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DATE PURPOSE fabia Minanda S. CUE COLUMN NOTES P= (746 50 lb DCL 4 Ay A Ax D F BD A 3 F FCD y + FEF y B G / 2 2 P=214,92 lb 2 4 FCD triangulo FCD =0 = FCD = EFy E = - 4/5 FCD Ay=D = =0 (4) - 214,92 (2) FEF (4) = - 429.84 + =0 FEF = 107,46 KN - 3/5 FCD (4) - 107,46 2 4 92(4)- 4FBD - 214,92 + 859,68 4FBD FBD= 161,19 EF3=0 A3 FEF + - SUMMARY A3= -53,73 KN de em FCD= + oj - (53,73 KN) FEF : 107, 46 XN FBD = 161, 19 KN