of Beams and the Whirling of Shafts. 83 



§ 3. If it is assumed that y is of the form wcos (pt -fe), 



e have 



dx 

 and, putting t—^- = K , 





rZ 4 « 



The solution of this equation is 



u — a cos K^ + 5 cosh K<r -f c sin Ka* f d sinh Kir, 



where a, 6, c, d are constants to be determined by the end 

 conditions and the conditions at the intermediate supports. 



The differential equation for u is satisfied at all points of 

 the beam, but the constants of the solution will be different 

 for each bay. 



§ 4. Consider the case of a continuous beam of n bays 

 simply supported at the ends and at (n — 1) intermediate 

 points. Let Z l5 l 2 -..l n be the lengths of the bays. Take the 

 origin successively at the left-hand end of each bay. The 

 equation giving the value of u for the rth bay will be of the 

 form 



a — « r (cos K# — cosh K#) -f c r sin Kx + d r sinh Ka?, 



since a r + b r = 0, because u = when x = 0. 



The further equations expressing the facts that u = Q and 



-Y~ , -7-1 > are continuous at the supports are of the form 

 ax dx z l L 



a r (oosKZ r — cosh Kl r )-t-e r sin Kl r + d r sinh KZ r = . . . (1) 



— a r (sin Kl r + sinh K/ r ) + c r cos Kl r + d r cosh K/,.= c r+ i + d r+ i. (2) 



a r (cos K/ r + cosh KZ,.) + c r sin K/ r — rf r sinh KZ r =2a f .+i. . . («*) 

 From (1) and (3) by addition and subtraction 



a r cos K/ r 4- c r sin KZr — a r cosh KZ r — t? r sinh K/,. = a r+ i, . . (4) 

 whence, if sin KZ r is not zero, 



<c r + d r =a r (cotli K/ r — cot K.l r ) — a r+ i(cosech K/,.— cosec K/,.) . (5 ) 



Let cothK/,.— cot K/ r = <£(K /,.) = </> 



cosechK/,.— cosec K( r = ylr(Kl r ) = \jr r . 



G 2 



