ASEN5022: Dynamics of Aerospace StructuresLecture 7: Vibration of String, Bar and ShaftThis lecture covers:• Begin with Hamilton’s Principle.• Introduce the kinematics of cables for deriving the strain energy of a cable.• Perform variations of Hamilton’s principle to obtain the the governing equations of motion andboundary conditions. Euler-Lagrange’s equations do not provide the boundary conditions.• Solve the homogeneous equtions of motion by satisfying the boundary conditions to obtainthe natura lfrequencies (modes) and mode shapes (vibration shapes).1Symbols used in this lecture:Transverse displacement: w(x, t) (m)Time derivative:˙w =dwdt(m/s)Spatial derivtive: wx=dwdx(m/m)String tension: T (x) (N)Mass per unit string length: ρ(x) (kg/m)String length: ds (m)End spring : K (N/m)2Formulation via Hamilton’s principle:Zt2t1[δL + δ¯Wnoncons]dt = 0, L = T − VT =ZL012ρ(x)˙w2dxV =ZL0T (x)(ds − dx) +12Kw2(L, t)(7.1)From the preceding figure, we findds = [1 + w2x(x, t)]12dx ≈ [1 +12w2x(x, t)]dx (7.2)Substituting ds in the above expression into (7.1), we obtainV =ZL012T (x) w2x(x, t) dx +12Kw2(L, t) (7.3)3Fig. 1. String in transverse vibrationSince there are important differences between the variations of discrete model vs. continuum model,4let’s perform the variation of the Lagrangian, δL, below.δL = δT − δVδT = δ{ZL012ρ(x) ˙w2(x, t) dx}δV = δ{ZL012T (x) w2x(x, t) dx}+ δ{12Kw2(L, t)}(7.4)Evaluation ofRt2t1δT dtδT = δ{ZL012ρ(x) ˙w2(x, t) dx}=ZL012ρ(x)δ{˙w2(x, t)} dx}=ZL0ρ(x){˙w(x, t) δ ˙w(x, t)} dx(7.5)5Zt2t1δT dt =Zt2t1ZL0ρ(x){˙w(x, t) δ˙w(x, t)} dxdt=ZL0{Zt2t1ρ(x) ˙w(x, t) δ ˙w(x, t) dt}dx=ZL0{ [ρ(x)˙w(x, t) δw(x, t)]t2t1− [Zt2t1ρ(x)¨w(x, t) δw(x, t) dt] }dx(7.6)Since w(x, t1) and w(x, t2) are considered known everywhere over the spatial domain 0 ≤ x ≤ L,we haveδw(x, t1) = δw(x, t2) = 0 (7.7)so that (Rt2t1δT dt) reduces toZt2t1δT dt = −Zt2t1ZL0ρ(x) ¨w(x, t) δw(x, t)dxdt (7.8)6Evaluation of δVSince the potential energy V does not involve time derivative of w(x, t), it is adequate to carry outsimply δV , viz.:δV = δZL012T (x) w2x(x, t) dx + δ12Kw2(L, t)=ZL0T (x) wx(x, t)δwx(x, t) dx+ Kw(L, t)δw(L, t)(7.9)The first term in the previous equation becomes:ZL0T (x) wx(x, t)δwx(x, t) dx = [T (x) wx(x, t)δw(x, t)]x=L− [T (x) wx(x, t)δw(x, t)]x=0−ZL0T (x) wxx(x, t)δw(x, t) dx(7.10)7Virtual work due to nonconservative force f(x, t)δ¯Wnoncons=ZL0f(x, t)δw(x, t)dx (7.11)Substituting the virtual work due to the kinetic energy δT (7.8), the potential energy δV (7.9)together with (7.10), the nonconservative force δWnoncons(7.11), into Hamilton’s principle (7.1), oneobtainsZt2t1hZL0{[−ρ(x)¨w(x, t)+ T (x) wxx(x, t) + f(x, t)] δw(x, t)}dx− [(T (x) wx(x, t) + Kw(x, t))δw(x, t)]x=L+ [T (x) wx(x, t)δw(x, t)]x=0i dt = 0(7.12)8The governing equations of motion plus the boundary conditionsIn order for the above variational expression to hold, first the term inside the brace {.