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The equations and the characteristic equation for the vibration analysis of a cable with a middle support. The analysis is based on the principle of virtual work and considers the effects of tension, mass, and spring forces. The document also includes the boundary and constraint conditions for the solution.

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Download Analysis of a Cable with a Middle Support: Equations and Characteristic Equation and more Assignments Aerospace Engineering in PDF only on Docsity! Solutions of Homework #3 (ASEN5022, Spring 2004) Problem : A cable with a flexible support at the middle. A construction company has been designing a cable that will run between two towers in a mountain. After a preliminary analysis and a scale model test, the management came to the conclusion that a third flexible tower needs to be placed. 3.1 Formulate the equations of motion for this new cable system complete with the appropriate boundary conditions. Assume that the original two supporting towers are substantially stiffer than the new one to be placed at the middle of the original cable. The design team has concluded that they should model not only the flexibility but also the inertia effect of the middle tower. 3.2 If the new system is to have its fundamental frequency about 50% higher than the original system (which was the reason why a third tower is needed), how are you going to accomplish this requirement? Observe that the cable tension, T , of the original system could not be increased any further as that will result in an unacceptable bending on the original two towers. w(x,t) z x f(x,t) O K0 K L ML M0 O Km Mm x z _ L /2 L /2 Figure 1 Cable with a middle support 1. Selection of the Coordinate System. There are two ways of selecting the coordinate system. Once is to use one coordinate system. In the present solution, we employ two independent coordinate systems, one for the left cable and another for the right cable as shown in Figure 1 above. 2. Hamilton’s principle for this problem. Introduce w(x, t) and w(x̄, t) whose domains are defined by { For the left cable: w(x, t), 0 ≤ x ≤ For the right cable: w(x̄, t), ≤ x̄ ≤ L , = L/2 (1) Using these distinct displacements, we have ∫ t2 t1 [δL + δW̄noncons]dt = 0, L = T − V T = ∫ 0 1 2ρ(x)ẇ 2(x, t) dx + ∫ L 1 2ρ(x̄)ẇ 2(x̄, t) dx̄ + 12 M0 ẇ2(0, t) + 12 ML ẇ2(L , t) + 12 Mm ẇ2(, t) V = ∫ 0 1 2 T (x) w 2 x (x, t) dx + ∫ L 1 2 T (x̄) w 2 x (x̄, t) dx̄ + 12 K0 w2(0, t) + 12 KL w2(L , t) + + 12 Km w2(, t) (2) Evaluation of ∫ t2 t1 δT dt ∫ t2 t1 δT dt = − ∫ t2 t1 {[ ∫ 0 ρ(x)ẅ(x, t) δw(x, t)dx ] + [ ∫ L ρ(x̄)ẅ(x̄, t) δw(x̄, t)dx̄ ] + M0ẅ(0, t) δw(0, t) + MLẅ(L , t) δw(L , t) + Mmẅ(, t) δw(, t)} dt (3) Evaluation of δV δV = − ∫ 0 T (x) wxx (x, t)δw(x, t) dx − ∫ L T (x̄) wx̄ x̄ (x̄, t)δw(x̄, t) dx̄ + { [T (x) wx (x, t)δw(x, t)]x= − [T (x) wx (x, t)δw(x, t)]x=0 } + { [T (x̄) wx̄ (x̄, t)δw(x̄, t)]x̄=L − [T (x̄) wx̄ (x̄, t)δw(x̄, t)]x̄= } + [K0 w(x, t)δw(x, t)]x=0 + [KL w(x̄, t)δw(x̄, t)]x̄=L + [Km w(x, t)δw(x, t)]x= (4) Virtual work due to nonconservative force f (x, t) δW̄noncons = 0 (5) due to free vibration problem of interest. Substituting (5), (4) and (3) into (1), one obtains Hamilton’s principle: W (0) = 0 [T W (x)x − T W (x̄)x̄ + Km W (x) − ω2 Mm W (x)]x= = 0, = L/2 W (L) = W (2) = 0 [W (x)|x= − W (x̄)]x̄= = 0 (16) Let’s focus on the second equation of the above equation set. First, dividing by the tension T and multiplying by , we have [ W (x)x − W (x̄)x̄ + Km T W (x) − ω 2 Mm T W (x)]x= = 0 (17) Second, we note that with β2 = ω2ρT ω2 Mm T = Mm ω 2 T = Mm 2 ρ ω2ρ T = Mm 2 ρ β2 = Mm ρ (β)2 = µm β̄2 µm = Mm ρ , β̄ = β, = L/2 (18) Third, introducing κm = Km T (19) equation (17) can be written as [ W (x)x − W (x̄)x̄ + κm W (x) − µm β̄2W (x)]x= = 