126 ON VIBRATIONS OF BEAMS OF VARIABLE CROSS-SECTION. 



be considered here) ; thus a particle, falling from a certain height, in the beginning 

 of motion has no kinetic energy, and at the end, when just reaching the floor, has no 

 potential energy. Similarly a pendulum bob, when in one of its extreme positions, 

 has no kinetic energy.; while passing through its neutral position it has no potential 

 energy. It can easily be shown that, in any state of free vibratory motion, the 

 energy, on the whole, is equally distributed into kinetic and potential energies.* 



By potential energy more specifically is meant the definite integral, of the virtual 

 work of forces, whose upper limit is determined by some fixed position of reference, 

 and whose lower limit is given by the instantaneous position of the system. In prob- 

 lems involving elastic bodies, this simply means work of deformation, and the energy 

 is often referred to as strain energy, or potential energy of the strain. 



Example i. Tension or Compression. — As is well known from strength of 



materials : — 



P l-L r- ^1. ^ • unit Stress . „. 



— ^i — -1 = E, that IS, — -. — zr-pi -. — = constant, 



a I unit deflection 



where F is the force acting on the area a ; / and k are the initial length and the 

 stressed length ; E is Young's modulus. Hence 



I I- ^^^ 



F F^l 

 and if the load is gradually applied, so that its average value is — , we have V= ^, 



which is the required expression of potential energy. 



Example 2. Potential Energy of Bending. — Let us consider two sections of the 



EJ 



beam, separated by the small length (ix. The bending moment, as we know, is M= , 



where / is the moment of inertia of the cross-section, and p is the radius of curva- 



ture, equal very nearly to i / — -^ ; let us put — = <^ ; then the angle a of the two 



sections will be, of course, such that ap = dx, in other words, a = ^dx. 



Let us now bend the beam a little more, so that the angle a will become a + da] 

 this will require the following work : — 



Mda = EJ^da = EJdx^d<p; 



in other words, the total work required for securing the desired curvature, starting 

 with a straight bar and reaching a certain value p, characterized by a corresponding 

 4), will be — 



dV=EJdx f'rpd^ = EJdx^ =. ■^•^'^^ 



So that the whole potential energy will be — 



f 



9 



*Ibbetson, "Elasticity," p. 283. 



