Fuerzas internas
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Fuerzas internas
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SoluciónReacciones de los soportesMiembro AB+→∑ F x= 0; N B– 266.7kN = 0N B= 266.7kN+↑∑ F y= 0;200kN – 200kN - V B= 0V B= 0∑ M B= 0; M B– 200kN(4m) – 200kN(2m) = 0M B= 400kN.m