lec19 (1)

# Lec19(1) - General solution procedure Elasticity condition(no dissipation d ψ = δ W reflecting that dD =(this is the result from analyzing the TD

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Unformatted text preview: General solution procedure Elasticity condition (no dissipation): d ψ = δ W reflecting that dD = (this is the result from analyzing the TD as done in class) ∂ψ ∂ψ ∂ψ • Step 1 : Express d ψ ( x , x ,..) = dx + dx + ... = dx 1 2 1 2 i ∂ x 1 ∂ x 2 ∂ x i ∂ F ∂ F ∂ψ • Step 2 : Express δ W ( ξ 1 , ξ 2 ,..) = d ξ 1 + d ξ 2 + ... = d ξ j ∂ξ 1 ∂ξ 2 ∂ξ j ∂ψ ∂ψ • Step 3 : Solve equations ∂ x i dx i = ∂ξ j d ξ j ∀ dx i , ∀ d ξ j Collect all terms dx i and d ξ j and set the entire expression to zero. In EQ, the expression must be satisfied for all displacement changes dx i , d ξ j Example II: Truss structure (1) Problem statement: Structure of three trusses with applied force F d : Forces in each truss 1 1 ξ F d δ 1 , N 1 δ 2 , N 2 Distance L =1 between the trusses N = k δ 1 1 N = k δ δ 3 , N 3 Trusses behave 2 2 like springs Goal: Calculate displacements δ i , ξ F d N 3 = k δ 3 Example II: Truss structure (2) Rigid bar: If two displacements...
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## This note was uploaded on 11/29/2011 for the course CIVIL 1.018j taught by Professor Markusbuehler during the Fall '08 term at MIT.

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Lec19(1) - General solution procedure Elasticity condition(no dissipation d ψ = δ W reflecting that dD =(this is the result from analyzing the TD

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