2 Beams internal forces

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Aerospace Structure – Exercises – 1 – Exercises #2 Beams: internal forces Aerospace Structure – Exercises – 2 – Beams: internal forces – Preliminary Notes Internal forces will be evaluated by applying equilibrium equations. The stress system on beam sections will be equivalent to the internal forces, while the internal forces will equilibrate the external load. To determine the sign of internal forces it can be considered a reference coordinate with Z axes directed as the normal outgoing from the section. On section A, internal forces will be positive if their directi ons are coincident with the right-handed coordinate system that is assigned to the section itself. On se ction B positive internal forces are opposite to those in section A. Moments will follow the right hand rule. Equal and opposite on the two sections. SIGN MUST BE DEFINED             A yzxzz A zzy A zzx A zzy A xzx A zzz dAxyM xdAM ydAM dAT dAT dAT       A B Aerospace Structure – Exercises – 3 – 0 dzrdTTT yyyy 0 y y rdz dT    0 0 y z yy TdzzrzT   Equilibrium of an infinitesimal element of the beam (length dz) leads to the development of very effective equations for the evaluation of internal forces. A generic beam element is in equilibrium under the action of th e internal forces that act a z=0, the internal forces that act at z=z and distributed load per unit l ength r z, rx, ry, depending on the loads direction. A differential relation can be obtained by considering an eleme nt of infinitesimal length dz: a differential equilibrium equation.   02  dzdzrdzdTTMdMM yyyxxx 0 y x Tdz dM    0 0 x z yx MdzzTzM   0 dzrdTTT xxxx 0 x x rdz dT    0 0 x z xx TdzzrzT    02  dzdzrdzdTTMdMM xxxyyy 0 y x Tdz dM    0 0 y z xy…