Determining Boundary Spring Constants of a Beam: Analyzing the Characteristic Equation, Assignments of Aerospace Engineering

Download Determining Boundary Spring Constants of a Beam: Analyzing the Characteristic Equation and more Assignments Aerospace Engineering in PDF only on Docsity! Solution of Homework #5 (ASEN5022, Spring 2005) 5.1 Vibrations of Beam Bending with Uncertain Boundary Conditions Vibration experiments were conducted on a beam in a laboratory setting and yielded the following results: • The measured first mode was: βL = 3.75. • The peak amplitude of its mode shape occurred at the beam span x = 0.54L , measured from the left end. Determine the boundary spring constants by modifying Matlab code provided. Hint: The first modes of classical boundary cases are given by: Simply supported at both ends: βL = π One end fixed and the other free (cantlievered beam): βL = 1.875 Both ends fixed: βL = 4.730 One end fixed and other end simply supported : βL = 3.937 Solution: Since the frequency (βL = 3.75) is higher than the case of simply supported at both ends (βL = 3.75) and lower than the fixed at one end and simply surpported at the other end (βL = 3.937), we conclude that the observed case would fall into partially moment-resisting ends. To this end, let’s recall the characteristicc equation (27) of the lecture notes (Lecture 13: 01 March) for this case: Simply Supported and Fixed-Fixed Beams (w(0) = w(L) = 0 or kw1 = kw2 → ∞)  0 −1 0 −1 −k̄ θ1 −β̄ −k̄θ1 β̄ − sin β̄ − cos β̄ − sinh β̄ − cosh β̄ β̄ sin β̄ β̄ cos β̄ −β̄ sinh β̄ −β̄ cosh β̄ −k̄ θ2 cos β̄ +k̄θ2 sin β̄ −k̄θ2 cosh β̄ −k̄θ2 sinh β̄     c1 c2 c3 c4   = 0 (1) which yields the following characteristic equation: 2β̄2 sin β̄ sinh β̄ + (k̄θ1 + k̄θ2) β̄ (sin β̄ cosh β̄ − cos β̄ sinh β̄) + k̄θ1 k̄θ2 (1 − cos β̄ cosh β̄) = 0 (2) In addition, at x = 0.54 the slope of the mode shape must be zero: Wx xm=0/54 = {(c1 cos βxm − c2 sin βxm + c3 cosh βxm + c4 sinh βxm)}|xm=0.54 = 0 (3) By solving for (c1, c2, c3) in terms of c4 from (1), we obtain (Mathematica here!):{ c1 c2 c3 } = 1 (sinh β̄ − sin β̄) kθ1 { 2β̄ sinh β̄ − (cos β̄ − cosh β̄) kθ1 −(sinh β̄ − sin β̄) kθ1 −2β̄ sin β̄ + (cos β − cosh β̄) kθ1 } c4 (4) If c4 = (sinh β̄ − sin β̄) kθ1, then we obtain:  c1 c2 c3 c4   =   2β̄ sinh β̄ − (cos β̄ − cosh β̄) kθ1 −(sinh β̄ − sin β̄) kθ1 −2β̄ sin β̄ + (cos β − cosh β̄) kθ1 (sinh β̄ − sin β̄) kθ1   (5) Substituting this into (3) and after some arrangements, kθ1 is found as denom2 = k1*(1.0-c*ch) + x*(s*ch-c*sh); k2 = -nom2/denom2; % determinant (root accuracy) test det_check = k2*k1*(1.0 -c*ch) +( k1 + k1)*x*(s*ch - c*sh) +2*x^2*s*sh; disp([’The left-end rotational spring, k_theta 1 = ’, num2str(k1,7)]); disp([’The right-end rotational spring, k_theta 2 = ’, num2str(k2,7)]); disp([’Characteristic root accuracy test, det(A) = ’, num2str(det_check,7)]); % plot the mode shape c1 = 2*x*sh + (ch-c)*k1; c2 = (s-sh)*k1; c3 = -2*x*s -(ch-c)*k1; c4 = (sh-s)*k1; ell = 0:0.01:1; W = c1*sin(x*ell) + c2*cos(x*ell) + c3*sinh(x*ell) + c4*cosh(x*ell); W = W/ max((W)); figure(1); plot(ell, W); xlabel(’Beam Span’); ylabel(’Modal Amplitude’); legend([’ Rotational end springs’, ’ k_theta1 = ’, num2str(k1,7), ... ’k_theta2 = ’, num2str(k2,7)]); grid on; *************************************************************************** The results are: κθ1 = 8.9232414 κθ2 = 0.8943559 (8) These values together with κw1 = κw2 = 108 were used to confirm the βL value as well as the mode shape by running a general Matlab codehanded out. They indeed match the theoretical values. