ASEN 5022 - Spring 2005Dynamics of Aerospace StructuresLecture 12: 24 FebruaryBeams with Uncertain Boundary ConditionsConsider a beam whose two ends are constrained by transversesprings (kw1, kw2) as well as rotational springs (kθ1, kθ2) as shownin the figure below.kw1kw2kθ1kθ2EI, m(x)xzwLBeam with unknown boundary conditionsFig. 1 Modeling of unknown boundary conditionsBefore we proceed further, it should be noted that the mathemati-cally knownboundaryconditionscanberealizedbytakingthelimitvalues of the four spring constants as follows:Simply supported : kw1= kw2→∞ and kθ1= kθ2= 0Cantilever beam x = 0:kw1= kθ1→∞ and kw2= kθ2= 0Fixed-fixed : kw1= kθ1→∞ and kw2= kθ2→∞Free-free : kw1= kθ1= 0 and kw2= kθ2= 0(1)For intermediate values of the unknown springs a convenient wayof identifying the boundary conditions is to invoke a variationalformulation. This is because the appropriate boundary conditions,both natural and essential, are determined as part of variationalprocess.Kinetic energy:T =12L0m(x)w(x, t)2tdx,w(x, t)t=∂w(x, t)∂t(2)Potential energy of the beam:Vb=12L0{ EI(x)w(x, t)2xx+ P(x)w(x, t)2x} dxw(x, t)xx=∂2w(x, t)∂x2,w(x, t)x=∂w(x, t)∂x(3)where EI(x) is the bending rigidity and P(x) is the pre-stressedaxial force.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)External energy due to distributed applied force f(x, t), shear forcesQ(1,2)and moments M(1,2)applied at both ends:δW =L0f(x, t)δw(x, t) dx + Q1δw(0, t)+ M1δw(0, t)x+ Q2δw(L, t)+ M2δw(L, t)x(5)Q1= Q(0, t), Q2= Q(L, t)M1= M(0, t), M2= M(L, t)(6)Hamilton’s principlet2t1[δT − δVb− δVs+ δW ] dt = 0 (7)Kinetic energyt2t1δTdt=−L0t2t1m(x)w(x, t)ttδw(x, t) dt dx(8)Variation of the internal energy of the beam VbδVb= [EI w(x, t)xxδw(x, t)x]|L0−{[EI w(x, t)xx]xδw(x, t)}|L0+L0{ [EI w(x, t)xx]xx− [P w(x, t)x]x}δw(x, t) dx(9)Variation of the potential energy of the unknown boundary forcesand momentsδVs={kw1w(0, t)δw(0, t) + kθ1w(0, t)xδw(0, t)x+ kw2w( L, t)δw(L, t) + kθ2w( L, t)xδw(L, t)x(10)t2t1[δT − δVb− δVs+ δW ] dt = 0 (11)t2t1[δT − δVb− δVs+ δW ] dt =−t2t1{−[EI w( L, t)xx]x+ kw2w( L, t) − Q2} δw(L, t)dt+t2t1{−[EI w(0, t)xx]x− kw1w(0, t) + Q1} δw(0, t)dt−t2t1{[EI w( L, t)xx] + kθ2w( L, t)x− M2} δw(L, t)xdt+t2t1{[EI w(0, t)xx] − kθ1w(0, t)x+ M1} δw(0, t)xdt−t2t1L0{ m(x)w(x, t)tt+ [EI w(x, t)xx]xx− f (x, t) } δw(x, t) dxdt = 0(12)The governing equation of motion:m(x)w(x, t)tt+ [EI w(x, t)xx]xx− f (x, t) = 0 (13)The boundary conditions:{−[EI w( L, t)xx]x+ kw2w( L, t) − Q2} δw(L, t) = 0{−[EI w(0, t)xx]x− kw1w(0, t) + Q1} δw(0, t) = 0{[EI w( L, t)xx] + kθ2w( L, t)x− M2} δw(L, t)x= 