Question

Problem 1. The natural frequencies wn of free vibration of a cantilever beam are determined from the roots of the equation: ET Cantilever beam Wn = (k~L)2 VPALA in which E = 2.0 x 1011 N/m is the elastic modulus, L = 0.45 m is the beam length, 1 = 4.5 x 10-11 m is the moment of inertia, A = 6.0 x 10-5 mº is the cross-sectional area, and p = 6850 kg/m' is the density per unit length. kn L = p can be estimated by solving the following single nonlinear equation: cos(u) cosh() = -1 Find the first natural frequency of the beam applying (a) bisection and (b) false-position method. Proceed iteration assuming es = 6.5% (stop at fifth iteration if it did not reach the tolerance), compare and explain the obtained results. For both cases, use H = 1.0 and Hu = 2.5. Plot the graph of the function to verify the results. Present detailed calculations for the first two iterations and tabulate results for the rest iterations.

Answer 1

f(u) =_cos (u) cos(u) + L0 a Bisection Method :- 11 Iteration : 10 fu- 1.83 (+) the = 2.5 file) -3.91(-) checking convergence criteris fulex Huu): -1,135 < Sarified M = Me + lu 1.0 +2.5 = 1.75 Plus) - $61.15) - 0.471 Cure : for next iteration, replacing y by the - Me - the 9.75 , the 2.5 "To Satisfy convergence criteria, foux H ) < 2 Therahon 2 :- - , = 1,15_ 25 J.15 +2.5 - 2.125 - Metlu Plu) = -1.234 (-ve) → For next iteration L, 15 A, E 1.12 El. py- xloo 2.125-1.75 X 100 2.125 = 19.66% > E. (6.5 ) Hence Continuing calculations, next iterations are summorized in .عاطمة

Iteration I of our p 1.0 1.75 1.15 1.75 4.86215 flue) the 1.832.5 0.67 2.5 Oca 2.125 0.47 1 .93% 0.26 4 .9375 flue) u fan & -3. 9 1.25 D4L -3.91 2.125 1.236 136 -1.934 4.9975 -0.24 9.614 -0,27 4.84335 0.126 5.0 -0.274,84062 -0.065 3:413%. Bisection meet his = 4.890625, with a = 2.679% False-position Method Iteration 1 - De fruw) 1.63373 Chw) = -3.9128446 4x(um) - Wexflue) = 1.4166645 f ) - fut flu) = 1.2123263 (tre) > Por next iteration, = 1.2122263 to satisfy fe) x fluu) co .2.5 Texation 2 te = 1.2123263 - 2.5 My = 1.72024342 My x flu) - lux flue) flu) - fluw Pau) - 0.5708212. (Eve) for next iteration, l = 2, en My Mix 100 = 16.0944 the = llu Ug7

Pace Dne Summarizing hext iterations in table Iteration le 1.4786495 ozu 1.7202434 1,819516 fan Mu 1.83373 2.5 1.2123263_3.5 0.5708272 2.5 0.22072 205 feu) un Fen) Ea +86495 1.2123253 -3.9128446 2093 Side 2 -3.9128446 1.720243|| Oos Fotost2 14.064 -3.9128446 1.819516 0.22072_5656:24 -3,9128646 1.85586_0.018314_1-9594 - False-position root, Ma = 1.85586 ceith Eo = 109.58%

Matlab Code to plot function

f = @(x)cos(x)*cosh(x)+1;

fplot(f,[1,2.5]);
yline(0);
grid on
xlabel("\mu");
ylabel("f(\mu)");

15

Natural frequency [I] For Bisection - KaL = M = 1.890625 con (K, L² EI V SAL - - (1.890625), (2.0x10") x64.5 x 10") 6850 XC60x10)x0.45)4 win = 82.6012 radis for False-position KAL- l = 1.85586 m = (85586) (20x10") x (usx 10'). 6850 x (6.0x10") 0.45) = 79.5913 rad/s W False-position noot is more accurate than bisection root because (Ea) Ep = 1.950 % < (Ea) Bis - 2.479%