Vibrations of a Rigid Beam

Copyright 2000 by James V. Herod. All rights reserved.

David L. Powers in his book, Boundary Value Problems , Third Edition (Published by Harcourt Brace Jovanovich Publishers) gives a problem concerning the vibrations of a rigid beam. The text states that the displacement u(t, x) of a uniform thin beam satisfies

= , for 0 < x < L and t > 0.

The ends of the beam are simply supported , which produces boundary conditions

u(t, 0) = u(t, L) = 0 and (t, 0) = (t,L) = 0.

A derivation of this equation can be found in many texts on undergraduate partial differential equations. See for example Donald W. Trim's Applied Partial Differential Equations published by PWS KENT Publishing Company.

It is natural to ask what is the difference between the transverse vibrations of a string and of a thin beam. An over simplified response would be that the beam offers resistance to bending. This resistance is responsible for changing the wave equation to the fourth order beam equation above. Hereafter we write this equation as

= .

The constant c incorporates the rigidity and the linear density of the beam. As for the boundary conditions, simply fastened is usually taken to mean that the ends of the beam are held stationary, but the slopes at the end points can move. One describes the remaining boundary conditions in terms of the bending moment of the beam. A simply fastened beam should have zero bending moments at the end.

All that remains now is to have the initial conditions:

u(0, x) = f(x) and (0, x) = g(x).

Derivative of the general solution

We expect separation of variables to lead to solutions of the form

u(t, x) = X(x) [ A cos( t) + B sin( t) ].

That is, we expect vibrations in the time variable t.

In this problem, separation of variables will lead an equation to

X''''/X = = - T ''/ T.

For the X function:

X '''' - X = 0.

We seek the general solution of this equation.

> restart; assume(lambda>0);

> dsolve(diff(X(x),x,x,x,x)-lambda^2*X(x)=0,X(x),method=laplace);

>

We get combinations of sines, cosines, hyperbolic sines, and hyperbolic cosines of . For simplicity, we write

X(x) = C cos( ) + D sin( ) +

E cosh( ) + F sinh( ).

As usual, to determine these constants, we apply the boundary conditions.

u(t, 0) = 0 implies that X(0) = 0, which implies that C + E = 0.

(t, 0) = 0 implies that X''(0) = 0, which implies that - C + E = 0.

From this, we conclude that C = E = 0.

We see what are the implications from the other end boundary conditions.

u(t, L) = 0 implies X(L) = 0, which implies that

D sin( ) + F sinh( ) = 0.

(t, L) = 0 implies that X ''(L) = 0, which implies that

- D sin( ) + F sinh( ) = 0.

We conclude that F sinh( ) = 0, so that F = 0, and

D sin( ) = 0, so that sin( ) = 0.

We know everywhere the sine function is zero: = .

This means that there is an infinity of solutions for the X equation and they all have the form

X(x) = sin( x).

The equation in T is easier. First, there are no boundary conditions on the T equation, and second, it is only second order. The equation is

= - T ''/ T.

Thus, T '' + T = 0.

We solve this equation.

> dsolve(diff(T(t),t,t)+c^2*(n*Pi/L)^4*T(t)=0,T(t));

>

We can now write down the general solution for the partial differential equation:

u(t, x) = .

Here is a check.

> u:=(t,x)->sum((a[n]*cos(c*(n*Pi/L)^2*t)+
b[n]*sin(c*(n*Pi/L)^2*t))*sin(n*Pi/L*x),
n= 1..3);

> simplify(diff(u(t,x),t,t)+
c^2*diff(u(t,x),x,x,x,x));

We check the boundary conditions.

> u(t,0);u(t,L);

> eval(subs(x=0,diff(u(t,x),x,x)));
eval(subs(x=L,diff(u(t,x),x,x)));

>

Details for an Example