The end conditions on the beam are modelled by assuming the following boundary conditions on the function u: u(0, t) = u x (0, t) = 0;

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1 M344 - ADVANCED ENGINEERING MATHEMATICS Lecture 1: Motion of a Deflected Beam The deflection of a cantilever beam that is fixed at one end and free to move at the other can be shown to satisfy a fourth-order partial differential equation of the form u tt = c u xxxx, where u(x, t) is the deflection at the point x along the length of the beam, at time t. The positive constant c is EI, where A is the cross-sectional area of ρa the beam, ρ is its density, E is the modulus of elasticity, and I is the moment of inertia. u 1 x 1 The end conditions on the beam are modelled by assuming the following boundary conditions on the function u: u(, t) = u x (, t) = ; u xx (L, t) = u xxx (L, t) =, for all t >. Using the method of Separation of Variables, let u(x, t) = X(x)T (t). Then T XT c XT = X T c c XT X β4 for some constant β 4. This results in the two ordinary differential equations c T = X X β 4 X = and T + c β 4 T =. The solution of the second-order equation in T will have the form T n (t) = a n cos(cβ nt) + b n sin(cβ nt), where the constants β 4 n are the eigenvalues of the equation in X. 1

2 Solution of X β 4 X =, X() = X () = X (L) = X (L) = The characteristic polynomial for this fourth-order boundary-value problem is r 4 β 4 =, and it has the four roots r = ±β, ±βi; therefore, the general solution of the differential equation can be written as X(x) = A cos(βx) + B sin(βx) + C cosh(βx) + D sinh(βx), and differentiating with respect to x, X (x) = βa sin(βx) + βb cos(βx) + βc sinh(βx) + βd cosh(βx). Using the boundary conditions at x =, X() = A + C = C = A; and X () = βb + βd = β(b + D) = D = B. We can now write X(x) = A(cos(βx) cosh(βx)) + B(sin(βx) sinh(βx)) X (x) = βa( sin(βx) sinh(βx)) + βb(cos(βx) cosh(βx)) X (x) = β A( cos(βx) cosh(βx)) + β B( sin(βx) sinh(βx)) X (x) = β 3 A(sin(βx) sinh(βx)) + β 3 B( cos(βx) cosh(βx)) The final two boundary conditions, at x = L, give X (L) = β A( cos(βl) cosh(βl)) + β B( sin(βl) sinh(βl)) = X (L) = β 3 A(sin(βL) sinh(βl)) + β 3 B( cos(βl) cosh(βl)) = and these two equations can be written in matrix form as ( cos(βl) + cosh(βl) sin(βl) + sinh(βl) sin(βl) + sinh(βl) cos(βl) + cosh(βl) ) ( A B ) = ( ). (1) This system of linear equations in A and B will have a unique solution A = B = unless the determinant of the matrix is zero. This means that there will be non-zero solutions if, and only if,

3 (cos(βl)+cosh(βl))(cos(βl)+cosh(βl))+(sin(βl)+sinh(βl))(sin(βl) sinh(βl)) =. Multiplying out and using the two identities (sin(x)) + (cos(x)) 1 and (cosh(x)) (sinh(x)) 1, results in the equation (cos(βl)) + cosh(βl) cos(βl) + (cosh(βl)) + (sin(βl)) (sinh(βl)) = + cosh(βl) cos(βl) =. (Checkthis!) This means that the eigenvalues β n must satisfy the condition cosh(β n L) cos(β n L) = 1. There will be an infinite sequence of positive solutions Lβ 1 < Lβ <, tending to, which can be found by using the following MAPLE instructions: for n from 1 to 15 do fsolve(cosh(x)*cos(x)=-1., x=(n-1)*pi..n*pi); od; For n equal 1,, and 3 the solutions are Lβ , Lβ , Lβ ; and as n, the solution Lβ n approaches the value (n 1) π. The corresponding eigenfunctions X n (x) are X n (x) = A n (cos(β n x) cosh(β n x)) + B n (sin(β n x) sinh(β n x)), and Equation (1) implies that B n and A n are related by ( ) cos(βn L) + cosh(β n L) B n = A n. sin(β n L) + sinh(β n L) Letting A n = 1, we have a set of eigenfunctions of the form ( ) cos(βn L) + cosh(β n L) φ n (x) = (cos(β n x) cosh(β n x)) (sin(β n x) sinh(β n x)). sin(β n L) + sinh(β n L) () The general solution of the partial differential equation can now be written in the form u(x, t) = φ n (x)t n (t) = φ n (x)(a n cos(cβnt) + b n sin(cβnt)). Two initial conditions on u(x, t) are needed to determine the coefficients a n and b n. These conditions will be given by specifying the initial deflection 3

