examen de analisis 2
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analisisTRANSCRIPT
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EXAMEN DE ANALISIS ESCTRUTURAL II
UNIDADESmkg
b (cm) = 0.3h(cm) = 0.6
Io = 0.0054 E= 218819.788867 210 E= 218819.788867f`c= 3 klb/in2 I= 0.0108 I= 0.0081
L= 6 L= 8MEPi= -24 MEPi= -117.187500MEPj= 36 MEPj= 70.312500
100 kg20kg 3m 5m 20kg
2 Io 1.5 Io 2.5 IoA B C D
6m 8m 5mK1 = 2EI/L K2 = 2EI/L K3= 2EI/L
MATRIZ DE RIGIDEZ GIROS 2° MIEMBRO2K1 K1 θA -(MEPAB)K1 2(K1 + K2) K2 θB -(MEPBA + MEPBC)
K2 2(K2 + K3) K3 θC -(MEPCB + MEPCD)K3 2(K3 + K4) K4 θD -(MEPDC + MEPDE)
K4 2K4 θE -(MEPED)
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MATRIZ DE RIGIDEZ GIROS 2° MIEMBRO1575.502480 787.751240 0 0 0 θA 24787.7512399 2461.722625 443.110072 0 0 θB 81.187500
0 443.1100725 3249.473865 1181.626860 0 θC -28.6458330 0 1181.626860 3781.205952 708.9761159 θD 208.3333333333330 0 0 708.976115931 1417.952232 θE -250
RESOLVIENDO EL SISTEMA DE ECUACIONES CON METODOS ITERATIVOS - METODOS DE SOBRERELAJACION
ϐ = 1.09
K θA ΘB ΘC Θd Θe0 0 0 0 0 01 0.016604 0.030157 - 0.014091 0.064856 -0.227525 2 -0.001325 0.036461 - 0.039467 0.114162 -0.233920 3 -0.003148 0.041508 - 0.057476 0.117166 -0.234981 4 -0.005734 0.045489 - 0.057638 0.117168 -0.234887 5 -0.007671 0.045838 - 0.057676 0.117162 -0.234892 6 -0.007687 0.045820 - 0.057667 0.117160 -0.234891 7 -0.007676 0.045816 - 0.057667 0.117160 -0.234891 8 -0.007675 0.045816 - 0.057667 0.117160 -0.234891 9 -0.007675 0.045816 - 0.057667 0.117160 -0.234891 10 -0.007675 0.045816 - 0.057667 0.117160 -0.234891
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METODO DE CROSS:
A) RIGIDEZ DEL ELEMENTO
TRAMO I L KAB 0.0108 6 0.00135BC 0.0081 8 0.0010125CD 0.0135 5 0.0027DE 0.0162 10 0.001215
B) FACTOR DE DISTRIBUCION
TRAMO K FDAB 0.00135 0.57BC 0.0010125 0.43
0.0023625
BC 0.0010125 0.27
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CD 0.0027 0.730.0037125
CD 0.0027 0.69DE 0.001215 0.31
0.003915
C) CALCULO DE LOS MOMENTOS DE EMPOTRAMIENTO
TRAMO MEAB -24BA 36
BC -117.187500CB 70.312500
CD -41.666667DC 41.666667
DE -250ED 250
0.57 0.43 0.27 0.73 0.69 0.31-24 36 -117.187500 70.312500 -41.666667 41.666667 -250 250
-81.1875 28.645833 -208.33333381.1875 -28.6458333333333 208.333333333333
46.39285714 34.79464286 -7.8125 -20.8333333 143.6781609 64.655172413793123.19642857 -3.90625 17.3973214286 71.83908046 -10.4166667 32.3275862
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E= 218819.788867 E= 218819.788867I= 0.0135 I= 0.0162L= 5 L= 10
MEPi= -41.666666666667 MEPi= -250MEPj= 41.6666666666667 MEPj= 250
200 kg5m 5m
3 Io E
10mK4 = 2EI/L
M EC. PEND. DEFLEX.AB 2EI/L(2 θA + θB - 3Ψ) + MEP
BA 2EI/L( θA + 2θB - 3Ψ) + MEP
BC 2EI/L(2 θB + θC - 3Ψ) + MEP
CB 2EI/L( θB + 2θC - 3Ψ) + MEP
CD 2EI/L( 2θC+ θD - 3Ψ) + MEP
DC 2EI/L( θC+ 2θD - 3Ψ) + MEP
DE 2EI/L( 2θD+ θE - 3Ψ) + MEP
ED 2EI/L( θD+ 2θE - 3Ψ) + MEP
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GIROSθA -0.057667
θB 0.045816
θC -0.057667
θD 0.117160
θE -0.234891
MATRIZ DE MOMENTO (debido a giros) M finales1575.502480 787.751240 0 0 0 AB787.7512399 1575.502480 0 0 0 BA
0 886.220144913215 443.110072 0 0 BC0 443.110072456608 886.220144913 0 0 CB0 0 2363.25371977 1181.6268599 0 CD0 0 1181.626860 2363.253720 0 DC0 0 0 1417.9522319 708.976115931 DE0 0 0 708.976116 1417.952232 E
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ASLO PS SOLO FALTA REMPLAZAR EN LA FORMULAA DE EC. PENDIENTE ES EL CUADRO QUE ESTA DE AMARILLO LOS GIROS ESTAN BIEN LOS COMPROBEE
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ASLO PS SOLO FALTA REMPLAZAR EN LA FORMULAA DE EC. PENDIENTE ES EL CUADRO QUE ESTA DE AMARILLO LOS GIROS ESTAN BIEN LOS COMPROBEE