0 (20) Therefore, (16) simplifies to W (0) = 0 [ W (x)x − W (x̄)x̄ + κm W (x) − µm β̄2W (x)]x= = 0 W (L) = W (2) = 0 [W (x)|x= − W (x̄)]x̄= = 0 (21) Finally, using W (x) = C1 sin(βx) + C2 cos(βx), 0 ≤ x ≤ W (x̄) = C3 sin(βx) + C4 cos(βx), ≤ x̄ ≤ L (22) the preceding boundary and constraint set (21) leads to the following characteristic equation: 0 1 0 0[β̄ cos(β̄) + (κm − µm β̄2) sin(β̄)] [−β̄ sin(β̄) + (κm − µm β̄2) cos(β̄)] −β̄ cos(β̄) β̄ sin(β̄) 0 0 sin(2β̄) cos(2β̄) sin(β̄) cos(β̄) − sin(β̄) − cos(β) C1 C2 C3 C4 = 0 0 0 0 (23) By successive manipulations of the third and fourth columns, it can be shown that the above equation simplifies to 0 1 0 0[β̄ cos(β̄) + (κm − µm β̄2) sin(β̄)] [−β̄ sin(β̄) + (κm − µm β̄2) cos(β̄)] −β̄ 0 0 0 sin(β̄) cos(β̄) sin(β̄) cos(β̄) 0 −1 C1 C2 C3 C4 = 0 0 0 0 ⇓ AC = 0 (24) This simplification is equivalent to taking the left and right cable coordinates independently for both cases as (0 ≤ x ≤ , 0 ≤ x̄ ≤ ). Setting det(A) = 0, one finds with µm = Mmρ : κm = Km T = (β)2 − 2(β) cos(β) sin(β) (25) Since the desired frequency is βL = 1.5π ⇒ β = 1.5π/2 (26) we find the spring constant to be κm = Km T = 10.26404145599745 (27) The fundamental mode shape is plotted in Figure 2 using the computation routines listed below. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 Beam span M od e sh ap e Two cables supported by a flexible middle support middle spring (km) =10.264 beta*L = 4.7124 Figure 2 Fundamental mode of a cable with a middle support % Computations of two cable vibrations % using a general characteristic equation % clear all; % % Km determined from the det(C_cable) % km=10.26404145599745; % % compute mode shape for this frequency % beta_ell = 1.5*pi/2; % target frequency % compute C_cable (4x4) matrix Ccable = CmatrixTwoCable(km, beta_ell); % it is assumed that the minor Cbeam(3x3) matrix is non-singular Cminor = Ccable(2:4, 2:4); b=-Ccable(2:4,1); % mode shape coefficients assuming c_4 =1 coef = Cminor\b; xL_coord=zeros(1,0); yL_coord =zeros(1,0); xR_coord=zeros(1,0); yR_coord =zeros(1,0); Step 4: Form the model equation set given by: Mẍ1 + K x1 − K 4 xm = f1 Mẍ2 + K x2 − K 4 xm = f2 Mm ẍm + (Km + K 2 )xm − K 4 (x1 + x2) = fm (30) whose characteristic equation is given by det (K − M ω2) 0 − K 4 0 (K − M ω2) − K4 − K4 − K4 (Km + K2 − Mm ω2) = 0 (31) For computational expediency, divide each of the three rows of the above characteristic matrix by M and using ωn = √ K/M , we have det (ω2n − ω2) 0 −ω 2 n 4 0 (ω2n − ω2) −ω 2 n 4 −ω2n4 −ω 2 n 4 (Km/M + ω 2 n 2 − Mm/M ω2) = 0 (32) Introducing the following non-dimensional parameters, Km M = Km K K M = κmω2n, µm = Mm M (33) equation(32) becomes det (ω2n − ω2) 0 −ω 2 n 4 0 (ω2n − ω2) −ω 2 n 4 −ω2n4 −ω 2 n 4 (κmω 2 n + ω 2 n 2 − µm ω2) = 0 (34) For this characteristic equation, one finds κm for the case of ω = 1.5π , with µm = 1, to be κm = 13.747 (35) which, when compared with the exact solution given by (27), yields about 34% off. One possible improvement for estimating the middle support spring is to distribute the spring constant Km as shown in Figure 4. Km O K M K M Mm (1-a)(1-a) K Ka a R ig h t ca b le Le ft c ab le xx x 1 2 m Km O K M K M Mm (1-a)(1-a) K Ka a R ig h t ca b le Le ft c ab le xx x 1 2 m (1-b) Km Possible Improvement of estimating the middle support spring Km by distributing Km into three ways as shown on the right in the above figure. b_ 2 Figure 4 Possible Improvement in modeling the support spring Whether this conjecture will be valid or not is left as an exercise to curious minds. Better yet, one may employ each half of the cable by a two-DOF model, instead of the one-DOF model I used herein. In any case, you now have a glimpse of ”old-fashioned” modeling approaches. Let me know if you would like to learn more about old-fashioned modeling practices.