Project 10.3 Vibrating Beams and Diving Boards

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Rudolf Hutchinson
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1 Project 10.3 Vibratig Beams ad Divig Boards I this project you are to ivestigate further the vibratios of a elastic bar or beam of legth L whose positio fuctio y(x,t) satisfies the partial differetial equatio ρ 2 y 2 t 4 y + EI = 0 (0 < x < L) (1) 4 x ad the iitial coditios y(x,0) = f(x), y t (x,0) = 0. First separate the variables (as i Example 3 of Sectio 10.3 i the text) to derive the formal series solutio yxt (, ) = c X( x)cos = 1 β 2 2 at 2 L (2) 4 where a = EI/ ρ, the ;@ c are the appropriate eigefuctio expasio coefficiets of the iitial positio fuctio f (x), ad the ; values ad ; X ( x )@ eigefuctios are determied by the ed coditios imposed o the bar. I a particular case, oe wats to fid both the frequecy equatio whose positive roots are the ; ad the explicit eigefuctios ; X ( x I this sectio we saw that the frequecy equatio for the fixed-fixed case (with y( 0) = y ( 0) = y( L) = y ( L) = 0) is cosh x cos x = 1, that is, sech x = cos x (3) The solutios of this equatio are located approximately by the graph below, where we see that the th positive solutio is give approximately by β ( 2 + 1) π / y = sech x -1 y = cos x pi/2 3pi/2 5pi/2 7pi/2 9pi/2 Project

2 Case 1: Higed at each ed The edpoit coditios are y( 0) = y ( 0) = y( L) = y ( L) = 0. (4) Accordig to Problem 8 i Sectio 10.3 of the text, the frequecy equatio is si x = 0, so β = π ad X( x) = siπ x for = 1, 2, 3,.... Suppose that the bar is made of steel (with desity δ = 7.75 g/cm 3 ad Youg's modulus E = dy/cm 2 ), ad is 19 iches log with square cross sectio of edge w = 1 i. (so its momet of iertia is I = 12 1 w 4 ). Determie its first few atural frequecies of vibratio (i Hz). How does this bar soud whe it vibrates? Case 2: Free at each ed The edpoit coditios are y ( 0) = y ( 0) = y ( L) = y ( L) = 0. (5) This case models, for example, a weightless bar suspeded i space i a orbitig spacecraft. Now show that the frequecy equatio is agai sech x = cos x as i (3), although the eigefuctios i this free-free case differ from those i the fixed-fixed case discussed i the text. From the figure above we see that the th positive solutio is give approximately by β ( 2 + 1) π / 2. Use the umerical methods of Project 10.2 to approximate the first several atural frequecies of free-free vibratio of the same physical bar cosidered i case 1. How does it soud ow? Case 3: Fixed at x = 0, free at x = L Now the boudary coditios are y(0) = y (0) = y (L) = y (L) = 0. (6) This is a catilever like the divig board illustrated i Fig of the text. Accordig to Problem 15 there, the frequecy equatio is cosh x cos x = 1. (7) From the figure at the top of the ext page, showig the graphs y = sech x ad y = cos x, we see that β (2 1)π /2 for large. Use the umerical methods of Project 10.2 to approximate the first several atural frequecies of vibratio (i Hz) of the particular divig board described i Problem 15 it is 4 meters log ad is made of steel (with desity δ = 7.75 g/cm 3 ad Youg's modulus E = dy/cm 2 ); its cross sectio is a rectagle with width a = 30 cm ad thickess b = 2 cm, so its momet of iertia about a horizotal axis of symmetry is I = 12 1 ab 3. Thus fid the frequecies at which a diver o this board should bouce up ad dow at the free ed for maximal resoat effect. 282 Chapter 10

3 1 y = sech x 0-1 y = -cos x pi/2 3pi/2 5pi/2 7pi/2 9pi/2 Usig Maple We eter the eigevalue equatio eq := sech(x) = cos(x): i (3) ad proceed to solve it for the first N = 10 of the ; 1 values, as follows: N := 10: beta := array(1..n): for from 1 to N do beta[] := fsolve(eq, x=(2*+1)*pi/2): od: seq(beta[], =1..N); , , , , , , , , , Note how we make use of the fact that β is approximately ( 2 + 1) π / 2. It is iterestig to check successive differeces of eigevalues: seq(beta[]-beta[-1], =2..N); , , , , , , , , We see that, for 6-place accuracy, it suffices to calculate the first half-doze β -values ad the add successive multiples of π. Project

