problema 27

2
Problema 27 1.1 1.2 F Z = 0 R AZ + D RZ = 354 + 428.68 F Y = 0 R AY + R DY = 145 + 147.5 M A = 0 M Y = 0 345 (0.05) + 428.68 (0.1) + R DZ (0.15) = 0 M x = 0 145 (0.05) + 145.5 (0.1) = R DY (0.15) R DZ = -403.78 N R AZ + R DZ = 782.58

Upload: gabriel-gallardo

Post on 12-Aug-2015

25 views

Category:

Engineering


5 download

TRANSCRIPT

Page 1: Problema 27

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 )

Page 2: Problema 27

1.3 Torque en el punto ``C’’

MC + TC = MB

15.32 + TC =17.7TC = 2.38

EN SENTIDO