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Statically Indeterminant Truss B P C L A D L Overall Equilibrium P P P P We can find all the external reactions but the truss is still statically indeterminant since there are too many members to solve for all the forces Take out, for example, BD B C P R B D A P D R P P Then for the remaining truss we can solve for all the forces in terms of P and R . R is our redundant force B C P R R A P D P P ∑F For example, at B x =0 R 2 2 ∑ Fy = 0 FBC = − FBC R FAB = − FAB R 2 2 Now, for the entire truss Fi 2 Li U =∑ i =1 2 Ai Ei 6 where we will take From the Principle of Least Work 6 F L ∂F ∂U =∑ i i i =0 ∂R i =1 AE ∂R Ai = A Ei = E 6 F L ∂F ∂U =∑ i i i =0 ∂R i =1 AE ∂R member ∂Fi / ∂R Fi Li ∂Fi / ∂R −R 2 / 2 − 2/2 RL / 2 L −R 2 / 2 − 2/2 RL / 2 L − P+R 2/2 − 2/2 L P 2 /2+ R/2 L −R 2 / 2 Li AB Fi L BC CD DA AC 2L BD 2L ( ) − 2/2 P 2+R 1 R 1 ( RL / 2 ( L 2 R+P 2 RL 2 Fi Li ∂Fi ( 4.83RL + 2.707 PL ) = =0 ∑ AE i =1 AE ∂R R = −0.56 P 6 ) ) Now, suppose that in assembling the truss, member BD was slightly short so that there was a lack of fit ∆R B C R R A P D P P In this case R does complimentary virtual work and we have 6 F L ∂F ∂U δU = δ R = ∑ i i i δ R = δ R∆ R ∂R i =1 AE ∂R which must be satrisfied for all δR so Fi Li ∂Fi = ∆R ∑ i =1 AE ∂R 6 ( 4.83RL + 2.707 PL ) = ∆ AE R = −0.56 P + R AE ∆R 4.83L so the force in BD is less compressive, which is what we expect since we have to pre-tension BD to get it to fit.