problema 27
Post on 12-Aug-2015
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Problema 271.1
1.2
∑ FZ = 0 RAZ + DRZ = 354 + 428.68∑ FY = 0 RAY + RDY = 145 + 147.5
∑ MA = 0
MY = 0 345 (0.05) + 428.68 (0.1) + RDZ (0.15) = 0Mx = 0 145 (0.05) + 145.5 (0.1) = RDY (0.15)
RDZ = -403.78 N → RAZ + RDZ = 782.58RDY = 146.47 N → RAY + RDY = 292.5
RAZ = 378.9RAY = 145.83
RD = ( RX ; RY ; RZ )RD = ( 0 ; -146.67 ; 403.78 )
1.3 Torque en el punto ``C’’
MC + TC = MB
15.32 + TC =17.7TC = 2.38
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