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 ; β @ 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 ; β @ 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 / 2. 1 0 y = sech x -1 y = cos x pi/2 3pi/2 5pi/2 7pi/2 9pi/2 Project 10.3 281
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 = 2 10 12 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. 10.3.8 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 = 2 10 12 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
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 ; β @ 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); 4.730040745, 7.853204624, 10.99560784, 14.13716549, 17.27875966, 20.42035225, 23.56194490, 26.70353756, 29.84513021, 32.98672286 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); 3.123163879, 3.142403216, 3.14155765, 3.14159417, 3.14159259, 3.14159265, 3.14159266, 3.14159265, 3.14159265 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 10.3 283
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); [121.6925898, 335.4503097, 657.6167321, 1087.073862, 1623.900808, 2268.092846, 3019.650238, 3878.572976, 4844.861051, 5918.514469] 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 ; β @ 1 values, as follows: M = 10; solutios = Table[ FidRoot[ eq, {x,(2*+1)*pi/2}], {, 1, M}]; beta = x /. solutios {4.7300, 7.8532, 10.9956, 14.1372, 17.2788, 20.4204, 23.5619, 26.7035, 29.8451, 32.9867} 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
{3.12316, 3.1424, 3.14156, 3.14159, 3.14159, 3.14159, 3.14159, 3.14159, 3.14159} 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) {569.485, 1569.81, 3077.45, 5087.18, 7599.37, 10614., 14131.1, 18150.6, 22672.5, 27696.9} Whereas the fudametal frequecy of the higed-higed bar of case 1 is (Pi^2/L^2) Sqrt[E0*I0/rho]/(2 Pi) 251.219 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 10.3 285
whose successive positive zeros we seek. The we proceed to fid the first N = 10 of the values as follows: ; β @ 1 N = 10; beta = 1 : N; for = 1 : N beta() = fzero(f, (2*-1)*pi/2); ed reshape(beta,5,2)' as = 1.8751 4.6941 7.8548 10.9955 14.1372 17.2788 20.4204 23.5619 26.7035 29.8451 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 = 1.0258 6.4285 17.9999 35.2725 58.3081 87.1021 121.6550 161.9668 208.0373 259.8667 Thus the divig board's fudametal frequecy is about 1.0258 cycles/sec, so for maximal resoat effect the diver should bouce up ad dow just about oce per secod. 286 Chapter 10