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CU-Boulder ASEN 5022 - Finite Element Modeling of Vibration of Bars

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ASEN 5022 - Spring 2005Dynamics of Aerospace StructuresLecture 14: 04 MarchFinite Element Modeling of Vibration of BarsLet’s consider the modeling of a bar with general boundary condi-tions by the finite element modeling.EA, m(x)LBar with unknown boundary conditionsM0LMk01L1kk02L2kTheoretical Basis: Same as the continuum bar model!Kinetic energy of the contimuun bar:Tbar=12L0m(x) ut(x, t)2dx, ut(x, t) =∂u(x, t)∂t(1)Potential energy of the continuum bar:Vbar=12L0{ EA(x) ux(x, t)2} dxux(x, t) =∂u(x, t)∂x(2)External energy due to distributed applied force f(x, t):δWbar=L0f(x, t)δu(x, t) dx(3)Kinetic energy of the discrete masses :Ts=12M0˙u20(t) +12ML˙u2L(t)(4)Potential energy of four springs:Vs=12{ k01u20(t) + k02[u0(t) − u(0, t)]2+ kL1u2L(t) + kL2[uL(t) − u(L, t)]2(5)Hamilton’s principlet2t1[δTtotal−δVbar−δVs+δWbar] dt = 0, Ttotal= Tbar+Ts(6)Key departure in FEM modeling of beams from continuum modeling:1. Instead of carrying out the variation (the δ-process of Hamilton’sprinciple(6) in terms of the continuum variable u(x, t), the energyexpressions, (T, V, W, etc.), are approximated on a completely freebar with an arbitrary length ,areaA, and mass density ρ = m(x).2. Ideally, the interpolation of the axial displacement u(x, t) overthe completely free bar segment, 0 ≤ x ≤ , is chosen to satisfy thehomogeneous differential beam equation, i.e.,EAu(x, t)xx= 0 (7)if possible. If not, it is chosen such that as the length of the bargetssmallerandsmaller,itapproximatelysatisfiesthehomogeneousequation.We now consider a completely free, isolated, small segment of a bar.EA, m(x)LBar with unknown boundary conditionsM0LMk01L1kk02L2k1212xA completely free element of length taken out from the continuum baran element isolatedu2u1lxApproximation of u(x, t) over the element segment, <<L:u(x, t) = c0(t) + c1(t)x +c2(t)x2+ ...+cn(t)xn(8)Class of elements:1.Linear Element or Constant Strain Element : Retain the first two termsin (8), viz., (c0, c1) to result in:u(x, t) = c0(t) + c1(t)x (9)2.Quadratic Element or Linear Strain Element : Retain the first threeterms in (8), viz., (c0, c1, c2) to result in:u(x, t) = c0(t) + c1(t)x +c2(t)x2(10)3. Cubic Element or Quadratic Strain Element : Retain the first four termsin (8), viz., (c0, c1, c2, c3) to result in:u(x, t) = c0(t) + c1(t)x +c2(t)x2+ c3(t)x3(11)How does one determine the coefficients,(c0, c1, c2, c3, etc.)