128 ON VIBRATIONS OF BEAMS OF VARIABLE CROSS-SECTION. 



ditions, the coefficient of the displacement must necessarily be = o; and this leads 

 to the desired relations between forces, etc., whence the solution of the problem. In 

 other words, the conception itself of equilibrium means that all forces are quite inert 

 to any small displacement that can be imagined. In a broader sense, and with more 

 special reference to problems of elastic systems, we might say that for equilibrium 

 the variation of the difference V — W must vanish, so that h {V — W) = o, where 

 V is the potential energy and VV the work done in storing it up. The variation sign 

 h implies that it is absolutely immaterial which way we try to deflect, slightly, such a 

 system — the work done will always be exactly the same as the change of energy, 

 and therefore the system is inert to any such change, — the true definition of 

 equilibrium. 



For our immediate purpose we shall, by way of example, take the following 

 problem : — 



A beam, AB, is clamped at its right end B and is resting on a continuous, yield- 

 ing foundation. The beam is loaded with a distributed load, varying, according to 

 the law of triangle, from left to right. The loading is really due to the weight of 

 the beam itself, whose breadth is = i, and whose depth is variable, h = ^ ho, where 

 ho is the initial value at A, and ^ is a variable coefficient, changing according to the 

 straight line law, ^ = x/l ; the origin is at A and the beam length is=/. Then the vari- 

 able distributed load will be g ^ ^ qo] and the elastic resistance of the foundation 

 will be ^ = I ^0- (This example is a prototype of a more elaborate problem of find- 

 ing the elastic line of a ship in drydock.) In order to apply the principle of virtual 

 work, let V be the potential energy of the bent beam plus that of the yielding founda- 

 tion ; and let W be the work done by the loading in causing the bending of the beam. 

 Then 5 {V — W)^o. In other words, the increment of the function in the 

 brackets, for any possible small deflection from the neutral position of equilibrium, 

 must always be = o. 



Now the potential energy will, in our case, consist of two components — that 

 due to the strain energy of the beam itself, and that due to the elasticity of the foun- 

 dation; the latter will consist of such elemental expressions as k. y. dx.^. So that 



2 



the total potential energy will be — 



The work of the external forces will be, for the deflection of the beam from its orig- 

 inal form to that which it will assume under the load, 



IF = J qydx. 



o 



Remembering that — 



x = ^l; k = ^k^\ 9 = ^9o\ ^1^° t^^^ -^ = —2-^ , 



