General solution procedure Elasticity condition (no dissipation): dψ=δW reflecting that dD = 0 (this is the result from analyzing the TD as done in class) ∂ψ∂ψ ∂ψ • Step 1: Express dψ(x , x ,..) = dx + dx + ... = dx1 2 1 2 i∂x1 ∂x2 ∂xi ∂F ∂F ∂ψ • Step 2: Express δW (ξ1,ξ2,..) = dξ1 + dξ2 + ... = dξj∂ξ1 ∂ξ2 ∂ξj ∂ψ ∂ψ • Step 3: Solve equations ∂xi dxi = ∂ξjdξj ∀dxi ,∀dξj Collect all terms dxi and dξj and set the entire expression to zero. In EQ, the expression must be satisfied for all displacement changes dxi,dξjExample II: Truss structure (1) Problem statement: Structure of three trusses with applied force Fd : Forces in each truss 11 ξ0Fd δ1, N1 δ2, N2 Distance L=1 between the trusses N = kδ1 1 N = kδ δ3, N3 Trusses behave 2 2 like springsGoal: Calculate displacements δi ,ξ0 for given Fd N3 = kδ3Example II: Truss structure (2) Rigid bar: If two displacements δ1,δ2 are specified can calculate the other displacements (kinematic constraint): Deformation δ11 δ21 δ3ξ0Therefore: δ −δ 3 1δ3 =δ1 + 22 1 = 2δ2 −δ1 ξ0 =δ2 −δ11 2 2Example II: Truss structure (3) Solution procedure: Elasticity condition (no dissipation): dψ=δW • Step 1: dψ(δ1,δ2) = 1 k[(4δ1 − 4δ2 )dδ1 + (− 4δ1 + 10δ2 )dδ2 ]2 • Step 2: δW (ξ1) = Fd ⎢⎣⎡− 12 dδ1 + 23 dδ2 ⎥⎦⎤ • Step 3: Solve equations dψ=δW ∀dxi ,∀dξj 12 k[(4δ1 − 4δ2 )dδ1 + (− 4δ1 +10δ2 )dδ2 ]= Fd ⎡⎢⎣− 12 dδ1 + 23 dδ2 ⎤⎥⎦ ⎡⎛⎜2kδ− 2kδ+ 1 Fd ⎞⎟dδ + ⎛⎜− 2kδ+ 5kδ− 3 Fd ⎞⎟dδ⎤⎥ = 0⎢1 2 1 1 2 2⎣⎝ 2 ⎠ ⎝ 2 ⎠⎦ ! !=0 =0 for elastic EQExample II: Truss structure (4) This results in linear system of equations: 1 ⎛−⎜⎜⎝ δδ2 ⎞⎟⎟⎠ ⎞ ⎟⎟⎠ ⎛ ⎞ ⎞ − d ⎟⎟⎠ − ⎛ ⎜⎜⎝ 1 − ad bc = ⎜⎜⎝ ⎟⎟⎠ − − ⎛ ⎜⎜⎝ 1 − ⎞ ⎛ ⎟⎟⎠ ⎜⎜⎝ /2 /2 d d F 3F = b 2 k 2 k M Solve for the unknown variables δ1,δ2,.. Note that (forming the inverse of a 2x2 matrix): 2k 5k a b c d c a This can be used to calculate M −1 to solve for δ1,δ2Example II: Truss structure (5) This results in: d 2 ⎛− M −1 Solve for the other unknown variables (utilize kinematic relationships and the ⎞⎜⎟⎟⎜2⎠⎝ spring equations): δ1 δ2 d 5 ⎟= ⎞⎟ ⎠ F /2 ⎛ ⎞ ⎛ ⎛ F 1 = ⎜⎜⎝ ⎟⎟⎠1/12⎞ ⎜⎜⎝1/3 ⎜⎜⎝ ⎟⎟⎠ d k k 2 6 3F /2 δ3= 7/12 Fd k 1/12 Fd N = iδN k = 1 i 1/3 Fd N == 2 Fd ξ0= 11/24 k N 3 7/12
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