Hoja1PROBLEMA 2.-3.- Calculo de momentos de empotramiento perfecto

MAB=-10.67MBA=10.67MBC=-7.50MCB=7.50MCD=-9.00MDC=9.001.- Calculo de las rigidez:2.- Factor de distribucion:KAB=0.188FDBA=0.429KBC=0.25FDBC=0.572KCD=0.167FDCB=0.600FDCD=0.4004.-Ejecucion del metodo0.4290.5720.60.4M-10.6710.67-7.57.5-991ra distr.0-1.22045-1.951.06050.439501er transp.-0.6100.53025-0.9700.219752da distr.0-0.20414625-0.330.690.2902do transp.-0.100.34-0.1600.143ra distri.0-0.13-0.210.115277680.047774203er transp.-0.0700.05763884-0.1100.02388714ta distrib.0-0.02219095-0.040.070.0304to transp.-0.0100.04-0.0200.025ta distrib.0-0.01-0.020.012530830.0051931205to transp.-0.0100.00626541-0.0100.002596566ta distrib.0-0.0024121800.01006to transp.0000007ma distrib.0000.001362120.00056450MF-11.479.07-9.078.19-8.199.41

5.- Obtencion de fuerza cortante y momento flexionante

VIGA AB:12.502

VA=6.49VB1=9.51Mi=16

VIGA BC:12.05157.38

VB2=8.67VC1=6.33Mi=15

VIGA CD:24.185.703

VC2=8.65VD=9.35Mi=13.50

6228 A

B

B

C

C

D

2268

Hoja2

Hoja3