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8 Displacement based approaches for discrete systems

Divisi per argomento di Strutture Aerospaziali per il corso di Aerospace Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Divisi per argomento di Strutture Aerospaziali per il corso di Aerospace Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Aerospace Structure – Exercises 8 1 Exercises #9 Displacement-based approaches for discrete systems Aerospace Structure – Exercises 8 2 Displacement-based approaches for discrete systems - Problem #1 For the systems of cart that is presented in the following figure: a) evaluate the stiffness matrix of the externally unconstrained system by means of direct stiffness method, if k1 = k2 = k3 = k b) add a constraint U 3 = 0 by means of Lagrange multiplier method and solve the system SOLUTION a) UNCONSTRAINED STIFFNESS MATRIX THROUGH DIRECT ST IFFNESS METHOD i) Expanded matrices of the elements The application of direct stiffness method consists in the evaluation of the expanded matrix, which contain the forces acting on spring member at the i-th node due to a unit variation of the j-th d.o.f. Expanded matrix for the first spring element The first spring member k 1, connect the d.o.f. U 1 and U 2. The term 1,1 of the expanded stiffness matrix K e (1) will represent the force F 1 transmitted to the spring at the node 1, for a unit variation of U 1. All the other d.o.f. can be considered frozen. Force is positive in the direction consistent with the positive sign of U 1. Hence: K e (1) (1,1) = k (the spring is compressed by U 1 = 1, the force acting on the spring is in the positive direction) The term 1,2 will represent the force F 2 that is transmitted to the spring at the node 1 for a positive unit variation of U 2, all the other d.o.f’s being frozen. If U 2 = 1, the spring is tensed and the force transmitted at the spring at node 1 will be negative. Hence: K e (1) (1,2) = -k The term 1,3 is zero, because if U 1 and U 2 are frozen, the first spring member is not loaded. The term 2,1 is the force that is transmitted to the spring at node 2, for a unit variation of U 1. In…

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