4 u(x, ) = f(x) and the initial velocity u t (x, ) = g(x) at each point x along the beam. In order to use these functions to obtain the coefficients a n and b n, it must first be shown that the functions φ n (x) form an orthogonal set on the interval [, L]. Theorem 1 The set of functions {φ n (x)} 1 defined in equation () is an orthogonal family on [, L]; that is φ m(x)φ n (x)dx = if m n. Proof: We first use integration by parts to show that if u and v are any sufficiently differentiable functions on x L, then u vdx = v(x)u (x) L = v(x)u (x) L ( v (x)u (x) L u v dx ) u v dx and integrating by parts two more times = v(x)u (x) L v (x)u (x) L + v (x)u (x) L v (x)u(x) L + If the right-hand term is subtracted from both sides, u vdx uv dx, uv dx = (v(x)u (x) v (x)u (x)+v (x)u (x) v (x)u(x)) L. (3) Now assume u = φ m and v = φ n where φ m and φ n are eigenfunctions of X β 4 X = for two different eigenvalues β n and β m. Then φ m = βmφ 4 m and φ n = βnφ 4 n, and equation (3) shows that φ m φ n dx φ m φ n dx = = (β 4 m β 4 n) β 4 mφ m φ n dx φ m φ n dx = φ m β 4 nφ n dx (φ n (x)φ m(x) φ n(x)φ m(x) + φ n(x)φ m(x) φ n (x)φ m (x)) L When the right-hand side of the equation is evaluated at the two endpoints x = and x = L, the boundary conditions on φ m and φ n can be seen to imply that every product has one term equal to zero; therefore, (βm 4 βn) 4 L φ mφ n dx =, and since β m β n, this means that φ mφ n dx = as required. 4

5 The values of the coefficients a n and b n can now be found by using the initial conditions. Since u(x, ) f(x) = φ n (x)(a n cos() + b n sin()) = a n φ n (x), the a n are the coefficients in an orthogonal series for f(x); therefore, a n = f(x)φ n(x)dx (φ n(x)) dx. Similarly, u t (x, t) = φ n (x)cβn( a n sin(cβnt) + b n cos(cβnt)) and u t (x, ) g(x) = cβnb n φ n (x) b n = 1 g(x)φ n(x)dx cβn (φ n(x)) dx. Example 1 Assume that c = 1, and the initial displacement function is f(x) =.5x on x 1m. The initial velocity is g(x). Find the displacement u(x, t) and plot the position of the beam at time increments of 1 1 of the period of the function T 1(t)..6 u(x,t) Position at t=k*period/1 k=6.4 k=5...4 k=4 k= x k= k=1 k= 5

6 In the figure above, the position of the beam at time k period is labelled 1 by k =, 1,, 6. The period of T 1 (t) = a 1 cos(cβ1) + b 1 sin(cβ1), in this particular example, is π cβ 1. With c = 1 and Lβ , the period is π and the curves labelled k =, 1,, 6 correspond to times (.18751) t =, 14.9, 9.8,, In the figure on page 1, the initial beam position was set equal to the first eigenfunction φ 1 (x). In this case, a 1 is the only non-zero coefficient, and the beam will oscillate precisely with period π. For the straight-line initial cβ1 position in the example, the oscillation is no longer exactly periodic because the periods of the functions T, T 3, are not constant multiples of the period of T 1, and none of the coefficients a n are zero for this straight-line initial position function. Exercises: 1. * Determine the second eigenfunction φ (x) and plot a graph for x 1 (assume that the length of the beam is 1).. * Redo Example 1 and find u(x, t), assuming that the initial displacement function is f(x) = φ (x) and the initial velocity function g(x). Find the period of this function u(x, t) and plot a graph showing the position of the beam at several different times during its period. 6