Kinetic Energy and Potential Energy - Dynamics of Aerospace Structures | ASEN 5022, Study notes of Aerospace Engineering

Download Kinetic Energy and Potential Energy - Dynamics of Aerospace Structures | ASEN 5022 and more Study notes Aerospace Engineering in PDF only on Docsity! ASEN 5022 - Spring 2005 Dynamics of Aerospace Structures Lecture 15: March 10 Finite Element Modeling of Beam Bending Vibration Let’s consider the modeling of a beam under general boundary conditions by the finite element modeling. kw1 kw2 kµ1 kµ2EI, m(x) x z w L Beam with unknown boundary conditions Potential energy of four unknown springs: Vs = 12{ kw1 w(0, t)2 + kθ1 w(0, t)2x + kw2 w(L , t)2 + kθ2 w(L , t)2x (4) Hamilton’s principle ∫ t2 t1 [δT − δVb − δVs + δW ] dt = 0 (5) From here on, FEM modeling differs from continuum modeling! Key departure in FEM modeling of beams from continuum modeling: 1. Instead of carrying out the variation (the δ-process of Hamilton’s principle(5) in terms of the continuum variable w(x, t), the energy expressions, (T, V, W, etc.), are approximated on a completely free beam with an arbitrary length , area A, the bending rigidity E I and mass density ρ. 2. The interpolation of the transverse displacement w(x, t) over the completely free beam segment 0 ≤ x ≤ is chosen to satisfy the homogeneous differential beam equation, i.e., E Iw(x, t)xxxx = 0 (6) if possible. If not, it is chosen such that as the length of the beam gets smaller and smaller, it approximately satisfies the homogeneous equation. The FEM beam bending element 1. Observing that w(x, t) and w(x, t)x are linearly independent from the result of variational formulation, we specify them at nodes 1 and 2 in the previous figure. Namely, two discrete variables at node 1: w(x1, t) and w(x1, t)x two discrete variables at node 2: w(x2, t) and w(x2, t)x (8) 2. For convenience, we introduce a local coordinate system or ele- mental coordinate system, (x, y) and set x1 = 0 and x2 = . Substituting (8) into the interpolation function(7), we obtain w1(t) = w(x1 = 0, t) = c0(t) + c1(t)0 + c2(t)02 + c3(t)03 θ1(t) = w(x1 = 0, t)x = c1(t) + 2c2(t)0 + 3c3(t)02 w2(t) = w(x2 = , t) = c0(t) + c1(t) + c2(t) 2 + c3(t) 3 θ2(t) = w(x2 = , t)x = c1(t) + c2(t)2 + c3(t)3 2 (9) Solving for the coefficients and backsubstituting into (7), one obtains w(x, t) = H1(ξ) w1(t) + H2(ξ) θ1(t), ξ = x/ + H3(ξ) w2(t) + H4(ξ) θ2(t) H1(ξ) = (1 − 3ξ2 + 2ξ3), H2(ξ) = (ξ − 2ξ2 + ξ3) H3(ξ) = (3ξ2 − 2ξ3), H4(ξ) = (−ξ2 + ξ3) (10) For notational clarity, we express (10) in the form w(x, t) = N(ξ) q(t) N(ξ) = 〈 H1 H2 H3 H4 〉 q(t)T = 〈 w1(t) θ1(t) w2(t) θ2(t) 〉 (11) 1 2 θ θl 2 w w1 2 l A completely free beam element and its nodal degrees of freedom Discretization of Elemental Strain (Potential) Energy V elb = 12 ∫ 1 0 E I (ξ) 3 w(ξ, t)2ξξ dξ = 12 ∫ 1 0 E I (ξ) 3 q(t)T [NTξξ · Nξξ ] q(t)dξ = 12q(t)T [ ∫ 1 0 E I 3 NTξξ · Nξξ dξ ] q(t) = 12q(t)T [kel] q(t) (16) where we utilized: w(ξ, t)ξξ = N(ξ)ξξ q(t) (17) Elemental mass matrix, mel m = ρ A   13 35 11 210 9 70 −13 420 11 210 2 105 13 420 − 2 140 9 70 13 420 13 35 −11 210 −13 420 − 2 140 −11 210 2 105   (18) Elemental stiffness matrix, kel k = E I 3   12 6 −12 6 6 4 2 −6 2 2 −12 −6 12 −6 6 2 2 −6 4 2   (19) Approximation of elemental external energy (3) δW el = ∫ 0 f(x, t) δw(x, t) dx(= ∫ 1 0 f(ξ, t) N(ξ) δq(t) dξ) + Q1 δw(0, t) + Q2 δw(L , t) + M1 δw(0, t)x + M2 δw(L , t)x (3) Let’s utilize the approximation(10): w(x, t) = N(ξ) q(t) w(x, t)x = 1 N(ξ)ξ q(t) (20) and observe that w(0, t) = q1(t), w( , t) = q3(t), w(0, t)x = q2(t), w( , t)x = q4(t) (21) so that we have δW el = δq(t)T { ∫ 0 NT f(x, t) dξ} + Q1 δq1(t) + Q2 δq3(t) + M1 δq2(t) + M2 δq4(t) = δq(t)T {fel }, fel = { ∫ 0 NT f(x, t) dξ} + 〈 Q1 M1 Q2 M2 〉T (22) 1.A Partition a beam into many elements. element element 3element 2 global node 1 2 3 4 5 element 1 (2)(1) (3) (4) (. . . ) Partition a beam into finite elements elelental node global node 1 2 3 42 3 1 2 1 2 1 2 element 3element 2element 1 u1 u2 u3 u4u2 u3 q1 (1) q2 (1) q1 (2) q2 (2) q1 (3) q2 (3) Partitioning and establishing the Boolean relation between the global degrees of freedom, u, and elemental degrees of freedom, q 1.B Establish the relation between the elemental and assembled(global) degrees of freedom. For element 1: q(1) = {q(1)1 q(1)2 } = {u1 u2 } For element 2: q(2) = {q(2)1 q(2)2 } = {u2 u3 } For element 3: q(2) = {q(3)1 q(3)2 } = {u3 u4 } ⇓  q(1) q(2) q(3)   =   I 0 0 0 0 I 0 0 0 I 0 0 0 0 I 0 0 0 I 0 0 0 0 I     u1 u2 u3 u4   ⇒ For general case: q = L ug (24) 3.A Sum up the total system kinetic energy. T total = n∑ el=1 T el 1 2{q̇(1)}T [m(1)]{q̇(1)} + 12{q̇(1)}T [m(1)]{q̇(1)}+ ... + 12{q̇(n)}T [m(n)]{q̇(n)} =   q̇(1) q̇(2) .. q̇(n)   T   m(1) 0 ... 0 0 m(2) ... 0 0 0 ... 0 0 0 .. m(n)     q̇(1) q̇(2) .. q̇(n)   T T total = 12 q̇ m q̇ (25) • Total assembled system external work: δW total = n∑ el=1 {δq(t)(el)}T {fel} = δqT f = δuTb {LT f } δW total = δuTg fg (28) Euler-Lagrange’s equations of motion Lagrangian: L = T total − V total = 12 u̇gMg u̇g − 12ugKg ug (29) From the generic equation of motion d dt ∂L ∂k − ∂L ∂qk = Qk (30) we obtain the FEM equation of motion from (29) and (30): Mg üg + Kg ug = fg (31)