SOLUCIÓN

4
 SOLUCIÓN – PRÁCTICA ALUMNO: RAMOS GOMEZ NEFER VIGA Y LOSA – INFINIT AMENTE RÍGIDA DATOS: E= 2000000.000 m1 = 10.000  Tn*Seg^2/ m m2 = 10.000  Tn*Seg^2/ m m3 = 10.000  Tn*Seg^2/ m 3 5 3 5 3 5 5 5 5 0.6 0.4

description

SAETGRSZHTSFDGSD

Transcript of SOLUCIÓN

SOLUCIN PRCTICA

ALUMNO: RAMOS GOMEZ NEFER

VIGA Y LOSA INFINITAMENTE RGIDA

35

35

3

5555

0.6060

0.40

DATOS: E= 2000000.000m1=10.000Tn*Seg^2/m

m2=10.000Tn*Seg^2/m

m3=10.000Tn*Seg^2/m

Hallar. T, F y W.Solucin: Hallamos la inercia 1,2 y 3F`cEI1I2I3H1H2H3

280.0002000000.0000.0070.0070.0073.0003.0003.000

Hallamos el K1, K2, Y K3. PISO 1PISO 2PISO 3

N COL.K1N COL.K2N COL.K3

9.00057600.0009.00057600.0006.00038400.000

K1=57600.000Tn/m

K2=57600.000Tn/m

K3=38400.000Tn/m

HALLAMOS, T, F, y W.Frecuencia

f(1) = 12.079

f(2) =12.079

f(3) =9.862

Periodo

T(1) =0.083

T(2) =0.083

T(3) =0.101

P(t)1 = 50.000

P(t)2 = 20.000

P(t)3 = 10.000

w^2 (1) = 16898.124

w^2 (2) = 6982.047

w^2 (3) =1079.829

Matriz rigidez.

w1=129.993

-168981.24057600.0000.000-0.000650.000

57600.000-72981.240-38400.000X-0.0008=20.000

0.000-38400.000-130581.2400.000210.000

X = N1*sen(w*t)

x(1) =0.0006v(1) =0.017a(1) =-9.408

x(2) =0.0008v(2) =0.024a(2) =-13.330

x(3) =0.0001v(3) =0.017a(3) =-1.576

w2=83.559

-69820.47057600.0000.000-0.000450.000

57600.00026179.530-38400.000X0.0004=20.000

0.000-38400.000-31420.470-0.000810.000

X = N1*sen(w*t)

x(1) =-0.0002v(1) =-0.026a(1) =1.576

x(2) =0.0002v(2) =0.027a(2) =-1.682

x(3) =-0.0007v(3) =0.039a(3) =4.636

w3=32.861

-10798.29057600.0000.0000.000150.000

57600.00085201.710-38400.000X0.0009=20.000

0.000-38400.00027601.7100.001610.000

X = N1*sen(w*t)

x(1) =0.0000v(1) =-0.003a(1) =-0.044

x(2) =0.0004v(2) =-0.027a(2) =-0.391

x(3) =-0.0003v(3) =-0.051a(3) =0.326