243 



At tlie same time, tlie forces, defined by %,, are in equilibrium. 



If tlie elastic ground were loaded with kij^, it would obtain the 

 the deflexion y,. In this case the beam and the ground would have 

 the same shape. However the load on the ground can only arise 

 from the beam. The deflexion y on the ground therefore involves 

 necessarily a reaction-load — ky^ on the beam. 



This latter load gives rise to another deflexion y, of the beam, 

 defined by : 



Ely^'" — -ky,=kt,{x) 



Hence 



y, = k' j ƒ ^/^ y-> 0) <I»+C,x + D, 







If we require again tiiat the load ky,, which follows from ?/,, 

 is in equilibrium, we find that: 



y, = - A' X, (X). 



From this, we deduce y, = — ^'" XiO^') ^"d so on. Therefore, the 

 terms of the series : 



y = - ^^^— - J'-, i'') - ^' K (*) - ^" ^. (•^) • • • 



represent elastic curves of a beam, which is loaded in a well- 

 defined manner. 



6. Fig. 1 illustrates the described construction in the case: 

 /=200cm., b = breath of the beam — 25 cm., I = 5000 cmS 

 E = 100000 kg/cm'; Ef = 5 X ^^' kg.cm', 1= 5 kg/cm', 

 A =3 6^' =: 125 kg/cm'. The load diagram has a parabolic form; the 

 specific load at the ends of the beam is 7* of its value at the 

 middle. The total load is 15000 kg. The scale of length in horizontal 

 direction is n := 5 (1 cm ^ — ► means 5 cm « — >). 



The deflexion are 25 times enlarged; 1 cm. T represents 7,4 cm . 



The linear load q„ which is statically equivalent to the given 

 load q, will give a sinking down to the beam, which is: 



_ 15000 KG _ 



^° ~ 125 KG/cm' X 200 cm ■" ' ^™' 



This sinking down is represented in figure la by 25 x 0,6 cm. 

 = 15 cm.; and gives rise to the straight line v,. This line also 