4 The physical parameters of the fixed-fixed bridge discussed i Sectio 10.3 of the text are give by delta := 7.75: A := 40: rho := delta*a: L := 120*30.48: I0 := 9000: E := 2*10^12: # volumetric desity # cross-sectioal area # liear desity # legth of bridge # momet of iertia # Youg's modulus 2 2 Its atural frequecies of vibratio, ω = ( β / L ) EI0 / ρ, are give i radias/sec by b := beta: w := map(b->(b^2/l^2)*sqrt(e*i0/rho), b): ad the i hertz (cycles/mi) by map(w->evalf(60*w/(2*pi)), w); [ , , , , , , , , , ] Thus the first "critical cadece" for marchig soldiers is about 122 steps per miute. Usig Mathematica We eter the eigevalue equatio eq = Sech[x] == Cos[x]; i (3) ad proceed to solve it for the first N = 10 of the ; 1 values, as follows: M = 10; solutios = Table[ FidRoot[ eq, {x,(2*+1)*pi/2}], {, 1, M}]; beta = x /. solutios {4.7300, , , , , , , , , } Note how we make use of the fact that β is approximately ( 2 + 1) π / 2. It is iterestig to check successive differeces of eigevalues: Table[beta[[]]-beta[[-1]], {,2,10}] 284 Chapter 10

5 { , , , , , , , , } We see that, for 5-place accuracy, it suffices to calculate the first half-doze β -values ad the add successive multiples of π. The physical parameters of the free-free steel bar of case 2 are give by delta = 7.75; (* volumetric desity *) A = 2.54^2; (* cross-sectioal area *) rho = delta*a; (* liear desity *) L = 19*2.54; (* legth of bridge *) I0 = (2.54^4)/12; (* momet of iertia *) E0 = 2*10^12; (* Youg's modulus *) 2 2 Its atural frequecies of vibratio, ω = ( β / L ) EI0 / ρ, are give i radias/sec by b = beta; w = (b^2/l^2) Sqrt[E0*I0/rho]; ad the i cycles per secod by w = w/(2 Pi) { , , , , , , , , , } Whereas the fudametal frequecy of the higed-higed bar of case 1 is (Pi^2/L^2) Sqrt[E0*I0/rho]/(2 Pi) cycles/sec, that is, about middle C, we see that the fudametal frequecy of the correspodig free-free bar of case 2 is above high C. Usig MATLAB First we defie the eigevalue equatio cosh x cos x = 1 i (7), for the fixed-free catilever of case 3, i the form of the fuctio f = ilie('cosh(x)*cos(x)+1','x') f = Ilie fuctio: f(x) = cosh(x)*cos(x)+1 Project

6 whose successive positive zeros we seek. The we proceed to fid the first N = 10 of the values as follows: ; 1 N = 10; beta = 1 : N; for = 1 : N beta() = fzero(f, (2*-1)*pi/2); ed reshape(beta,5,2)' as = Note how we make use of the fact that β is approximately ( 2 1) π / 2. The physical parameters of the divig board described i Problem 15 i the text are give by delta = 7.75; a = 30; b = 2; A = a*b; rho = delta*a; L = 400; I0 = a*b^3/12; E = 2*10^12; % volumetric desity % width of divig board % its thickess % cross-sectioal area % liear desity % legth of divig board % momet of iertia % Youg's modulus 2 2 Its atural frequecies of vibratio, ω = ( β / L ) EI0 / ρ, are give i radias/sec by b = beta; w = (b.^2/l^2)*sqrt(e*i0/rho); ad the i cycles per secod by w = w/(2*pi) w = Thus the divig board's fudametal frequecy is about cycles/sec, so for maximal resoat effect the diver should bouce up ad dow just about oce per secod. 286 Chapter 10

Application 10.5B Rectangular Membrane Vibrations

Application 10.5B Rectangular Membrane Vibrations Applicatio.5B Rectagular Membrae Vibratios Here we ivestigate the vibratios of a flexible membrae whose equilibrium positio is the rectagle x a, y b. Suppose it is released from rest with give iitial displacemet,