0{[EI w(0, t)xx] − kθ1w(0, t)x+ M1} δw(0, t)x= 0(14)Free vibrations of a beamThe free vibration of beams can be modeled from (13) and (14) bysettingP = Q1= Q2= M1= M2= f (x, t) = 0 (15)so that (13) and (14) are simplified tom(x)w(x, t)tt+ [EI w(x, t)xx]xx= 0{−[EI w( L, t)xx]x+ kw2w( L, t) } δw(L, t) = 0{−[EI w(0, t)xx]x− kw1w(0, t) } δw(0, t) = 0{[EI w( L, t)xx] + kθ2w( L, t)x} δw(L, t)x= 0{[EI w(0, t)xx] − kθ1w(0, t)x} δw(0, t)x= 0(16)Free Vibration Governing Equation and Four Natural BoundaryConditions:m(x)w(x, t)tt+ [EI w(x, t)xx]xx= 0{−[EI w( L, t)xx]x+ kw2w( L, t) }=0{−[EI w(0, t)xx]x− kw1w(0, t) }=0{[EI w( L, t)xx] + kθ2w( L, t)x}=0{[EI w(0, t)xx] − kθ1w(0, t)x}=0(17)Four natural boundary conditions - cont’dw(x, t) = W(x)ejωt(18)which, when substituted into (17), yields−ω2mEIW(x) + W(x)xxxx= 0, 0 ≤ x ≤ L−W(0)xxx−kw1EIW(0) = 0w(0)xx−kθ1EIW(0)x= 0−W(L)xxx+kw2EIW(L) = 0w( L)xx+kθ2EIW(L)x= 0(19)The frequency equation of vibrations of a uniform beam can beobtained by assuming the solution in form ofW = c1sinβx + c2cosβx + c3sinhβx + c4coshβxβ4=ω2mEIWx= β(c1cosβx − c2sinβx + c3coshβx + c4sinhβx)Wxx= β2(−c1sinβx − c2cosβx + c3sinhβx + c4coshβx)Wxxx= β3(−c1cosβx + c2sinβx + c3coshβx + c4sinhβx)(20)¯β3−¯kw1−¯β3−¯kw1−¯kθ1−¯β −¯kθ1¯β−¯β3cos¯β¯β3sin¯β¯β3cosh¯β¯β3sinh¯β−¯kw2sin¯β −¯kw2cos¯β −¯kw2sinh¯β −¯kw2cosh¯β¯β sin¯β¯β cos¯β −¯β sinh¯β −¯β cosh¯β−¯kθ2cos¯β +¯kθ2sin¯β −¯kθ2cosh¯β −¯kθ2sinh¯βc1c2c3c4= 0 (21)where¯β = β L,¯kθi= kθi/(EI/L),¯kwi= kwi/(EI/L3)(22)Free-free beam (kw1= kw2= kθ1= kθ2= 0)¯β30 −¯β300 −¯β 0¯β−¯β3cos¯β¯β3sin¯β¯β3cosh¯β¯β3sinh¯β¯β sin¯β¯β cos¯β −¯β sinh¯β −¯β cosh¯βc1c2c3c4= 0⇓det10−100 −10 1−cos¯β sin¯β cosh¯β sinh¯βsin¯β cos¯β −sinh¯β −cosh¯β= 0(23)Cantilever beam (kw2= kθ2= 0, kw1→∞, kθ1→∞.)det¯β3−¯kw1−¯β3−¯kw1−¯kθ1−¯β −¯kθ1¯β−cos¯β sin¯β cosh¯β sinh¯βsin¯β cos¯β −sinh¯β −cosh¯β= 0 (24)det0 −10 −1−10 −11−cos¯β sin¯β cosh¯β sinh¯βsin¯β cos¯β −sinh¯β −cosh¯β= 0⇓(1 + cos¯β cosh¯β) = 0(25)which gives¯βk={1.875, 4.694, 7.855, ...} (26)1 2 3 4 5 6 7 8 9 10020406080100120 Fixed end rotational spring parameter for cantilever beam rotational spring, k (theta) /(EI/L) Error in fundamental frequency, percentEffect of fixed end rotational spring on fundamental frequency of a cantileverbeamSimply Supported and Fixed-Fixed Beams(w(0) = w( L) = 0 or kw1= kw2→∞)det0 −10 −1−¯kθ1−¯β −¯kθ1¯β−sin¯β −cos¯β −sinh¯β −cosh¯β¯β sin¯β¯β cos¯β −¯β sinh¯β −¯β cosh¯β−¯kθ2cos¯β +¯kθ2sin¯β −¯kθ2cosh¯β −¯kθ2sinh¯βc1c2c3c4= 0⇓¯β2sin¯β sinh¯β +¯kθ1¯β(sin¯β cosh¯β − cos¯β sinh¯β)−¯kθ2¯β cos¯β sinh¯β +¯kθ1¯kθ2(1 − cos¯β cosh¯β) =