?uu12loxx=- /2lx= /2lNode 1Node 2Linear (2-Node) Elementuu12loxx=- /2lx= /2lNode 1Node 2Quadratic (3-Node) Elementu3x=0Bar ElementsThe linear bar element1. We specify the two discrete nodes, viz., nodes 1 and 2 in theprevious figure. Namely,At node 1: u(−/2, t) = u1(t)At node 2: u(/2, t) = u2(t)(12)so that we have the following two equations:u1(t) = c0(t) + c1(t) ∗ (−/2)u2(t) = c0(t) + c1(t) ∗ (/2)⇓u(x, t) = (12− x/) u1(t) + (12+ x/) u2(t)u(ξ, t) =12(1 − ξ) u1(t) +12(1 + ξ) u2(t), ξ = 2x/⇓(13)The Standard form of Linear (Two-Noded) Bar Elementu(ξ, t) = N1(ξ) u1(t) + N2(ξ) u2(t)N1(ξ) =12(1 − ξ)N2(ξ) =12(1 + ξ)− 1 ≤ ξ ≤ 1(14)For notational clarity, we express (14) in the formu(ξ, t) = N(ξ) q(t)N(ξ) =N1(ξ) N2(ξ)q(t)T=u1(t) u2(t)(15)Construction of Linear Bar ElementElemental kinetic energy:Telbar=1212−12m(x) ˙u2(x, t)dx=121−1m(ξ) ˙u2(ξ, t)(12dξ), x =12ξ(16)Elemental potential energy:Velbar=12/2−/2EA(x) u2x(x, t) dx=121−1EA(ξ) (2)2u2ξ(ξ, t)(12dξ)(17)Construction of Linear Bar Element – Cont’dTelbar=121−1m(ξ) ˙u2(ξ, t)(12dξ)=121−112ρ A˙q(t)T[NT· N]˙q(t)dξ=12˙qT[1−112ρ A NT· N dξ]˙q(t)=12˙q(t)T[mel]˙q(t)(18)where we utilized (15), viz.:u(ξ, t) = N(ξ) q(t)(19)Construction of Linear Bar Element – Cont’dVelbar=121−1EA(ξ) (2)2u2ξ(ξ, t)(12dξ)=121−12EA(ξ)q(t)T[NTξ· Nξ] q(t)dξ=12q(t)T[1−12EANTξ· Nξdξ] q(t)=12q(t)T[kel] q(t)(20)Elemental mass matrix, melbarm =ρ A6 2112(21)Elemental stiffness matrix, kelbark =EA 1 −1−11(22)Approximation of elemental external energy (3)δWel=/2−/2f(x, t)δu(x, t) dx= δq(t)T{1−1NTf(x, t)12dξ}= δq(t)T{fel}fel={1−1NTf(x, t)12dξ}(23)Summary of elemental energy expressions:Telbar=12{˙q(t)(el)}T[mel] {˙q(t)(el)}Velbar=12{q(t)(el)}T[kel] {q(t)(el)}δWelbar={δq(t)(el)}T{fel}(24)Question: How do we model a long bar with bar elements?Answer:(1) Partition the long bar into small elements. This needs to beexplained.(2) Generate the elemental energy expression. This is done (see(24)).(3) Sum up the elemental energy to form the total system energy.(4) Perform the variation of the total system energy to obtain theequations of motion!Let’s now work on items (1), (3) and (4).1.A Partition a bar into many elements.element element 3element 2global node 12345element 1(2)(1)(3)(4)(. . . )Partition a bar into finite elementselelental nodeglobal node123423121212element 3element 2element 1u1u2u3u4u2u3q1(1)q2(1)q1(2)q2(2)q1(3)q2(3)Partitioning and establishing the Boolean relation between the globaldegrees of freedom, u, and elemental degrees of freedom, q1.B Establishtherelationbetweentheelemental andassembled(global)degrees of freedom.For element 1: q(1)=q(1)1q(1)2=u1u2For element 2: q(2)=q(2)1q(2)2=u2u3For element 3: q(2)=q(3)1q(3)2=u3u4⇓q(1)q(2)q(3)=I 0000 I 000 I 0000I 000I 0000Iu1u2u3u4⇒For a general case:q= Lcuc(25)where the subscript c designates the continuum ddomain.3.A Sum up the total system kinetic energy.Ttotalbar=nel=1Telbar12{˙q(1)}T[m(1)]{˙q(1)}+12{˙q(1)}T[m(1)]{˙q(1)}+... +12{˙q(n)}T[m(n)]{˙q(n)}=˙q(1)˙q(2)..˙q(n)Tm(1)0 ... 00 m(2)... 000... 000.. m(n)˙q(1)˙q(2)..˙q(n)T⇓Ttotalbar=12{˙q}T[m] {˙q}(26)3.B Sum up the total system strain energy.Vtotalbar=nel=1Velbarb12{q(1)}T[k(1)]{q(1)}+12{q(1)}T[k(1)]{q(1)}+... +12{q(n)}T[k(n)]{q(n)}=q(1)q(2)..q(n)Tk(1)0 ... 00 k(2)... 000... 000.. k(n)q(1)q(2)..q(n)T⇓Vtotalbar=12{q}T[k] {q}(27)3.C Assemble the partitioned elements back into a bar.This corresponds to


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