} must vanish,which yields the governing equation of motion:ρ(x)¨w(x, t) = T (x) wxx(x, t) + f(x, t) (7.13)Similarly, the two terms in the bracket must also vanish:[(T (x) wx(x, t) + Kw(x, t))δw(x, t)]x=L= 0 (7.14)[T (x) wx(x, t)δw(x, t)]x=0= 0 (7.15)9Boundary conditions for continuum stringTwo types of boundary conditions:Essential (or geometric) boundary conditions: when displacements are specified.Natural (or force) boundary conditions: when the boundary conditions come from force(moment) balance considerations.Boundary conditions at x = L:From (7.14) we findT (L) wx(L, t) + Kw(L, t) = 0 (7.16)or w(L, t)] = 0 (7.17)Since the right end is left to move, (7.17) is not appropriate. Hence, the correct boundary conditionis given by (7.17), which is a natural boundary condition.10Boundary conditions at x = 0:From (7.16) we findT (0) wx(0, t) = 0 (7.18)or w(0, t) = 0 (7.19)Since the right end is fixed, (7.18) is not appropriate. Hence, the correct boundary condition is givenby (7.19), which is an essential or geometric boundary condition.Observe that the variational equation for string, viz., Hamilton’s principle (7.12) provides theinformation to determine the boundary conditions, depending on the end configuration orconstraint condition.This is a distinct advantage of the variational formulation of continuum models as opposed toNewton’s approach.11Application of the cable model to torsional shaft and axial rod problems Thederivation of the governing equation and the associated boundary conditions for continuum stringcan be used for the modeling of continuum shaft and axial bar. This is summarized in the Table 8.1Variables for String, Shaft and Axial BarModel String Shaft Axial barVariable w(x, t) (m) θ(x, t) (rad) u(x, t) (m)Stiffness T (x) (N) GJ(x) (Nm2) EA(x) (N)Table 8.1 Conversion Variables for String, Shaft and Axial BarHence, the equations of motion for bar, shaft and axial bar can be derived by appropriate variableand parameter changes.Transverse vibration of stringsEquation (7.13) with f(x, t) = 0 leads to continuum vibration problem for strings together with theboundary conditions (7.16)-(7.19). To this end, we seek the solution of the homogeneous equation of12(7.13) in the form ofw(x, t) = W (x) F (t) (7.20)by employing the separation of variable used for the solution of partial differential equations. Sub-stituting (7.20) into (7.13), (7.16) and (7.19), we obtainddx[T (x)dW (x)dx]F (t)= ρ(x)W (x)d2F (t)dt2, 0 < x < L[T (x)dW (x)dx+ KW (x)]|x=LF (t) = 0[W (x)]|x=0F (t) = 0(7.21)Applying the temporal part F(t) in the form ofF (t) =¯F ejωt,¯F 6= 0 (7.22)we obtain the following boundary-value problem set:ddx[T (x)dW (x)dx] + ω2ρ(x)W (x) = 0, 0 < x < L (7.23)[T (x)dW (x)dx+ KW (x)]|x=L= 0 (7.24)13[W (x)]|x=0= 0 (7.25)When the tension T (x) and mass ρ(x) are constant along the string length, equation (7.23) can berearranged asd2W (x)dx2+ β2W (x) = 0,β2= ω2ρ/T, 0 < x < L(7.26)whose solution has the form ofW (x) = C1sin βx + C2cos βx (7.27)Substituting the preceding solution into the (x = L)-boundary condition (7.24) yields:T β{C1cos βL − C2sin βL}+ K{C1sin βL + C2cos βL} = 0(7.28)Similarly, the (x = 0)-boundary condition (7.25) yields:C1sin(β ∗ 0) + C2cos(β ∗ 0) = 0 ⇒ C2= 0 (7.29)14With C2= 0, (7.28) reduces toT β cos βL + K sin βL = 0 (7.30)Fixed end at x = L:When K → ∞, viz.,T βK→ 0, we havesin βL = 0 ⇒ β =ωqTρ=nπL, n = 1, 2, ... (7.31)So that we obtain the