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Beam Span M od al A m pl itu de Rotational end springs for beta L =3.75 0.8943559k_theta 2 = 8.923241k_theta 1 = Veam with partially constrained support conditions 5.2 FEM Modeling of Vibration of Bars Consider a bar whose ends are constrained by springs and masses shown below as discussed in the class. For your convenience, you may utilize the non-dimensional parameters given by κ01 = k01 L/E A, κ02 = k02 L/E A, κL1 = kL1 L/E A, κL2 = kL2 L/E A µ0 = M0/ρL , µL = ML/ρL M0 LM k01 L1 kk02 L2 k m0 k0( ) mL kL( )Mc Kc( ) Unassembled Model (hardware) Unassembled Model (software) Figure for Problem 5.2 5.2.1 Perform an FEM vibration analysis by gradually increasing the number of bar elements for the case of free-free bar. This condition can be realized in your FEM code by setting all of the end masses and springs to be zero. Determine the number elements needed to obtain the FEM computed two lowest frequencies to be within 0.1 % of the continuum results. 5.2.2 Repeat Problem 5.2.1 with your choice of soft support conditions by selecting appropriate end masses and strpings. 5.2.3 Repeat Problem 5.2.1 with your choice of hard support conditions by selecting appropriate end masses and strpings. Discuss your FEM analysis results as compared to the analytical results (the continuum solution). What have you learned? The governing equation of motion for the continuum bar: ρ(x)ü(x, t) = E A(x) uxx (x, t) + f(t) (16) The governing equation of motion for the left-side spring mass: M0 ü0(t) + K01 u0(t) + K02 [u0(t) − u(0, t)] = 0 (17) The governing equation of motion for the right-side spring mass: ML üL (t) + KL2 u0(t) + KL1 [uL (t) − u(L , t)] = 0 (18) The boundary conditions: E A(L) ux (L , t) + KL1 [u(L , t) − uL (t)] = 0 − E A(0) ux (0, t) + K02 [u(0, t) − u0(t)] = 0 (19) Derivation of Characteristic Equation With f (t) = 0, the general solution of the continuum equation (16) assumes the form of u(x, t) = {C1 sin βx + C2 cos βx} F(t), F(t) = e jωt , β = ω √ ρ E A (20) Similary, the discrete displacements, (u0(t), uL (t)), for (17) and (18) are assumed to have their solutions in the form: u0(t) = ū0 e jωt uL (t) = ūL e jωt (21) Substituting the preceding two solution forms into the two discrete mass equations (17) and (18), and the two boundary conditions (19), we obtain the following characteristic equation [ (K01 + K02 − ω2 M0) 0 0 −K02 0 (KL1 + KL2 − ω2 ML ) −KL1 sin βL −KL1 cos βL 0 −KL1 (KL1 sin βL + E Aβ cos βL) (KL1 cos βL − E Aβ sin βL) −K02 0 −E Aβ K02 ]{ u0(t) uL (t) C1(t) C2(t) } = 0 (22) For the ease of algebra, we multiply all the rows of the above equation by (L/E A) and simply by introducing the following non-dimensional parameters: κ01 = K01 L E A , κ02 = K02 L E A , κL1 = KL1 L E A , κL2 = KL2 L E A , ω2 M0 L/E A = (β2 E A ρ )(M0 L/E A) = (βL)2 M0 ρL = (βL)2 µ0, µ0 = M0 ρL ω2 ML L/E A = (β2 E A ρ )(ML L/E A) = (βL)2 ML ρL = (βL)2 µL , µL = ML ρL (23) Utilizing these nondimensionalization, (22) becomes:   (κ01 + κ02 − (βL)2µ0) 0 0 −κ020 (κL1 + κL2 − (βL)2µL ) −κL1 sin βL −κL1 cos βL 0 −κL1 (κL1 sin βL + (βL) cos βL) (κL1 cos βL − (βL) sin βL)−κ02 0 −(βL) κ02   { u0(t) uL (t) C1(t) C2(t) } = 0 (24) The characteristic equation of the above equation (after relying on Mathematica!) is obtained by setting the determinant of its solution matrix to be zero:  = ((−k2L1) ((x ((k01 + k02 − µ0 x2)) cos[x] + k02 ((k01 − µ0 x2)) sin[x])) + ((kL1 + kL2 − muL x2)) ((k02 ((k01 − µ0 x2)) ((x cos[x] + kL1 sin[x])) − x ((k01 + k02 − µ0 x2)) (((−kL1) Cos[x] + x sin[x]))))) (25) Alternatively, one may eliminate (u0(t), uL (t) ) (17) and (18) to obtain u0(t) = (K01 + K02 − ω2 M0)−1 K02 u(0, t) uL (t) = (KL1 + KL2 − ω2 ML )−1 KL1 u(L , t) (26) Eliminating the discrete displacements, (u0(t), uL (t) ), via the preceding relation in the two boundary conditions (19), we obtain: E A(L) ux (L , t) + KL1 [ 1 − (KL1 + KL2 − ω2 ML )−1 KL1] u(L , t) = 0 − E A(0) ux (0, t) + K02 [1 − (K01 + K02 − ω2 M0)−1 K02 ]u(0, t) = 0 (27) Rearranging the the above equation by multiplying (KL1 + KL2 − ω2 ML ) and (K01 + K02 − ω2 M0) yields: (KL1 + KL2 − ω2 ML ) E A(L) ux (L , t) + KL1 [ (KL1 + KL2 − ω2 ML ) − KL1} u(L , t) = 0 − (K01 + K02 − ω2 M0) E A(0) ux (0, t) + K02 [(K01 + K02 − ω2 M0) − K02 ] u(0, t) = 0 (28)