 00 1000 0 400 0 1000 0 666 .67 1000 0 1000
0 1000 1000 0 1000 2000
7) Hallamos [ K ]
{ Q }=[ K ] { D }
{ } [ ]{ } {Q1 } {Q2 } = {Q3 } {Q4 }
¿ ¿ ¿ ¿
¿ ¿ ¿ ¿
{ D1 } { D2 } { D3 } {D 4 }
[ K ¿ ] =[ K 11 ]1+ [ K 11 ] 2+ [ K 11 ] 3
[ K ¿ ] =¿
[
][
][
400 0 0 144 0 360 400 0 0 + + 0 666 .67 1000 0 240 0 0 666 .67 −1000 0 1000 2000 360 0 1200 0 −1000 2000
[ K ¿ ] =¿
[
944 0 360 0 1573 .34 0 360 0 5200
]
]
]
8) Ensamblaje del vector de cargas de fijación
{Q} = {ℝ} + {F} BARRA 1: Mº Y 12=
−120∗1.72∗1.3 =−50.09 9
120∗1.32∗1.7 Mº Y 21= =38.31 9 Z º 12=
120∗1.7+50.09−38.31 =71.93 3
Zº 21=
120∗1.3−50.09+38.31 =48.07 3
Mº X 12=Mº X 21=0
BARRA 2: Mº X 13=−Mº X 31= Zº 13=Zº 31=
−32∗25 =−66.67 12
32∗5 =80 2
Mº Y 13=Mº Y 31=0
BARRA 3: Mº Y 41=Mº Y 14=0 Zº 41=Zº 14=0 Mº X 41=Mº X 14 =0
9) Cálculo de [ r ' ] BARRA 1:
de cada barra:
{ }
0 −50.09 {r ' 1 } 71.93 [ r ' ] 1= = {r ' 2 } 0 38.31 48.07
{ }
BARRA 2:
{ }
−66.67 0 {r ' 1 } 80 [ r ' ] 2= = {r ' 3 } 66.67 0 80
{ }
En la Barra 2 no coinciden los sistemas de coordenadas, por lo tanto: {r }2=[B]{r ' }2
{ }[
]{ }
{r ' 1 } {r 1 } = [R] [0] {r ' 3 } [0] [ R] {r 3 }
∝=90 º Falta ¿
[ ]{ }
1 0 [ r ] 1= 0 0 1 0
0 0 0 0 −50.09 1 0 ¯ 50.09 0 1 71.93 = 71.93 = {r 1 } 0 {r 2 } ¯1 0 0 38.31 0 0 38.31 48.07 0 1 48.07
{ }
[ ]{ }{ }
1 0 [ r ] 2= 0 ¯1 0 0
0 0 −66.67 −66.67 0 0 −50.09 1 0 0 80 71.93 0 1 80 = = −66.67 −38.31 0 0 66.67 0 0 ¯1 0 0 80 48.07 0 1 80
10) Ensamblamos [R ]
{ }{
R1 {r 1 }1+{r 1 }2 +{r 1 }3 R {r }1 { R }= 2 = R3 {r }2 R4 {r }3
{ R }=
}
{ }{ } { }{ { { {
}{} { } { } { } {
0 −66.67 0 + + 0 −50.09 0 71.93 80 0
−66.67 −50.09 151.93
0 −50.09 71.93
0 −50.09 71.93
−66.67 0 80 −66.67 0 80
=
−66.67 0 80 −66.67 0 80
} } } }
Como {F}, entonces: { ℝ } = {R} - {F}
{} { { { {
−66.67 −50.09 151.93
{ R }={ R }−{F }=
0 −50.09 71.93 −66.67 0 80 −66.67 0 80
11) Hallamos
} } } }
{ }
{Q1 } {Q2 } {Q }= {Q3 } {Q4 }
{} { { { {
} } } }
66.67 50.09 −151.93
{ Q }=−{R }=
0 50.09 −71.93 66.67 0 −80 66.67 0 −80
12) De los valores anteriores tomo : {Q L } {Q L }=[ K ¿ ] +{D L }
Para el problema: {Q1 }=[ K 11 ] +{D1 }
{ } { } {} { } D1 x 66.67 50.09 =[ K 11 ] D1 y −151.93 θ
