Analisis Matricial: b (m) h (m)
ANALISIS MATRICIAL 10 Tn 2.5 m 2.0 Tn/m B A D C 5.0 m 6.0 m 4.0 m f'c= 210 E= 2.17E+06 I= 0.003125 EI= 6792.83 E
Views 59 Downloads 4 File size 234KB
Recommend stories
- Author / Uploaded
- Elio Paucar
Citation preview
ANALISIS MATRICIAL 10 Tn 2.5 m
2.0 Tn/m B
A
D
C
5.0 m
6.0 m
4.0 m
f'c= 210 E= 2.17E+06 I= 0.003125 EI= 6792.83 EA= ∞
b (m) 0.30
kg/cm2 Tn/m2 m4 Tn*m2
h (m) 0.50
1.- Definimos el sistema Q-D Y q-d 2
1
3
1
4
2
3
Q-D 4GDL 1
2
3
4
5
6
q-d 2.- Generamos la matriz de transformacion de desplazamientos [A] D1=1
Di =0 ,
i ≠1
[C1] =
D2=1
Di =0 ,
i ≠2
[C2] =
D3=1
Di =0 ,
Di =0 ,
[A] =
D2=1 0 1 1 0 0 0
D3=1 0 0 0 1 1 0
D4=1 0 0 0 0 0 1
[A1] [A2] [A3]
3.- Calculamos el [k'] matriz de rigidez de barra en el sistema local) barra 1: [k'1]=
5434.27 2717.13
2717.13 5434.27
4528.56 2264.28
2264.28 4528.56
6792.83 3396.42
3396.42 6792.83
barra 2: [k'2]=
barra 3: [k'3]=
0 0 0 1 1 0
i ≠4
[C4] =
D1=1 1 0 0 0 0 0
0 1 1 0 0 0
i ≠3
[C3] =
D4=1
1 0 0 0 0 0
[� ′]=[■8(4𝐸𝐼⁄𝐿&2𝐸𝐼⁄𝐿@2𝐸 𝐼⁄𝐿&4𝐸𝐼⁄𝐿)]
0 0 0 0 0 1
4.- Transformando el [k'] al sistema global [�_𝑖 ]=[𝐴_𝑖 ]^𝑇∗[ 〖�′〗 _𝑖 ]∗[𝐴_𝑖 ]
Ai^t
barra 1 [k1]=
5434.266 2717.133 0.000 0.000
2717.133 5434.266 0.000 0.000
0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000
barra 2 [k2]=
0 0 0 0
0 4529 2264 0
0 2264 4529 0
0 0 0 0
barra 3 [k3]=
0 0 0 0
0 0 0 0
0 0 6793 3396
0 0 3396 6793
5434.27 2717.13 0.00 0.00
2717.13 5434.27 0.00 0.00
0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00
0.00 0.00 6792.83 3396.42
0.00 0.00 3396.42 6792.83
ki*Ai
1 0 0 0
0 1 0 0
5434.26628 2717.13314 2717.13314 5434.26628
0 1 0 0
0 0 1 0
0 0
0 0 1 0
0 0 0 1
0 0
0.00 0.00 0.00 0.00
0.00 4528.56 2264.28 0.00
0.00 2264.28 4528.56 0.00
0.00 0.00 0.00 0.00
5434 2717 0 0
2717 9963 2264 0
0 2264 11321 3396
0 0 3396 6793
0 0
4528.55523 2264.27762 2264.27762 4528.55523
6792.83285 3396.41642 3396.41642 6792.83285
0 0 0 0
0 0 -2.66666667 2.66666667
[𝐾]=∑▒[�_𝑖 ]
+
+
=
6.- Calculo de {R} por ensamblaje 10 Tn
2.00 Tn/m 6.25 Tn-m 2.67 Tn-m
6.25 Tn-m 5 Tn
5 Tn
2.67 Tn-m 4 Tn
barra 1 {r1}=
-6.25 6.25
Tn-m
{r3}=
-2.67 2.67
Tn-m
4 Tn
barra 2 {r2}=
0.00 0.00
Tn-m
barra 3
transformando {ri} al sistema global {𝑅}=∑▒ 〖 [𝐴_𝑖 ]^𝑇∗[𝑟_𝑖 ] 〗 {R}=
+
{R}=
-6.25 6.25 -2.67 2.67
1 0 0 0
0 1 0 0
-6.25 6.25
0 0 1 0
0 0 0 1
-2.67 2.67
+
0 1 0 0
0 0 1 0
-6.25 6.25 0 0
0 0
Tn-m
7.- Calculo de {D} de los desplazamientos en el sistema global {𝑄}=[𝐾]∗{𝐷}
{𝐷}=[𝐾]^(−1)∗{𝑄} {Q}= - {R}
K^-1 =
2.15E-04 -6.19E-05 1.46E-05 -7.29E-06
{Q}=
-6.19E-05 1.24E-04 -2.92E-05 1.46E-05
6.25 -6.25 2.67 -2.67 1.46E-05 -2.92E-05 1.11E-04 -5.54E-05
Tn-m
-7.29E-06 1.46E-05 -5.54E-05 1.75E-04
1
2
3
4 1 2 3 4
0 0
0 0
5.- Ensamblamos la matriz de rigidez
[K]=
0 0
0.002 -0.001 0.001 -0.001
{D}=
rad rad rad rad
8.- Determinanos los esfuerzos y desplazamientos en el sistema local
Desplazaminetos
{𝑑_𝑖 }=[𝐴_𝑖 ]∗{𝐷}
- barra 1 {d1}=
1 0
0 1
0 0
0 0
0.002 -0.001 0.001 -0.001
=
0.00178916 -0.0012781
rad rad
{d2}=
0 0
1 0
0 1
0 0
0.002 -0.001 0.001 -0.001
=
-0.0012781 0.00071639
rad rad
{d3}=
0 0
0 0
1 0
0 1
0.002 -0.001 0.001 -0.001
=
0.00071639 -0.00075077
rad rad
- barra 2
- barra 3
Esfuerzos
{𝑞_𝑖 }={𝑞_𝑖 }𝑝+{𝑞_𝑖 }�
donde:{�_� }�=[ 〖�′〗 _� ]∗{�_� }
{𝑞_𝑖 }={�_� }𝑝+[ 〖�′〗 _𝑖 ]∗{𝑑_� }
entonces: barra 1
{q1}=
-6.25 + 6.25
5434.26627982 2717.13314 2717.13313991 5434.26628
1.8E-03 -1.3E-03
=
0.00 4.17
T-m T-m
barra 2 {q2}= barra 3 {q3}=
0.00 + 0.00
4528.555 2264.278
2264.278 4528.555
-1.3E-03 7.2E-04
=
-4.17 T-m 0.35 T-m
-2.67 + 2.67
6792.833 3396.416
3396.416 6792.833
7.2E-04 -7.5E-04
=
-0.35 T-m 0.00 T-m
CALCULO DE LOS CORTANTES Y MOMENTOS MAXIMOS {V}i = {V}isos - {(Mi + Mj)/Lij} {V}j = {V}isos+ {(Mi + Mj)/Lij} 5.00 5.00 0.00 0.00 4.00 4.00
{V}isos =
DMF (Tn-m)
i
Tn
{V}T =
j
4.17 5.83 0.64 -0.64 4.09 3.91
Tn
-4.17 -0.35
+ 10.42
DFC (Tn)
3.827
4.09
4.17 0.64
0.64
+ x= 2.04 m -5.83
-3.91