ASEN 5022 - Spring 2004Dynamics of Aerospace StructuresLecture 14: March 02FiniteElementModelingofBeamBendingVibrationLet’s consider the modeling of a beam under generalboundary conditions by the finite element modeling.kw1kw2kµ1kµ2EI, m(x)xzwLBeam with unknown boundary conditionsTheoretical Basis: Same as the continuum beam model!Kinetic energy:T =12L0m(x)w(x, t)2tdx,w(x, t)t=∂w(x, t)∂t(1)Potential energy of the beam:Vb=12L0{ EI(x)w(x, t)2xx} dxw(x, t)xx=∂2w(x, t)∂x2,w(x, t)x=∂w(x, t)∂x(2)External energy due to distributed applied force f(x, t),shear forces Q(1,2)and moments M(1,2)applied at bothends:δW =L0f(x, t)δw(x, t) dx+ Q1δw(0, t) + Q2δw(L, t)+ M1δw(0, t)x+ M2δw(L, t)xQ1= Q(0, t), Q2= Q(L, t)M1= M(0, t), M2= M(L, t)(3)Potential energy of four unknown springs:Vs=12{ kw1w(0, t)2+ kθ1w(0, t)2x+ kw2w( L, t)2+ kθ2w( L, t)2x(4)Hamilton’s principlet2t1[δT − δVb− δVs+ δW ] dt = 0 (5)From here on, FEM modeling differs from continuummodeling!Key departure in FEM modeling of beams from continuummodeling:1. Instead of carrying out the variation (the δ-process ofHamilton’s principle(5) in terms of the continuum vari-able w(x, t), the energy expressions, (T, V, W, etc.),areapproximated on a completely free beam with an arbi-trary length ,areaA, the bending rigidity EI and massdensity ρ.2. The interpolation of the transverse displacementw(x, t) overthecompletelyfreebeamsegment0 ≤ x ≤ is chosen to satisfy the homogeneous differential beamequation, i.e.,EIw(x, t)xxxx= 0 (6)if possible. If not, it is chosen such that as the lengthof the beam gets smaller and smaller, it approximatelysatisfies the homogeneous equation.We now consider a completely free isolated small segmentof a beam.kw1kw2k2EI, m(x)xzwL12wx12k1θθlA completely free element of length taken out from the beaman element isolatedApproximation of w(x, t) over the element segment,<<L:w(x, t) = c0(t) + c1(t)x + c2(t)x2+ c3(t)x3(7)Observe that the above interpolation satisfies the homoge-neous differential equation of beam (6).How does one determine the coefficients,(c0, c1, c2, c3)?The answer comes from the finite element method.The FEM beam bending element1. Observing that w(x, t) and w(x, t)xare linearly inde-pendent from the result of variational formulation, wespecify them at nodes 1 and 2 in the previous figure.Namely,two discrete variables at node 1: w(x1, t) and w(x1, t)xtwo discrete variables at node 2: w(x2, t) and w(x2, t)x(8)2. For convenience, we introduce a local coordinate sys-temorelementalcoordinatesystem, (x, y) andset x1= 0and x2= .Substituting (8) into the interpolation function(7),weobtainw1(t) = w(x1= 0, t) = c0(t) + c1(t)0 + c2(t)02+ c3(t)03θ1(t) = w(x1= 0, t)x= c1(t) + 2c2(t)0 + 3c3(t)02w2(t) = w(x2= , t) = c0(t) + c1(t) + c2(t)2+ c3(t)3θ2(t) = w(x2= , t)x= c1(t) + c2(t)2 + c3(t)32(9)Solvingforthecoefficientsandbacksubstitutinginto(7),one obtainsw(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 formw(x, t) = N(ξ ) q(t)N(ξ) =H1H2H3H4q(t)T=w1(t)θ1(t)w2(t)θ2(t)(11)12θθl2ww12lA completely free beam element and its nodal degrees of freedomElement DiscretizationElemental kinetic energy:Tel=120m(x)w(x, t)2tdx=1210m(ξ) w(ξ, t)2tdξ, x = ξ(12)Elemental potential energy:Velb=120EI(x)w(x, t)2xxdx=1210EI(ξ)3w(ξ, t)2ξξdξ(13)Discretization of Elemental Kinetic EnergyTel=1210m(ξ) w(ξ, t)2tdξ=1210ρ A˙q(t)T[NT· N]˙q(t)dξ=12˙qT[10ρ A NT· N dξ ]˙q(t)=12˙q(t)T[mel]˙q(t)(14)where we utilized (11), viz.:w(x, t) =3 i=0ci(t)xi= N(ξ ) q(t)(15)Discretization of Elemental Strain (Potential) EnergyVelb=1210EI(ξ)3w(ξ, t)2ξξdξ=1210EI(ξ)3q(t)T[NTξξ· Nξξ] q(t)dξ=12q(t)T[10EI3NTξξ· Nξξdξ ] q(t)=12q(t)T[kel] q(t)(16)where we utilized:w(ξ, t)ξξ= N(ξ )ξξq(t)(17)Elemental mass matrix, melm = ρ A133511210970−1342011210210513420−2140970134201335−11210−13420−2140−112102105(18)Elemental stiffness matrix, kelk =EI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)δWel=0f(x, t)δw(x, t) dx(=10f(ξ, t) N(ξ ) δq(t)dξ)+ Q1δw(0, t) + Q2δw(L, t) + M1δw(0, t)x+ M2δw(L, t)x(3)Let’s utilize the approximation(10):w(x, t) =N(ξ) q(t)w(x, t)x=1N(ξ)ξq(t)(20)and observe thatw(0, t) = q1(t), w(, t) = q3(t), w(0, t)x= q2(t), w(, t)x= q4(t)(21)so that we haveδWel= δq(t)T{0NTf(x, t)dξ}+Q1δq1(t) + Q2δq3(t) + M1δq2(t) + M2δq4(t)= δq(t)T{fel}, fel={0NTf(x, t)dξ}+Q1M1Q2M2T(22)Summary of elemental energy expressions:Tel=12{˙q(t)(el)}T[mel] {˙q(t)(el)}Velb=12{q(t)(el)}T[kel] {q(t)(el)}δWel={δq(t)(el)}T{fel}(23)Question: How do we model a long beam with beam elements?Answer:(1) Partition the long beam into small elements. This needs to beexplained.(2) Generate the elemental energy expression. This is done (see (23)).(3) Sum up the elemental energy to form the total system energy.(4) Perform the variation of the total system energy to obtain the equa-tions of motion!Let’s now work on items (1), (3) and (4).1.A Partition a beam into many elements.element element 3element 2global node 12345element 1(2)(1)(3)(4)(. . . )Partition a beam 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 betweenthe global degrees of freedom, u, and elemental degrees offreedom, q1.B Establish the relation between the elemental and as-sembled(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)=I0000I 000I 0000I 000I 0000Iu1u2u3u4⇒For general case:q= Lug(24)3.A Sum up the total system kinetic energy.Ttotal=n el=1Tel12{˙q(1)}T[m(1)]{˙q(1)}+12{˙q(1)}T[m(1)]{˙q(1)}+...