D1 x 66.67 −1 D 1 y =[ K 11 ] 50.09 −151.93 θ
{ } { } } { }{ D1 x 66.67 −1 D 1 y =❑ 50.09 −151.93 θ
D1 x 0.08398 m. = D1 y 0.03184 m. −0.03503 rad . θ
13) Cálculo de fuerzas de extremo de barras: BARRA 1: {d }=
{ }{ } d1 d2
{ D1 } { D2 }
Hallamos {d } :
{ q ' } =[ k ' ] +{d ' } {d ' }1=[ B ] ⸆ 1 +{d }1
]{ }
[
{d 1 }1 {d ' }1= [R ]⸆ [0 ] [0 ] [R] ⸆ {d 2 }1
[ ]{
1 0 0 0.08398 ' {d 1 }= 0 1 0 0.03184 0 0 1 −0.03503
{
0.08398 ' {d 1 }= 0.03184 −0.03503
}
}
BARRA 2: {d ' }2=[ B ] ⸆ 2 +{d }2
[
]{ }
{d 1 }2 {d ' }2= [R ]⸆ [0 ] [0 ] [R] ⸆ {d 3 }2
[
]{
{
}
0 1 0 0.08398 ' {d 2 }= −1 0 0 0.03184 0 0 1 −0.03503 0.03184 {d ' 1 }= −0.08398 −0.03503
}
BARRA 3: {d ' }3=[ B ] ⸆3 +{d }3
[
]{ }
[
]{
{d 1 }3 {d ' }3= [R ]⸆ [0 ] [0] [R ]⸆ {d 4 }3
−1 0 0 0.08398 {d ' 2 }= 0 −1 0 0.03184 0 0 −1 −0.03503
{
−0.08398 {d ' 1 }= −0.03184 −0.03503
}
}
14) Empleamos: {q ' }n =[ k ' ] n+ {d ' }n
n = 1, 2, 3
BARRA 1:
{ }[
q '1 [ k ' 11 ] [k ' 12] = q ' 2 1 [ k ' 21 ] [k ' 22]
{q ' 1 }1=[ k ' 11 ]1 {d' 1 }1
]{ } 1
d '1 d '2
1
[
400 0 0 {q ' 1 }1= 0 666.67 1000 0 1000 2000
]{ 1
}{ }
0.08398 33.592 = 0.03184 −13.803 −0.03503 1 −38.22
1
{q ' 2 }1=[ k ' 21 ]1 {d ' 1 }1
[
{q ' 2 }1=
−400 0 0 0 −666.67 −1000 0 −1000 1000
]{ 1
}{ }
0.08398 −33.592 = 0.03184 13.803 −0.03503 1 −38.22
BARRA 2:
{ }[
]{ }
q '1 [ k ' 11 ] [k ' 12 ] d ' 1 = q ' 3 2 [ k ' 21 ] [k ' 22 ] 2 d ' 3
2
{q ' 1 }2=[ k ' 11 ]2 {d' 1 }2
[
240 0 0 ' {q 1 }2= 0 144 360 0 360 1200
]{ 2
}{
0.03184 7.6416 = −0.08398 −24.70392 −0.03503 2 −72.2688
}
2
{q ' 3 }2=[ k '21 ]2 {d ' 1 }2
{q ' 3 }2=
[
−240 0 0 0 −144 −360 0 −360 600
]{ 2
0.03184 −7.6416 = −0.08398 24.70392 −0.03503 2 9.2148
}{
}
}{
}
3
2
BARRA 3:
{ }[
q '1 [k ' 11 ] [k ' 12 ] = q ' 4 3 [k ' 21] [k ' 22 ]
]{ } 3
d'1 d'4
3
{q ' 1 }3=[ k '11 ]3 {d '1 }3
[
400 0 0 {q ' 1 }3= 0 666.67 1000 0 1000 2000
]{ 3
−0.08398 −33.592 = −0.03184 −56.2568 −0.03503 3 −101.9
1
{q ' 2 }1=[ k ' 21 ]1 {d ' 1 }1
[
−400 0 0 {q 4 }3 = 0 −666.67 −1000 0 −1000 1000 '
]{ 3
}{ }
0.08398 33.592 0.03184 = 56.2568 −0.03503 3 3.19
GRAFICAS
DIAGRAMA DE CORTANTE
DIAGRAMA DE MOMENTO FLECTOR
3
DIAGRAMA DE TORSION
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