3 Beams statically indeterminate systems

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Aerospace Structure – Exercise 3 1 Exercises #3 Statically Indeterminate Systems Aerospace Structure – Exercise 3 2 Statically Indeterminate Systems - EX. 1 Evaluate the indeterminate reaction in the A support of the following structure: DATA L = 2000 mm L 1 = 500 mm r = 2.0 N/mm E=200000 Mpa J = 500000 mm 4 SOLUTION a) Problem set-up b) True System Internal forces in the domain 0 < z 1 < L 1 ( ) ( ) ( ) 11 2 10 11111 1 0 111 0 2 1 1 Lzper rz dz zTzM rz rdz zT z x z y << − == − =− = ∫ ∫ Internal forces in the domain 0 < z 2 < L 2 A L L1 r rL 1+Y 0 < z 2 < L 2 (1/2)rL 1 + z y Ty Mx L2 L1 r Y z1 y1 z2 y2 + z y Ty Mx Aerospace Structure – Exercise 3 3 ( ) ( ) ( ) 2 2 2 2 2 1 2 1 0 2 2 2 2 2 2 2 1 2 2 1 2 1 0 0 1 1 1 2 2 2 z y z f T z rdz Y rL rz Y rL per z L M T z dz rL rz Yz rL z rL = − − − = − + + < <   = − = − + + +     ∫ ∫ c) Unit dummy system Expression of internal forces: 11 1 1 00 0 Lzper M T x y <<=′ =′ Lzper zM T x y <<⋅−=′ −=′ 2 22 2 01 1 d) PVWC External Work: The external work for the variation that corresponds to the dummy systems is zero, as the application point of the reaction has a null displacement. 0* =eWδ Complementary deformation work: Only flexural deformation work is considered, in bo th beam stretches. The work in the first domain is zero. 2 0 22 2 0 22 1 0 11* 221 dz EJ MMdz EJ MMdz EJ MMW L xx L xx L xx d ∫∫∫ ′=′+′=δ PCVW states ** de WW δδ = ⇒      +++=+++= =     +++=     +++ ∫∫ 2 2 2 1 3 2 1 3 2 4 2 0 2 2 2 1 3 2 1 3 2 4 2 0 22 2 1 2 2 1 2 2 3 222 0 2 12 12 2 2 4 1 3 1 3 1 8 11 4 1 3 1 3 1 8 11 2 1 2 11 2 1 2 11 2 22 LrL LrL YL rL EJ zrL zrL Yz rz EJ dz zrL zrL Yz rz EJ dz zrL zrL Yz rz EJ L zL The equation can be solved to find the reaction Y. By substituting the numerical data, considering L 2=L-L1=1500 mm, the…