QUINTA PRÁCTICA CALIFICADAANÁLISIS ESTRUCTURAL I

PROBLEMA 01.- En la estructura cargada como se muestra por el método de Kani con desplazamiento horizontal, determinar:

a) Los momentos en las barras

b) Las reacciones en los apoyos

c) Los diagramas en el centro de las vigas

d) Las deflexiones en el centro de las vigas

A.- CALCULO DE MOMENTOS EN LOS EXTREMOS DE LAS BARRAS.

1. MOMENTOS DE EMPOTRAMIENTO

ME12= -10.331 T-mME21= 10.331 T-mME34= -15.496 T-mME43= 15.496 T-mME45= -12.397 T-mME54= 12.397 T-m

2. MOMENTOS DE PISO

MPISO12=3.05∗103

=10.167

MPISO35=3.66∗(10+15)

3=30.500

3. COEFICIENTES DE GIRO

μij=

−12

∗K ij

∑ K

NUDO 1

K12=30∗553

12∗520=799.880 μ12=−0.392

K13=30∗303

12∗305=221.312 μ13=−0.108

∑ K=1021.192

NUDO 2

K21=30∗553

12∗520=799.880 μ21=−0.392

K24=30∗303

12∗305=221.312μ24=−0.108

∑ K=1021.192

NUDO 3

K31=30∗303

12∗305=221.311 μ31=−0.042

K34=40∗703

12∗520=2198.718μ34=−0.412

K36=40∗303

12∗366=245.902μ36=−0.046

∑ K=2665.932

NUDO 4

K42=30∗303

12∗305=221.312μ42=−0.021

K45=35∗653

12∗520=1540.365μ45=−0.151

K47=40∗503

12∗366=1138.434 μ47=−0.112

K43=40∗703

12∗520=2198.718μ43=−0.216

∑ K=5098.829

NUDO 5

K54=35∗653

12∗520=1540.365μ54=−0.456

K58=

34∗40∗283

12∗366=149.945μ58=−0.044

4. COEFICIENTES DE DESPLAZAMIENTO

γ ij=

−32

∗K ij

∑ K

K13=30∗303

12∗305=221.312 γ13=−0.750

K24=30∗303

12∗305=221.312 γ24=−0.750

∑ K=442.624

K36=40∗303

12∗366=245.902 γ36=−0.240

K47=40∗503

12∗366=1138.434 γ 47=−1.113

K58=

34∗40∗283

12∗366=149.945 γ 58=−0.147

∑ K=1534.281

4 3 -10.331 -0.392 -10.331 10.331 -0.392 10.331 -0.108 -6.515 -4.050 -0.108 1.483 6.118 1.798 -1.116 1.685 8.776 -2.505 0.495 2.418 9.246 -3.752 -0.690 2.547 9.308 -4.056 -1.034 2.564 9.311 -4.119 -1.117 2.565 9.310 -4.130 -1.135 2.565 9.309 -4.131 -1.138 2.565 9.309 -4.131 -1.138 2.565 9.309 -4.131 -1.138

2.565 -1.138

-0.750 10.167 -0.750

-8.346 0.651 -8.346 -0.057-10.686 1.273 -10.686 0.628-10.480 1.134 -10.480 0.945-10.280 0.976 -10.280 1.050-10.210 0.922 -10.210 1.078-10.192 0.907 -10.192 1.085-10.188 0.904 -10.188 1.086-10.187 0.903 -10.187 1.087-10.187 0.903 -10.187 1.087-10.187 0.903 -10.187 1.087

1 -0.042 5 -0.021 2 -15.496 -0.412 -15.496 15.496 -0.216 3.099 -0.151 -12.397 12.397 -0.456 12.397 -0.046 6.384 -0.586 -0.112 -0.410 -5.653 -0.044 0.713 12.483 6.459 -0.304 4.515 -3.537 -0.545 1.394 11.122 9.722 3.349 6.797 -6.315 -0.341 1.242 9.579 10.801 5.041 7.550 -7.513 -0.609 1.069 9.045 11.090 5.600 7.753 -7.915 -0.725 1.010 8.900 11.159 5.751 7.801 -8.025 -0.764 0.994 8.866 11.174 5.786 7.812 -8.052 -0.774 0.990 8.859 11.178 5.794 7.814 -8.058 -0.777 0.989 8.857 11.178 5.796 7.814 -8.059 -0.777 0.989 8.857 11.178 5.796 7.814 -8.059 -0.778

0.989 5.796 -0.778

-0.240 30.500 -1.113 -0.147-7.353 -34.098 -4.503 0.273-8.958 -41.542 -5.487 2.422-9.413 -43.654 -5.766 3.048-9.526 -44.175 -5.834 3.245-9.551 -44.293 -5.850 3.299-9.556 -44.317 -5.853 3.312-9.557 -44.322 -5.854 3.315-9.558 -44.323 -5.854 3.316-9.558 -44.323 -5.854 3.316-9.558 -44.323 -5.854 3.316

6 -0.500 0.000 0.000 0.000

LOS MOMENTOS FINALES EN LOS EXTREÑOS DE LAS BARRAS SON:

M12 = -10.331 + 2 * 9.309 + -4.131 = 4.155 TmM21 = 10.331 + 2* -4.131 + 9.309 = 11.377 TmM34 = -15.496 + 2* 8.857 + 11.178 = 13.397 TmM43 = 15.496 + 2* 11.178 + 8.857 = 46.710 TmM45 = -12.397 + 2* 7.814 + -8.059 = -4.827 TmM54 = 12.397 + 2* -8.059 + 7.814 = 4.094 TmM13 = 0.000 + 2* 2.565 + 0.903 + -10.187 = -4.155 TmM31 = 0.000 + 2* 0.903 + 2.565 + -10.187 = -5.817 TmM24 = 0.000 + 2* -1.138 + 1.087 + -10.187 = -11.377 TmM42 = 0.000 + 2* 1.087 + -1.138 + -10.187 = -9.152 TmM36 = 0.000 + 2* 5.796 + -9.558 + 0.000 = -7.580 TmM63 = 0.000 + 2* 0.000 + 0.989 + -9.558 = -8.569 TmM47 = 0.000 + 2* 5.796 + -44.323 + 0.000 = -32.731 TmM74 = 0.000 + 2* 0.000 + 5.796 + -44.323 = -38.527 TmM58 = 0.000 + 2* -0.778 + 3.316 + -5.854 = -4.093 TmM85 = 0.000 + 2* 3.316 + -0.778 + -5.854 = 0.000 Tm