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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